Oswaal CBSE Chapterwise & Topicwise Question Bank Class 12 Chemistry Book (For 2022-23 Exam) 9789355955593

Table of contents :
Cover
Title Cover
Copyright Page
Contents
Syllabus
Sample Question Paper, 2021-22
Solved Paper, 2021-2022
Toppers Answers 2020
Chapter 1- The Solid State
Chapter 2- Solutions
Chapter 3- Electrochemistry
Chapter 4- Chemical Kinetics
Chapter 5- Surface Chemistry
Chapter 6- General Principles and Processes of Isolation of Elements
Chapter 7- p-Block Elements
Chapter 8- ‘d’ and ‘f’ Block Elements
Chapter 9- Coordination Compounds
Chapter 10- Haloalkanes and Haloarenes
Chapter 11- Alcohols, Phenols and Ethers
Chapter 12- Aldehydes, Ketones and Carboxylic Acids
Chapter 13- Amines
Chapter 14- Biomolecules
Chapter 15- Polymers
Chapter 16- Chemistry in Everyday life

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“978-93-5595-559-3”

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YEAR 2022-23

SYLLABUS COVERED

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TABLE OF CONTENTS Latest CBSE Syllabus released on 21st April 2022 for Academic Year 2022-2023 (CBSE Cir. No. Acad 48/2022) l Latest Term-I & Term-II Board Papers 2021-22 - Fully solved l Latest Topper’s Answers 2020





l

8 - 15 17 - 33 34 - 48

In each chapter, for better understanding, questions have been classified the typology issued by CBSE as : R - Remembering, A - Applying,

U AE

- Understanding, - Analysing & Evaluating

*1. The Solid State 1 - 27 Topic - 1 : Classification of Solids Based on Different Binding Forces, Crystal Lattices, Unit Cells, Packing in Solids Topic - 2 : Voids, Packing Efficiency, Voids, Calculations Related to Unit Cell Dimensions Topic - 3 : Defects in Solids, Electrical and Magnetic Properties, Band Theory of Metals



l

Self-Assessment Test-1 28 - 29 Solutions 30 - 64 Topic - 1 : Types of Solutions, Expression of Concentration of Solutions and Solubility Topic - 2 : Vapour Pressure, Raoult's Law, Ideal and Non-ideal Solutions Topic - 3 : Colligative Properties, Determination of Molecular Mass, Abnormal molecular mass, van't Hoff Factor Self-Assessment Test-2 65 - 66 Electrochemistry 67 - 95 Topic - 1 : Electrolytic Conductivity, Electrolytes and Kohlrausch’s Law Topic - 2 : Redox Reactions and Electrochemical Cells, Electrode Potential and Nernst Equation Topic - 3 : Electrolysis, Laws of Electrolysis, Batteries, Fuel Cells and Corrosion 96 - 97 l Self-Assessment Test-3 4. Chemical Kinetics 98 - 124 Topic - 1 : Rate of a Chemical Reaction and Factors Affecting Rate of Reactions



2. l 3.



C

- Creating.

Topic - 2 : Order of a Reaction, Integrated Rate Equations and Half life of a Reaction Topic - 3 : Concept of Collision Theory, Activation Energy and Arrhenius Equation 125 - 126 l Self-Assessment Test-4 *5. Surface Chemistry

Topic - 1 : Adsorption and its Types, Factors Affecting Adsorption Topic - 2 : Catalysis and its Types,



Enzyme Catalysis



Topic - 3 : Colloids, Types of Colloids,

l

127 - 149

Characteristics and Preparation of Colloids Self-Assessment Test-5

150 - 151

*6. General Principles and Processes

of Isolation of Elements



Topic - 1 : Principles and Methods



of extraction



Topic - 2 : Principles of Extraction of



Aluminium Copper, Zinc

l

and Iron Self-Assessment Test-6

*7. p-Block Elements

Topic - 1 : Group-15 Elements,



Properties and Some



Important Compounds



Topic - 2 : Group-16 Elements,



Properties and Some



Important Compounds

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152 - 172

173 - 174 175 - 212

l

Topic - 3 : Group-17 Elements, Properties and Some Important Compounds Topic - 4 : Group-18 Elements, Properties and Some Important Compounds

Self-Assessment Test-10 11. Alcohols, Phenols and Ethers Topic - 1 : Methods of Preparation and Properties of Alcohols and Phenols Topic - 2 : Methods of Preparation and Properties of Ethers l Self-Assessment Test-11 12. Aldehydes, Ketones and Carboxylic Acids Topic - 1 : Aldehydes and Ketones Topic - 2 : Carboxylic Acids l Self-Assessment Test-12 13. Amines l Self-Assessment Test-13 14. Biomolecules Topic - 1 : Carbohydrates, their Classification and Importance Topic - 2 : Proteins, Hormones, Vitamins and Nucleic Acids *15. Polymers l Self-Assessment Test-14 *16.  Chemistry in Everyday life l





TABLE OF CONTENTS

Self-Assessment Test-7 213 - 214 8. ‘d’ and ‘f’ Block Elements 215 - 243 Topic - 1 : d-Block Elements, their Properties and Compounds Topic - 2 : f-Block Elements, Lanthanoids and Actinoids l Self-Assessment Test-8 244 - 245 9. Coordination Compounds 249 - 268 Topic - 1 : Coordination Compounds: Properties and IUPAC Name Topic - 2 : Werner’s Theory Bonding in Coordination Compound, VBT, CFT l Self-Assessment Test-9 269 - 270 10. Haloalkanes and Haloarenes 271 - 308 Topic - 1 : Haloalkanes and their Properties Topic - 2 : Haloarenes and Polyhalogen Compounds



*Kindly note that this Chapter/Topic has been deleted from the Latest CBSE Syllabus for Academic Year 2022-23, Hence it is optional to study it.

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348 - 349 350 - 397

398 - 399 400 - 428 429 - 430 431 - 459

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CBSE CIRCULAR 2022-23 dsUæh; ek/;fed f'k{kk cksMZ

CENTRAL BOARD OF SECONDARY EDUCATION

April 21, 2022



CBSE/Acad/2022

Cir No Acad-48/2022 All Heads of Institutions affiliated to CBSE

Subject : CBSE - Secondary and Senior School Curriculum 2022-23

1. CBSE annually provides curriculum for classes IX to XII containing academic content, syllabus for examinations with learning outcomes, pedagogical practices and assessment guidelines.

2. Considering the feedback of stakeholders and other prevailing conditions, the Board will conduct the annual scheme of assessment at the end of the Academic Session 2022-23 and the curriculum has been designed accordingly. Details are available at the link https://cbseacademic.nic.in/ curriculum_2023.html

3. It is important that schools ensure curriculum transaction as per the directions given in the initial pages of the curriculum document. The subjects should be taught as per the curriculum released by the Board with the help of suitable teaching-learning strategies such as Art-Integrated Education, Experiential Learning, and Pedagogical Plans etc. wherever possible.

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(Dr. Joseph Emmanuel) Director (Academics)

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SYLLABUS Latest Syllabus issued by CBSE for Academic Year 2022-23 CHEMISTRY (Code No. 043) CLASS-XII (Theory) Time : 3 Hours

Title No. of Periods Marks Solutions 15 7 Electrochemistry 18 9 Chemical Kinetics 15 7 d- and f-Block Elements 18 7 Co-ordination Compounds 18 7 Haloalkanes and Haloarenes 15 6 Alcohols, Phenols and Ethers 14 6 Aldehydes, Ketones and Carboxylic Acids 15 8 Amines 14 6 Biomolecules 18 7 Total 160 70 Unit II : Solutions 15 Periods Types of solutions, expression of concentration of solutions of solids in liquids, solubility of gases in liquids, solid solutions, Raoult’s law, colligative properties - relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, osmotic pressure, determination of molecular masses using colligative properties, abnormal molecular mass, Van’t Hoff factor. Unit III : Electrochemistry 18 Periods Redox reactions, EMF of a cell, standard electrode potential, Nernst equation and its application to chemical cells, Relation between Gibbs energy change and EMF of a cell, conductance in electrolytic solutions, specific and molar conductivity, variations of conductivity with concentration, Kohlrausch’s Law, electrolysis and law of electrolysis (elementary idea), dry cellelectrolytic cells and Galvanic cells, lead accumulator, fuel cells, corrosion. Unit IV : Chemical Kinetics 15 Periods Rate of a reaction (Average and instantaneous), factors affecting rate of reaction: concentration, temperature, catalyst; order and molecularity of a reaction, rate law and specific rate constant, integrated rate equations and half-life (only for zero and first order reactions), concept of collision theory (elementary idea, no mathematical treatment), activation energy, Arrhenius equation. Unit VIII : d and f Block Elements 18 Periods General introduction, electronic configuration, occurrence and characteristics of transition metals, general trends in properties of the first-row transition metals – metallic character, ionization enthalpy, oxidation states, ionic radii, colour, catalytic property, magnetic properties, interstitial compounds, alloy formation, preparation and properties of K2Cr2O7 and KMnO4. Lanthanoids – Electronic configuration, oxidation states, chemical reactivity and lanthanoid contraction and its consequences. Actinoids - Electronic configuration, oxidation states and comparison with lanthanoids. Unit IX : Coordination Compounds 18 Periods



































S. No. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.





Total Periods (Theory 160 + Practical 60) 70 Marks

Coordination compounds - Introduction, ligands, coordination number, colour, magnetic properties and shapes, IUPAC nomenclature of mononuclear coordination compounds. Bonding,

(9)

SYLLABUS Werner’s theory, VBT, and CFT ; structure and stereoisomerism, importance of coordination compounds (in qualitative analysis, extraction of metals and biological system).



Haloalkanes : Nomenclature, nature of C–X bond, physical and chemical properties, optical rotation mechanism of substitution reactions.



Haloarenes : Nature of C–X bond, substitution reactions (Directive influence of halogen in monosubstituted compounds only).



Alcohols : Nomenclature, methods of preparation, physical and chemical properties (of primary alcohols only), identification of primary, secondary and tertiary alcohols, mechanism of dehydration, uses with special reference to methanol and ethanol. Phenols : Nomenclature, methods of preparation, physical and chemical properties, acidic nature of phenol, electrophillic substitution reactions, uses of phenols. Ethers : Nomenclature, methods of preparation, physical and chemical properties, uses.





Unit XII : Aldehydes, Ketones and Carboxylic Acids





14 Periods





Uses and environmental effects of - dichloromethane, trichloromethane, tetrachloromethane, iodoform, freons, DDT .

Unit XI : Alcohols, Phenols and Ethers



15 Periods





Unit X : Haloalkanes and Haloarenes

Aldehydes and Ketones : Nomenclature, nature of carbonyl group, methods of preparation, physical and chemical properties, mechanism of nucleophilic addition, reactivity of alpha hydrogen in aldehydes,uses. Carboxylic Acids : Nomenclature, acidic nature, methods of preparation, physical and chemical properties;uses. 14 Periods

Amines : Nomenclature, classification, structure, methods of preparation, physical and chemical properties, uses, identification of primary, secondary and tertiary amines. Diazonium salts: Preparation, chemical reactions and importance in synthetic organic chemistry.











Unit XIV : Biomolecules

18 Periods









Unit XIII : Amines



15 Periods

Carbohydrates - Classification (aldoses and ketoses), monosaccahrides (glucose and fructose), D-L configuration oligosaccharides (sucrose, lactose, maltose), polysaccharides (starch, cellulose, glycogen); Importance of carbohydrates. Proteins - Elementary idea of - amino acids, peptide bond, polypeptides, proteins, structure of proteins - primary, secondary, tertiary structure and quaternary structures (qualitative idea only), denaturation of proteins; enzymes. Hormones - Elementary idea excluding structure. Vitamins : Classification and functions. Nucleic Acids : DNA and RNA. Note : The content indicated in NCERT textbooks as excluded for the year 2022-23 is not to be tested by schools.

( 10 )

SYLLABUS PRACTICALS 3 Hours/30 Marks Evaluation Scheme for Examination

Marks

Volumetric Analysis

08

Salt Analysis

08

Content Based Experiment

06

Project Work

04

Class record and Viva

04 Total

PRACTICALS SYLLABUS

30













60 Periods Micro-chemical methods are available for several of the practical experiments. Wherever possible, such techniques should be used. A. Surface Chemistry (a) Preparation of one lyophilic and one lyophobic sol Lyophilic sol - starch, egg albumin and gum Lyophobic sol - aluminium hydroxide, ferric hydroxide, arsenous sulphide. (b) Dialysis of sol-prepared in (a) above. (c) Study of the role of emulsifying agents in stabilizing the emulsion of different oils. B. Chemical Kinetics (a) Effect of concentration and temperature on the rate of reaction between Sodium Thiosulphate and Hydrochloric acid. (b) Study of reaction rates of any one of the following: (i) Reaction of Iodide ion with Hydrogen Peroxide at room temperature using different concentration of Iodide ions. (ii) Reaction between Potassium Iodate, (KIO3) and Sodium Sulphite: (Na2SO3) using starch solution as indicator (clock reaction). C. Thermochemistry : Any one of the following experiments (a) Enthalpy of dissolution of Copper Sulphate or Potassium Nitrate. (b) Enthalpy of neutralization of strong acid (HCI) and strong base (NaOH). (c) Determination of enthaply change during interaction (Hydrogen bond formation) between Acetone and Chloroform. D. Electrochemistry : Variation of cell potential in Zn/Zn2+|| Cu2+/Cu with change in concentration of electrolytes (CuSO4 or ZnSO4) at room temperature. E. Chromatography : (a) Separation of pigments from extracts of leaves and flowers by paper chromatography and determination of Rf values. (b) Separation of constituents present in an inorganic mixture containing two cations only (constituents having large difference in Rf values to be provided). F. Preparation of Inorganic Compounds : Preparation of double salt of Ferrous Ammonium Sulphate or Potash Alum. Preparation of Potassium Ferric Oxalate. ( 11 )

SYLLABUS

G.

Preparation of Organic Compounds :



Preparation of any one of the following compounds Acetanilide





(i)





(ii) Di -benzal Acetone



(iii) p-Nitroacetanilide



(iv) Aniline yellow or 2 - Naphthol Aniline dye.

H.

Tests for the functional groups present in organic compounds :



Unsaturation, alcoholic, phenolic, aldehydic, ketonic, carboxylic and amino (Primary) groups. Characteristic tests of carbohydrates, fats and proteins in pure samples and their detection in given foodstuffs.

J.

Determination of concentration/ molarity of KMnO4 solution by titrating it against a standard solution of :



(i)

Oxalic acid,







I.





(ii) Ferrous Ammonium Sulphate



Qualitative analysis Determination of one cation and one anion in a given salt. Cation :Pb2+, Cu2+ As3+, A 3+, Fe3+, Mn2+, Zn2+, Ni2+, Ca2+, Sr2+, Ba2+, Mg2+, NH4+  

Anions: (CO3)2-, S2-, (SO3)2-, (NO2)-, (SO4)2-, C -, Br -, I-, PO3-4, (C2O4)2-, CH3COO -,NO3ℓ









K.

ℓ



(Students will be required to prepare standard solutions by weighing themselves).

(Note: Insoluble salts excluded)





INVESTIGATORY PROJECT Scientific investigations involving laboratory testing and collecting information from other sources. A few suggested Projects. l Study of the presence of oxalate ions in guava fruit at different stages of ripening.



l Study of quantity of casein present in different samples of milk.



l Preparation of soybean milk and its comparison with the natural milk with respect to curd formation, effect of temperature, etc.



l Study of the effect of Potassium Bisulphate as food preservative under various conditions (temperature, concentration, time, etc.)



l Study of digestion of starch by salivary amylase and effect of pH and temperature on it.



l Comparative study of the rate of fermentation of following materials: wheat flour, gram flour, potato juice, carrot juice, etc.



l Extraction of essential oils present in Saunf (aniseed), Ajwain (carum), Illaichi (cardamom).



l Study of common food adulterants in fat, oil, butter, sugar, turmeric power, chilli powder and pepper.





Note : Any other investigatory project, which involves about 10 periods of work, can be chosen with the approval of the teacher.

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SYLLABUS Practical Examination for Visually Impaired Students of Classes XI and XII Evaluation Scheme Time Allowed : Two hours

Max. Marks : 30 Topic

Marks

Identification/ Familiarity with the apparatus

5

Written test (based on given/ prescribed practicals)

10

Practical Record

5

Viva

10 Total

30

General Guidelines • The practical examination will be of two hour duration. • A separate list of ten experiments is included here. • The written examination in practicals for these students will be conducted at the time of practical examination of all other students. • The written test will be of 30 minutes duration. • The question paper given to the students should be legibly typed. It should contain a total of 15 practical skill based very short answer type questions. A student would be required to answer any 10 questions. • A writer may be allowed to such students as per CBSE examination rules. • All questions included in the question papers should be related to the listed practicals. Every question should require about two minutes to be answered. • These students are also required to maintain a practical file. A student is expected to record at least five of the listed experiments as per the specific instructions for each subject. These practicals should be duly checked and signed by the internal examiner. • The format of writing any experiment in the practical file should include aim, apparatus required, simple theory, procedure, related practical skills, precautions, etc. • Questions may be generated jointly by the external/internal examiners and used for assessment.





1.



• The viva questions may include questions based on basic theory/principle/concept, apparatus/materials/ chemicals required, procedure, precautions, sources of error, etc. Items for Identification/Familiarity of the apparatus for assessment in practicals (All experiments) Beaker, glass rod, tripod stand, wire gauze, Bunsen burner, Whatman filter paper, gas jar, capillary tube, pestle and mortar, test tubes, tongs, test tube holder, test tube stand, burette, pipette, conical flask, standard flask, clamp stand, funnel, filter paper Hands-on Assessment



• Identification/familiarity with the apparatus



• Odour detection in qualitative analysis

2. List of Practicals

The experiments have been divided into two sections:



Section A and Section B.



The experiments mentioned in Section B are mandatory.

( 13 )

SYLLABUS SECTION - A A. Surface Chemistry (1) Preparation of one lyophilic sol Lyophilic sol - starch, egg albumin and gum (2) Preparation of one lyophobic sol Lyophobic sol – Ferric hydroxide B. Chromatography Separation of pigments from extracts of leaves and flowers by paper chromatography and determination of Rf values (distance values may be provided). C. Tests for the functional groups present in organic compounds: (1) Alcoholic and Carboxylic groups. (2) Aldehydic and Ketonic D. Characteristic tests of carbohydrates and proteins in the given foodstuffs. E. Preparation of Inorganic Compounds- Potash Alum

SECTION - B (Mandatory) F. Quantitative analysis (1) (a) Preparation of the standard solution of Oxalic acid of a given volume (b) Determination of molarity of KMnO4 solution by titrating it against a standard solution of Oxalic acid. (2) The above exercise [F 1 (a) and (b)] to be conducted using Ferrous ammonium sulphate (Mohr’s salt) G. Qualitative analysis : (1) Determination of one cation and one anion in a given salt. Cations- NH4+ – – – Anions – (CO3)2 , S2 , (SO3)2 , Cl–, CH3COO– (Note: Insoluble salts excluded) Note: The above practicals may be carried out in an experiential manner rather than recording observations. Prescribed Books: 1. Chemistry Part -I, Class-XII, Published by NCERT. 2. Chemistry Part -II, Class-XII, Published by NCERT. 3. Laboratory Manual of Chemistry, Class XI Published by NCERT. 4. Other related books and manuals of NCERT including multimedia and online sources.

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SYLLABUS QUESTION PAPER DESIGN Classes –XI and XII (2022-23)

S.No.

Domains

Marks

%

1.

Remembering and Understanding: Exhibit memory of previously learned material by recalling facts, terms, basic concepts and answers. Demonstrate understanding of facts and ideas by organizing, comparing, translating, interpreting, giving descriptions and stating main ideas.

28

40

2.

Applying : Solve problems to new situations by applying acquired knowledge, facts, techniques and rules in a different way.

21

30

3.

Analysing, Evaluating and Creating: Examine and break information into parts by identifying motives or causes. Make inferences and find evidence to support generalizations. Present and defend opinions by making judgments about information, validity of ideas or quality of work based on a set of criteria. Compile information together in a different way by combining elements in a new pattern or proposing alternative solutions.

21

30

For more details kindly refer to Sample Question Paper of class XII for the year 2022-23 to be published by CBSE at its website. qq

( 15 )

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( 16 )

Sample Question Paper, 2021-22 (Issued by CBSE Board on 14th January, 2022) TERM-II

CHEMISTRY SOLVED Time allowed : 2 Hours

Max. Marks : 35



General Instructions : (i) There are 12 questions in this question paper with internal choice. (ii) SECTION A - Q. No. 1 to 3 are very short answer questions carrying 2 marks each. (iii) SECTION B - Q. No. 4 to 11 are short answer questions carrying 3 marks each. (iv) SECTION C- Q. No. 12 is case based question carrying 5 marks. (v) All questions are compulsory. (vi) Use of log tables and calculators is not allowed

OR

1. Arrange the following in the increasing order of their property indicated (any two): [1×2=2]





Section - A



(a) Benzoic acid, Phenol, Picric acid, Salicylic acid (pka values).



(b) Acetaldehyde, Acetone, Methyl tert butyl ketone (reactivity towards NH2OH).



(c) ethanol, ethanoic acid, benzoic acid (boiling point)

2. Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The Lm of ‘B’ increases 1.5 times while that of A increases 25 times. Which of the two is a strong electrolyte? Justify your answer. Graphically show the behaviour of ‘A’ and ‘B’. [2] 3. Give reasons to support the answer:

[2]



(a) Presence of alpha hydrogen in aldehydes and ketones is essential for aldol condensation.



(b) 3–Hydroxy pentan-2-one Tollen’s test.

shows

positive



4. Account for the following:





Section - B (a) Aniline cannot be prepared by the ammonolysis of chlorobenzene under normal conditions.



(b) N-ethylethanamine boils at 329.3K and butanamine boils at 350.8K, although both are isomeric in nature.



(c) Acylation of aniline is carried out in the presence of pyridine. [1×3=3]

Convert the following:

(a) Phenol to N-phenylethanamide.



(b) Chloroethane to methanamine.



(c) Propanenitrile to ethanal.

[1×3=3]

5. Answer the following questions:

(a) [Ni(H2O)6] 2+ (aq) is green in colour whereas [Ni(H2O)4 (en)]2+ (aq) is blue in colour, give reason in support of your answer .



(b) Write the formula and hybridization of the following compound:



tris(ethane-1,2–diamine) cobalt(III) sulphate [1+2] OR

In a coordination entity, the electronic configuration of the central metal ion is t2g3eg1

(a) Is the coordination compound a high spin or low spin complex?



(b) Draw the crystal field splitting diagram for the above complex. [1+2]

6. Account for the following:

(a) Ti(IV) is more stable than Ti (II) or Ti(III).



(b) In case of transition elements, ions of the same charge in a given series show progressive decrease in radius with increasing atomic number.



(c) Zinc is comparatively a soft metal, iron and chromium are typically hard. [1×3=3]

18 ]



Oswaal CBSE Chapterwise & Topicwise Question Bank, CHEMISTRY, Class – XII

7. An alkene ‘A’ (Mol. formula C5H10) on ozonolysis gives a mixture of two compounds ‘B’ and ‘C’. Compound ‘B’ gives positive Fehling’s test and also forms iodoform on treatment with I2 and NaOH. Compound ‘C’ does not give Fehling’s test but forms iodoform. Identify the compounds A, B and C. Write the reaction for ozonolysis and formation of iodoform from B and C. [3]



(c) It has been observed that first ionization energy of 5d series of transition elements are higher than that of 3d and 4d series, explain why? [1×3=3] OR



On the basis of the figure given below, answer the following questions:



8. Observe the figure given below and answer the questions that follow: [3]



(b) What is the application of this process?



(c) Can the same process occur without applying electric field? Why is the electric field applied?

9. What happens when reactions: (a) N-ethylethanamine reacts with benzenesulphonyl chloride.



(b) Benzylchloride is treated with ammonia followed by the reaction with Chloromethane.



(c) Aniline reacts with chloroform in the presence of alcoholic potassium hydroxide. [1×3=3]



(b) Why do transition metals of 3d series have lower melting points as compared to 4d series?



(c) In the third transition series, identify and name the metal with the highest melting point. [1×3=3]

Section - C 12. Read the passage given below and answer the questions that follow.



OR

(a) Write the IUPAC name for the following organic compound:

(b) Complete the following:





[1×3=3]









10. Represent the cell in which the following reaction takes place. The value of E˚ for the cell is 1.260 V. What is the value of Ecell?

2Al(s) + 3Cd2+ (0.1M) ® 3Cd (s) + 2Al3+ (0.01M) [3] 11. (a) Why are fluorides of transition metals more stable in their higher oxidation state as compared to the lower oxidation state?

(a) Why manganese has lower melting point than chromium?



(a) Which process is represented in the figure?

(b) Which one of the following would feel attraction when placed in magnetic field: Co2+, Ag+, Ti4+, Zn2+?

Are there nuclear reactions going on in our bodies?

There are nuclear reactions constantly occurring in our bodies, but there are very few of them compared to the chemical reactions, and they do not affect our bodies much. All of the physical processes that take place to keep a human body running are chemical processes. Nuclear reactions can lead to chemical damage, which the body may notice and try to fix. The nuclear reaction occurring in our bodies is radioactive decay. This is the change of a less stable nucleus to a more stable nucleus. Every atom has either a stable nucleus or an unstable nucleus, depending on how big it is and on the ratio of protons to neutrons. The ratio of neutrons to protons in a stable nucleus is thus around 1:1 for small nuclei (Z < 20). Nuclei with too many neutrons, too few neutrons, or that are simply too big are unstable. They eventually transform to a stable form through radioactive decay. Wherever there are atoms with unstable nuclei (radioactive atoms), there are nuclear reactions occurring naturally. The interesting thing is that there are small amounts of radioactive atoms everywhere: in your chair, in the ground, in the food you eat, and yes, in your body.

[ 19



The most common natural radioactive isotopes in humans are carbon-14 and potassium-40. Chemically, these isotopes behave exactly like stable carbon and potassium. For this reason, the body uses carbon-14 and potassium-40 just like it does normal carbon and potassium; building them into the different parts of the cells, without knowing that they are radioactive. In time, carbon-14 atoms decay to stable nitrogen atoms and potassium-40 atoms decay to stable calcium atoms. Chemicals in the body that relied on having a carbon-14 atom or potassium-40 atom in a certain spot will suddenly have a nitrogen or calcium atom. Such a change damages the chemical. Normally, such changes are so rare, that the body can repair the damage or filter away the damaged chemicals. The natural occurrence of carbon-14 decay in the body is the core principle behind carbon dating. As long as a person is alive and still eating, every carbon-14 atom that decays into a nitrogen atom is replaced on average with a new carbon-14 atom. But once a person dies, he stops replacing the decaying carbon-14 atoms. Slowly the carbon-14 atoms decay to nitrogen without being replaced, so that there is less and less carbon-14 in a dead body. The rate at which carbon-14 decays is constant and follows first order kinetics. It has a half - life of nearly 6000 years, so by measuring the relative amount of carbon-14 in a bone, archeologists can calculate when the person died. All living organisms consume

carbon, so carbon dating can be used to date any living organism, and any object made from a living organism. Bones, wood, leather, and even paper can be accurately dated, as long as they first existed within the last 60,000 years. This is all because of the fact that nuclear reactions naturally occur in living organisms.





CBSE SAMPLE QUESTION PAPER 2021-22 (Term-II)

(Source: The textbook Chemistry: The Practical Science by Paul B. Kelter, Michael D. Mosher and Andrew Scott states)



(a) Why is Carbon -14 radioactive while Carbon -12 not? (Atomic number of Carbon: 6)



(b) Researchers have uncovered the youngest known dinosaur bone, dating around 65 million years ago. How was the age of this fossil estimated?



(c) Which are the two most common radioactive decays happening in human body?



(d) Suppose an organism has 20 g of Carbon -14 at its time of death. Approximately how much Carbon -14 remains after 10,320 years? (Given antilog 0.517 = 3.289) OR



(d) Approximately how old is a fossil with 12 g of Carbon -14 if it initially possessed 32 g of Carbon-14? (Given log 2.667 = 0.4260) [1+1+1+2]

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2.



Picric acid < salicylic acid < benzoic acid III>II (B) II>III>I (C) II>I>III (D) III>I>II 18. Which of the following forms strong pp – pp bonding? (A) S8 (B) Se6 (C) Te4 (D) O2 19. F2 acts as a strong oxidising agent due to (A) low Dbond H° and low Dhyd H°



































(C)

















(A) AgCl (B) NaCl (C) FeO (D) ZnO 12. Which of the following acids reacts with acetic anhydride to form a compound Aspirin? (A) Benzoic acid (B) Salicylic acid (C) Phthalic acid (D) Acetic acid 13. Which of the following statements is wrong? (A) Oxygen shows pπ-pπ bonding. (B) Sulphur shows little tendency of catenation. (C) Oxygen is diatomic whereas sulphur is polyatomic. (D) O–O bond is stronger than S–S bond. 14. Amino acids which cannot be synthesized in the body and must be obtained through diet are known as (A) Acidic amino acids (B) Essential amino acids (C) Basic amino acids (D) Non-essential amino acids 15. Which one of the following halides contains Csp2 – X bond? (A) Allyl halide (B) Alkyl halide (C) Benzyl halide (D) Vinyl halide 16. On mixing 20 ml of acetone with 30 ml of chloroform. The total volume of the solution is (A) 50 ml (D) =10 ml 17. Consider the following compounds

CH3 – CH = CH – CH2 – OH ¾PCC ¾¾¾ ®

the product formed is: (A) CH3 – CHO and CH3CH2OH (B) CH3 – CH = CH – COOH (C) CH3 – CH = CH – CHO (D) CH3 – CH2 – CH2 – CHO 29. Enantiomers differ only in (A) boiling point (B) rotation of polarized light (C) melting point (D) solubility 30. The number of lone pairs of electrons in XeF4 is (A) zero (B) one (C) two (D) three 31. Sulphuric acid is used to prepare more volatile acids from their corresponding salts due to its (A) strong acidic nature (B) low volatility (C) strong affinity for water (D) ability to act as a dehydrating agent 32. An element with density 6 g cm–3 forms a fcc lattice with edge length of 4 × 10–8 cm. The molar mass of the element is (NA = 6 × 1023 mol–1) (A) 57.6 g mol–1 (B) 28.8 g mol–1 –1 (C) 82.6 g mol (D) 62 g mol–1

26 ]



compound 'Y' is

(B) CH3CH2CH2CH2CH3 (C)

(D) All of the above Given below are the questions (45-49) labelled as Assertion (A) and Reason (R). Select the most appropriate answer from the options given below: (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true. 45. Assertion (A): A raw mango placed in a saline solution loses water and shrivel into pickle. Reason (R): Through the process of reverse osmosis, raw mango shrivel into pickle. 46. Assertion (A): H2S is less acidic than H2Te. Reason (R): H-S bond has more Dbond H° than H–Te bond. 47. Assertion (A): Chlorobenzene is less reactive towards nucleophilic substitution reaction. Reason (R): Nitro group in chlorobenzene increases its reactivity towards nucleophilic substitution reaction. 48. Assertion (A): Due to schottky defect, there is no effect on the density of a solid. Reason (R): Equal number of cations and anions are missing from their normal sites in Schottky defect. 49. Assertion (A): Fluorine forms only one oxoacid HOF. Reason (R): Fluorine atom is highly electronegative.































Section-C





















(A) Boiling point of solution (B) Freezing point of solvent (C) Boiling point of solvent (D) Freezing point of solution 39. XeF6 on reaction with NaF gives (A) Na+ [XeF7]– (B) [NaF2]– [XeF5]+

















(C) SO2 (g) (D) N2O (g) 38. In the following diagram point, ‘X’ represents





























(C) AsH3 (D) BiH3 35. The boiling point of a 0.2 m solution of a nonelectrolyte in water is (Kb for water = 0.52 K kg mol–1) (A) 100°C (B) 100.52°C (C) 100.104°C (D) 100.26°C 36. Nucleic acids are polymer of (A) amino acids (B) nucleosides (C) nucleotides (D) glucose 37. Which of the following gas dimerises to become stable? (A) CO2 (g) (B) NO2 (g)





I

(ii) (iii)

Long range order …ABC ABC ABC….

(iv)

Number of atoms per unit cell=2 Metal excess defect due to anionic vacancies





Stoichiometric defects (a)















(i)



(v)











Crystalline solids

(b) F- centres (c) Schottky and Frenkel defects (d) fcc structure

Which of the following is the best matched options? (A) (i)- (d), (ii)-(a), (iii)-(b), (iv)-(c) (B) (i)-(c), (ii)-(a), (iii)-(d), (v)-(b) (C) (i)-(c ), (ii)-(a), (iii)-(d), (iv)-(b) (D) (i)-(a), (ii)-(b), (v)-(c), (iv)-(d)

















(C) C2H5–Br (D) C2H5–I 44. Which of the following isomer of pentane (C5H12) will give three isomeric monochlorides on photochemical chlorination?

II









This section consists of 6 multiple choice questions with an overall choice to attempt any 5. In case more than desirable number of questions are attempted, only first 5 will be considered for evaluation. 50. Match the following:











(C) Na+[XeF6]– (D) [NaF2]+ [XeF5]– 40. Glucose on reaction with Br2 water gives (A) Saccharic acid (B) Hexanoic acid (C) Gluconic acid (D) Salicylic acid 41. Which of the following is optically inactive? (A) (+) – Butan–2–ol (B) (–) – Butan–2–ol (C) (±) –Butan–2–ol (D) (+)–2–Bromobutane 42. Which of the following is not a correct statement? (A) Halogens are strong oxidising agents. (B) Halogens are more reactive than interhalogens. (C) All halogens are coloured. (D) Halogens have maximum negative electron gain enthalpy. 43. Which of the following has highest boiling point? (A) C2H5–F (B) C2H5–Cl















































































34. Which of the following is the weakest reducing agent in group 15? (A) NH3 (B) PH3













(D)





(C)







(B)





(A)









(A)



33. In the reaction





Oswaal CBSE Chapterwise & Topicwise Question Bank, CHEMISTRY, Class – XII

[ 27









































(C) CH3 — CH2 — OH (D) CH3 — OH 54. Which of the following group increases the acidic character of– phenol? – (A) CH3O (B) CH3 – (C) NO2 (D) All of these 55. Consider the following reaction

the products X and Y are (A)























51. Which of the following analogies is correct? (A) XeF2: linear:: XeF6: square planar (B) moist SO2: Reducing agent:: Cl2: bleaching agent (C) N2: Highly reactive gas:: F2: inert at room temperature (D) NH3: strong base:: HI: weak acid 52. Complete the following analogy: Curdling of milk : A :: a-helix : B (A) A: Primary structure B: Secondary structure (B) A: Denatured protein B: Primary structure (C) A: Secondary structure B: Denatured protein (D) A: Denatured protein B: Secondary structure Case: Read the passage given below and answer the following questions (53-55). Alcohols and phenols are acidic in nature. Electron withdrawing groups in phenol increase its acidic strength and electron donating groups decrease. Alcohols undergo nucleophilic substitution with hydrogen halides to give alkyl halides. On oxidation primary alcohols yield aldehydes with mild oxidising agents and carboxylic acids with strong oxidising agents while secondary alcohols yield ketones. The presence of –OH groups in phenols activates the ring towards electrophilic substitution. Various important products are obtained from phenol like salicylaldehyde, salicylic acid, picric acid, etc. 53. Which of the following alcohols is resistant to oxidation? (A) (B)





SOLVED PAPER - 2021-22 (Term-I)





(B)





(C)







(D)







SOLUTIONS Section-A

Explanation: H2S2O7 is pyrosulphuric acid.

2. (D) All of the above



3. (C)



Explanation: Glass, plastic, rubber all are amorphous solids as they are non-crystalline solids in which the atoms and molecules are not organized in a definite lattice pattern.

4. (B) slightly less than 109°28’

Explanation: n-hexane and n-heptane will form an ideal solution over entire range of concentrations as their intermolecular interactions (solutesolvent) after forming solution are similar to their intermolecular attractions (solute-solute, solventsolvent) before mixing the components.



1. (C) n-hexane and n-heptane

Explanation: C-O-H bond angle in alcohol is slightly less than 109°28’. The oxygen atom is sp3 hybridised but because of the mutual repulsion of 2 lone pairs of electrons on it the resultant bond angle C-O-H is slightly less than the tetrahedral angle.

28 ]

7. (A) O



5. (A)

Explanation: Group 16 elements (chalcogens) have six valence electrons each with general electronic configuration of ns2np4. Highly electronegative oxygen shows -2 oxidation state as it accepts 2 electrons to complete its octet. However, down the group from S, Se, Te, Po the stability of –2 oxidation state decreases with decrease in the electronegativity of elements. They show +2,+4, +6 oxidation states.

Explanation: Propene yields two products, however only one predominates as per Markovnikov’s rule i.e., 2-bromopropane which on heating with aq. KOH gives secondary alcohol. Aq. KOH is alkaline in nature so it gives hydroxide ion which is a nucleophile. It replaces halide(bromide in this case) ion and form alcohols.

8. (D) Isotropic in nature.



Oswaal CBSE Chapterwise & Topicwise Question Bank, CHEMISTRY, Class – XII



9. (D) psolute = p°solvent.xsolvent Explanation: According to Raoult’s law, the vapour pressure of a solution of a non-volatile solute is equal to the vapour pressure of the pure solvent at that temperature multiplied by its mole fraction.

6. (C) a nitrogenous base and a pentose sugar

N N

N

O

Explanation: An azeotropic mixture that has a boiling point lesser than its constituents is known as minimum boiling azeotropes and show positive deviation from Raoult’s law.

N

11. (D) ZnO

HO

10. (A) shows a positive deviation from Raoult’s law

Explanation: Nucleosides are composed of a nitrogenous base and a pentose sugar. NH2









Explanation: Crystalline solids are anisotropic in nature i.e., they possess different properties in different directions. However, they have a sharp melting point, regular geometry and are true solids.

Explanation: In metal excess defect due to cations, heating the compound releases extra cations which occupy the interstitial sites in crystals and the same number of electrons goes to neighbouring interstitial sites. E.g. ZnO.

12. (B) Salicylic acid



OH



15. (D) Vinyl halide

13. (B) Sulphur shows little tendency of catenation



Explanation:

Explanation: (A) Vinylic halides These are the compounds in which the halogen atom is bonded to a sp2-hybridised carbon atom of a carbon-carbon double bond (C = C).

14. (B) Essential amino acids 16. (B) = 50 ml

Explanation: Amino acids which cannot be synthesized in the body are obtained through diet are called essential amino acids.







Explanation: Oxygen is small in size therefore the lone pairs of electrons repel the O-O bond greatly as compared to lone pairs of electrons in S-S bond. Hence, S-S bond is stronger than the O-O bond and the sulphur shows greater tendency for catenation than oxygen.

[ 29

24. (B) NaCl and NaClO



Tert-halide with strong base favours elimination reaction not the substitution reaction.

Explanation: Chlorine reacts with cold and dilute NaOH to produce a mixture of sodium chloride (NaCl) and sodium hypochloride (NaOCl). 2NaOH + Cl2 ® NaCl + NaOCl +H2O

18. (D) O2



19. (B) low Dbond H° and high Dhyd H°

25. (C) Molar mass of solute (M)

Explanation: Oxygen molecule has strong pp – pp bonding.







Explanation:

17. (B) II>III>I Explanation: Chlorine exhibits –I effect, pulling electrons towards itself, thus creating a slight positive charge on the carbon attached. In compound 2, –I effect of chlorine is most pronounced making the compound reactive towards SN2 reaction. In compound 3, –I effect of chlorine is strong but is countered by the resonance effect making substitution possible at o-, p- positions. In compound 1, –I effect of chlorine is balanced by the +I effect of the cycloalkyl group.



23. (D)





Explanation: There will be no change in volume on mixing chloroform and acetone. It will be: 20 + 30 = 50 ml





SOLVED PAPER - 2021-22 (Term-I)

Explanation: Boiling Point Elevation

20. (A) Glucose



Explanation: Flourine molecule has high hydration energy and low bond dissociation energy. That's why it acts as a strong oxidising agent.

21. (C) NO



Explanation: Glucose is known as dextrose.

Explanation:

...(i)

Molality

x m= ×1000 W

Molality = m Number of mole of solute = x Weight of solvent in grams = W ...(ii) Moles of solute = mass of solute / molar mass of solute ...(iii) From eq. (i), (ii), (iii), we derive that elevation of boiling point is inversely proportional to the molar mass of the solute.

22. (B) SiO2

SECTION-B 26. (A) 0.08 Explanation: Mole fraction of gas X in solution = 0.04 Pressure = 2.5 bar Let p1 = P0 X1 2.5 = 0.04 P0 ...(i) Let pressure be doubled, then p2 5.0 = X2 P0 ...(ii) Dividing Eqn ii by eqn I, we get 5.0/2.5 = X2/0.04 2 × 0.04 = X2 X2 = 0.08













Three dimensional structure of SiO2















Explanation: SiO2 is a network solid. A network solid or covalent network solid (also called atomic crystalline solids) is a chemical compound (or element) in which the atoms are bonded by covalent bonds in a continuous network extending throughout the material.





3Cu(s) + 2HNO3(aq) ® 3Cu(NO3)2(aq) + 2NO(g) + H2O(l)





Where,

30 ]



Oswaal CBSE Chapterwise & Topicwise Question Bank, CHEMISTRY, Class – XII

27. (D) Thymine

28. (C) CH3 – CH = CH – CHO

Explanation: Sulphuric acid has high boiling point so has low volatility hence, used to prepare more volatile acids.

Where PCC : Pyridinium chloro oxochromate. PCC oxidises primary alcohols to aldehydes

32. (A) 57.6 g mol–1

da3 N A z

M = 6 × (4 × 10–8)3 × 6 × 1023/4 = 576 × 10–1 = 57.6 g mol–1

33. (C)



Explanation: The XeF4 (xenon tetrafluoride) molecule is hypervalent with six electron pairs around the central xenon (Xe) atom. Out of these, four are bonding pairs, and two are lone pairs.

M=

For an FCC, z = 4





30. (C) two



Explanation: Enantiomers are a pair of molecules that exist in two forms that are mirror images of one another but cannot be superimposed one upon the other. However, they differ in direction in which they rotate polarized light, either dextro (d or +) or levo (l or -), when dissolved in solution.



Explanation: Density of element d = 6 g cm–3 Edge length of FCC lattice, a = 4 × 10–8 cm NA = 6 × 1023 mol–1 z = no. of effective constituent particles in one unit cell Density of cubic unit cell zM d= 3 a NA

29. (B) rotation of polarised light



31. (B) low volatility



PCC ® Explanation: CH3 – CH = CH – CH2 – OH ¾¾¾¾







Explanation: Instead of Thymine, RNA has Uracil as nitrogenous base.

36. (C) nucleotides



Explanation: NH3 is the weakest reducing agent in the hydrides of group 15. The reducing character of hydrides of group 15 elements increases from top to bottom because the Z-H strength (Here Z= N, P, As, Sb, or Br) bond decreases down the group due to an increase in the size of the central atom.

Explanation: Nucleic acids are polymers of nucleotides.

35. (C) 100.104°C





37. (B) NO2 (g)

Explanation: Molality of solution = 0.2 m Kb of water = 0.52 K kg mol–1 Boiling Point Elevation







For most non-electrolytes dissolved in water, the van't Hoff factor is essentially 1. Hence, Elevation in Boiling point = 0.52 × 0.2 = 0.104C° Therefore, Boiling point = 100 + 0.104 = 100.104 C°

34. (A) NH3



Explanation: Bromocyclohexane gives Grignard reagent on treatment with Mg and dry ether, which on hydrolysis yields cyclohexane.

Explanation: In NO2 structure, there is an unpaired electron on nitrogen atom, so it is reactive, hence it dimerises to pair up its electron and gains stability as N2O4. NO2(g) is an odd electron molecule Hence, it undergoes dimerisation to form stable molecule with even number of electrons.

[ 31



47. (B) Both A and R are true but R is not the correct explanation of A. Explanation: Presence of nitro group on ortho or para position in the ring makes the ring more electron deficient and activated towards nucleophilic substitution reaction as compared to chlorobenzene.

38. (A) Boiling point of solution

39. (A) Na+[XeF7] Explanation: XeF6 + NaF ® Na+[XeF7]



Explanation: Equal number of cations and anions are missing from their normal sites in Schottky defect which leads to considerable decrease in density.

–

40. (C) Gluconic acid

49. (B) Both A and R are true but R is not the correct explanation of A. Explanation: Absence of d-orbitals in fluorine is the reason it doesn't form the oxoacids having higher oxidation states such as +3, +5, or +7. So, the +1 oxidation state is shown by fluorine only with the element oxygen. Thus, it forms only one oxoacid, HOF.

41. (C) (±) Butan-2-ol







Explanation: Gluconic acid.





48. (D) A is false but R is true.







Explanation: Point X represents the boiling point of the solution.











SOLVED PAPER - 2021-22 (Term-I)



51. (B) moist SO2: Reducing agent:: Cl2: bleaching agent Explanation: The reducing character of sulphur dioxide is due to evolution of nascent hydrogen when it is moist. Chlorine is a good bleaching agent, due to its oxidising properties because it produces nascent oxygen.



52. (D) A: Denatured protein

44. (B) CH2 CH2 CH2 CH2 CH3

53. (A)



Explanation: When alkanes larger than ethane are halogenated, isomeric products are formed.

45. (C) A is true but R is false. Explanation: A raw mango placed in saline solution loses water and shrivels into pickle. Mango looses water due to osmosis and turns into a pickle.

Explanation: Tert –alcohols are resistant to oxidation.



Explanation: H2S is less acidic than H2Te as we move down the group the bond dissociation enthalpy decreases and it is easier to remove H+.

54. (C) NO2–





46. (A) Both A and R are true and R is the correct Explanation of A.



B: Secondary structure

Explanation: Curdling of milk is denaturation of protein while alpha helix is a secondary structure of protein.





Explanation: Schottky and Frenkel defects are stoichiometric defects. Crystalline solids have long range order. FCC structure is made up of layers of octahedral, -type planes. These stack in a sequence ABC ABC. Metal excess defects due to anionic vacancies is called F-centres.

43. (D) C2H5–I Explanation: For the same alkyl group the boiling points of haloalkanes are in the order of RF < RCl< RBr < RI as with the increase in size of halogen atom the magnitude of van der Waals forces of attraction increases, resulting in higher boiling points.



50. (B) (i)-(c), (ii)-(a), (iii)-(d), (v)-(b)





Explanation: Interhalogen compounds are more reactive than all halogens. Halogens due to their tiny size and effective nuclear charge, are highly electronegative with low dissociation energies and high negative electron gain enthalpies. Therefore, they have a high tendency to gain an electron and act as strong oxidizing agents. Halogens absorb different quanta of radiations that lie in the visible region. This typically results from the excitation of outer electrons to higher energy levels, resulting in different colours.

SECTION-C



42. (B) Halogens are more reactive than inter halogens.



Explanation: (+-) Butan-2-ol is a racemic mixture so, it is optically inactive.

Explanation: NO2 is an electron withdrawing group, hence it increases the acidic character of phenol.

32 ]



Oswaal CBSE Chapterwise & Topicwise Question Bank, CHEMISTRY, Class – XII

55. (D)





Explanation:





SOLVED PAPER - 2021-22 (Term-I)

[ 33

C.B.S.E.

Topper's* Answers

2020

Class–XII

Chemistry

Delhi & Outside Delhi *Note : This paper is solely for reference purpose. The pattern of the paper has been changed for the academic year 2022-23.

MM: 70



Maximum Time: 3 hours

General Instructions: Read the following instructions very carefully and strictly follow them: (i) Question paper comprises four sections – A, B, C and D. (ii) There are 37 questions in the questions paper. All questions ae compulsory. (iii) Section – A : Q. No. 1 to 20 are very short answer type questions carrying one mark each. Answer these questions in one word or one sentence. (iv) Section B : Q. No. 21 to 27 are short answer type questions carrying two marks each. (v) Section – C : Q. No. 28 to 34 are long answer type-I questions carrying three marks each. (vi) Section – D : Q. No. 35 to 37 are long answer type-II questions carrying five marks each. (vii) There is NO overall choice in the question paper. However, an internal choice has been provided in 2 questions of two marks, 2 questions of three marks and all the 3 questions of five marks. You have to attempt only one of the choices in such questions. (viii) However, separate instructions are given with each section and question, wherever necessary. (ix) Use of calculators and log tables is NOT permitted.

SECTION - A

Read the given passage and answer the questions 1 to 5 that follow : The halogens have the smallest atomic radii in their respective periods. The atomic radius of fluorine is extremely small. All halogens exhibit – 1 oxidation state. They are strong oxidising agents and have maximum negative electron gain enthalpy. Among halogens, fluorine shows anomalous behaviour in many properties. For example electro negativity and ionisation enthalpy are higher for fluorine than expected whereas bond dissociation enthalpy, m.p. and b.p. and electron gain enthalpy are quite lower than expected. Halogens react with hydrogen to give hydrogen halides (HX) and combine amongst themselves to form a number of compounds of the type XX′, XX′3, XX′5 and XX′7 called inter-halogens.

1. Why halogens have maximum negative electron gain enthalpy ? Ans.



2.







TOPPER'S ANSWERS - 2020

Why fluorine shows anomalous behaviour as compared to other halogens ?

3.







Ans.

Arrange the hydrogen halides (HF to HI) in the decreasing order of their reducing character.





Ans.

4. Why fluorine is a stronger oxidizing agent than chlorine ? Ans.

5. What are the sizes of X and X′ in the interhalogen compounds ? Ans.

Question 6 to 10 are one word answers. 6. Name the cell used in hearing aids and watches. Ans.

7. How much charge in terms of Faraday is required to reduce one mol of MnO4– to Mn2+ ? Ans.

[ 35

36 ]



Oswaal CBSE Chapterwise & Topicwise Question Bank, CHEMISTRY, Class – XII

8. Write the slope value obtained in the plot of log [Ro]/ [R] Vs. time for a first order reaction. Ans.

9. Name the sweetening agent used in the cooking of sweets for a diabetic patient. Ans.



10. Name the polymer which is used for making electrical switches and combs. Ans.



Questions 11 to 15 are multiple choice questions. 11. In the Mond’s process the gas used for the refining of a metal is

(A) H 2

Ans.

(B) CO2

(C) CO

(D) N2



12. The conversion of an alkyl halide into an alcohol by aqueous NaOH is classified as

(A) a dehydrohalogenation reaction (C) an addition reaction

(B) a substitution reaction (D) a dehydration reaction

Ans.

13. CH3CONH2 on reaction with NaOH and Br2 in alcoholic medium gives (A) CH3CH2NH2 (B) CH3CH2Br (C) CH3NH2 (D) CH3COONa Ans.

14. The oxidation state of Ni in [Ni(CO)4] is (A) 0 (B) 2

(C) 3

(D) 4

Ans. 15. Amino acids are (A) acidic (B) basic (C) amphoteric (D) neutral Ans. Questions 16 to 20. (A) Both Assertion (A) and Reason (R) are correct statements and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are correct statements, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is correct, but Reason (R) is wrong statement. (D) Assertion (A) is wrong, but Reason (R) is correct statement. 16. Assertion (A) : Conductivity of an electrolyte increases with decrease in concentration. Reason (R) : Number of ions per unit volume decreases on dilution. Ans.

17. Assertion (A) : The C—O—C bond angle in ethers is slightly less than tetrahedral angle. Reason (R) : Due to the repulsive interaction between the two alkyl groups in ethers.

Ans.

18. Assertion (A) : Low spin tetrahedral complexes are rarely observed. Reason (R) : Crystal field splitting energy is less than pairing energy for tetrahedral complexes.

[ 37



TOPPER'S ANSWERS - 2020

Ans.



Assertion (A) : Elevation in boiling point is a colligative property. Reason (R) : Elevation in boiling point is directly proportional to molarity.



19.







Ans.



20. Assertion (A) : Oxidation of ketones is easier than aldehydes. Reason (R) : C—C bond of ketones is stronger than C—H bond of aldehydes.

Ans.

20 × 1 = 20



State Raoult’s law for a solution containing volatile components. What is the similarity between Raoult’s law and Henry’s law ? 2

21.





SECTION - B



22.







Ans.

Ans.





Write the role of (a) Dilute NaCN in the extraction of Gold. (b) CO in the extraction of Iron.

1+1=2 OR How is leaching carried out in the case of low grade copper ores ? Name the method used for refining of copper metal. 2

38 ]



Oswaal CBSE Chapterwise & Topicwise Question Bank, CHEMISTRY, Class – XII

23. Define adsorption with an example. What is the role of adsorption in heterogeneous catalysis ? OR

2

Define Brownian movement. What is the cause of Brownian movement in colloidal particles ? How is it responsible for the stability of Colloidal Sol ? 2

Ans.

24. (a) Write the IUPAC name and hybridisation of the complex [Fe(CN)6]3–.





(b) What is the difference between an ambidentate ligand and a chelating ligand ?

Ans.







(Given : Atomic number of Fe = 26) 1+1=2

[ 39

How do antiseptics differ from disinfectants ? Name a substance which can be used as a disinfectant as well as

26.





Ans.

2



an antiseptic.



25.







TOPPER'S ANSWERS - 2020

Identify the monomers in the following polymers :

(i)

Ans.



27.





Ans.

1+1=2





(ii)







Draw the structures of the following : (i) H2S2O8

(ii) XeF6

1+1=2

40 ]



Oswaal CBSE Chapterwise & Topicwise Question Bank, CHEMISTRY, Class – XII



A 0.01 m aqueous solution of AlCl3 freezes at – 0.068 °C. Calculate the percentage of dissociation. [Given : Kf for Water = 1.86 K kg mol–1] 3

28.





SECTION - C





Ans.

29. When a steady current of 2A was passed through two electrolytic cells A and B containing electrolytes ZnSO4 and CuSO4 connected in series, 2 g of Cu were deposited at the cathode of cell B. How long did the current flow ? What mass of Zn was deposited at cathode of cell A ? Ans.

[Atomic mass : Cu = 63.5 g mol–1, Zn = 65 g mol–1; 1F = 96500 C mol–1]

3

[ 41





TOPPER'S ANSWERS - 2020

30. Differentiate between following :

(i) Amylose and Amylopectin



(ii) Globular protein and Fibrous protein



(iii) Nucleotide and Nucleoside

1+1+1=3

Ans.







31. Identify A, B, C, D, E and F in the following :

6×½=3

42 ]



Oswaal CBSE Chapterwise & Topicwise Question Bank, CHEMISTRY, Class – XII

Ans.



32.

Give the structures of final products expected from the following reactions : (i) Hydroboration of propene followed by oxidation with H2O2 in alkaline medium.





(ii) Dehydration of (CH3)3 C—OH by heating it with 20% H3PO4 at 358 K.



(iii)







3×1=3

OR

How can you convert the following ?



(i) Phenol to o-hydroxybenzaldehyde.



(ii) Methanal to ethanol.



(iii) Phenol to phenyl ethanoate.

1+1+1=3

Ans.



Ans.





33.

Give reasons : (i) Aniline does not undergo Friedal-Crafts reaction. (ii) Aromatic primary amines cannot be prepared by Gabriel’s phthalimide synthesis. (iii) Aliphatic amines are stronger bases than ammonia.

3×1=3





TOPPER'S ANSWERS - 2020









[ 43

44 ]



Oswaal CBSE Chapterwise & Topicwise Question Bank, CHEMISTRY, Class – XII

34. Write three differences between lyophobic sol and lyophilic Ans. OR Define the following terms : (i) Protective colloid (ii) Zeta potential (iii) Emulsifying agent

3

1+1+1=3

Ans.

Ans.

(a) (b)

Give reasons : (i) Transition metals and their compounds show catalytic activities. (ii) Separation of a mixture of Lanthanoid elements is difficult. (iii) Zn, Cd and Hg are soft and have low melting point. Write the preparation of the following : (i) Na2Cr2O7 from Na2CrO4 (ii) K2MnO4 from MnO2 OR (a) Account for the following : (i) Ti3+ is coloured whereas Sc3+ is colourless in aqueous solution. (ii) Cr2+ is a strong reducing agent. (b) Write two similarties between chemistry of lanthanoids and actinoids. (c) Complete the following ionic equation : 3 MnO42– + 4H+ —→



35.





SECTION - D

3+2=5

2+2+1=5

[ 45





(a) Write the products formed when benzaldehyde reacts with the following reagents :



36.









TOPPER'S ANSWERS - 2020



(i) CH3CHO in presence of dilute NaOH



(ii)



(iii) Conc. NaOH



(b) Distinguish between following :





(i) CH3—CH = CH—CO—CH3 and CH3—CH2—CO—CH = CH2





(ii) Benzaldehyde and Benzoic acid.

3 + (1 + 1) = 5 OR



(a) Write the final products in the following :



(ii)



(iii) CH2 = CH—CH2—CN







(i)

(a) DIBAL − H (b) H 3O+

(b) Arrange the following in the increasing order of their reactivity towards nucleophilic addition reaction :









Ans.

(c) Draw the structure of 2, 4 DNP derivative of acetaldehyde.

3+1+1=5

46 ]



Oswaal CBSE Chapterwise & Topicwise Question Bank, CHEMISTRY, Class – XII





37. (a) A first order reaction is 25% complete in 40 minutes. Calculate the value of rate constant. In what time will the reaction be 80% completed ?

(b) Define order of reaction. Write the condition under which a bimolecular reaction follows first order kinetics. 3+2=5 OR



(a) A first order reaction is 50% complete in 30 minutes at 300 K and in 10 minutes at 320 K. Calculate activation energy (Ea) for the reaction.







(b) Write the two conditions for collisions to be effective collisions.



(c) How order of reaction and molecularity differ towards a complex reaction ?





Ans.

(R = 8.314 J K–1 mol–1)

[Given : log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021, log 5 = 0.6991]

3+1+1=5





TOPPER'S ANSWERS - 2020





[ 47

48 ]



Oswaal CBSE Chapterwise & Topicwise Question Bank, CHEMISTRY, Class – XII









Metal Excess: Due to anionic vacanci es (LiCl) and presence of extra cations (ZnO)

olids

Co va

ic

n efi

id s

Magnetic Properties

Metal Deficiency: Metal shows variable vacancy (Fe 2+, Fe3+)

Im pu ri t

yf

or

ei g

Po in At t de a p fec t oi nt s

Imperfections

The Solid State

n

nd uc to r

Molecules held by dipole-dipole interactions  (HCl, SO 2)

Atoms /molecules held by weak dispersion forces /London forces  (Ar, He)

O O O O

O O O O

O B O O

O O O O

disturb

re

iom etr y

Schottky: Equal number of ions missing (NaCl, KCl)

First Level

Second Level

Trace the Mind Map Third Level

Interstitial : Particles occupy interstitial site (non -ionic solids)

Vacancy : Lattice sites vacant (non-ionic solids)

Frenkel :Smaller ion dislocated to interstitial site (ZnS, AgCl)

Stoic h

Ferromagnetic: domains unequal  (Fe3O4, MgFe2O4)

Antiferromagnetic: domain opposite and equal  (MnO)

Ferromagnetic: strongly attracted  domains in same direction  (Fe, Co)

Diamagnetic: weakly repelled  Paired electrons  (H2O, NaCl)

Paramagnetic: weakly attracted  Unpaired electrons  (O2, Cu2+)

p-type (positive charge)

(Si, Ge)  Very large gap between conduction and valence band

(Polyethylene, clay)  Small energy gap between conduction and valence band

conduction and valence band

(Cu, Al)  Overlapping / very small gap between

Molecules held by hydrogen bonding  (H2O(ice)

r

n-type (negative charge) O O O O O O As O O O O O O O O O

Semico

pe sha ite

so l

lids

Electrical Properties

lw

t so

d ith

len

sa om at

Six spheres at vertices of octahedron  0.414 – 0.732

Trigo nal

talli cs

:

Four spheres at the vertices of tetrahedron  0.225 – 0.414

Three spheres in contact  0.155 – 0.225

Voids : Empty space between spheres

Packing Efficiency

Crystal Lattice

Smallest Unit

 ccp/fjcc  ABCABC….type  74%  (Cu, Ag)  bcc  square close packing  68%  (Li, Na)

 hcp  ABABAB…..type  74%  (Mg,Zn)

 scp  AAA….type  52.4%  (Po)

Body centred: One particle at its body centre and at its cor n. Face centred: One particle at centre of each face and at its corner. End centred: One particle at centre of any two opposite faces and at its corner.

Me

n Io

Triclinic : (K2Cr 2O 7, H3BO 3)   b  c      90°; primitive

zM Density = 3 a NA

Metal atoms held by metallic bond  (Fe, Cu)

s 

Non-metals held by covalent bond  (SiC, C)

Molecules held by intermolecular forces

ed 

Ions held by strong coulombic forces  (NaCl, MgO) d bon

of sm

ge n Hydro

a l l cry s ta

la Po 

Cubic : (NaCl, Cu) a=b=c; primitive, bc, fc; === 90° Tetragonal : (SnO 2, T iO2) a=b  c; === 90°; primitive, BCT Orthorhombic :  (KNO 3, BaSO 4) a  b  c === 90°; primitive, BCO, FCO, Hexagonal :  (ZnO, CdS) a=b  c; == 90° y =120°; primitive Rhombohedral or Trigonal : (CaCO 3, HgS) a=b=c; ==90°;primitive Monoclinic : (Monoclinic sulphur, Na2SO4.10H2O) == 90°   90°; primitive, ec

THE SOLID STATE

[ 1

1

CHAPTER @PROCBSE JOIN NOW

Syllabus

THE SOLID STATE

Classification of solids based on different binding forces : molecular, ionic, covalent and metallic solids, amorphous and crystalline solids (elementary idea). Unit cell in two dimensional and three dimensional lattices, calculation of density of unit cell, packing in solids, packing efficiency, voids, number of atoms per unit cell in a cubic unit cell, point defects. Electrical and magnetic properties. Band theory of metals, conductors, semiconductors and insulators and n and p type semi conductors.

Trend Analysis List of Concepts Numericals related to density of solids, unit cells Defects in solids, properties of solids

2018 D/OD 1Q (3 Marks) 1Q (1 Mark)

2019 D 1Q (3 Marks) 1Q (1 Mark)

OD 1Q (3 Marks)

2020 D/OD Was not in Syllabus Was not in Syllabus

TOPIC-1

Classification of Solids, Unit Cells and Packing in Solids

Revision Notes  Solids

 Solids are chemical substances which are characterised by definite mass, shape and volume, rigidity, high density, low compressibility.

TOPIC - 1 Classification of Solids, Unit Cells and Packing in Solids .... P. 2

TOPIC - 2   The constituent particles (atoms, molecules or ions) are closely packed and Voids, Packing Efficiency, held together by strong intermolecular forces. Calculations Related to Unit  General characteristics of solids are given below : Cell Dimensions .... P. 9 (i) Solids have definite mass, shape and volume. (ii) The intermolecular distance is minimum in solids and intermolecular forces TOPIC - 3 are strong. Defects in Solids, Electrical and Magnetic Properties, (iii) The constituent particles i.e., atoms, molecules or ions have fixed positions. Band Theory of Metals (iv) Solids cannot be compressed except foam, rubber, sponge, etc.  .... P. 18 (v) Most of them have high melting and boiling point. (vi) Solids are rigid.  Types of solids : There are two types of solids. (a) Crystalline solids : The solids in which the constituent particles have an Scan to know more about ordered arrangement (long range order) are crystalline solids. For example, this topic Sodium chloride, Diamond, Iodine, etc.  Characteristics of Crystalline solids (i) Constituent particles are regularly arranged. (ii) They possess strong melting point. Amorphous and (iii) Their outer surface also show a regular arrangement during the formation of crystals. crystalline solids

[ 3

THE SOLID STATE

(iv) These are anisotropic in nature.

(v) Crystalline solids have a definite geometric shape with flat faces and sharp edges.



(vi) Crystalline solids have long range order of arrangement of constituent particles.



 Classification of Crystalline solids : On the basis of binding forces, crystalline solids are classified as follows : (i) Molecular solids (ii) Ionic solids (iii) Metallic solids (iv) Covalent solids (b) Amorphous solids : The solids in which constituent particles do not have ordered arrangement (short range order) are amorphous solids. These have a range of melting point. For example, plastic, glass, etc. Characteristics of Amorphous solids : (i) Constituent particles are not arranged regularly and are irregular. (ii) These solids do not show sharp melting point. (iii) Outer surface does not show regular arrangement during the formation of crystals. (iv) These are isotropic in nature. (v) They have short range order of arrangement of constituent particles.  Isotropy and Anisotropy :  Solids whose value of any physical property would be same along any direction. This property is called isotropy.  Amorphous solids are isotropic in nature.   The solids whose some of the physical properties like electrical resistance or refractive index show different values when measured in different directions in the same crystal, this property is called anisotropy.  Crystalline solids are anisotropic in nature.

Fig. 1 : Two - dimensional representation of structure of crystalline solid (SiO2) Quartz

Fig. 2 : Two - dimensional representation of structure of amorphous solid (SiO2) Quartz Glass

 Different Types of Solids : S. No. 1.

Type of solids

Constituent Bonding/AtParticles tractive Forces

Examples

Physical

Electrical

Nature

Conductivity

Molecular solids (i) Non-polar Molecules

Dispersion or Ar, CCl4, H2, Soft London forces I2, CO2

Insulator

Very low

(ii) Polar

Dipole-dipole HCl, SO2 interactions

Soft

Insulator

Low

(iii) Hydrogen bonded

Hydrogen

Hard

Insulator

Low

H2O (ice)

bonding

2.

Ionic solids

Ions (positive and negative)

3.

Metallic solids

Positive ions Metallic bond- All metals are Hard but in a sea of ing alloys malleable delocalized and ductile electrons

4.

Melting Point

Covalent or Network solids

Atoms

Coulombic or NaCl, MgO, Hard but brittle Electrostatic ZnS, CaF2

Covalent bond- SiO2 (Quartz), Hard SiC, C (Diaing mond), AlN C (Graphite)

Soft

Insulators in solid High state but conductors in molten state and in aqueous solutions Conductors in solid state as well as in

Fairly high

molten state Insulators

Conductor (exception)

Very high

4 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



Z  Crystal lattice : The regular three dimensional arrangement of the constituent particles in a crystal in which each particle is represented by a point is called a crystal lattice and points are called lattice points. c  Unit cell : It is the smallest, fundamental repeated three  dimensional unit of a crystal lattice.  X a  Characteristics of unit cell : A unit cell is characterized by the b  following parameters : (i) Axial lengths a, b and c of the three edges along the three axis i.e., axial distances. Y (ii) Axial angles a, b and γ between the pairs of edges (b, c), Fig. 3 : Characteristic parameters of a unit cell (c, a) and (a, b) respectively. Thus, a unit cell is characterized by six parameters edges : a, b, c and parameters : a, b and γ. Scan to know  Types of unit cell : There are two types of unit cells namely, primitive unit cell and centred unit more about cell. this topic  Primitive unit cell : In the primitive unit cell, constituent particles are present only on the corner positions of a unit cell. There are seven types of primitive unit cells.  Centred unit cell or non-primitive unit cell : In this type of unit cell, particles (or points) are present not only at the corners but also at some other positions. Unit cell and These are of three types : Crystal lattices (i) Face-centred cubic (fcc) : Particles (or points) are located at the corners and also at the centre of each face. (ii) Body-centred cubic (bcc) : Particles (or points) are located at the corners and also at the centre within the body. (iii) End-centred cubic (ecc) : Particles (or points) are located at the corners and also at the centres of the two opposite end faces.

Primitive

Face-centred

Body-centred

End-centred

Fig. 4 : Different types of unit cells  Bravais lattices : These are the arrangement of lattice points in three dimensional space of crystal shown by relative distance and facial angles along the three axis. There are 14 Bravais lattices, as shown in table below : S. No.

Unit cell

1. 2. 3.

Primitive Body-centred Face-centred

4. 5.

Crystal system basic

Relative axial distances

Axial angles

Symmetry

Bravais lattice

Examples

Cubic

a=b=c

α = β = γ = 90°

9 planes, 13 axis

3

NaCl, KCl, ZnS, Diamond

Primitive Body-centred

Tetragonal

a=b≠c

α = β = γ = 90°

5 planes, 5 axis

2

6. 7. 8. 9. 10.

Primitive Body-centred Face-centred End-centred Primitive

Orthorhombic

a≠b≠c

α = β = γ = 90°

3 planes, 3 axis

4

TiO2, SnO2, PbSO4, NH4Br KNO3, BaSO4, K2SO4

Hexagonal

a=b≠c

ZnO, CdS

Primitive

a=b=c

1

NaNO3, HgS

12. 13.

Primitive End-centred

Trigonal or Rhombohedral Monoclinic

a≠b≠c

α = γ = 90° β ≠ 90°

7 planes, 7 axis 7 planes, 7 axis 7 planes, 10 axis

1

11.

α = β = 90° γ = 120° α = β = γ ≠ 90°

2

14.

Primitive

Triclinic

a≠b≠c

α ≠ β ≠ γ ≠ 90°

Monoclinic sulphur, PbCrO4 CuSO4.5H2O, K2Cr2O7

No planes, No axis

1

[ 5

THE SOLID STATE

 Number of atoms in a unit cell : Crystal lattice which includes number of unit cells and constituent particles are represented by lattice points. Number of atoms in a unit cell (Z) : for simple cubic Z = 1, for bcc Z = 2 for fcc Z = 4 and for ecc Z = 2. Number of atoms per unit cell Type of cell

Number of atoms at corner

Number of atoms at faces

Number of atoms at centre of cube

Total

Simple cubic Crystal (scc)

8´

1 =1 8

0

0

1

Body-centred cubic (bcc)

8´

1 =1 8

0

1

2

Face-centred cubic (fcc)

8´

1 =1 8

1 =3 2

0

4

6´

 Closed-packed structures : The constituent particles are closely packed in solids and there is minimum space between particles. These structures are called close packed structures.  Types of close packing : (a) Close packing in one dimension : When the spheres representing particles are touching each other in a row, it is called close packing. (b) Closed packing in two dimensions : This type of packing is obtained by placing the rows of close-packed spheres. This can be done in two ways Fig. 5 : Square close packing of : spheres in two dimensions (i) Square close packing : The particle in the adjacent rows may show a horizontal as well as vertical alignment forming squares. A central sphere is surrounded by four other spheres in two dimensions. (ii) Hexagonal close packing : The second row (particles) may be placed above the first row. It is repeated in the next row. Each sphere is in contact with six other spheres.

Fig. 6 : Hexagonal close packing of spheres in two dimensions Scan to know (c) Close-packing in three dimensions : All solids exist in three dimensional structures. These more about structures can be obtained by placing two dimensional layers one above other. They this topic can be of two types : (i) Square close packed layers : In this arrangement, spheres of both the layers are perfectly aligned horizontally as well vertically. If the arrangement of spheres at the first layer is called ‘A’ type. All the layers have the same arrangement. Thus, this lattice has AAA Number of Atoms type pattern. The co-ordination number is 6 in three dimensions. in a Unit cell (ii) Hexagonal close packed layers : It is more efficient and leaves less space unoccupied by spheres. The central sphere is in contact with six other spheres in two dimensions.  Co-ordination number : The number of closest neighbours of any constituent particle is called its co-ordination number. Co-ordination number of hcp and ccp is 12 while in bcc, it is 8.  Atomic radius : It is defined as half of the distance between neighbouring atom in a crystal. It Scan to know is expressed in terms of the edge (a) of unit cell of the crystal. more about a (i) Simple cubic structure (sc) : Radius of atom ‘r’ = , as atoms touch along the edges. this topic 2



(ii) Body-centred cubic structure (bcc) : Radius of atom ‘r’ =



(iii) Face centred cubic structure (fcc) : Radius of atom ‘r’ =

3 a. 4

a 2 2

.

Packing, closed packed structures, packing efficiency

6 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Mnemonics • Concept: Types of solid state. • Mnemonic: She ate cake. • Interpretation: She ® solid state Ate ® Amorphous solid Cate ® Crystalline solid • Concept: Primitive unit cell • Mnemonic: Class Test Of Hindi Takes More Time • Interpretation: *Cubic *Orthorhombic *Trigonal *Triclinic

*Tetragonal *Hexagonal *Mono Clinic

• Concept: Types of Crystalline solid • Mnemonic: Make a Cup of Ice-cream Milk • Interpretation: Metallic, Covalent, Ionic, Molecular Solid • Concept: Void • Mnemonic: Try To Offer • Interpretation: Trigonal Voids, Tetrahedral Voids, Octahedral Voids

• Concept: Centered • Mnemonic: Family Beats Enemy • Interpretation: Face, Body and End centered

[ 7

THE SOLID STATE

Know the Terms

 Order : It designates the presence or absence of some symmetry or correlation in a many-particle system.  Ionic crystal : A crystal structure that grows from chemical bonds between two oppositely charged atoms which are held together by electrostatic attraction.  Fluidity : The physical property of a substance that enables it to flow.

Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS: Q. 1. Which of the following is not true about the ionic solids? (a) Bigger ions form the close packed structure. (b) Smaller ions occupy either the tetrahedral or the octahedral voids depending upon their size. (c) Occupation of all the voids is not necessary. (d) The fraction of octahedral or tetrahedral voids occupied depends upon the radii of the ions occupying the voids. Ans. Correct option : (d) Explanation : In ionic solids, smaller ions occupy the voids, and this depends on stoichiometry of the compounds not on the radius of ions. Q. 2. Solid A is very hard electrical insulator in solid as well as in molten state and melts at an extremely high temperature. What type of solid is it? (a) Ionic solid (b) Molecular solid (c) Covalent solid (d) Metallic solid Ans. Correct option : (c) Explanation : Covalent solids are generally hard, act as insulators and melting points of such solids are extremely high. Q. 3. Which of the following is a network solid? (a) SO2 (solid) (b) I2 (c) Diamond (d) H2O (ice) Ans. Correct option : (c) Explanation : A network solid consists of a network atoms of same or different elements connected to each other by covalent bonds. So, diamond is a network solid as it contains network of carbon atoms . [B] ASSERTIONS AND REASONS: (a) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion. (b) Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion. (c) Assertion is correct, but reason is wrong statement. (d) Assertion is wrong, but reason is correct statement.

(1 mark each) Q. 1. Assertion : Most of the solids possess high melting point. Reason : They have strong intermolecular forces of attraction. Ans. Correct option : (a) Explanation : Most of the solids possess high melting point due to the presence of intermolecular forces of attraction between their particles. Q. 2. Assertion : Amorphous solids possess a longrange order in the arrangement of their particles. Reason : The formation of amorphous solids involves very rapid cooling. Ans. Correct option : (d) Explanation : Amorphous solids do not possess a long-range order in the arrangement of their particles because their formation involves rapid cooling. Q. 3. Assertion (A) : Crystalline solids are anisotropic in nature. Reason (R) : Some of their physical properties show same electrical and optical properties in different directions in the same crystal. Ans. Correct option : (c) Explanation : Crystalline solids are anisotropic in nature as some of their physical properties show different electrical and optical properties in different directions in the same crystal. [C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1. Define crystalline solids. Ans. The solids in which the constituent particles have an ordered arrangement are called crystalline solids.  Q. 2. Calculate number of atoms in simple cubic unit cell. 1 Ans. Number of atoms at corner = 8 ´ = 1 8 Number of atoms at faces = 0 Number of atoms at centre = 0 Total number of atoms = 1+0+0 = 1 Q. 3. What is meant by close packing in one dimension? Ans. When the spheres representing particles are touching each other in a row, it is called close packing in one dimension.

Short Answer Type Questions-I Q. 1. Classify the following as amorphous or crystalline solids : Polyurethane, naphthalene, benzoic acid, teflon, potassium nitrate, cellophane, polyvinyl chloride, fibre glass, copper.

(2 marks each)

Ans. Amorphous solids : Polyurethane, teflon, cellophane, polyvinyl chloride, fibre glass. Crystalline solids : Naphthalene, benzoic acid, potassium nitrate, copper.

8 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q. 2. Under which situations can an amorphous substance change to crystalline form? Ans. An amorphous solid on heating at same temperature may become crystalline. Slow heating and cooling over a long period makes an amorphous solid to have some crystalline character. Q. 3. Define the term ‘amorphous’. Give a few examples of amorphous solids.

Ans. In amorphous solids, constituent particles are of irregular shapes and have short-range order. These solids are isotropic in nature and melt over a range of temperature. Therefore, amorphous solids are sometimes called pseudo solids or super cooled liquids. They do not have definite heat of fusion. When cut with a sharp-edged tool, they cut into two pieces with irregular surfaces. Examples of amorphous solids include glass, rubber, and plastic.

Short Answer Type Questions-II Q. 1. Discuss the types of Unit cells. Ans. (i) Face-centred cubic (fcc) : Particles (or points) are located at the corners and also at the centre of each face. (ii) Body-centred cubic (bcc) : Particles (or points) are located at the corners and also at the centre within the body. (iii) End-centred cubic (ecc) : Particles (or points) are located at the corners and also at the centres of [1×3=3] the two opposite end faces. Q. 2. Give a brief account on the following : (i) Square close packing (ii) Hexagonal close packing Ans. (i) Square close packing : The particle in the adjacent rows may show a horizontal as well as vertical alignment forming squares. A central sphere is surrounded by four other spheres in two dimensions.



(3 marks each)

(ii) Hexagonal close packing : The second row (particles) may be placed above the first row. It is repeated in the next row. Each sphere is in contact with six other spheres.

Hexagonal close packing of spheres in two dimensions Q. 3. Give the general characteristics of solids. Ans. (i) Solids have definite mass, shape and volume.

(ii) The intermolecular distance is minimum in solids and intermolecular forces are strong. (iii) The constituent particles i.e., atoms, molecules or ions have fixed positions.

(iv) Solids cannot be compressed except foam, rubber, sponge etc. (v) Most of them have high melting and boiling point.

Square close packing of spheres in two dimensions

(vi) Solids are rigid.

Long Answer Type Questions Q. 1. Explain (a) The basis of similarities and differences between metallic and ionic crystals. (b) Ionic solids are hard and brittle. Ans. (a) Similarity between metallic and ionic crystals : The basis of similarities between metallic and ionic crystals are that both the crystals are held by the electrostatic forces of attraction. In metallic crystals, the electrostatic force acts between the positive ions and the electrons. In ionic crystals, it acts between the oppositely-charged ions. Hence, both have high melting points.  [1½] Differences between metallic and ionic crystals : The basis of differences between metallic and ionic crystals are that in metallic crystals, the electrons are free to move and so, metallic crystals can conduct electricity. However, in ionic crystals, the ions are not free to move. As a result, they cannot

(5 marks each)

conduct electricity. However, in molten state or in aqueous solution, they conduct electricity.[1½] (b) The constituent particles of ionic crystals are ions. These ions are held together in threedimensional arrangements by the electrostatic force of attraction. Since the electrostatic force of attraction is very strong, the charged ions are held in fixed positions. That is why ionic crystals are hard and brittle.[2] Q. 2. (a) Ionic solids conduct electricity in molten state but not in solid state. Explain. (b)  Classify the following solids in different categories based on the nature of intermolecular forces operating in them :  Potassium Sulphate, Tin, Benzene, Urea, Ammonia, Water, Zinc Sulphide, Graphite, Rubidium, Argon, Silicon Carbide.

[ 9

THE SOLID STATE

Ans. (a)  In ionic compounds, electricity is conducted through ions. In solid state, ions are held together by strong electrostatic forces and are not free to move about within the solid. Hence, ionic solids do not conduct electricity in solid state. However, in molten state or in solution form, the ions are free to move and conduct electricity.  [2] (b) Potassium sulphate (K2SO4) → Ionic solid Tin (Sn) → Metallic solid Benzene (C6H6) → Molecular solid (non-polar) Urea (NH2CONH2) → Polar molecular solid Ammonia (NH3) → Polar molecular solid Water (H2O) → Hydrogen bonded molecular solid Zinc sulphide (ZnS) → Ionic solid Graphite → Covalent or network solid Rubidium (Rb) → Metallic solid Argon (Ar) → Non-polar molecular solid Silicon carbide (SiC) → Covalent or network solid[3]

Commonly Made Error  Some students can not identify the class of solids correctly. Answering Tip  Learn and understand the attractive forces and properties of a particular solid. Q. 3. (a) Give characteristics of crystalline solids. (b) Give two examples of covalent solids. Ans. (a) (i) Constituent particles are regularly arranged.   (ii) They possess sharp melting point.  (iii) T heir outer surface also show a regular arrangement during the formation of crystals.   (iv) These are anisotropic in nature.     (v) Crystalline solids have a definite geometrical shape with flat faces and sharp edges.   (vi) Crystalline solids have long range order of arrangement of constituent particles. [3] (b) Quartz and Diamond

[2]

TOPIC-2 Voids, Packing Efficiency, Calculations Related to Unit Cell Dimensions

Revision Notes  Voids or holes :  The holes left in the close packing of spheres are called as interstitial sites or voids or holes.  Trigonal voids :  The void enclosed by three spheres in contact is called a trigonal void. Tetrahedral  There are 24 voids around each sphere. Void  There are eight trigonal voids per atom in a crystal. Fig. 7 : Tetrahedral void  Radius ratio is 0.155.  Tetrahedral voids :  The void surrounded by four sphere lying at the vertices of a regular tetrahedron is called tetrahedral void.  There are 8 tetrahedral voids around each sphere and two voids per atom of crystal.  So, number of tetrahedral voids = 2 × Number of close packed spheres = 2N.   The radius of a tetrahedral void in a closest packed arrangement is 22.5% of the sphere involved in this arrangement. r Thus, void ==0.0.225 255 rsphere 

Octahedral voids :  The void surrounded by six spheres lying at the vertices of a regular octahedron is called octahedral void.  There are 6 octahedral voids around such sphere.  There is one void per atom in a crystal. So, Number of octahedral voids = 1 × Number of close packed spheres = N

Fig. 8 : Octahedral void



10 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

The radius of an octahedral void in a close packed arrangement is 41.4% of the sphere involved in this arrangement. Thus, rvoid = 0.414 rsphere

 Packing efficiency :  It is the percentage of entire space which is covered by the particles.  Calculation of packing efficiency of various types of structures is as follows : (i) Packing efficiency of hcp and ccp structures : Both are equally efficient.

Packing efficiency of hcp or ccp =





Volume occupied by four spheres in the unit cell Total volum me of the unit cell ( a3 )

 

4 4 × πr 3 × 100 3 = = 74% 3  4r     2

× 100

  4r   for ccp, a =   2  

 4r  (ii) Packing efficiency of bcc structure : For bcc structure a is  · .  3





Packing efficiency of bcc =

Volume occupied by two spheres in the unit cell Total volumee of the unit cell ( a3 )

´ 100

4 2 × πr 3 × 100 3 = = 68% 3  4r     3

(iii) Packing efficiency of simple cubic lattice : It has less packing efficiency as compared to hcc and bcc. For simple cubic lattice a = 2r.



Packing efficiency of simple cubic lattice =

Volume of one atom or sphere present in the unit cell Total volume of unit cell

4 3 pr ´ 100 = 3 = 52.4% ( 2r )3

 Calculations of density involving unit cell dimensions :

Density of unit cell ‘d’ =

Mass of the unit cell Volume of the unit cell

Mass of the unit cell = Number of atoms in the unit cell × Mass of each atom = Z × m where, ‘Z’ is the number of atoms in one unit cell and ‘m’ is mass of each atom. Mass of each atom, m = m =

Atomic mass Avogadro's number M NA

Volume of cube = a3, where ‘a’ is the edge length of the cube Density of unit cell ‘d’ =

Scan to know more about this topic

M × Z N A × a3

It is also the relation between the density ‘d’ and the edge length ‘a’ of unit cell.

Calculation of density of unit cell

[ 11

THE SOLID STATE

Mnemonics • Concept: Imperfections • Mnemonic: Indian Friends are Very Special • Interpretation: Interstitial, Frenkel, Vacancy, Schottky

Know the Formulae 

Density of the unit cell (d) =

Z ×M Mass of the unit cell = N A × a3 Volume of the unit cell

 Various parameters of cubic system : Unit cell

No. of atoms per unit cell

Distance between nearest neighbour (d)

Coordination Number

Radius (r)

Simple cubic

1

a

6

a 2

Face-centred cubic

4

2

12

2 2

Body-centred cubic

2

3 a 2

8

3 a 4



a

Packing efficiency =

a

Volume occupied by atoms in unit cell × 100 Total volume of the un nit cell

 Packing efficiency of different crystals : Crystal system

Packing efficiency

Simple cubic Body-centred cubic Face-centred cubic Hexagonal close-packed

52.4% 68% 74% 74%

Radius of the cation r+ = − Radius of the anion r  Structural arrangement of different radius ratio of ionic solids : 

Radius ratio =

 r+   r −

Radius ratio  

Possible C.N.

Structural arrangement

Examples

0.155 – 0.225

3

Trigonal planar

B2O3

0.225 – 0.414

4

Tetrahedral

ZnS, SiO44–

0.414 – 0.732

6

Octahedral

NaCl

0.732 – 1

8

Body-centred cubic

CsCl

12 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q. An element crystallizes in a bcc lattice with cell edge of 500 pm. The density of the element is 7.5 g cm–3. How many atoms are present in 300g of the element? Solution: STEP I: Z = 2 STEP II:

d=

Z ´M a3 ´ NA

STEP III:

NA =

Z ´M d ´ a3

STEP IV:

NA =

2 ´ 300 [7.5 ´ (5 ´ 10-8 cm)3 ]

STEP V: = 6.4 × 1023 atoms

d=              

Z ´M a3 ´ NA

Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS : Q. 1. A compound is formed by two elements M and N. The element N forms ccp lattice and atoms of M 1 rd of tetrahedral occupy two atoms an Mercury 3 voids. What is the formula of the compound (a) MN2 (b) M2N3 (c) M3N2 (d) M2N2 Ans. Correct option : (b) Explanation : Suppose the atoms N in the ccp = a ∴ No. of tetrahedral voids = 3a 2a No. of atoms M = :a=2:3 3 Hence compound as M2 : N3 Q. 2. Silver crystallises in for Lattice. It edge length of the unit cells is 4.07 × 10–8 cm density and is 10.5g cm–3. Calculate the atomic mass of silver. (a) 144 g/mol (b) 106.8 g/mol (c) 106.6 g/mol (d) 213 g/mol Ans. Correct option : (c) Explanation : Z´M d= 3 a ´ NA M=

d ´ N A ´ a3 Z

=





(

10.5 ´ 6.022 ´ 10 ´ 4.07 ´ 10 23

= 106.6 g mol -1

4

)

-8 3

[½] 

(1 mark each) Q. 3. The correct order of the packing efficiency in different types of unit cells is ________. (a) fcc < bcc < simple cubic (b) fcc > bcc > simple cubic (c) fcc < bcc > simple cubic (d) bcc < fcc = simple cubic Ans. Correct option : (b) Explanation : The correct order of the packing efficiency in different types of unit cells is given below : Unit Cell

Packing efficiency

fcc

74%

bcc

68%

Simple cubic 52.4% fcc > bcc > simple cubic Q. 4. The total number of tetrahedral voids in the face centred unit cell is _____. (a) 6 (b) 8 (c) 10 (d) 12 Ans. Correct option : (b) Explanation : The total number of tetrahedral voids in the face centred unit cell is 8.

[B] ASSERTIONS AND REASONS : g cm

-3



(a)

Both assertion and reason are correct statements, and reason is the correct explanation of the assertion.

[ 13

THE SOLID STATE

(b)



Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion. Assertion is correct, but reason is wrong statement.

(c)

(d)

Assertion is wrong, but reason is correct statement.

Q. 1. Assertion (A) : Total number of octahedral voids present in unit cell of cubic packing including the one that is present at the body centre is four.

R  eason (R) : Besides the body centre these is one octahedral void present at the centre of each of the 4 faces of the unit cell and each of which is shared between two adjacent unit cells. Ans. Correct option : (c) Explanation : Total number of octahedral voids present in unit cell of cubic packing including the one that is present at the body centre, is four. Q. 2. Assertion (A) : A two dimensional arrangement where each sphere is in contact with four of its neighbour and has a 2-D coordination number 4 is square close packing arrangement. Reason (R) : In such arrangement, if centres of the neighbouring spheres are joined, a square is formed. Ans. Correct option : (a) Explanation : In square close packing arrangement in two dimension, each sphere is in contact with 4 spheres and if the centres of neighbouring spheres are joined a square is formed. Q. 3. Assertion : The packing efficiency of simple cubic lattice is 52.4%. Reason : The number of atoms per unit cell is 2.

Ans. Correct option : (c) Explanation : For simple cubic lattice a = 2r Number of atoms per unit cell = 1 [C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1. What is the formula of a compound in which the element P forms ccp lattice and atoms of occupy 2/3rd of tetrahedral voids?  U [CBSE Foreign Set 1 2017] Ans. P3Q4

[CBSE Marking Scheme, 2017] Q. 2. What is the formula of a compound in which the element P forms hcp lattice and atoms of Q occupy 2/3rd of octahedral voids?  U [CBSE Foreign Set 2 2017] Ans. P3Q2

[CBSE Marking Scheme, 2017] Q. 3. What is the formula of a compound in which the element P forms ccp lattice and atoms of Q occupy 1/3rd of tetrahedral voids?  U [CBSE Foreign Set 3 2017] Ans. P3Q2

[CBSE Marking Scheme, 2017] Q. 4. A metallic element crystallises into a lattice having a ABC ABC … pattern and packing of spheres leaves out voids in the lattice. What type of structure is formed by this arrangement.  U [CBSE Comptt Set 2 2017] Ans. ccp/fcc

[CBSE Marking Scheme, 2017] Q. 5. A metallic element crystallises into a lattice having a pattern of AB AB … and packing of spheres leaves out voids in the lattice. What type of structure is formed by this arrangement?  U [CBSE Comptt Delhi Set 1 2017] Ans. hcp

[CBSE Marking Scheme, 2017]

Short Answer Type Questions-I Q. 1. Atoms of a element P form ccp lattice and those of the element Q occupy 1/3rd of tetrahedral voids and all octahedral voids. What is the formula of the compound formed by the elements P and Q? [CBSE, SQP, 2020-21] Ans.





Let no. of Atoms of elements P be x No. of tetrahedral voids = 2x[½] No. of octahedral voids = x Atoms of Q = 1/3 (2x) + x = 5x + 3[½] P3Q2x[1] P3Q2 [CBSE SQP Marking Scheme 2020]

Q. 2. An element with density 11.2 g cm–3 forms a fcc lattice with edge length of 4 × 10–8 cm. Calculate the atomic mass of the element. (Given : NA = 6.022 × 1023 mol–1) U [CBSE Delhi 2014]

(2 marks each)

Ans. d = 11.2 g/cm3, Z = 4, a = 4 × 10–8 cm Z´M ⇒ d = N A ´ a3 ⇒

11.2 =

4´M 6.022 ´ 10 23 ´ ( 4 ´ 10 -8 )3

[½]

23 -8 -8 -8 ⇒ M = 11.2 ´ 6.022 ´ 10 ´ 4 ´ 10 ´ 4 ´ 10 ´ 4 ´ 10 4 [1]

⇒

M = 11.2 × 6.022 × 16 × 10–1 ∴ M = 107.9 g mol–1 or 107.9 u[½] [CBSE Marking Scheme 2014]

Commonly Made Error  Sometimes, students are unable to find the atomic mass of element for a crystal lattice.

14 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

d × N A × a3 Z 7.2 × 6.022 × 10 23 × ( 288 × 10 −10 )3 = g/mol 2

Answering Tip  Students must remember the formula for calculation and also the number of atoms per unit cell. Q. 3. An element ‘X’ (At. mass= 40 g mol-1) having f.c.c structure, has unit cell edge length of 400 pm. Calculate the density of ‘X’ and the number of unit cells in 4 g of ‘X’. (NA = 6.022 × 1023 mol-1)3  U [CBSE, Delhi Set 1, 2018] Ans.    d =







=

ZM [½] a3 N A

(

4 ´ 40

)

-8 3

23

[½]

´ 6.022 ´ 10 4 ´ 10  = 4.15 g/cm3[½] No. of unit cells = total no. of atoms /4[½]

é4 ù = ê ´ 6.022 ´ 1023 ú / 4 [½] 40 û ë = 1.5 ´ 1022 [½]

 

(or any other correct method) [CBSE Marking Scheme, 2018] –

Q. 4. An element ‘X’ (At. mass = 40 g mol 1) having fcc structure, has unit cell edge length of 400 pm. Calculate the density of ‘X’ and the number of unit – cells in 4 g of ‘X’. (NA = 6.022 × 1023 mol 1) Ans.    d =

ZM [½] a3 N A

M=

[1] = 51.7869 g/mol Q. 6. An element crystallises in fcc lattice with cell edge of 400 pm. Calculate its density if 250 g of this element contain 2.5 × 1024 atoms. [CBSE Comptt. Delhi Set 2, 3 2017] Ans. In fcc,

Z = 4;        d = (Z × M)/a3 × NA   



No. of atoms =



24   2.5 × 10 =

(i) [½]

w × NA M

250 g × NA  M M = [250 × NA]/2.5 × 1024



[1] (ii)

Putting values of M in equation (i) d = 4 × 250 g × NA /[2.5 × 1024 atoms × (400 × 10-10 cm3) × NA][½] d = 6.25 g/cm3 (or any other correct method)[1] [CBSE Marking Scheme, 2017]



Q. 7. An element crystallises in fcc lattice with cell edge of 400 pm. Calculate its density if 250 g of this element contain 2.5 × 1024 atoms. [CBSE Comptt. Delhi Set 2, 3 2017] Ans. In fcc,

Z = 4;

= [½] -8 3 ´ 6.022 ´ 10 23 4 ´ 10  = 4.15 g/cm3[½] No. of unit cells = total no. of atoms /4[½] é4 ù = ê ´ 6.022 ´ 10 23 ú / 4 [½] 40 ë û



 d = (Z × M)/a3 × NA   

= 1.5 ´ 1022 [½] (Or any other correct method) [CBSE Marking Scheme, 2018]



Q. 5. An element exists in bcc lattice with a cell edge of 288 pm. Calculate its molar mass if its density is 7.2 g/cm3. A [CBSE Comptt. OD Set 1, 2, 3 2017]



Ans. In bcc, Z = 2; d = (Z × M)/a3 × NA

Q. 8. An element crystallizes in a fcc lattice with cell edge of 400 pm. The density of the element is 7 g cm-3. How many atoms are present in 280 g of the element?  [CBSE OD Set 2 2016]





 

(

4 ´ 40

)

(i)



Putting values of M in equation (i)

 

M = 7.2g/cm3 × (288 × 10-10 cm)3 NA/2

[1] [1]

= 51.8 g/mol (or any other correct method)[1]

[CBSE Marking Scheme 2017] Detailed Answer : a = 288 pm = 288 × 10-10 cm d = 7.2 g/cm3 In bcc, Z = 2 NA = 6.022 × 1023 Z×M d= [1] N A × a3   

No. of atoms =



 

2.5 × 10 24 =

(i) [½]

w × NA M

250 g × NA M 

[1]

M = [250 × NA]/2.5 × 1024 (ii) Putting values of M in equation (i)  d = 4 × 250 g × NA /[2.5 × 1024 atoms × (400 × 10-10 cm3) × NA] [½]  d = 6.25 g/cm3 (or any other correct method) [1] [CBSE Marking Scheme, 2017]

Ans. Volume of the unit cell a3 = (400 pm)3 = (4 × 10–8 cm)3 = 64 × 10–24 cm3







Volume of 280 g of the element =

mass density

=

280 cm3= 40 cm3 7

Number of unit cells in this volume

[ 15

THE SOLID STATE



=

40 64 ×10

-24

Commonly Made Error

= 6.25 × 1023 unit cells

 Some students get confused to calculate the total number of atoms in given mass of an element.



Since = 24 Therefore, total no. of atoms in 280 g = 4 ×6.25 ×1023 = 2.5 ×104 atoms   (or any other correct method) [CBSE Marking Scheme 2016]

Answering Tip  Be precise in answer and illustrate all the steps for the answer. OR

( 300 × 6.022 × 10 23 ) 2 × 10 24 [Topper’s Answer 2016] = 90.3 g Z×M d= 3 a NA 4 × 90.3 = ( 250 × 10 −10 )3 × N 0



=



Q. 9. An element crystallizes in a fcc lattice with cell edge of 250 pm. Calculate the density if 300g of this element contains 2 × 1024 atoms.  U [CBSE, Delhi Set 1,2016] Ans. 2 × 1024 atoms weigh = 300g 6.022 × 1023 atoms weigh ( 300 × 6.022 × 10 23 ) = 2 × 10 24   = 90.3 g Z×M d= 3 a NA 4 × 90.3 = ( 250 × 10 −10 )3 × N 0 Q. 1. An element crystallizes bcc−3 lattice with cell = 38.in 4 gacm edge of 500 pm. The density of the element is 7.5 g cm–3. How many atoms are present in 300 g of the element ? A [CBSE OD Set 1, 2016]

= 38.4 g cm −3



(or any other correct method) 1 [CBSE Marking Scheme 2016]

Short Answer Type Questions-II Ans.

(3 marks each) Z = 2 d =

Z×M a3 × N A



16 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



NA =



NA =

Z×M d×a

2 × 300 [7.5 × ( 5 × 10 −8 cm )3 ]

 = 6.4 × 1023 atoms Z×M d = 3 a × NA  OR Z×M 7.5 = ( 500 )3 × 10 −30 × 6.022 × 10 23





[½]



 Z = 2 Z×M d = 3 a × NA



NA =



NA =





Z×M d × a3

6.022 × 10 23 × 300 = 6.4 ×1023 atoms 282.3 [3] [CBSE Marking Scheme 2016]

Q. 3.



Silver crystallizes in fcc lattice. If edge length of the unit cell is 4.07 × 10–8 cm and density is 10.5 g cm–3, calculate the atomic mass of silver.

A [CBSE Comptt. OD 2016; KVS TBQ, NCERT] Ans.

d =

Z×M a3 × N A

d × N A × a3 M = [1] Z



10.5 × 6.022 × 10 23 × ( 4.07 × 10 −8 )3 g cm −3 [1] 4  = 106.6 g mol–1 [1] [CBSE Marking Scheme 2016]



=

Q. 4. An element with molar mass 27 g mol–1 forms a cubic unit cell with edge length 4.05 × 10–8 cm. If its density is 2.7 g cm–3, what is the nature of A [CBSE Delhi 2015] the cubic unit cell ? Ans.

 2 × 300

[7.5 × ( 5 × 10 −8 cm )3 ]

= 6.4 × 1023 atoms Z×M d= 3 a × NA              OR Z×M 7.5 = ( 500 )3 × 10 −30 × 6.022 × 10 23      

  300 g =



7.5 × 125 × 10 −24 × 6.022 × 10 23 M = 2 = 282.3 g/mol 282.3 g = 6.022 × 1023 atoms 6.022 × 10 23 × 300 300 g = 282.3     = 6.4 ×1023 atoms[3] [CBSE Marking Scheme 2016] Q. 2. An element crystallizes in a bcc lattice with cell edge of 500 pm. The density of the element is 7.5 g cm–3. How many atoms are present in 300 g of the element ? A [CBSE OD Set 1, 2016] Ans.

7.5 × 125 × 10 −24 × 6.022 × 10 23       M =  2    = 282.3 g/mol 282.3 g = 6.022 × 1023 atoms

  

3





Z ×M a3 × N A d a3 N0 Z= . M         

d=

[½]

2.7 g cm −3 × 6.022 × 10 23 mol −1

[3]

× ( 4.05 × 10 −8 cm )3  [1] Z=   M = 3.999 ≈ 4  [½] Face centred cubic cell / fcc[1]  [CBSE Marking Scheme, 2015] 1

OR





[Topper’s Answer 2015]

[ 17

THE SOLID STATE

Q.5.

Silver crystallizes in fcc lattice. If edge length of the unit cell is 4.077 × 10–8 cm, then calculate the radius of silver atom.

A [CBSE Comptt. OD 2015]

 Ans.

r =

2. a 4

1.414 × 4.077 × 10 −8 cm 4 = 1.44 × 10–8 cm [CBSE Marking Scheme, 2015] [3]

⇒ ∴ 

Commonly Made Error  Some students get confused to calculate radius of an atom for the lattice. Answering Tip  Remember the formula for radius of a particular lattice.

=

Long Answer Type Questions Q. 1. (i) Calculate packing efficiency in ccp structure. (ii) Express the relationship between atomic radius (r) and edge length (a) in the bcc unit cell.  R+ U (iii) A  metal crystallises into two cube system face centred cubic (fcc) and body centred cubic (bcc) whose unit cell length are 3.5 and 30 A respectively. Calculate the ratio of densities of fcc and bcc. U [CBSE, SQP, 2020-21] Ans. (i) For ccp structure 4r a = 2 2 r or r = a = [½] 2 2 2 Packing efficiency Volume occupied by four spheres in the [½] unit cell × 100% = Total volume of the unit cell 4 4 ´ pr 3 ´ 100 [½] 3 = % ( 2 2r )3 16 3 pr ´ 100 = 3 % = 74% [½] 16 2r 3 3a [1] (ii) r =

2

(iii) Since,   d = ZMN0a3 For fcc,   Z = 4 Therefore d = 4 ´ M / Na (3.5 ´ 10 2 )3 g / cm 3 

[½] [1]

For bcc, Z = 2 Therefore 2 3 3   d = 2 ´ M / Na (3.0 ´ 10 ) g / cm 

[1] d = 4(3.5 ´ 102 )3 / 2(3.0 ´ 102 )3 = 317.1 [1] [CBSE SQP Marking Scheme 2020]

Q. 2. (i) Silver metal crystallizes with a face-centred cubic lattice. The length of unit cell is found to be 4.077 × 10–8 cm. Calculate atomic radius and density of silver. (Atomic mass of Ag = 108 u, NA = 6.02 × 1023 mol–1) A

(5 marks each)



(ii) An element occurs in the bcc structure with cell edge of 288 pm. The density of the element is 7.2 g cm–3. How many atoms of the elements does 208 g of the element contain ? Ans. (i) G  iven : a = 4.077 × 10–8 cm, Z = 4, M = 108 g mol–1 NA = 6.023 × 1023 [½] Z´M d = 3 a ´ NA

=

( 4.077 ´ 10 ) ´ 6.023 ´ 10 23

= 10.58 g/cm3



4 ´ 108 -8 3

For fcc,

r =

a

=

[½] -8

4.077 ´ 10 cm

[½] 2 2 = 1.44 × [½]   10–8 cm. (ii) For the bcc structure, Z = 2 Z´M Density     d = 3 [½] a ´ NA 7.2 cm–3 =

2 2

( 288 × 10

−10

2×M

cm ) × ( 6.022 × 10 23 mol −1 ) 3

[½] or M = 51.8 g mol–1 [½] By mole concept, 51.8 g of the element contains 6.022 × 1023 atoms [½] 208 g of the element will contain 6.022 ´ 10 23 ´ 208 atoms [½] 51.8 = 24.17 × 1023 atoms. [½] Q. 3. (i)  What is the formula of a compound in which element B forms ccp lattice and atoms A surrounds 2/3 part of tetrahedral voids? (ii) The formula of a crystalline solid is XY2O4 in which oxide ion forms ccp lattice and cation X occupies in all tetrahedral voids and cation Y occupies in octahedral voids, state that (a)  Calculate % part of cation X occupied in tetrahedral voids. (b)  Calculate % part of cation Y occupied in octahedral voids. U

18 ] ns. A



Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(i) In ccp lattice, Number of molecules  = N Number of tetrahedral voids = 2N Number of atoms of element 2 A = ´ 2N 3 4N = 3  

Formula of compound = A4B3 [2] (ii) In ccp lattice, Each oxide ion has two tetrahedral voids and one octahedral void.





















Therefore, for 4 oxide ions, Number of tetrahedral voids = 8 Number of octahedral voids = 4 1 tetrahedral void out of 8 occupied by ion X and 2 octahedral voids out of 4 occupied by ion Y.[1] Hence, % of tetrahedral void occupied by 1 X = ´ 100 = 12.5% [1] 8 % of tetrahedral void occupied by 2 Y = ´ 100 = 50% 4

TOPIC-3

Defects in Solids, Electrical and Magnetic Properties, Band Theory of Metals

Revision Notes Defects in solids :  It is also known imperfection in solids.  Ideal crystal is that which has definite repeating arrangement of particles of atoms in crystal.  Any departure from perfectly ordered arrangement of atom in crystal is called defect or imperfection. Mainly there are two types of defects  Point defect.  Line defect. Point defect : This defect is also known as atomic imperfections. When deviations exist from the regular arrangement around a point or an atom in a crystalline substance, the defect is called point defect. Types of point defects : Point defects are classified into three types : (a) Stoichiometric defects (b) Impurity defects (c) Non-stoichiometric defects (a) Stoichiometric defects : Defects which do not change the stoichiometry of solids are called stoichiometric defects. These are also intrinsic or thermodynamic defects. There are four types of this defect : (i) Vacancy defect : When some lattice sites in a crystalline solid are vacant, then crystal is said to have vacancy defect. This defect arises on heating. As a Vacancy defect Vacancy defect result, the density of the solid decreases. (ii) Interstitial defect : When some extra constituent Fig. 9 : Crystalline solid showing vacancy defect particles occupy the interstitial site in crystal, defect is known as interstitial defect. It arises by applying high pressure on the crystal. This defect increases the density of the crystal. Both these defects are shown by non-ionic – + – – + – + solids. (iii) Frenkel defects : This defect arises when an + – + – + – + – ion leaves its fixed position and occupies an – + – + + – + interstitial space. It creates vacancy at lattice point. There is no change in number of ions. + – + – + – + – This defect is shown by the crystal having – + – + – + – + lower co-ordination number. The ionic crystals Vacancy which have large difference in the size of ions, + – + – + – + – show this defect. This defect does not affect – + + – + – + the density of the solid. This defect is found in ZnS, AgCl, AgBr, AgI, etc. Also called dislo+ – + – + – – cation defect, it creates vacancy defect at the original site and interstitial defect at the new Fig. 10 : Frenkel defect site.  (I)

[ 19

THE SOLID STATE

Consequences of Frenkel defect :















l Crystal becomes conductor of electricity. l Stability of crystal decreases. l Dielectric constant increases.

Cation vacancy (iv) Schottky defect : The Schottky defect is often visually demonstrated using the following layout of anions and cations. Fig. 11 : Schottky Defect Positive symbols represents cations (i.e., Na+) and the negative symbols – represents anions (i.e., Cl ). This defect causes vacancy of equal numbers of cations and anions. In addition, this layout is applicable only for ionic crystal compounds in which cations are of almost similar sizes. For example NaCl, FeO, FeS, etc. AgBr shows both Frenkel and Schottky defects. Consequences of Schottky defect : l Density of crystal decreases. l Lattice energy of crystal decreases. l Stability of crystal decreases. l Crystal becomes conductor of electricity. (b) Impurity defects : SrCl2 or CaCl2 is added to molten NaCl and it is crystallized. Some of the Na+ ions are replaced by Sr2+ or Ca2+ ions. Each Sr2+ or Ca2+ ions replaces two Na+ ions. It occupies the site of one of the ions and other site remains vacant. The number of cationic vacancies are equal to the number of bivalent cations added. Similar example of impurity defect is addition of CdCl2 to AgCl. (c) Non-stoichiometric defects : Those defects which lead to change in composition of solids are called non-stoichiometric defects. These defects are of two types : (i) Metal excess defect, (ii) Metal deficiency defect. (i) Metal excess defect : This defect arises due to anionic vacancies or due to the presence of extra cation in the interstitial sites. The anionic sites occupied by unpaired electrons. This defect occur in the following ways : l Metal excess defect due to anion vacancies : In this defect, negative ion from the crystal lattice may be missing from its lattice site leaving a hole or vacancy which is occupied by the electron originally associated with anion. In this way crystal remains neutral. Alkali halides like NaCl and KCl show this type of defect. l F-Centres : These are anionic sites occupied by unpaired electrons. F– centres impart colour to crystals. The colour results by the excitation of electrons when they absorb energy from the visible light falling on the crystal. For example NaCl becomes yellow in colour when heated with sodium vapours. LiCl becomes pink, KCl becomes violet. l Metal excess defect due to interstitial cation : In this defect, an extra cation occupies interstitial position in the lattice and the free electron is trapped in the vacancy (vicinity) of this interstitial cation, but crystal remains neutral. For example : Zinc oxide on heating loses oxygen and turn yellow.





Cation occupying interstitial site

1 heat ZnO ¾¾¾® Zn 2 + + O2 + 2e 2

The excess of Zn2+ ions move to interstitial sites and the electrons to neighbouring interstitial site. (ii) Metal deficiency defect : In this, a cation is missing from its lattice site. To maintain electrical neutrality, one of the nearest metal ion acquires two positive charges. This type of defect occurs in compounds where metal can exhibit variable valency. e.g., transition metal compounds like NiO, FeO, FeS, etc.  Compounds which act as semiconductors : (i) 13–15 compounds : When the solid state materials are produced by combination of elements of groups 13 and 15, the compounds obtained are called 13–15. For example, GaAs, AlP. (ii) 12–16 compounds : Combination of elements of groups 12 and 16 yield some solid compounds are called 12–16 compounds. For example, ZnS, CdS.  Line Defects : Deviations or irregularities from the ideal arrangement in entire row of lattice point in a crystalline solid is known as line defects.  Types of solids on the basis of electrical conductivity : Solids are classified into three groups : (i) Conductors : The solids which permit maximum flow of electricity are known as conductors. Their conductivity order is 104 to 107 ohm–1 m–1. In metals, conduction takes place due to electrons while in ionic solids due to ions. e.g., all metals, aqueous solution of NaCl etc. (ii) Semi-conductors : The solids which permit less flow of electricity as compared to conductors are known as semiconductors. The conductivity order ranges from 10–6 to 104 ohm–1 m–1. As the temperature rises their conductivity value also rises because electrons from the valence band jump to conduction band, e.g., Si, Ge.

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



(iii) Insulators : The solids which are unable to conduct the electricity are known as insulators. Their conductivity value ranges from (10–20 to 10–10) ohm–1 m–1. For example : Sulphur, Phosphorus etc.  Band theory : Band theory explains the conductivities of conductors, insulators and semiconductors. The interaction between the valence band and the conduction band decide the conductivity of a particular material. If the energy gap between the two shells is negligible, the electrons can jump easily from valance band to conduction band, they behave as good conductors. When the energy gap is small then the electrons can jump from valence band to conduction band on providing little energy then these type of solids behave as semiconductors. When this gap is large enough for the electrons to excite themselves from valence band to conduction band then they act as insulators. This energy gap between the conduction band and valence band in insulators is known as forbidden zone.  Doping : The mixing of impurity in a solid crystal is known as doping. Due to doping, the conductance of semiconductors can be increased forming extrinsic semiconductors. The number of electrons or cation vacancies can be increased by doping.  n-type of semiconductors : Silicon and germanium (Group-14) doped with donor impurity (like P or As of Group-15) are called n-type semiconductor. Here n means negative, which in fact is nature of charge on electron, doping of phosphorus (P) in silicon (Si). Silicon and phosphorus have 4 and 5 valence electrons respectively. The extra electron of phosphorus increases conductivity of the crystal.  p-type of semiconductors : Silicon or germanium (Group-14) doped with acceptor impurity (Like B, Al or Ga of Group-13) is called p-type semiconductor. Here ‘p’ indicates positive e.g., doping of boron in silicon. Silicon and boron contain 4 and 3 valence electrons respectively. Due to deficiency of one electron in boron (B), positively charged hole increases conductance.  Classification of substances on the basis of magnetic properties : (i) Paramagnetic substances : The substances which are attracted in external magnetic field are known as paramagnetic substances. In such type of substances, atoms or molecules have unpaired electrons. In magnetic field, the alignment of magnetic moments develops magnetic property. Thus, substances acquire temporary magnetism in the magnetic field. e.g. Cu2+, O2 etc. (ii) Diamagnetic substances : The substances which are repelled in magnetic field are known as diamagnetic substances. The atoms or molecules of these substances have paired electrons only. e.g., Zn, TiO2, NaCl etc. (iii) Ferromagnetic substances : The substances which are attracted most easily in magnetic field are known as ferromagnetic substances. These compounds acquire permanent alignment of magnetic moments in magnetic field. e.g., Fe, Co, Ni, CrO2 etc. (iv) Anti-ferromagnetic substances : Those paramagnetic substances whose resultant magnetic moment due to alignment of magnetic moments in the magnetic field is zero are known as anti-ferromagnetic substances. e.g., MnO. The alignment of magnetic moment is found (a) opposite to each other in equal number. (v) Ferrimagnetic substances : Paramagnetic substances in (b) which resultant magnetic moment due to alignment of ­ magnetic moments in magnetic field is less than that of ferromagnetic substances are known as ferrimagnetic (c) substances. In such substances, magnetic moments alignment in parallel way is more than in opposite Fig. : 11. Schematic alignment of magnetic moments in side. e.g. Fe3O4, MgFe2O4, CuFe2O4 and ZnFe2O4. (a) ferromagnetic, (b) anti-ferromagnetic and



(c) ferrimagnetic substances

(1 mark each)



Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS : Q. 1. The crystal showing defect is :





[ 21

THE SOLID STATE



Ans. (a)

[CBSE, SQP, 2020-21] [CBSE Marking Scheme 2021]

Explanation : The Frenkel defect is formed when an atom or smaller ion (usually cation) leaves its place in the crystal lattice, creating a vacancy, and becomes an interstitial by lodging in a nearby location. Q. 2. Which stoichiometric defect does not change the density of the crystal? (a) Frenkel defect (b) Schottky defect (c) Interstitial defect (d) F-centres Ans. Correct option : (a) Explanation : In Frenkel defect, one of the ion is missing from its lattice site and occupies an interstitial site. So, density of the crystal does not change. Q. 3. Which of the following detects is also known as dislocation defect? (a) Frenkel defect (b) Schottky defect (c) Non-stoichiometric defect (d) Simple interstitial defect Ans. Correct option : (a) Explanation : Frenkel defect is also known as dislocation defect because in this defect one of the ion is missing from its lattice site and occupies an interstitial site. Q. 4. The lattice site in a pure crystal cannot be occupied by………………….. (a) Molecule (b) Ion (c) Electron (d) Atom Ans. Correct option : (c) Explanation : The lattice site in a pure crystal cannot be occupied by an electron because in case of pure crystal, no ion is missing or added. Q. 5. Examine the given defective crystal A+ B– A+ B– A+ B– 0 B – A+ B– A+ B– A+ 0 A+ – + – + B A B A B– How is the density of the crystal affected by this defect? (a) Density increases (b) Density decreases (c) No effect on density (d) Density first increases then decreases Ans. Correct option : (b) Explanation : The given defective crystal shows that there is missing of one cation and one anion from

their lattice positions which is Schottky defect. Due to missing of ions, density of the crystal decreases. Q. 6. Interstitial multi- compounds are format when small atoms are dropped under the curved lattice of metals. Whether the following is not the characteristics property of interstitial compounds? (a) They have high melting points in compound to pure metals (b) They are very hard (c) They retain metallic conductive (d) They are chemically very conductivity Ans. Correct option : (d) Explanation : Interstitial compounds are usually non-stoichiometric and are neither typically ionic nor covalent. Hence, interstitial compounds are chemically inert. [B] ASSERTION AND REASON : (a)  Both assertion and reason are correct statements, and reason is the correct explanation of the assertion. (b)  Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion. (c) Assertion is correct, but reason is wrong statement. (d) Assertion is wrong, but reason is correct statement. Q. 1. Assertion : Density of the crystal decreases in Frenkel defect.  Reason : In this defect, one of the ion is missing from lattice position and occupies interstitial site. Ans. Correct option : (a) Explanation : Density of the crystal decreases in Frenkel defect as no ion is missing from the crystal. Q. 2.  Assertion : Diamagnetic substances are attracted in external magnetic field.  Reason : These substances do not possess unpaired electrons Ans. Correct option : (d) Explanation : Diamagnetic substances are repelled in external magnetic field because these substances do not possess unpaired electrons. Q. 3. Assertion (A) : Schottky defect arises when a non-ionic solid is heated.  Reason (R) : It happens because some of the lattice sites are vacant in the crystal. Ans. Correct option : (d) Explanation : Vacancy defect arises when a non-ionic solid is heated. It happens because some of the lattice sites are vacant in the crystal. [C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1. What type of stoichiometric defect is shown by ZnS and why? [CBSE, Delhi Set 1, 2019]

Ans.  Frenkel defect due to large difference in size of ions. [CBSE Marking Scheme 2019] Q. 2. Analysis shows that FeO has a non-stoichiometric composition with formula Fe0.95O. Give reason.  [CBSE, Delhi set 1, 2018]

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Ans.  Shows metal deficiency defect / It is a mixture of Fe2+ and Fe3+ /Some Fe2+ ions are replaced by Fe3+ /Some of the ferrous ions get oxidized to ferric ions. [CBSE Marking Scheme 2018]

Ans. AgBr 

Q. 3. What type of stoichiometric defect is shown by ZnS ? R [CBSE Comptt. OD Set I 2017]

Ans. Frenkel defect 

Ans. Frenkel defect 

Q. 7. What type of magnetism is shown by a substance if magnetic moments of domains are arranged in same direction? R [CBSE Delhi Set 2016]

Q. 6. What type of stoichiometric defect is shown by AgCl ? R [CBSE Comptt. OD Set I 2017]

[CBSE Marking Scheme 2017]

Q. 4. What type of stoichiometric defect is shown by NaCl ? R [CBSE Comptt. OD Set I 2017] Ans. Schottky defect 

[CBSE Marking Scheme 2017]

Q. 5. Which ionic compound shows both Frenkel and Schottky defects? [CBSE Comptt. OD set 2017]

[CBSE Marking Scheme 2017]

[CBSE Marking Scheme 2017]

Ans. Ferromagnetism [CBSE Marking Scheme 2017] Detailed Answer : Ferromagnetism, magnetic moments are oriented in same direction.

Short Answer Type Questions-I Q. 1. (i) Give reasons : In stoichiometric defects, NaCl exhibits Schottky defect and not Frenkel defect. (ii) A metal crystallizes in a body centred cubic structure. If ‘a’ is the edge length of its unit cell, ‘r’ is the radius of the sphere. What is the relationship between ‘r’ and ‘a’?

(2 marks each) electrically neutral because the number of cations and anions remains same even after the missing of ion from its lattice position.

Ans. (i) Due to comparable size of cation and anion/ large size of sodium ion.[1] (ii) r = 0.414 R[1] [CBSE Marking Scheme 2017] Schottky defect is shown by the ionic solids having significantly small difference in their cationic and anionic radius whereas Frenkel defect is shown by ionic solids having large difference in their cationic and anionic radius. In NaCl, the radius of both Na+ and Cl– have very small difference in their radius. As a result of it, NaCl shows Schottky defect.

Q. 2. Explain the following terms with suitable examples : (i) Frenkel defect (ii) F-centres R [CBSE Comptt. OD 2016] Ans. (i) Frenkel defect : The defect in which the smaller ion/cation is dislocated to interstitial site. ½ Example : Silver halides, ZnS. (Any one) ½ (ii) F-centres : The anion vacancy occupied by an electron. ½ Example : NaCl, KCl, LiCl (Any one) ½ [CBSE Marking Scheme 2016] Detailed Answer : Frenkel defect : This defect is created in the crystal when one of the ion (cation or anion) is missing from its lattice position and occupies the vacant interstitial site. The crystal remains







Detailed Answer :







Example : AgCl, Agl, ZnS, etc. [½] F-centres : In metal exces defect due to anion vacancies the ‘holes’ occupied by electrons are called F-centres or clour centres. These centres are responsible for the colour of the compound.  [½] Example : KCl, LiCl [½]

Commonly Made Error  Some students get confused between Schottky defect and Frenkel defect. Answering Tip  Learn and understand point defects. Q. 3. (i)  Name the non-stoichiometric point defect responsible for colour in alkali metal halides.  (ii) ZnO appears as yellow on beating. Why? A  Ans. (i) Metal excess or anionic vacancies or F-centres.  [1] (ii) Z  nO loses oxygen on heating and anion vacancies are occupied by electrons which disperse complementary colour by absorbing light in visible region.[1]

[ 23

THE SOLID STATE

Short Answer Type Questions-II Q. 1. (i) Based on the nature of intermolecular forces, classify the following solids : Silicon carbide, Argon (ii) ZnO turns yellow on heating. Why? (iii) What is meant by groups 12-16 compounds? Give an example?  R [CBSE OD Set 1 2017] Ans. (i) Covalent solid/network solid, molecular solid [½+½] 1 O2 + 2e − 2 2+ Because excess Zn ions move to interstitial sites and the electrons move to neighbouring voids[1]



(ii) ZnO Heating → Zn 2 + +





(iii) C ompounds prepared by combination of groups 12 and 16 behave like semiconductors. For e.g., ZnS, CdS, CdSe, HgTe  (Any one) [½+½] [CBSE Marking Scheme 2017] Detailed Answer : (a) Silicon carbide : Covalent or network solid Argon : Non-polar molecular solid[1]













(b) When ZnO is heated, it loses oxygen. In ZnO, Zn2+ ions occupy the interstitial sites and electrons are trapped in the interstitial sites for the neutralization. Due to presence of electrons in the interstitial voids, the colour is yellow. 1 Heating ® Zn 2 + + O2 + 2e - [1] ZnO ¾¾¾¾ 2

(c) Compounds formed between group-12 elements with elements of group-16 are called 12 – 16 compounds. For example : ZnS, CdS. [1] Q. 2. (i) Based on the nature of intermolecular forces, classify the following solids : Sodium sulphate, Hydrogen  (ii)  What happens when CdCl2 is doped with AgCl?  (iii) Why do ferrimagnetic substances show better magnetism than anti-ferromagnetic substances? R [CBSE OD Set 3 2017] Ans. (i) Na2SO4 : Ionic, H2 : Molecular [½+½] (ii) Impurity defect/Schottky defect [1] (iii) In ferrimagnetism, domains/magnetic moments are aligned in opposite direction in unequal numbers while in antiferromagnetic the domains align in opposite direction in equal numbers so they cancel magnetic moments completely, net magnetism is zero/ [1] diagrammatic explanation. [CBSE Marking Scheme 2017]

(3 marks each)

Detailed Answer : (a) Sodium sulphate : Ionic solid Hydrogen : Molecular solid[1] (b) Addition of CdCl2 to AgCl crystal causes impurity defect wherein one Cd2+ ion replaces two Ag+ ions from the crystal. One site is occupied by one Cd2+ ion and the other remains vacant. [1] (c) In ferrimagnetic substances, magnetic domains are aligned in parallel and antiparallel directions in unequal numbers resulting in non-zero net magnetic moment. Whereas in antiferromagnetic substances, the domains are oppositely oriented and cancel out magnetic moment of each other. Thus ferrimagnetic substances show better magnetism than antiferromagnetic substances. [1] Q. 3. (a) Based on the nature of intermolecular forces, classify the following solids : Benzene, Silver (b) AgCl shows Frenkel defect while NaCl does not. Give reason. (c) What type of semiconductor is formed when Ge is doped with Al? R [CBSE OD Set 2 2017] Ans.

(a) Benzene - molecular solid [½] Silver - metallic solid [½] (b) Size of Ag+ ion is smaller than Na+ ion.  [1] (c) p-type  [1] [CBSE Marking Scheme 2017]

Detailed Answer : (b) AgCl shows frenkel defect while NaCl does not because Frenkel defect is shown by ionic solids in which difference in size of cations and anions is large. This condition is rightly applicable to AgCl.[1] (c) p-type extrinsic semiconductor.[1] Q. 4. Define the following :

(i) Schottky defect



(ii) Frenkel defect



(iii) F-centre

R [CBSE Comptt. Delhi 2015]

Ans. (i) The defect in which equal number of cations and anions are missing from the lattice.  [1] (ii) Due to dislocation of smaller ion from its normal site to an interstitial site.  [1] (iii) Anionic vacancies are occupied by unpaired electron. [CBSE Marking Scheme 2015]  [1] Q. 5. Examine the given defective crystal : X+ Y– X+ Y– X+ Y–

O Y– X+ Y–

X+

Y– X+ O X+

–

Y

X+ Y– X+ Y–

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



Answer the following questions : (i)  Is the above defect stoichiometric or nonstoichiometric ? (ii) Write the term used for this type of defect. Give an example of the compound which shows this type of defect. (iii) How does this defect affect the density of the A [CBSE OD 2015] crystal ? Ans. (i) Stoichiometric defect. [1]



(ii) Schottky defect NaCl (or any other example).[1]



(iii) Density of crystal decreases. Commonly Made Error

 Some students are unable to identify the defect from the structure of crystal. Answering Tip  Learn and understand point defects with structures.

Long Answer Type Questions

(i) d = (Z × M)/a3 × Na [½]   11.5 = Z × 93/[(300 × 10-10)3 × 6.02 × 1023][1]    Z = 2.0[½] Body centred cubic (bcc)[1]



(ii) Amorphous solids

Crystalline solids

Short range order

Long range order

Isotropic

Anisotropic



[1 + 1] [CBSE Marking Scheme 2017]



OR

(i) n = given mass/molar mass = 8.1/27 mol [½] 8.1 23 Number of atoms = × 6.022 × 10 27  [½] Number of atoms in one cell unit = 4 (fcc) [½] Number of unit cells = é 8.1 ´ 6.022 ´ 10 23 ù / 4 ë 27 û = 4.5 ´ 10 22







[½] OR 27 g of Al contains = 6.022 × 1023 atoms [½] 8.1 g of Al contains = (6.022 × 1023/27) × 8.1 [½] No. of unit cells = total no of atoms/4 [½] =  827.1 × 6.022 × 10 23  / 4  



22    = 4.5 × 10 [½] (ii) (a)  Due to comparable size of cation and anion/large size of sodium ion. [1]    (b) Phosphorous has 5 valence eg, an extra electron results in the formation of n-type semiconductor. [1]

   (c)  In ferrimagnetism domains/magnetic moments are aligned in opposite direction in unequal numbers while in antiferromagnetic the domains align in opposite direction in equal numbers so they cancel magnetic moments completely, net magnetism is zero/diagrammatic representation [1] [CBSE Marking Scheme 2017]

etailed Answer : D (i) M = 93 g mol-1 d = 11.5 g cm-3 a = 300 pm = 300 × 10-10 cm = 3 × 10-8 cm. We know that[1] Z×M d= N A × a3 d × N A × a3 M 11.5 × 6.022 × 10 23 × ( 3 × 10 −8 )3 = 93  [1] = 2.01(appro ox) As the number of atoms present in the given unit cell is nearly equal to 2, hence it is body centred cubic unit cell (BCC). [1] Z=

(or any other correct difference)

(5 marks each)



Q. 1. (i) An element has atomic mass 93 g mol–1 and density 11.5 g cm–3. If the edge length of its unit cell is 300 pm, identify the type of unit cell. (ii) Write any two differences between amorphous solids and crystalline solids. Ans.

[1]



Crystalline solids (ii) Amorphous solids (a) They have short (a) They have long range order i.e., there is range order i.e., there is a regular arrangement a regular arrangement of constituent particles of constituent particles over a short distance. over a long distance. (b) They are isotropic (b) They are anisoi.e., show identical tropic i.e., show electrical and optical different electrical properties in all and optical properties directions. in all directions. [1 + 1] Q. 2. (i) Calculate the number of unit cells in 8.1 g of aluminium if it crystallizes in a fcc structure. (Atomic mass of Al = 27 g mol–1)  (ii) Give reasons : (a) In stoichiometric defects, NaCl exhibits Schottky defect and not Frenkel defect. (b) Silicon on doping with Phosphorous forms n-type semiconductor. (c) Ferrimagnetic substances show better magnetism than antiferromagnetic substances. A [CBSE Delhi Set 1, 2, 3 2017] (i) 27g of Al contains 6.022 × 1023 atoms [½] 8.1g of Al contains (6.022 × 1023/27) × 8.1 [½]

[ 25

THE SOLID STATE

No. of unit cells = Total number of atoms/4





 8.1  = × 6.022 × 10 23  / 4  2.7  22

= 4.5 × 10 [1] (ii) (a) Schottky defect is shown by the ionic solids having significantly small difference in their cationic and anionic radius whereas Frenkel defect is shown by ionic solids having large difference in their cationic and anionic radius. In NaCl, the radius of both Na+ and Cl– have very small difference in their radius. As a result of it, NaCl shows Schottky defect.[1]  (b) Phosphorous is an electron rich element of group 15. On doping with element of group 14, silicon, an n-type semiconductor will form because here the conductivity is due to the presence of extra electrons.[1]  (c)  Ferrimagnetic substances have magnetic moments of the domains aligned in parallel and anti-parallel directions in unequal numbers whereas antiferromagnetic substances domains oppositely oriented and cancel out each other’s magnetic moment. Therefore, ferromagnetic substances show better magnetism than antiferromagnetic substances.[1] Q. 3. (i) Following is the schematic alignment of magnetic moments : Identify the type of magnetism. What happens when these substances are heated?  (ii) If the radius of the octahedral void is ‘r’ and radius of the atoms in close packing is ‘R’. What is the relation between ‘r’ and ‘R’?  (iii) T  ungsten crystallizes in body centred cubic unit cell. If the edge of the unit cell is 316.5 pm. What is the radius of Tungsten atom?



Ans. (i) Ferrimagnetism. These substances lose ferrimagnetism heating and become paramagnetic. (ii) r = 0.414 R 3 (iii) r= a 4

[1] on [1] [1] [1]

3 × 316.5 4 r = 136.88 pm [½ + ½] r=





Q. 4.

(i) Identify the type of defect shown in the figure : –

+

–

+

–

–

+

+

+

–

–



+





(ii)

r=



(iii)

ρ= 8.92 =





3 a 4 [1] ZM a3 N A

[½]

 Z × 63

( 3.608 × 10 −8 )3 6.022 × 10 23

[1]

    Z = 4. So it is face centred cubic lattice [½] [CBSE Marking Scheme 2017] Write the type of magnetism observed Q. 5.(a) (i)  when the magnetic moments are oppositely aligned and cancel out each other.  (ii) Which stoichiometric defect does not change the density of the crystal ? (b) Examine the given defective crystal A

+

B

–

A

+

B

–

A

–

A

+







(ii) A metal crystallizes in a body centred cubic structure. If ‘a’ is the edge length of its unit cell, ‘r’ is the radius of the sphere. What is the relationship between ‘r’ and ‘a’? (iii) An element with molar mass 63 g/mol forms a cubic unit cell with edge length of 360.8 pm. If its density is 8.92 g/cm3. What is the nature of the cubic unit cell? R [CBSE SQP 2017] Ans. (i) Schottky defect[1] It is shown by ionic substances in which the cation and anion are of almost similar sizes.[1]

+ –

–

+

+

–

– –

–

+

+ –

+

+ +

– –

+ + –

+

+

–

–

+

+ – + –

+

–

+

–

+

–

+ –

–

+

+ –

+

+ –

–

B

–

A B

+

0

B

B

A

–

A

B

+

B

+ 0

A

A

– +

B–

– + – + Answer the following questions : (i) What type of stoichiometric defect is shown by the crystal ? (ii) How is the density of the crystal affected by this defect ? (iii)  What type of ionic substances show such defect ? U [CBSE OD 2014]



Ans. (a) (i) Anti ferromagnetism.[1]  (ii) Frenkel defect. [1]

(b)  (i) Schottky defect. [1]   (ii) Decreases[½] (iii) Alkali metal halides/ionic substances having almost similar size of cations and anions (NaCl/KCl).[CBSE Marking Scheme 2014]

Detailed Answer (a) (i) When the magnetic moments are oppositely aligned in equal number and cancel out each other then such type of magnetism is called Anti-ferromagnetism.

–

+ –

What type of substances show this defect?

Fig. Alignment of magnetic moment in Antiferromagnetic substance [1]

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(ii) In Frenkel defect, density of the crystal does not change because in this defect, one of the ions is missing from its lattice position and occupies interstitial site and number of cations and anions remains same.[1] (b) (i) A+ B– A+ B– A+ B– 0 B – A+ B– + – A B A+ 0 A+ B – A+ B– A+ B–











(ii) The density of this crystal decreases due to missing of cations and anions. [1]





  (iii) Ionic substances having almost similar size of cations and anions show such defect.[1]

In this defect, equal number of cation and anion is missing from their lattice positions and cation and anion vacancies are created. So, this defect is called Schottky defect. [1]

Visual Case Based Questions Q. 1. Read the passage given below and answer the following questions : (1 × 4 = 4) All real structures are three dimensional structures. They can be obtained by stacking two dimensional layers one above the other while placing the second square close packed layer above the first we follow the same rule that was followed when one row was placed adjacent to the other. The second layer is placed over the first layer such that the spheres of the upper layer are exactly above there of the first layer. In his arrangement spheres of both the layers are perfectly aligned horizontally as well as vertically. A metallic element crystallise into a lattice having a ABC ABC pattern and packing of spheres leaves out voids in the lattice. The following questions are multiple choice questions. Choose the most appropriate answer :  (i)  What type of structure is formed by this arrangement?   (a) ccp (b) hcp  (c) ccp/fcc (d) none of the above   (ii)  Name the non-stoichiometric point defect responsible for colour in alkali metal halides (a) Frenkel defect (b) Interstitial defect (c) Schottky defect (d) F-centres (iii) What is the total volume of atoms in a face centred cubic unit cell of a metal? (r is atomic radius) (a) 103 pr3 (b) 8 pr3 3 (c) 83 pr (d) 13 pr3 (iv) Which of the following statements not true for the amorphous and crystalline solids? (a) Amorphous solids are isotropic and crystalline solids are anisotropic. (b) Amorphous solids are short range order and crystalline solids are long range order. (c) Amorphous solids melt at characteristic temperature while crystalline solids melt over a range of temperature. (d) Amorphous solids have irregular shape and crystalline solids have a geometrical shape.

(4 marks each) OR





Which of the following statements is not true for unit cell?

(a) Each cubic unit cell has 8 atoms on its corners the total number of atoms in one unit cell is 1. (b) A unit cell is characterized by its dimensions along the three edges a, b, c. (c) Each body centred cube cell has 2 atoms in one unit cell. (d) Each face centred cubic cell contains only one constituent particle present at the centre of each face. Ans. (i) Correct option : (c)



(ii) Correct option : (d)





(iii) Correct option : (a)





Number of atoms per unit cell in fcc = 4





∴ Total volume of atoms present in fcc unit cell









=4´

4 3 16 3 pr = pr 3 3

(iv) Correct option : (c) Explanation : Crystalline solids melt at characteristic temperature while amorphous solids melt over a range of temperature. OR Correct option : (d) Explanation : Each fcc unit cell contains one constituent particle present at the centre of each face, besides the ones that are at its corners. Q. 2. Study the diagram given below and answer the following questions : (1 × 4 = 4)

[ 27

THE SOLID STATE







(i)





In these questions, a statement of Assertion followed by a statement of Reason is given. Choose the correct answer out of the following choices. (a)  Assertion and reason both are correct statements and reason is correct explanation for assertion. (b)  Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c)  Assertion is correct statement but reason is wrong statement. (d)  Assertion is wrong statement but reason is correct statement. Assertion : The diagram shows Schottky defect. Reason : Schottky defect occurs in ionic solids.

(ii)

Assertion : LiCl Crystal is pink.



Reason : Pink colour of LiCl crystal is due to excess Lithium.



(iii)

Assertion : The crystal lattice density increases due to the defect shown in the diagram.



Reason : Tetrahedral voids are surrounded by 4 constituent particles.



(iv)

Assertion : AgCl shows Frenkel defect while NaCl does not.



Reason : Frenkel defect is shown when anionic vacancies are occupied by unpaired electrons.



Ans.



























(i) Correct option : (b) Explanation : The diagram shows Schottky defect as it has equal number of cationic and anionic vacancies. (ii) Correct option : (a)  xplanation : LiCl crystal is pink due to excess E Lithium. It is caused by metal excess defect caused by a anionic vacancies(F-centres). (iii) Correct option : (d) Explanation : The crystal lattice density decreases due to the defect shown in the diagram-Schottky defect. (iv) Correct option : (b) Explanation : AgCl shows Frenkel defect while NaCl does not as the size of Ag+ ion is smaller than Na+ ion. Frenkel defect occurs when the smaller ion dislocates from its normal site to an interstitial site.

ll

28 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Self Assessment Test-1 Time : 1 Hour [A] OBJECTIVE TYPE QUESTIONS :

Max. Marks : 25 (1 mark each)

1.  Read the passage given below and answer the following questions : (1 × 4 = 4) As we know that free electrons are responsible for the conduction of electricity in metals. Electrons are occupied in atomic orbital. The atomic orbital forms molecular orbital in metallic crystal. Molecular orbitals are very close in energy and referred to as bands. To be a conductor, there must be some electrons in conduction band. On the basis of conductivity, solids are classified in three types – Conductor, Semiconductor and insulator. In these questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not the correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. (i) Assertion : In a metallic crystal, molecular orbitals are very close in energy and referred to as bonds. Reason : There are number of atoms in metallic crystals whose atomic orbitals overlap together to form molecular orbitals. (ii) Assertion : Metals are good conductor of electricity. Reason : They contain free electrons to move. (iii) Assertion : In semi-conductors, some of the electrons may jump from valence band to the conduction band. Reason : An overlap zone is present between them. (iv) Assertion : Insulators can conduct electricity. Reason : A large energy gap is present between valence band and the conduction band in insulators. OR Assertion : Valence band is unoccupied level. Reason : Electrons are filled in bonding molecular orbitals due to low energy. The following questions (No. 2 to 5) are Multiple Choice Questions carrying 1 mark each : 2. The expression for density of crystal is : ZM a3 N A (a) 3 (b) a NA ZM

a3 M NAM (c) (d) ZN A Za 3

3. The number of atoms in bcc unit cell of a mono atomic elementary substance is (a) 1 (b) 2 (c) 3 (d) 4 4.  Copper is crystallised in fcc structure with edge length of unit cell 361 pm, the atomic radius of copper is (a) 108 pm (b) 127 pm (c) 157 pm (d) 181 pm 5. The number of octahedral voids present per atom in cubic close packing structure is (a) 1 (b) 2 (c) 3 (d) 4 In the following questions (No. 6 & 7), a statement of Assertion followed by a Statement of Reason is given. Choose the correct answer out of the following choices.

(a)

Assertion and reason both are correct statements and reason is correct explanation for assertion.

(b)

Assertion and reason both are correct statements but reason is not the correct explanation for assertion.



(c)

Assertion is correct statement but reason is wrong statement.

(d)

Assertion is wrong statement but reason is correct statement.

6.  Assertion : The substances which are strongly attracted by magnetic field are said to exhibit ferro magnetism.  Reason : Ferro magnetism arises due to spontaneous alignment of magnetic moments of ions or atoms in the same direction. 7. Assertion : The density remains unchanged in the crystals exhibiting Frenkel defect.  Reason : Doping of group 14 element with suitable element of group 13 produces p-type of semiconductors. The following questions (No. 8 & 9) are Short Answer Type-I and carry 2 marks each. 8. A compound formed by elements A and B crystallises in the cubic structure where A atoms are at the corners of the cube and B atoms are at the centre of the cube. What is the formula of the compound? 9. Sodium metal crystallizes in body centred 1 attice with edge length (a) = 4.29 Å. What is the radius of sodium atom?

[ 29

SELF-ASSESSMENT TEST

The following question (No. 10 & 11) are Short Answer Type II carrying 3 marks each. 10. (i) Explain why Frenkel defect is not found in pure alkali metal halides? (ii) What are F-centres? Give their significance. 11. If there are 2.5 × 1024 atoms in 290 g of the element and it crystallizes in a body centred cubic structure with a density of 7.13 g/cm3, what is its atomic radius? Question No. 12 is Long Answer Type Question carrying 5 marks. 12. (i) Define face centred cubic unit cell. (ii)  Show by simple calculation that % of space

occupied by spheres in a face centred cubic unit cell is 74 %. (i) Which type of solids exhibit anisotropic nature? (ii) A substance forms face centred cubic lattice. The edge length od unit cell is 630 pm and density of substance is 1.984 g cm–3, then calculate the molar mass of substance. (iii) The structure of lithium metal crystal is bcc. The density of crystal is 1.53 g cm–3 and its molar mass is 6.94 g mol–1. Calculate the volume of its unit cell.

 

Finished Solving the Paper ? Time to evaluate yourself !

OR SCAN THE CODE

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For elaborate Solutions

ll

Volume of solution in mL)

Volume of component  100 Total volume of solution

Mole fraction

uid

Types

Normality: Number of gram equivalents of the solute dissolved in one litre of solution

Equivalent weight =

ou se

s

First Level

Second Level

Trace the Mind Map Third Level

No. of moles of solute Volume of solution in litres

Molarity : Number of moles of solute in 1L solution

Gas – Solid  O2 in Pd Liquid – Solid  Amalgam of Hg with Na Solid – Solid  Cu dissolved in gold

Gas – Liquid  O2 dissolved in water Liquid – Liquid  Ethanol dissolved in water Solid – Liquid  Glucose dissolved in water

Gas – Gas  Mixture of O2 and N 2 Liquid – Gas  Chloroform with N2 Solid – Gas  Camphor in N2

Exothermic sol H < 0, Solubility decreases

Endothermic sol H > 0, Solubility increases

Not significant

No. of moles of solute Mass of solvent in kg

Molecular mass Valency

Gram Equivalents of solute Mass of solute = Equivalent weight

G a

Molality: Number of moles of solute per kilogram of the solvent

No. of gram equivalentof solute×100 Volume of solution

No. of moles of component Total no. of moles of all components

No. of parts of components×10 6 Total no. of parts of components of solution

Mass of component in solution  100 Total mass of solution

Solutions

Solubility

Different methods to express concentration of solution

Raoult’s Law

Parts per million : For trace quantities

Mass percentage (w/w)

Volume percentage (v/v)

 100

Mass by volume p ercentage (w/v)

(Mass of solute(g)

Maximum boiling azeotrope

H mix = negative Vmix = negative

Vmix = positive H mix = positive

Increases with increase in pressure

Abnormal molecular mass: molecular mass different from expected value

Colligative properties

For any solution, the partial vapour pressure of each volatile component is directly proportional to its mole fraction.

Non-ideal solution  (Mixture of chloroform and acetone)

Ideal solution  (n-hexane and n-heptane)

Minimum boiling azeotrope

W2 ×M 1 P°1 - P1  M 2 ×W1 P1

K b ×1000 × W2 M 2 × W1

K f × W2 ×1000 M 2 ×W1

 Relative lowering of vapour pressure 

 Elevation of boiling point  Tb =

 Depression in freezing point  Tf =

 Osmotic pressure   = CRT

Normal molar mass = Abnormal molar mass



Gas in L iq

Increases with decrease in temperature



Partial pressure of gas in vapour phase is proportional to the mole fraction of gas in the solution. p =KHx



30 ] Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

2

CHAPTER

Syllabus

SOLUTIONS

Types of solutions, expression of concentration of solutions of solids in liquids, solubility of gases in liquids, solid solutions, Raoult’s law, colligative properties - relative lowering of vapour pressure, elevation of boiling point, depression of freezing point, osmotic pressure, determination of molecular masses using colligative properties, abnormal molecular mass, van’t Hoff factor.

Trend Analysis List of Concepts

2018 OD

Henry’s Law, Raoult’s Law Ideal and Non-ideal Solution Colligative Properties Numericals based on Colligative Properties

1Q (3 marks) 1Q (2 marks)

2019 D OD 2Q (2 marks) 1Q 2Q (2 marks) (2 marks) 1Q (2 marks) 1Q 1Q (3 marks) (3 marks)

2020 D OD 2Q (2 marks) 1Q (1 mark) 2Q (1 mark) 1Q 1Q (3 marks) (3 marks)

TOPIC-1

Types of Solutions, Expression of Concentration of Solutions and Solubility

Revision Notes  Solution : A homogeneous mixture of two or more pure substances is known as solution.  If the constituents of the solution are two, it is called binary, if three then ternary, if four then quaternary and so on.  Two constituents of the solution are :

(i) Solute : A substance that is dissolved in another substance in lesser amount, forming a solution. For example : Sugar, salt, etc.



(ii) Solvent : A substance in which another substance is dissolved in larger amount forming a solution. For example : Water, milk, etc.



Note : Solvent determines the physical state of the solution.

TOPIC - 1 Types of Solutions, Expression of Concentration of Solutions and Solubility .... P. 31 TOPIC - 2 Vapour Pressure, Raoult’s Law, Ideal and Non-ideal Solutions .... P. 39 TOPIC - 3 Colligative Properties, Determination of Molecular Mass Abnormal molecular mass, van’t

.... P. 48 Hoff Factor  Types of Solutions : Any state of matter (solid, liquid or gas) can act both as a solvent and as a solute during the formation of a solution. Therefore, depending upon the physical states of solute and solvent, we can have following nine different types of solutions :

32 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

S. No.

Types of Solution

Solute

Solvent

Examples

1.

Solid – Solid

Solid

Solid

Alloys like brass, bronze, etc.

2.

Solid – Liquid

Solid

Liquid

Solution of sugar, salt, urea etc. in water.

3.

Solid – Gas

Solid

Gas

Sublimation of substances like iodine, camphor, etc, into air, dust or smoke particles in air.

4.

Liquid – Solid

Liquid

Solid

Hydrated salts, mercury in amalgamated zinc, etc.

5.

Liquid – Liquid

Liquid

Liquid

Alcohol in water, benzene in toluene.

6.

Liquid – Gas

Liquid

Gas

Aerosol, water vapour in air.

7.

Gas – Solid

Gas

Solid

Hydrogen adsorbed in palladium.

8.

Gas – Liquid

Gas

Liquid

Aerated drinks.

9.

Gas – Gas

Gas

Gas

Mixture of gases, etc.

 Aqueous solution : A solution containing water as solvent is known as aqueous solution. For example, sugar solution.  Non-aqueous solution : A solution containing solvent other than water is known as non-aqueous solution. For example, iodine dissolved in alcohol.  Saturated solution : A solution in which no more solute can be dissolved at the same temperature is known as saturated solution.  Unsaturated solution : A solution in which more amount of solute can be dissolved at the same temperature is known as unsaturated solution.  Solubility : Solubility can be defined as the maximum amount of solute that can be dissolved in 100 g of solvent to form a saturated solution at a given temperature. l Causes of Solubility : (i) Inter ionic attraction in the solute molecules : Molecules are stabilised in the lattice due to electrostatic forces and the energy released is known as lattice energy.  (ii) Inter molecular attraction between solvent molecules. (iii) Solvation : It denotes the force of attraction between solute and solvent molecules.  (iv) Temperature. l Factors affecting Solubility : (i) Nature of Solute and Solvent : “Like dissolves like” i.e., polar solvents like water and ammonia can dissolve polar solute or ionic solute while non-polar solvents can dissolve non-polar organic solutes.  (ii) Temperature : Solubility increases with increase in temperature. It increases for endothermic reaction while it decreases for exothermic reaction. (iii) Pressure : The solubility of solid in liquid is not affected significantly by pressure because solids and liquids cannot be compressed.  (iv) Hydration Energy : It is the amount of energy released when ions formed by 1 mole of ion get hydrated. It is an exothermic process.  Method of expressing Concentration of Solution : The concentration of solution is the amount of solute present in the given quantity of solute or solvent. It can be expressed in any of the following types : (i) Mass percentage w : It is the amount of solute in grams dissolved per 100 gm of solution. W





Mass% of a solute =

Mass of solute in the solution × 100 Total mass of the solution

(ii) Volume percentage v : It is defined as volume of a solute dissolved per 100 ml of solution. V Volume% of a solute =

Volume of solute × 100 Total volume of the solution

(iii) Mass by volume percentage v : It is defined as mass of solute dissolved per 100 ml of solution. It is V commonly used in medicine and pharmacy.

[ 33

SOLUTIONS

Mass of solute × 100 Volume of solution



Mass by volume % of solute =

(iv) Parts per million (ppm) : It can be defined as the parts of a component per million (106) parts of the solution. It is used to express the concentration of a solute present in trace quantities.



ppm (A) =

Parts per million can be expressed in three ways : (a) Mass to mass Mass of a component ppm (mass to mass) = × 10 6 Total mass of solution (b) Volume to volume



Number of the parts of the component (A) × 10 6 Total number of parrts of all the components of the solution

ppm (volume to volume) =

Volume of a component Total volume of solution

× 10 6

(c) Mass to volume



ppm (mass to volume) =

Mass of a component Volume of solution

× 10 6

(v) Mole Fraction : It is the ratio of number of moles of a particular component to the total number of moles of all the components. e.g., mole fraction of component A. nA χA = , n A + nB



where nA is the number of moles of component ‘A’ and nB is the number of moles of component ‘B’. nB Similarly, χB = nA + nB Sum of mole fractions of all the components is always one.

χA + χB = 1 (vi) Molarity (M) : It is defined as the number of moles of solute per litre of solution. Number of moles of solute     Molarity = Volume of solution (in Litres)      M =

WB × 1000 MB × V



where, WB = Weight of solute, V = Volume of solution in ml, MB = Molar mass of solute. Unit is mol L–1 or M (molar).



And

Weight of solute ( WB ) = Moles of solute Molar mass of solute ( M B )

(vii) Molality (m) : It is defined as the number of moles of solute per 1000 g or 1 kg of solvent. Number of moles of solute Molality = Mass of solvent in kg



m =

WB × 1000 MB × W

where, WB = Weight of the solute, MB = Molar mass of solute, W = Mass of solvent in g Unit is mol kg–1 or molal (m). Molality and mole fraction do not change with change in temperature. (viii) Normality (N) : It is defined as number of gram equivalents of solute dissolved per litre of solution.

Normality =



N =

Number of gram equivalent of solute Volume of solution in Liitre WB × 1000 EB × V

where, WB = Mass of solute, EB = Equivalent weight of solute, V = Volume of solution in ml  Relationship between Molarity (M) and Molality (m) :

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

  



34 ]

d M 1 − B = M 1000 m

Scan to know more about this topic

Partial pressure (p)

where, m = Molality of solution, M = Molarity of solution, MB = Molar mass of solute, d = Density of solution in g ml–1  Relationship between Mole fraction of solute (χB) and Molality (m) : Solubility of χB × 1000 gases m= (1 − χ B ) × M A    where χB is mole fraction of solute, m is molality and MA is molar mass of solvent. The relationship between pressure and solubility is guided by Henry’s Law. According to this law, ‘‘The mass of a gas dissolved in given volume of the liquid at a constant temperature depends upon the pressure applied.’’ It can also be stated as the partial pressure of the gas (p) in vapour phase is proportional to the mole fraction of the gas (χ) in the solution. p = KHχ, where KH = Henry’s constant.

 (mole fraction)

Fig. 1 : The slope of the line in Henry’s constant, (KH) l Applications of Henry’s law : (i) To increase the solubility of CO2 in soda water and soft drinks, the bottle is sealed under high pressure. (ii) To avoid the toxic effects of high concentration of nitrogen in blood, the tanks used by scuba divers are filled with air diluted with helium (11.7%), nitrogen (56.2%) and oxygen (32.1%). (iii) At high altitudes, low blood oxygen causes climber to become weak and make them unable to think clearly, which are symptoms of a condition known as anoxia. l Limitations of Henry’s law : This law is applicable only when : (i) The pressure of gas is not too high and temperature is not too low. (ii) The gas should not undergo any chemical change. (iii) The gas should not undergo association or dissociation in the solution.

Know the Formulae 



Number of moles of the component Mole fraction of a component = Total number of moles of alll the components n1 n2 x1 = , x2 = ( x1 + x 2 = 1) n1 + n2 n1 + n2



Molarity (M) =

Number of moles of solute Volume of solution in Litre



Molality (m) =

Number of moles of solute Mass of solvent in kg



Normality (N) =

    

Number of gram equivalent of solute Volume of solution in Liitre

wö Mass of solute in the solution Mass percentage æ × 1000 çW ÷ = Total mass of the solution è ø

æv ö Volume of solute × 100 Volume percentage ç ÷ = Total volume of the solution èV ø

æw ö Mass of solute Mass by volume percentage ç ÷ = × 100 Volume of solution èV ø p = KH. x where KH = Henry’s law constant p = partial pressure of the gas in vapour phase x = mole fraction of the gas in the solution ppm of component A =

Mass of component A × 10 6 Total mass of solution

[ 35

SOLUTIONS

Q. Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride. Solution: STEP-1: Let the total mass of the solution be 100 g and the mass of benzene be 30 g. \ Mass of carbon tetrachloride  = (100 − 30) g      = 70 g [½] Molar mass of benzene (C6H6) = (6 × 12 + 6 × 1) g mol–1 = 78 g mol–1 30 ∵ Number of moles of C6H6 = mol 78 [1]

STEP-2: Molar mass of carbon tetrachloride (CCl4)        = 1 × 12 + 4 × 35.5          = 154 g mol–1 \ Number of moles of CCl4 70 = mol 154





= 0.3846 mol

= 0.4545 mol [1] STEP-3: Thus, the mole fraction of C6H6 is given as : Number of moles of C6H6 Number of moles of C6H6 + Number of moles of CCl4

Objective Type Questions Q. 1. A molar solution is one that contains one mole of a solute in (a) 1000 g of the solvent (b) one litre of the solvent (c) one litre of the solution (d) 22.4 litre of the solution R Ans. Correct option : (c) Explanation : A molar solution is one that contains one mole of a solute in one litre of the solution. Number of moles of solute Molarity ( M ) =   Volume of solution in L Q. 2. In which mode of expression, the concentration of a solution remains independent of temperature? (a) Molarity (b) Normality (c) Formality (d) Molality R Ans. Correct option : (d) Explanation : The molality of a solution does not change with temperature. Q. 3. The increase in the temperature of the aqueous solution will result in its

0.3846 0.3846 + 0.4545

= 0.458[½]

(1 mark each) (a) Molarity to increase (b) Molarity to decrease (c) Mole fraction to increase (d) Mass % to increase R Ans. Correct option : (b) Explanation : An increase in temperature increase the volume of solution and therefore it will result in its molarity to decrease. Q. 4. KH value for Ar(g), CO2(g), HCHO(g) and CH4(g) are 4.039, 1.67, 1.83 × 10–5, and 0.143, respectively. Arrange these gases in the order of their increasing solubility (a) HCHO < CH4 < CO2 < Ar (b) HCHO < CO2 < CH4 < Ar (c) Ar < CO2 < CH4 < HCHO (d) Ar < CH4 < CO2 < HCHO Ans. Correct option : (c) Explanation : According to Henry’s law, P = KH C 1 KH µ C

[A] MULTIPLE CHOICE QUESTIONS :

=

36 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Where P = Partial pressure of gas C = Concentration of gas KH = Henry’s constant It implies that as the value of KH increases, mole fraction of gas solute in solvent decreases. Hence, higher the KH value, lower is the solubility of gas. The order of increasing solubility of gases in : Ar < CO2 < CH4 < HCHO





[B] ASSERTION AND REASON TYPE QUESTIONS : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Q. 1. Assertion : A molar solution is more concentrated than molal solution. Reason : A molar solution contains one mole of solute in 1000 mL of solution. Ans. Correct option : (a) Explanation : A molar solution is more concentrated than molal solution because 1 molar solution contains 1 mole of solute in 1 litre of the solution which include both solute and solvent. Q. 2. Assertion : Molarity of 0.1 N solution of HCl is 0.1 M. Reason : Normality and molarity of a solution are always equal.

Ans. Correct option : (c) Explanation : Normality and molarity of a solution are not always equal. Normality depends on chemical equivalent of the substance while molarity depends on molecular mass of the substance. Q. 3. Assertion : Molarity of a solution changes with temperature. Reason : Molarity is dependent on volume of solution. Ans. Correct option : (a) Explanation : As molarity is dependent on volume of solution and volume rises with increase in temperature. Molarity is inversely proportional to temperature. So, as temperature increases, volume increases and molarity decreases. [C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1. Define Normality of a solution. Ans. It is defined as the number of gram equivalents of the solute dissolved per litre of the solution. Q. 2. What is the Normality of 0.2 M aqueous solution of a tribasic acid? Ans. Normality = molarity × basicity of acid = 0.2 × 3 = 0.6 N Q. 3. Give the type of solution in Aerated drinks. Ans. Gas – Liquid (Solute = Gas, Solvent = Liquid) Q. 4. Why soda bottle kept at room temperature fizzes on opening? Ans. Carbon dioxide gas is filled in the soda bottle under pressure. It leads to higher amount of gas to dissolve in the water, which is released as bubbles or fizz when the bottle is kept open at room temperature.

Short Answer Type Questions-I Q. 1. State Henry’s law. Calculate the solubility of CO2 in water at 298K under 760 mm Hg. (KH for CO2 in water at 298 K is 1.25 × 106 mm Hg)  R + U [CBSE Outside Delhi Set-1, 2020] Ans. Henry’s law : The mass of a gas dissolved in a given volume of the liquid at a constant temperature depends upon the pressure which is applied. KH for CO2 = 1.25 × 106 mm Hg



Partial pressure of CO2 K H for CO2 760 mm Hg = 1.25×10 6 mm Hg.

xCO2 =

= 608 ´ 10 -6







Mole fraction represents the solubility of CO2 in water. [2] Q. 2. State Henry’s law and write its two applications. R [CBSE Delhi Set-3 2019]

(2 marks each)

Ans. Henry’s law states that “the partial pressure of the gas (p) in vapour phase is proportional to the mole fraction of the gas (c) in the solution”. [1] • To increase the solubility of CO2 in soft drinks • At high altitudes the partial pressure of oxygen is less than that at the ground level. This leads to low concentrations of oxygen in the blood and tissues of people living at high altitudes or climbers. • Scuba divers must cope with high concentrations of dissolved gases while breathing air at high pressure underwater. Increased pressure increases the solubility of atmospheric gases in blood.  (Any two) [½ + ½] [CBSE Marking Scheme, 2019] Q. 3.

Calculate the molality of ethanol solution in which the mole fraction of water is 0.88. U

Ans. Mole fraction of water, XH2O = 0.88 Mole fraction of ethanol,  XC2H5OH = 1 – 0.88

[ 37  





SOLUTIONS



n1 =



...(1)





1000 = 55.5 moles 18 

  Substituting the value of n1 in equation (1)



[½]

n2 = number of moles of ethanol. n1 = number of moles of water. Molality of ethanol means the number of moles of ethanol present in 1000 g of water.





n2 XC2H5OH = n1 + n2





  = 0.12

[½] n2 = 7.57 moles [½] Molality of ethanol (C2H5OH) = 7.57 m Alternatively,   Mole fraction of water = 0.88 [½]    Mole fraction of ethanol = 1 – 0.88 = 0.12 [½] Therefore 0.12 moles of ethanol are present in 0.88 moles of water. Mass of water = 0.88 × 18 =15.84 g of water.[½] Molality = number of moles of solute (ethanol) present in 1000 g of solvent (water)

= 0.12 × 1000 / 15.84 = 7.57 m [½] Molality of ethanol (C2H5OH) = 7.57 m [CBSE Marking Scheme 2018] Commonly Made Error

 Students get confused between the terms molarity and molality.

[½]

n2 = 0.12 55.5 + n2



Answering Tip  Students should remember that molarity is volume based concept and molality is mass based concept. Q. 4.

Calculate the molarity of NaOH solution obtained by dissolving 2g of NaOH in 50 ml of its solution. U Ans. ∵ 50 ml of NaOH solution contains = 2g of NaOH  ∴ 100 ml of NaOH solution will contain 2 ´ 1000 = 50 = 40 g of NaOH [1] Molecular mass of NaOH = 23 + 16 + 1 = 40 Mass Number of moles of NaOH = molecular mass 40 = 1 mol = 40  ∴ Molarity of the solution = 1 M [1]

Short Answer Type Questions-II



Ans.







(i) W  hat is the relationship between Molarity and Normality? R (ii) One litre of Water at N.T.P. dissolves 0.08 g of nitrogen. Calculate the amount of nitrogen that can be dissolved in four litres of water at 0oC and at a pressure of 1520 mm. (i) Molarity (M) × Molecular mass of solute = Normality (N) × Equivalent mass of solute  [1] (ii) Solubility of gas (C1) = 0.08 g/litre P1 = 760 mm P2 = 1520 mm Solubility of gas (C2) at pressure P2 = ? By Henry’s law, C1 P1 = C 2 P2 C ×P 0.08 ´ 1520 \ C2 = 1 2 = = 0.16 g/litre P1 760 ∴ Solubility of nitrogen in 4 litres of water = 0.16 × 4 = 0.64 g [1]

Commonly Made Error  Some students get confused in using correct formula for calculation of solubility of gas. Answering Tip  Students must understand Henry’s law.

Q. 2.



Ans.















Q. 1.



(3 marks each)

8.0575 × 10–2 kg of Glaubers’s salt is dissolved in water to obtain 1 dm3 of a solution of density 1077.2 kg m–3. Calculate the molarity, molality and mole fraction of Na2SO4 in the solution. Mass of Glauber’s salt = 8.0575 × 10–2 kg = 8.0575 × 10–2 × 103 g = 80.575 g Molecular mass of Glauber’s salt (Na2SO4.10 H2O) = 322 Number of moles Glauber’s salt 80.575 = = 0.25 322 Mass of solution per dm3 = 1077.2 g m–3  = 1077.2 × 103 g m–3 = 1077.2 × 103 × 10–3 g dm–3 = 1077.2 g Mass of water = 1077.2 – 80.575 = 996.625g 0.25 = 0.25 Molarity = 1 dm 3 0.25 mol ´ 1000 Molality = = 0.2508 996.625 Molarity [1] mass of water Molarity + molecular mass of water 0.25 = 4.49 ´ 10 -3 = 996.625 0.25 + 18

mole fraction =



38 ] Q.3.

Ans.











Q. 4.

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Calculate the concentration of a solution that is obtained by mixing 300 g of 25% solution NH4NO3 with 150 g of 40% solution of NH4NO3.  U Total mass of solution = 300 + 150 = 450 g Amount of solute present in 300 g of 25% solution 25 = 300 ´ = 75 g  [1] 100 Similarly, Amount of solute present of 150 g of 40% solution 40 [1] = 150 ´ = 60 g  100 Total mass of solute  = 75 + 60 = 135 g Concentration of solution (in %) mass of solute in gm = ´ 100 mass of solution in gm 135 = ´ 100 450 [1] = 30% (a) Define mole fraction (b) Explain the following phenomena with the help of Henry’s law :



(i) Painful condition known as bends.



(ii) Feeling of weakness and discomfort in breathing at high altitude.  R

Ans.

(a) It may be defined as the ratio of number of moles of one component to the total number of moles of all the components (solvent and solute) present in the solution. (b) (i) Deep sea divers depend upon compressed air for breathing at high pressure under water. The compressed air contains N in addition to O2, which are not very soluble in blood at normal pressure. However, at depths when the diver breathes in compressed air from the supply tank, more N2 dissolves in the blood and other body fluids because the pressure at that depth is far greater than the surface atmospheric pressure. When the diver comes towards the surface, the pressure decreases, N2 comes out of the body quickly forming bubbles in the blood stream. These bubbles restrict blood flow, affect the transmission of nerve impulses. The bubbles can even burst the capillaries or block them and starve the tissues of O2. This condition is called the bends, which are painful and lifethreatening.[1] (ii) At high altitudes, the partial pressure of O2 is less than that at the ground level. This results in low concentration of oxygen in the blood and tissues of the people living at high altitudes or climbers. The low blood oxygen causes climbers to become weak and unable to think clearly known as anoxia.[1]

Long Answer Type Questions (i) Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN. (ii) C  ommercially available concentrated hydro chloric acid contains 38% HCl by mass and has density 1.19g cm–3. What is the molarity of this solution? U+ Ans.(i) 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN. Then, total mass of the solution = (6.5 + 450) g = 456.5 g Therefore, mass percentage of C9H8O4

Q. 1.

6.5 ´ 100% 456.5 [2] = 1.424% =

(ii)















Density of HCl = 1.19 g cm–3 Mass of 1000 cm3 of commercial HCl  = 1000 × 1.19 = 1190 g [1] Mass of HCl in 1190 g (1000 cm3) of the solution 1190 ´ 38 = 452.2 g    = [1] 100 Molar mass of HCl      = 1 + 35.5 = 36.5 g 452.2  452.2 g of HCl = moles of HCl 36.5  =12.4 moles of HCl ∴ 1.0L of the commercial HCl solution contains 12.4 moles of HCl. Molarity of the commercial HCl solution = 12.4 M [1]

(5 marks each)

Q. 2. Concentration terms such as mass percentage, ppm, mole fraction and molality are independent of temperature, however molarity is a function of temperature. Explain.  R Ans. (a) Mass Percentage (w/W) : It is defined as the ratio of mass of the component of the solution and total mass of the solution multiplied by 100%. (b) Parts per million : It is defined as the ratio of number of parts of the component to the total number of parts of all components of the solution multiplied by 106 and it is used to express concentration of a solution where solutes are present in traces. It is denoted by the alphabet x and subscript written on the right hand side of x denotes the component of which mole fraction is being calculated. (c) Mole Fraction : It is defined as the ratio of number of moles of the component in the solution to the total number of moles of all components in the solution. (d) Molality : It can be defined as the ratio of number of moles of solute to the mass of solvent in kg or it can also be defined as the number of moles of solute present in unit kilogram of solvent. Moles of solute Molality ( m ) = Mass of solvent in kilogram

[ 39

SOLUTIONS

(e)

Q.3. Ans.



































Molarity : It is defined as the ratio of number of moles of solute to the volume of solution in litre or it can also be defined as the number of moles of solute present in unit litre or cubic decimetre of solution. It is function of temperature due to the dependence of volume on temperature whereas Mass %, ppm, mole fraction and molality are independent of temperature because mass does not depend on temperature. [5] 4.0 g of NaOH are contained in one decilitre of solution. Calculate the following : (i) Molality fraction of NaOH (ii) Molarity of NaOH (iii) Molality of NaOH Density of solution = 1.038 g/cm3 U (i) Density of solution = 1.038 g/cm3  Mass of 100 cm3 of solution = 1.038 × 100  = 103.8 g  Mass of NaOH in 100 cm3 solution = 4g  Mass of water = 103.8 – 4.0 = 99.8 g [1] 4.0  Number of moles of NaOH (nA) = 40 = 0.1 mol  Number of moles of water (nB) = 99.8 18   = 5.54 mol nA  ∴ Mole fraction of NaOH XA = n A+ n B





(ii)  (iii)

























=

0.1 = 0.018 0.1 + 5.54

[1]



Molarity of NaOH solution Mass of NaOH per litre of solution 40 = = = 1M Molecular mass of NaOH 40 Mass of NaOH in 99.8 g of water = 4.0 g ∴ Mass of NaOH in 1000 g of water 4.0 ´ 1000 = 99.8 = 40.08 g [1]  Molality of NaOH solution Mass of NaOH per 100 g of water = Molecular mass of NaOH   40.08 = 40 = 1.002 mol kg–1 [1]

Commonly Made Error  Sometimes, students are unable to calculate the concentration of solution from given data. Answering Tip  Understand the formulae to calculate concentration of solution in different ways.

the

TOPIC-2

Vapour Pressure, Raoult’s Law, Ideal and Non-ideal Solutions

Revision Notes  Vapour pressure is the pressure exerted by vapours over a liquid at equilibrium state at constant temperature.  Vapour pressure depends on the following factors : (i) Nature of the liquid. (ii) Temperature : Vapour pressure of a liquid increases with increase in temperature. Scan to know more about  Raoult’s law for a solution of volatile liquids : It states that for a solution of volatile liquids, the this topic partial vapour pressure of each component of the solution is directly proportional to its mole fraction in solution. Suppose a solution is prepared by mixing two volatile liquids A and B. Let χA and χB respectively be their mole fractions, and let pA and pB be their partial vapour pressures respectively in the Raoult”s law solution at a particular temperature.

o o If pA and pB are their vapour pressures in the pure state respectively, then according to Raoult’s law :





o pA = pA χA





o pB = pB χB



Considering Dalton’s law of partial pressure, Substituting values of pA and pB,

ptotal = pA + pB





o o o o ptotal = χA pA + χB pB = (1 – χB) pA + χB pB



o o o = pA + ( pB – pA )χB

40 ]



Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

The composition of the vapour phase in equilibrium with the solution can be determined from the partial pressure of the two components. If ϒA and ϒB are the mole fractions of components A and B respectively in the vapour phase, then pA = γAptotal and pB = γBptotal In general pi = γi ptotal  Raoult’s law as a special case of Henry’s law : According to Raoult’s law, the vapour pressure of volatile component (A) in a given solution is given as :



o pA = pA χA According to Henry’s law, in the solution of a gas in a liquid, the gaseous component is normally so volatile that it exists as a gas and solubility depends upon Henry’s law to which : pA = KHχA

o On comparing both expressions pA is equal to KH.  Raoult’s law for non-volatile solute : For a solution containing non-volatile solute present in a volatile solvent, Raoult’s law may be stated as the relative lowering of vapour pressure for a solution is equal to the mole fraction of solute. 0 χB = pA − pA , pA0



 χB = Mole fraction of solute,

where, pAo

– pA = Lowering of vapour pressure.  Ideal solution : A solution which obeys Raoult’s law over a wide range of concentration at specific temperature is called ideal solution. 

o o (i) Raoult’s law is obeyed. pA = pA χA, pB = pB χB (ii) ∆mixH = 0, (iii) ∆mixV = 0, Ideal solutions (iv) The force of attraction between A-A and B-B is nearly equal to A-B. and azeotropes Some examples of ideal solutions are : (i) n-hexane and n-heptane, (ii) Ethyl bromide and ethyl chloride, (iii) Benzene and toluene, (iv) Chlorobenzene and bromobenzene. Non-ideal solution : A solution which does not obey Raoult’s law for all the concentrations is called a non-ideal solution.



o o (i) Raoult’s law is not obeyed, i.e., pA ≠ pA χA and pB ≠ pB χB (ii) ∆ mixH ≠ 0,



(iii) ∆ mixV ≠ 0,



(iv) The force of attraction between A-A and B-B is not equal to A-B.

 



Scan to know more about this topic

Some examples of non-ideal solutions are : (i) Water and ethanol (ii) Chloroform and acetone (iii) Ethanol and cyclohexane A non-ideal solution can show either positive or negative deviation from Raoult’s law. Positive deviation from Raoult’s law : In this type of deviation, A-B interactions are weaker than the interaction between A-A or B-B and leads to increase in vapour pressure. Some examples are : (i) Water and ethanol, (ii) Chloroform and water, (iii) Ethanol and CCl4, (iv) Methanol and chloroform, (v) Benzene and methanol, (vi) Acetic acid and toluene. Negative deviation from Raoult’s law : In this type of deviation in non-ideal solutions, the intermolecular attractive forces between A-A and B-B are weaker than those between A-B and leads to decrease in vapour pressure. Some examples are : (i) Chloroform and acetone, (ii) Chloroform and methyl acetate, (iii) H2O and HCl, (iv) H2O and HNO3, (v) Acetic acid and pyridine, (vi) Chloroform and benzene.

[ 41

SOLUTIONS

p

p

1

2

1 = 0 Mole fraction 1 2 = 1

1 = 1 2 = 0

Vapour pressure of solution

Vapour pressure

Vapour pressure

Vapour pressure of solution

p

2

p

1

1 = 0 Mole fraction 1 2 = 1

2

1 = 1 2 = 0

2

(b)

(a)

Fig. 2 : The vapour pressures of two component systems as a function of composition : (a) A solution that shows positive deviation from Raoult’s law, and (b) A solution that shows negative deviation from Raoult’s law.  Azeotropes : Liquid mixtures which distil over without change in composition are called constant boiling mixtures or azeotropes or azeotropic mixtures.  Minimum boiling azeotropes : Non-ideal solutions showing large positive deviation from Raoult’s law form minimum boiling azeotropes at a specific composition. e.g, water and benzene, chloroform and methanol.  Maximum boiling azeotropes : Non-ideal solutions showing large negative deviation from Raoult’s law form maximum boiling azeotropes which boil at temperature higher than the boiling points of its components. e.g. mixture of HNO3 and H2O.

Know the Formulae  Raoult’s law for a solution of volatile solute in volatile solvent : o pA = pA χA



o pB = pB χB  Raoult’s law for a solution of non-volatile solute and volatile solvent :

p oA − p A

p oA

= iχ B = i

nB W × MA (for dilute solution) =i B nA WA × MB

Mnemonics • Concept: Raoult’s law for Non-volatile Solute • Mnemonic: R.L. is Very Poor Student = Most Failure Student • Interpretation: Relative Lowering Of Vapour Pressure For A Solution Is Equal To The Mole Fraction Of Solute. • Concept: Ideal solution • Mnemonic: ISRaeL • Interpretation: Ideal Solution Obeys Raoult’s Law • Concept: Non-Ideal solution • Mnemonic: Nano Scale Device Research Labaratory • Interpretation: Non-Ideal Solution Does Not Obey Raoult’s Law

42 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q. The vapour pressure of benzene and toluene at 20oC are 75 mm Hg and 22 mm Hg respectively. 23.4 g of benzene and 64.4 g of toluene are mixed. If the two form an ideal solution, calculate the mole fraction of benzene in the vapour phase if the vapours are in equilibrium with the liquid mixture at the same temperature. Solution: STEP-1: Mass of benzene  = 23.4 g Molecular mass of benzene (C6H6) = 12 × 6 + 1 × 6 = 78 \ Number of moles of benzene in solution 23.4 [½] = = 0.3 78  Mass of toluene = 64.4 g Molecular mass of toluene (C7H8) = 12 × 7 + 1 × 8 = 92 \ Number of moles of toluene in solution

64.4 = 0.7 [½] 92  STEP-2: Mole fraction of benzene 0.3 = 0.3 [½] = 0.3 + 0.7  Mole fraction of toluene 0.7 = = 0.7  [½] 0.3 + 0.7   Partial pressure of benzene vapour = 75 × 0.3 = 22.5 mm [1] Partial pressure of toluene vapour = 22 × 0.7 = 15.4 mm [1] Total vapour pressure = 22.5 + 15.4 = 37.9 mm STEP-3: Mole fraction of benzene in vapour phase Partial vapour pressure = Total vapour pressure =

=

Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS : Q. 1. When 1 mole of benzene is mixed with 1 mole of toluene the vapour will contain : (Given : vapour of benzene = 12.8kPa and vapour pressure of toluene = 3.85 kPa). [CBSE, SQP, 2020-2021] (a)  equal amount of benzene and toluene as it forms an ideal solution (b) unequal amount of benzene and toluene as it forms a non ideal solution (c) higher percentage of benzene (d) higher percentage of toluene Ans. Correct option : (c) Explanation : When 1 mole of benzene is mixed with 1 mole of toluene the vapour will contain higher percentage of benzene. As it is an ideal solution, it follows Raoult’s law.

22.5 = 0.59 37.9 

[1]

(1 mark each) The vapour pressure of the solution depends on the mole fraction of the solvent. P =c P0 soln solvent solvent Psoln is the vapour pressure of the solution χ is the mole fraction of the solvent 0 Psolvent is the vapour pressure of the pure solvent Since the mole fraction of both the components is same, but the vapour pressure of benzene is higher than the toluene, its percentage will be greater in the vapour of the solution. Q. 2. If two liquids A and B form minimum boiling azeotrope at some specific composition then _________. (a) A–B interactions are stronger than those between A–A or B–B.



[ 43

SOLUTIONS

(b) Vapour pressure of solution increases because more number of molecules of liquids A and B can escape from the solution. (c) Vapour pressure of solution decreases because less number of molecules of only one of the liquids escape from the solution. (d)  A–B interactions are weaker than those between A–A or B–B. R Ans. Correct Option : (d) Explanation : When solute-solvent or A-B interactions are weaker than the A-A or B-B interactions, molecules of A or B will find it easier to escape than in pure state. This will increase the vapour pressure and result in positive deviation from Raoult’s law. Such solutions are called minimum boiling azeotropes. Q. 3. For a dilute solution, Raoult’s law states that (a) The lowering of vapour pressure is equal to the mole fraction of solute. (b) The relative lowering of vapour pressure is equal to the mole fraction of solute. (c)  The relative lowering of vapour pressure is proportional to the amount of solute in solution. (d) The vapour pressure of the solution is equal to the mole fraction of the solute. R Ans. Correct option : (b) Explanation : According to Raoult’s law, for a dilute solution, the relative lowering of vapour pressure is equal to the mole fraction of solute. PA 00 - PA PA -0 PA = X B PA 0 = X B P Where 0 A PA 0 - PA PA -0 PA = Relative lowering of vapour pressure PA = Relative lowering of vapour pressure PA 0 X B = mole fraction of solute  X B = mole fraction of solute [B] ASSERTIONS AND REASONS: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.

(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Q. 1. Assertion : An ideal solution obeys Henry’s law. Reason : In an ideal solution, solute-solute as well as solvent-solvent interactions are similar to solutesolvent interaction. [CBSE Delhi Set-3, 2020] Ans. Correct option : (d) Explanation : An ideal solution obeys Raoult’s law. Q. 2. Assertion : Dimethyl ether is less volatile than ethyl alcohol.  Reason : Dimethyl ether has greater vapour pressure than ethyl alcohol. Ans. Correct option : (d) Explanation : Dimethyl ether is more volatile than ethyl alcohol. Q. 3. Assertion : Vapour pressure increase with increase in temperature. Reason : With increase in temperature, more molecules of the liquid can go into vapour phase. Ans. Correct option : (a) Explanation : Vapour pressure increase with increase in temperature because more molecules of the liquid can go into vapour phase with increase in temperature. [C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1. What is an ideal solution? Ans. The solution which obeys Raoult’s law is called ideal solution. Q. 2. Define azeotropic mixture. Ans. A solution which distilled without change in composition at a particular temperature is called azeotropic mixture. Q. 3. What do you meant by minimum boiling azeotropes? Ans. The azeotropes which are formed by those liquid pairs which show positive deviation from ideal behaviour.

Short Answer Type Questions-I

(2 marks each)

Q. 2. State Raoult’s law for a solution containing volatile components. Write two characteristics of the solution which obey Raoult’s law at all concentrations. [CBSE, Delhi Set-1, 2019]

Ans. Raoult’s law for a solution containing volatile components states that the partial pressure of a volatile component present in a solution is directly proportional to the mole fraction of that component at a given temperature. PA a cA or PA = KcA Raoult’s law and Henry’s law are similar as both gives equation to find partial pressure of gases.

Ans. For a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution. [1]



PA = K χ A (Raoult's law)

PA = K H χ A [2] (Henry's law)



Q. 1. State Raoult’s law for a solution containing volatile components. What is the similarity between Raoult’s law and Henry’s law?  [CBSE, Delhi Set 1 & 2, 2020]

(i) ∆mix H = 0 (ii) ∆mix V = 0 (iii) The components have nearly same intermolecular force of attraction       [½, ½] 

(any two) [CBSE Marking Scheme, 2019]

44 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Detailed Answer : Raoult’s law for a solution containing volatile components - The partial vapour pressure of each volatile components of the solution is directly proportional to mole fraction present in solution. The ideal solutions have characteristics to obey Raoult’s law for the following conditions. (i)  The enthalpy of mixing of the pure components to form the solution is zero, ∆mix H = 0. That means no heat is absorbed or released during the mixing of the two pure components. (ii) The volume of the mixing also be zero, ∆mix V = 0. That means the total volume of the solution is equal to the sum of the volume of the two components.



(iii) In pure components, A and B, the intermolecular attractions between solute-solute interactions and solvent-solvent interaction are almost similar to the solute-solvent interaction. [2] Q. 3. Write two differences between ideal solutions and non-ideal solutions. [CBSE, Delhi Set 2, 2019] Ans. Ideal solution

Non-ideal solution

Obeys Raoult’s law at all Does not obey range of concentrations. Δmix H = 0, Δmix V = 0

Δmix H ≠ 0, Δmix V ≠ 0 (or any other difference)

[1 + 1] [CBSE Marking Scheme, 2019]

Detailed Answer : Ideal solution

Non-ideal solution

(i) Raoult’s Law

Solutions which obey Raoult’s law over Do not obey Raoult’s law over the entire range the entire range of concentrations. of concentrations.

(ii) D ü  H

The enthalpy of mixing of the pure The enthalpy of mixing of pure components to components to form a solution is zero. form a solution is not zero. D mix  H > 0 D mix  H = 0

(iii) D mix  V

The volume of the mixing is also zero, The volume of the mixing is not zero, D mix  V  ¹ 0 . D mix  V = 0. The total volume of the solution is not equal to That means the total volume of the the sum of the volume of the two components. solution is equal to the sum of the volume of the two components.

(iv) In pure components, A and B, the Intermolecular intermolecular attractions between soluteinteractions solute interactions and solvent-solvent interactions are almost similar to the solute-solvent interaction.

[2] Also, the freezing point of water is higher than when water contains KCl. [2]

 

Ans. (a)  Ethanol-acetone interaction is weaker than pure ethanol or acetone interactions. (b) On adding KCl, vapour pressure of the solution decreases[1 + 1]  [CBSE Marking Scheme 2019] Detailed Answer : (a) When ethanol is mixed with acetone, it shows positive deviation from Raoult’s law and acetone molecules get in between the host molecules and break some hydrogen bonds, which requires higher energy than energy released in the formation of new hydrogen bonds. This results fall in temperature. (b) According to Raoult’s law, when a non-volatile solid is added to the solvent, its vapour pressure decreases resulting in decrease in freezing point.

Q. 5. Give reasons : (a) An increase in temperature is observed on mixing chloroform and acetone. (b) Aquatic animals are more comfortable in cold water than in warm water. [2] A [CBSE, Outside Delhi Set 3, 2019]



Ans. (a) Due to stronger interaction between chloroform and acetone than pure chloroform or acetone interactions.[1] (b) Because of high solubility of oxygen gas /low KH value in cold water than in warm water. [1]  [CBSE Marking Scheme 2019] Detailed Answer : (a) On addition of chloroform and acetone, chloroform forms strong hydrogen bonding with acetone.

Q. 4. Give reasons : (a)  A decrease in temperature is observed on mixing ethanol and acetone. (b) P  otassium chloride solution freezes at a lower temperature than water. [2] A [CBSE, Outside Delhi, 2019]

In pure components, A and B, the intermolecular attractions between solute-solute interactions and solvent-solvent interaction are not similar to the solute-solvent interaction.

CH3

Cl C O

CH3

H C

Cl Cl

[ 45

SOLUTIONS



 his results in release of energy due to increase T in attractive forces. Hence, the dissolution is an exothermic process. (b) At a given pressure, the solubility of oxygen in water increases with decrease in temperature. Therefore, the concentration of oxygen in sea is more in cold water and thus, the presence of more oxygen at lower temperature makes the aquatic animals more comfortable in cold water.  [2] Q. 6. Define the following terms : (i) Ideal solution (ii) Molarity (M) [2]  [CBSE, Delhi Set 2, 2017] Ans. (i) The solution that obeys Raoult's Law over the entire range of concentration. [1] (ii) Number of moles of solute dissolved per litre wb × 1000 of solution or M = . [1] M b × V( mL)



[CBSE Marking Scheme, 2017] Detailed Answer : (i) An ideal solution which obeys Raoult’s law over entire range of concentration. The necessary condition to reach ideal solution is : Enthalpy of mixing of the pure components to form the solution, ΔHmix = 0 and volume of mixing, ΔVmix = 0 [1] (ii) Molarity is defined as the number of moles of solute present in 1000 mL of the solution. Molarity is represented by M.

OR Azeotropes : A liquid mixture which distills at constant temperature without undergoing any change in composition is called azeotrope. [1] Minimum boiling azeotrope is formed by positive deviation from Raoult’s law.  [½] Example : (i) Water and benzene (ii) Chloroform and methanol [½] Q. 8. The experimentally determined molar mass for what type of substances is always lower than the true value when water is used as solvent. Explain. Give one example of such a substance which does not show a large variation from the true value. R Ans. When there is dissociation of solute into ions in dilute solutions (ignoring interionic attractions) the number of particles increases. As the value of colligative properties depends on the number of particles of the solute, the experimentally observed value of colligative property will be higher than the true value, therefore the experimentally determined (observed) molar mass is always lower than the true value. [1] For KCl (electrolyte), the experimentally determined molar mass is always lower than the true value when water is used as solvent. [½] Glucose (non-electrolyte) does not show a large variation from the true value. [½] Q. 9. Why a mixture of carbon disulphide and acetone shows positive deviation from Raoult’s law? What type of azeotrope is formed by this mixture ? Ans. Intermolecular forces of attraction between carbon disulphide and acetone are weaker than the pure components.[1] Minimum boiling azeotrope at a specific composition.[1] Q. 10.  State Raoult’s law for a solution containing volatile components. What is the similarity between Raoult’s law and Henry’s law ? R Ans. Raoult’s law for a solution containing volatile components states that the partial pressure of a volatile component present in a solution is directly proportional to the mole fraction of that component at a given temperature. PA a cA or PA = KcA [1] Raoult’s law and Henry’s law are similar as both gives equation to find partial pressure of gases. PA = KcA PA = KHcA [1] (Raoult’s law) (Henry’s law)



M =

Number of moles of solute × 1000 [1] Volume of solution in mL

Q. 7. What is meant by positive deviations from Raoult’s law ? Give an example. What is the sign of DmixH for positive deviation ? OR Define azeotropes. What type of azeotrope is formed by positive deviation from Raoult’s law ? Give an example. R [CBSE Delhi 2015] Ans. In case of positive deviation from Raoult’s law, the intermolecular attractive forces between the solutesolvent molecules are weaker than those between the solute-solute and solvent-solvent molecules. Example : Mixture of ethanol and acetone. Sign for ∆mixH is positive [1+½+½]

Short Answer Type Questions-II Q. 1.





(a) Define vapour pressure of the liquid. (b) (i) Gas (A) is more soluble in water than Gas (B) at the same temperature. Which one of the two gases will have the higher value of KH (Henry’s constant) and why ? (ii) In non-ideal solution, what type of deviation shows the formation of maximum boiling azeotropes ?  R+

Ans.





(3 marks each)

(a) The pressure exerted by the vapours above the liquid surface in equilibrium with the liquid at a given temperature is called vapour pressure of the liquid.[1] (b) (i) Gas B will have the higher value of KH (Henry’s constant) as lower is the solubility of the gas in the liquid higher is the value of KH. [1]

46 ]



Q. 2.

Ans.





Q. 3.

Ans.











Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(ii) In non-ideal solution, negative deviation shows the formation of maximum boiling azeotropes. [1] (i) Give the factors affecting the vapour pressure of a liquid. (ii) Suggest the most important type of intermolecular attractive interaction in the following pairs : (a) n-hexane and n-octane (b) I2 and CCl4 R+









(i) Nature of liquid : Liquids having week intermolecular forces are volatile and possess higher vapour pressure. Temperature : Vapour pressure increases with increase in temperature. [½ + ½] (ii) (a) van der Waals forces of attraction. (b) van der Waals forces of attraction.[1] In aqueous solution containing 20% by weight of liquid ‘A’ (Mol.wt = 140) has vapour pressure of 160 mm at 57o C. Find the vapour pressure of pure A, if that of water is 150 mm at this temperature. U 20 Number of moles of A dissolved = = 0.143 140  [½]

Ans. ptotal = p1o + (p2o - p1o)x2 [1] 600 = 450 + (700 - 450) x2 [1] x2 = 0.6 [½] x1 = 1 - 0.6 = 0.4 [½] [CBSE Marking Scheme 2017]





Detailed Answer : o

 pAo = 450 mm Hg, pB = 450 mm Hg  pTotal = 600 mm Hg According to Raoult’s law,



 pA = cA × pAo o  pB = cB × pB     pTotal = pA + pB



o o = (1 - cB) pAo + cB pB = pAo + ( pB - pAo ) cB   600 = 450 + (700 - 450)cB 600 - 450 = 250cB 0.143  150 = 250cB = 0.031 Mole farction of A  (cA) = 0.143 + 4.44  [½]  c = 150 = 0.6 B 4.44 250 Mole farction of B (cB) = = 0.969 0.143 + 4.44 ⇒    cA = 1 - 0.6 = 0.4  [½]

Number of moles of

  B = 80 = 4.44  18

[½]

Q. 1. (a) 30 g of urea (M = 60 g mol–1) is dissolved in 846 g of water. Calculate the vapour pressure of water for this solution if vapour pressure of pure water at 298 K is 23.8 mm Hg. (b) Write two differences between ideal solutions and non-ideal solutions. U + R [CBSE OD Set-1, 2 & 3, 2017]

0 0 Ans. (a) (i) (PA – PA)/PA = (wB × MA)/(MB × wA) 23.8 − PA = (30 × 18)/60 × 846 23.8



160 = PA ´ 0.031 + 150 ´ 0.969

 \ PA = 472.58 mm [1] Q. 4. The vapour pressure of pure liquids A and B at 400 K are 450 and 700 mm Hg respectively. Find out the composition of liquid mixture if total pressure at this temperature is 600 mm Hg. [CBSE Comptt. Delhi 2017]

Long Answer Type Questions



 Total pressure = 160mm = PA ´ c A + PB ´ c B

23.8 – PA = 23.8 × [(30 × 18)/60 × 846] 23.8 – PA = 0.2532

1

(5 marks each)

PA = 23.9 – 0.2532 = 23.55 mmHg (b) Ideal solution Non-ideal solution

1

(a) It obeys Raoult's law (a) Does not obeys Raoult's over the entire range law over the entire range of concentration. of concentration. (b) Dmix H = 0 (b) Dmix H is not equal to 0. (c) Dmix V = 0 (c) Dmix V is not equal to 0. 1 + 1 (Any two correct difference) [CBSE Marking Scheme, 2017]

½ ½ OR

[ 47

SOLUTIONS



 

[Topper’s Answer 2017] 5 Commonly Made Error

Q. 2.

 Some students can not calculate the vapour pressure of water for the solution from given data. Answering Tip Ans.

 Students must understand the relationship between relative of vapour pressure and mole fraction.

The vapour pressures of ethanol and methanol are 44.5 mm and 88.7 mm Hg respectively. An ideal solution is formed at the same temperature by mixing 60 g of ethanol with 40 g of methanol. Calculate the total vapour pressure of the solution and the mole fraction of methanol in the vapour. Mass of ethanol = 60 g Molecular mass of ethanol (C2H5OH) = 12 ´ 2 + 1 ´ 5 + 16 + 1 = 46 



































Q. 3.

Ans.

1.304 = 0.51 Mole fraction of ethanol = 1.304 + 1.250  [½] Mole fraction of methanol = 1 – 0.51 = 0.49 [½] Vapour pressure of pure ethanol PoC2 H5 OH = 44.5 mm Vapour pressure of pure methanol o   P CH3 OH = 88.7 mm Vapour pressure due to ethanol P = PoC2 H5 OH ´ 0.51    C2 H5 OH   = 44.5 mm × 0.51 = 22.69 mm [½] Vapour pressure due to methanol P = PoCH3 OH ´ 0.49   CH3 OH   = 88.7 mm × 0.49 = 43.46 mm [½] Total Vapour pressure =PC2 H5 OH + PCH3 OH

Vapour pressure of solution

p

Vapour pressure of solution

p

1

2

1 = 0 Mole fraction 1 2 = 1

1 = 1 2 = 0

p

2

1 = 0 Mole fraction 1 2 = 1





Vapour pressure of a two-component solution showing positive deviation from Raoult’s law : In the case of an ideal solution, the enthalpy of the mixing of the pure components for forming the solution is zero. ΔsolH = 0

Vapour pressure of solution



p

2

1 = 1 2 = 0

2

(b)

(a)



p

1

[1]

2

 = 22.69 + 43.46 = 66.15 mm [1] Mole fraction of methanol in the vapour state PCH3 OH 43.46 = =  [1]p Total vapour pressure 66.15 1 = 0.657 What is meant by positive and negative deviations from Raoult’s law and how is the 1 = 0 Mole fraction sign of ∆mixH related to positive and negative 1 1 2 = R deviations from Raoult’s law?  2 According to Raoult’s law, the partial vapour (a) pressure of each volatile component in any solution is directly proportional to its mole fraction. The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. The solutions that do not obey Raoult’s law (non-ideal solutions) have vapour pressures either higher or lower than that predicted by Raoult’s law (given in figure). [1]

I f the vapour pressure is higher, then the solution is said to exhibit positive deviation, and if it is lower, then the solution is said to exhibit negative deviation from Raoult’s law.

Vapour pressure





Vapour pressure



∴ Number of moles of ethanol = 60 = 1.304 [½] 46 Mass of methanol = 40 g Molecular mass of methanol (CH3OH) = 12 ´ 1 + 1 ´ 3 + 16 + 1 = 32  ∴ Number of moles of methanol = 40 = 1.250 32  [½]

Vapour pressure of solution

Vapour pressure



Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Vapour pressure





48 ]

p

2

p

1

1 = 0 Mole fraction 1 2 = 1

1 = 1 2 = 0

2

[1]

(b)

In the case of solutions showing positive deviations, absorption of heat takes place. so, ΔsolH = Positive [fig (a)] In the case of solutions showing negative deviations, evolution of heat takes place. so, ΔsolH = Negative [fig (b)]

[2]

TOPIC-3

Colligative Properties, Determination of Molecular Mass, Abnormal Molecular Mass, van’t Hoff Factor

Revision Notes  Colligative properties : Certain properties of solutions depend only on the number of particles of the solute (molecules or ions) and do not depend on the nature of solute, such properties are called colligative properties. These are : (i) Relative lowering of vapour pressure, (ii) Depression in freezing point, (iii) Elevation of boiling point, (iv) Osmotic pressure of the solution.  Relative lowering of vapour pressure : The relative lowering of vapour pressure is the ratio of lowering of vapour pressure and vapour pressure of pure solvent which is equal to the mole fraction of solute.

1 =

2 =

[ 49

SOLUTIONS



o Vapour pressure of pure solvent = pA



o Lowering of vapour pressure  = pA – pA Relative lowering of vapour pressure

pAo − pA n = χsolute = pAo N+n

where n and N are the number of moles of solute and solvent respectively.  Elevation of the boiling point : The difference in boiling point of solution and pure solvent is called elevation of the boiling point. Boiling point of pure solvent = Tb° Boiling point of Boiling point of solution = Tb Solvent Solution Increase in boiling point ∆Tb = Tb – Tb° is known as 1 Atm elevation of boiling point for dilute solution.

∆Tb ∝

∆p = xB p°

Vapour pressure



∆Tb = KχB ∆Tb =

K b × 1000 × w2 M2 × w1

t en lv n So tio lu So

0

∆Tb

Tb Tb

Where, w2 = weight of solute in g Fig. 3 : The vapour pressure curve for solution M2 = Molar mass of solute lies below the curve for pure water. The diagram w1 = weight of solvent in g shows that ∆Tb denotes the elevation of boiling point ∆Tb = Kbm of a solvent in solution. Kb = Boiling point elevation constant or molal elevation constant or Ebullioscopic constant.  Depression of freezing point : According to Raoult’s law, when a non-volatile solid is added to the solvent its vapour pressure decreases and it would become equal to that of solid solvent at lower temperature. Thus, the difference in the freezing point of pure solvent and that of the solution is known as depression in freezing point.

Temperature (K)

o The freezing point of pure solvent = T f The freezing point when non-volatile solute is dissolved in it = Tf (Freezing point of solution)

o The decrease in freezing point ∆Tf = T f – Tf is known as depression in freezing point. For dilute solution, ∆Tf = KχB W M ∆Tf = K B × A M B WA

∆Tf = K f m.

We know,

WB × 1000 = molality M B × WA (ii) K × MA = Kf Kf = Freezing point depression constant or molal depression constant or Cryoscopic constant. (i)

Vapour pressure



en oz Fr

lv so

t en

t ven sol uid q i L on uti Sol

f 

f

f Temperature/K





Fig. 4 : Diagram showing ∆Tf , depression of the freezing point of a solvent in a solution.  Osmosis : The process in which there is net flow of solvent to the solution by a semipermeable membrane is called osmosis.  Osmotic pressure : The extra pressure that is applied to stop the flow of solvent to solution across a semipermeable membrane is called osmotic pressure of the solution.

50 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII P atm + 

Patm

SPM

Solution



Solvent

Fig. 5 : The excess pressure equal to the osmotic pressure must be applied on the solution side to prevent osmosis. For dilute solution, osmotic pressure is proportional to the molar concentration (C) of the solution at a given temperature T. Thus π = CRT as π is the osmotic pressure and R is the gas constant. n (n is the number of moles, V is the volume of solution) π= VRT  Reverse osmosis : The direction of osmosis can be reversed, if a pressure larger than the osmotic pressure is applied to the solution side. Now the pure solvent flows out of the solution through the semipermeable membrane. This phenomenon is called reverse osmosis.

Piston Pressure >

Salt water

Fresh water





SPM

Fig. 6 : Reverse osmosis occurs when a pressure larger than the osmotic pressure is applied to the solution.   Abnormal molecular mass : When the molecular mass calculated with the help of colligative property is different from theoretical molecular mass, it is called abnormal molecular mass.  van’t Hoff factor(i) : The ratio of the observed (experimental) value of a colligative property to the normal (calculated) value of the same property is called as van’t Hoff factor. Mathematically, Observed (exp erimental ) value of a colligative property i= Normal ( calculated) value of the same colligative property ∆ i = obs Or, ∆ cal



Water outlet



where Δobs and Δcal respectively represent the observed and calculated value of a colligative property. Thus,



(i) for lowering of vapour pressure,

i=

( ∆p )obs ( ∆p )cal



(ii) for elevation of boiling point,

i=

( ∆Tb )obs ; ( ∆Tb )cal



(iii) for depression of freezing point,

i=



(iv) for osmotic pressure,

i=



Since a colligative property is propotional to number of particles of solute. Normal molecular mass i= Observed molecular mass

( ∆Tf )obs ( ∆Tf )cal π obs ; π cal

;

[ 51

SOLUTIONS



Normal molecular mass = i × Calculated molecular mass. i=

Total number of moles of particle after association/dissocciation n/dissociation Total number of moles of particle before association

 Hypertonic solution : A solution is called hypertonic, if its concentration is higher than that of the solution separating it by a semipermeable membrane.  Hypotonic solution : A solution is called hypotonic, if its concentration is lower than that of the solution separating it by a semipermeable membrane.  Isotonic solution : Two solutions are called isotonic, if they exert the same osmotic pressure at a given temperature. Isotonic solutions have same molar concentration. When such solutions are separated by semipermeable membrane no osmosis occurs between them.

Mnemonics • Concept: Different Colligative properties • Mnemonic: RLVP_DFP_EBP_OP • Interpretation: l Relative Lowering Of Vapour Pressure l Depression In Freezing Point l Elevation Of Boiling Point l Osmotic Pressure

Know the Formulae  Modified equations for colligative properties  : (i) Relative lowering of vapour pressure of solvent pAo − pA n = o p N +n A (ii) Elevation of boiling point ∆Tb = iKbm (iii) Depression of freezing point ∆T f = iK f m (iv) Osmotic pressure of solution inRT π = V é m ù ê V = C ú û ë  Determination of molecular mass using colligative properties : (i) Relative lowering of vapour pressure : nB pAo − pA nB = χB = = o n p n + n A A A B WB × M A pAo − pA = WA × M B pAo or

π = i CRT

MB =

pAo

pAo

− pA

×

WB × M A WA

(ii) Elevation of boiling point : ∆Tb = Kb × m W 1000 ⇒ ∆Tb = K × B × . b M B WA in gms MB =



K b × WB × 1000 ∆Tb × WA

52 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(iii) Depression of freezing point : ∆Tf = Kf × m W 1000 ⇒ ∆Tf = Kf × solute × M solute Wsolvent MB =

1000 × WB × K f ∆Tf × WA



(iv) Osmotic pressure :



π = CRT,

⇒

π =

No. of moles ×R×T Volume of solution



π =

WB RT × , here MB = WB×RT MB V π×V

 Degree of dissociation (α) :

α =

i −1 , here i → van’t Hoff factor n −1

n = No. of ions produced per formula of the compound

 Degree of association (α) :

α =

1−i 1 1− n

 Strength = Molarity × Mol. wt.

= Normality × Eq. wt.

 Kb = 0.512 K kg/mol for water  Kf = 1.86 K kg/mol

Q. The freezing point of benzene decreases by 2.12 K when 2.5 g of benzoic acid (C6H5COOH) is dissolved in 25 g of benzene. If benzoic acid forms a dimer in benzene, calculate the van’t Hoff factor and the percentage association of benzoic acid. (Kf = for benzene = 5.12 K kg mol–1) Solution: STEP-1: DTf = iKfm

2.12 K = i.

5.12 Kkgmol-1 ´ 2.5 g ´ 1000 gKg -1 122 gmol-1 ´ 25 g



i = 0.505

[1]

S TEP-2: for association a 2 a = 0.99

[½]

i = 1-





[½]

ercentage association of benzoic acid P 99.0% [½]

[ 53

SOLUTIONS

Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS : Q. 1. The values of van’t Hoff factors for KCl, NaCl and K2SO4, respectively are ____________ . (a) 2, 2 and 2 (b) 2, 2 and 3 (c) 1, 1 and 2 (d) 1, 1 and 1 Ans. Correct option : (b) Explanation : KCl (K+ + Cl–) and NaCl (Na+ + Cl–) ionize to give 2 ions and K2SO4 (K+ + SO2– 4 ) ionize to form 3 ions. So, van’t Hoff factors for KCl, NaCl, and K2SO4 are 2, 2 and 3 respectively Q. 2. We have three aqueous solutions of NaCl labelled as ‘A’, ‘B’ and ‘C’ with concentrations 0.1 M, 0.01 M and 0.001 M, respectively. The value of van’t Hoff factor for these solutions will be in the order (a) iA < iB < iC (b) iA > iB > iC (c) iA = iB = iC (d) iA < iB > iC Ans. Correct option : (c) Explanation : The value of van’t Hoff factor will be i. It is due to complete dissociation of strong electrolyte (NaCl) in dilute solutions and on complete dissociation value of i for NaCl is 2. Q. 3. Consider the figure and mark the correct option. Piston (A) Piston (B) 8PM

Fresh water (A)

Concentrated sodium chloride solution in water (B)

(a)  Water will move from side (A) to side (B) if pressure lower than osmotic pressure is applied on piston (B). (b) Water will move from side (B) to side (A) if pressure greater than osmotic pressure is applied on piston (B). (c) Water will move from side (B) to side (A) if pressure equal to osmotic pressure is applied on piston (B). (d) Water will move from side (A) to side (B) if pressure equal to osmotic pressure is applied on piston (A). Ans. Correct option : (b) Explanation : Water will move from side (b) to side (a) if a pressure greater than osmotic pressure is applied on piston (b). This is a process of reverse osmosis. Q. 4. Which of the following solutions in water has highest boiling point? (a) 1 M NaCl (b) 1 M MgCl2 (c) 1 M urea (d) 1 M glucose Ans. Correct option : (b) Explanation : 1 M MgCl2 in aqueous solution gives maximum number of ions than other solutions. So, it has highest boiling point.

(1 mark each) Q. 5. Which of the following 0.1 m aqueous solution will have the lowest freezing point? (a) Al2(SO4)3 (b) C2H10O5 (c) KI (d) C12H22O11 Ans. Correct option : (a) Explanation : Since depression in freezing point is a colligative property so aqueous solution giving maximum number of ions i.e., Al2(SO4)3 has the lowest freezing point. [B] ASSERTIONS AND REASONS : In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Q. 1. Assertion : Elevation in boiling point is a colligative property. Reason : Elevation in boiling point is directly proportional to molarity.  [CBSE Delhi Set 1 & 2, 2020] Ans.  Correct option : (a)   Explanation : Elevation in boiling point is a colligative property. It is directly proportional to molarity. DTb = K b ´ m Q. 2. Assertion : 0.1 M solution of KCl has great osmotic pressure than 0.1 M solution of glucose at same temperature. Reason : In solution KCl dissociates to produce more number of particles.  [CBSE Delhi Set 2, 2020] Ans.  Correct option : (a)   Explanation : KCl is ionic compound, hence dissociates into ions but glucose is a covalent compound which doesnot dissociate into ions. Q. 3. Assertion : When NaCl is added to water a depression in freezing point is observed. Reason : The lowering of vapour pressure of a solution causes depression in the freezing point. Ans.  Correct option : (a)   Explanation : When a non-volatile solute is added to water, freezing point lowers due to lowering of vapour pressure. Q. 4. Assertion : When methyl alcohol is added to water, boiling point of water decreases. Reason : When a volatile solute is added to a volatile solvent, elevation in boiling point is observed. Ans.  Correct option : (c)   Explanation : When methyl alcohol (volatile) is added to water, boiling point of water decreases because vapour pressure increases when volatile solute is added to volatile solvent.

54 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

[C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1. What happens when a pressure greater than osmotic pressure is applied on the solution side separated from solvent by a semi-permeable membrane?  [CBSE, Outside Delhi, 2020] Ans. Reverse osmosis will take place and the level of solution will decrease. Q.2. Give one practical application of depression of freezing point in automobiles.

Ans. Use of antifreeze (i.e. solution of ethylene glycol) in radiators. Q.3. How is the value of colligative property related to molecular mass of the non-volatile solute? R Ans. colligative property µ 1 molecular mass of the non - volatile solute Q.4. What is the effect of temperature on osmotic pressure? R Ans. osmotic pressure ∝ temperature

Short Answer Type Questions-I



Q. 1. A glucose solution which boils at 101.04°C at 1 atm. What will be relative lowering of vapour pressure of an aqueous solution of urea which is equimolal to given glucose solution? (Given : Kb for water is 0.52 K kg mol–1 )  Ans. ΔTb = Kb m ΔTb = 101.04 – 100 = 1.04 °C or m = 1.04 /0.52 = 2 m [1] 2 m solution means 2 moles of solute in 1 kg of solvent. 2 m aq solution of urea means 2 moles of urea in 1kg of water. No. of moles of water = 1000/18 = 55.5 Relative lowering of vapour pressure = x2 (where x2 is mole fraction of solute) [½] Relative lowering of vapour pressure = n2/n1+n2 (n2 is no. of moles of solute , n1 is no. of moles of solvent) = 2/2 + 55.5 = 2/57.5 = 0.034 [½] Q. 2. Give reasons : (a) Cooking is faster in pressure cooker than in cooking pan. (b) Red Blood Cells (RBC) shrink when placed in saline water but swell in distilled water.

Ans. (a) In case of liquids, due to increase of pressure inside the cooker, the boiling point of water increases leading to faster cooking than in pan. (b) RBC shrink in saline water due to loss of water owing to exosmosis. In distilled water they swell due to endosmosis as the water enters the RBC. [1+1] Q. 3.  Calculate the freezing point of a solution containing 60 g of glucose (Molar mass = 180 g mol 1) in 250 g of water. (Kf of water = 1.86 K kg mol 1) U [CBSE D/OD, 2018] Ans. ΔTf = Kf m w ´ 1000 = Kf × 2 M2 ´ w1 

1.86 ´ 60 ´ 1000 180 ´ 250



=



= 2.48 K



DTf =T fo -Tf   2.48 = 273.15 – Tf Tf = 273.15 – 2.48 = 270.67 K OR

[2]





U [CBSE OD Set-1 2019]

(2 marks each)



[Topper’s Answer 2018]

[ 55

SOLUTIONS

Commonly Made Error  Students do not write applied formula or sometimes miss the next step of value assignment to all the entities in it. Answering Tip

=

Observed colligative property Calculated colligative propertty



Value of i is less than 1 in case of association. Value of i is greater than 1 in case of dissociation. Value of i is equal to 1 in case of no association or dissociation. [1] Q. 6. Explain why on addition of 1 mol glucose to 1 litre water the boiling point of water increases. Ans. Vapour pressure of the solvent decreases in the presence of non-volatile solute (glucose) as the mole fraction of water decreases hence boiling point increases. [2] Q. 7. A 1.00 molar aqueous solution of trichloroacetic acid (CCl3COOH) is heated to its boiling point. The solution has the boiling point of 100.18°C. Determine the van’t Hoff factor for trichloroacetic acid. (Kb for water = 0.512 K kg mol–1). Ans. Tb – 100.18 °C = 373.18 K T°b – 100°C = 373K ΔTb = iKbm [1] (373.18 – 373) K = i × 0.512 K kg mol–1 × 1 m  0.18K = i × 0.512 K kg mol–1 × 1m,  i = 0.35 [1] Q. 8. Explain why on addition of 1 mol of NaCl to 1 litre of water, the boiling point of water increases, while addition of 1 mol of methyl alcohol to one litre of water decreases its boiling point. Ans. Sodium chloride (NaCl) is a non-volatile solute, therefore, addition of NaCl to water lowers the vapour pressure of water. As a result, boiling point of water increases. Methyl alcohol, on the other hand, is more volatile than water, hence, its addition increases the total vapour pressure of the solution and in decrease in boiling point of water results. [2] Q. 9. Which of the following solutions has higher freezing point ? 0.05 M Al2(SO4)3, 0.1 M K3[Fe(CN)6] Justify. Ans. 0.05 M Al2(SO4)3 has higher freezing point. [1] 0.05 M Al2(SO4)3 : i = 5, DT ∝ No. of particles; f DTf = i × concentration    = 5 × 0.05 = 0.25 moles of ions 0.1 M K3[Fe(CN)6] : i = 4, [½] = 4 × 0.1 = 0.4 moles of ions [½] Q. 10. 18 g of glucose, C6H12O6 (Molar Mass = 180 g mol–1) is dissolved in 1 kg of water in a saucepan. At what temperature will this solution boil ? (Kb for water = 0.52 K kg mol–1, boiling point of pure water = 373.15 K) Ans.  DTb = Kb × m [½] 18 g 1 ´ Tb – Tb° = 0.52 K kg mol–1 × 180 g mol -1 1 kg

 Always state the formula applied before starting the calculations. Write each step clearly as it carries marks.





Q. 4. Define the following terms : (i) Colligative properties (ii) Molality (m)

R [CBSE D/OD, 2017]

Ans. (i) Properties that are independent of nature of solute and depend on number of moles of solute only.[1] (ii) Number of moles of solute dissolved per kg of the solvent.[1] [CBSE Marking Scheme, 2017] Detailed Answer : (i) Colligative properties are those properties of the solutions which depend upon the number of solute particles present in the solution irrespective of their nature and are relative to the total number of particles present in the solution. Some colligative properties are elevation of boiling point of solvent, depression of freezing point of solvent, etc. [1] (ii) Molality is the number of moles of solute dissolved in 100 g of a solvent. It is represented by m and is used to express concentration of a solution. It can be calculated as : [1] Number of moles of solute m = × 1000. Weight of solvent in grams Q. 5. Define the following terms : (i) Abnormal molar mass (ii) van’t Hoff factor (i) R [CBSE Delhi set 3, 2017] Ans. (i) If the molar mass calculated by using any of the colligative properties to be different than theoretically expected molar mass.[1] (ii) Extent of dissociation or association or ratio of the observed colligative property to calculated colligative property. [1] [CBSE Marking Scheme, 2017] Detailed Answer : (i) Abnormal molar mass – There are certain cases where due to association or dissociation of molecules, the molar mass of a substance calculated from its colligative property is either lower or higher than the expected or normal value. Such molar mass is called abnormal molar mass. [1] (ii) van’t Hoff factor – To account for the extent of dissociation or association, van’t Hoff introduced a factor i, known as the van’t Hoff factor. Normal molar mass i = Abnormal molar mass





[½]

0.52 ⇒ Tb – 373.15 = 10 ⇒ Tb – 373.15 = 0.052 K ⇒ Tb = 0.052 + 373.15 ∴  Tb = 373.202 K

[½] [½]

56 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q. 11. What are colligative properties? Write the colligative property which is used to find the molecular mass of macromolecules. R Ans. Properties that depend on the number of solute particles irrespective of their nature relative to the total number of particles present in the solution.[1] Osmotic pressure[1] Q. 12. Define the following terms : (a) Ideal solution (b) Osmotic pressure. R Ans. (a) The solution which obeys Raoult’s law over the entire range of concentration. [1]

(b) It is the excess pressure that must be applied to a solution to prevent osmosis. [1] Q. 13. Calculate the boiling point elevation for a solution prepared by adding 10 g CaCl2 to 200 g of water, assuming that CaCl2 is completely dissociated. (Kb for water = 0.512 K kg mol-1; Molar mass of CaCl2 = 111 g mol-1) U Ans. DTb = iKbm Here, m = wB × 1000/MB × wA DTb = [3 × 0.512 K kg mol-1 × 1000 × 10 g/ [1] [111 g mol-1 × 200 g] [1] = 0.69 K

Short Answer Type Questions-II Q. 1. A 0.01 m aqueous solution of AlCl3 freezes at – 0.068oC. Calculate the percentage of dissociation. [Given : Kf for Water = 1.68 K kg mol–1]  U [CBSE Delhi Set 1 & 2, 2020] Ans. Given, m = 0.01 m     DTf (s) = –0.068°C   Kf (aq) = 1.86 K kg mol–1     DTf = i Kf m ∆Tf i= ×m Kf      0.068 [1] i= × 0.01 m = 3.65 1.86     



→ Al3+ + 3Cl–

1 + 3α =    1     3.65 = 1 + 3a



MB =

[1]

 Some student get confused to calculate percentage of dissociation. Answering Tip  To calculate percentage of dissociation, first find van’t Hoff Factor (i). Q. 2. The freezing point of a solution containing 5g of benzoic acid (M = 122 g mol–1) in 35g of benzene is depressed by 2.94 K. What is the percentage association of benzoic acid if it forms a dimer in solution ? (Kf for benzene = 4.9 K kg mol–1)  U [CBSE Outside Delhi Set 1, 2020]

K f ´ WB WA ´ DTf

    WB= 5 gm WA = 0.035 kg

4.9 ´ 5 24.5 MB = = 0.035 2.94 0.1029 ´ Kf = 4.9 K kgmol–1 –1     = 238 g mol  DTf = 2.94 Normal molar mass of C6H5COOH         = 122 g mol–1 normal molar mass 122 i= = = 0.513 observed molar mass 238

% of association of acid (a) 2C 6 H 5COOH  (C 6 H 5COOH )2

0.513 - 1 i -1 ( -0.487) = = = 0.974[5] 1 1 / n-1 ( -0.5) -1 2 Percentage association of acid = 0.974 × 100 = 97.4% Q. 3. A 4% solution w/W of sucrose (M = 342g mol–1) in water has a freezing point of 271.15K. Calculate the freezing point of 5% glucose (M = 180 g mol–1) in water. (Given : Freezing point of pure water = 273.15 K)   U [CBSE Delhi Set 1, 2019] n=2a=





Commonly Made Error







initial 1 mol 0 0 At equilibrium 1–a a 3a Total number of moles at equilibrium     = 1 – a + a + 3a = 1 + 3a[1] Total no. of moles at equilibrium i=      Initial no. of moles

3.65 − 1       α = 3 Percentage dissociation = 0.88%.

Ans. Observed molar mass of benzoic acid :

Ans.

  DTf = K f m



 





AlCl3

(3 marks each)

K f = DTf ´

2 ´ 342 ´ 96 4 ´ 1000

  = 16.4 Km–1



DTf = K f m '



=



  

      

M2 ´ w1 w2 ´ 1000

K f w2 ´ 1000 M2 ´ w1

=

16.4 ´ 5 ´ 1000 95 ´ 180

= 4.8 K DTf = Tf° - Tf 4.8 = 273.15 – Tf Tf = 268.35 K [CBSE Marking Scheme, 2019]

[ 57

SOLUTIONS





OR







[Topper’s Answer 2019]

58 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



   Freezing point of solution = 271.15 K Freezing point of pure water = 273.15 K          Glucose solution = 5% 

M = 180 g  mol -1   To calculate : Freezing point of 5% glucose solution. Formula : DTf = i ´ K f ´ m  

moles of solute Kg of solvent mass moles = molar mass  Sucrose solution is 4% (w/w) which means there is 4.0 grams of sucrose dissolved in 100g of solution. Mass of solution = mass of solute + mass of solvent 100.0 g = 4.0g + mass of solvent. Hence, mass of solvent = 100.0 – 4.0 = 96.0 g 1 kg 96 g ´ = 0.096 kg   1000 g ½ Moles of solute 1 mass 4.0g  = moles = -1 molar mass  342 g  mol   = 0.011695 moles Molality of solution : moles of solute ½ m= kg of solvent 0.011695 moles    = 0.096 kg water = 0.1218 m   Sucrose is a non - electrolyte, hence i – 1 D = Freezing point of solvent      Freezing point of solution       



m=

 DTf = 273.15 K - 271. 15 K = 2.00 K  DTf = i ´ K f ´ m

2.00 K = i ´ K f ´ 0.1218 -1   K f =  16.42 Km For glucose solution, Glucose solution is 5% (w/w) which means there is 5.0 grams of glucose dissolved in 100g of solution. Mass of solution = mass of solute + mass of solvent 100.0 g = 5.0g + mass of solvent. Hence, mass of solvent = 100.0 – 5.0 = 95.0 g 1 kg 95.0 g ´ 1000 g = 0.095 kg Moles of solute mass molar mass  5.0g  = = 0.0277 moles   180 g  mol -1   Molality of solution : moles =



moles of solute kg of solvent 0.0277 moles = = 0.2923 m   0.095 kg water

m=



Glucose is a non - electrolyte, hence i – 1



  DTf = i ´ K f ´ m



  DTf = 1 ´ 16.42 ´ 0.2923



Km -1   DTf =  4.801 K DTf = Freezing point of solvent - Freezing point of solution



    



  4.801 = 273.15 K -  Freezing point of solution

Freezing point of solution =  273.15 K - 4.801 K   =  268.35 K Thus, the freezing point of the glucose solution is 268.35 K. [3] Q. 4. A solution containing 1.9 g per 100 mL of KCI (M = 74.5 g mol–1) is isotonic with a solution containing 3 g per 100 mL of urea (M = 60 g mol–1). Calculate the degree of dissociation of KCI solution. Assume that both the solutions have same temperature. [CBSE OD Set 2, 2019]



Ans.

π1 (urea)=π 2 (KCl) π1 (urea)=π 2 (KCl) C1RT=i C 2 RT C1RT=i C 2 RT n1 n2 (V1 = V2 ) n1 n=2 i V V = i1 (V2 1 = V2 ) V1 V 30 2 1.9 30 60 = 1i .× 9 74.5 =i× 60 5 i =74 1..96 i = 1.96 i −1 αi −=1 n −1 α= n − 11.96 − 1 = −1 1.96 2 −1 = 2=−01.96 or 96% ½ = 0.96 or 96%   [CBSE Marking Scheme 2019]

Detailed Answer : Given : Sucrose solution =  4% ( w / W )   M  =  342 g  mol -1

Detailed Answer :

pKCl = pUrea

i ´ 0.19 ´ 0.0821 ´ 300 0.3 ´ 0.0821 ´ 300 = 74.56 ´ 1 60 ´ 1 (As temperature is same)   i ´ 0.19 = 0.3 74.5 60 i=

0.3 ´ 74.5 = 1.96 60 ´ 0.19

 K + + Cl KCl ionizes as : KCl 

Initial conc.      1  



Conc. at equilibrium  1 – α   α   α

0    0

Number of effective particles after dissociation =1–α+α+α=1+α van’t Hoff factor Number of moles after dissociation = Number of moles before dissociation

[ 59

SOLUTIONS

Ans. (a) As compared to other colligative properties, its magnitude is large even for very dilute solutions / macromolecules are generally not stable at higher temperatures and polymers have poor solubility / pressure measurement is around the room temperature and the molarity of the solution is used instead of molality. 1 (b) Because oxygen is more soluble in cold water or at low temperature. 1 (c) Due to dissociation of KCl / KCl (aq) ® 1 K+ + Cl-, i is nearly equal to 2.  [CBSE Marking Scheme, 2018]







   =



1+a = 1+ a 1 1.96 = 1 + α ; α = 1.96 – 1 = 0.96 α = 0.96 × 100 = 96% Q. 5. Give reasons for the following : (a) Measurement of osmotic pressure method is preferred for the determination of molar masses of macromolecules such as proteins and polymers. (b) Aquatic animals are more comfortable in cold water than in warm water. (c) Elevation of boiling point of 1 M KCl solution is nearly double than that of 1 M sugar solution. [CBSE Delhi/OD 2018] OR



[Topper’s Answer 2018]

60 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q. 6. A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water if the freezing point of pure water is 273.15 K. Given : (Molar mass of sucrose = 342 g mol-1) (Molar mass of glucose = 180 g mol-1) [CBSE Delhi/OD2017]



Ans.   DTf = Kf m 1 Here,    m = w2 × 1000/M2 × W1 273.15 - 269.15 = Kf × 10 × 1000/342 × 90 1     Kf = 12.3 K kg/mol 1    DTf = Kf m = 12.3 × 10 × 1000/180 × 90  = 7.6 K  Tf = 273.15 - 7.6 = 265.55 K (or any other correct method) 1 [CBSE Marking Scheme 2017] 1 Detailed Answer : ΔTt = (273.15 – 269.15) K = 4K Molar mass of sucrose (C12H22O11)  = (12 × 12) + (22 × 1) + (11 × 16)  = 342 g mol–1 10% solution of sucrose in water means 10 g of sucrose is present in (100 – 10)g of water. 10 Number of moles of sucrose = = 0.0292 mol 342 Therefore, molality of the solution

Ans.



=

0.0292 ´ 1000 = 0.3244 mol kg–1 90

We know that ΔTt = Kf × m

∆Tt 4 = = 12.33 K Kg mol–1 m 0.3244 Molar mass of glucose (C6H12O6) = (6 × 12) + (12 × 1) + (6 × 16) = 180 g mol–1 10% solution of glucose in water means 10 g of glucose is present in (100 – 10) g of water. 10 Number of moles of glucose = = 0.0555 mol 180 Therefore,molality of the solution 0.0555 ´ 1000  = 90 ⇒

 Kf =



 = 0.6166 mol kg–1 W e know that ΔTt = Kf × m ⇒ ΔTt = 12.33 × 0.6166 = 7.60 K

So, the freezing point of 10% glucose solution in water is (273.15 – 7.60) K = 265.55 K 3 Q. 7. Calculate the boiling point of solution when 4 g of MgSO4 (M = 120 g mol–1) was dissolved in 100 g of water, assuming MgSO4 undergoes complete ionization. (Kb for water = 0.52 K kg mol–1)  U



DTb = iKb.m i = 2 = i ´ Kb ´

½ W2 ´ 1000 M ´ W1

= 2 × 0.52 K kg mol–1 4g ´ 1000 g / kg 1 × 120 g / mol ´ 100g

= 2 ´ 0.52 = 0.346 K ½ 3 Boiling point of water 373.15 K = 373 K   0 Tb = Tb + DTb = 373.15K + 0.346 K or 373 K + 0.346 K = 373.496 K or 373.346 K1  [CBSE Marking Scheme, 2016] Q. 8. A solution of glucose (molar mass = 180 g mol-1) in water has a boiling point of 100.20C. Calculate the freezing point of the same solution. Molal constants for water Kf and Kb are 1.86 K kg mol-1 and 0.512 K kg mol-1 respectively. [CBSE Foreign Set-1, 2, 3 2017]

Ans. Given : Tb of glucose solution = 100.20°C ∆Tb = Kb. m m = 0.20/0.512 m = 0.390 mol/kg

[1]

∆Tf = Kf . m [½] ∆Tf = 1.86 K kg/mol × 0.390 mol/kg [½] ∆Tf = 0.725 K Freezing point of solution = 273.15 - 0.725 = 272.425 K[1]  [CBSE Marking Scheme 2017] Q. 10. Calculate the boiling point of solution when 2 g of Na2SO4 (M = 142 g mol–1) was dissolved in 50 g of water, assuming Na2SO4 undergoes complete ionization.  [CBSE OD Set-2 2017]

[ 61



SOLUTIONS

Ans. (i)

∆Tf = Kf m

[½]

⇒

∆Tf = Kf WB ×1000 M B ×WA

[½]

⇒ (ii) ⇒ ⇒

∆Tf =

1.86 K kg mol −1 × 45g × 1000 g kg −1 60 g mol −1 × 600 g

∆Tf = 2.325 K or 2.325 °C [1] Tf° – Tf = 2.325 °C 0 °C – Tf = 2.325 °C Tf = – 2.325 °C or 270.675 K [1] [CBSE Marking Scheme 2015]

Commonly Made Error  Students often forget to mention the formula applied. Formula carries marks. Answering Tip  Write working formula followed by data in the working formula. Q. 12. A solution is prepared by dissolving 10 g of nonvolatile solute in 200 g of water. It has a vapour pressure of 31.84 mm Hg at 308 K. Calculate the molar mass of the solute. (Vapour pressure of pure water at 308 K = 32 mm Hg) [CBSE OD 2015] Ans.

Ws × M solvent p0 − p = , s = solute 0 M s × Wsolvent p

[1]

⇒(32 – 31.84)/32 = 10 × 18/Ms × 200 [1] Ms = 180 g/mol [1] [CBSE Marking Scheme 2015]

Q. 11. 45 g of ethylene glycol (C2H4O2) is mixed with 600 g of water. Calculate (i) the freezing point depression and (ii) the freezing point of the solution (Given : Kf of water = 1.86 K kg mol–1) [CBSE Comptt. Delhi 2015]

Q. 13. 3.9 g of benzoic acid dissolved in 49 g of benzene shows a depression in freezing point of 1.62 K. Calculate the van’t Hoff factor and predict the nature of solute(associated or dissociated). (Given : Molar mass of benzoic acid= 122 g mol–1, Kf for benzene = 4.9 K kg mol–1)  [CBSE Delhi, 2015] Ans. ∆Tf = iK f m m=

ΔTf = Depression in freezing point = 1.62 i = van’t Hoff factor Kf = constant = 4.9 × 1000 m == 3.9 1000 == 00..65 65 m 122 × 122 49 × 49 ∆T Tf ∆ f ii == K m K ff m 11..62 62 == 4.9 × 0.65 == 00..50 50 4.9 × 0.65 50 ∴ ii == 00..50

As the value of i < 1, the solute is associated.



Q. 1. (i) A 10% solution (by mass) of sucrose in water has a freezing point of 269.15 K. Calculate the freezing point of 10% glucose in water if the freezing point of pure water is 273.15 K.  Given : (Molar mass of sucrose = 342 g mol-1) (Molar mass of glucose = 180 g mol-1) [CBSE Delhi Set-1, 2, 3 2017] (ii) Define the term : Molality (m) [CBSE OD Set-1, 2, 3 2017]

Ans. (i) Here,

DTf = Kf m m = w2 × 1000/M2 × W1

[1]

3

Commonly Made Error  Students forgot to put the value of i in ionic solids Answering Tip  Remember to put the value of i in questions where complete dissociation is mentioned for ionic solids.

Long Answer Type Questions

WB × 1000 MB × WB

(5 marks each)

273.15 - 269.15 = Kf × 10 × 1000/342 × 90[1] Kf = 12.3 K kg/mol[1] DTf = Kf m = 12.3 × 10 × 1000/180 × 90 = 7.6 K Tf = 273.15 - 7.6 = 265.55 K [½]  (or any other correct method)[1] (ii) Number of moles of solute dissolved in per kilo gram of the solvent. [1] [CBSE Marking Scheme 2017] [1]

62 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q. 2. (a) Calculate the freezing point of solution when 1.9 g of MgCl2 (M = 95 g mol–1) was dissolved in 50 g of water, assuming MgCl2 undergoes complete ionization. (Kf for water = 1.86 K kg mol–1) (b) (i) Out of 1 M glucose and 2 M glucose, which one has a higher boiling point and why? (ii)  What happens when the external pressure applied becomes more than the osmotic pressure of solution ?

 (b) (i)

M b × Wa

[1]

 1.86 × 1.9  DTf = 3 ×  95 × 50  × 1000  



[1] = 2.23 K Tf – DTf = 273.15 – 2.23 or 273 – 2.23 Tf = 270.92 K or 270.77 K[1] 2 M glucose has a higher boiling point because higher the number of particles, lesser is the vapour pressure. [½ + ½]

(ii) Reverse osmosis.  [1] [CBSE Marking Scheme 2016] Detailed Answer : (a) WB (solution) = 1.9 g ws W(H2O) = 50 g M(MgCl2) = 95 g/mol i = 3(for MgCl2)



m =

WB in kg Molar mass × mass of solvent



DTf = iKfm



DTf = 3 × 1.86 ×



DTf = 2.232 K



DTf = DT°f – Tf



Also, DTf = DT°f – Tf



(b)  Blood cells are isotonic with 0.9% sodium chloride solution. What happens if we place blood cells in a solution containing; (i) 1.2% sodium chloride solution ?



(ii) 0.4% sodium chloride solution ?

Ans. (a)

    



DTf = i

Ans. (a)

K f Wb × 1000

Q. 3. (a) When 2.56 g of sulphur was dissolved in 100 g of CS2, the freezing point lowered by 0.383 K. Calculate the formula of sulphur (SX). (Kf for CS2 = 3.83 K kg mol–1, Atomic mass of Sulphur = 32 g mol–1).

1.9 1000 × ( kg ) 95 50

Tf = T°f – DTf = 273.15 – 2.232 = 270.918 K  [3] (b) (i) 2M glucose, because more will be the concentration molality, more will be the elevation in boiling point. (ii) Reverse osmosis takes place. [½ × 2]





[1]

[1] [1] [CBSE Marking Scheme, 2016]

(a)

DTf = iKfm



DTf = K f ×



MB =



=



WB 1 × ( Kg ) WA ( Kg ) WA Kf

∆Tf × WA ( Kg ) 3.83 × 2.56 × 1000 0.383 × 100( kg )

= 256 g/mol MB = n × Atomic mass MB n = Atomic mass





M b × Wa

[1]



M = 256 S × x = 256 32 × x = 256[1] x = 8

(b) (i) Shrinks (ii) Swells  Detailed Answer :



K f Wb × 1000

 3.83 × 2.56     0.383 =  × 1000  M × 100 



=

256 =8 32

\ S8 [3] (b) (i) Water moves out from blood cell, hence will shrink. (ii) Water will enter into blood cell, hence will swell. [1 × 2]

Visual Case–Based Questions Q. 1. Read the passage given below and answer the following questions : (1 × 4 = 4) Scuba apparatus includes a tank of compressed air toted by the diver on his or her back, a hose for carrying air to a mouthpiece, a face mask that covers the eyes and nose, regulators that control air flow, and gauges that indicate depth and how much air remains in the tank.

DTf =

(1 mark each)



 diver who stays down too long, swims too deep, A or comes up too fast can end up with a condition called “the bends.” In this case, bubbles of gas in the blood can cause intense pain, even death.



In these following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

[ 63

SOLUTIONS



(b)  Assertion and Reason both are correct statements but Reason is not correct explanation for Assertion.



(c) Assertion is correct statement but Reason is wrong statement.



(d) Assertion is wrong statement but Reason is correct statement. (i) Assertion : Scuba divers may face a medical condition called ‘bends’. Reason : ‘Bends’ can be explained with the help of Henry’s law as it links the partial pressure of gas to that of its mole fraction. (ii) Assertion : Bends is caused due to formation of nitrogen bubbles in the blood of scuba divers which blocks the capillaries. Reason : Underwater high pressure increases solubility of gases in blood, while as pressure gradually decreases moving towards the surface, gases are released and nitrogen bubbles are formed in blood. (iii) Assertion : Soft drinks and soda water bottles are sealed under high pressure.  Reason : High pressure maintains the taste and texture of the soft drinks. (iv) Assertion : Anoxia is a condition experienced by climbers which makes them suddenly agile and unable to think clearly.  Reason : At high altitudes the partial pressure of oxygen is less than that at the ground level. OR  Assertion : Solubility of gases in liquids decreases with rise in temperature. Reason : As dissolution is an exothermic process, the solubility should decrease with increase of temperature. Ans. (i) Correct option : (a) Explanation : Henry’s law explains some biological phenomena like the ‘bends’ experienced by the scuba divers. Since mole fraction of a gas in the solution is a measure of its solubility. [1] (ii) Correct option : (a) Explanation : Scuba divers must cope with high concentrations of dissolved gases while breathing air at high pressure underwater. Increased pressure increases the solubility of atmospheric gases in blood. When the divers come towards surface, the pressure gradually decreases. This releases the dissolved gases and leads to the formation of bubbles of nitrogen in the blood. This blocks capillaries and creates a medical condition known as bends. [1] (iii) Correct option : (c) Explanation : The bottle is sealed under high pressure to increase the solubility of CO2 in soft drinks and soda water. [1]



(iv) Correct option : (d) Explanation : At high altitudes the partial pressure of oxygen is less than that at the ground level. This leads to low concentrations of oxygen in the blood and tissues of people living at high altitudes or climbers. This leads to a condition called anoxia caused due to low oxygen in blood, making the climbers to become weak and unable to think clearly. [1] OR Correct option : (a) Explanation : Solubility of gases in liquids decreases with rise in temperature. As dissolution is an exothermic process, the solubility should decrease with increase of temperature. [1] Q. 2. Read the passage given below and answer the following questions :  (1 × 4 = 4) Raoult’s law states that for a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution. Dalton’s law of partial pressure states that the total pressure (ptotal) over the solution phase in the container will be the sum of the partial pressures of the components of the solution and is given as : Ptotal = P1 +P2 Vapour pressure of solution

Vapour pressure

(a)  Assertion and Reason both are correct statements and Reason is correct explanation for Assertion.

p2

k 1=0 k 2=1

p1

Mole fraction k1

k 1 =1 k 2 =1

k2



(i) What type of deviation from Raoult’s law does the above graph represent ?



(a) First positive then negative



(b) Negative deviation



(c) Positive deviation



(d) First negative then positive (ii) In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M MgCl2 solution is _____________.



(a) the same



(b) about twice



(c) about three times



(d) about six times

64 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(iii) A solution of two liquids boils at a temperature more than the boiling point of either of them. What type of deviation will be shown by the solution formed in terms of Raoult’s law ?



(a) Negative deviation



(b) Positive deviation



(c) First positive then negative



such as freeze concentration of liquid food and to find the molar mass of an unknown solute in the solution. Freeze concentration is a high quality liquid food concentration method where water is removed by forming ice crystals. This is done by cooling the liquid food below the freezing point of the solution. The freezing point depression is referred as a colligative property and it is proportional to the molar concentration of the solution (m), along with vapor pressure lowering, boiling point elevation, and osmotic pressure. These are physical characteristics of solutions that depend only on the identity of the solvent and the concentration of the solute. The characters are not depending on the solute’s identity. (Jayawardena, J. A. E. C., Vanniarachchi, M. P. G., & Wansapala, M. A. J. (2017). Freezing point depression of different Sucrose solutions and coconut water.)

(d) First negative then positive (iv) Which of the following aqueous solutions should have the highest boiling point ?

(a) 1.0 M NaOH (b) 1.0 M Na2SO4 (c) 1.0 M NH4NO3 (d) 1.0 M KNO3 Ans. (i) Correct option : (b) Negative deviation (ii) Correct option : (c) Explanation :

[1]

ΔTf =iKf​m,  where i=1 for glucose. glucose​ ΔTf =

1 × Kf​× 0.01 In case of MgCl2 → Mg2+ + 2Cl−, where i = 3, MgCl 2

∆Tf

= 3 × 0.01 × Kf

MgCl 2

⇒ ∆Tf

glucose​

= 3 × ΔTf

Hence, the depression in freezing point of MgCl2​  is three times that of glucose.[1] (iii) Correct option : (a) Explanation : Since the Boiling point of the solution is more than the b.p. of the individual components in the solution, it indicates that the vapour pressure exerted by the solution is less than the expected, as boiling starts when vapour pressure equals the atmospheric pressure. Hence, the solution shows a negative deviation from the Raoult’s law.[1] (iv) Correct option : (b) Explanation : Na2SO4 will release 3 moles of ions/ moles of Na2SO4 in the aqueous solution, and b.p being a colligative property, the b.p of this solution will be the highest as other solutions release only 2 ions each. [1] Q. 3. Read the passage given below and answer the following questions: Boiling point or freezing point of liquid solution would be affected by the dissolved solids in the liquid phase. A soluble solid in solution has the effect of raising its boiling point and depressing its freezing point. The addition of non-volatile substances to a solvent decreases the vapor pressure and the added solute particles affect the formation of pure solvent crystals. According to many researches the decrease in freezing point directly correlated to the concentration of solutes dissolved in the solvent. This phenomenon is expressed as freezing point depression and it is useful for several applications

(i) When a non volatile solid is added to pure water it will: (a) boil above 100°C and freeze above 0°C (b) boil below 100°C and freeze above 0°C (c) boil above 100°C and freeze below 0°C (d) boil below 100°C and freeze below 0°C (ii) Colligative properties are: (a) dependent only on the concentration of the solute and independent of the solvent’s and solute’s identity. (b) dependent only on the identity of the solute and the concentration of the solute and independent of the solvent’s identity. (c) dependent on the identity of the solvent and solute and thus on the concentration of the solute. (d) dependent only on the identity of the solvent and the concentration of the solute and independent of the solute’s identity. (iii) Assume three samples of juices A, B and C have glucose as the only sugar present in them. The concentration of sample A, B and C are 0.1M, .5M and 0.2 M respectively. Freezing point will be highest for the fruit juice: (a) A

(b) B

(c) C (d) All have same freezing point (iv) Identify which of the following is a colligative property : (a) freezing point (b) boiling point (c) osmotic pressure (d) all of the above Ans. 1. (b) 2. (d)

3. (a)

4. (c) ll

[ 65

SELF-ASSESSMENT TEST

Self-Asessment Test - 2 Time : 1 Hour 1.  Read the passage given below and answer the following questions : (1 × 4 = 4)



 ost of the gases are soluble in water to some extent. M The solubility of gas in water generally depends upon nature of the gas, temperature and pressure. In general, the gases which are easily liquefiable are more soluble in water. The dissolution of gas in water is exothermic process. Hence, the solubility of gas decreases with rise in temperature. The effect of pressure on the solubility of a gas is given by Henry’s Law which states that mass of the gas dissolved per unit volume of a liquid at particular temperature is directly proportional to the pressure of the gas above liquid at equilibrium. (i) The solubility of gas in water depends upon:

(a) Nature of the gas (b) Temperature (c) Pressure (d) All of the above

Max. Marks : 25 (c) H2O + HCl (d) H2O + C2H5OH 4. Which of the following is a colligative property? (a) Osmotic pressure (b) Building point (c) Vapour pressure (d) Electrical conductivity 5. The number of moles of solute present in 1000 g of the solvent is known as (a) Molarity (b) Molality (c) Normality (d) Mole fraction In the following questions (no. 6 & 7), a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a)

Assertion and reason both are correct statements and reason is correct explanation for assertion.

(b)

Assertion and reason both are correct statements but reason is not the correct explanation for assertion.

(d) None of these



(c)

(iii) T  he solubility of gas __________ with rise in temperature:

Assertion is correct statement but reason is wrong statement.

(d)

Assertion is wrong statement but reason is correct statement.

(ii) The dissolution of gas in water is: (a) Endothermic process (b) Exothermic process



(c) Both (a) and (b)

(a) Increases (b) Decreases (c) Remains some (d) First increases and then decreases (iv) T  he effect of pressure on the solubility of a gas given by: (a) Raoult’s Law (b) Henry’s Law (c) Boyle’s Law (d) Charle’s Law The following questions (No. 2 to 5) are Multiple Choice Questions carrying 1 mark each. 2. Which of the following is not correct for an ideal solution? (a) It should obey Raoult’s Law (b) ∆Hmix = 0 (c) ∆Hmix ≠ 0 (d) ∆Vmix = 0 3.  Select the non-ideal solution showing positive deviation from Raoult’s Law. (a) CHCl3 + C6H6 (b) (CH3)2CO + C6H5NH2

6. Assertion : Cooking time is reduced in pressure cooker. Reason : Boiling point of water inside the pressure cooker is elevated. 7. Assertion : Two solution having same osmotic pressures are called isotonic solutions. Reason : Lowering of vapour pressure is not a colligative property. The following questions (No. 8 & 9), are Short Answer Type-I and carry 2 marks each. 8. 10 g glucose is dissolved in 90 g of water then what will be the mass % of glucose? 9. The freezing point of solution of 0.1 g week monatomic acid dissolved in 22g water in 272.817 k. Calculate molar mass of acid. (Kf = 1.86 k kg mol–1) The following questions (No. 10 & 11) are Short Answer Type-II carrying 3 marks each. 10. 10.8 g sucrose is dissolved in 100 g of water. At which temperature, this solution will boil at 1.013 bar pressure? The value of Kb for water is 0.52 k kg mol–1

66 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

11. A substance X (molecular mass = 94) associates as 2X  X2 when dissolved in CCl4. If 10g of X is dissolved in 2 kg of ccl4, the freezing point is lowered by 1.08oC, Kf for ccl4 is 31.8 k kg mol–1. Calculate the degree of dissociation of X. Q.No 12 is a Long Answer Type Question carrying 5 marks each.

(molar mass = 180 g mol–1) in water. Calculate the mass of glucose present in one litre of its solution. OR

(a)

What will be the value of van’t Hoff factor for dilute solution of Al2(SO4)3 in water?

(b)

Among 1 m glucose, 1m KCl and 1 m K2SO4, which will have minimum freezing point and why?



A solution of glycerol (C3H8O3) is formed by dissolving some glycerol in 500 g water. The boiling point of this solution is 100.42oC. How much quantity of glycerol was dissolved to form this solution? (For water, Kb = 0.512 k kg mol–1)

Q.12. (i) Define the following terms :

(a) Molarity



(b) Molal elevation constant (Kb)

(ii) A solution containing 15 g urea (molar mass = 60 g mol–1) per litre of solution in water has the same osmotic pressure (isotonic) as a solution of glucose

(c)

 

Finished Solving the Paper ? Time to evaluate yourself !

OR SCAN THE CODE

SCAN

For elaborate Solutions

ll

Rusting of iron, tarnishing of silver

rG° = –n FE°cell

rG° = –RT InK

Second Level

Trace the Mind Map

First Level

Electrode : Pt coated with Pt black, electrolyte : acidic solution pressure 1 bar Pt(s) |H2(g)| H+(aq)

Third Level

V Unit : Ohm () I

Temperature)

1 A Unit : Siemens (s) = R l or ohm Increases on A = l dilution

Strong electrolyte m = m – Ac½ (KCl) Limiting molar conductivity of an electrolyte can be represented as sum of individual contribution of anions and cations of the electrolyte °m = v+°++v–°–

Kohlrausch law of independent migration of ions

C=

Electrolytic (Ionic) Conductance: depends on: Nature of electrolyte added,  Size of ions solvation  Nature of solvent and its viscosity,  Concentration of electrolyte,  Temperature (increases with increase in

Electronic conductance: Depends on:- Nature and structure of metal, -Number of valence electrons per atom, -Temperature (Decreases with increase in temperature)

 nE° cell  KC = Antilog    0.0591  2.303 RT log K C E°cell = nF

Weak electrolyte Ù  = m (acetic acid) Ù°m

A  m; = V l Limiting molar conductivity :If molar conductivity reaches a limiting value when concentration approaches zero. C  0, m = m

m =

Conductivity Cell

l A  = Resistivity Unit : Ohm – Meter

R= 

(Lead storage battery, Ni-Cd cell)

Leclanche cell (Dry cell)

Mercury cell

2.303RT [M] log n+ Ecell = E° cell – nT [M ] 0.059 1 log n+ Ecell = E°cell – n [M ]

e r e us ed ) cannot b

Mn+(aq) + ne–  M(s)

Electrical resistance

R=

Resistance

Two copper strips dipped in an aqueous solution of CuSO4  Anode : Cu  Cu2+ + 2e– cathode : Cu2+ + 2e–  Cu

1. Calculate °m for any electrolyte from ° of individual ions 2. Determine value of dissociation constant for weak electrolytes

and o n ce

Nernst equation

Batteries

Products of electrolysis depends upon

Faraday’s laws of electrolysis

 2nd Law  Amount of different subst ances liberated by same quantity of electricity passing through electrolytic solution are proportional to their chemical equivalent weights  …… W1 W2 W3   E1 E2 E3

Electrochemistry

Fuel cells

Cathode  Reduction takes place Anode  Oxidation takes place

Types of Cells

A chemical compound that dissociates U shaped inverted into ions and tube connecting conducts electric two electrolytic current solution

A series of half–cells arranged in increasing standard oxidation potentials.

Negative E°Stronger reducing agent than H+/H2

Positive E° Weaker reducing agent than H+/H2

Electrode potential when concentration of all species in half cell is unity.

Potential difference between electrode and electrolyte.

Half–cell  two portions of cell

Corrosion

Relation between cell potential and Gibbs energy

Galvanic cell that converts energy of combustion of fuels like H 2, CH4 directly into electrical energy

 1st Law  Amount of substance in a chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through electrolyte  W = Zit

ls

Daniell Cell: cathode: Copper, Anode : Z inc; Salt bridge : Agar agar; electrolyte : KCN/KNO3  Reduction : Cu2+ + 2e  Cu;Zn(s) Oxidation : Zn2++2e– ;Zn|Zn2+(C1) || Cu 2+(C2)|Cu



eta

Electrochemical phenomenon in which metal oxide of metal forms coating on metal surface.



m

Painting, barrier protection, rust solutions



ELECTROCHEMISTRY

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3

CHAPTER

Syllabus

ELECTROCHEMISTRY

¾¾ Redox reactions, EMF of a cell, standard electrode potential, Nernst equation and its application to chemical cells, Relation between Gibb’s energy change and EMF of a cell, conductance in electrolytic solutions, specific and molar conductivity, variations of conductivity with concentration, Kohlrausch’s Law, electrolysis and law of electrolysis (elementary idea), dry cell-electrolytic cells and Galvanic cells, lead accumulator, fuel cells, corrosion.

Trend Analysis List of Concept names

2018 D/OD

D

Electrolytic & molar conductivity and Numericals

1Q (2 marks)

1Q (2 marks)

Electrochemical cell, Nernst equation and Numericals

2Q (3 marks) 1Q (5 marks)

1Q (5 marks)

-

1Q (3 marks)

-

1Q (2 marks)

-

1Q (1 mark)

-

-

-

Electrolysis and Numericals Primary and Secondary cells, Fuel cell Corrosion and its protection

TOPIC-1

2019

Electrolytic Conductivity, Electrolytes and Kohlrausch’s Law

Revision Notes

2020

OD 1Q (2 marks) 1Q (3 marks) 1Q (1 mark) 1Q (3 marks)

D 2Q (1 mark)

1Q (1 mark) 1Q (3 marks) 1Q (1 mark) 1Q (1 mark)

OD 1Q (2 marks) 1Q (3 marks) 1Q (1 mark) 1Q (3 marks) 1Q (2 marks) 1Q (1 mark) -

TOPIC - 1 Electrolytic Conductivity, Electrolytes and Kohlrausch’s Law .... P. 68 TOPIC - 2 Redox Reactions and Electrochemical Cells, Electrode Potential and Nernst Equation .... P. 76 TOPIC - 3 Electrolysis, Law of Electrolysis, Batteries, Fuel Cells and Corrosion .... P. 87

 Electrochemistry is the branch of chemistry which deals with the study of the production of electricity from energy released during spontaneous chemical reactions and the use of electrical energy to result in non-spontaneous chemical transformations.  Electrolytic conduction : The flow of electric current through an electrolytic solution is called electrolytic conduction.

[ 69





Electrolyte : A substance that dissociates in solution to produce ions and hence conducts electricity in dissolved state or molten state. Weak electrolyte – H 2CO3, CH 3COOH, HCN, MgCl 2 . Strong electrolyte – NaCl, HCl, NaOH.  Degree of ionisation : It is the ratio of number of ions produced to the total number of molecules in electrolyte.  Resistance is defined as the property of given substance to obstruct the flow of charge. It is directly proportional to the length (l) and inversely proportional to its area of cross-section (A). l l Rµ or; R = r A A r = Resistivity or specific resistance.  Resistivity : If a solution is placed in between two parallel electrodes having cross sectional area ‘A’ and distance ‘l’ apart, then l R=r , A 



ELECTROCHEMISTRY

where r = resistivity and its SI unit is Ohm-m also Ohm-cm is used as a unit.

 Conductance : The ease with which the current flows through a conductor is called conductance. It is reciprocal of resistance. i.e.,

C =



The SI unit of conductance is Siemens (S).





R=r

1 A A = =k R l ρl

l , A

1S = 1 Ohm–1 = 1 W–1  Conductivity : It is reciprocal of resistivity and is denoted by k (Greek Kappa).



k = C×

Scan to know more about this topic

l , A



where C = Conductance of the solution





l = Distance/ length





A = Area of cross section



Its SI unit is S m–1. Also expressed as S cm–1



It depends upon the :



(i) Nature of the material



(ii) Temperature



(iii) The number of valence electrons per atom or size of the ions produced and their solvation(electrolytes)

Molar conductance

 Metallic conductance is the electrical conductance in metals that occurs due to the movement of electrons. It depends upon the :

(i) Nature and structure of the metal



(ii) Number of valence electrons per atom



(iii) Temperature

 Electrolytic or ionic conductance is the conductance of electricity that occurs due to ions present in the solution. It depends upon the :

(i) Nature of electrolyte or interionic attractions



(ii) Solvation of ions



(iii) Nature of solvent and its viscosity



(iv) Temperature

 Wheatstone bridge helps us to measure R4.  Cell constant (G) : It is the ratio of distance between electrodes to the cross-sectional area between electrodes.



Cell constant (G) =



It depends on the :



(i) Distance between the electrodes



(ii) Area of cross section.

l in cm–1 or m–1 A

70 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII







Molar conductivity : It is defined as the conducting power of all the ions produced by one gram mole of an electrolyte in a solution. It is denoted by Lm. κ Lm = × 1000 S cm 2 mol-1 , C where k = Conductivity C = Concentration of solution. SI unit of molar conductivity is S m2 mol–1.  Debye Huckel Onsager equation : It is applicable for strong electrolyte : L = L° – A C , where L° = Limiting molar conductivity, L = Molar conductivity, A = Constant and C = Concentration of solution.  Kohlrausch’s law of independent migration of ions : According to this law, limiting molar conductivity of an electrolyte, at infinite dilution, can be expressed as the sum of contributions ∞ conductivity of the cations is denoted by λ+ and that of from its individual ions. If the molar ∞ the anions by λ − then the law of independent migration of ions is

∞ Λ ∞m = v+ λ+ + v– λ ∞− or L0 = v+ λ+0 + v– λ−0 , where, v+ and v– are the number of cations and anions per formula of electrolyte. Applications of Kohlrausch’s law (i) Calculation of molar conductivities of weak electrolyte at infinite dilution. (ii) Calculation of degree of dissociation (a) of weak electrolytes :







(iii) Determination of dissociation constant (K) of weak electrolytes :







(iv) Determination of solubility of sparingly soluble salts : K × 1000 Solubility = Λ°m



Degree of dissociation (a) =

Ka =

c Λm

Λ°m cα 2 c Λm2 = (1 − α ) Λ m0 ( Λ m0 − Λ m 2 )

Know the Formulae Potential difference (V) Resistance ( R )



Current (I) =



Resistance (R) = ρ



Conductance (C) = K





 





l A A l

Specific conductivity (k) = C× Cell constant (G) =ρ

l Cell constant = R A

l A

For strong electrolyte, Lm = L°m – A C L° = n+L°+ + n–L°– Degree of dissociation (a) =



K =



Solubility =

c Λm

Λ°m cα 2 (1 − α )

=

K × 1000 Λ°m

c( Λ °m )

Λ °m ( Λ °m

− Λm )

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Molar conductivity and Kohlrausch’s law

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ELECTROCHEMISTRY

Mnemonics • Concept: Electrolyte • Mnemonic: SEDC • Interpretation: Strong Electrolytes Dissociate Completely. • Mnemonic: WED Prime • Interpretation: Weak Electrolytes Dissociate Partially.

Know the Terms

 Superconductors : Materials with a zero resistance.  Limiting molar conductivity : Molar conductivity when concentration approaches zero.  Electrolyte : Substance that dissociates into electrically conducting ions.  Over voltage : It is the difference between the potential required for the evolution of a gas and its standard reduction potential.

Q. Calculate the molar conductivity and degree of dissociation. Conductivity of 2.5 × 10–4 M methanoic acid is 5.25 × 10–5 S cm–1. (Given : Lo(H+) = 349.5 S cm2 mol–1 and Lo 2 –1 (HCOO–) = 50.5 S cm mol .) Solution:

1000 ´ k S cm2 mol-1 STEP-1: Lm = C

[1]

STEP-2: Lm

1000×5.25×10−5 S cm2mol−1 = −4 2.5×10 = 210 S cm2mol–1 STEP-3:

= LoHCOO– + Lo H+ = (50.5 + 349.5)S cm2mol–1 = 400 S cm2mol–1 a = Lm / L°m a = 210 / 400 = 0.525 [1]

Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS :

∧ = ∧o − A C Which of the following equality holds?



∧ = ∧ o as C  →∞ (a) ∧ = ∧ o as C  → A (b)



(c) ∧ = ∧ o as C  → 0 (d) ∧ = ∧ o as C  →1



(1 mark each) Ans. Correct option : (b)

Q. 1. Debye-Huckel Onsager equation for strong electrolytes:

[1]

Ù0m HCOOH



[CBSE, Delhi Set-2, 2020]



Explanation : When c → ∞



Then ∧ = ∧o Q. 2. Which of the following option will be the limiting molar conductivity of CH3COOH if the limiting molar conductivity of CH3COONa is 91 Scm2mol–1? Limiting molar conductivity for individual ions are given in the following table.

72 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

S.No

Ions

limiting molar conductivity / Scm2mol–1

1

H+

349.6

2

+

Na

+

3

K

4

OH

50.1 73.5 199.1

(a) 350 Scm2mol–1 (b) 375.3 Scm2mol–1 2 –1 (c) 390.5 Scm mol (d) 340.4 Scm2mol–1  U [CBSE, SQP 2020-21] Ans. Correct option : (c) Explanation : The limiting molar conductivity (∧mo) for strong and weak electrolyte can be determined by using Kohlrausch’s law which states that “the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.” ∧CH3COONa = ∧CH3COO− + ∧Na+ 91 Scm2mol−1 = ∧CH3COO− + 50.1 Scm2mol−1 ⇒ ∧CH3COO− = 40.9 Scm2mol−1 For acetic acid, ∧CH3COOH = ∧CH3COO− + ∧H+ = 40.9 Scm2mol−1 + 349.6 Scm2mol−1 = 390.5 Scm2mol−1 Q. 3. Which of the statements about solutions of electrolytes is not correct ? (a) Conductivity of solution depends upon size of ions. (b) Conductivity depends upon viscosity of solution. (c) Conductivity does not depend upon solvation of ions present in solution. (d) Conductivity of solution increases with temperature. R Ans. Correct option : (c) Explanation : Conductivity depends upon solvation of ions present in solution. Greater the solvation of ions of an electrolyte, lesser will be the electrical conductivity of the solution. Q. 4. When 0.1 mol CoCl3(NH3)5 is treated with excess of AgNO3, 0.2 mol of AgCl are obtained. The conductivity of solution will correspond to (a) 1 : 3 electrolyte

(b) 1 : 2 electrolyte

A (c) 1 : 1 electrolyte (d) 3 : 1 electrolyte Ans. Correct option : (b) Explanation : When 0.1 mole of CoCl3(NH3)5 was reacted with excess of AgNO3, we get 0.2 moles of AgCl. So, there are two chloride ions that are free and not part of the complex. The formula for complex has to be [Co(NH3)5Cl]Cl2. [Co(NH3)Cl]Cl2 → [Co(NH3)5Cl]2+ + 2Cl− Therefore, the conductivity of the solution will be 1 : 2 electrolyte.

[B] ASSERTION AND REASON TYPE QUESTIONS : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Q. 1. Assertion : Conductivity of an electrolyte increases with decrease in concentration.   Reason : Number of ions per unit volume decreases on dilution. Ans. Correct option : (d) Explanation : Conductivity of an electrolyte decreases with decrease in concentration because of ions per unit volume decreases on dilution. Q. 2. Assertion : Λm for weak electrolytes shows a sharp increase when the electrolytic solution is diluted. Reason : For weak electrolytes degree of dissociation increases with dilution of solution. Ans. Correct option : (a) Explanation : Weak electrolytes dissociate partially in concentrated solution. On dilution, their degree of dissociation increases hence, their Λm increases sharply. Q. 3. Assertion : Electrolytic conduction increases with increase in temperature. Reason : Increase in temperature cause the electronic movement more rapid Ans. Correct Option : (c) Explanation : As the temperature of electrolytic solution is increased, the kinetic energy of the ion increases. This results in the increase of electrical conductance of electrolytic solutions. [C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1. What is electrolyte? R Ans. Substance that dissociates into electrically conducting ions in molten state or aqueous solution. Q. 2. State Kohlrausch’s law. R Ans.  The molar conductivity of an electrolyte at infinite dilution is equal to sum of the conductivities of the individual ions. Ù ¥m = n + l ¥+ + n - l ¥Q. 3. What is meant by the term of infinite dilution? A  Ans.  The term ‘infinite dilution’ means a solution that is so dilute that it has a maximum or limiting molar conductivity which does not increase on further dilution.

[ 73

ELECTROCHEMISTRY

Short Answer Type Questions-I Q. 1. Calculate the degree of dissociation (a) of acetic acid if its molar conductivity (Lm) is 39.05 S cm2 mol-1. Given L˚(H+) = 349.6 S cm2 mol-1 and L0(CH3COO-) = 40.9 S cm2 mol-1. A [CBSE Delhi Set-1, 2, 3 2017] Ans. L°CH COOH = L°CH COO- + L°H+ [½] 3 3 = 40.9 + 349.6 = 390.5 S cm2/mol [½] Now, a = Lm/L0m [½] = 39.05/390.5 = 0.1 [½] [CBSE Marking Scheme 2017]

Commonly Made Error  Students tend to forget specifically mentioning the formula and start the calculations or do not mention all the steps or specific units.

Answering Tip  Write the working formula in each step followed by value assignment for each entity. Give appropriate unit alongwith the answer. Q. 2. The conductivity of a 0.01 M solution of acetic acid at 298 K is 1.65 × 10–4 S cm–1. Calculate molar conductivity (Lm) of the solution. A [CBSE Comptt. Delhi/OD 2018] Ans.     Lm =

1000κ C

[½]

1.65 × 10−4 S cm −1 × 1000 cm 3 L−1    Lm = [½] 0.01 mol L−1        = 16.5 S cm2 mol–1 [1] [CBSE Marking Scheme 2018] -3

Q. 3. The conductivity of 10 mol/L acetic acid at 25°C is 4.1 × 10–5 S cm–1. Calculate its degree of dissociation if L0m for acetic acid at 25°C is A 390.5 S cm2 mol-1. 1000κ Ans. ∧m = [½] C 1000 cm 3 L−1 × 4.1 × 10−5 S cm −1 ∧m = [½] 10−3 mol L−1

= 41 S cm2 mol–1 c ∧m [½] ∧0m



α =



41 = 0.105 [½] α = 390.5

(2 marks each)

Q. 4. Why on dilution the Lm of CH3COOH increases drastically, while that of CH3COONa increases gradually? U Ans. In case of CH3COOH which is a weak electrolyte, the number of ions increases on dilution due to an increase in degree of dissociation resulting in drastic increase in Lm. [1] CH 3COOH+H 2O → CH 3COO−+H 3O+  In the case of CH3COONa which is a strong electrolyte, the number of ions remains the same but the inter-ionic attraction decreases resulting in gradual increase in L m. [1] Q. 5. Solutions of two electrolytes ‘A’ and ‘B’ are diluted. The limiting molar conductivity of ‘B’ increases 1.5 times while that of ‘A’ increases 25 times. Which of the two is a strong electrolyte ? Justify A your answer. Ans. ‘B’ is a strong electrolyte. [1]  A is a strong electrolyte which is completely dissociated into ions, but on dilution interionic forces overcome and ions are free to move. So there is slight increase in molar conductivity on dilution.  [1] Q. 6. The resistivity of a 0.8M solution of electrolyte is 5 × 10–3 Ωcm. Calculate its molar conductivity. A Ans.  Resistivity (ρ) = 5 × 10–3 Ωcm Conductivity of solution(k) 1 Resistivity 1 = 5 ´ 10 -3 Wcm =

= 0.2 ´ 10 -3 W -1cm -1 = 200 W -1cm -1

[1] Molar conductivity 1000 ´ k Ùm = M 1000 ´ 200 = = 2.5 ´ 10 5 W -1cm 2 mol -1 [1] 0.8

Commonly Made Error  Some students get confused in conductivity and molar conductivity and calculate conductivity rather than molar conductivity

Answering Tip  Learn and understand the difference between conductivity and molar conductivity and also formula for their calculation.

74 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Short Answer Type Questions-II

Λm / 5 cm2 mol–1

Q. 1. The following curve is obtained when molar conductivity (Lm) is plotted against the square root of concentration, c½ for two electrolytes A and B:



Where ∧m = Molar conductivity Ù ¥m = molar conductivity at infinite dilution b = constant c = Concentration of solution

[1]

(ii) C  onductivity of NaCl decreases on dilution as the number of ions per unit volume decreases.  [1] Whereas molar conductivity of NaCl increases on dilution as on dilution the interionic interactions overcome and ions are free to move. [1]

400

200 A

Q. 3. The electrical resistance of a column of 0.05 M KOH solution of diameter 1 cm and length 45.5 cm is 4.55 × 103 ohm. Calculate its molar conductivity.A [CBSE Foreign Set-1, 2, 3 2017]

B 0.2 0.4 C½ (mol L–1)½



(i) How do you account for the increase in the molar conductivity of the electrolyte A on dilution ? (ii) As seen from the graph, the value of limiting molar conductivity (L°m) for electrolyte B cannot be obtained graphically. How can this value be obtained ? (iii) Define limiting molar conductivity. U+ R Ans. (i) As seen from the graph, electrolyte A is a strong electrolyte which is completely ionised in solution. With dilution, the ions are far apart from each other and hence the molar conductivity increases. [1] (ii)  To determine the value of limiting molar conductivity for electrolyte B, indirect method based upon Kohlrausch law of independent migration of ions is used.  [1] (iii) W  hen concentration approaches zero, the molar conductivity is known as limiting molar conductivity.  [1] Q. 2. (i) Give Debye Huckel Onsager equation for strong electrolyte. R (ii) G  iven are the conductivity and molar conductivity of NaCl solutions at 298K at different concentrations : Concentration

Conductivity

Molar conductivity

(M)

(S cm–1)

(S cm2 mol-1)

0.100

106.74 × 10–4

106.7

0.05



(3 marks each)

55.53 × 10

–4

111.1

0.02 23.15 × 10–4 115.8 Compare the variation of conductivity and molar conductivity of NaCl solutions on dilution. Give reason. A Ans. (i) Debye Huckel Onsager equation for strong electrolyte is : ¥ Ùm = Ùm -b c

Ans. A = pr2 = 3.14 × 0.5 × 0.5 cm2 = 0.785 cm2 [½] l = 45.5 cm G* = l/A = 45.5 cm/0.785 cm2 = 57.96 cm–1 [½] k = G*/R [½] = 57.96 cm–1/4.55 × 103 Ω = 1.27 × 10–2 S cm–1  [½] Lm = k × 1000/C [½] –2 –1 3 = [1.27 × 10 S cm ] × 1000/0.05 mol/cm = 254.77 S cm2 mol–1 [½] [CBSE Marking Scheme 2017]

Commonly Made Error  Students often convert centimeter into meter.

Answering Tip  Check the compatibility of units. Q. 5. (i)  State the law which helps to determine the limiting molar conductivity of weak electrolyte. (ii) Calculate limiting molar conductivity of CaSO4 (limiting molar conductivity of calcium and sulphate ions are 119.0 and 160.0 S cm2 mol–1 respectively) R+A Ans. Kohlrausch law of independent migration of ions : (i) The limiting molar conductivity of an electrolyte can be represented as the sum of the individual contribution of the anions and cations of the electrolyte. [1] (ii)

Λ om (CaSO4) = LoCa2+ + Lo SO24

[1] = 119.0 S cm2 mol–1 + 160.0 S cm2 mol–1 = 279.0 S cm2 mol–1 [1]

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ELECTROCHEMISTRY

Commonly Made Error

Answering Tips

 Students only write the mathematical expression.

 Write the law as stated. Stick to the statement as the marks are alloted to that only.  Do not forget to mention the units.

Long Answer Type Questions Q. 1.

(a) The electrical resistance of a column of 0.05 M KOH solution of length 50 cm and area of cross-section 0.625 cm2 is 5 × 103 ohm. Calculate its resistivity, conductivity and molar conductivity.   (b)  Predict the products of electrolysis of an aqueous solution of CuCl2 with platinum electrodes. (Given : E

= + 0.34 V, E ½ Cl

o Cu 2 + / Cu

EHo + / H

2 ( g),

o

(

2

/ Cl-

)

= + 1.36V ;

= 0.00 V, E(o½ O2 / H2 O) = + 1.23V)

Pt



[CBSE, OD Set-1, 2020] Ans.  (a)  Given : A = 0.625 cm2, l = 50 cm

R = 5 × 103 ohm, r = ?



m = 0.05 m, K = ?



∧m = ?



Cell constant =



Resistivity =



⇒

50 cm  = = 80 cm −1 A 0.625 cm 2

R R×A or cell constant l

5 × 10 3 × 0.625 50













1 l × Conductivity = Resistivity A

1 50 50 × = 5 × 10 3 0.625 5 × 10 3 × 625 × 10 −3 10 = = 0.016 scm −1 625

10 3 K Molar conductivity ( Λ m ) = M 10 × 1000 K = 320 sm2 mol [3] Λm = × 1000 = M 625 × 0.05 (b)  Given : E° 2+ = +0.34 V Cu / Cu



° E(½ Cl

2

/ Cl − )

E°H+ / H

2 (g)

= +1.36 V

° , Pt = 0.00 V , E(½ O2 / H 2 O ) = +1.23 V

3

Cu

−

+ 2e  → Cu( s); E° = 0.34

H(+aq ) + e - ¾ ¾®

U

¥ l (CH = 40.9 ohm -1cm 2 mol -1 COO- ) 3

¥ ¥ l m(CH = l (CH + l (¥H+ ) COO- ) 3 COOH) 3

= 40.9 + 349.8







= 390.7 ohm -1cm 2 mol -1

At C = 0.1 M Degree of dissociation Ù a= m Ù ¥m 5.20 = = 0.013 i.e.1.3%  390.7 At C = 0.001 M Ù a= m Ù ¥m 49.2 = = 0.125 i.e. 12.5% 390.7 



[2]

[1½]

[1½]

Commonly Made Error  Sometimes students get confused to calculate degree of dissociation of acid.

Answering Tip

At cathode : 2+ ( aq )

2H 2O( l )  → O2 (g) + 4 H + ( aq ) + 4 e − ; E° = +1.23 V

and 40.9 ohm-1cm2mol-1 respectively. ns. Given A l (¥H+ ) = 349.8 ohm -1cm 2 mol -1

5 × 10 ohm ⇒ ⇒ 62.5 ohm cm. 80 cm −1 ⇒ 62.5 ohm cm.

The reaction with a higher value of E° takes place at the cathode, so deposition of copper will take place at the cathode. At anode : The oxidation reactions are possible at the anode. 1 Cl (−aq )  → Cl 2 ( g ) + e − ; E° = 1.36 V 2

At the anode the reaction with a lower value of E° is preferred. But due to the over potential of oxygen, Cl– gets oxidised at anode to produce Cl2 gas. [2] . 2. The molar conductivities of acetic acid at 298 K Q at the concentrations of 0.1 M and 0.001 M are 5.20 and 49.2 S cm2mol-1 respectively. Calculate the degree of dissociation of acetic acid at these concentration. Given l (¥CH+1 ) and l (¥CH COO- ) are 349.8

3



(5 marks each)

1 H 2 ( g ); E°= 0 V 2

 Learn and understand the formula to calculate molar conductivity of solution at infinite dilution and degree of dissociation.

76 ] Q. 3.

Ans.

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

The resistance of 0.01 M acetic solution when measured in a conductivity cell of cell constant 0.366 cm-1, is found to be 2220 Ω. Calculate degree of dissociation of acetic acid at this concentration. Also find the dissociation constant of acetic acid. Given that value of ¥ -1 cm2mol-1 l ¥H+ and l CH - as 349.1 and 40.9 Ω 3 COO respectively. Conductivity(k) of 0.01 M acetic acid

1 k = ´ (cell constant) R 1 = ´ 0.366 cm -1 2220 W

¥ Ù ¥m(CH3 COOH) = l ¥H+ + l CH COO3

= 349.1 + 40.9



Initial conc (mol/L)

k ´ 1000 M 1.648 ´ 10 -4 ´ 1000 = 0.01 = 16.48 W -1cm 2 mol -1 [1]

Ùm =



Degree of dissociation of acetic acid Ù a= m Ù ¥m 16.48 = = 0.0422 [1] 390 CH 3COOH  CH 3COO - + H +



= 1.648 ´ 10 -4 W -1cm -1 [1] Molar conductivity

[1]

= 390 W -1cm 2 mol -1 

Equilibrium conc (mol/L) Dissociation constant [CH 3COO- ][H + ] K= [CH 3COOH] C µ ´C µ = C-Cµ C µ2 = ( 1- µ) 0.01 ´ (0.0422)2 1 - 0.0422 = 1.86 ´ 10 -5

C

0

0

C − Cα

Cα

Cα

=

Molar conductivity at infinite dilution

[1]

TOPIC-2

Redox Reactions and Electrochemical Cells, Electrode Potential and Nernst Equation Scan to know more about this topic

Revision Notes

 Redox reaction : A chemical reaction in which oxidation and reduction both processes take place is known as redox reaction. Oxidation is a process in which any substance loses one or more electrons while reduction is the process in which one or more electrons are gained by another substance. Nernst equation, Gibbs energy  Galvanic cell : A device in which the redox reaction is carried indirectly and chemical energy is and conductance converted to electrical energy. It is also called galvanic cell or voltaic cell.  Redox couple : It is defined as having together the oxidised and reduced form of a substance taking part in an oxidation or reduction half reaction.  Concept : Electrochemical cell Interpretation : Electrochemical cell converts chemical energy into electrical energy or vice-versa.  Concept : Redox reaction Scan to know more about Mnemonic : eRROR this topic Interpretation : Redox reaction involves both oxidation and reduction  Galvanic cell or Voltaic cell : It consists of two metallic electrodes dipped in electrolytic solutions. Electrical energy is produced as a result of chemical reaction which takes place in this cell.  Daniell cell : It is a type of galvanic cell which consist of two electrodes (Zn & Cu) in contact Electrochemistrywith the solution of its own ion i.e., ZnSo4 & CuSo4 respectively. Voltaic cell

Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s)

Cell is represented as, Zn(s) |Zn2+(aq) (C1) || Cu2+(aq) (C2)| Cu(s)  Salt Bridge and its function : It is an inverted U-shaped glass tube which contains a suitable salt in the form of a thick paste made in agar-agar. It performs following functions : (i) It completes inner cell circuit. (ii) It prevents transference of electrolyte from one half-cell to the other. (iii) It maintains the electrical neutrality of the electrolytes in the two half-cells.

Scan to know more about this topic

Electrochemical cells

[ 77

ELECTROCHEMISTRY

 Electrode Potential : It is the potential developed by the electrode with respect to the standard reference electrode. By convention, the reference electrode is standard hydrogen electrode which have a potential of zero volt.  Standard Electrode Potential : Electrode potential at 25°C, 1 bar pressure and 1 M solution is known as standard electrode potential (E°). The standard electrode potential of any electrode can be measured by connecting it to Standard Hydrogen Electrode (SHE). SHE has a standard potential at all temperatures. It consists of a platinum foil coated with platinum black dipped into an aqueous solution in which the H+ = 1 M at 25°C and 1 bar pressure. The potential difference between the two electrodes of a galvanic cell is called the cell potential (measured in volts). It is also called the emf of the cell when no current is flowing through the circuit.  EMF of the cell : It is the sum of electric potential differences produced by separation of charges that occur at each phase boundary in the cell. Ecell = Ecathode – Eanode In terms of standard oxidation electrode potential : E°cell = E°cathode – E°anode, where E°cathode = standard electrode potential of cathode and E°anode = standard electrode potential of anode  Standard oxidation potential : It is the potential difference when given electrode is in contact with its ions having 1 molar concentration, undergoes oxidation when coupled with standard hydrogen electrode.  Electrochemical series : It is the arrangement of the element in order of their increasing electrode potential values. The series has been established by measuring the potential of various electrodes occurs SHE.  Nernst equation : If the concentration of species in the electrode reaction is not equal to 1 versus M, then we use Nernst equation. For a general electrode, Mn+(aq) + ne– ® M(s) the Nernst equation can be written as E

(Mn+ / M)

[M] RT 0 − ln = E n+ M / M) nF  M n+( aq )  (  

where E° = Standard electrode potential, R = Gas constant (8.31 JK –1 mol–1), T = Temperature (K), n = Number of moles of electrons and F = Faraday (96500 C).

At equilibrium,



E°cell = 0.059 log K c n



Kc = Equilibrium constant



M Kc =   [Mn + ]

For the cell with the net reaction,

ne -

aA + bB → mM + nN

the Nernst equation at 298 K can be written as

[M] [N] 0.059 log n [A ]a [B]b m



Ecell = E°cell

-

n

where E°cell = E°cathode – E°anode  Gibbs energy : DG° = nFE°cell For cell reaction to be spontaneous, DG must be negative, Calculations of DrG° and DrG : DrG° = – nF E°cell and DrG = – nF Ecell We also know that, Gibbs energy change is equal to the useful work done. For cell reaction to be spontaneous, DG must be negative. DG° = – 2.303 RT log K.

78 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Mnemonics • Concept: Nernst equation • Interpretation: OPIIEc • Interpretation: Oxidising Power Increases With Increase In E° Value.

Know the Formulae  Ecell = Ecathode – Eanode  E°cell = E°cathode – E°anode  Nernst equation :

Ecell = E°cell –

2.303 RT [C]c [ D]d log nF [ A]a [ B]b



Ecell = E°cell –

0.0591 [C]c [ D]d log at 298 K n [ A]a [ B]b

 DrG° = – nFE°cell  DrG° = – 2.303 RT log KC.

Objective Type Questions

(1 mark each)



EMg2+/ Mg

(b) log[Mg2+]

(a)

The graph of E Mg 2+ /Mg vs. log [Mg 2+ ] is EMg2+/ Mg





log[Mg2+]

(d) log[Mg2+]



EMg2+/ Mg

(c)



log[Mg2+]

U





Q. 1. An electrochemical cell behaves like an electrolytic cell when (a) Ecell = Eexternal (b) Ecell = 0 (c) Eexternal > Ecell (d) Eexternal < Ecell  R [CBSE Outside Delhi Set-2, 2020] Ans. Correct option : (c) Explanation : If an external opposite potential is applied on the electrochemical cell, the reaction continues to take place till the opposite voltage reaches the value 1.1V. At this stage, no current flow through the cell and if there is any further increase in the external potential(Eexternal), then reaction starts functioning in opposite direction i.e. an electrochemical cell behaves like an electrolytic cell. Eexternal > Ecell Q. 2. Electrode potential for Mg electrode varies according to the equation : 0.059 1 E Mg 2 + /Mg =E°Mg 2+ /Mg log . 2 [Mg 2+ ]

EMg2+/ Mg

[A] MULTIPLE CHOICE QUESTIONS :

Ans. Correct option : (b) Explanation :

E Mg 2+ /Mg = E0 Mg 2+ / Mg +

0.059 2+ log é ëMg ù û 2

Compare this equation with the equation of straight line y = mx + c.

The graph of

E Mg 2+ /Mg vs. log [Mg2+] is a

straight line with a positive slope and intercept

E Mg 2+ /Mg . Q. 3. In an electrochemical process, a salt bridge is used                 (a) as a reducing agent             (b) as an oxidizing agent (c) to complete the circuit so that current can flow    (d) None of these R Ans. Correct option : (c) Explanation : In an electrochemical cell, a salt bridge is used to complete the circuit so that current can flow. Q. 4. Following reactions occur at cathode during the electrolysis of aqueous silver chloride solution : Ag+(aq) + e– → Ag(s) E° = +0.80 V

[ 79

ELECTROCHEMISTRY



H+(aq) + e– →

1 H 2 (g) 2

Ans. Correct option : (c) Explanation :

E° = 0.00 V



On the basis of their standard reduction electrode potential (E°) values, which reaction is feasible at the cathode ?



(a) Ag+(aq) + e– → Ag(s)



1 (b) H (aq) + e → H 2 (g) 2 +

–

E° = +0.80 V E° = 0.00 V

(c) Both reactions are feasible (d) None of the above U Ans. Correct option : (a) Explanation : Ag+(aq) + e– ® Ag(s); E° = + 0.80 V. 1 H+(aq) + e– ® H 2 (g) ; E° = 0.00 V. 2 On the basis of their standard reduction potential (E°) values, cathode reaction is given by the one with higher E° values. Thus, Ag+(aq) + e– ® Ag(s) reaction will be more feasible at cathode.











0 ECell = ECell −

[ Mg 2+ ] 0.059 log n [Cu 2+ ]

= 2.71V −

0.1 0.059 log 2 0.001

= 2.71V −

0.059 log 10 2 2

ECell = 2.651 V Q. 8. Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidising power. ClO4–

Ion

Reduction EQ=1.19V Q potential E /V (a) ClO–4 > IO–4 > BrO– 4 (c) BrO–4> IO–4 > ClO– 4 

IO4–

BrO4–

EQ=1.65V

EQ=1.74V

(b) IO–4 > BrO–4 > ClO–4 (d) BrO–4 > ClO–4 > IO–4

Ans. Correct option : (a) Explanation : Higher the reduction potential, higher is its tendency to get reduced. Hence, the order of oxidising power is  : BrO–4> IO–4 > ClO–4 Q. 9. Using the data given below find strongest reduction agent. E − Cr O2 − /Cr3+ =1.33 V, E − Cl /C1− =1.36 V

Depict the galvanic cell in which the given reaction takes place.    (a) Cu2+ (aq)|Cu(s) ||Ag+(aq)|Ag(s)    (b) Cu(s) | Cu2+(aq) || Ag+ (aq) | Ag(s) (c) Ag+(aq)|Ag(s)||Cu2+ (aq)|Cu(s)    (d) Ag(s)|Ag+(aq)||Cu2+ (aq)|Cu(s) Ans. Correct option : (b) Explanation : Oxidation half reaction

2

2

4 (a) Cl- (b) Cr (c) Cr3+ (d) Mn2+ Ans. Correct option : (b) Explanation : The negative value of standard reduction potential for Cr3+ to Cr means that the redox couple is a stronger reducing agent.

Cu(s) ® Cu 2 + (aq) + 2e Reduction half reaction

7

E − MnO- /Mn2+ =1.51V, E − Cr3+ /Cr = − 0.74 V







Ag + (aq) + e - ® Ag(s) Cu(s) | Cu2+(aq) || Ag+ (aq) | Ag(s) Anode salt Cathode (Oxidation) bridge (Reduction) Q. 6. Which of the following statements is not correct ? (a) Copper liberates hydrogen from acids. (b) In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine. (c) Mn3+ and Co3+ are oxidising agents in aqueous solution. (d) Ti2+ and Cr2+ are reducing agents in aqueous solution. Ans. Correct option : (a) Explanation : Copper does not liberate hydrogen from acids because copper lies below hydrogen in electrochemical series. So, copper does not have sufficient electrode potential to liberate elemental hydrogen form compounds in which oxidation state of hydrogen is +1. Q. 7. Calculate the emf of the following cell at 298 K : Mg(s)|Mg2+ (0.1 M)||Cu2+ (1.0 × 10–3 M)|Cu(s) [Given = E°Cell = 2.71 V] (a) 1.426 V (b) 2.503 V (c) 2.651 V (d) 1.8 V 





Q. 5. Consider the following reaction : Cu(s) + 2Ag+(aq) → 2Ag(s) + Cu2+(aq)



U

[B] ASSERTIONS AND REASONS: In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Q. 1. Assertion : Ecell should have a positive value for the cell to function. Reason : Ecathode < Eanode. Ans. Correct option : (c) Explanation : Ecell = Ecathode – Eanode. To have positive value of Ecell, Ecathode should be greater than Eanode. Q. 2. 0Assertion : Cu is less reactive than hydrogen.

Reason : E°Cu2+ / Cu is negative.

80 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Cu

/cu

Q. 3. Assertion : Copper sulphate can be stored in zinc vessel. Reason : Zinc is more reactive than copper. Ans. Correct option : (d) Explanation : Zinc will get dissolved in CuSO4 solution, since, zinc is more reactive than copper. E+ Ag increases with increase with increase in in + increases / Ag / Ag increases with increase in Q. 4. AssertionE: AgE Ag+ / Ag



+ + + ions. concentration of Ag Ag concentration of of Ag ions. concentration ions. E+ Ag has a positive value. has a positive value. Reason : E AgE + / Ag / Ag has a positive value.

Ag+ / Ag Ans. Correct option : (b) Explanation : E Ag + / Ag = E

Ag + / Ag



On increasing [Ag+], EAg+/ has a positive value.

Ag

[C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1. Can we store copper sulphate in a iron vessel? Ans. No, we can not store copper sulphate in an iron vessel because copper having higher reduction potential will accept the electron and will change into solid copper. So, iron will react with copper sulphate. Q. 2. Define electrochemical series.

Ans. Free energy decreases in Galvanic cell, so its value is negative, i.e. ΔG < 0.

RT 1 − log nF  Ag + 

Ans. (i) Zinc to silver [1] (ii) Concentration of Zn2+ ions will increase and Ag+ ions will decrease. [1] [CBSE Marking Scheme 2017] Q.2.  Calculate the emf of the following cell at 298 K Cr(s)/Cr3+ (0.1M)//Fe2+ (0.01M)/Fe(s) [Given : E°cell = + 0.30 V] Ans. 2Cr(s) + 3 Fe2+(aq.) → 3Fe(s) + 2Cr3+ (aq.) [½] n = 6 2



Cr 3 +  2.303RT   − log 3 nF Fe2+    

Q.3. Explain redox potential. Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidising power.



Commonly Made Error  Sometimes students get confused to write correct formula for calculation of emf of the cell.

Answering Tip  Learn and understand Nernst equation for an electrochemical cell.

Reduction potential EQ/V

EQ=1.19V

IO4-

BrO4-

EQ=1.65V EQ=1.74V

Ans. Redox potential (also known as reduction / oxidation potential) is a measure of the tendency of a chemical species to acquire electrons from or lose electrons to an electrode and thereby be reduced or oxidised respectively. Redox potential is measured in volts (V), or millivolts (mV). The more positive the reduction potential of a species, the greater the species’ affinity for electrons and tendency to be reduced.[1] The higher the reduction potential, the higher is its tendency to get reduced. Hence, the order of oxidising power is : BrO–4> IO–4 > ClO–4 [1] Q. 4. Following reactions can occur at cathode during the electrolysis of aqueous silver nitrate solution using Pt electrodes : Ag+(aq) + e– Ag (s); E0=0.80 V 1 H+ (aq)+ e–  E0=0.00 V 2 H2 (g) ;

[½]

[½]

ClO4-

On the basis of their standard electrode potential values, which reaction is feasible at cathode and why ? Ans. Ag+(aq) + e− → Ag(s) Because it has higher reduction potential. [1] Detailed Answer : As reaction with higher value of standard electrode potential occurs at cathode, Ag gets reduced. So, the reaction occurring at cathode is



Ion



2

10−1  0.059   ECell = 0.30 V − V log 3 − 2 6 10     ECell = 0.26 V

(2 marks each)



Q. 1.  In a galvanic cell, the following cell reaction occurs: Zn(s) + 2Ag+(aq) ® Zn2+(aq) + 2Ag(s) Eocell = +1.56 V (i) Is the direction of flow of electrons from zinc to silver or silver to zinc? (ii) How will concentration of Zn2+ ions and Ag+ ions be affected when the cell functions? A&E [CBSE Foreign Set-1, 2, 3 2017]

ECell =

R

Ans. The arrangement of various elements in the order of increasing values of standard reduction potentials is called electrochemical series.

Short Answer Type Questions-I

o ECell

will increase and it

Q. 3. What is the charge of free energy for a galvanic cell? R

Ag + e − → Ag , Θ





Ans. Correct option : (c) Explanation : Cu is less reactive than hydrogen because EQ 2+ is positive.





Ag+(aq) + e− → Ag(s)

[2]

0.059 [Prod.] log n [React.] [Prod.] 0.059 log g + n [React.]

o Ecell = Ecell o Ecell = Ecell

ELECTROCHEMISTRY

Q. 5. Calculate E°cell for the following reaction at 298 K : 2+

2Cr(s) + 3Fe

= 0.261 +

3+

(0.01M) → 2Cr (0.01M) + 3Fe(s)

Given : Ecell = 0.261 V Ans. Nernst Equation :

0.059 [Prod.] log n [React.] [Prod.] 0.059 log g + n [React.]

= 0.261 +

o Ecell = Ecell -



o Ecell = Ecell



0.059 [Cr 3+ ]2 = 0.261 + log 6 [Fe 2+ ]3



-2 2

0.059 (10 ) log 6 (10 -2 )3 0.059 = 0.261 + log 10 2 6 0.059 ´ 2 = 0.261 + 6 = 0.261 + 0.01966 = 0.28068 V » 0.281V



= 0.261 +

0.059 [Cr 3+ ]2 log 6 [Fe 2+ ]3

0.059 (10 -2 )2 log 6 (10 -2 )3 0.059 = 0.261 + log 10 2 6 0.059 ´ 2 = 0.261 + 6 = 0.261 + 0.01966 = 0.28068 V » 0.281V 

U

[ 81





[2]

OR







[Topper’s Answer] [2] sulphate while electrode potential of Pt is less than Cu. Due to this reason, Pt cannot displace Cu from copper sulphate. [2]

Q.6. Iron displaces copper from copper sulphate solution but Pt does not why? Ans. Electrode potential of Fe is more than electrode potential of Cu. So, Fe displaces Cu from copper

Short Answer Type Questions-II

Q. 1.  Consider the following reaction : Cu(s) + 2Ag+(aq) ® 2Ag(s) + Cu2+(aq)



(i)

(ii)

(3 marks each)

Depict the galvanic cell in which the given reaction takes place. Give the direction of flow of current.

82 ] (iii) 

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Write the half-cell reactions taking place at cathode and anode. U [CBSE Comptt. Delhi/OD 2018]

Ans. (i) Cu(s) | Cu2+(aq) || Ag+ (aq) | Ag(s) [1] (ii) Current will flow from silver to copper electrode in the external circuit. [1] (iii) Cathode: 2Ag+(aq) + 2e– → 2Ag(s) Anode: Cu(s) → Cu2+ (aq) + 2e– [1] [CBSE Marking Scheme 2018]

Q. 2.  (a) The cell in which the following reaction occurs : 2 Fe3+ (aq) + 2 I– (aq) → 2 Fe2+ (aq) + I2 (s) ° has Ecell = 0.236 V at 298 K. Calculate the standard Gibb’s energy of the cell reaction. (Given : 1 F = 96,500 C mol–1)      (b) How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours ? (Given : 1 F = 96,500 C mol–1) 3 Ans. (a)

0 ΔG0 = – nFEcell [½] n = 2 ΔG0 = – 2 × 96500 C /mol × 0.236 V [½] = – 45548 J/mol = – 45.548 kJ/mol [½]

Ans.

Commonly Made Error  Students forget to mention the working formula and start the calculations or do not mention all the steps or specific units. Marks are allotted for formulas too.

Answering Tip  Always write the working formula followed by the value substitution for each entity. Do not forget to mention units wherever required. Q. 4.  Calculate e.m.f. of the following cell at 298 K :

2Cr(s) + 3Fe2+ (0.1M) ® 2Cr3+ (0.01M) + 3Fe(s)



E°(Cr3+ / Cr) = – 0.74 V



E° (Fe2+ / Fe) = – 0.44 V.



2I– → I2 + 2e–



For the given cell reaction, n = 2.

(b) I = 0.5A t = 2 hours = 2 × 60 × 60 s = 7200s [1½] Q = It = 0.5 × 7200 = 3600 coulombs A flow of 96500 C is equal to flow of 1 mole of electrons is 6.023 × 1023 electrons. ∴ 3600 c is equivalent to of electrons =

6.023 ´ 10 23 × 3600 96500

= 2.246 × 1022 electrons [1½] Q. 3. Calculate DrG0 and log Kc for the following reaction at 298 K.



ΔG° = – n FE cell = – 2 × 96500 × 0.236 = – 45548 J mol–1 = – 45.55 kJ mol–1







°



2Cr( s) + 3Fe2+( aq) → 2Cr 3+( aq) + 3Fe( s) [(E°cell = 0.30 V), 1F = 96500 C mol-1] A [CBSE Comptt. OD Set-2 2017]

U [CBSE Delhi Set-1, 2 & 3, 2016]

Ans.

(b) Q = I t = 0.5 × 2 × 60 × 60 [½] = 3600 C 96500 C = 6.023 × 1023 electrons 3600 C = 2.25 × 1022 electrons [1]  [CBSE Marking Scheme, 2017] Detailed Answer : (a) 2 Fe3+ + 2e– → 2 Fe2+

DrG0 = -nFE°cell, n = 6 [½] = -6 × 96500 C/mol × 0.30 V = -173700 J/mol = -173.7 kJ/mol[1] E°cell = 0.059V/n × log Kc [½] log Kc = 0.30 V × 6/0.059V = 30.5 [1] [CBSE Marking Scheme 2017]

E°cell = E°cathode – E°anode = (– 0.44) – (– 0.74) V = 0.30 V E = E° –

[½]

0.0591 [ Products] log n [ Reactants]



= 0.30 –

0.0591 [Cr 3+ ]2 log 6 [ Fe 2+ ]3

[½]



= 0.30 –

0.0591 [0.01]2 log 6 [0.1]3

[1]









 − 0.0591  = 0.30 –   = 0.3098 V [1] 6   



[CBSE Marking Scheme 2016]

Q. 5.  Calculate the emf of the following cell at 25° C :     Fe | Fe2+ (0.001 M) || H+ (0.01 M) | H2(g) (1bar) | Pt(s)     E° (Fe2+ / Fe) = – 0.44 V E° (H+ / H2) = 0.00 V A [CBSE Delhi 2015] Ans. Cell reaction is Fe(s) + 2H+(aq) ® Fe2+ (aq) + H2(g) E°cell = 0.00 – (– 0.44) = 0.44 V 0.0591 [ Fe 2 + ] log + 2 2 [H ]



Ecell = E°cell –



= 0.44 V –





0.0591 0.001 V log 2 ( 0.01)2

= 0.44 V – 0.02955 V Ecell = 0.41045 V

[3]

[ 83

ELECTROCHEMISTRY

Q. 6.

Ans.

A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn(NO3)2 solution and metallic plate of lead in 0.02 M Pb(NO3)2 solution. Calculate the emf of the cell. Write the chemical equation for the electrode reactions and represent the cell. (Given : E°Zn2+/Zn = – 0.76 V; E°Pb2+/Pb = – 0.13V)

Anode reaction : Zn(s) ® Zn2+(aq) + 2e– Cathode reaction : Pb2+(aq) + 2e– ® Pb(s) Cell representation : Zn(s)|Zn2+(aq)||Pb2+(aq)|Pb(s) According to Nernst equation :  Zn 2+  0.0591   − log E = E° cell cell  Pb2+  2  



A

[½] [½] [½]

0.0591 0.1 log 2 0.02 = 0.63 – 0.02955 × log 5 = 0.63 – 0.02955 × 0.6990 = 0.63 – 0.0206 = 0.6094 V

Ecell = [– 0.13 – (– 0.76)] -

[½]



[½]

Commonly Made Errors  The cell representation is given incorrectly by many candidates.  The calculation of emf of the cell by using Nernst equation is incorrect, in some cases.

Answering Tip [½]

 Do more practice of cell representation and numeri-

cal based on Nernst equation.

Long Answer Type Questions

o Zn 2 + / Zn

= -0.76 V, E

Given : E



[Given : log 10 = 1]

o Ag + / Ag

(b) X and Y are two electrolytes. On dilution molar conductivity of ‘X’ increases 2.5 times while that Y increases 25 times. Which of the two is a weak electrolyte and why?



E°

Zn 2+ / Zn

E° + Ag / Ag

= −0.76 V

= + 0.80 V

Ecell = E°cell −

emf = ?

[ Anode] 0.0591 log n [Cathode]

E°cell = E°cathode − E°anode



= E°Ag / Ag − E°



= 0.80 – (–0.76) = 1.56 V

Zn

Ecell = 1.56 −

= 1.56 −

2+

weak electrolyte

c  

[2]

Commonly Made Error  Some students get confused to find Eocell correctly

Answering Tip to identify the Eocathode and Eoanode from given standard reduction potentials.

 Understand

Q. 2.   Eocell for the given redox reaction is 2.71 V.  Mg + Cu2+(0.01 M) Mg2+(0.001 M) + Cu Calculate Ecell for the reaction. Write the direction of flow of current when an external opposite potential applied is (i) Less than 2.71 V

Ans.



m 

= + 0.80 V

 [CBSE Outside Delhi Set-1, 2020] Ans. (a) Zn(s)/ Zn2+ (0.1 M) || (0.01M) Ag+/Ag(s)

strong electrolyte

Q. 1. (a) Calculate e.m.f. of the following cell : Zn(s)/Zn2+ (0.1 M) || (0.01 M) Ag+/Ag(s)



(5 marks each)

/ Zn

Ecell =E° cellS

=E° cellS

2+

0.0591 [Zn ] log 2 [Ag+ ]2

0.059  logKc n 

[1]

0.059  10 -3 log -2 =2.71 + 0.0295[1] 2 10

Ecell = 2.7395 V (i) Cu to Mg / Cathode to anode / Same direction[1] (ii) Mg to Cu / Anode to cathode / Opposite direction  [CBSE Marking Scheme, 2019] [1]

0.0591 [0.1] log 2 [0.01]2

= 1.56 – 0.0295 log 1000 = 1.56 – 3 (0.0295)= 1.56 – 0.09 = 1.4715[3] (b) Y is a weak electrolyte as n dilution complete dissociation of weak electrolyte takes place and thus a sharp increase in molar conductivity while in strong electrolyte it has already dissociated completely. So on dilution molar conductivity does not rises much.



(ii) Greater than 2.71 V[CBSE Delhi Set-1, 2019]

Detailed Answer :



o Ecell = Ecell -

é Mg 2 + ù 0.059 log ë 2 + û n éëCu ùû

Ecell = 2.71 -

0.059 [0.001] log 2 [0.01]

Ecell = 2.71 - ( -0.0295) = 2.74 V 

[3]

84 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(i) When external opposite applied voltage is less than 2.71, it is less than E

o cell

, therefore, the

(ii)  When external opposite applied potential is greater than 2.71, it is greater than Eocell , therefore, the reaction will be reversed, and the current will flow from anode to cathode. [1]



electrons will flow from the anode to the cathode, and current will flow from cathode (copper electrode) to anode (magnesium electrode). [1] OR









[ 85



ELECTROCHEMISTRY

[Topper's Answer 2016]

rises. This increase leads to increase in the number of ions in the solution. Thus, the molar conductivity rises sharply of a weak electrolyte at low concentration. The molar conductivity of strong electrolyte decreases a bit with an increase in concentration. This is due to increase in interionic attraction due to higher number of ions per unit volume. On dilution, ions move apart, weakening interionic attractions and thus conductance increases. Limiting molar conductivity for weak electrolytes is obtained by using Kohlrausch law of independent migration of ions. [2]

Q. 3. (a) Represent the cell in which the following reaction takes place :

2Al ( s ) + 3Ni 2 + ( 0.1M ) ¾¾ ® 2Al 3 + ( 0.01M ) + 3Ni ( s ) Calculate its emf if Eocell = 1.41V. (b)  How does molar conductivity vary with increase in concentration for strong electrolyte and weak electrolyte? How can you obtain limiting molar conductivity

(Ù )

for weak



electrolyte? [CBSE Outside Delhi Set-2, 2019]



Ans. (a) Al(s) | Al3+(0.01M) || Ni2+ (0.1 M) | Ni(s)[1]

[ Al 3 + ] 0.059 log 6 [Cu 2 + ]3 2

Ecell = Eocell −

[½]

[0.01] 0.059 log 6 [0.1]3 = 1.4198 V or Ecell = 1.42 V 2

Ecell = 1.41V − Ecell

[1] [½]

(b) Λm decreases with increase in concentration for both strong & weak electrolyte. Λ0m can be obtained for weak electrolyte by applying Kohlrausch law Λ0m = V+l0+ + V–l0–[1 + 1] [CBSE Marking Scheme 2019]



Q.4. (a) Write the cell reaction and calculate the e.m.f. of the following cell at 298 K : Sn (s) | Sn2+ (0.004 M) || H+ (0.020 M) | H2 (g) (1 bar) | Pt (s) (Given : E°Sn2+/ Sn = - 0.14V) (b) Give reasons : (i) On the basis of E° values, O2 gas should be liberated at anode but it is Cl2 gas which is liberated in the electrolysis of aqueous NaCl. (ii) Conductivity of CH3COOH decreases on dilution. [CBSE Delhi/Outside Delhi, 2018]





Detailed Answer :

(a) 2Al (s) + 3Ni 2 + (0.1M ) ¾¾ ® 2Al 3 + (0.01M ) + 3Ni (s)



Al (s)|Al 3 + ||Ni 2 + |Ni (s) ¾¾ ® Cel l reaction





[0.01] 0.0591 log 6 [0.1]3

E = Eo –

2

Ecell = E -

  = 1.41 - 0.0591 log 0.1 6 0.0591   = 1.41 + log 10 6

0.0591 6  = 1.41 + 0.00985 = 1.42 V [3] (b)  When the concentration of weak electrolyte becomes very low, its degree of ionization

Ans. (a) Sn + 2H+ ® Sn2+ + H2 (Equation must be balanced)[1]

  = 1.41 +



éSn 2 + ù 0.059 û log ë 2 2 éH + ù ë û 

0= [0 – (– 0.14) ] – 0.0295 log

[½]

(0.004 ) (0.02 )2 

[½]

= 0.14 – 0.0295 log 10 = 0.11 V / 0.1105 V[1]

(b) (i) Due to over potential/ overvoltage of O2 [1]

(ii)  The number of ions per unit volume decreases.[1] [CBSE Marking Scheme, 2018]

86 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Detailed Answer :

(a) Sn(s)| Sn 2 + (0.004 M)||H + (0.020 M)



| H 2 (g)(1 bar)| Pt(s) E°cell

=

E° + (H /H 2 )

- E° 2 + (Sn /Sn)

= 0.00 - ( - 0.14) = + 0.14V



0 Ecell =

Sn(s) + 2H + (aq) ® Sn 2 + (aq) + H 2 (g) 0.0591 [Sn 2 + ] log n [ H + ]2

H 2O  H + + OH The value of E° of O2 is higher than Cl2 but O2 is evolved from H2O only when the higher voltage is applied. So, because of this Cl2 is evolved instead of O2.

(ii)  Conductivity varies with the change in the concentration of the electrolyte. The number of ions per unit volume decreases on dilution. So, conductivity decreases with decrease in concentration. Therefore, conductivity of CH3COOH decreases on dilution. [3+2] Q. 5.  (a) Calculate E°cell for the following reaction at 298K: 2Al(s) + 3Cu2+ (0.01M) → 2Al3+ (0.01M) + 3Cu(s) Given : E°cell = 1.98 V  (b)  Using the E° values of A and B, predict which is better for coating the surface of iron [E°(Fe2+/Fe) = – 0.44V] to prevent corrosion and why ? Given : E°(A2+/A) = – 2.37V : E°(B2+/B) = – 0.14V  [CBSE Outside Delhi Set-1, 2 & 3, 2018]



0 Ecell = Ecell +

0.0591 [ Al 3 + ]2 log n [Cu 2 + ]3





[1]

0.0591 [Al 3 + ]2 log n [Cu 2 + ]3

0 Ecell = 1.98 V +

0.0591 (0.01)2 [1] log 6 (0.01)3 

0 Ecell = 1.98 V +

0.0591 log 10 2 6



 Students often make error identifying oxidation reaction and reduction reaction from a cell representation.

 Practice numericals with oxidation and reduction reactions.

(b) (i) NaCl ® Na + + Cl -

0 Ans. (a) Ecell = Ecell −

Commonly Made Error

Answering Tip

( 4 ´ 10 -3 ) 0.0591 log 2 ( 2 ´ 10 -2 )2 = 0.14 - 0.0295 log 10 = 0.14 - 0.0295 = 0.1105 V

= 0.14 -



[Q log 10 = 1] 0.0591V ´2 1.98 V + 6

0 Ecell = 1.9997 V [1] (b) A, because its E0 value is more negative[1 + 1] [CBSE Marking Scheme, 2016]



2H + (aq) + 2e - ® H 2 (g)

Ecell = E°cell -

0.0591 ´ 2 ´ log 10 6

0 Ecell = 1.98 V + 0.0197 V

Sn(s) ® Sn 2 + (aq) + 2e -



0 Ecell = 1.98 V +

Q. 6. (a) The conductivity of 0.001 mol L–1 solution of CH3COOH is 3.905 × 10–5 S cm–1. Calculate its molar conductivity and degree of dissociation (α). Given λ0 (H+) = 349.6 S cm2 mol–1 and λ0 (CH3COO–) = 40.9 S cm2 mol–1 (b) Define electrochemical cell. What happens if external potential applied becomes greater than E°cell of electrochemical cell ?  [CBSE Outside Delhi Set-1, 2 & 3, 2016] Ans. (a)

Λm =k × 1000/C

=

3.905 ´ 10 –5 ´ 1000 0.001

= 39.05 S cm2/mol  CH3 COOH → CH3COO– + H+ Λ0CH3COOH =l0CH3COO– + l0H+ = 40.9 + 349.6  Λ0CH3COOH = 390.5 S cm2/mol a =

Lm L 0m

=

39.05 = 0.1  390.5

[1]

[1]

[1]

(b) Device used for the production of electricity from energy released during spontaneous chemical reaction and the use of electrical energy to bring about a chemical change. [1] The reaction gets reversed / It starts acting as an electrolytic cell & vice – versa.  [1]  [CBSE Marking Scheme, 2016]

Commonly Made Error  Students often make mistakes in writing half-cell reactions. Also, in some cases calculation errors are seen while solving Nernst equation as the students miss out the power of concentration terms.

Answering Tip  Practice numericals with Nernst equation, by writing every step clearly.

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ELECTROCHEMISTRY

TOPIC-3

Electrolysis, Law of Electrolysis, Batteries, Fuel Cells and Corrosion

Revision Notes  Electrolysis is the process of decomposition of an electrolyte when electric current is passed through either its aqueous solution or molten (fused) state. This process takes place in electrolytic cell.  Faraday’s first law of electrolysis : The amount of chemical reaction which occurs at any electrode during electrolysis is proportional to the quantity of electricity passed through the electrolyte. m = Z × I × t, where Z = Electrochemical equivalent   Faraday’s second law of electrolysis: Amount of various substances liberated by the same quantity of electricity passed through the electrolytic solution is proportional to their chemical equivalent weights. w1 w2 = E1 E1  Products of electrolysis depend on (i) Physical state of material. (ii) Types of electrode being used.  Battery is a combination of galvanic cells in series and used as a source of electrical energy. Types of batteries : (i) Primary batteries are nonchargeable batteries such as Lechlanche cell and Dry cell. (ii) S  econdary batteries are chargeable cells involving reversible reaction. Example, Lead storage battery and Nickel-cadmium cells.  Dry cell (Lechlanche cell) : The anode consists of a zinc container and the cathode is a graphite electrode surrounded by powdered MnO2 and C. The space is filled with paste of NH4Cl and ZnCl2.

Scan to know more about this topic

Faraday’s law of electrolysis

Fig 1 : Dry Cell 2+

–



At anode: Zn(s) → Zn (aq) + Ze



At cathode: MnO2(s) + NH4+(aq)+ 2e– → MnO(OH) + NH3



The net reaction: Zn + NH4+(aq) + MnO2 → Zn2+ + MnO(OH) +NH3

 Lead storage battery :

Anode - Spongy lead



Cathode - Lead packed with lead dioxide



Electrolyte -Aqueous solution of H2SO4

Fig 2 : Lead storage battery

88 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Discharge reaction of cell : At anode: Following reaction takes place at anode: Pb(s) +SO42–(aq) → PbSO4(s) +2e– Reaction at cathode: PbO2 filled in lead grid gets reduced to Pb2+ ions which combines with SO42– ions to form PbSO4(s). Complete cathode reaction is as follows : PbO2(s) + 4H+(aq) + SO42–(aq) + 2e– → PbSO4(s) + 2H2O(l) Complete cell reaction: Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l) Recharge reaction of cell : It changes the direction of electrode reaction. PbSO4 accumulated at cathode gets reduced to Pb. At cathode, PbSO4(s) +2e– → Pb(s) + SO42–(aq) At anode, PbSO4 gets oxidised to PbO2. Scan to know more about PbSO4(s) + 2H2O → PbO2(s) + 4H+(aq) + SO42–(aq) + 2e– this topic Complete cell reaction would be as follows: PbSO4(s) + 2H2O(l) charge → Pb(s) + PbO2(s) + 2H2SO4(aq)  Conventions for representing the galvanic cell : (i) Double vertical line is used for salt bridge. Left hand side of the double line is anode and Corrosion the cathode is on the right hand side. (ii) A single vertical line is used to separate metal and the electrolytic solution. (iii) If there is no metallic surface involved, we write Pt. Example: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) We use inert electrode Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)  Fuel cells : Electrical cells that are designated to convert the energy from the combustion of fuels such as hydrogen, carbon monoxide or methane directly into electrical energy are called fuel cells. In the cell: Anode: [H2(g) + 2OH–(aq) → 2H2O (l) + 2e–] × 2 Cathode : O2(g) + 2H2O(l) + 4e– → 4OH–(aq) Net reaction: 2H2(g) + O2 → 2H2O (l).

Fig 3 : Fuel cell using H2 and O2 produces electricity  Corrosion : The process of slow conversion of metals into their undesirable compounds (usually oxide) by reaction with moisture and other gases present in the atmosphere.

Rusting of iron : Fe( s) + 2 H + ( aq ) +

1 O2 ( aq ) ® Fe 2 + (aq) + H 2O(l) 2

1 O2 ( g ) + 2 H 2O(l) ® Fe 2O3 ( s) + 4 H + 2 Fe 2O3 + xH 2O ® Fe 2O3 xH 2O ( Rust )

2 Fe 2 + ( s) +



Prevention of Corrosion : (i) Barrier protection: By covering the surface with paint or a thin film of grease or by electroplating. (ii) Sacrificial protection: By galvanization. (iii) Alloying.

[ 89

ELECTROCHEMISTRY

Mnemonics • Concept: Fuel Cell • Interpretation: FCCEE • Interpretation: Fuel Cell Converts Chemical Energy of a Fuel Into Electrical Energy

Know the Formulae  Q = It  m = ZIt

Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS: Q. 1. The amount of electricity required to produce one mole of Zn from ZnSO4 solution will be (a) 3F (b) 2F (c) 1F (d) 4F [CBSE Delhi Set-3, 2020] Ans. Correct option : (b) Explanation : ZnSO4  Zn 2 + + SO4 2 1mol

1mol

Zn 2 + + 2e - ® Zn  Number of electrons gain = 2 \ The amount of electricity required to produce one mole of Zn from ZnSO4 Solution = 2F

  Q. 2. Zinc is coated over iron to prevent rusting of iron because (a) Eo Zn2+ / Zn = Eo Fe2+ / Fe (b) E Eo Zn2+ / Zn > Eo Fe2+ / Fe (c) Ans.

o Zn 2 + / Zn

< Eo Fe2+ / Fe

(d) None of these [CBSE Delhi Set-3, 2020]

Correct option : (b) Explanation : Zinc is coated over iron to prevent rusting of iron because standard reduction potential of Zn is lesser than Fe.

Eo Zn 2+ / Zn < Eo Fe2+ / Fe  Q. 3.  In a lead storage battery : (a) PbO2 is reduced to PbSO4 at the cathode. (b) Pb is oxidised to PbSO4 at the anode. (c) Both electrodes are immersed in the same aqueous solution of H2SO4. (d) All the above are true.  [CBSE Outside Delhi Set 1, 2020] Ans. Correct option : (b) Explanation : In a lead storage battery, • At Cathode, PbO 2 +2H 2O + 2e - ® Pb2 + + 4OH



Pb 2 + + SO4 2 - ® PbSO 4

(1 mark each)

• At Anode, Pb ® Pb2+ + 2e -

Pb 2+ + SO4 2 - ® PbSO4 • Both electrodes are immersed in the same aqueous solution of H2SO4. . 4. Which of the following is correct for spontaneity Q of a cell ?

(a) DG = –ve, Eo = +ve

(b) DG = +ve, Eo = 0

(c) DG = –ve, Eo = 0 (d) DG = +ve, Eo = –ve Ans. Correct option : (a) Explanation : For spontaneity of cell, DG = - ve E =0 [B] ASSERTIONS & REASONS: In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Q.1.  Assertion : Mercury cell does not give steady potential.    Reason : In the cell reaction, ions are not involved in solution. Ans.  Correct option : (d)      Explanation : Mercury cell gives a steady potential because in the cell reaction, ions are not involved in the solution.  Q.2. Assertion : Electrolysis of NaCl solution gives chlorine at anode instead of O2.      Reason : Formation of oxygen at anode requires over voltage. o

90 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Ans.  Correct option: (a)      Explanation : Formation of oxygen has lower value of E° than formation of chlorine even then it is not formed because it requires over voltage. Q. 3.  Assertion : Zinc protects the iron better than tin even after the cracks.      Reason : Oxidation potential of Zn > Fe but oxidation potential of Sn < Fe. Ans.  Correct option : (a)      Exaplanation : Oxidation potential of Zn > Fe and Sn < Fe      ∴ Zinc protects the iron better than tin even after the cracks.

Ans. At anode, water should get oxidised in preference to Cl-, but due to over voltage/ over potential Cl- is oxidised in preference to water.

[C] VERY SHORT ANSWER TYPE QUESTIONS:

Ans. H2–O2 fuel cell

Q. 1. The products of electrolysis of aqueous NaCl at the respective electrodes are : Cathode : H2 Anode : Cl2 and not O2. Explain. A

Q.4.  Name the cell used in hearing aids and watches  [CBSE, Delhi Set-1, 2020]

Q.2. Which electrolyte is used in fuel cell? What is the relation which expresses thermodynamic efficiency of cell? R Ans. Aqueous sodium hydroxide is used in fuel cell. [½]

h=

DG - nFE [½] = DH DH

Q.3.  Give an example of fuel cell.

Ans. Primary cell.

Short Answer Type Questions-I



Q. 1. Write the name of the cell which is generally used in hearing aids. Write the reactions taking place at the anode and the cathode of this cell. U + R [CBSE Outside Dehi Set-1, 2017]

ns. Mercury cell[1] A Anode : Zn(Hg) + 2OH– → ZnO(s) + H2O + 2e– [½] Cathode : HgO + H2O + 2e– → Hg(l) + 2OH– [½]  [CBSE Marking Scheme, 2017] Detailed Answer : Mercury cell is generally used in hearing aids.[1] At anode : Zn (Hg) + 2 OH– → ZnO(s) + H2O + 2e– At cathode : HgO(s) + H2O + 2e– → Hg(l) + 2OH– Overall reaction : Zn(Hg) + HgO(s) → ZnO(s)+ Hg(l)[1] Q. 2. Write the name of the cell which is generally used in transistors. Write the reactions taking place at the anode and the cathode of this cell.  U + R [CBSE Outside Dehi Set-3, 2017] ns. Dry Cell/Leclanche cell[1] A Anode : Zn(s) → Zn2+ + 2e–[½] Cathode : MnO2 + NH4+ + e– → MnO(OH) + NH3 [½] [CBSE Marking Scheme, 2017] Detailed Answer : The cell which is used in the transistors is Dry cell. At anode : Zn(s) → Zn2+ + 2e–

(2 marks each)

At cathode : MnO2 + NH+4 + e– → MnO(OH) + NH3 Ammonia produced in the reaction forms a complex with Zn2+ ion. Zn2+ + 4NH3 → [Zn(NH3)4]2+ [2] Q. 3. From the given cells : Lead storage cell, Mercury cell, Fuel cell and Dry cell. Answer the following : (i) Which cell is used in hearing aids?   (ii)  Which cell was used in Apollo Space Programme?   (iii)  Which cell is used in automobiles and inverters? (iv) Which cell does not have long life?  R [CBSE Dehi Set-1,2 & 3 2016] ns. (i) Mercury cell A (ii) Fuel cell [½ + ½] (iii) Lead storage cell (iv) Dry cell [½ + ½]  [CBSE Marking Scheme, 2016] Q.4. Write the electrode reactions for a H2–O2 fuel cell.  R Ans. At anode, éë H 2 (g) + 2OH - (aq) ® 2H 2O(l) + 2e - ùû ´ 2 [½]

At cathode, O2 (g) + 2H 2O(l) + 4e - ® 4OH - (aq) [½] Net reaction, 2H 2 (g) + O2 (g) ® 2H 2O(l) [1]

Short Answer Type Questions-II Q. 1. When a steady current of 2A was passed through two electrolytic cells A and B containing electrolytes ZnSO4 and CuSO4 connected in series, 2 g of Cu were deposited at the cathode of

R



(3 marks each)

cell B. How long did the current flow? What mass of Zn was deposited at cathode of cell A? [Atomic mass : Cu = 63.5 g mol–1, Zn = 65 g mol–1; 1F = 96500 C mol–1]

[ 91

ELECTROCHEMISTRY

Ans. Zn 2 + ( aq ) + 2e −  → Zn(s) 2 mol 1 mol

Cu 2 + + 2e −  → Cu( s) 2 mol

1 mol



(2 gm given)

The charge Q on a mole of electrons, Q = nF

Calculation of time for the flow of current : n = 1 mol Q = 1 × 96500 C mol–1 = 96500 C Molar mass of Cu = 63.5 gm mol–1



∵  63.5 gm of Cu is deposited by electric charge = 96500C ∴ 2 gm of Cu is deposited by electric charge

=

96500 × 2 = 3039.37 C 63.5

[1]



Let 2 A of current be passed for time t, quantity of electricity used = 2A × t = 3039.37 C 3039.37C or, t = = 1519.68 s 2



= 25 min. 33 s Calculation of mass of Zn deposited : W1 E1 Mass of Zn = = W2 E 2 Mass of Cu Molar mass of Zn / Charge on Cu = Molar mass of Cu / Charge on Cu

[1]

(b) Fuel cell is the name given to the galvanic cells which are designed to convert the energy of combustion of fuels like hydrogen, methane, methanol, etc. directly into electrical energy.  [1½] Q.3. The electrolysis of a metal salt solution was carried out by passing a current of 4 A for 45 minutes. It resulted in deposition of 2.977 g of a metal. If atomic mass of the metal in 106.4 g mol–1, calculate the charge on the metal cation. Ans. Let the charge on the metal ion = n+ Reduction half-reaction,

[½]

Quantity of electricity required for depositing 106.4 g of metal = n × 96500 C [1]  Quantity of electricity required for depositing n ´ 96500 ´ 2.977 2.977 g of metal = = n ´ 2700 [1] 106.4 Quantity of electricity passed = 4 × 45 × 60 C = 10800 C 10800 = n × 2700 10800 =4 n= 2700

[1]



[1]

Q.2.  (a) Calculate the mass of Ag deposited at cathode when a current of 2 amperes was passed through a solution of AgNO3 for 15 minutes. (Given : Molar mass of Ag = 108 g mol–1, 1 F = 96500 C mol–1) (b) Define fuel cell. Ans. (a) m = Zit 108 ´ 2 ´ 15 ´ 60 = 1 ´ 96500

(or any other correct method) [½] (b) Cells that convert the energy of combustion of fuels directly into electrical energy. [CBSE Marking Scheme, 2017] [1]  Detailed Answer : (a) t = 900 s Charge = Current × Time = 2 × 900 = 1800 C According to the reaction Ag+ (aq) +e– → Ag(s) We require 1 F to deposit 1 mol or 108 g of Ag For 1800 C, the mass of Ag deposited will 108 ´ 1800 be = = 2.0145 g [1½] 1 ´ 96500

M n + + ne - ® M (1mol) (n mol) (106.4 g)

Amount of Zn deposited :

65 = 2 × 2 = 2.0472 g 635 2

= 2.01 g

Charge on metal ion = +4[1]

Long Answer Type Questions



Q.1. (a) A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeSO4 and ZnSO4 until 2.8 g of Fe deposited at the cathode of cell X. How long did the current flow? Calculate the mass (Molar mass: Fe = 56 g mol-1, Zn = 65.3 g mol-1, 1 F = 96500 C mol-1) (b) In the plot of molar conductivity L m vs. square root of concentration (C½), following curve obtained for two electrolytes A and B:

(5 marks each)

∧m

C

92 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Answer the following : (i) Predict the nature of electrolytes A and B:

(ii)  As concentration of strong electrolyte approaches zero, the molar conductivity of the plot intercepts the molar conductivity axis, giving the limiting value of molar conductivity



(ii)  What happens on extrapolation of L m to concentration approaching zero for electrolytes A and B? [U] [CBSE Delhi Set 1, 2019]



Ans. (a)

m =z I t 56 ´ 2 ´ t  2.8 = 2 ´ 96500 t = 4825 s / 80.417 min



m1 E 1 =  m2 E2

E0m . The plot of molar conductivity of weak

electrolyte tends to infinity as its concentration approaches zero; it does not intersect the molar conductivity axis. [1]

[½] [½]

Q.2. (a)  The conductivity of 0.001 mol L–1 acetic acid is 4.95 × 10–5 S cm–1. Calculate o

Ùm the dissociation constant if for acetic acid is 390.5 S cm2 mol–1. (b) Write Nernst equation for the reaction at 25°C :

[½]

2.8 56 2 = ´ mZn 2 65.3 

[½]



2 Al ( S ) + 3Cu 2 + ( aq ) ¾¾ ® 2 Al3 + ( aq ) + 3Cu ( s )

[1]

(c) What are secondary batteries? Give an example.  [5]

(b) (i) A- Strong electrolyte, B-weak electrolyte [1] (ii) Ù o m for weak electrolytes cannot be obtained by extrapolation while Ù o m for strong electrolytes can be obtained as intercept.[1]  [CBSE Marking Scheme, 2019] Detailed Answer :

96500 C = 9650 C 0.05 ´  2 = 0.1 F = 0.1 F ´ F The quantity of charge is related to current as

Q = It Therefore, the time needed to deposit 2.8 g Fe is :





t=

Q 9650 C = = 4825 s I 2 A

So, the current flowed through the cells for 4825 seconds. The amount of Zn deposited in cell Y can be calculated using Faraday’s second law : mass of Zn Eq.wt of Zn = mass of  Fe Eq wt of Fe molar mass of Zn / charge on zinc ion = molar mass of  Fe / charge on iron ion mass of Zn = 2.8 g ×



Ans. (a) Λ m =

65.3 g / 2 = 3.265 g ≈ 3.3g 56 g / 2

Therefore, the mass of Zn deposited in cell Y in the same time is 3.3 g.[3] (b) (i) Molar conductivity of strong electrolytes increases linearly as the square root of the concentration decreases; therefore, electrolyte A is a strong electrolyte. Molar conductivity of weak electrolytes increases non-linearly as square root of concentration decreases; therefore, electrolyte B is a weak electrolyte. [1]

k 4.95 × 10 −5 S cm −1 1000 cm 3 = × 0.001 mol L−1 L c = 49.5 S cm 2 mol −1

(a) Charge required to deposite 2.8 g Fe:

mass 2.8 g mol Fe =   = = 0.05 mol molar mass 56 g.mol -1 2 F charge is required to discharge 1 mol of Fe2+ ions as Fe, therefore deposition of 0.05 mol Fe will need







mZn=3.265

α=

49.5 S cm 2 mol −1 Λm = = 0.126  Λ m 390.5 S cm 2 mol −1

K=

c α2 0.001mol L−1 × ( 0.126 )2 = 1 − 0.126 (1 − α ) = 1.8 × 10 −5 mol L−1



(If K= cα2 , then K= 1.6 × 10−5 mol L–1)

0.059 [ Al 3 + ]2 Θ (b) E [1] log (cell) = E ( cell ) − 6 [cu 2 + ]3 (c) Batteries which are rechargeable Example- Lead storage, Ni-Cd batteries (Or any other one example ) [½ + ½]  [CBSE Marking Scheme 2019] Commonly Made Error  Some student do not write correct Nernst equation for the given cell reaction. Answering Tip  Do practice to write Nernst equation for the cell reaction. Q.3. (a) For the reaction + 2AgCl (s) + H2 (g) (1 atm) 2Ag(s)+2H (0.1 M)+2Cl (0.1 M), ΔG°= – 43600 J at 25°C. Calculate the e.m.f. of the cell. –n [log 10 = –n] (b) Define fuel cell and write its two advantages. Ans. (a) ΔGo = – nFEo[½] –43600 = – 2 × 96500 ×Eo Eo = 0.226 V  E = Eo – 0.059/2 log ([H+]2 [Cl–]2 / [H2])[½] = 0.226 – 0.059/2 log[ (0.1)2 × (0.1)2 ] / 1[½] = 0.226 – 0.059 /2 log 10-4[½]

No. of electrons (n) = 2 F = 96500 C  DGo = -nFEo ELECTROCHEMISTRY



= 0.226 + 0.118 = 0.344 V (Deduct half mark if unit is wrong or not written)[1] (b) Cells that convert the energy of combustion of fuels (like hydrogen, methane, methanol etc.) directly into electrical energy are called fuel cells.[1] Advantages : High efficiency, non polluting (or any other suitable advantage)[½+½] [CBSE Marking Scheme, 2018] Detailed Answer (a)



DGo = -43600 J No. of electrons (n) = 2 F = 96500 C  DGo = -nFEo -43600 = -2 ´ 96500 ´ Eo

-43600 \ Eo = = 0.226V -2 ´ 96500 For the reaction, 2AgCl(s) + H 2 (g) (1 atm)

-43600 = -2 ´ 96500 ´ Eo -43600 \ Eo = = 0.226V -2 ´ 96500 For the reaction, 2AgCl(s) + H 2 (g) (1 atm) ® 2Ag(s) + 2H+ (0.1M) + 2Cl - (0.1M) From Nernst equation E=Eo -

0.059 [H + ]2 [Cl - ]2 log 2 [H 2 ] (  Concentration of solids are taken as unity)

0.059 (0.1)2 (0.1)2 log 2 1 0.059 4 = 0.226 log10 2 [3] = 0.226 + 0.118 = 0.344V = 0.226 -

(b) Fuel Cell : The cell which converts chemical energy of a fuel directly into electrical energy is called fuel cell. [1] Adavntages of fuel cell : • Pollution free working • High efficiency [1]

® 2Ag(s) + 2H+ (0.1M) + 2Cl - (0.1M) From Nernst equation

Visual Case Based Questions

0.059 [H + ]2 [Cl - ]2 o - the log Q. E=E 1. Read passage given below and answer the 2 [H ] following questions2 : (1 × 4 = 4) ( Concentration of solids are determined taken as unity)  The cell constant is usually by 2 measuring resistance 0.059 the(0.1) (0.1)2 of the cell containing a = 0.226 log solution is already known. 2 whose conductivity 1 For this purpose, we generally use KCl solutions 0.059 = 0.226 - conductivity log10 4 whose is known accurately at various 2 concentrations and at different temperatures. = 0.226 + 0.118 = 0.344V Consider the resistance of a conductivity cell filled with 0.1 M KCl solution is 200 Ohm. If the resistance of the same cell when filled with 0.02 M KCl solution is 420 Ohm. (Conductivity of 0.1 M KCl solution is 1.29 S m–1.) The following questions are Multiple Choice Questions. Choose the most appropriate answer: (i) What is the conductivity of 0.02 M KCl solution ? (a) 0.452 S m–1 (b) 0.215 S m–1 (c) 0.614 S m–1 (d) 0.433 S m–1 (ii) What will happen to the conductivity of the cell with the dilution ? (a) First decreases then increases (b) Increases (c) First increases then decreases (d) Decreases (iii) The cell constant of a conductivity cell ________. (a) Changes with change of electrolyte. (b) Changes with change of concentration of electrolyte. (c) Changes with temperature of electrolyte. (d) Remains constant for a cell. (iv) SI unit for conductivity of a solution is (a) S m–1 (b) S m2 mol–1

[ 93

(4 marks each)

(c) mol cm–3 (d) S cm2 mol–1 OR Which of the following is not true? The conductivity of solutions of different electrolytes in the same solvent and at a given temperature differs due to (a) size of the ions in which they dissociate (b) concentration of ions (c) charge of the ions in which they dissociate (d) is independent of ions movement under a potential gradient Ans. (i) Correct option : (c)

 xplanation : Conductivity of 0.02 mol L–1 KCl E Solution= Cell constant resistance

258 420 = 0.614 Sm −1 =

(ii) Correct option : (d) Explanation : The conductivity decreases with dilution. (iii) Correct option : (d) Explanation : The cell constant of a conductivity cell remains constant for a cell. (iv) Correct option: (a) Explanation: SI unit for conductivity of a solution is S m–1. OR Correct option: (d) Explanation: The conductivity of solutions of different electrolytes in the same solvent and at a

94 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

given temperature differs due to size and charge of the ions in which they dissociate, concentration of ions, ease with which the ions move under a potential gradient. Q.2.  Read the passage given below and answer the following questions : (1 × 4 = 4) A galvanic cell consists of a metallic zinc plate immersed in 0.1 M Zn(NO3)2 solution and metallic plate of lead in 0.02 M Pb(NO3)2 solution. The following questions are multiple choice questions. Choose the most appropriate answer: (i) How will the cell be represented ? (a) Zn(s)| Zn2+(aq)|| Pb2+(aq)|Pb(s) (b) Zn2+(s)| Zn(aq)|| Pb2+(aq)|Pb(s) (c) Pb2+(aq)|Pb(s)||Zn2+(s)| Zn(aq) (d) Pb(s)|Pb2+(aq)||Zn2+(s)| Zn(aq) (ii) Calculate the emf of the cell. (a) 6.01 V (b) 0.412 V (c) 0.609 V (d) 4.12 V (iii) What product is obtained at cathode ? (a) Zn (b) Pb (c) Zn2+ (d) Pb2+ (iv) Which of the following statement is not correct about an inert electrode in a cell ? (a) It does not participate in the cell reaction. (b) It provides surface either for oxidation or for reduction reaction. (c)  It provides surface for conduction of electrons. (d) It provides surface for redox reaction. Ans. (i) Correct option : (a) Cell representation :



Zn(s)|Zn2+(aq)||Pb2+(aq)|Pb(s) (ii) Correct option : (c) According to Nernst equation :  Zn 2+  0.0591   Ecell = E°cell − log  2+   2 Pb   

[1]

[½]

Ecell = [– 0.13 – (– 0.76)] -



= 0.63 – 0.02955 × log 5 = 0.63 – 0.02955 × 0.6990 = 0.63 – 0.0206 = 0.6094 V

 The cell representation is given incorrectly by many

candidates.  The calculation of emf of the cell by using Nernst equation is incorrect, in some cases.

Answering Tip  Do more practice of cell representation and numerical based on Nernst equation. (iii) Correct option : (b) Anode reaction : Zn(s) → Zn2+(aq) + 2e– Cathode reaction : Pb2+(aq) + 2e– → Pb(s)



In a galvanic cell, the following cell reaction occurs : Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s) E°cell = +1.56 V The following questions are multiple choice questions. Choose the most appropriate answer : (i) What is the direction of the flow of electrons ? (a) First from silver to zinc, then the direction reverses (b) Silver to zinc (c) First from zinc to silver, then the direction reverses (d) Zinc to silver (ii) How will concentration of Zn2+ ions and Ag+ ions be affected when the cell functions ?



Commonly Made Errors



The potential difference between the two electrodes of a galvanic cell is called the cell potential and is measured in volts. The cell potential is the difference between the electrode potentials (reduction potentials) of the cathode and anode. It is called the cell electromotive force (emf) of the cell when no current is drawn through the cell. It is now an accepted convention that we keep the anode on the left and the cathode on the right while representing the galvanic cell. A galvanic cell is generally represented by putting a vertical line between metal and electrolyte solution and putting a double vertical line between the two electrolytes connected by a salt bridge.



[½]

[1]

(iv) Correct option : (a) Explanation : Inert electrode does not participate in redox reaction and acts only as source or sink for electrons. It provides surface either for oxidation or for reduction reaction. [1]

Q.3.  Read the passage given below and answer the following questions : (1 x 4 = 4)



0.1 0.0591 log 2 0.02







(a) Concentration of both Zn2+ and Ag+ ions increase (b) Concentration of Zn2+ increases and Ag+ ions decreases (c) Concentration of Zn2+ decreases and Ag+ ions increases (d) Concentration of both Zn2+ and Ag+ ions decreases (iii) Name the cell which is generally used in inverters ?



(a)  Mercury cell (b)  Leclanche cell (c)  Lead storage battery (d)  Lithium ion battery OR Which cell uses a 38% solution of sulphuric acid as an electrolyte ? (a)  Lead storage cell (b)  Leclanche cell (c)  Lithium ion battery (d)  Fuel cell

[ 95

ELECTROCHEMISTRY







(iv)  The standard electrode potential for Daniell cell is 1.1V. Calculate the standard Gibbs energy for the reaction : Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s). (a)  215.36 kJ mol–1 (b)  –212. 27 kJ mol–1 (c)  212.27 kJ mol–1 (d)  –218 kJ mol–1

ns. (i) Correct option : (d) A Zinc to silver is the flow of electrons.[1] (ii) Correct option : (b) Concentration of Zn2+ increases and Ag+ ions decreases[1] (iii) Correct option : (c) Lead storage battery is used in inverters. [1] OR Correct option : (a) Explanation : Lead storage battery is the most common secondary cell. It consists of a lead anode and a grid of lead packed with lead oxide (PbO2) as cathode. A 38% solution of sulphuric acid is used as electrolyte.[1] (iv) Correct option : (b) o ∆rGo = –nFEcell n in the above equation is 2, o F = 96487 C mol–1 and Ecell = 1.1 V o Therefore, ∆rG = –2 × 1.1V × 96487 C mol–1 r = – 21227 J mol–1 = – 212.27 kJ mol–1 Q.4.  Products of electrolysis depend on the nature of material being electrolysed and the type of electrodes being used. If the electrode is inert (e.g., platinum or gold), it does not participate in the chemical reaction and acts only as source or sink for electrons. On the other hand, if the electrode is reactive, it participates in the electrode reaction. Thus, the products of electrolysis may be different for reactive and inert electrodes. Aqueous copper sulphate solution and aqueous silver nitrate solution are electrolysed by 1 ampere current for 10 minutes in separate electrolytic cells. In these questions, a statement of assertion followed by a statement of reason. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion.



(b)  Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. (i)  Assertion : The mass of copper and silver, deposited on the cathode be same. Reason : Copper and silver have different equivalent masses. (ii)  Assertion : At equilibrium condition Ecell = 0 or ΔrG = 0. Reason : Ecell is zero when both electrodes of the cell are of the same metal. (iii)  Assertion : The negative sign in the expression EZn2+/Zn = – 0.76V means Zn2+ cannot be oxidised to Zn. Reason : (iv)  Assertion : In a galvanic cell, chemical energy is converted into electrical energy. Reason : Redox reactions provide the chemical energy to the cell. OR Assertion : Copper sulphate cannot be stored in zinc vessel. Reason : Zinc is less reactive than copper. Ans. (i) Correct option : (d) Explanation : W = itE/96500 = 1 × 10 × 60 × 31.8/96500 for copper. It will be different for silver since the equivalent weight of silver is different.[1] (ii) Correct option : (b)  At equilibrium, condition of Ecell = 0, ∆rG = 0 [1] (iii) Correct option : (a) It shows that the reduced form of (Zn) is not stable. It is difficult to reduce Zn2+ to Zn. Rather the reverse reaction i.e Zn can get oxidised to Zn2+ and H+ will get reduced as it is stabler among both the reduced species.[1] (iv) Correct option : (a) Explanation: The redox reactions provide the chemical energy to the galvanic cell which is converted into electrical energy. [1] OR Correct option: (c) Explanation: Copper sulphate cannot be stored in zinc vessel as zinc is more reactive than copper. [1] ll

96 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Self Assessment Test-3 Time : 1 Hour

Max. Marks : 25

1.  Read the passage given below and answer the following questions : (1 × 4 = 4) Products of electrolysis depend on the nature of material being electrolysed and the type of electrodes being used. If the electrode is inert (e.g., Platinum or gold), it does not participate in the chemical reaction and acts only as source or sink for electrons. On the other hand, if the electrode is reactive, it participates in the electrode reaction. Thus, the products of electrolysis may be different for reactive and inert electrodes. The products of electrolysis depend on the different oxidising and reducing species present in the electrolytic cell and their standard electrode potentials. The following questions are Multiple Choice Questions. Choose the most appropriate answer: (i) On the basis of the standard electrode potential values stated for acid solutions, predict whether Ti4+ species may be used to oxidise Fe(II) to Fe(III)

Ti4+ + e– → Ti3+ Fe3+ + e– → Fe2+

E° = +0.01V  E° = +0.77V

(a) Ti4+ can oxidise Fe(II) to Fe(III)

(b) Ti4+ cannot oxidise Fe(II) to Fe(III)



(c) Ti4+ can oxidise Fe(III) to Fe(II)



(d) Ti4+ can reduce Fe(II) to Fe(III)



(ii) Based on the data, arrange Fe2+, Mn2+ and Cr2+ in the increasing order of stability of +2 oxidation state. (Give a brief reason)

E°Cr3+/Cr2+ = –0.4V E°Mn3+/Mn2+ = +1.5V

E°Fe3+/Fe2+ = +0.8V (a) Fe2+ < Mn2+ < Cr2+

(b) Mn2+ < Cr2+ < Fe2+



(c) Cr2+ < Fe2+ < Mn2+



(d) Mn2+ < Cr2+ < Fe2+



(iii) Name the cell which is generally used in hearing aids.



(a) Daniell cell



(b) Mercury cell



(c) Lead storage battery



(d) Lithium cell



(iv) Which of the following statement is not true.



(a) Graphite is used as an inert electrode.



(b) An active electrode may affect the outcome of the electrode reaction in the cell.



(c) Products of electrolysis is independent of the electrode potential of the species participating.



(d) The electrolysis products are dependent on electrolytic solution and electrodes.

OR Which of the following statement is not correct about an inert electrode in a cell? (a) It does not participate in the cell reaction (b) It provides surface either for oxidation or for reduction reaction (c) It provides surface for conduction of electrons (d) It provides surface for redox reaction The following questions (No. 2 to 5) are Multiple Choice Questions carrying 1 mark each. 2. Galvanised iron is formed by coating of zinc on iron, while its reverse is not possible. Its reason is: (a) negative electrode potential of zinc is more than iron. (b) Zinc is lighter than iron. (c) the melting point of zinc is lesser than iron. (d) negative electrode potential of zinc is lesser than iron. 3. A device which converts chemical energy of fuel like hydrogen and methane into electric energy directly is called directly is called: (a) electrolytic cell (b) dynamo (c) Ni-Cd cell (d) Fuel cell R 4. Standard reduction potentials of half-cell reaction are given below: F2(g) + 2e– → 2F-(aq); Eo = +2.85V Cl2(g) + 2e– → 2Cl–(aq); Eo = +1.36V Br2(g) + 2e– → 2Br–(aq); Eo = +1.06V I2(g) + 2e– → 2I–(aq); Eo = +0.53V Statement oxidant and reductant are respectively (a) F, I (b) Cl, F (c) Br, I (d) F, Br U 5.  The conductivity of aqueous solution of an electrolyte depends on (a) charge present on ions (b) movement of ions (c) number of ions (d) all of the above R In the following questions (No. 6 & 7), a statement of assertion is followed by a statement of Reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not the correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. 6.  Assertion : Zinc displaces copper from copper sulphate solution. Reason : Zinc is more reactive than copper. 7. Assertion : Following reaction is possible 2Ag + H2SO4 → Ag2SO4 + H2 Reason : The tendency to oxidize Ag in less than Hz.

[ 97

SELF-ASSESSMENT TEST

The following questions (No. 8 & 9), are Short Answer Type-I and carry 2 marks each. 8.  Why on dilution the Lm of CH3COOH increases drastically, while that of CH3COONa increases gradually? A&E 9.  (i) Calculate ΔrG° for the reaction   Mg(s) + Cu2+(aq) → Mg2+(aq) + Cu(s) 0         Given : Ecell = + 2.71 V, 1 F = 96500 C mol–1 (ii) Name the type of cell which was used in Apollo space programme for providing electric power.  + R Q.No. 10 & 11 are Short Answer Type-II carrying 3 marks each. 10.  What is corrosion? Explain the electrochemical theory of rusting of iron and write the reactions involved in rusting of iron. 11. (i) The cell in which following reaction occurs : 2 Fe3+(aq) + 2I(aq) → 2Fe2+(aq) + I2(s) 0 has Ecell = 0.236 V at 298 K. Calculate the standard gibbs energy of the cell reaction. (Given 1F = 96,500 C mol–1) (ii) How many electrons flow through a metallic wire if a current of 0.5 A passed for 2 hours. (Given 1F = 96,500 C mol–1)

Q.No 12 is a Long Answer Type Question carrying 5 marks each. Q.12. (i) Define the following terms :

  (a) Molar conductivity (Λm)



  (b) Secondary batteries



  (c) fuel cell



 (ii) State the following laws :



  (a ) Faraday first law of electrolysis



  (b) Kohlrausch’s law of independent migration of ions R OR



(i) Define the term degree of dissociation. Write an expression that relates the molar conductivity of a weak electrolyte to its degree of dissociation.

(ii) For the cell reaction

   Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s)



   Calculate the equilibrium constant at 25°C.



   How much mnimum work would be obtained by operation of the cell?



EoNi 2 / Ni  0.25 V and EoAg  / Ag  0.80 V 

R+ U

 

Finished Solving the Paper ? Time to evaluate yourself !

OR SCAN THE CODE

SCAN

For elaborate Solutions

ll

Effective Collision

Rate = PZAB e–Ea/RT P is Steric or Probability factor

[ [

Unit (k) mol L-1 s-1 s-1

[R] vs t

In[R] vs t

kt =[R]0-[R]

kt =In {[R]-0/[R]}

First Level

Second Level

Trace the Mind Map Third Level

reaction depends on the collision frequency and effective collisions.

0.693 t1/2 = k

First Order

[R]0 t1/2 = 2k

Zero Order

Time in which the concentration of a reactant is reduced to one half of its initial concentration.

Half-life of A Reaction

Pseudo First Order Reaction

Chemical Kinetics

2

1

0

Order

Rate of Chemical Reaction

A

nf l ea uen c ct ion ing

–1

Unit

mol–1L s–1

s–1

mol L s –1

=

Rate of disappearance of R Decease in concentration of R – [R] = Time taken t Rate of appearance of P Increase in concentration of P + [P] = Time taken t

=

Rate of change in concentration of reactant/product at a particular time rinst= –d[R] =+d[P] dt dt

Concentration : Higher the concentration of reactants, faster is the rate of reaction. Temperature : Increases with increase in temperature. Becomes almost double with 10°C rise. Presence of Catalyst : Increases with a catalyst. Surface Area : Greater is the surface area, faster is the rate of reaction. Activation Energy : Lower the activation energy, faster is the reaction.

Change in concentration of reactants or products in unit time; Unit : mol L-1 s-1 or atm s-1

Expression in which reaction rate in given in terms of molar concentration of reactants with each term raised to power which may or may not be same as stoichiometric coefficient of reactants in a balanced chemical equation. aA + bB  cC + dD d [R] a Rate = k [A] [B]b = dt c d [C] [D] k= [A]a–[B]b

Order of a Reaction

Molecularity of a Reaction

Appearance of products or disappearance of reactants over a long time interval. d [P] rav = dt = slope –d [R] = –slope rav = dt

Sum of powers of concentration of the reactants in the rate law.

Number of reacting species taking part in an elementary reaction colliding to bring out a reaction.

Collision Theory: Rate of

Integrated Rate Equation

Straight Plot

Integrated Rate Law

Number of collisions per unit volume of reaction mixture Rate = ZAB e–Ea/RT

u eq Fr n o lisi Col

Arrhenius Equation

Not truly of first order but under certain conditions behave, as first order reaction. In such reaction, one reactants is in excess.  Acid hydrolysis of ethyl acetate  Inversion of sugar

1

0

Order

Integration of differential rate equation to give a relation between concentrations at different times and rate constant.

Ea 2.303RT Ea T2 –T1 log k2 = k1 2.303RT T1T2

k = Ae–Ea/RT log k = log A –

Activation Energy, Ea : Energy required to form an intermediate called activated complex (C) Ea= Threshold energy - Average Kinetic energy of reacting molecules

Collision in which molecules collide with sufficient kinetic energy and proper orientation for breaking of bonds and formation of new bonds.



en cy

Ea and proper orientation of the molecules determine the criteria for an effective collision.



ate

ve ra ge R I rs r cto of Fa ate r



98 ] Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

4

CHAPTER

Syllabus

CHEMICAL KINETICS

 Rate of a reaction (Average and instantaneous), factors affecting rate of reaction: concentration, temperature, catalyst; order and molecularity of a reaction, rate law and specific rate constant, integrated rate equations and half-life (only for zero and first order reactions), concept of collision theory (elementary idea, non mathematical treatment). Activation energy, Arrhenius equation.

Trend Analysis List of Concept names Rate of reaction, Rate law, Rate constant and Numericals

2018 D/OD 1Q (2 marks)

2019 D 1Q (2 marks)

OD -

Order of reaction, Integrated rate equation, Half Life and Numericals

-

-

1Q (2 marks) 1Q (3 marks)

Collision theory, Arrhenius equation and Numericals

1Q (3 marks)

-

-

TOPIC-1

Rate of a Chemical Reaction and Factors Affecting Rate of Reactions

 Chemical Kinetics : It is the branch of physical chemistry which deals with study of the rate of chemical reaction and the mechanism by which the reaction occurs.  Rate of Reaction : The rate of reaction is the change of concentration of any reactant or product with time, for a reaction. A+B→C Decrease in concentration of A −∆[ A ] Rate of reaction, A = = Time taken ∆t

TOPIC - 1 Rate of a Chemical Reaction and Factors Affecting Rate of Reactions .... P. 99 TOPIC - 2 Order of a Reaction, Integrated Rate Equations and Half-life of a Reaction

Revision Notes

Similarly,

2020 D OD 1Q (1 mark) 2Q (1 mark) 3Q 1Q (1 mark) (5 marks) 1Q 1Q (1 mark) (5 marks) 1Q (3 marks)

B =

.... P. 106 TOPIC - 3 Concept of Collision Theory, Activation Energy and Arhenius equation .... P. 117

−∆[ B ] ∆t

D [C ] Dt where, [A], [B] and [C] are molar concentrations of the reactants and the product respectively.  Unit of rate of reaction : mol L–1 s–1 or mol L–1 min–1 (in liquid), atm s–1 or atm min–1 (in gaseous form)  Instantaneous rate of reaction : Instantaneous rate is defined as the rate of change in concentration of any one of the reactant or product at a particular time. and for product

C =

100 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII





-d[ B] -d[ A] dx +d[C ] = = = dt dt dt dt  Average rate of reaction : The rate of reaction measured over a long time interval is called

Instantaneous rate =

average rate of a reaction. Average rate =

Scan to know more about this topic

Dx , where, Dx = change in concentration in given Dt

time and Dt = time taken.

Rate of reaction

Concentration (moles/litre)

 Factors affecting the rate of a chemical reaction :  (i) Concentration of reactants : Rate of reaction is directly proportional to the concentration of the reactants. Thus, to increase the rate of a reaction the concentration of the reactants has to be increased. (ii) Temperature : The rate of a reaction increases with the increase in temperature. Increase in temperature increases the kinetic energy of the molecules which results in the increase in rate of reaction. (iii) Pressure : Pressure affects the rate of only gaseous reactions. Increase in pressure decreases volume and increases concentration. Increase in concentration increases the rate of reaction.  (iv) Presence of catalyst : The rate of many reactions is greatly affected by the presence of a catalyst. In the presence of a catalyst, the activation energy of a reaction decreases due to which the reaction proceeds at a faster rate.  (v) Nature of the reactants : In a chemical reaction, some bonds are broken while some new bonds are formed. Thus, if the molecules are simpler, then less bonds will rupture and the rate of reaction becomes fast while in complex molecules, more bonds will rupture and consequently the rate of reaction decreases. (vi) Surface area of the reactants : In some heterogeneous reactions, the reaction takes place at the surface of the reactant. Thus in such reactions, the reaction rate is greatly affected by the surface area. Marble powder reacts faster than marble chips. (vii) Effect of radiations : The reactions which are initiated by the radiations of particular wavelengths are termed as photochemical reactions. These reactions generally proceed at a faster rate than normal thermal reactions. Product

Reactant

Time ( t )







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Factors affecting Rate of a reaction

(viii) Effect of physical state : Rate of reaction depends upon physical state of the reactant, e.g., I2(g) reacts faster than I2(s). AgNO3(aq) reacts with NaCl but AgNO3(s) does not react with NaCl.  Rate Law : Rate law or rate equation is the expression which relates the rate of reaction with concentration of the reactants. The constant of proportionality ‘k’ is known as rate constant. The rate law states that the rate of reaction is directly proportional to the product of molar concentration of reactants and each concentration is raised to some power which may or may not be equal to stoichiometric coefficients of reacting species. Rate = k[A]m [B]n  Rate Constant : Rate constant is also called specific reaction rate. When concentration of both reactants are unity (one), then the rate of reaction is known as rate constant. It is denoted by ‘k’.  Molecularity : Total number of atoms, ions or molecules of the reactants involved in the reaction is termed as molecularity. It is always a whole number. It is never more than three. It cannot be zero. Example : NH4NO2 → N2+2H2O (Unimolecular reaction) 2HI → H2 + I2 (Bimolecular reaction)  2NO+ O2 → 2NO2 (Trimolecular reaction)  Elementary Reaction : An elementary reaction is a chemical reaction in which one or more of the chemical species react directly to form products in a single reaction step and with a single transition state.  For a complex reaction generally, molecularity of the slowest step is same as the order of the overall reaction.  Initial rate of reaction : The rate at the beginning of the reaction when the concentrations have not changed appreciably is called initial rate of reaction.

 Rate Determining Step : The slowest step in the reaction mechanism is called rate determining step.

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CHEMICAL KINETICS

Objective Type Questions

(1 mark each) 0.2 ( 0.6 ) = 0.1 ( 0.3)y

[A] MULTIPLE CHOICE QUESTIONS :

y

r = k[A]x [B]y dx = k[A]x [B]y dt mol L-1s -1 = k (mol L-1 )x (mol L-1 )y





∴ x=1 Rate law Rate = k[A][B]2 Q. 3. Consider the reaction A  B. The concentration of both the reactants and the products varies exponentially with time. Which of the following figures correctly describes the change in concentration of reactants and products with time?

(a)

mol L-1s -1 k= (mol L-1 )x (mol L-1 )y

Initial Initial Initial Expericoncentration concentration concentration of ment of [A]/mol L−1 of [B]/mol L−1 [C]/mol L−1 1.

0.30

0.30

0.10

2.

0.30

0.60

0.40

3.

0.60

0.30

0.20

(c)

(d)

(a) Rate = k [A] [B] (b) Rate = k [A][B] (c) Rate = k [A][B] (d) Rate = k [A]2[B]0 Ans. Correct option : (b) Explanation : Suppose order with respect to A and B are x and y respectively. Rate = k[A]x[B]y For experiment 1, 0.1 = k(0.3)x(0.3)y…(i) For experiment 2, 0.4 = k(0.3)x(0.6)y…(ii) For experiment 3, 0.2 = k(0.6)x(0.3)y…(iii) Dividing equation (ii) by (i) y 0.6)y 0.4 ( 0.6 0.4 == y 0.1 (0.3 0.1 0.3)y ∴ yy == 22 Dividing equation (iii) by (i)



A&E

[B]

[A] Time →

[B]

[A] Time →

2

Concentration →

(b)

Concentration →

Concentration of either ‘A’ or ‘B’ were changed keeping the concentration of one of the reactants constant and rates were measured as function of initial concentration. Following results were obtained. Choose the Correct option for this reaction. A&E

2

[A]

Concentration Concentration → →

Where (x + y) = order of the reaction Q. 2. Compounds ‘A’ and ‘B’ react according to the following chemical equation  : A(g) + 2B(g) → 2C(g)

[B]

Time →

= (mol L-1 )1 - ( x + y ) s-1



Concentration →

Q. 1. The unit of rate constant depends upon the (a) molecularity of the reaction (b) activation energy of the reaction (c) order of the reaction (d) temperature of the reaction  U [CBSE, Delhi Set-3, 2020] Ans. Correct option : (c) Explanation : For the reaction, xA + yB ® Product

[B]

[A]

[B] [A] Time →

Ans. Correct option : (b) Explanation : As the reactant A’s concentration decreases with time, so the product B’s concentration increases. Also since the reaction is reversible, the increase and decrease in concentration with respect to time is similar. [B] ASSERTION AND REASON TYPE QUESTIONS : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct state­ ments and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement.

102 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Rate1 = k[A 1 ] [A 2 ] = [2A 1 ] Rate 2 = k[2A 1 ] Rate 2 = k ´ 2 Rate1



Ans. The molecularity of a reaction is the number of total molecules taking part in elementary step of reaction. So, minimum one molecule is required for a reaction to occur. Hence, the value of molecularity is never zero. Q. 2. Write the formula for expressing rate of the reaction. N2 + 3H2 → 2NH3 U Ans. N 2 + 3H 2 ® 2NH 3

Q. 1. Assertion : Rate of reaction doubles when concentration of reactant is doubled if it is a first order reaction. Reason : Rate constant also doubles. Ans. Correct option : (c) Explanation : For first order reaction

(Reactants)

Rate of reaction =

For a given reaction, rate constant is constant and independent of the concentration of reactant.



Q. 2. Assertion : The rate of reaction increases with increase in temperature. Reason : The reactant molecules collide less frequently. Ans. Correct option : (c) Explanation : As the temperature of a reaction is increased, the rate of the reaction increases because the reactant molecules collide more frequently and with greater energy per collision.



Q. 3. Assertion : Dust particles suspended in the air inside unheated gain electrons can sometimes react explosively. Reason : The dust particles have large surface area for the reaction. Ans. Correct option : (a)

(Product)

Explanation : Since the dust particles have large surface area for the reaction so these particles suspended in the air inside unheated gain electrons can sometimes react explosively.

[C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1. Why the molecularity of a reaction can not be zero?

=

- d[N 2 ] 1 æ - d[H 2 ] ö = ç ÷ dt 3 è dt ø 1 æ d[NH 3 ] ö ç ÷ 2 è dt ø

Q. 3. The rate of reaction decreases with the progress of reaction. Why? U Ans. The rate of reaction depends on the concentration of reactants. Since, the concentration of reactants decreases with time, so rate of reaction also decreases. Q. 4. Reactions having molecularity more than three occur rarely. Why? R Ans.  It is because of the fact that collision of more than three molecules is not possible at a time. Q. 5. If the rate equation is given below :

Rate = k[A]2[B]



then what will be the unit of its rate and rate constant?

Ans. Unit of rate = mol L–1s–1

[½] Unit of rate Unit of rate constant (k) = Unit of [A 2 ] ´ Unit of [B]



=

= mol -2 L2 s -1







−

I 2H 2O 2 alkaline  → 2H 2O + O 2 medium







The proposed mechanism is given below :





(1) H 2O 2 + I - ® H 2O + IO -  slow ( )







 (i) Write rate law of the reaction.





 (ii) Write overall order of reaction.





(iii) O  ut of step (1) and (2), which one is rate determining step? U [CBSE Delhi Set-1, 2019]



Ans.  (i) Rate = k [H2O2] [ I¯] (ii) Order = 2 

[½]

(2 marks each)

(iii) Step 1 [½] [CBSE Marking Scheme, 2019]

Detailed Answer :  (i) The rate law for the reaction is

(2) H 2O 2 + IO - ® H 2O + I - + O 2  fast ( )









Short Answer Type Questions-I Q. 1. For a reaction,

mol L-1s -1 (mol L-1 )2 (mol L-1 )

[1] [½]





rate = −

d [ H 2O2 ] dt

= k [ H 2O2 ]  l − 

(ii) This reaction is first order with respect to both H2O2 and I–. The overall order of the reaction is bimolecular, 2. The order of the reaction is determined from the slowest step of the reaction ­ mechanism. (iii) The first reaction is slow, so this is the rate determining step. [2]

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CHEMICAL KINETICS

OR



[Topper’s Answer 2019] [2]



Commonly Made Error

Answering Tip

 Some students are not able to write correct expression for the rate of the reaction.

 Learn and understand the rate law.

Q. 2. For the reaction 2N2O5 (g) 4NO2 (g) + O2 (g), – – the rate of formation of NO2 (g) is 2.8 × 10 3 M s 1. Calculate the rate of disappearance of N2O5 (g). [CBSE, Delhi/Outside Delhi Set, 2018] U,



Ans.



Rate =

1 ∆( NO2 ) −1 ∆( N 2O5 ) = ½ ∆t ∆t 4 2

1 −1 ∆( N 2O5 ) ( 2.8 × 10−3 ) = 4 2 ∆t

 − ∆( N 2O5 )   Rate of disappearance of N2O5    ∆t





= 1.4×10–3 M/s 1 (Deduct half mark if unit is wrong or not written) [CBSE Marking Scheme 2018]

½ OR



Ans.

[Topper’s Answer 2018]



Commonly Made Error



Students tend to forget specifically mentioning the formula and start the calculations or do not mention all the steps or specific units.

Answering Tip



Pay special attention to writing of units in the answer as missing out units in the last step leads to deduction of marks.

Q. 3. For a reaction the rate law expression is represented as follows: Rate = k [A][B]1/2 (i) Interpret whether the reaction is elementary or complex. Give reason to support your answer.

 (ii) Write the unit of rate constant for this reaction if concentration of A and B is expressed in moles/L.  U [CBSE SQP, 2020-21]

Ans.

(i) This is a complex reaction.

[½]

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



 rder of reaction is 1.5. Molecularity cannot O be 1.5, it has no meaning for this reaction. The reaction occurs in steps, so it is a complex reaction. [½] (ii) Rate = k[A][B][½] Unit of Rate constant (k) Unit of rate = Unit of [A] ´ Unit of [B]½ -1 -1

mol L s = -1 -1 ½ (mol L )(mol L ) 1 -1 [1] = mol 2 L 2 s1

Q.4. The following results have been obtained during the kinetic studies for the reaction : P + 2Q → R + 2S

Exp. 1 2 3 4

Initial P(mol/L) 0.10 0.30 0.10 0.20

Initial Q (mol/L) 0.10 0.30 0.30 0.40

Init. Rate of Formation of R (M min-1) 3.0 × 10−4 9.0 × 10−4 3.0 × 10−4 6.0 × 10−4

Determine the rate law expression for the reaction.  [CBSE SQP, 2020-21] Ans. Let the rate law expression be Rate = k [P]x [Q]y from the table we know that Rate 1 = 3.0 × 10–4 = k (0.10)x (0.10)y Rate 2 = 9.0 × 10–4 = k (0.30)x (0.30)y

Rate 3 = 3.0 × 10–4 = k (0.10)x (0.30)y

Rate 1/ Rate 3 = (1/3)y or 1 = (1/3)y [½]

So y = 0 Rate 2/ Rate 3 = (3)x or 3 = (3)y So x = 1

[½]

Rate = k [P]

[1]

hv Q.5. For a reaction : H2 + Cl2 → 2HCl



Rate = k



  (i)  Write the order and molecularity of this reaction.

(ii) Write the unit of k.  [CBSE, Outside Delhi Set 1, 2016] Ans. (i) Zero order reaction, Molecularity is 2 / bimolecular reaction. [[½] + [½]] –1 –1 (ii) mol L s . [1]  [CBSE Marking Scheme, 2016]

Commonly Made Error  Some students get confused to write correct unit of rate constant for the reaction.

Answering Tip  Understand the unit of rate constant for different orders of reaction.

Short Answer Type Questions-II



é 0.02 - 0.03 ù = -ê ú 25 ë û é -0.01 ù = -ê ú ë 25 û Q. 2.

= 4 ´ 10 -4 mol L-1 min -1 

4 ´ 10 -4 = mol L-1s -1 60 = 6.66 ´ 10 -6 mol L-1s -1 For the reaction 2A + B → A2B

Rate = k[A] [B]2, k = 2.0 × 10–6 mol–2L2s–1 Calculate the initial rate of the reaction when [A] = 0.1 mol L–1 and [B] = 0.2 mol L–1 . Calculate the rate of reaction after [A] is reduced to 0.06 mol L–1. Ans. 2A + B → A2B Initial rate = k[A] [B]2   = 2.0 × 10−6 × 0.1 × (0.2)2   = 8.0 × 10−9 mol L–1 [1] When concentration of [A] is reduced from 0.10 mol L–1 to 0.06 mol L–1 i.e. 0.04 mol L–1, concentration of A has been used in the reaction. Therefore, the concentration of B





For the reaction R → P, the concentration of a reactant changes from 0.03 to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds.  Ans. R→P  [R1] = 0.03 M  [R2] = 0.02 M  ∆t = 25 min D[R] Average rate = Dt [R ] - [R1 ] ù é = -ê 2 ú Dt ë û Q. 1.

(3 marks each)

1 ´ 0.04 2 = 0.02 mol L-1  =

[2]



[1]



[½]

Hence,   [B] = 0.2 – 0.02   

= 0.18 mol L–1

[½]

  Rate = k[A][B]2



  = 2.0 × 10−6 × 0.06 × (0.18)2



  = 3.89 × 10−9 mol L–1 s–1

[1]

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CHEMICAL KINETICS

Q.3.

The decomposition of NH3 on platinum surface is a zero order reaction. What will be rate of production of N2 and H2 when the value of k is



2.5 × 10−9 mol L–1 s–1? Ans.





Rate of production of H2



∵ Rate of reaction =



Rate of production of N2



∵ Rate of reaction = Rate of production of N2

 = 3 × 2.5 × 10−4



1 d[H 2 ] 3 dt

  = k = 2.5 ´ 10 -4 mol L-1s -1 

1 ´ Rate of production of H2 3 ∴ Rate of production of H2 = 3 ×Rate of reaction



1 æ d[NH 3 ] ö d[N 2 ] Rate of reaction = - ç ÷= 2 è dt ø dt

 = 7.5 × 10−4mol L–1 s–1 [1]



Commonly Made Error

∵ Reaction is zero order, so, the rate of reaction = k[NH 3 ]0



∴ Rate of production of N2 = 2.5 × 10−9 mol L–1s–1[1]



2NH 3 ® N 2 + 3H 2

=



 Some students find it difficult to calculate rate of production of the reaction products. [1]

Answering Tip  Students must be clear about the concept of rate of reaction.

Long Answer Type Questions Q. 1.











A reaction is of first order with respect to A and second order with respect to B (i) Write differential rate equation. (ii) How is the rate effected when the concentration of B is tripled? (iii) Howistherateaffectedwhentheconcentration of both A and B is doubled? U+ Ans. (i) Differential rate equation dx = k[A][B]2 [1] dt (ii) Let [A] = a, [B] = b If [B] increases three times  [B] = 3b ∴ Rate = k[A][B]2 Rate1 = k × a × b2…(i) Rate2 = k × a × (3b)2…(ii) From eq (i) and (ii) Rate2 k ´ a ´ (3b )2 = Rate1 k ´ a´ b      Rate2 = 9 ´ Rate1

(5 marks each)

∴ The rate becomes nine times when the concentration of B is tripled. [2]    (iii) If [A] and [B] is doubled then [A] = 2a [B] = 2b    Rate1 = k ´ a ´ b  …(i)



Q.2.

…(ii) Rate2 = k ´ (2 a) ´ (2 b) From eq. (i) & (ii) Rate 2 k ´ (2 a) ´ (2 b)2 = Rate1 k ´ a ´ b2 =8 Rate 2 = 8Rate1    ∴ The rate becomes eight times when the concentration of both A and B is doubled.[2] (a) Reaction 2A + B + C → D + 2E shows first order with respect to A, second order with











respect to B and zero order with respect to C. Determine : (i) Rate law of reaction     (ii)  What will be the rate of reaction on doubling of concentration of A, B and C? (b) For the reaction, 1 N2O5 ® 2NO 2 + O 2 2  he rate of dissociation of N2O5 is 5.65 × 10–5 T mol L–1 s–1. Determine : (i) Rate of reaction (ii) Rate of formation of NO2  (iii) Rate of formation of O2 U

Ans. (a) (i) 2A + B + C → D + 2E

Rate Law



Rate (r) = k[A]1 [B]2[C]0 [1]   (ii) If concentration of A, B & C is doubled



    r1 = k[A]1 [B]2[C]0…(i)



          r2 = k[2A]1 [2B]2[2C]0



From equation (i) & (ii)

r1 k[A]1[B]2 [C]0 1 = = r2 k[2A]1[2B]2 [2C]0 8

   ∴ r2 = 8 × r1 Hence rate of reaction becomes eight times.[1] 1 (b) (i) N 2O5 ® 2NO2 + O2 2



…(ii)

Rate of reaction =



−d[ N 2O5 ] 1  d[ NO2 ]  =   dt 2  dt 

 d[O2 ]  = 2  dt    Rate of reaction = Rate of dissociation of N2O5   = 5.65 × 10–5 mol L–1s–1[1]

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



of dissociation dissociation of of N N 2O O5  (ii) Rate Rateof 2 5 1 1 Rateof of formation formation of of NO NO2 == 2 ´´ Rate 2 2 Rate of dissociation of NO Rate of dissociation of NO22 Rateof of formation formation of of N N 2O O5 ==22 ´´ Rate 2 5 5.65 ´´10 10--55 ==22 ´´ 5.65 11.3 ´´10 10--55mol molLL--11ss--11 ==11.3



[1]



(iii) Rate of dissociation of N2O5 = 2 × Rate of formation of O2



Rate of formation of O2 1 = ´ Rate of dissociation of N 2O5 2 1 = ´ 5.65 ´ 10 -5 2  = 2.825 ´ 10 -5 mol L-1s -1

[1]

TOPIC-2

Order of a Reaction, Integrated Rate Equations and Half-life of a Reaction Revision Notes  Order of reaction : Order is defined as the sum of powers of concentration of the reactants in the experimentally derived rate equation or rate law expression. Order of reaction is experimentally determined and is not written from the balanced chemical equation. Order of reaction can be whole number, zero or fractional.  Zero order reaction : The rate of reaction does not change with the concentration of the reactants. i.e. Rate = k [A]0,

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Kinetics- Initial rates and Integrated Rate laws

[ A]0 - [ A] , where ‘k’ is rate constant, [A] is initial concentration of reactant. k= 0 t   Unit of the rate constant k is mol L–1 s–1. This reaction will be zero order reaction. Decomposition of gaseous ammonia on hot platinum, thermal decomposition of HI on gold surface and photochemical reaction between hydrogen and chlorine are examples of zero order reaction.  First order reaction : The rate of reaction is directly proportional to the first power of the concentration of reacting substance. Rate constant of the first order reaction is 2.303 a log k = t a x



k = 2.303 log [A 0 ] ,

⇒

t



[ A]



where ‘a’ is initial concentration and (a – x) is the concentration after time ‘t’. Unit of ‘k’ is s–1 or min–1. Decomposition of N2O5 and N2O are examples.  Pseudo first order reaction : If a reaction is not truly of the first order but under certain conditions becomes reaction of first order is called pseudo first order reaction, e.g., acidic hydrolysis of ester (ethyl acetate). H+   CH3COOC2H5 + H2O    CH3COOH + C2H5OH

 Second order reaction : The reaction in which sum of powers of concentration terms in rate law or rate equation is equal to 2. dx \ = k[A] [B] dt

Unit of rate constant is mol–1 L s–1 or M–1 s–1, where M is molarity. Reaction

Order

Unit of rate constant

Zero order

0

mol–1 L–1 s–1

First order

1

s–1

Pseudo first order

1

s–1

Second order

2

mol–1 L s–1

Example Sunlight

H2 + Cl2 → 2HCl 2N2O5 → 4NO2 + O2 H+

→ C6H12O6 + C6H12O6 C12H22O11 + H2O  H2 + I2 → 2HI

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CHEMICAL KINETICS

 Equation for typical first order gas phase reaction : A(g) → B(g) + C(g) p 2.303 k = log i t pA 2.303 pi or k = log t (2pi − pt ) where pi is the initial pressure of A at time, t = 0 and pt is the total pressure at time t.  Half-life of a reaction : The time taken for a reaction when half of the initial value has reacted is called half-life of a reaction. [A]0 For zero order reaction, t1/2 = , 2k 0 where [A] is initial and last concentration of reaction it means there is no change in concentration and ‘k’ is rate constant. 0.693 For 1st order reaction, t1/2 = k  nth order reaction : In general for nth order reaction of the type dx A → products, where, = k[A]n dt  1 1  1 k =   − n − 1 n t(n − 1) [ A] [ A]0 n−1  where [A]0 is initial concentration, [ A] is final concentration after time t and n can have all the values except 1.  Half-life of a reaction of nth order : 1 t1/2 µ 1 [ A]n0 t1/2 µ [A] for zero order t1/2 is independent of [A] for 1st order t µ 1 for 2nd order 1/2 [A] t1/2 µ

1 [A]2

for 3rd order

Amount of substances left after n half-lives = [A]0 2n

 Integrated rate laws for the reactions of zero and first order : Order 0

1

Reaction

Differential rate law

Integrated rate law

Straight line plot

Half Life

Units of k

A→P

d[ A] =–k dt

kt = [A]0 – [A]

[A] vs. t

[ A]0 2k

conc. time–1

A→P

d[ A] = – k[A] dt

type

[A] = [A]0 e–kt kt =

ln[ A]0 [ A]

ln [A] vs. t

ln

2 k

time–1

 Life time : The time in which 98% of the reaction is completed is called life time.

Mnemonics • Concept: Zero order • Mnemonics : ZOR don’t CCR • Interpretation : In zero order reaction, the rate of reaction does not change with concentration of the reactants.



108 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Know the Formulae  Integrated Rate Equations :

(i) For a zero order reaction :







(ii) For a first order reaction :





t =

t =

 Temperature coefficient =

[ R]0 − [ R] [ R]0 and t[½] = k 2k 0.693 [ R] 2.303 log 0 and t[½] = k k [ R]

k ( T + 10 ) k ( T)

Objective Type Questions Q. 1. In a chemical reaction X → Y, it is found that the rate of reaction doubles when the concentration of X is increased four times. The order of the reaction with respect to X is [CBSE, Delhi Set-2, 2020] (a) 1 (b) 0 (c) 2 (d) 1/2 Ans. Correct option : (d) Explanation : X → Y Rate(r) ∝ [X]n [Where n = Order of reaction] If the concentration X is increased by 4 times  X’ = 4X Then, Rate(r’) ∝ [X’]n r ' [ 4X ] = =2  r [X ]n r’ is new rate, X’ is a new concentration [4]n = 2 1 \n = 2 1 Order of reaction =  2 Q. 2. The half-life period for a zero order reaction is equal to (b) 2k 0.693 (a) [R ]0 k 2.303 [R ]0 (c) (d) k 2k (where [R]0 is initial concentration of reactant and k is rate constant).  R [CBSE Outside Delhi Set-2, 2020] Ans. Correct option : (d) Explanation : Half life period of a zero order [R]0 reaction = 2k Where [R]0 = initial concentration of reactant k = Rate constant Q. 3. For a zero order reaction, the slope in the plot of [R] vs. time is

-k (b) –k 2.303 +k (c) (d) +k 2.303 (where [R] is the final concentration of reactant)

(a)



[A] MULTIPLE CHOICE QUESTIONS :

(1 mark each)

 R [CBSE Outside Delhi Set-3, 2020] ns. Correct option : (b) A Explanation :



n



Q. 4. A first order reaction is 50% completed in 1.26 × 1014 s. How much time would it take for 100% completion ? U (a) 1.26 × 1015 s (b) 2.52 × 1014 s (c) 2.52 × 1028 s (d) Infinite Ans. Correct option : (d) Explanation : The reaction will be 100% complete only after infinite time. Q. 5. Consider a first order gas phase decomposition reaction given below : A(g) B(g) + C(g) The initial pressure of the system before decomposition of A was ‘pi’. After lapse of time ‘t’, total pressure of the system increased by x units and became ‘pt’. The rate constant k for the reaction is given as : (a) k=

pi 2.303 log pi - x t

(b) k =

pi 2.303 log t 2 pi - pt

(c) k=

pi 2.303 log 2 pi + pt t

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CHEMICAL KINETICS



(d) k = 2.303 log

t

pi pi + x

Ans. Correct option : (b) Explanation : Let us consider a first order gas phase decomposition reaction : A(g) ® B(g) + C(g) The initial pressure of the system before decomposition of A is ‘Pi.’ After lapse of time ‘t’, total pressure of the system increased by x units and became ‘Pt’. Hence, the pressure of A decreased by x atom. Initial pressure :    Pi atom  0       0 Pressure after time t : (Pi – x) x atm   x atm Pt = (Pi – x) + x + x = Pi + x atm. x = Pt – Pi = Pi – (Pt + Pi) PA = 2Pi – Pt

k=



=

[A]0 2.303 log t [A] Pi 2.303 log 2 Pi − Pt t



Q. 6. For the reaction A → B, the rate of reaction becomes three times when the concentration of A is increased by nine times. What is the order of reaction ?  (a) 1 (b) 2 (c) 1/2 (d) 0 Ans. Correct option : (c) Explanation : A → B Rate of reaction r ∝ [A]n…(i) If concentration of A is increased by nine times, then rate of reaction becomes three times, r ' = 3r A ' = 9A r ' ∝ [A ']n



3r ' ∝ [9A]n …(ii) From eq. (i) and (ii) [A] n r = 3r [9A] n 1  1 =  3  3 1 = 2n n=1 2

2n

∴ Order of reaction = [½]

[B] ASSERTION AND REASON TYPE QUESTIONS : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.



(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Q. 1. Assertion : Hydrolysis of an ester follows first order kinetics. Reason : Concentration of water remains nearly constant during the course of the reaction.  [CBSE, Outside Delhi Set 1, 2020] Ans. Correct option : (a)

Explanation :

CH 3COOC 2 H 5 + H 2O → CH 3COOH + C 2 H 5OH ester

Hydrolysis of an ester follows first order kinetics as [H2O] remains nearly constant during the course of the reaction. It is pseudo first order reaction.

Q. 2. Assertion : For complex reactions molecularity and order are not same. Reason : Order of reaction may be zero. Ans. Correct option : (b) Explanation : For a complex reaction, Order of overall reaction = molecularity of slowest step As rate of overall reaction depends upon total number of molecules involved in slowest step of the reaction. Hence, for complex reaction, molecularity and order are not same. Q. 3. Assertion : Order of the reaction can be zero or fractional. Reason : We cannot determine order from balanced chemical equation. Ans. Correct option : (b) Explanation : Order of a reaction may be zero or fractional. It can be determined through the rate law expression by sum of power of reactants.  [1] [C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1. Write the slope value obtained in the plot of log [R0] / [R] vs. time for a first order reaction.  R [CBSE, Delhi Set 1, 2020] Ans. K Slope  R0  2.303 R 

− K (t) Time 2.303 Q. 2. For which reaction, the rate of reaction does not decrease with time? U Ans. For zero order reaction the rate of reaction does not decrease with time because it does not depend on concentration of reactants Q. 3. What is the order of photochemical reaction? R Ans. Zero Order reaction

Slope value =

110 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Short Answer Type Questions-I Q. 1. Define order of reaction. Predict the order of reaction in the given graphs: (a)

t1/2





(ii) Half-life period of a reaction (t1/2) : Half-life of a reaction is the time in which the concentration of a reactant is reduced to half of its original value. [1] Q. 4. Write units of rate constants for zero order and for the second order reactions if the concentration is expressed in mol L–1 and time in second. U [CBSE Comptt. OD 2015]

Ans. Zero order : mol L–1s–1[1] Second order : L mol–1s–1[1] [CBSE Marking Scheme 2015]

[R]o (b)

(2 marks each)

Commonly Made Error t1/2



[R]o

Where [R]0 is the initial concentration of reactant and t1/2 is half life. U [CBSE, Outside Delhi set 2, 2019]



Ans. Order of reaction is defined as the sum of powers to which the concentration terms are raised in the rate law equation. [1] (a) First order (b) zero order [½]+[½]  [CBSE Marking Scheme 2019]



Learn the units of various order reactions and also try and understand the underlying concept.

Q. 5. (i)  What is the order of the reaction whose rate constant has same units as the rate of reaction ? (ii) For a reaction A + H2O → B; Rate ∝ [A]. What is the order of this reaction?  A [CBSE Comptt. OD Set-3 2017] Ans. (i) Zero Order[1] (ii) Pseudo-first Order[1] [CBSE Marking Scheme 2017]

 Sometimes students can not predict the order of reaction from the given graph.

 Do practice to interpret the graph between initial concentration and half life for a particular order of reaction

The concept of rate law and determination of order of reaction is not clear to many students. Hence, students make errors in identifying order of reaction.

Answering Tip

Commonly Made Error

Answering Tip



Q. 6. For a chemical reaction R ® P, variation in ln [R] vs time (t) plot is given below: ln[Ro]

Pt

Q. 2. For a reaction : 2NH3 (g) → N2(g) + 3H2(g); Rate = k ; (i) Write the order and molecularity of this reaction. (ii) Write the unit of k.  U [CBSE, Delhi Set 1, 2016]

Ans. (i) Zero order, bimolecular / unimolecular. [½], [½] (ii) mol L–1 s–1.  [1]  [CBSE Marking Scheme, 2016] Q. 3. Explain the following terms : (i) Rate constant (k) (ii) Half-life period of reaction (t1/2). R [CBSE Comptt. OD 2015] Ans. (i) Rate constant (k) : Rate constant is rate of the reaction when the concentration of reactants is unity.  [1]

ln[R]





t

For this reaction : (i) Predict the order of reaction. (ii) What is the unit of rate constant (k)? A [CBSE Comptt. Delhi Set-1, 2 2017] Ans. (i) First order.[1] (ii) s-1/time-1[1] [CBSE Marking Scheme 2017]

Commonly Made Error  Students often misinterpret graph either in hurry or overlook the details.

[ 111

CHEMICAL KINETICS

Answering Tip  Read the concept, understand and practice the interpretation of graphs of various orders.

log[Ro]/[R]

Q. 7. For a chemical reaction R ® P, variation in log [Ro]/[R] vs time plot is given below:



Time For this reaction : (i) Predict the order of reaction (ii) What is the unit of rate constant (k)? A [CBSE Comptt. Delhi Set-3 2017]

(ii) s-1/time-1[1]



[CBSE Marking Scheme 2017]

Q. 8. For a certain chemical reaction variation in concentration [A] vs. time (s) plot is given below: (i) Predict the order of the given reaction?



(ii) What does the slope of the line and intercept indicate? A (iii) What is the unit of rate constant k?  Ans. (i) Zero order reaction  [½] (ii) Slope represents –k ; Intercept represents [R]0 [½] +[½] (iii) mol L–1 s–1



[½]

Q. 9. (i) Explain why H2 and O2 do not react at room temperature.

(ii) Write the rate equation for the reaction A2+3B2 ® 2C, if the overall order of the A&E+A reaction is zero.

Ans. (i) Due to high activation energy[1] (ii) Rate = k [A2]0[B2]0

[1]

Q. 10. Derive integrated rate equation for rate constant of a first order reaction.



[CBSE Comptt. OD Set-1, 2 2017]

Ans.

or

  

t =

C 2.303 log o  k Ct

[½]

Let Co = 1 and Ct = 3/10 so Co/Ct = 1/ (3/10) = 10/3

2.303 10 ´ 5730 log  0.693 3

[½]

t = 19042 × (1-0.4771) = 9957 years

[½]

t =

Q. 13. (i) What is the order of the reaction whose rate constant has same units as the rate of reaction ? (ii) For a reaction A + H2O → B; Rate ∝ [A]. What is the order of this reaction ? U Ans. (i) Zero order (ii) Pseudo-first order

[1] [1]

Commonly Made Error

R→P Rate = −

Q. 11. State a condition under which a bimolecular reaction is kinetically first order reaction. C Ans. Let us take a bimolecular reaction : A + B ® Product Rate = k [A] [B] When concentration of [B] is taken in excess then rate law will become : Rate = k [A] where, k = constant The rate depends only on one of the reactant as there is negligible change in its concentration so it is bimolecular but is of first order. [2] Q. 12. The C-14 content of an ancient piece of wood was found to have three tenths of that in living trees. How old is that piece of wood? (log 3= 0.4771, log 7 = 0.8540 , Half-life of C-14 = 5730 years) A Ans. k = 0.693/t1/2 k = 0.693/5730 years–1 [½]



Ans. (i) First order.[1]

Integrating this equation, we get In [R] = –kt + I …(i) When t = 0, R = [R]0, where [R]0 is the initial concentration of the reactant. Therefore, equation (i) can be written as In [R]0 = –k × 0 + I In [R]0 = I Substituting the value of I in equation (i) In [R] = –kt + ln[R]0 Rearranging this equation [R]   In  [1] = −kt [R]0 1 [R] k = In 0 t [R] [R] 2.303 k= log 0 t [R] [CBSE Marking Scheme 2017]  [1]

d[ R] = k[ R ] dt

d[ R] = −kdt [ R]

 Student get confused in order of reactions. Answering Tip  Learn the units of various order reactions and also try and understand the underlying concept.

112 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q. 14. A first-order reaction has a rate constant 1.15 × 10−3 s−1. How long will 5 g of this reactant take to reduce to 3 g ? A Ans.

t=



[R ] 2.303 log initial k [R final ]

2.303 5 log −3 1.15 × 10 3 2.303 × 0.2219 = 1.15 × 10 −3

Initial amount = 5 g Final concentration = 3 g Rate constant = 1.15 × 10–3 s–1 We know that for a First order reaction,

=

= 444.379 s[2]

Short Answer Type Questions-II Q. 1.

The following data were obtained for the reaction : A + 2B → C



Experiment

[A]/M

[B]/M

Initial rate of formation of C/M min-1

1

0.2

0.3

4.2 × 10–2

2

0.1

0.1

6.0 × 10–3

3

0.4

0.3

1.68 × 10–1

4

0.1

0.4

2.40 × 10–2

(a) Find the order of reaction with respect to A and B. (b) Write the rate law and overall order of reaction. (c) Calculate the rate constant (k). [CBSE, Outside Delhi Set-2, 2019]



Ans. Rate = k[A]p[B]q On solving (a) Order with respect to A=2 , B=1 [½] + [½] (b) Rate = k[A]2[B]1 ; overall order = 3 [½] + [½] (c) Experiment 1 : 4.2 × 10−2 = k (0.2)2 (0.3); k=3.5

(3 marks each)



4=

[0.4 ]y [0.1]y



( 4 )1 = ( 4 )y



y=1



Dividing equation (i) by (ii), we get



4.2 ´ 10 -2 k [0.2 ] [0.3] = x y -1 1.68 ´ 10 k [0.4 ] [0.3] x



0.25 =

y

[0.1]x [0.2 ]x



  (0.25) = (0.5)x



(0.5)2 = (0.5)x

x=2



(i) S  o the rate of reaction with respect to A is 2 and with respect to B is 1.



(ii) Rate law = k[A]2[B]



Overall order of reaction is 3. (iii) Rate constant, K =

Rate

[ A ]2 [B] 6.0 ´ 10 -3

Experiment 2 : 6.0 × 10−3 = k (0.1)2 (0.1) ; k=6  [1] (Full marks may be awarded for any one correct answer)



         =



         = 6.0 mol -2 L2 min -1 



Q. 2.

[CBSE Marking Scheme 2019]

Detailed Answer :

Let the order of reaction with respect to A be x and with respect to B be y.

∴ Rate of reaction = k [ A ]x [ B]y According to details given ,



4.2 × 10–2 = k[0.2]x [0.3]y…(i)



6.0 × 10–3 = k[0.1]x [0.1]y…(ii)



1.68 × 10–1 = k[0.4]x [0.3]y…(iii) –2

x

y



2.40 × 10 = k[0.1] [0.4] …(iv)



Dividing equation (iv) by (ii), we get



2.40 ´ 10 -2 k [0.1] [0.4 ] = x y  6.0 ´ 10 -3 k [0.1] [0.1] x

y

[3]

Following data are obtained for the reaction : N2O5 → 2NO2 + ½O2 t/s

[N2O5]/mol L–1

0

300

1.6 × 10–2 0.8 × 10–2

600 0.4 × 10–2

(a) Show that it follows first order reaction. (b) Calculate the half-life. (Given log 2 = 0.3010, log 4 = 0.6021) [3]  [CBSE, Delhi set 1, 2017] Ans. (a)



( 0.1)2 ( 0.1)



k =

[ A]o 2.303 log [ A] t

[½]

=

2.303 1.6 ´ 10 -2 log 300 0.8 ´ 10 - 2

=

2.303 log 2 = 2.31 × 10–3 s–1 [½] 300

[ 113

CHEMICAL KINETICS



At 600 s,

k =

[ A]o 2.303 log t [ A]

[½]



2.303 1.6 ´ 10 -2 log 0.4 ´ 10 -2 = 600



= 2.31 × 10–3 s–1



k is constant when using first order equation therefore it follows first order kinetics. [½] OR



In equal time interval, half of the reactant gets converted into product and the rate of reaction is independent of concentration of reactant, so it is a first order reaction.

(b) t1/2 = 0.693/k = 0.693/2.31 × 10–3 = 300 s

(If student writes directly that half life is 300 s, award full marks). [1]



[CBSE Marking Scheme, 2017]

Detailed Answer : (a) For first order reaction the integral rate law is :

 kt =ln

reaction from the given data of a reaction.

Answering Tip  Do practice to solve the numericals and be precise in your answer. Q. 3. A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed. (Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021) A [CBSE OD Set-1, 2, 3 2017] Ans.

For t = 600 s, at = 0.4 × 10–2 mol L–1



Using first set of data in the rate law,



1.6 ´ 10

-2

0.8 ´ 10

-2





 k × 600 = ln



 k = 0.00231 s



–1

The value of k is consistent, therefore it follows first order reaction. [1½]







OR







[½]



...(i) [½] ...(ii) [½]

2.303 100 log 20 k 75 = 2.303 100 t log k 25 log 4 / 3  log 4  20/t = 0.1250/0.6021

 =

1.6 ´ 10 -2 0.4 ´ 10 -2

2.303 100 log k 75

2.303 100 log k 25    Divide (i) equation by (ii)

  

Using second set of data in the rate law,



 

2.303 [ A]0 log k A 

t=

  k = 0.00231 s–1





t=

–1



 k × 300 = ln

 Some students get confused to show the order of

20 min =

 Given, a0 = 1.6 × 10 mol L For t = 300 s, at = 0.8 × 10–2 mol L–1



Commonly Made Error

a0 a1 –2



(b) The half-life of first order reaction is given by the following equation : log 2 ln2 t[½] = = 2.303 × k k log 2  ∴ t[½] = 2.303 × = 300.08 s. [1½] 0.00231

[½]

 t = 96.3 min [1] (or any other correct procedure) [CBSE Marking Scheme 2017]

114 ]



Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

[3] 



[Topper’s Answer 2017]

 Q. 4. For the first order thermal decomposition reaction, the following data were obtained :

Time / sec



C2H5Cl(g) → C2H4(g) + HCl(g) Total pressure / atm

0

0.30

300

0.50



Calculate the rate constant



(Given : log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021) [CBSE, Outside Delhi Set 1, 2016]

ns. Given : Initial pressure, A P0 = 0.30 atm Pt = 0.50 atm t = 300 s P0 2.303 Rate constant, k = log t 2P0 - Pt

 2.303 0.30 = log 300 s 2 ´ 0.30 - 0.50 =

2.303 0.30 log 300 s 0.60 - 0.50

2.303 0.30 = log 300 s 0.10



2.303 = log 3 300 s 



=



1.099 = 300 s



= 0.0036s–1

[1]

2.303 ´ 0.4771 300 s

= 3.6 × 10–3 s–1 (deduct [½] mark if unit is not written) [1] [CBSE Marking Scheme, 2016]



Commonly Made Error [1]





Students tend to overlook or incorrectly copy the data.

Answering Tip



Read the question carefully to write the correct values of concentration. Write working formula followed by substitution of values. Do not forget to mention unit with the answer.

[ 115

CHEMICAL KINETICS

Long Answer Type Questions Q. 1.

(a) A first order reaction is 25% complete in 40 minutes. Calculate the value of rate constant. In what time will the reaction be 80% completed?





  (b) Define order of reaction. Write the condition under which a bimolecular reaction follows first order kinetics. [CBSE, Delhi set 1, 2020]

Ans.

(a) t =



[ A]0 2.303 log [ A] k

Dividing (i) by (ii)



40 2.303 100 2.303 100 = log / log t k 75 k 20



40 0.1250 = t 0.6991



[CBSE Comptt. OD Set-1, 2 2017] (i) For a first order reaction [ R] 2.303 log 0 , k = [ R] t



k=

where [R]0 = initial concentration, [R] = conc. after time t



When half of the reaction is completed, [R] = [R]0/2. Representing, the time taken for half of the reaction to be completed, by t1/2, equation becomes : [ R]0 2.303 log t1 / 2 [ R]0 / 2 

[½]

⇒  t1/2 =

2.303 log 2 k 

[½]



⇒  t1/2 =

2.303 × 0.3010 k



⇒  t1/2 =

0.693  k



 k =



[½]



0.6691 × 40 = 223.712 min. t= 0.1250

 

Ans.



2.303 4 log 40 k 3 = t 2.303 log 5 40 0.6021 / 4.771 = t 0.6991



  (ii) F  or a first order reaction, show that the time required for 99% completion of a first order reaction is twice the time required for the completion of 90%.

2.303 100 ...(ii) log k 20









2.303 100 (i) 40 min = log ...(i) k 75 (ii) t =

(5 marks each)



2.303 100 2.303 100 = log log t 100 − 0.25 40 75

 he above equation shows that half-life of first T order reaction is independent of the initial concentration of the reactant. [½] (ii) For a first order reaction



t =

2.303 a log k a-x

[½]



t99% =

2.303 100 log k 1

[½]





[3]



=

2.303 × 2 k

(b) Order of reaction : The sum of the coefficients of the reacting species that are involved in the rate equation for the reaction, is called order of reaction.



=

4.606 k

[½]



and

2.303 100 log 10  k

[½]

2.303 2.303 log 10 = k k

[½]



    

2.303 = (log 4 − log 3) 40



2.303 = ( 0.6021 − 0.4771) 40



2.303 = × 0.125 = 0.007196 40



Q. 2.

 

 = 7.196 × 10–3 min–1 

The condition under which a bimolecular reaction follows first order kinetics is when one of the reactants is taken in large excess that its concentration hardly changes.[2] (i) Write the rate law for a first order reaction. Justify the statement that half life for a first order reaction is independent of the initial concentration of the reactant.

=

t90% =



= t99%

2.303 log 100 k

=2



t90%



t99% = 2 × t90%

[½]



116 ] Q. 3.

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(i) A first order reaction is 75% completed in 40 minutes. Calculate its t1/2.



 (ii) Predict the order of the reaction in the given plots:

(a) t1/2

(b) t1/2

[R]o





where [R]0 is the initial concentration of reactant.



(Given : log 2 = 0.3010, log 4 = 0.6021) [CBSE Foreign Set-1, 2, 3 2017]

2.303 [ A]0 Ans. (i) k = log t [ A] 

 

[½]

2.303 100 log 40 25 

[½]



[½]

t1 / 2 =

0.693 k 

t1 / 2 =

0.693 = 19.98 min = 20 min 0.0347 min−1 

[½] [1]

2 NO + O2 → 2 NO2 Initial rate of Experiment





[NO]/M [O]2/M

formation of NO2/M min-1

1

0.3

0.2

7.2 × 10-2

2

0.1

0.1

6.0 × 10-3

3

0.3

0.4

2.88 × 10-1

4

0.4

0.1

2.40 × 10-2



(i) Find the order of reaction with respect to NO and O2. (ii) Write the rate law and overall order of reaction. (iii) Calculate the rate constant (k).



2.40 × 10−2 k[0.4]x [0.1]y = 6.0 × 10−3 k[0.1]x [0.1]y x=1  Dividing eqn 3 by eqn 1

 [1] order w.r.t. NO = 1, order w.r.t. O2 is 2. [½]+[½] (ii) Rate law Rate = k[NO]1[O2]2, overall order of the reaction is 3. [½]+[½] (iii) Rate constant k =

rate [NO]1[O2 ]2

k = 6.0 mol

−2

=

7.2 × 10−2 0.3 × ( 0.2 )2

2

L min−1



[1]



Rate = k[A] [B]2



(a) How is the rate of reaction affected if the concentration of B is doubled ?



Ans.



(b) What is the overall order of reaction if A is present in large excess ? (ii) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction. (log 2 = 0.3010) U [Question for Green board] (i) A + B → P Rate = k[A] [B]2 (a)  When concentration of B is doubled it means concentration of B becomes 2 times.  Thus, Rate = k[A]1 [2B]2 = k[A] [4B2]  So, the rate becomes 4 times. [1] (b)  Order of reaction is the number of molecules whose concentration alters after the reaction. If A is present in excess i.e., its concentration is unaffected. So, rate depends only on the concentration of B. as k = [B]2 Thus, the reaction is of second order. [1] (ii) STEP 1 : For the 1st order reaction :



A [CBSE Foreign Set-1, 2, 3 2017]

Ans. (i)

[1]

2.88 × 10−1 k[0.3]x [0.4]y = 7.2 × 10−2 k[0.3]x [0.2]y y=2

[1]

(b) Zero order reaction [1] [CBSE Marking Scheme 2017]

Eqn (4)

Dividing eqn 4 by eqn 2

[CBSE Marking Scheme 2017]

Q. 4. The following data were obtained for the reaction:

Eqn (3)

y

Q. 5. (i) For a reaction A + B → P, the rate is given by

(ii) (a) First order reaction

x

2.40 × 10 = k[0.4] [0.1]



2.303 100 = log 4 40 25   2.303 = × 0.6021 40   –1     k = 0.0347 min 



-2

[R]o



=

2.88 × 10-1 = k[0.3]x[0.4]y



Rate = k[NO]x[O2]y



7.2 × 10-2 = k[0.3]x[0.2]y

Eqn (1)



6.0 × 10-3 = k[0.1]x[0.1]y

Eqn (2)



k =

2.303 a log t a−x

t = 30 × 60 = 1800 s

2.303 100 log 1800 100 − 50 2.303 = log 2 1800

k =

[ 117

CHEMICAL KINETICS

STEP 2 : and



2.303 × 0.3010 1800 2.303 a log t = k (a − x) =

=

[1]

STEP 3 :

2.303 100 log k 100 − 90

=

2.303 2.303 log 10 = [1] k k

By putting the value of k here, we get ⇒



t =

2.303 × 1800 2.303 × 0.3010

= 5.98 × 103 s

[1]

TOPIC-3

Concept of Collision Theory, Activation Energy and Arrhenius Equation

Revision Notes



 The rate of reaction is dependent on temperature. This is expressed in terms of temperature coefficient. Rate constant at 308 K Temperature coefficient = nt at 298 K Rate constan It is observed that for a chemical reaction with rise in temperature by 10°C, the rate constant is nearly doubled.  Activation energy : It is an extra energy which must be possessed by reactant molecules so that collision between reactant molecules is effective and leads to the formation of product molecules.  Activation energy (Ea) for n reaction can not be zero. It is not possible that every collision between molecules will be effective. Ea can not have negative value.  Threshold energy : The minimum energy that the reacting species must possess in order to undergo effective collision to form product molecules is called threshold energy. Activation theory (Ea) = Threshold energy (ET) – Average energy of the reactions (ER)



 Arrhenius equation : Activated complex is defined as unstable intermediate formed between reacting molecules. It is highly unstable and readily changes into product. Arrhenius equation gives the relation between rate of reaction and temperature. k = Ae − Ea / RT where k =Rate constant A = Frequency factor (Arrhenius factor) Ea = Activation Energy. R =Gas constant T = Temperature in Kelvin ln k = ln A – Ea / RT Ea log k = log A − 2.303 RT Ea A plot of log k with 1/T gives a straight line with slope = − 2.303 R If k2 and k1 are rate constants at temperature T2 and T1 respectively, then T2 − T1 k Ea log 2 = − 2.303 R T1 − T2 k1 log k

1

118 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

 Those collisions which lead to the formation of product molecules are called effective collisions. Rate of reaction = f × z, where ’z’ is collision frequency and ‘f’ is fraction of collisions which are effective.  The number of collisions that take place per second per unit volume of the reaction mixture is called collision frequency. It is represented by ‘Z’.  Activated complex is defined as unstable intermediate formed between reacting molecules. It is highly unstable and readily changes into product.  According to the collision theory, rate of reaction depends on the collision frequency and effective collisions.

Rate = Z AB e − Ea / RT ,







where ZAB represents the collision frequency of reactants A and B. e − Ea / RT represents the fraction of molecules with energies equal to or greater than Ea.

 According to collision theory, another factor P which is called steric factor refers to the orientation of molecules which collide, is important and contributes to effective collision, k = PZ AB e − Ea / RT  Catalyst : A catalyst is a substance that alters the rate of reaction without itself undergoing any chemical change at the end of reaction. Intermediate complex theory : A+B + X →A B → A→B+ X Reactants Product Catalyst Catalyst



X         Intermediate complex Characteristics of catalyst : (i) Catalyzes only the spontaneous reaction. (ii) Does not change the equilibrium constant. (iii) Catalyzes both the forward and backward reactions. (iv) Does not alter the free energy change (ΔG) of a reaction. (v) A small amount of the catalyst can catalyse large amount of reactions.

Mnemonics • Concept: Effect of Collision • Mnemonics : ECFPM • Interpretation : Effective collisions lead to formation of product molecules. • Concept: Catalyst • Mnemonics : CAR • Interpretation : A catalyst alters the reaction

Know the Formulae  Temperature coefficient =

Rate constant at (T+10)° Rate constant at T°

 Rate of reaction = f × Z where Z is the collision frequency, f is the fraction of collisions.  Arrhenius equation : k = Ae − Ea / RT  Rate = PZe − Ea / RT

[ 119

CHEMICAL KINETICS



Q. (i) For the reaction A → B, the rate of reaction becomes twenty seven times when the concentraion of A is increased three times. What is the order of the reaction ? (ii) The activation energy of a reaction is 75.2 kJ mol–1 in the absence of a catalyst and it lowers to 50.14 kJ mol–1 with a catalyst. How many times will the rate of reaction grow in the presence of a catalyst if the reaction proceeds at 25°C? Solution: STEP - I : (i) r = k[R]n When concentration is increased three times, R]] = 3a a [[R =3 27rr = = kk((3 3a a))nn 27  [½] 27 kk((3 3a a))nn 27 or 27 27 = = or ka nn rr ka 3nn or or 3 333 = 3nn =3 =3 = n= 3 =3 n

(ii) STEP - I : According to equation,



log k = log A -

Ea 2.303 RT



For uncatalysed reaction Ea (1) log k1 = log A …(i) [½] 2.303 RT  STEP - II : For catalysed reaction Ea (2) …(ii) [½] log k2 = log A 2 . 303 RT 



STEP - III : A is equal for both the reactions. Subtracting equation (i) from equation (ii) k2 Ea (1) - Ea (2) log = k1 2.303 RT

(75.2 - 50.14)kJ mol-1 k2 = k1 2.303 ´ 8.314 JK -1mol-1 ´ 298 K k log 2 = 4.39 k1 [½] k2 = antilog(4.39) k1 [1] [1] = 22.45 .45 ´× 110 044 Arrhenius = Rate of reaction increases by 2.45 × 104 times. [1] log

Objective Type Questions Q. 1. The slope in the plot of ln[R] vs. time gives

(a) +k (b)

+k 2.303



(c) –k

−k 2.303



(d)

(where [R] is the final concentration of reactant.) [CBSE, Outside Delhi Set 1, 2020]

Ans. Correct option : (c) Explanation :



[A] MULTIPLE CHOICE QUESTIONS :

(1 mark each)

↑

↑

↑

↑

120 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



Q. 2. According to Arrhenius equation, rate constant k is equal to Ae–Ea /RT. Which of the following options represent the graph of ln k versus 1/T ? (a)

(b)

ln k

ln k 1/T →

1/T →

(c)

(d) ln k

ln k

1/T →

1/T →

Ans. Correct option : (a) Explanation : According to Arrhenius equation, − E / RT

a k = A.e Taking log on both sides

ln k = ln A −

Ea −E 1 = − a × + In A RT R T

This equation can be related to equation of straight line, y = mx + c. It is evident that slope of the plot can be given as slope =

Ea R

Intercept = ln A.  Q. 3. Consider the following figure and mark the Correct option :

Energy →

Activated complex

Products Reactants







Reaction coordinate →

(a) Activation energy of forward reaction is E1 + E2 and product is less stable than reactant.

(b) Activation energy of forward reaction is E1 + E2 and product is more is stable than reactant. (c) Activation energy of both forward and backward reaction is E1 + E2 and reactant is more stable than product. (d) Activation energy of backward reaction is E1 and product is more stable than reactant. U Ans. Correct option : (c) Explanation : Activation energy of forward reaction is E1 + E2 and their product is less stable than reactants. Ea (forward) = E1 + E2 As energy of reactants is less than products and the product is less stable than the reactant. [B] ASSERTION AND REASON TYPE QUESTIONS : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Q. 1. Assertion : Rate constants determined from Arrhenius equation are fairly accurate for simple as well as complex molecules.  Reason : Reactant molecules undergo chemical change irrespective of their orientation during collision. Ans. Correct option : (c)  Explanation : According to Arrhenius equation, the rate constant determined is closely accurate for single as well as complex reaction because the reactant molecules with proper orientation and sufficient kinetic energy undergo a chemical change. Q. 2. Assertion : The rate of reaction on also increase with its product if one of the product act as catalyst. Reason: A catalyst lowers the activation energy of reaction. Ans. Correct option: (a) Explanation : A catalyst increases the rate of reaction by decreasing activation energy. Q. 3. Assertion : There is no reaction known for which ΔG is positive, yet it is spontaneous. Reason : For photochemical reaction, ΔG may or may not be negative. Ans. Correct option: (a)  Explanation : For photochemical reaction, ΔG may or may not be negative. While for spontaneous reaction, ΔG is negative. [C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1. Define activation energy. R Ans. The minimum extra energy over and above the average potential energy of the reactants which must be supplied to the reactants to enable them to cross over the energy barrier between reactants and product is called activation energy. Q. 2. All collisions taking place between reactants do not form products. Why? U Ans.  Only those collisions are effective for a reaction which help reactant molecules to get threshold energy level. Q. 3. What is the effect of (i) catalyst (ii) increase in temperature on activation energy? R Ans. (i) A catalyst decreases the activation energy.[½] (ii) A  n increase in temperature decreases the activation energy.[½]

[ 121

CHEMICAL KINETICS

Short Answer Type Questions-I Q. 1. What is the effect of adding a catalyst on (i) Activation energy (Ea), and

(2 marks each) log x =

209.5 ´ 10 3 J mol -1 2.303 ´ 8.314 JK -1mol -1 ´ 581K = -18.8323

(ii) Gibb’s energy (∆G) of a reaction ?



=-

[CBSE, Delhi Set 3/ Outside Delhi Set 1, 2017]

ns. (i) Decreases A (ii) No effect

[½] [½] [CBSE Marking Scheme, 2017]

Detailed Answer :





     X = Antilog (–18.8323)



   = Antilog 19.1677    = 1.471 × 10–19 [2] Q. 3. The rate of a particular reaction quadruples on increasing temperature from 293K to 313K. For this reaction, calculate activation energy. Ans. T1 = 293 K T2 = 313 K k2 =4 k1

(i) Adding of a catalyst lowers the activation energy (Ea) of reactants. [½]

- Ea 2.303 RT

(ii) Adding of a catalyst has no effect on Gibb’s free [½]

energy. Q. 2. The activation energy for the reaction,



2HI(g) → H2(g) + I2(g)













According to Arhenius equation, k Ea é T2 - T1 ù log 2 = ê ú k1 2.303 R ë T1T2 û

-1



is 209.5 KJ mol at 581K. Calculate the fraction of molecules of reactants having energy equal to or

greater than activation energy. n Ans.  x = = e - Ea Rt N   Ea = 209.5 ×103 J mol–1 T = 581 K R = 8.314 J K–1 mol–1

A&E

Ea é 313 - 293 ù 2.303 ´ 8.314 êë 293 ´ 313 úû 20 Ea ´ 0.6020 = 2.303 ´ 8.314 91709 Ea = 528545 J mol -1 log 4 =

= 52.854 K J mol -1

Short Answer Type Questions-II Q. 1.

Calculate Ea for this reaction and rate constant if its half-life period be 300 minutes. [CBSE, Outside Delhi Set 1, 2020]

 Ans.

The rate constant for the first order decomposition of N2O5 is given by the following equation : k = (2.5 × 1014 s–1) e(–25000K)/T



 k = (2.5 × 1014 l–1) e(–25000K)/T  t1/2 = 300 minutes Ea = 25000 K R  Ea = 25000 × R × K  = 25000 × 8.314 J K–1 mol–1 × K  = 207850 J mol–1 = 207.850 KJ mol–1 0.693 0.693 t½ = ÞK= min -1 K 300

(3 marks each)

Ans.

k2 = 0.693/20,



k1 = 0.693/40



[2]

log

k2 Ea é 1 1 ù = ê - ú k1 2.303R ë T1 T2 û

k2/k1 = 2 log 2 =

é 320 - 300 ù Ea ê ú 2.303 ´ 8.314 ë 320 ´ 300 û [3]

Ea = 27663.8 J/mol or 27.66 kJ/ mol

Q.3. Two reactions of the same order have equal pre exponential factors but their activation energies differ by 24.9 kJ mol–1. Calculate the ratio between the rate constants of these reactions at 27°C. (Gas constant R = 8.314 J K–1 mol–1) U

[2]

 = 0.231 × 10–2  [1]  = 2.31 × 10–3 min–1 Q. 2. A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction. (Given : log 2 = 0.3010, log 4 = 0.6021, R = 8.314 JK 1mol 1) [CBSE, Outside Delhi, 2018]

Ans.

The Arrhenius equation : k = Ae–Ea/RT





Taking log on both sides : log k = log A −





For reaction (i) : log k1 = log A −





For reaction (ii) : log k 2 = log A −

Ea 2.303 RT

Ea (1)  2.303RT

[1]

Ea ( 2 )  2.303RT

[1]

122 ]







Subtracting (i) from (ii) log k1 Ea (1) − Ea ( 2 ) = log k 2 2.303 RT log k1 24.9 × 1000 = log k 2 2.303 × 8.3 × 300 = 4.342 k1 = antilog(4.342) = 2.198 × 10 4  k2

 [1]

Commonly Made Error  Students often substitute incorrect data in the working formula leading to wrong result. Answering Tip  List the given data and carefully identify the values to be computed in the formula. Q.4. The rate constant for the first order decomposition of H2O2 is given by the following equation :

Ans.

1.0 × 10 4 log k = 14.2 − K; T  Calculate Ea for this reaction and rate constant k if its half-life period be 200 minutes. (Given : R = 8.314 J K–1 mol–1) U [CBSE, Delhi Set 1, 2016]

Ea 2.303 RT



At 320 K , t½ = 10 min t½ =

0.693 0.693 or k = k t½

0.693 = 0.0693 min −1 10 According to Arrhenius equation : k2 =

log

k2 Ea  1 1  =  −  k1 2.303R  T1 T2 



(5 marks each) = log

Ea  T2 − T1  0.0693 =   0.0231 2.303R  T1T2 

= log

Ea 0.0693  320 − 300  = 0.0231 2.303 × 8.314  300 × 320   R = 8.314 JK −1 mol −1 

= log 3 =



Ea × 20 2.303 × 8.314 × 300 × 320

log 3 × 2.303 × 8.314 × 300 × 320 20 [∵ log 3 = 0.4771] 0.4771 × 2.303 × 8.314 × 300 × 320 = 20 = 43848.5 J mol–1 = 43.85 k J mol–1  [2] (b) Two conditions for collisions to be effective collision are : (i) The reactant molecules must have attained sufficient energy to break chemical bonds (ii) The reactant molecules must have the proper orientation.  [2] (c) The number of the reacting species that collide simultaneously in a chemical reaction is called as molecularity of a reaction. The sum of the coefficients of the reacting species is the order of reaction. For complex reactions, molecularity has no significance while the order of reaction is applicable.  [1]

Visual Case-Based Questions Q. 1. Read the passage given below and answer the following questions : 1×4=4

[½]

Ea = 1.0 × 104 × 2.303 × 8.314 = 191471.4 J/mol 1 t1 0.693 [½] = 2 k  0.693 min k = 200 = 0.0034 min–1 1  [CBSE Marking Scheme, 2016]

Long Answer Type Questions Q. 1. (a) A first order reaction is 50% completed in 30 minutes at 300 K and in 10 minutes at 320 K. Calculate activation energy (Ea) for the reaction. (R = 8.314 J K–1 mol–1)  (b) Write the two conditions for collisions to be effective collisions.  (c) How order of reaction and molecularity differ towards a complex reaction? [Given : log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021, log 5 = 0.6991] + U [CBSE, Delhi Set-1, 2 & 3, 2020]  Ans. (a) At 300K, t1/2 = 30 min. 0.693 0.693 t½ = or k = k t½ 0.693 k1 = = 0.0231 min −1 30

log A - Ea 2.303RT  1.0 ´ 10 4 K = T

log k =





Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



Ea =

(4 marks each) The rate of a reaction, which may also be called its velocity or speed, can be defined with relation to the

[ 123

CHEMICAL KINETICS

concentration of any of the reacting substances, or to that of any product of the reaction. If the species chosen is a reactant which has a concentration c at time t the rate is - dc/dt, while the rate with reference to a product having a concentration x at time t is dx/dt. Any concentration units may be used for expressing the rate; thus, if moles per liter are employed for concentration and seconds for the time, the units for the rate are moles litre–1sec–1. For gas reactions pressure units are sometimes used in place of concentrations, so that legitimate units for the rate would be (mm. Hg) sec–1 and atm. sec–1 The order of a reaction concerns the dependence of the rate upon the concentrations of reacting substances; thus, if the rate is found experimentally to be proportional to the ath power of the concentration of one of the reactants A, to the bth power of the concentration of a second reactant B, and so forth, via., rate = k CAa CAb the over-all order of the reaction is simply n = a + b + ----- (2) Such a reaction is said to be of the ath order with respect to the substance A, the bth order with respect to B and so on... (Laidler, K. J., & Glasstone, S. (1948). Rate, order and molecularity in chemical kinetics. Journal of Chemical Education, 25(7), 383.) In the following questions, a statement of Assertion followed by a statement of Reason is given. Choose the correct answer out of the following choices on the basis of the above passage. (CBSE QB 2021) (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Assertion: Rate of reaction is a measure of change (i) in concentration of reactant with respect to time. Reason: Rate of reaction is a measure of change in concentration of product with respect to time. (ii) Assertion: For a reaction: P + 2Q → Products, Rate = k [P]1/2[Q]1 so the order of reaction is 1.5 Reason: Order of reaction is the sum of stoichiometric coefficients of the reactants. (iii) Assertion: The unit of k is independent of order of reaction. Reason: The unit of k is moles L–1s–1. (iv) Assertion: Reactions can occur at different speeds. Reason: Rate of reaction is also called speed of reaction. Ans. (i) Correction option is (b) (ii) Correction option is (c) (iii) Correction option is (d) (iv) Correction option is (b) Q. 2. Read the passage given below and answer the following questions : (1 × 4 = 4)



The rate of the reaction is proportional to the concentration of the reactant. Hydrogenation of ethene results in the formation of ethane. The rate constant, k for the reaction was found to be 2.5 × 10–15s–1. The concentration of the reactant reduces to one-third of the initial concentration in 5 minutes. The following questions are multiple choice questions. Choose the most appropriate answer: (i) Find the order of reaction : (a) Zero order (b) First order (c) Second order (d) Fractional order (ii) The rate law equation is : (a) Rate = k [C2H6] (b) Rate = k [C2H4]2 (c) Rate = k [C2H4] (d) Rate = k [C2H4]2 (iii) The half-life for the reaction is: (a) 2.772 × 10–24 s (b) 2.772 × 10–12 s (c) 1.386 × 10–24 s (d) 1.386 × 10–12 s (iv) The rate constant of the reaction after 5 minutes is: (a) 0.4290 min–1 (b) 0.1297 min–1 –1 (c) 0.2197 min (d) 0.6591 min–1 OR The slope of the curve in the reaction is : (a) K (b) –K (c) 2K (d) –2K Ans. (i) Correct option: (b) Explanation : Since the rate of the reaction is proportional to the concentration for the reactant i.e. ethene so, it is first order reaction. [1] (ii) Correct option : (c) Explanation : Catalyst

(iii)

C 2 H 4 + H 2  → C2 H6 Ethene Ethane

Rate law equation, Rate = K [C2H4] [1] Correct option : (a) Explanation : For first order reaction, 0.693 t1 2 = K 0.693 = 2.5 × 10 −15 s−1 −24 = 2.772 × 10 s [1] (iv) Correct option : (c) Explanation : t = 5 min [R]0 =3 [R]



For first order reaction, [R] 2.303 K log t [R] 2.303 log 3

=



2.303

× 0.4771

0.2197 min

[1]

124 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

OR Correct option : (b) Slope = –k [1] Q. 3. Concentration dependence of rate is called differential rate equation. Integrated differential equations give relation between directly measured experimental data i.e. concentration at different times and rate constant. The integrated rate equations are different for the reactions of different reaction orders. The first-order reaction has a rate constant 1.15 × 10–3s–1. The following questions are multiple choice questions. Choose the most appropriate answer:  (1 × 4 = 4) (i) How long will 5g of this reactant take to reduce to 3g? (a) 222.189 s (b) 444.379 s (c) 111.095 s (d) 888.789 s (ii) When the rate constant has same units as the rate of reaction, the order of the reaction is : (a) Zero order (b) First order (c) Second order (d) Fractional order (iii) Under which condition a bimolecular reaction is kinetically first order reaction: (a) When two reactants are involved. (b) When one of the reactants is in excess. (c) When one of the reactants does not involve in reaction. (d) None of these. (iv) For a reaction, A + H2O → B Rate ∝ [A] The order of the reaction is: (a) Zero order (b) Fractional order (c) Pseudo first order (d) Second order OR The integrated rate equation for rate constant of a first order reaction is : [R] 2.303 log 0 (a) k = t [R]

1é 1 1 ù (b) k = ê ú t ë [R] [R]0 û

1 é 1 1 ù ê 2ú 2t ë [R] [R]02 û (d) None of these Ans. (i) Correct option: (b) Explanation : Initial amount= 5 g Final concentration = 3 g Rate constant= 1.15 × 10–3 s–1 We know that for a First order reaction [R ] 2.303 t= log initial [ R final ] k 2.303 5 log = 1.15 × 10 −3 3 2.303 × 0.2219 = 1.15 × 10 −3 = 444.379 s  (c) k =

[1] (ii) Correct option: (a) E  xplanation : In case of zero order reaction, the rate constant has same units as the rate of reaction. r = k [A]0 r=k Unit of rate = mol L–1s–1 Unit of k = mol L–1s–1 [1] (iii) Correct option: (b) Explanation : When one of the reactants is in excess, a bimolecular reaction is kinetically first order reaction.[1] (iv) Correct option: (c) Explanation : A + H2O → B r ∝ [A] (∵ [H2O] = excess) It is called pseudo first order reaction. [1] OR Correct option: (a) Explanation : A → products For first order reaction, Integrated rate equation is, [R]0 2.303 k=  [1] log t [R] 

[ 125

SELF-ASSESSMENT TEST

Self Assessment Test - 4 Time : 1 Hour 1. Read the passage given below and answer the following questions : (1 × 4 = 4) The formation of product is possible only if the molecules of reactants collide. For the colliding molecules, the total number of collisions which occur among the reacting species per second per unit volume is called collision frequency. All the collisions in a reaction mixture which are effective in bringing about the chemical reaction are called effective collisions. The number of effective collisions is very less as compared to total number of collisions. A certain minimum energy is required for effective collisions between reactant molecules in a chemical reaction. This certain minimum energy is called threshold energy. Only those molecular collisions are effective whose energy is either equal to or more than threshold energy. For effective collision, we have to provide certain extra amount of energy to them, this extra amount of energy is called activation energy. In the chemical reaction, the colliding molecules must have proper orientation at the time of collision so that the old bonds breaks and the new bonds are formed.  In these questions, a statement of Assertion is followed by a statement of Reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and the reason is correct explanation for assertion. (b) Assertion and reason both are correct statements and the reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. (i) Assertion : The formation of product is possible only if the molecules of reactants collide. Reason : In a chemical reaction, chemical bonds in the reactants are broken and new bonds in the products are formed. (ii) Assertion : All the collisions in a reaction mixture are effective in bringing about the chemical reaction. Reason : The collisions which bring about the chemical reaction is called effective collisions. (iii) Assertion : The number of effective collisions is very less than total collisions. Reason : First molecules come in activated state and only those molecules form products whose energy is equal to or more than threshold energy. (iv) Assertion : Greater than concentration of reactants, smaller is the rate of the reaction.

Max. Marks : 25











Reason : The number of collisions between the reacting molecules increases on increasing concentration of reactants. OR Assertion : In some cases, the reactant molecules having equal to or greater than threshold energy do not form products. Reason : They posses proper orientation.

The following question (No. 2 to 5) are Multiple Choice Questions carrying 1 mark each. 2.  Which of the following quantity changes by addition of catalyst in a chemical reaction? (a) Entropy (b) Internal energy (c) Enthalpy (d) Activation energy R 3.  The initial concentration of the reactant becomes twice in a reaction, then half-life period of that reaction is not affected. The order of that reaction is (a) first (b) second (c) more than zero but less than one (d) zero U 4.  The unit of rate constant for the zero order reaction is (a) L s–1 (b) L mol–1s–1 –1 –1 (c) mol L s (d) mol s–1 R 5.  The temperature dependence of the rate constant is given by the Arrhenius equation: E E (a) ln k = ln A − a (b) ln A = ln k − a RT RT Ea − (c) ln k = A (d) ln A = RTln Ea–ln k RT  R

The following questions (No. 6 & 7), a statement of Assertion is followed by a statement of Reason. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not the correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. 6. Assertion : Half-life period is always independent of initial concentration. Reason : Half-life period is inversely proportional to rate constant. 7. Assertion : Order of a reaction cannot be written from the balanced equation of a reaction. Reason: Rate law can provide some clue about the reaction mechanism.

126 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

The following questions (No. 8 & 9), are Short Answer Type-I and carry 2 marks each. 8. The rate of a reaction depends upon the temperature and is quantitatively expressed as : (i) If a graph is plotted between log k and 1/T, write the expression for the slope of the reaction? (ii) If at under different conditions Ea1 and Ea2 are the activation energy of two reactions. If Ea1 = 40J/mol and Ea2 = 80 J/mol. Which of the two has a larger value of the rate constant? 9. (i)  Rate law r = k[A]1/2[B]2 is given for the reaction A + B → P. What will be the order of the reaction? (ii) The rate constant (k) is 5.5 × 10–14s–1 for a first order reaction. Calculate the half-time period of this reaction. Q.No. 10 & 11 are Short Answer Type-II carrying 3 marks each. 10. For the reaction A + B → products, the following initial rates were obtained at various given initial concentrations :



S.No.

[A] mol/L

[B] mol/L

Initial rate M/s

1.

0.1

0.1

0.05

2.

0.2

0.1

0.10

3

0.1

0.2

0.05

Determine the half-life period.

11. A first order reaction is 50% complete in 50 minutes at 300 K and the same reaction is again 50% complete in 25 minutes at 350 K. Calculate activation energy of the reaction. A&E Q.No 12 is a Long Answer Type Question carrying 5 marks. 12. F  or the hydrolysis of methyl acetate in aqueous solution, the following results were obtained :



t/s

0

300

600

[CH3COOCH3] / mol L-1

0.60

0.30

0.15

(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.

(ii) Calculate the average rate of reaction between the time interval 30 lo 60 seconds. 

(Given log 2 = 0.3010, log 4 = 0.6021) OR

(i) For a reaction A + B → P, the rate is given by Rate = k[A] [B]2 (a) How is the rate of reaction affocted if the concentration of B is doubled ? (b)  What is the overall order of reaction if A is present in large excess ? (ii)  A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reactions. (log 2 = 0.3010)

U



 

Finished Solving the Paper ? Time to evaluate yourself !

OR SCAN THE CODE

SCAN

For elaborate Solutions

ll

SURFACE CHEMISTRY

[ 127

5

CHAPTER

Syllabus

SURFACE CHEMISTRY

¾¾ Adsorption - physisorption and chemisorption, factors affecting adsorption of gases on solids, catalysis: homogeneous and heterogeneous, activity and selectivity of solid catalysts; enzyme catalysis, colloidal state: distinction between true solutions, colloids and suspension; lyophilic, lyophobic, multi-molecular and macromolecular colloids; properties of colloids; Tyndall effect, Brownian movement, electrophoresis, coagulation, emulsion - types of emulsions.

Trend Analysis List of Concepts

2018 D/OD

D

2019

1Q (1 mark) 1Q (3 marks) 1Q (3 marks)

1Q (1 mark) 1Q (1 mark) 1Q (3 marks)

Adsorption, absorption Colloids Differences between types of colloids, catalysis and emulsion Miscellaneous type

OD 1Q (3 marks) 1Q (1 mark)

2020 D 1Q (2 marks)

OD

1Q (5 marks) 1Q (3 marks)

TOPIC-1

Adsorption and its Types, Factors Affecting Adsorption

Revision Notes  Surface chemistry : The branch of chemistry which deals with the phenomenon that occurs at the surfaces is called surface TOPIC - 1 chemistry. This phenomenon is studied with the help of Adsorption and its Types, Factors Affecting Adsorption .... P. 128 adsorption and colloidal state.  Adsorption is a surface phenomenon in which the substance TOPIC - 2 gets accumulated on the surface of a solid rather than in the Catalysis and its Types, Enzyme Catalysis bulk of a solid or liquid. The surface that adsorbs is called .... P. 133 adsorbent and the one that gets adsorbed is called an adsorbate. TOPIC - 3 For example : Pt can adsorb large amount of hydrogen gas. Colloids, Types of Colloids, Characteristics and  Types of adsorption : Preparation of Colloids .... P. 137 (i) Physisorption : Physisorption is also called physical adsorption. If the adsorbate is held on a surface of adsorbent by weak van der Waals forces, the adsorption is called physical adsorption or physisorption. (ii) Chemisorption : Chemisorption is also called chemical adsorption. If the forces holding the adsorbate are as strong as in chemical bonds, the adsorption process is known as chemical adsorption or chemisorption.

[ 129







SURFACE CHEMISTRY

 





Desorption : The process of removing an adsorbed substance from a surface on which it is adsorbed is called desorption. In absorption, molecules of a substance are uniformly distributed throughout the body of solid or liquid. For example : Ammonia absorbed by water, water absorbed by anhydrous CaCl2. Sorption : When adsorption and absorption take place simultaneously, it is called sorption, e.g., dying of cotton fabrics. The dye is adsorbed initially on the surface of cotton fibre, but later the fibre has uniform concentration of dye throughout. Enthalpy of adsorption : Adsorption generally occurs with release in energy, i.e., it is exothermic in nature. The enthalpy change for the adsorption of one mole of an adsorbate on the surface of adsorbent is called enthalpy or heat of adsorption. Differences between Adsorption and Absorption : S. No.

Adsorption

Absorption

(i)

It is a surface phenomenon. Adsorbate molecules Absorption occurs in the bulk of absorbing are held at the surface of adsorbent. substance.

(ii)

The concentration of the adsorbate at the Absorbed material is uniformly distributed adsorbent is much more than that in the bulk. throughout the bulk. Thus, concentration is same throughout.

(iii)

Initially rate of adsorption is rapid. It decreases Absorption occurs with uniform rate. slowly till equilibrium is attained.

(iv)

Example : Water vapours are adsorbed on silica Example : Water vapours are absorbed by anhygel. drous CaCl2.



 Factors affecting adsorption of gases on solids : (i) Nature of gas (ii) Nature of adsorbent (iii) Specific area of the solid (iv) Pressure of the gas (v) Effect of temperature (vi) Activation of adsorbent  Adsorption isobar : A plot of extent of adsorption (x/m) vs. temperature at constant pressure is called adsorption isobar.  Adsorption isotherm : The plot of extent of adsorption (x/m) vs. pressure (p) at constant temperature is called adsorption isotherm, where, ‘x’ is the quantity of gas adsorbed by unit mass ‘m’ of the solid ‘adsorbent’. It is called adsorption isotherm.  Freundlich Adsorption Isotherm : It gives the relationship between magnitude of adsorption (x/m) and pressure at a constant temperature. It can be given by mathematical equation. x = kp1/n ....(i) m

log

x 1 = log k + log p ...(ii) m n

x 1 A plot of log vs. log p gives a straight line with slope m n and y intercept = log k.

In case of solution, the isotherm takes the form x x 1 = k(C)1/n ; log = log k + log C m m n

where x is the amount of adsorbate adsorbed on m gram of adsorbent at pressure p or concentration C of the adsorbate, k and n are constants, n > 1. Scan to know  Applications of Adsorption : more about this topic (i) In removing colouring matter from solution (vii) In curing diseases (ii) In gas masks (viii) In dehumidifiers (iii) In separating noble gases (ix) In adsorption analysis (iv) In dyeing of cloth (x) In creating high vacuum Freundlich”s (v) In chromatography (xi) In ion-exchange resins Adsorption Isotherm (vi) In froth flotation process (xii) Heterogeneous catalysis

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Know the Terms  Saturation pressure : The pressure beyond which adsorption becomes independent of pressure is called as saturation pressure.  Competitive adsorption : A strong adsorbate gets adsorbed more efficiently as compared to weak adsorbate. That strong adsorbate can displace already adsorbed one from the surface of the adsorbent. This is known as competitive adsorption or preferential adsorption.

Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS : Q. 1. Which one of the following is not applicable to the phenomenon of adsorption U (a) ΔH > 0 (b) ΔG < 0 (c) ΔS < 0 (d) ΔH < 0 Ans. Correct option : (a) Explanation : Adsorption is an exothermic process, so the ΔH of adsorption is always negative. ΔH < 0 ΔG = ΔH – TΔS ΔG = Change in Gibbs free energy ΔH = Change in enthalpy T = Temperature in Kelvin ΔG = Change in entropy Since, adsorption is a spontaneous process, the thermodynamic requirement is at constant temperature and pressure, ΔG must be negative. So the enthalpy ΔH as well as entropy ΔS of the system is negative. Q. 2. On the basis of data given below predict which of the following gases shows least adsorption on a definite amount of charcoal? Gas CO2 SO2 CH4 H2 Critical temp./K 304 630 190 33  (a) CO2 (b) SO2 (c) CH4 (d) H2 Ans. Correct option : (d) Explanation : Higher critical temperatures indicate easily liquefiable gases which are readily adsorbed as the van der Waal’s forces responsible for adsorption of gases on solid surfaces are stronger near critical temperatures. Q. 3. Which of the following is an example of absorption? U (a) Water on silica gel (b) Water on calcium chloride (c) Hydrogen on highly divided nickel (d) Oxygen on metal surface Ans. Correct option : (b) Explanation : Water on calcium chloride is an example of absorption, rest of all are examples of adsorption. [B] ASSERTIONS AND REASONS:

In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

(1 mark each)

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.



(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.



(c) Assertion is correct statement but reason is wrong statement.



(d) Assertion is wrong statement but reason is correct statement. Q. 1. Assertion : Extent of adsorption of adsorbate from solution phase increases with increase in amount of adsorbate in solution. Reason : Freundlich’s equation describes the behaviour of adsorption from a solution. Ans. Correct option : (a) Explanation : Freundlich’s equation describes the behaviour of adsorption from a solution. 1 x = kC n m

It also written as x 1 log = log k + log C m n

x = mass of adsorbate m = mass of adsorbent p = equilibrium pressure of the gaseous adsorbate in case of experiments made in gas phase (gas/solid interaction with gaseous species/adsorbed species). C = equilibrium constant of adsorbate in case of experiments made with an aqueous solution in contact with dispersed solid phase (dissolved species/adsorbed species). k and n are constants of a given adsorbate and adsorbent at a given temperature (from there, the item isotherm needed to avoid significant gas pressure fluctuations due to uncontrolled temperature variations in the case of adsorption experiments of a gas onto a solid phase). Q. 2. Assertion : At the equilibrium position in the process of adsorption ∆H = T∆S. Reason : Adsorption is accompanied by decrease in surface energy. Ans. Correct option : (b) Explanation : ∆G = ∆H – T∆S Gibb’s Free energy equation Adsorption is a spontaneous process so ∆G is negative. At equilibrium it becomes zero and ∆H = T∆S is attained.

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Ans. Both are surface phenomenon / both increase with increase in surface area (or any other correct similarity). [CBSE Marking Scheme, 2017]

Q. 2. Why is chemisorption referred to as activated adsorption? U Ans. Chemisorption involves formation of bonds between the solid surface (adsorbent) and gaseous atoms/molecules (reactants). Since formation of chemical bonds requires high activation energy, it is known as activated adsorption. Q. 3. Freundlich adsorption isotherm is given by the expression x/m = kp1/n what conclusions can be drawn from this expression? Ans. Following conclusions can be drawn from the equation: (a) When 1/ n = 0, the adsorption is independent of pressure. (b) When n = 0, x/ m vs p graph is a line parallel to x-axis. [½+ ½] (or any other correct conclusion)

Q. 3. Assertion : Extent of physisorption of a gas increases with decrease in temperature. Reason : It is due to decrease in strength of van der Waals forces. Ans. Correct option : (c) Explanation : In physisorption, phenomenon particles are held to the surface by weak van der Waals’ force of attraction which get destroyed due to increase in temperature. [C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1. Write one similarity between Physiorption and Chemisorption. R [CBSE D/OD 2017]

Short Answer Type Questions-I

Ans.

A graph drawn between extent of adsorption and the pressure of the gas at constant temperature is called adsorption isotherm. [1]





 relationship between the amount adsorbed A (x/m) and the equilibrium pressure (p) can be given by

x = kp1 / n , m           where, n is a positive integer and k is the constant. This is known as Freundlich adsorption isotherm.[1]

[1]



Q. 1. Define adsorption with an example. What is the role of adsorption in heterogeneous catalysis? Ans. The phenomenon of higher concentration of the molecular species on the surface of the solid in the bulk is called adsorption. e.g. Activated charcoal adsorbs a number of gases like ammonia, phosgene etc. In heterogeneous catalysis, reactants are gaseous and the catalyst is solid. The concentration of the reactant molecules on the surface of catalyst increases and hence the rate of reaction increases.[2] Q. 2. Define Brownian movement. What is the cause of Brownian movement in colloidal particles? How is it responsible for the stability of Colloidal sol? R [CBSE Delhi Set-1 2020] Brownian movement : The particles of the dispersed phase of colloidal solution execute a continuous zig-zag motion. This phenomenon is called Brownian movement. Cause of Brownian movement : The dispersed phase particles of the colloidal solution are constantly collided by the molecules of the dispersion medium. Brownian movement opposes the force of gravity on the colloidal particles. So these particles always remain in a state of motion and do not settle. [2] Q. 3. What do you understand by an adsorption isotherm? Give a brief description of Freundlich’s isotherm. R

(2 marks each)

Commonly Made Error  Students often plot the graph and do not write the variables.

Answering Tip  While plotting a graph, both the independent and dependent variables must be written. Q.4.

Ans.

With the help of an example, explain how physisorption changes to chemisorption with rise in temperature. Give the reason for change. U  Adsorption of hydrogen gas on finely divided nickel is physisorption at low temperatures as it involves weak van der Waals forces. With the rise in temperature, hydrogen molecules dissociate to form hydrogen atoms which are held on the surface by chemisorption. [2]

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Short Answer Type Questions-II Q. 1. Write any three differences between physisorption and chemisorption.[CBSE OD 2015] OR Write three differences between Physisorption and Chemisorption. [CBSE Foreign Set-1, 2 2017] Ans. S. No. (i)

Physisorption Chemisorption Due to van der Waals Due to chemical bond forces formation

(ii)

Reversible

(iii)

Enthalpy of adsorption Enthalpy of adsorption is low (20-40 kJ/mol. is high (80-240 kJ/mol.



x 1 = log k + log p m n At high pressure x µ p ° m

(c) log

Irreversible

(or any other correct difference)

[1+1+1] [CBSE Marking Scheme 2017]

Detailed Answer: S. No. Physisorption (i) It arises by weak van der Waals forces. (ii) It usually takes place at a low temperature and decreases with increasing temperature.

Chemisorption It arises by strong forces like chemical bond. It takes place at a high temperature and increases with increase in temperature.

(iii)

It is reversible.

It is irreversible.

(iv)

It depends on the ease of liquification of the gas.

The extent of adsorption is not related to liquification of the gas.

(v)

It is not very specific.

It is highly specific.

(vi)

It forms multimolecular layers.

It forms monomolecular layers.

(vii)

It does not require any It requires activation activation energy. energy. (Any three) [3] Q. 2. (a) Write the dispersed phase and dispersion medium of milk.   (b) Why is adsorption exothermic in nature?   (c) Write Freundlich adsorption isotherm for gases at high pressure. U [CBSE OD Set-2 2019]

Ans. (a) Dispersed phase = liquid; Dispersion medium = liquid   (b) Due to the formation of new bonds / force of attraction between adsorbate and adsorbent. x   (c) = kp0 = k[1 × 3 = 3] m [CBSE Marking Scheme, 2019] Detailed Answer : (a) Dispersed phase in milk - Liquid/oil Dispersion medium in milk - Liquid/water. (b)  Because adsorption leads to a decrease in the residual forces on the surface of the adsorbent. This causes a decrease in the surface energy of the adsorbent. Therefore, adsorption is always exothermic.

Commonly Made Error

[3] 





 Sometimes, students do not write correct equation for Freundlich adsorption isotherm at high pressure. They often use concentration term in place of pressure.

Answering Tip  Learn and isotherm.

understand

Freundlich

adsorption

Q. 3. (a) Write the dispersed phase and dispersion medium of butter. (b) Why does physisorption decrease with increase in temperature? (c) A colloidal sol is prepared by the method given in the figure. What is the charge on AgI colloidal particles formed in the test tube? How is this sol represented? AgNO3 solution

KI solution 



(3 marks each)

U [CBSE OD Set-3 2019]

Ans. (a) Dispersed phase = Liquid   Dispersion medium = Solid/liquid [1]   (b) Because physisorption is exothermic in nature. / With rise in temperature, desorption starts.[1]   (c) Negatively charged , AgI/I– [½ + ½] [CBSE Marking Scheme, 2019] Detailed Answer : (a) Dispersed phase : Liquid Dispersion medium : Solid (b) Because it is an exothermic reaction. (c) When AgNO3 solution is added to KI solution, negatively charged sol of AgI is formed due to elective adsorption of I– ion from the dispersion medium. It is represented as given below. AgI/I– Negatively charged. [3]

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TOPIC-2

Catalysis and its Types, Enzyme catalysis

Revision Notes  Catalysis : The process in which catalyst is used to increase the rate of reaction without changing itself is called catalysis. MnO

2  → 2KCl + 3O2 2KClO3 473 − 633 K



The catalyst remains unchanged with respect to mass and composition. Catalyst does not affect ∆H, ∆S, ∆G and equilibrium constant k.

 Promoters : Those substances which increase the activity of catalyst are called promoters. e.g., Mo is promoter whereas Fe is catalyst in Haber’s process. Fe   N2(g) + 3H2(g)    2NH3(g) Mo

 Poisons or Inhibitors : Those substances which decrease the activity of catalyst are called catalytic poisons or inhibitors, e.g., arsenic acts as catalytic poison in the manufacture of sulphuric acid by contact process.  Types of Catalysis : (i) Homogeneous Catalysis : When the catalyst mixes homogeneously with the reactant and forms a single phase, the catalyst is said to be homogeneous and this kind of catalysis is known as homogeneous catalysis. e.g., catalytic oxidation of SO2 to SO3 in presence of NO is an example of homogeneous catalysis. (g) 2SO2(g) + O2(g) NO  → 2SO3(g) (ii) Heterogeneous Catalysis : When the catalyst forms a separate phase (usually a solid phase), it is said to be heterogeneous catalysis. In heterogeneous catalysis, the reactants are gases and reaction starts from the surface of the solid catalyst. This is the reason why heterogeneous catalysis is also called ‘surface catalysis’. e.g.



Cu (Solid) CO(g) + 2H2(g) ZnO  → CH3OH(l) / Cr O 2

3

 Adsorption theory of Heterogeneous Catalysis : According to modern adsorption theory, there are free valencies on the surface of solid catalyst and mechanism involves following five steps : (i) Diffusion of reactant molecules on the surface of the catalyst. (ii) Adsorption of the reactant molecules on the surface of the catalyst by forming loose bonds with the catalyst due to presence of free valencies. (iii) Occurrence of a chemical reaction forming an intermediate on the surface. (iv) Desorption of the product molecules from the surface.

(v) Diffusion of product molecules away from the surface of the catalyst.

Fig. 1: Adsorption of reacting molecules, formation of intermediate and desorption of products.  Important features of solid catalysts : (i) Activity : It is the ability of catalyst to increase the rate of a reaction. (ii) Selectivity : It is the ability of a catalyst to direct a reaction to yield a particular product.







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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Shape-selective catalysis by zeolites : It is the catalytic reaction that depends upon the pore structure of the catalyst and the size of the reactant and product molecules. Zeolites are the shape selective catalyst having honey comb structures. Zeolite catalyst, ZSM-5 is used in petroleum industry to convert alcohols into gasoline by dehydration.

 Enzyme Catalysis : In enzyme catalysis, enzymes are biological catalysts which catalyse specific biochemical reactions. They are globular proteins having high molecular mass. e.g., NH2CONH2 + H2O Urease  → 2NH3 + CO2 (Urea)  Characteristics of enzymes : (i) Enzymes form a colloidal solution in water and hence they are very active catalysts. (ii) Like inorganic catalyst they cannot disturb the final state of equilibrium of a reversible reaction. (iii) They are highly specific in nature i.e., one catalyst cannot catalyse more than one reaction. (iv) They are highly specific to temperature. The optimum temperature of their activity is 25°C – 35°C. They are deactivated at 70°C. (v) A small quantity of enzyme catalyst is sufficient for a large change. (vi) They are destroyed by UV rays. (vii) Their efficiency is decreased in the presence of electrolytes.  Some examples of Enzyme catalyzed reactions.





(i) C12H22O11 Invertase → C6H12O6 + C6H12O6 (Inversion of cane sugar) (Cane sugar) (Glucose) (Fructose)



(ii) Maltose Maltase  → Glucose + Glucose



Lactase → Glucose + Galactose (iii) Lactose 



Zymase → 2C2H5OH + 2CO2 (iv) C6H12O6  (Glucose) (Ethyl alcohol)



→ Peptides (v) Proteins 

Pepsin

Lactobacilli

→ Curd (vi) Milk  enzyme  Mechanism of Enzyme catalyzed reaction : (i) Binding of enzyme to substrate to form an activated complex. E + S → ES (ii) Decomposition of the activated complex to form the product. ES → E + P where E is enzyme, S is substrate and P is product.  Co-enzymes : Certain substances, which can increase the activity of enzymes are known as Co-enzymes.

Know the Terms





Zeolites : Aluminosilicate with three dimensional network containing Al—O—Si network which acts as a good shape-selective catalyst.  Enzymes : Complex nitrogenous organic compounds which are produced by living plants and animals.  Optimum pH : Particular pH at which the rate of an enzyme catalysed reaction is maximum.

Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS : Q. 1. In which of the following reactions, heterogeneous catalysis is involved ? (i) 2SO2(g) + O2 (g) → 2SO3 (g) NO ( g )



(ii) 2SO2 (g)  → 2SO3 (g) Pt ( s )

( s) (iii) N2 (g) + 3H2 (g) Fe → 2NH3 (g)

(1 mark each)

(I) (iv) CH3COOCH3 (l) + H2O (l) HCI  → CH3COOH (aq) + CH3OH (aq)

(a) (c)  Ans.

(ii), (iii) (i), (ii), (iii)

(b) (ii), (iii), (iv) (d) (iv) A

Correct option : (a) Explanation : In reactions (ii) and (iii), catalysts are in solid state, and reactants and products are in gaseous states.

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Q. 2. Which of the following statements are correct about solid catalyst ? (a) Same reactants may give products by using different catalysis. (b) Catalyst is required in large quantities to catalyze reactions. (c) Catalytic activity of a solid catalyst does not depend on the strength of chemisorption. (d) None of the above Ans. Correct option : (a) Explanation : Same reactants may give different products by using different catalysts. e.g., CO(g) + 3H2(g) Ni → CH4(g) + H2O(g). Cu CO(g) + H2(g) → HCHO(g) Q. 3. In a reaction, catalyst changes (a) No change (b) qualitatively (c) Chemically (d) quantitatively Ans. Correct option : (b) Explanation : Catalyst changes qualitatively in a reaction. Since it does not participate in reaction, it does not change chemically, quantitatively. [B] ASSERTION AND REASON TYPE QUESTIONS : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Q. 1. Assertion : Zeolites are good shape - selective catalysts. Reason : Zeolites have honeycomb like structures. Ans. Correct option : (a) Explanation : Zeolites are microporous with honeycomb like structures and are good shapeselective catalysts. Q. 2. Assertion : Inhibitors or poisons completely destroy the catalytic activity of the enzymes.

Reason : Inhibitors or poisons interact with the active functional groups on the enzyme surface. Ans. Correct option : (a) Explanation : Inhibitors or poisons completely destroy the catalytic activity of the enzymes by interacting with the active functional groups on the enzyme surface. Q. 3. Assertion : The gas molecules are held up on the catalyst’s surface by loose chemical combination. Reason: Catalyst remains unchanged at the end of reaction. Ans. Correct option : (b) Explanation : The gas molecules are held up on the catalyst’s surface by loose chemical combination. Different molecules are adsorbed side by side, reacting with each other, resulting in formation of new products. [C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1. What is meant by ‘shape selective’ catalysis ?  R Ans. Shape selective catalysis is a chemical reaction in which the rate depends on the pore size of the catalyst, and also on the shape and size of the reactant and product molecules. Zeolite acts as shape selective catalyst. Q. 2. What is the role of desorption in the process of catalysis? R [CBSE Foreign Set-1, 2, 3 2017] Ans. To make the surface available again for more reaction to occur/To remove the product formed from the surface of the catalyst. [CBSE Marking Scheme 2017] Q. 3. CO(g) and H2(g) react to give different products in the presence of different catalysts. Which ability of the catalyst is shown by these reactions ? Ans.

[Topper’s Answer 2018]

Short Answer Type Questions-I Q. 1. Define the following terms: (i) Desorption (ii) Critical micelle concentration (iii) Shape selective catalysis R [CBSE Comptt. Delhi Set-1 2017] Ans. (i)  The process of removing an adsorbed substance from a surface on which it is adsorbed. [1] (ii) The formation of micelles takes place only above a particular concentration called CMC.  [1]



(2 marks each)

(iii) The catalytic reaction that depends upon the pore structure of the catalyst and size of the reactant and product molecules. [1]  [CBSE Marking Scheme 2017]

Commonly Made Error  Students often write irrelevant information thus losing time. Concise answer saves time.

Answering Tip  As it is a 3 marks question, just write the definition of each.

136 ]



Q. 2. (i) Write the expression for Freundlich’s equation to describe the behaviour of absorption from solution.  (ii) What causes charge on sol particles? (iii) Name the promoter used in the Haber’s process for the manufacture of ammonia. U Ans. (i)

x 1 = kC n m 

[1]

  (ii) The charge on the sol particles is due to  (a)  Electron capture by sol particles during electrodispersion.  (b)  Preferential adsorption of ions from [1] solution.    (c) Formulation of electrical double layer.  (Any one reason)   (iii) Molybdenum acts as a promoter for iron. [1]

Q. 3. Write one difference in each of the following:   (i) Lyophobic sol and lyophilic sol (ii) Solution and colloid (iii) Homogeneous catalysis and heterogeneous catalysis U [CBSE Delhi Set-1, 2017] Ans. (i) Lyophobic sols are liquid (dispersion medium)hating and lyophillic sols are liquid (dispersion medium)-loving colloids.[1] (ii)  Solution is a homogeneous mixture while colloid is heterogeneous mixture/does not show Tyndall effect – shows Tyndall effect.[1] (iii) Homogeneous catalysis : Reactants and catalyst are in same phase.  Heterogeneous catalysis : Reactants and catalyst are not in same phase.[1] (or any other correct difference) [CBSE Marking Scheme 2017]

Detailed Answer : (i) Lyophobic sol

These are unstable These are quite stable and can be easily and cannot be easily coagulated. coagulated.

 (ii)

Solution

Colloid

These are homogeneous. These are heterogeneous. In the solutions, the In the colloid, the particles are ions and particles are either molecules. single macromolecules or aggregates of molecules or atoms.

 (iii)

Homogeneous catalysis Heterogeneous catalysis The catalyst dissolves in The catalyst does not the reaction medium. dissolve in the reaction medium. The catalyst and the The catalyst and the reactants constitute a reactants are in different single phase. phases.



Commonly Made Error  Students write more than one point of difference, which is not required, as each part carries just one mark.

Answering Tip  Write short, concise one difference only as each subquestion carries only 1 mark.

Long Answer Type Questions Q.1. Describe the role of adsorption in heterogenous catalysis giving the step wise mechanism.  R Ans: Reactants which are in the gaseous state or in solution phase are adsorbed on the surface of the solid catalyst and the chemical reaction occurs on the surface of catalyst. The mechanism of catalysis involves the following steps: (i) Diffusion of reactants to the surface of the catalyst. (ii) Adsorption of reactants on the surface of the solid catalyst. (iii) Chemical reaction on the catalyst’s surface  through formation of an intermediate. (iv) Desorption of products from the catalyst surface thus making it available again for reaction. (v) Diffusion of reaction products away from the catalyst’s surface.[5]

Lyophilic sol

These are also called These are also called irreversible sols. reversible sols.





Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q.2. Ans:

 

 

(5 marks each) What do you understand by shape selective catalysis? Why are zeolites good shape selective catalysts? List an important use of zeolites. U The catalytic reaction which depends on the pore structure of catalyst and the size of reactant and product molecules is called shape-selective catalysis. Zeolites have honey comb like structures. Their pores and cavities provide sites for reaction to reactant molecules depending on their size and shape. Zeolites are microporous aluminosilicates with three- dimensional network of silicates where some Si atoms are replaced by Al atoms giving Al-O-Si frame. Zeolites are used as catalysts in petrochemical industries for cracking of hydrocarbons and isomerisation. [5]

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Q.3. Ans:

List five characteristics of enzyme catalysis. R Characteristics of enzyme catalysis are as follows: (i) It occurs with high efficiency. One enzyme  molecule transforms millions of reactant molecules per minute. (ii) It is highly specific in nature. Each enzyme can catalyse only a specific reaction.

(iii) It is highly active under optimum temperature and pH. The optimum temperature is between 298-310K and optimum pH is between 5-7. (iv) Activators and co-enzymes increase the activity of the biocatalytic enzymes. (v) Inhibitors and poisons interact with the active functional groups of the enzymes and reduce or destroy the catalytic activity.[5]

TOPIC-3

Colloids, Types of Colloids, Characteristics and Preparation of Colloids

Revision Notes  Colloids : A colloid is a heterogeneous system in which one substance is dispersed (dispersed phase) as very fine particles in another substance called dispersion medium. The size of colloidal particles is in the range 1–1000 nm.  Colloidal solution : A colloidal solution is a heterogeneous system in which a definite substance is distributed in the form of very small particles as dispersed phase in another substance called dispersion medium, e.g., glue, ink, smoke, etc.  Dispersed phase : Dispersed phase is the component present in small proportion like solute in the solution.  Dispersion medium : The medium in which the colloidal particles are dispersed is called dispersion medium.  Crystalloids : The substances whose aqueous solution can pass through the semi-permeable membrane are called crystalloids.  Types of Colloidal solutions : S. No.

Dispersed phase

Dispersion medium

Name

Examples

1.

Solid

Gas

Aerosol

Smoke, dust particles.

2.

Solid

Liquid

Sol

As2S3, Gold sol, starch, gum, muddy water.

3.

Solid

Solid

Solid sol

Coloured gem stones, alloys, pearls, ruby glass.

4.

Liquid

Solid

Gel

Jellies, cheese, butter.

5.

Liquid

Liquid

Emulsion

Milk, hair cream, cod-liver oil.

6.

Liquid

Gas

Aerosol

Fog, mist, cloud.

7.

Gas

Solid

Solid foam

Pumice stone, foam rubber, cork.

8.

Gas

Liquid

Foam

Whipped lather.

cream,

froth,

some

soap

 Classification based on nature of interaction between dispersed phase and dispersion medium : (i) Lyophilic colloids : Lyophilic means ‘‘solvent loving’’. Those substances which when mixed with a suitable solvent as the dispersion medium directly form the colloidal solution are called lyophilic substances and the solution thus formed is called lyophilic solution. They are also known as intrinsic colloids. For example : gum, gelatin, starch, rubber, etc. (ii) Lyophobic colloids : Lyophobic means ‘‘solvent hating’’. These are the substances, when mixed with dispersion medium do not form colloidal solution. Their solution can be prepared only by special method, such substances are called lyophobic and the solutions formed by them are called lyophobic solutions. They are also known as extrinsic colloids. For example : metals and their sulphides. Scan to know more about  Classification based on the type of particles of the dispersal phase : Multimolecular, this topic Macromolecular and Associated colloids : (i) Multimolecular Colloids : Multimolecular colloids contain dispersed particles less than 1 nm made of aggregates of many molecules. These are lyophobic colloids. In multimolecular colloids, particles are held together by weak van der Waals forces. For example, sulphur sol, Multi, Macro gold sol etc. and Associated colloids

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(ii) Macromolecular Colloids : Macromolecular colloids are molecularly dissolved solutions of a polymer with particle size of colloidal range and are lyophilic colloids. In macromolecular colloids, particles are held by chemical bonds. For example : (a) Naturally occurring macromolecular colloids (starch, cellulose, proteins and enzymes). (b) Man-made macromolecular colloids (polythene, nylon). (iii) Associated Colloids (Micelles) Micelles : Those colloids which behave as normal strong electrolytes at low concentrations, show colloidal properties at higher concentrations due to the formation of aggregated particles of colloidal dimensions. Such compounds are also referred to as associated colloids. Surface active agents like soaps and synthetic detergents belong to this class. They also form ions. Micelles may contain 100 molecules or more. Mechanism of micelle formation : Micelles are generally formed by the specific type of molecules which have lyophilic as well as lyophobic ends. Such molecules are known as surface active molecules or surfactant molecules. Sodium oleate, C17H33COO–Na+ (one of the soaps) is a typical example of such type of molecule. The long hydrocarbon part of oleate radical (C17H33–) is lyophobic end while COO– part is lyophilic end. When the concentration of the solution is below its CMC (3 × 10–3 mol L–1), sodium oleate behaves as normal electrolyte and ionises to give Na+ and C17H33COO– ions. As the concentration exceeds CMC, the lyophobic part starts preceding away from the solvent and are made to approach each other. However, the polar — COO– part tend to interact with the solvent. This ultimately leads to the formation of the cluster having the dimensions of the colloidal particles. In each such cluster, a large number of (usually 100 or more) oleate groups are clumped together in a spherical manner so that their hydrocarbon parts interact with one another but —COO– parts remain projected in water. Hence, the mechanism of micelles formation is same as that of soap.

 Preparation of colloidal solution : Colloidal solutions can be prepared by following methods :

(i) Mechanical dispersion : A suspension of coarsely ground particles prepared in dispersion medium is fed into a colloidal mill and speed of rotation is adjusted so as to get the particles of colloidal size.



(ii) Electrodispersion : (Bredig’s arc method): In this method, two rods of the metal (Au, Cu, Pt, Ag) to be dispersed are kept immersed in cold and a direct electric arc is struck between them. In this way, vapours of metal are formed which then immediately condense to form particles of colloidal size.

Ice

Electrodes Arc

Water

Bredig’s arc method

(iii) Chemical methods : l Oxidation : Solutions of non-metals are prepared by this method; e.g., colloidal solution of sulphur. HNO ( conc. )

3 → 2S + 2H2O 2H2S + 2(O) 

l Reduction : Metal sols can be prepared by this method; e.g., gold sol 2AuCl3 + 3SnCl2 → 2Au + 3SnCl4

2AuCl3+3HCHO+3H2O → 2Au (sol) + 3HCOOH + 6HCl

l Hydrolysis : Hydroxides sols are prepared by this method; e.g., Fe(OH)3, Al(OH)3 FeCl3 + 3H2O → Fe(OH)3 + 3HCl



l Double decomposition : This method is used to prepare colloids from inorganic salts; e.g.,

As2O3 + 3H2S → As2S3 + 3H2O l Exchange of solvent : Some substances which form true solution in one solvent, forms colloidal solution in another due to lowering of solubility, e.g., sulphur dissolved in alcohol forms colloidal solution in water and phenolphthalein dissolved in alcohol forms colloidal solution in water.

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l Excessive cooling : The colloidal solution of ice in CHCl3 or ether can be obtained by freezing a solution of water in solvent. The molecules of water combine to form particles of colloidal size.  Peptization : The conversion of precipitate into colloidal solution in presence of peptizing agent is called peptization. Peptizing agent is generally an electrolyte.  Purification methods of colloidal solutions : Colloidal solution can be purified by following methods : (i) Dialysis : In dialysis, particles of true solutions can pass through parchment paper or cellophane membrane. On the other hand, sol particles cannot pass through these membranes. A bag made up of such a membrane is filled with the colloidal solution and is then suspended in fresh water. The crystalloid particles pass out leaving behind the colloidal sol.



(a) Arrangement of oleate ions on the surface of water at low concentrations of soap.

(b) Arrangement of oleate ions inside the bulk of water (ionic micelle) at critical micelle concentration of soap.

(ii) Electrodialysis : Movement of ions across the membrane can be quickened by applying electric potential through two electrodes. This method is faster than simple dialysis and is known as electrodialysis. –

Impure sol Distilled water Water + Electrolyte

Cellophane bag (Dialysing membrane) Fig. 3.Electrodialysis Electr odialysis

(iii) Ultrafiltration : The process of separating colloidal particles by specially prepared filter papers whose pore size is reduced by dipping it in colloidal solution (e.g., 4% nitrocellulose in mixture of alcohol and ether) Scan to know  Properties of colloids : more about (i) Brownian movement : The zig-zag movement of colloidal particles when seen under this topic powerful microscope is called Brownian movement.

Scattering of Light



Brownian movement (ii) Tyndall effect : Scattering of light by colloidal particles is called Tyndall effect. Eye

Tyndall cone Light source

Microscope Scattered light Colloidal solution

Colloidal solution

Tyndall effect

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(iii) Electrophoresis : The movement of colloidal particles towards one of the electrodes on passage of electricity through colloidal solution is called electrophoresis.

 Hardy-Schulze rule :

(i) Opposite charged ions are effective for coagulation.



(ii) The coagulating power of electrolyte increases with increase in charge on the ions used for coagulation. e.g., Al3+ > Ba2+ > Na+ for negatively charged colloids.





[Fe(CN)6]4– > PO43– > SO42– > Cl – for positively charged colloids.





The reciprocal of coagulation value is called coagulating power. i.e., lower the coagulation value, higher will be coagulating power.

 Electrokinetic potential or Zeta potential : The potential difference between the fixed layer and diffused layer of opposite charges is called zeta potential.  Coagulation : Coagulation which can be reversed by shaking is called flocculation or coagulation.  Coagulating value : The minimum concentration of an electrolyte which is required to cause the coagulation or flocculation of a sol is known as coagulation value.

Coagulation value ∝



1 Coagulating power

 Gold number : The number of milligram of protective colloids which must be added to 10 ml of given gold sol to prevent it from coagulation by addition of 1 ml of 10% of NaCl solution.  Applications of colloids :

(i) Sewage disposal

(ii) Purification of drinking water



(iii) Smoke precipitation

(iv) Medicines



 (v) Tanning

(vi) Rubber industry.

 Distinction between True solution, Colloids and Suspension : S. No.

True Solution

Colloids

Suspension

1.

It is homogeneous.

It appears to be homogeneous but It is heterogeneous. actually it is heterogeneous.

2.

Its particle size is less than 1 nm. Its particle size is 1 nm to 1000 nm.

3.

It passes through filter paper.

It passes through ordinary filter It does not pass through filter paper. paper but not through ultrafilter.

4.

It does not show Tyndall effect.

It shows Tyndall effect.

5.

It has higher value of colligative It has low value of colligative It has very low value of colligative property. property. property.

6.

Its particles cannot be seen even Its particles can be seen by powerful Its particles can be seen even with with powerful microscope. microscope due to scattering of naked eye. light.

Its particles are larger than 1000 nm.

It does not show Tyndall effect appreciably.

Know the Terms  Crystalloids : The substances whose aqueous solution can pass through the semipermeable membrane.  CMC : Critical Micelle Concentration is the concentration of surfactant above which micelle formation takes place.  Streaming potential : When a liquid is forced through a porous material or a capillary tube, a potential difference is set up between the two sides. This is known as streaming potential.  Dorn potential : When a particle is forced to move through a resting liquid, a potential difference is set up. It is known as Dorn potential also called sedimentation potential.  U-number : The number of milligrams of a hydrophilic sol which is sufficient to produce the colour change from red to blue in 1 cc of gold sol.

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Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS : Q. 1. Which of the following is most effective in coagulating negatively charged hydrated ferric oxide sol? (i) NaNO3 (ii) MgSO4 (iii) AlCl3 U [CBSE Comptt. Delhi Set-1 2017] Ans. AlCl3/Al3+.

[CBSE Marking Scheme 2017]

Q. 2. Which of the following is most effective in coagulating positively charged hydrated ferric oxide sol? (i) NaNO3 (ii) Na2SO4 (iii) (NH4)3PO4 U [CBSE Comptt. Delhi Set-2 2017] Ans. (NH4)3PO4/PO43–.



[CBSE Marking Scheme 2017]

Q. 3.  Which of the following is most effective in coagulating positively charged methylene blue sol? (i) Na3PO4 (ii) K4 [Fe(CN)6] (iii) Na2SO4 U [CBSE Comptt. Delhi Set-3 2017]

Ans. K4[Fe(CN)6]/[Fe(CN)6]4–. [CBSE Marking Scheme 2017] [B] ASSERTIONS AND REASONS: In the following questions a statement of Assertion followed by a statement of Reason is given. Choose the correct answer out of the following choices.

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.



(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.



(c) Assertion is correct statement but reason is wrong statement.



(d) Assertion is wrong statement but reason is correct statement. Q. 1. Assertion : An ordinary filter paper impregnated with collodion solution stops the flow of colloidal particles. Reason : Pore size of the filter paper becomes more than the size of colloidal particle. U Ans. Correct option: (c) Explanation: Pore size of the filter paper becomes less than the size of colloidal particles hence colloidal particles do not flow through it. Q. 2. Assertion : Colloidal solutions do not show Brownian motion. Reason : Brownian motion is responsible for stability of sols.  U

(1 mark each) Ans. Correct option: (d) Explanation: Colloidal particles show Brownian movement and it is responsible for the stability of colloidal solution. Q. 3. Assertion : Detergents with low CMC are more economical to use. Reason : Cleansing action of detergents involves the formation of micelles. These are formed when the concentration of detergents becomes equal to CMC. Ans. Correct option: (a) Explanation: Cleansing of clothes takes place by micelles and their formation starts at CMC. The lesser is the CMC, the better and more economical is the detergent. Q. 4. Assertion : Coagulation power of Al3+ is more than Na+. Reason : Greater the valency of the flocculating ion added, greater is its power to cause precipitation (Hardy-Schulze rule). Ans. Correct option: (a) Explanation: According to Hardy-Schulze law, the greater is the valency of the coagulating ion, the more is the power to coagulate the colloidal solution. Thus, coagulation power of Al3+ is greater than that of Na+. Q. 5. Assertion : Colloidal solutions show colligative properties. Reason : Colloidal particles are large in size. Ans. Correct option: (b) Explanation: Colloidal particles are large in size and hence the number of particles is lesser than the true solution. The lesser number of particles results in lower colligative properties. Q. 6. Assertion : Colloids are stable. Reason : Brownian movement has a stirring effect, which does not allow the particles to settle Ans. Correct option: (a) Explanation: Colloids are stable due to Brownian movement which is the movement of colloidal particles striking against the dispersion medium. It prevents them from settling down. [C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1. Why are medicines more effective in colloidal state? OR What is the difference between an emulsion and a gel? R

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

[Topper’s Answer 2018] Detailed Answer :

Q. 3. What type of colloid is formed when a solid is dispersed in a gas? Give an example.

Due to large surface area these are easily assimilated or adsorbed. [CBSE Marking Scheme, 2019]

R [CBSE OD Set-3 2019]

Ans. Aerosol, Example- Smoke (Or any other correct example). [CBSE Marking Scheme, 2019] ½ + ½

OR Emulsion – both dispersed phase and dispersion medium are liquid Gel- Dispersed phase is liquid while dispersion medium is solid. [CBSE Marking Scheme, 2019]

Q. 4. Write the main reason for the stability of colloidal sols. U [CBSE D/OD 2016] Ans. Like charged particles cause repulsion/Brownian motion/solvation. 

Q. 2. What type of colloid is formed when a liquid is dispersed in a sol? Give an example. R [CBSE OD Set-2 2019]

Ans. Gel, Example- Cheese / butter. [CBSE Marking Scheme, 2019]

[CBSE Marking Scheme, 2016] Detailed Answer : The stability of colloidal sols is due to the presence of traces of electrolyte.

Short Answer Type Questions-I Q.1. How does the precipitation of colloidal smoke take place in Cottrell precipitator? R Ans. In Cottrell precipitator, charged smoke particles are passed through a chamber containing plates with charge opposite to the smoke particles. Smoke particles lose their charge on the plates and get precipitated. [2] Q.2. Explain how do the vital functions of the body like digestion get affected during fever. A&E Ans. The rate of an enzyme reaction is maximum at a particular temperature. On either side of the optimum temperature, the enzyme activity decreases. The optimum temperature range for enzymatic activity is 298-310 K. Normal human body temperature being 310 K is suited for enzymecatalysed reactions. If a person is suffering from

fever, the temperature will be over 310 K. This will adversely affect the enzymatic reactions. [2] Q.3. Explain the following: (i) Same substance can act both as colloids and crystalloids. (ii) Artificial rain is caused by spraying salt over clouds. A&E Ans. (i) T he nature of the substance whether colloid or crystalloid depends upon size of the solute particles. When the size of solute particles lies between 1 to 1000nm, it behaves as a colloid.[1] (ii) The colloid water particles of the clouds get neutralized and coagulated to bigger water drops by spraying salt over clouds and as a result artificial rain is caused.[1]



Short Answer Type Questions-II

Q. 1. Write three differences between lyophobic sol and lyophilic sol. Ans. Differences between lyophobic and lyophilic sol : S. No 1.

Lyophobic sol These are irreversible.

Lyophilic sol These are reversible.

(2 marks each)

(3 marks each)

2.

They exhibit Tyndall effect.

They do not exhibit Tyndall effect.

3.

These are unstable and hence require traces of stabilizers.

These are self stabilised.

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AgNO3 solution U [CBSE OD Set-3 2019]

Ans. (a) Dispersed phase = Liquid; Dispersion medium = Gas[1] (b) Due to weak van der Waal’s forces in physisorption whereas strong chemical forces in chemisorption. [1] (c) Positively charged , AgI/Ag+ [½ + ½] [CBSE Marking Scheme, 2019] Q. 4. Write one difference in each of the following : (a) Multimolecular colloid and associated colloid (b) Coagulation and peptization (c) Homogeneous catalysis and heterogeneous catalysis Associated colloid

Aggregation of large Aggregation of large number of small atoms or number of ions in conmolecules. centrated solutions.

(b) 

[1]

Coagulation

Peptization

Settling down of colloidal particles.

Conversion of precipitate into collodial soil by adding small amount of electrolyte. [1]

Homogenous catalysis

Heterogeneous catalysis

Reactants and catalyst are in same phase.

Reactants and catalyst are in different phases.

Detailed Answer : (a) Multimolecular colloids are formed by the aggregate of a large number of molecules which generally have diameter less than nm. e.g., sols of gold. Associated colloids are formed by aggregation of a large number of ions in concentrated solution. e.g., Soap sol. [1] (b) Peptization is the process of converting a fresh precipitate into colloidal particles by shaking it with the dispersion medium in the presence of a small amount of suitable electrolyte. Coagulation is a process of agitating together the colloidal particles so as to change them into large sized particles which ultimately settle as precipitate. [1] (c) Homogeneous catalysis is a kind of catalysis in which the catalyst is present in the same phase as reactant. e.g., synthesis of sulphur trioxide from sulphur dioxide and oxygen in the presence of nitric oxide as catalyst. Heterogeneous catalysis is a catalysis that involves the catalyst in the different phase from that of reactant. e.g., In the manufacturing ammonia, iron along with traces of K2O and Al2O3 is used as a catalyst. [1] Q. 5. (a) Write the dispersed phase and dispersion medium of milk. (b) Write one similarity between physisorption and chemisorption. (c) Write the chemical method by which Fe(OH)3 sol is prepared from FeCl3. (a) Dispersed phase-liquid, Dispersion mediumliquid. [1]



(b) Both are surface phenomena/both increase with increase in surface area (or any other correct similarity) [1]



(c) Hydrolysis/FeCl3+H2O

Ans. (a) Multimolecular colloid

(c)

[CBSE Marking Scheme, 2017] [1]



Q. 2. Define the following terms : (i) Protective colloid (ii) Zeta potential (iii) Emulsifying agent R [CBSE Delhi Set-1 2020] Ans. (i) Protective colloid : The process by which the lyophobic sols are protected from coagulation by electrolytes due to the previous addition of a lyophilic colloid is called protection and the colloids are called protective colloids. (ii) Zeta potential : It is the potential difference between the dispersion medium and the stationary layer of fluid attached to the dispersed particle. (iii) Emulsifying agent : The substances which are added to stabilize the emulsions are called emulsifiers or emulsifying agents. e.g. lecthin. Q. 3. (a)  Write the dispersed phase and dispersion medium of dust. (b)  Why does physisorption reversible whereas chemisorption is irreversible? (c)  A colloidal sol is prepared by the method given in the figure. What is the charge on AgI colloidal particles formed in the test tube? How is this sol represented? KI Solution



SURFACE CHEMISTRY



hydrolysis

 Fe(OH)3 (sol)+ 3HCl. [CBSE Marking Scheme, 2017] [1]

Detailed Answer : (a) Dispersed phase of milk : Liquid Dispersion medium of milk : Liquid [1] (b) Physisorption and chemisorption depend on the nature of the gas / both are exothermic processes. [1] (c) A red sol of Fe(OH)3 is obtained by adding some ferric chloride to a beaker of boiling water. FeCl3 + 3H2O → Fe(OH)3 + 3HCl  [1]

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q. 6. Write one difference between each of the following : (i) Multimolecular colloid and Macromolecular colloid (ii) Sol and Gel (iii) O/W emulsion and W/O emulsion Ans. (i) Multimolecular colloid : A large number of atoms or smaller molecules of a substance aggregate together to form species having size in the colloidal range. Macromolecular colloid : Large sized molecules whose particle size lies in the colloidal range. [1] (ii) Sol are solid dispersed in liquid while gel are liquid dispersed in solid. [1] (iii) In O/W emulsion, water acts as dispersion medium while in W/O emulsion, oil acts as dispersion medium. [1] [CBSE Marking Scheme, 2017]

Multimolecular colloid Macromolecular colloid These colloids are formed when small particles aggregate to form a particle having size in colloidal range.

These are substances, whose individual molecules have very large size with colloidal dimensions.

Example – Gold sol.

Example – Starch. [1]

(ii) Gel

A colloidal system in which dispersed phase is a solid and dispersion medium is a liquid.

A colloidal system in which dispersed phase is a liquid and dispersion medium is a solid.

Example – Paints.

Example – Cheese. [1]

(iii)



Ans. (i) S. No.

Detailed Answer : (i)

Sol

(ii)  Zeta potential : The potential difference between the fixed layer and the diffused layer of opposite charges is called the electrokinetic potential or zeta potential. [1] (iii) Associated colloids : There are some substances which at low concentrations behave as normal strong electrolytes, but at higher concentrations exhibit colloidal behaviour due to the formation of aggregates. The aggregated particles thus formed are called associated colloids or micelles.  [1] Q. 8. (i) Differentiate between adsorption and absorption. (ii) Out of MgCl2 and AlCl3, which one is more effective in causing coagulation of negatively charged sol and why ? (iii) Out of sulphur sol and proteins, which one forms multimolecular colloids ?  U [CBSE Delhi 2016]

O/W emulsion

W/O emulsion

In this type of emulsion, the dispersed phase is oil and the dispersion medium is water.

In this type of emulsion, the dispersed phase is water and the dispersion medium is oil.

Example – Milk.

Example – Butter and cold creams. [1]

Q. 7. Define the following terms : (i) Lyophilic colloid, (ii) Zeta potential, (iii) Associated colloids. R [CBSE OD 2016] Ans. (i) Lyophilic colloid : Liquid loving colloidal sols directly formed by mixing substances like gum, gelatin, starch, rubber, etc., with a suitable liquid (the dispersion medium) are called lyophilic sols. e.g., muddy water.  [1]

Adsorption

Absorption

(a)

It is a surface phenomenon.

It is a bulk phenomenon.

(b)

The accumulation of molecular species at the surface rather than in the bulk of a solid or liquid is called as adsorption.

The substance is uniformly distributed throughout the bulk of the solid essentially a bulk phenomenon.

[Any one difference]  [1] (ii) AlCl3 is more effective in causing coagulation of negatively charged sol because according to Hardy and Schulze rule, greater the valency of the flocculating ion, greater is its ability to bring coagulation. [1] (iii) Sulphur sol forms multimolecular colloids.[1] [CBSE Marking Scheme 2016] Q. 9. Give reasons for the following observations : (i) Leather gets hardened after tanning. (ii) Lyophilic sol is more stable than lyophobic sol. (iii) It is necessary to remove CO when ammonia is prepared by Haber’s process. A&E [CBSE Delhi 2015] Ans. (i) Mutual coagulation [1] (ii) Strong interaction between dispersed phase and dispersion medium or solvated layer [1] (iii) CO acts as a poison for catalyst [1] [CBSE Marking Scheme 2015] Detailed Answer: (i) Animal hides are colloidal in nature. When a hide which has positively charged particles, is soaked in tannin, which contains negatively charged colloidal particles, mutual coagulation takes place. This results in the hardening of leather. (ii) Lyophilic colloids have great affinity for the dispersion medium i.e., dispersed phase particles are solvated to a greater extent in case of lyophilic colloids. Hence, lyophilic sols are relatively more stable than lyophobic sols.  (iii) It is necessary to remove CO when ammonia is prepared by Haber’s process because it acts like a poison which reduces the activity of catalyst.

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Q. 10. (a)  A colloidal sol is prepared by the method in given figure. What is the charge of AgI colloidal particles in the test tube? How is the sol formed, represented? 3



(b) Explain how the phenomenon of adsorption finds application in Heterogeneous catalysis. (c) Which of the following electrolytes is the most effective for the coagulation of Fe(OH)3 sol which is a positively charged sol ? NaCl, Na2SO4, Na3PO4 U



Ans. (a) Negative charge is developed on the sol. [½] Sol is represented as AgI /I– [½] (b) Adsorption of reactants on the solid surface of the catalysts increases the rate of reaction. [1] (c) Na3PO4 [½] Hardy-Schulze rule [½] [CBSE SQP Marking Scheme 2018] Q. 11. Answer the following questions : (i) What happens when a freshly precipitated Fe(OH)3 is shaken with a little amount of dilute solution of FeCl3 ? (ii) Why are lyophilic colloidal sols more stable than lyophobic colloidal sols ? (iii) What form Freundlich adsorption equation will take at high pressure ? Ans. (i) A  reddish brown coloured colloidal solution is obtained. [1] (ii) Stability of lyophilic sols is due to : (a) same charge on all the colloidal particles. (b) solvation of the colloidal particles. [½ + ½] (iii) At high pressure amount of gas adsorbed (x/m) becomes independent of pressure (P). x = k × P0 [1] m

Q. 12. 

Explain the following phenomenon giving reasons: (i) Chemical adsorption increases with increase in temperature. (ii) Alum is applied on a cut to stop bleeding. (iii) Sky appears blue in colour. A&E [CBSE Comptt. OD Set-3 2017]

Ans. (i) High energy of activation is needed.  1 (ii)  Blood being a colloidal solution, gets coagulated by alum (an electrolyte). 1 (iii) Dust particles along with water suspended in air scatter blue light which reaches our eyes. 1  [CBSE Marking Scheme 2017]

Detailed Answer: (i) This happens initially as increase in temperature provides high energy of activation required in chemical adsorption. (ii) It is due to coagulation of blood. Alum acts as an electrolyte which coagulates blood because blood is a colloidal solution. (iii) Because of scattering and dispersion of light. The light gets incident on the atmospheric particles scattering the sunlight. Due to larger scattering of blue colour, sky appears blue in colour. Q. 13. Give reason for the following observations : (i) When silver nitrate solution is added to potassium iodide solution, a negatively charged colloidal solution is formed. (ii) Finely divided substance is more effective as an adsorbent. (iii) Lyophilic colloids are also called reversible sols.  A&E [CBSE Comptt. Delhi/OD 2018] Ans. (i) The precipitated silver iodide adsorbs iodide ions from the dispersion medium resulting in the negatively charged colloidal solution. [1] (ii) Due to large surface area. [1] (iii) If the dispersion medium is separated from the dispersed phase, the sol can be reconstituted by simply remixing with the dispersion medium. That is why these sols are also called reversible sols.  [1] [CBSE Marking Scheme 2018] Q. 14. What happens when (a) a freshly prepared precipitate of Fe(OH)3 is shaken with a small amount of FeCl3 solution ? (b) persistent dialysis of a colloidal solution is carried out ? (c) an emulsion is centrifuged ?  U [CBSE Delhi/OD 2018] Ans. (a) Peptization occurs / Colloidal solution of Fe(OH)3 is formed.  [1] (b) Coagulation occurs. [1] (c) Demulsification or breaks into constituent liquids. [1] [CBSE Marking Scheme 2018] Detailed Answer: (a) When a freshly prepared precipitate of Fe(OH)3 is shaken with a small amount of FeCl3 solution, peptization occurs by converting the Fe(OH)3 precipitate into colloidal solution of positively charged Fe(OH)3. (b) On persistent dialysis, the electrolyte present is completely removed resulting in the coagulation of the colloidal solution. (c) On centrifugation, an emulsion gets separated into its constituent liquids. Q. 15. (i) What is the role of activated charcoal in gas mask?

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(ii) A colloidal sol is prepared by the given method in figure. What is the charge on hydrated ferric oxide colloidal particles formed in the test tube? How is the sol represented?

(iii) How does chemisorption vary with temperature? U [CBSE Delhi Set-1 2019]

Ans. (i) Adsorption of toxic gases.

FeCl3 Solution

[1] –

(ii) Negative charge ; Fe2O3.xH2O/OH

[½ + ½]

(iii) Increases with increase in temperature/ First increases then decreases. [CBSE Marking Scheme, 2019] [1] NaOH Solution

OR Ans. (a)

(b)



(c)



Detailed Answer : (i) The activated charcoal adsorbs the poisonous gases rather than oxygen and provides fresh oxygen for inhaling. (ii) When ferric chloride is added to NaOH, a negatively charged sol is formed with adsorption of OH– ions. The sol is represented as : [Fe(OH)3]OH– (iii) Adsorption is an exothermic process. According to Le Chatelier’s principle, at low temperature, forward reaction is favourable. As temperature, increases enough energy is being provided for the molecules to reach the activation energy. Therefore, initially as temperature increases, chemisorption increases. As the high temperature helps in bond breaking, after certain temperature, chemisorption decreases.  Q. 16. Answer the following questions: (a) Which of the following electrolytes is most effective for the coagulation of AgI/Ag+ sol? MgCl 2 , K 2 SO4, K4[Fe(CN)6]



[Topper’s Answer 2018]

(b) What happens when a freshly precipitated Fe(OH)3 is shaken with a little amount of dilute solution of FeCl3. (c) Out of sulphur sol and proteins, which one forms macromolecular colloids?



Ans. (a) K4[Fe(CN)6]  [1] (b) Fe(OH)3 is converted into colloidal state by preferential adsorption of Fe3+ ions. [1] (c) Proteins  [1]  [CBSE SQP Marking Scheme 2020] Detailed Answer : (a) According to the Hardy-Schulze rule, greater the valence of the flocculating ion, greater is its ability to bring coagulation. Thus, K4[Fe(CN)6] is most effective in coagulation of AgI/ Ag+ sol. (c) Macromolecular colloids are molecularly dissolved solutions of a polymer with particle size of colloidal range. Proteins are polymers of amino acids and form macromolecular colloids.



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Long Answer Type Questions Q. 1. How do emulsifiers function to stabilise the emulsion? Give two examples of emulsifiers. U Ans. (i) It is believed that an emulsifier gets concentrated at the oil-water interface i.e., the surface at which oil and water come in contact with each other. It forms a protective coating around each drop of oil and thus, prevents the oil drop from coming in contact with one another. The oil drops remain suspended in water and are not coagulated. Emulsifier Oil





Water

(ii) A  ccording to an another view, the role of the emulsifier is the same as that of lubricant in a machine. Just as a lubricant reduces the friction in the various parts of machine, an emulsifier also tries to reduce the interfacial tension between oil and water by suitable means. Thus, oil and water remain in company of each other and do not get

separated. The commonly used emulsifying agents are soaps, detergents, lyophilic colloids, proteins, gums, gelatin, caesin, agar etc. 5 Q. 2.  Differentiate between multi-molecular and macro- molecular colloids. Name one example of each. How does associated colloids differ from these two colloids? U Ans. (i)  In multi-molecular colloids, the colloidal particles are an aggregate of atoms or small molecules with a diameter of less than 1nm. The molecules in the aggregate are held together by van der Waals forces of attraction. Examples of such colloids include gold sol and sulphur sol. [1] (ii) In macro-molecular colloids, the colloidal particles are large molecules having colloidal dimensions. These particles have a high molecular mass. When these particles are dissolved in a liquid, sol is obtained. For example : Starch, nylon, cellulose, etc. [1] (iii) Certain substances tend to behave like normal electrolytes at lower concentrations. [1] However, at higher concentrations, these substances behave as colloidal solutions due to the formation of aggregated particles. Such colloids are called aggregated colloids. [2]

Visual Case-Based Questions Q. 1. Read the passage given below and answer the following questions : U (1 × 4 = 4) The amount of moisture that leather adsorbs or loses is determined by temperature, relative humidity, degree of porosity, and the size of the pores. Moisture has great practical significance because its amount affects the durability of leather, and in articles such as shoes, gloves and other garments, the comfort of the wearer. High moisture content accelerates deterioration and promotes mildew action. On the other hand, a minimum amount of moisture is required to keep leather properly lubricated and thus prevent cracking. The study indicates that adsorption of moisture by leather is a multi- molecular process and is accompanied by low enthalpies of adsorption. Further at 75-percent relative humidity, the adsorption is a function of surface area alone. Untanned hide and chrome- tanned leathers have the largest surface areas. The leathers tanned with vegetable tanning materials have smaller surface areas since they are composed of less hide substance and the capillaries are reduced to smaller diameters, in some cases probably completely filled by tanning materials. This process of tanning occurs due to mutual coagulation of positively charged hide with negatively charged tanning material. The result of the study indicated that untanned hide and chrome-tanned leather adsorb the most water vapour. (Source:Kanagy, J. R. (1947). Adsorption of water vapor by untanned hide and various leathers at

(5 marks each)

(4 marks each)

100 F. Journal of Research of the National Bureau of [CBSE SQP, 2020-21] Standards, 38(1), 119-128.) In these questions a statement of Assertion followed by a statement of Reason is given. Choose the correct answer out of the following choices. (a)  Assertion and reason both are correct statements and reason is correct explanation for assertion. (b)  Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. (i) Assertion : Vegetable tanned leather cannot adsorb a large amount of moisture.  Reason : Porous materials have higher surface area. (ii) Assertion : Animal hide soaked in tannin results in hardening of leather.  Reason : Tanning occurs due to mutual coagulation. (iii) Assertion : Adsorption of moisture by leather is physisorption.  Reason : It is a multimolecular process and is accompanied by low enthalpies of adsorption. (iv) Assertion : Leathers tanned with vegetable tanning materials have smaller surface areas.  Reason : The capillaries present in leather are reduced to smaller diameters.

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

OR Assertion : Leather absorbs different amount of moisture.  Reason : Some moisture is necessary to prevent cracking of leather. Ans. (i) Correct option : (b)[1] Explanation : Vegetable tanned leather cannot adsorb a large amount of moisture as these leathers have smaller surface areas since they are composed of less hide substance and the capillaries are reduced to smaller diameters, in some cases probably completely filled by tanning materials. Whereas porous materials have higher surface area. (ii) Correct option : (a)[1] Explanation : Animal hide soaked in tannin results in hardening of leather as tanning occurs due to mutual coagulation. (iii) Correct option : (a)[1] Explanation : Adsorption of moisture by leather is physisorption. It is a multimolecular process and is accompanied by low enthalpies of adsorption.  (iv) Correct option : (a) [1] Explanation : Leathers tanned with vegetable tanning materials have smaller surface areas, the capillaries present in leather are reduced to smaller diameters. OR Correct option : (b)[1] Explanation : The amount of moisture that leather adsorbs or loses is determined by temperature, relative humidity, degree of porosity, and the size of the pores. A minimum amount of moisture is required to keep leather properly lubricated and thus prevent cracking. Q. 2. Read the given passage and answer the questions (i) to (v) that follow : Colloidal particles always carry an electric charge which may be either positive or negative. For example, when AgNO3 solution is added to KI solution, a negatively charged colloidal sol is obtained. The presence of equal and similar charges on colloidal particles provide stability to the colloidal sol and if, somehow, charge is removed, coagulation of sol occurs. Lyophobic sols are readily coagulated as compared to lyophilic sols. (i) What is the reason for the charge on sol particles ?  (ii) Why the presence of equal and similar charges on colloidal particles provide stability ? (ii) Why a negatively charged sol is obtained on adding AgNO3 solution to KI solution ? (iv) Name one method by which coagulation of lyophobic sol can be carried out. (v) Out of KI or K2SO4, which electrolyte is better in the coagulation of positive sol ?  [CBSE OD Set-1, 2020] Ans. (i) The charge on the colloidal sol particles depends upon the preferential adsorption of the ions from the electrolyte. When AgNO3 is added to KI solution the yellow precipitate of AgI absorbs I– ions from the electrolyte and form negatively charged sol. (ii) Presence of equal and similar charges on colloidal particles provides stability to colloids as repulsive forces between charge particle having same charge, prevent from colliding when they come closer to each other. Hence provide stability.



(iii) On adding AgNO3 to KI solution the yellow ppt. of AgI absorbs I– ions from the electrolyte KI, so negatively charged sol is obtained. AgI/I– (iv) By chemical method : (i) Oxidation, (ii) Reduction, (iii) Hydrolysis, (iv) Double decomposition (Any one) (v) K2SO4 because the coagulation power increases with increase in charge on the ions, SO42– > I– Q. 3. Read the passage given below and answer the following questions: U (1 × 4 = 4) Some colloids are stable by their nature, i.e., gels, alloys, and solid foams. Gelatin and jellies are two common examples of a gel. The solid and liquid phases in a gel are interdispersed with both phases being continuous. In most systems, the major factor influencing the stability is the charge on the colloidal particles. If a particular ion is preferentially adsorbed on the surface of the particles, the particles in suspension will repel each other, thereby preventing the formation of aggregates that are larger than colloidal dimensions. The ion can be either positive or negative depending on the particular colloidal system, i.e., air bubbles accumulate negative ions, sulphur particles have a net negative charge in a sulphur sol, and the particles in a metal hydroxide sol are positively charged. Accumulation of charge on a surface is not an unusual phenomenon-dust is attracted to furniture surfaces by electrostatic forces. When salts are added to lyophobic colloidal systems the colloidal particles begin to form larger aggregates and a sediment forms as they settle. This phenomenon is called flocculation, and the suspension can be referred to as flocculated, or colloidally unstable. If the salt is removed, the suspension can usually be restored to its original state; this process is called deflocculation or peptization. The original and restored colloidal systems are called deflocculated, peptized, or stable sols. Why does a small amount of salt have such a dramatic effect on the stability of a lyophobic colloidal system? The answer lies in an understanding of the attractive and repulsive forces that exist between colloidal particles. Van der Waals forces are responsible for the attractions, while the repulsive forces are due to the surface charge on the particles. In a stable colloid, the repulsive forces are of greater magnitude than the attractive forces. The magnitude of the electrical repulsion is diminished by addition of ionized salt, which allows the dispersed particles to aggregate and flocculate. River deltas provide an example of this behaviour. A delta is formed at the mouth of a river because the colloidal clay particles are flocculated when the freshwater mixes with the salt water of the ocean (source: Sarquis, J. (1980). Colloidal systems. Journal of Chemical Education, 57(8), 602. doi:10.1021/ ed057p602) 1. Gelatin is a ________________ colloidal system. A. Solid in solid B. Solid in gas C. Liquid in solid D. Liquid in gas 2. Colloidal solutions are stable due to: A. presence of charges on the colloidal particles

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SELF-ASSESSMENT PAPER

B. formation of aggregates by colloidal particles C. preferential adsorption on the surface D. preferential absorption on the surface 3. Settling down of colloidal particles to form a suspension is called: A. flocculation B. peptization C. aggregation D. deflocculation 4. When Van der Waals forces are greater than forces due to the surface charge on the particles, A. flocculation occurs. B. the colloid is stable. C. peptization takes place. D. deflocculation occurs. 5. The particles in suspension will repel each other, thereby preventing the formation of aggregates that are larger than colloidal dimensions. This statement explains: A. formation of delta B. river water is a colloidal of clay particles C. effect of salt on lyphobic colloid D. phenomenon of flocculation (Ans: 1C, 2C, 3A,4A,5B) Q. 4. Read the passage given below and answer the following questions: Industrially widely applied esterification reactions are commonly catalysed using mineral liquid acids, such as sulphuric acid and p-toluenesulphonic acid. The catalytic activity of homogeneous catalysts is high. They suffer, however, from several drawbacks, such as their corrosive nature, the existence of side reactions, and the fact that the catalyst cannot be easily separated from the reaction mixture. The use of solid acid catalysts offers an alternative and has received a lot of attention in the past years. Solid acid catalysts are not corrosive and, coated onto a support, they can be easily reused. Examples of solid acid catalysts used in esterification reactions include ion-exchange resins , zeolites and superacids like sulphated zirconia and niobium acid. Ion-exchange resins are the most common heterogeneous catalysts used and have proven to be effective in liquid phase esterification and etherification reactions. Because of their selective adsorption of reactants and swelling nature, these resins not only catalyse the esterification reaction but also affect the equilibrium conversion. Shortcomings include insufficient thermal resistance, which limits the reaction temperature to 120 °C, preventing widespread use in industry. Zeolites, like Y, X, BEA, ZSM-5 and MCM-41 offer an interesting alternative and have proven to be efficient catalysts for esterification reactions. Zeolites have found wide application in oil refining, petrochemistry and in the production of fine chemicals. Their success is based on the possibility to prepare zeolites with strong Brønsted acidity that can be controlled within a certain range, combined with a good resistance to high reaction temperatures. In this study, the activity of various commercial available solid acid catalysts is assessed with respect to the esterification of acetic acid with butanol. The ion-exchange resins Amberlyst 15 and Smopex-101, the acid zeolites H-ZSM-5, H-MOR, H-BETA and H-USY, and the solid superacids sulphated zirconia and niobium acid are selected. Comparative esterification experiments have been carried out using the homogeneous catalysts sulphuric acid,





ptoluenesulphuric acid and a heteropolyacid (HPA). The weight-based activity of the heterogeneous catalysts tested is maximum for Smopex101. The following table gives the activity of different catalysts in the esterification reaction between acetic acid and butanol at 750C.

Here: kobs: observed reaction rate constant ( m3mol-1 s-1) kc catalysed reaction rate constant (m3mol-1gcat-1 s-1) Please note: k c = k obs/ amount (in g) (source: PETERS, T., BENES, N., HOLMEN, A., & KEURENTJES, J. (2006). Comparison of commercial solid acid catalysts for the esterification of acetic acid with butanol. Applied Catalysis A: General, 297(2), 182–188. doi:10.1016/j.apcata.2005.09.00) 1. Which of the following are heterogeneous catalysts for esterifctaion reaction: A. sulphuric acid and p-toluenesulphonic acid B. sulphuric acid and niobium acid C. p-toluenesulphonic acid and niobium acid D. niobium acid and sulphated zirconia 2. Unit for observed rate constant for esterification reaction is m3mol-1s-1, so the reaction is: A. zero order B. first order C. second order D. third order 3. The catalytic activity of homogeneous catalysts is high. The weight based activity of HPA is less than which of the following heterogenous catalysts? A. Smopex-101 B. Amberlyst 15 C. sulphated ZrO2 D. H-USY-20 4. The weight-based activity of the heterogeneous catalysts tested decreases in the following order: A. Smopex-101 > Amberlyst 15 > sulphated ZrO2 > H-USY-20 >H-BETA-12.5 > HMOR-45 > Nb2O5 > H-ZSM-5-12 B. Smopex-101 > Amberlyst 15 > sulphated ZrO2 > H-USY-20 >H-BETA-12.5 > HMOR-45 > H-ZSM-5-12> Nb2O5 C. Smopex-101 > Amberlyst 15 > sulphated ZrO2 > H-USY-20 >H-BETA-12.5 > Nb2O5> H-MOR-45 > H-ZSM-5-12 D. Smopex-101 > sulphated ZrO2 > Amberlyst 15 > H-USY-20 >H-BETA-12.5 > HMOR-45 > H-ZSM-5-12> Nb2O5 5. Catalysts used in oil refining industry are: A. ion exchange resins B. superacids C. zeolites D. mineral liquid acids (Ans.: 1D, 2C, 3A, 4A, 5C) 

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Self Assessment Test - 5 Time : 1 Hour

Max. Marks : 25

Q. 1.  Read the passage given below and answer the following questions : (1 × 4 = 4)

a colloidal solution is comparatively small as compared to a true solution. Hence, the values of colligative properties (osmotic pressure, lowering in vapour pressure, depression in freezing point and elevation in boiling point) are of small order as compared to values shown by true solutions at same concentrations.

Pharmaceutical preparations for the treatment of skin conditions and skin care are usually supplied in the form of emulsions. This pharmaceutical dosage form is thermodynamically unstable and must be stabilized by the addition of emulsifying agent. Emulsified systems range from lotions having comparatively low viscosity to creams which are more viscous. There are two basic types of emulsions, that is, oil in water (O/W) and water in oil (W/O). In addition to these two types, a relatively complex emulsion, called multiple emulsions can also be formulated. Emulsions generally have certain advantages over other dosage forms as the drug solubilized may be more bioavailable. The following questions are multiple choice questions. Choose the most appropriate answer : (i) An emulsion cannot be broken by (a) Heating (b) Emulsifying agent (c) Freezing (d) None of above (ii) Which of the following is an example of oil dispersed in water? (a) Butter (b) Soap (c) Vanishing cream (d) All of the above (iii) Which of the following substances can be used to precipitate the negatively charged emulsions? (a) KCl (b) Glucose (c) Urea (d) All of the above (iv) Proteins, Gums, natural soaps are categorised as (a) O/W emulsifying agents (b) Electrolytes (c) W/O emulsifying agents (d) Emulsions Q.2. Read the passage given below and answer the following questions :  (1 × 4 = 4) Various properties exhibited by the colloidal solutions are described below: (i)  Colligative properties : Colloidal particles being bigger aggregates, the number of particles in



(ii) Tyndall effect : If a homogeneous solution placed in dark is observed in the direction of light, it appears clear and, if it is observed from a direction at right angles to the direction of light beam, it appears perfectly dark. Colloidal solutions viewed in the same way may also appear reasonably clear or translucent by the transmitted light but they show a mild to strong opalescence, when viewed at right angles to the passage of light.



Choose the correct answer out of the following choices. (a)  Assertion and reason both are correct statements and reason is correct explanation for assertion.



(b) A ssertion and reason both are correct statements but reason is not correct explanation for assertion.



(c) A  ssertion is correct statement but reason is wrong statement.



(d) A  ssertion is wrong statement but reason is correct statement.



(i) Assertion : Colloidal solutions show colligatlve properties.

Reason : Colloidal particles are large in size.



 (ii) Assertion : Colloidal particles always carry an electric charge.



Reason : The sol particles acquire positive or negative charge by preferential adsorption of positive or negative ions.



(iii) Assertion : Colloidal solutions show Tyndall effect, where the path of a beam of light is visible through the colloidal dispersion.



Reason : Colloidal particles scatter light in a straight line in the solution.  (iv) Assertion : The colour of colloidal solution is independent of the wavelength of light scattered by the dispersed particles. Reason : The colour of colloidal solution changes with the manner in which the observer receives the light.





[ 151

SELF-ASSESSMENT TEST

Q.3.  Which of the following is most effective in coagulating positively charged methylene blue sol ? (a) Na3PO4

(b) K4 [Fe(CN)6]



(d) NaNO3

(c) Na2SO4

Q.7. Write one difference in each of the following :

(i) Lyophobic sol and Lyophilic sol

    (ii) Solution and Colloid

Q.4. In the following question a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.



(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.



(c)  Assertion is correct statement but reason is wrong statement.



(d)  Assertion is wrong statement but reason is correct statement.



 (iii) Emulsion and Gel

[3]

Q.8. (a) Why does physisorption decrease with increase in temperature?

(b) A colloidal sol is prepared by the method given in the figure. What is the charge on AgI colloidal particles formed in the test tube? How is this sol represented?

AgNO3 solution

Assertion : Detergents with low CMC are more economical to use. Reason : Cleansing action of detergents involves the formation of micelles. These are formed when the concentration of detergents becomes equal to CMC. 

KI solution

Q.5. Write any two differences between Physisorption and Chemisorption. [2] Q.6. (a) Write the dispersed phase and dispersion medium of milk.

(b) Why is adsorption exothermic in nature ?

[2]

(c) Differentiate between adsorption and absorption.

Q.9. What is the difference between multi-molecular and macro-molecular colloids ? Give one example of each. How are associated colloids different from these two types of colloids ? [5]

 

Finished Solving the Paper ? Time to evaluate yourself !

OR SCAN THE CODE

SCAN

For elaborate Solutions

ll

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

6

CHAPTER

Syllabus

GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

¾¾ Principles and methods of extraction – concentration, oxidation, reduction–electrolytic method and refining; occurrence and principles of extraction of aluminium, copper, zinc and iron.

Trend Analysis List of Concept names Methods and Principles of Isolation of Elements Refining of Nickel, Extraction of Aluminum and Extraction of Iron Extraction of Gold and Role of NaCN and Zn

TOPIC-1

2018 D/OD

2019 D 1Q (3 marks)

2020 OD

1Q (3 marks) 1Q (3 marks)

Principles and Methods of Extraction

Revision Notes

D

OD

Chapter was not included in syllabus

TOPIC - 1 Principles and Methods of extraction. .... 153 TOPIC - 2 Principles of Extraction of Aluminium, Copper, Zinc and Iron .... 164

 Minerals : The naturally occurring chemical substances in the earth’s crust which are obtained by mining.  Ore : The mineral from which a particular metal can be extracted conveniently and economically. Scan to know  Gangue : The earthy materials associated with the ores. more about  Occurrence of metals : this topic In free state : Very few metals exist in the free or native state. Only metals like gold, platinum and mercury are occasionally found in the free state, i.e., in the pure form. In the combined state : The rest of the metals occur in the combined form of compounds such as oxides, carbonates, sulphides, sulphates, silicates, chlorides, nitrates, phosphates etc. Metallurgy Note : Copper and silver are two metals which occur in free as well as combined state as sulphides, oxides or halides ores.  Metallurgy : It is the entire scientific and technological process used to obtain the pure metal from its ore.  Flux : The substance which is added in the ore to convert non-fusible gangue to fusible compound is called flux. There are three types of flux : acidic flux (Silica borax), basic flux (Limestone) and neutral flux (Graphite).  Slag : The fusible compound formed by combination of flux and gangue is called slag.  The processes involved in the metallurgy : (i) Concentration of the ore (ii) Isolation of metal from its concentrated ore (iii) Refining or purification of metals



Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Metallurgy : The whole process of obtaining a pure metal from one of its ore is known as metallurgy. A general scheme of various metallurgical operations employed for the extraction of metals from ores is given below :





154 ]

Concentrate dore

concentrate dore



Concentration of ore : It is a process used for removing the gangue from the ore and increasing ore’s grade on the basis of difference between the physical or chemical properties of the gangue and the ore. The concentration of the ore is carried out by the following methods : (i) Crushing and Grinding : The huge lumps are first broken into small pieces in the jaw crushers and then powdered with the help of a ball mill or stamp mill. This process is termed as pulverisation. (ii) Levigation or Gravity separation : It is based on difference in densities (gravities) of ore and the gangue. In this process, ore is washed with stream of water under pressure, light impurities are washed away whereas heavy ore is left behind. e.g. Generally oxides and carbonates ores are concentrated by this method. (iii) Magnetic separation method : Ore and gangue are separated, if only one of them is magnetic in nature. Magnetic separation method is used to remove tungsten FeWO4–magnetic ore particles from cassiterite (non magnetic–SnO2). It is also used to concentrate magnetite (Fe3O4), chromite (FeCr2O4) and pyrolusite (MnO2) from unwanted gangue. Finely ground ore

Magnetic roller Magnetic particles Non-magnetic particles

Fig. 1 : Magnetic separation method (iv) Froth flotation process : This process makes use of the principle of preferential wetting of solid surfaces by various liquids. This process is used for the concentration of sulphide ores e.g., ores of lead, zinc and copper, because of the fact that metallic sulphides are more wetted by certain oils (pine oil) and less by water. The mixture is then agitated by passing a blast of air through it. The froth is formed which carries the ore particles along with it to the surface leaving the impurities behind. The froth is scummed off and in this way the ore is concentrated by froth flotation process.

[ 155

GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS Air

Froth containing mineral

Gangue

Fig. 2 : Froth flotation process

Scan to know

more about (v) Hydraulic washing : It is based on the difference in the gravities of the ore and the gangue this topic particles. An upward stream running water is used to wash the powdered ore. The lighter gangue particles are washed away and the heavier ores are left behind. (vi) Leaching : It is used if the ore is soluble in a suitable reagent which can selectively dissolve the ore but not the impurities. Extraction of  Conversion of ore into oxide : Following two methods are used to convert the ore into : metals (i) Calcination : It is a process in which ore is heated in the absence of air so as to convert carbonate ores into oxides. Process temperature is below the melting point of treated ores. In this process, the moisture and volatile impurities are removed. Thereby ore becomes porous.





FeCO3 Heat  → FeO + CO2 Siderite

Fe2O3.xH2O(s) Heat  → Fe2O3(s) + xH2O(g) (ii) Roasting : It is a process in which ore is heated in regular supply of air at a temperature below the melting point of the metal so as to convert the given ore into oxide ore. It is also used to remove impurities as volatile oxides. Sulphide ores are converted into oxide by roasting. e.g.,





2ZnS + 3O2 → 2ZnO + 2SO2 This process is done in reverberatory furnace. Charge hooper Hangers Tie rod Charge hooper

Air and oil Magnesite Furnace charge Silica

Fig. 3 : A section of a modern reverberatory furnace  Slag : The compound formed on reaction of gangue with flux is called slag. It is a fusible mass which floats over metal. FeO

+

SiO2

→

Iron (II) oxide Silica (Basic Gangue) (Acidic flux)

FeSiO3 (slag)

Iron (II) silicate

 Reduction of oxide to metal : Reduction of the metal oxide involves heating it with some other substances acting as a reducing agent. The common reducing agents used are carbon, carbon monoxide or any other metal like Al, Mg etc. Some common methods used for the reduction are given below : (i) Auto reduction : In this method, inactive metals can be reduced simply by heating the ore in air. Extraction of copper, lead, antimony, mercury etc, have been carried out by this process. e.g.,



2Cu2S + 3O2 → 2Cu2O + 2SO2 ↑­­





Cu2S + 2Cu2O → 6Cu + SO2 ↑­­

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



(ii) Smelting : In this process, metal oxide is reduced to metal with C or CO. e.g.,









> 1123 K

Fe2O3 + 3C → 2Fe + 3CO ↑­­ K Fe2O3 + 3CO 1123  → 2Fe + 3CO2 ↑­­

ZnO + C → Zn + CO ­ (iii) Aluminothermic reduction : The process of reduction of metal oxide by aluminium is known as aluminothermic reduction. Metals like manganese and chromium are extracted by thermite process. 3MnO4 + 8Al → 4Al2O3 + 3Mn Cr2O3 + 2Al → Al2O3 + 2Cr (iv) Reduction with hydrogen : It is an efficient reducing agent for metal oxides. For this purpose, the roasted ore is heated in a current of hydrogen and metal oxide is reduced to metal. For example, oxides of W, Mo, etc. are reduced with hydrogen. WO3 + 3H2 → W + 3H2O  Hydrometallurgy : The process of extraction of a metal by dissolving the ore in a suitable reagent followed by precipitation or displacement of the metal by a more electropositive metal is known as hydrometallurgy.  Refining or Purification of Metals : (i) Liquation : This method is based on the principle of difference in melting points of metal and impurity. It is the process of refining a low melting metal like tin which can be made to flow on a sloping surface. (ii) Zone refining : This method is particularly used when metals are required in high degree of purity. In this method, a metal rod is placed inside a small high frequency induction furnace. A narrow zone of metal is melted (Fig. 4). This furnace is now slowly moved along the rod. The pure metal recrystallizes out of the melt while impurities remain in the melt which moves along with the melted zone of the rod with the movement of the furnace. The process is repeated several times. The end of the rod where the impurities have collected is cut off. This method is employed for the purification of germanium, silicon, gallium, etc., which are used in semiconductors. Molten zone Pure metal Impure metal

Moving furnace

Fig. 4 : Zone refining (iii) Electrolytic refining : This method is based upon the phenomenon of electrolysis. The crude metal is made anode whereas the thin sheet of pure metal is made cathode. Electrolyte is the solution of same salt of the metal. On passing electricity, the metal from the anode goes into solution as ions due to oxidation, while pure metal gets deposited at the cathode due to reduction of metal ions. The less electropositive impurities settle down below the anode as anode mud. Reaction : At anode : M → Mn+ + ne– At cathode : Mn+ + ne– → M

Crude metal Anode mud

Solution of electrolyte Pure metal

Fig. 5 : Electrolytic refining (iv) Vapour phase refining : Vapour phase refining is illustrated by the following two methods : (a) Mond’s process : This method is applied for purification of nickel. Nickel metal when heated in a stream of carbon monoxide forms volatile nickel carbonyl [Ni(CO)4]. The impurities present in the impure nickel are left behind as solid. The vapour when heated at higher temperature (450–470 K) decomposes giving pure nickel and carbon monoxide.

330 − 350 K Ni + 4CO → Ni(CO)4

Impure nickel

− 470 K 450   → Ni + 4CO

Pure nickel

[ 157



GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

(b) Van Arkel method : Small amount of very pure titanium or zirconium metal can be prepared by this method. Impure metal is heated in an evacuated vessel with I2. TiI4 is formed which vaporizes leaving behind impurities. The gaseous MI4 is decomposed on a white hot tungsten filament. 1800K → ZrI4 Tungsten Zr + 2I2   → Zr + 2I2 filament

870K













Impure

Pure

523K → TiI4 Ti + 2I2 

Impure

1700K Tungsten  → Ti + 2I2 filament

Pure



 Chromatographic method : It is a method of separation or purification based on differential adsorption on an adsorbent. In column chromatography, Al2O3 is used as adsorbent. The mixture to be separated is taken in suitable solvent and applied on the column. They are then eluted out with suitable solvent (eluant). The weakly adsorbed component is eluted first, then the more strongly adsorbed and so on. This method is suitable for those elements which are available only in minute quantity and the impurities are not very much different in their chemical behaviour from the element to be purified.

Mixture of compounds

Coloured bands

Stationary phase

Fig. 6 : Column chromatography (Laboratory Method)  Thermodynamic principle of metallurgy : This principle helps in choosing a suitable reducing agent for the reduction of particular metal oxide to metal. For any process, at any specified temperature, Gibb’s free energy change (DG) is given by DG = DH – TDS where, DH is the enthalpy change and DS is the entropy change for any process. If DG is positive for any reaction, then to make such reaction spontaneous, it is coupled with another reaction of large negative DG value so that the sum of DG becomes negative. This is known as coupling reaction.  Ellingham diagram : This diagram was proposed by Ellingham to select the suitable reductant for the reduction of metal oxide. 200 + 0 4Ag

– 200

O2 

+ 4 Hg

O 2Ag 2

O2 

O 2 Hg 2

 2 F e + O2

– 600

2CO

2F eO

CO 2 2 + O2 ZnO 2

2Zn

G° /kJ mol

O  2 Cu2

C + O2  CO2

2C +

O2  2 CO

+ O2

–1

of O2

– 400

4 Cu + O2

l 2O 3

/3 A

– 800 4/3

– 1000

Al

2 + O2

+ 2Mg

O 2

2MgO

– 1200 0°C



400°C 800 °C 1200°C TEMPERATURE

1600°C 2000°C

Fig. 7 : Ellingham diagram for some oxides

158 ]



Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

In this diagram, graph is plotted between change in standard free energy (ΔG°) and absolute temperature (T) for the formation of oxide of elements. This diagram helps in predicting the feasibility of reduction of an ore. The criterion of feasibility of reduction is negative value of change in free energy. This diagram explains the following important facts : (i) Entropy decreases during formation of metal oxide from metal i.e., DS is negative. y x M(s) + O2(s) → MxOy(s) 2

(ii) Change in entropy (DS) increases on melting or boiling (change in state) of a substance. Hence, during the change in state, change in free energy takes place suddenly. (iii) Formation of carbon monoxide is the result of oxidation. It is due to positive change in entropy (DS). 2C(s) + O2(g) → 2CO(g)  Limitations of Ellingham diagram : (i) Ellingham diagram simply indicates the feasibility of a reduction process as it is based only on thermodynamic principles. It is unable to explain the kinetics of a reduction process. On the basis of Ellingham diagram, it cannot be predicted that how fast a reduction process will occur. (ii) Reactions are assumed at equilibrium in this diagram.

Know the Terms Refining : The process of purifying the impure metals is called refining. Froth stabilisers : Substances like cresol and aniline which stabilise the froth. Extraction : The process used to obtain metals in free state from the concentrated ore is called extraction. Ellingham diagram : The graphical representation of Gibbs energy.

Q. What chemical principle is involved in choosing a reducing agent for getting the metal from its oxide ore ? Consider the metal oxides, Al2O3 and FeO and justify the choice of reducing agent in each case.

ofO 2

0 – 100

–1

   

G° /kJ mol



– 400 – 500 – 600 – 700 – 800 – 900

2Cu 2O

4C u+O2

– 200 – 300

2F e+O2

2 CO

+ 2 Zn

– 1000 – 1100

2 C+

2C O 2

+O 2

O2

O 2 Zn

Si+ O2

O2

+O2

CO2

2CO

SiO

O2 Al + 4/3

2Mg

C

2F eO

+O2

2/3

Al 2O 3

2MgO

– 1200 0°C 273 K

400°C 800° C 1200°C 1600° C 2000° C 673 K 1073 K 1473 K 1873 K 2273 K TEMPERATURE

Solution: STEP-I: The feasibility of thermal reduction can be predicted on the basis of Ellingham diagram. Metals for which the standard free energy of formation (Δf G°) is more negative can reduce those metals for which Δf G° is less negative. At a given temperature, any metal will reduce the oxide of other metals which lie above it in the Ellingham diagram. STEP-2: Below the temperature approx. 1623 K, corresponding to the point of intersection of Al2O3 and MgO curves, Mg can reduce alumina. STEP-3: At temperatures below 1073 K, the CO, CO2 line lies below Fe, FeO line, thus CO is a better reducing agent. At temperatures above 1073 K, coke will reduce FeO and itself get oxidised to CO.

GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

Objective Type Questions Which of the following reactions is an example of auto-reduction?

(a) Fe3O 4 + 4CO → 3Fe+ 4CO 2 (b) Cu 2O+C → 2Cu + CO (c) Cu 2+(aq.)+Fe (s) → Cu (s)+Fe2+(aq.) 1 1 (d) Cu 2O + Cu 2S → 3Cu + SO 2 2 2

(1 marks each)

[A] MULTIPLE CHOICE QUESTIONS : Q. 1.

R

Ans.

Correct option : (d) Explanation : Reaction includes reduction of copper (I) oxide by copper (I) sulphide and in this process copper is reduced by itself. This process is called as auto-reduction. The solidified copper so obtained is known as blistered copper. Q. 2. Brine electrolysed by using inert electrodes. The reaction at anode is : 1 Θ = 1.36 V . (a) Cl− (aq.) → Cl2 (g)+e− ; ECell 2 + − Θ (b) 2H 2O (l ) → O 2 (g) + 4H + 4e ; ECell = 1.23 V . + − Θ (c) Na (aq.) + e → Na (s); ECell = 2.71 V . 1 Θ A (d) H+ (aq.) + e− → H 2 (g); ECell = 0.00 V . 2 Ans. Correct option : (a) Explanation : Electrolysis of brine solution: At anode: 2Cl– (aq) ® Cl2(g) + 2e– At cathode: 2H2O(l) + 2e– ® 2OH–(aq) + H2(g) Reaction: 2NaCl + 2H2O ® Cl2 + H2 + 2NaOH Q. 3. In the Mond’s process the gas used for the refining of a metal is (a) H2 (b) CO2 (c) CO R (d) N2 Ans. Correct option : (c) [B] ASSERTION AND REASON TYPE QUESTIONS: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation of assertion. (b) Assertion and reason both are correct statements but reason is not the correct explanation for assertion.

(c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Q. 1. Assertion : Zone refining method is very useful for producing semiconductors. Reason : Semiconductors are of high purity. R Ans. Correct option: (b) Explanation: The impurities of semiconductors are more soluble in molten zone and the ultrapure semiconductor crystallizes in zone refining method. Q. 2. Assertion : Zirconium can be purified by Van Arkel method. Reason : ZrI4 is volatile and decomposes at 1,800 K. R [NCERT Exemplar] Ans. Correct option: (a) Explanation: Zirconium is also purified by vapour phase refining method in which it is treated with iodine to form ZrI4 which on heating decomposes to give pure zirconium. Q. 3. Assertion : Nickel can be purified by Mond’s process. Reason : Ni(CO)4 is a volatile compound which decomposes at 460 K to give pure Ni. R Ans. Correct option: (a) Explanation: When nickel (Ni) is treated with carbon monoxide, it forms nickel tetra carbonyl Ni(CO)4, while impurities are left behind. When the vapour of Ni(CO)4 is heated at 460 K, it is decomposed to give pure nickel, while carbon monoxide is removed as gas. [C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1. Name the substance used as depressant in the separation of two sulphide ores in Froth flotation method. R [CBSE SQP 2020] Ans. Sodium cyanide. Q. 2. Name the depressant which is used to separate PbS and ZnS containing ore in froth flotation process. R [CBSE OD Set 2, 2020] Ans. NaCN: It selectively allows PbS form froth but prevents ZnS from coming to froth. Q. 3. Name the method used for the refining of Zinc.  R [CBSE OD Set 3, 2020] Ans. Electrolytic refining.

Short Answer Type Questions-I Q. 1.

[ 159

What factors should be considered while extraction of metals by electrochemical method?  R

Ans.

(2 marks each) The following factors should be considered while using electrochemical method for metal extraction.





160 ] Q.2.

Ans.

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(a) Reactivity of the metal (b) Suitability of the electrode How can two sulphide ores be separated by froth floatation method? Explain by giving an example.  U The separation of two sulphide ores can be done by adjusting the proportions of oil to water or can be also done by using depressants. In the case of an ore containing ZnS and PbS, the depressant used is NaCN. It forms complex with

Q. 3. Ans.

(a)

(b)

ZnS and prevents it from coming with froth but PbS remains with froth. Give two requirements for vapour phase refining.  C The two requirements for vapour phase refining are given below : The metal should form a volatile compound with available reagent. The volatile compound should be unstable and easily decomposable so that the recovery is easy.

Short Answer Type Questions-II

Commonly Made Error  Students often write lengthy answers to direct questions thus losing time.

Answering Tip  Write only the principle and not the process/steps involved in the process. Q. 2. (i) Write the name of the method used for the refining of the following metals: (a) Titanium (b) Germanium (c) Copper (ii) Write the name of the method of concentration applied for the following ores: (a) Zinc blende (b) Haematite (c) Bauxite R [CBSE Foreign Set-3 2017]

Ans. (i) (a) Vapour phase refining /van Arkel method[½] (b) Zone refining ½ (c) Electrolytic refining ½  (ii) (a) Froth floatation process ½ (b) Magnetic separation ½ (c) Leaching [½]  [CBSE Marking Scheme 2017] . 3. (i) Write the principle involved in the following: Q (a) Zone refining of metals (b) Electrolytic refining (ii) Name the metal refining by each of the following processes: (a) Mond Process (b) van Arkel Method  R [CBSE Comptt. OD Set-1, 3 2017]



Q. 4. (i) What is the principle behind ‘Zone refining’ of metal ? Name an element which is refined by this method. (ii) Write the name of the metal refined by each of the following processes: (a) Distillation (b) Liquation  R [CBSE Comptt. Delhi Set-3 2017] Ans. (i) Impurities are more soluble in the molten state than in the solid state of the metal. ½ Example: Ge/Si/ B (any other) ½ (ii) (a) Zn/Hg 1 (b) Sn [CBSE Marking Scheme 2017] 1

Ans. (i) Zone refining : Impurities are more soluble in the molten state than in the solid state of metal.[1]    (ii) Froth Floatation : Mineral particles are wetted by oils forming froth while gangue particles are wetted by water and settle down.[1]   (iii) Chromatography : Different components of a mixture are differently adsorbed on an adsorbent.  [1] [CBSE Marking Scheme 2017]

Ans. (i) (a) See SAT- II Q.1 (i)[1] (b) The more basic/reactive ones go the anode mud. [1]  (ii) (a) Ni (b) Ti/Zr [½ + ½]  [CBSE Marking Scheme 2017]

. 5. (i) Write the principle of electrolytic refining. Q (ii) Why does copper obtained in the extraction from copper pyrites have a blistered appearance? (iii) What is the role of depressants in the froth flotation process? R + A&E + U [CBSE OD Set-3 2017] Ans. (i) On passing current through the electrolytic cell, the pure metal gets deposited on the cathode.1 (ii) Evolution of SO2 gas 1 (iii) I t selectively prevents one of the sulphide ores from coming to the froth. 1  [CBSE Marking Scheme 2017]

. 1. Write the principle of the following : Q (i) Zone refining (ii) Froth flotation process (iii) Chromatography R [CBSE OD Set-1 2017]

(3 marks each)

Detailed Answer: (ii) Copper pyrites is concentrated by froth flotation process. The molten copper is poured and cooled. The sulphur dioxide evaluating from the melt gets trapped in the cooler parts of the surface giving a blistery appearance. Q. 6. (i) Write the principle of vapour phase refining. (ii) What is the role of depressant in froth flotation process? (iii) Write the name of reducing agent to obtain iron from Fe2O3 at high temperature.







R + U [CBSE Foreign Set-1 2017]

[ 161

GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS



Q. 8. (i) Write the principle of vapour phase refining . (ii) Write the role of dilute NaCN in the extraction of silver. (iii) What is the role of collectors in the froth flotation process? Give an example of a collector.

Q. 7. (i) Write the principle of Zone refining . (ii) What is the role of collectors in froth flotation process? Give an example of a collector. (iii) Write the name of a reducing agent to obtain Fe from Fe2O3 at low temperature. 

R + U [CBSE Foreign Set-2 2017]



Ans. (i) See SAT-(II) Q.1 (i) 1 (ii) C  ollectors enhance non-wettability of the mineral particles Ex. Pine oil /fatty acids. 1 (iii) Carbon monoxide (CO). 1  [CBSE Marking Scheme 2017]







R + U [CBSE OD Set-2 2017]

Ans. (i) Metal is converted into its volatile compound and collected else where. It is then decomposed at high temperature to give pure metal. 1 (ii) It acts as a leaching agent/forms soluble complex with Ag. 1 (iii) Enhance non-wettability of mineral particles. For e.g. Pine oil, Fatty acids, xanthates (Any one).  1  [CBSE Marking Scheme 2017]



Ans. (i) Refer Q.8 (i)[1] (ii) I t selectively prevents one of the sulphide ores from coming to the froth.[1] (iii) Coke. [CBSE Marking Scheme 2017] 1

. 9. (i) Name the method of refining which is based on the principle of adsorption . Q (ii) What is the role of depressant in froth flotation process? (iii) What is the role of limestone in the extraction of iron from its oxides? Ans.

3 [Topper’s Answer 2017] 







OR

R [CBSE OD Set-2 2016]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII







162 ]

3

[Topper’s Answer 2016] Q. 10. (i) Name the method of refining of nickel. (ii) What is the role of cryolite in the extraction of aluminium ? (iii) What is the role of limestone in the extraction of iron from its oxides ? R [CBSE OD Set-1 2016]



dilute NaCN and Zn in the process. R + U [CBSE Delhi/OD 2018]

OR



Ans. (i) Mond's process. [1] (ii) Cryolite acts as a solvent. The melting point of alumina is very high. It is dissolved in cryolite which lowers the melting point and brings conductivity. [1] (iii) Limestone is decomposed to CaO, which removes the silica impurity of the ore as slag. [1] [CBSE Marking Scheme 2016]



Write the chemical reactions involved in the process of extraction of Gold. Explain the role of dilute NaCN and Zn in this process.  R [CBSE D/OD 2018]



Ans. 4Au(s) + 8CN–(aq) 2H2O(aq) + O2(g) → 4[Au(CN)2]– (aq) + 4OH–(aq)[1] 2[Au(CN)2]–(aq) + Zn(s) → 2Au(s) +[Zn(CN)4]2–(aq) [1] (No marks will be deducted for not balancing) NaCN leaches gold/NaCN acts as a leaching agent / complexing agent [½] Zn acts as reducing agent / Zn displaces gold. [½] [CBSE Marking Scheme 2018]

Q. 11. Write the chemical reactions involved in the process of extraction of Gold. Explain the role of





OR

[Topper’s Answer 2018]

[ 163

GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

Detailed Answer : During extraction of gold, aqueous NaCN leaches gold from its ore in presence of air to form a complex [Au(CN)2]–. The reaction takes place is given as below : 4Au + 8CN– + 2H2O + O2 → 4[Au(Cn)2]– 4OH– Zinc reduces the complex formed to give pure gold as follows : 2[Au(CN)2]– + Zn → 2Au + [Zn(CN)4]2– Pure gold Role of Dil. NaCN : Dil. NaCN is used as complexing agent oxidising agent which oxidises Au to Au+. Role of Zn : Zinc is used as reducing agent which reduces Au+ to pure gold.

(b) Write chemical reactions taking place in the extraction of copper from Cu2S.



2Cu 2S + 3O2 → 2Cu 2O + 2SO2

. 12. Describe how the following steps can be carried out? Q (a) Recovery of Gold from leached gold metal complex. (b) Conversion of Zirconium iodide to pure Zirconium. (c) Formation of slag in the extraction of copper. (Write the chemical equations also for the reactions C involved) Ans. (a) Leached gold complex is treated with zinc and gold is recovered by displacement method. ½ 2Au[(CN)2]–(aq) + Zn(s) → 2Au(s)  +[Zn(CN)4]2–(aq) ½ (b) Zirconium iodide is decomposed on a tungsten filament; electrically heated to 1800 K. Pure Zr metal is deposited on the filament.  ½ ZrI4 → Zr + I2 ½ (c) Silica is added to the ore and heated. It helps to slag off iron oxide as iron silicate.  ½ FeO + SiO2 → FeSiO3 (slag) ½ Q. 13. Explain the use of the following: (a) NaCN in Froth Floatation Method. (b) Carbon monoxide in Mond’s process. (c) Coke in the extraction of Zinc from Zinc Oxide U Ans. (a) NaCN is used as depressant to separate two sulphide ores ( ZnS and PbS) in Froth flotation Method. 1 (b) Carbon monoxide forms a volatile complex of nickel, nickel tetracarbonyl. 1 (c) Coke is used as a reducing agent to reduce zinc oxide to zinc. 1 Q. 14.  Give reasons :



(If iron sulphide is present) :



2FeS + 3O2 → 2FeO + 2SO2



(ii) Removal of iron oxide as slag :



FeO + SiO2 → FeSiO3  slag ( )



(iii) Reduction of coper(I) oxide :



Cu 2O + C → Cu + CO



2Cu 2O + Cu 2S → 6Cu + SO2 



(a) Name the method of refining which is : (i) Used to obtain semiconductor of high purity, (ii) Used to obtain low boiling metal.

R [CBSE Delhi Set-1 2019]

 Ans. (a) (i) Zone refining

(a) (i) Zone refining

(ii) distillation

(b) (i) Roasting of the sulphide are:        



[3]

Q. 15. Write the principle of the following : (a) Hydraulic washing (b) Chromatography (c) Froth-floatation process R [CBSE OD, Set-3 2019]

Ans. (a) This is based on difference in gravities of the ore and gangue particles. 1

(b) Different compounds of a mixture are differently adsorbed on an adsorbent. 1 (c) The mineral particles become wet by oil while the gangue particles by water. 1 [CBSE Marking Scheme 2019]

Detailed Answer : (a) Hydraulic washing : It is based on the differences in gravities of the ore and the gangue particles. (b) Chromatography : Different components of a mixture are differently adsorbed on an adsorbent. (c) Froth-floatation process : This is based upon the preferential wetting of mineral/ ore particles by oil while the gangue particles by water.

Long Answer Type Question Q. 1. Explain the following: (a) Below 710 K, CO2 is a better reducing agent whereas above 710 K, CO is a better reducing agent. (b)  Sulphide ores are generally converted into oxides before reduction.

(ii) Distillation[½ + ½]

(b) 2Cu2S + 3O2 → 2Cu2O + 2SO2[1] 2Cu2O + Cu2S → 6Cu + SO2 [CBSE Marking Scheme, 2019] [1] Detailed Answer :



(5 marks each) (c) In the reverberatory furnace, silica is added to the sulphide ore of copper. (d) At high temperatures, carbon and hydrogen are not used as reducing agents. (e) For purification of Ti, vapour phase refining method is used.

A&E

164 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Ans. (a)  Ellingham diagram which relates Gibbs free energy and temperature at below 710 K. ∆G(C, CO2 ) < ∆G(C, CO) . Thus, CO2 is a better reducing agent than CO while above 710 K, CO becomes a very good reducing agent. [1] (b) Sulphide ores cannot be reduced easily but oxide ores can be easily reduced. So, sulphide ores are generally converted into oxides before reduction.  [1] (c) Copper pyrites contain iron sulphide in addition to copper sulphide. In the reverberatory furnace, copper ore is roasted to give oxides. FeO is removed by adding silica from the matte containing Cu2S and FeS. [1]





2FeS+ 3O2 ® 2FeO + 2SO2





FeO + SiO2 ® FeSiO3 (Slag)



(d) At the high temperature carbon and hydrogen react with metals to form carbides and hydrides respectively. Hence, they are not used as reducing agents. [1]

(e) Ti reacts with iodine to form TiI4 which is volatile and decomposes to give Ti at high temperature to give extra pure titanium. 530K , 800 K → TiI4 1 → Ti (Pure) Ti (Impure)+2I2 





+ 2I2 [1]



TOPIC-2

Principles of Extraction of Aluminium, Copper, Zinc and Iron Revision Notes  Chief Ores and Methods of Extraction of Some Common Metals : Metal Copper

Occurrence

Extraction Method

Copper pyrites, CuFeS2

Remark

Roasting of sulphide partially It is self reduction in a and reduction. specially designed converter.

Cuprite, Cu2O Malachite, CuCO3.Cu(OH)2 Copper glance, Cu2S Azurite, 2CuCO3.Cu(OH)2

2Cu2O+Cu2S → 6Cu+ SO2

Aluminium

Bauxite, Al2O3.xH2O Cryolite, Na3AlF6 Kaolinite, [Al2(OH)4Si2O5] Aluminosilicates

Electrolysis of Al2O3 dissolved A good source of electricity in molten cryolite or in Na3AlF6. is needed in the extraction of Al.

Zinc

Zinc blende or Sphalerite, ZnS Zincite, ZnO Calamine, ZnCO3

Roasting and then reduction The metal may be purified with C. by fractional distillation.

Iron

Haematite, Fe2O3

Reduction with the help of Limestone is added as CO and coke in blast furnace. flux which removes SiO2 Chemical reduction with CO. as calcium silicate (slag) floats over molten iron Calcination followed by and prevents its oxidation. reduction with CO. Temperature approaching Roasting followed by reduction. 2170 K is required. Chemical reduction with CO.

Magnetite, Fe3O4 Siderite, FeCO3 Iron pyrites, FeS2 Limonite, Fe2O3.3H2O

Sulphuric acid leaching is also employed.

 Flowchart for Extraction of Iron : Iron ore : Haematite (Fe2O3) and Magnetite (Fe3O4) ↓ Concentration of ore : Electromagnetic separation ↓ ↓ Calcination and roasting :  → Moisture, CO2, SO2 and As2O3 are removed Concentrated Ore + Air heat FeO is oxidised to Fe2O3

GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

[ 165

↓ Smelting : (Blast furnace) 2FeS + 3O2 → 2FeO + 2SO2 FeO + SiO2 → FeSiO3 (Slag) FeS + Cu2S → Matte ↓ Bessemerisation in Bessemer's converter lined with lime and magnesia 2FeS + 3O2 → 2FeO + 2SO2 FeO + SiO2 → FeSiO3 (Slag) 2Cu2S + 3O2 → 2Cu2O + 2SO2 self   → 6Cu + SO2 2Cu2O + Cu2S reduction

 Flowchart for the Extraction of Zinc : Zinc ore : Zinc blende (ZnS), Calamine (ZnCO3) and Zincite (ZnO)

↓



Concentration by Froth Flotation Process :

Powdered ore + Water + Pine oil + Foam (containing sulphide)

↓ Roasting in Reverberatory Furnace in the presence of air : 2ZnS + 3O2 → 2ZnO + 2SO2 ZnS + 2O2 → ZnSO4 2ZnSO4 → 2ZnO + 2SO2 + O2



↓ Reduction :



ZnO + C → Zn + CO ↓



Purification by Electrolytic Method Anode : Plates of impure Zn Cathode : Plates of pure Zn Electrolyte : Solution of ZnSO4 Deposition of pure zinc on cathode

 Flowchart for Extraction of Aluminium : Aluminium ore : Bauxite Al2O3.xH2O





↓ Concentration of ore : Leaching Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2Na[Al(OH)4](aq) 2Na[Al(OH)4](aq) + CO2(g) → Al2O3. xH2O(s) + 2NaHCO3(aq) Al2O3 xH2O(s) 1470K → Al2O3(s) + xH2O(g)





↓ Electrolytic reduction : (Hall and Heroult process) At cathode : Al3+(melt) + 3e– → Al(l) At anode : C(s) + O2–(melt) → CO(g) + 2e– C(s) + 2O2–(melt) → CO2(g) + 4e– 2Al2O3 + 3C → 4Al + 3CO2

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Extraction of Copper

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

 Flowchart for the Extraction of Copper : Ores : Copper pyrite (CuFeS2)

↓



Concentration of ore by Froth flotation method









↓





Roasting : 2CuFeS2 + O2 → Cu2S + 2FeS + SO2 2FeS + 3O2 → 2FeO + 2SO2 ↓ Smelting : FeO + SiO2 → FeSiO3 (Slag) ↓ Bessemerisation : 2Cu2S + 3O2 → 2Cu2O + 2SO2 2Cu2O + Cu2S → 6Cu + SO2 ↓ Purification : By electrolytic refining by taking impure Cu as anode, pure Cu as cathode and acidified CuSO4 solution as electrolyte : At anode : Cu → Cu2+ + 2e– At cathode : Cu2+ + 2e– → Cu

 Varieties of iron and their comparison : S. No.

Properties

Cast Iron

Wrought Iron

Steel

1.

Iron content

94 – 96%

98·5 – 98·8%

98·5 – 99·5%

2.

Carbon content

2·5 – 4·5%

0·12 – 0·25%

0·5– 1·5%

3.

Content of Si, P, S and Mn

1·5%

0·95 – 1·4%

—

4.

Hardness

Very hard

Soft

Hard

5.

Melting point

1200°C

1500°C

1300 °C

6.

Malleability

Brittle

Malleable

Malleable

7.

Welding

Not possible

May be done

Can be done but with difficulty

8.

Rust

Does not rust

Rusts

Does not rust

 Some important types of Ores : S. No. 1.

Ore type Native

Cu, Ag, Au, Hg, As, Bi, Sn, Pd, Pt

Example

2.

Oxides

Al2O3, Fe2O3, Fe3O4, SnO2, MnO2, TiO2, FeCr2O4, WO3, Cu2O, ZnO

3.

Carbonates

CaCO3, MgCO3, FeCO3, PbCO3, BaCO3, SrCO3, ZnCO3, MnCO3, CuCO3

4.

Sulphides

Ag2S, Cu2S, PbS, ZnS, HgS, FeS, Bi2S3, NiS,CaS, MoS2

5.

Halides

NaCl, KCl, AgCl, MgCl2.6H2O, NaCl and MgCl2 (in sea water)

6.

Sulphates

BaSO4, SrSO4, PbSO4, CuSO4, CaSO4. H2O

7.

Silicates

Be3Al2Si6O18, ZnSiO4, Sc2Si2O7, NiSiO3, MgSiO3

8.

Phosphates

CrPO4, LaPO4, Th3(PO4)4, LiF.AlPO4

Know the Terms  Complex ores : These are the mixtures of several minerals. For example : Lepidolite [K(Li, Al, Rb)2. (Al, Si)4O10 (F, OH)2], Triphylite [LiFePO4].  Native ores : These ores contain metals in their elemental form associated with alluvial impurities like clay, sand, etc.

[ 167

GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

Q. Answer the following : (i) What is the role of cryolite in the metallurgy of aluminium ? (ii) Differentiate between roasting and calcination. (iii) What is meant by the term ‘chromatography’ ? Solution: STEP-1: It lowers the melting point of alumina / acts as a solvent.  STEP-2: S. No. Roasting Calcination (a) Process of heating the ore below Process of heating the ore below its its melting point with excess of melting point in absence or limited air supply of air (b) Volatile impurities are removed as Water and inorganic impurities are oxides. removed. STEP-3: It is a process of separation of different components of a mixture which are differently adsorbed on a suitable adsorbent.

Objective Type Questions

(1 mark each)

[A] MULTIPLE CHOICE QUESTIONS :

Ans.

Correct option : (a)

Q. 1.





Explanation : Electrolysis of brine solution:





At anode: 2Cl– (aq) ® Cl2(g) + 2e–





(b) Cu 2O+C → 2Cu + CO

At cathode: 2H2O(l) + 2e– ® 2OH–(aq) + H2(g)





(c) Cu 2+(aq.)+Fe (s) → Cu (s)+Fe2+(aq.) 1 1 (d) Cu 2O + Cu 2S → 3Cu + SO 2 2 2

Reaction: 2NaCl + 2H2O ® Cl2 + H2 + 2NaOH

Q. 3.

Which of the following reactions is an example of auto-reduction?

(a) Fe3O 4 + 4CO → 3Fe+ 4CO 2

R

Ans.

Correct option : (d) Explanation : Reaction includes reduction of copper (I) oxide by copper (I) sulphide and in this process copper is reduced by itself. This process is called as auto-reduction. The solidified copper so obtained is known as blistered copper.

Q. 2.

Brine electrolysed by using inert electrodes. The reaction at anode is : 1 Θ = 1.36 V . (a) Cl− (aq.) → Cl2 (g)+e− ; ECell 2 + − Θ (b) 2H 2O (l ) → O 2 (g) + 4H + 4e ; ECell = 1.23 V . + − Θ (c) Na (aq.) + e → Na (s); ECell = 2.71 V . 1 Θ A (d) H+ (aq.) + e− → H 2 (g); ECell = 0.00 V . 2



When copper ore is mixed with silica in a reverberatory furnace, copper matte is produced. The copper matte contains _____________. (a) sulphides of copper (II) and iron (II). (b) sulphides of copper (II) and iron (III). (c) sulphides of copper (I) and iron (II). (d) sulphides of copper (I) and iron (III). R Ans. Correct option : (c)



Explanation : Copper ore when mixed with silica, iron oxide slags off as iron silicate and copper is produced in the form of copper matte which contains Cu2S (I) and FeS (II).

[B] ASSERTION & REASON TYPE QUESTIONS :



In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

168 ]

(b)



(c)

(d) Q. 1. Ans.

Q. 2.



Ans.

Both assertion and reason are correct statements, and reason is the correct explanation of the assertion. Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion. Assertion is correct, but reason is wrong statement. Assertion is wrong, but reason is correct statement. Assertion : Zone refining method is very useful for producing semiconductors. Reason : Semiconductors are of high purity. Correct option: (b) Explanation: The impurities of semiconductors are more soluble in molten zone and the ultrapure semiconductor crystallizes in zone refining method. Assertion : Zirconium can be purified by van Arkel method. Reason : ZrI4 is volatile and decomposes at 1,800 K. Correct option: (a) Explanation: Zirconium is also purified by vapour phase refining method in which it is treated with iodine to form ZrI4 which on heating decomposes to give pure zirconium.

Q. 3.

Assertion : Nickel can be purified by Mond’s process.



Reason : Ni(CO)4 is a volatile compound which decomposes at 460 K to give pure Ni.



Ans.

Correct option: (a)



Explanation: When nickel (Ni) is treated with carbon monoxide, it forms nickel tetracarbonyl Ni(CO)4, while impurities are left behind. When the vapour of Ni(CO)4 is heated at 460 K, it is decomposed to give pure nickel, while carbon monoxide is removed as gas.



[C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1.



(a)

Which reducing agent is employed to get copper from the leached low grade copper ore ? R

Ans. Hydrogen/Iron.

[CBSE Marking Scheme 2014]

Q. 2. What is the role of zinc metal in the extraction of silver ? R



Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



Ans. Zn acts as reducing agent. [CBSE Marking Scheme 2014]

Ans.

 ame the substance used as depressant in N the separation of two sulphide ores in Froth flotation method. R Sodium cyanide.

Short Answer Type Question-I

(2 marks each)

Q. 1. A mixture containing two compounds A and B is passed through a column of Al2O3 using alcohol as eluent. Compound A is eluted in preference to compound B, which compound is more readily adsorbed on the column?

A&E

Ans. As the mixture of compounds A and B is passed through a column of Al2O3 by using alcohol as eluent and compound A is eluted in preference to compound B, it indicates that compound B is more readily adsorbed on the column. [2] Q.2. How can the impurities like sulphur, silicon and phosphorus be removed from cast iron? Write a reaction used for the preparation of wrought iron (purest form of iron) from cast iron. Ans. The required reaction is given below : Fe2O3 + 3C → 2Fe + 3CO Limestone is added as flux and the impurities of sulphur, silicon and phosphorus are converted to their oxides and pass into slag. [2] Q. 3.  What is meant by Vapour phase refining? Write any one example of the process which illustrates this technique, giving the chemical equations involved. OR

Q. 3.



Write and explain the reactions involved in the extraction of gold. R

Ans. Vapour phase refining : It is a refining method in which the metal is converted into its volatile compound and collected elsewhere. It is then decomposed to give pure metal.  Example : Mond’s process for refining of Nickel / van Arkel method for refining of Zirconium. [1] Equations involved : 

330 − 350K Ni + 4CO  → Ni(CO)4 [½] 450 - 470K [½] Ni(CO)4 ¾¾¾¾ ® Ni + 4 CO  OR Extraction of gold involves leaching the metal with CN–. [½] Oxidation reaction :

4Au( s) + 8CN − ( aq ) + 2H 2 O( aq ) + O2 ( g ) → 4[Au(CN)2 ]− ( aq ) + 4OH − ( aq )  [½] The metal is recovered by displacement method :  2[Au(CN)2 ]- ( aq ) + Zn( s) ®

2 Au( s) + [Zn(CN)4 ]2 - ( aq ) Zinc acts as a reducing agent.

 [½] [½]

GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

Short Answer Type Questions-II Q. 1. Outline the principles of refining of metals by the following methods : (i) Distillation (ii) Zone refining R (iii) Electrolysis Ans. (i) The impurities are evaporated from volatile metals to obtain the pure metal as distillate.  [1] (ii) This method is based on the principle that the impurities are more soluble in the molten state than in the solid state of the metal.  [1] (iii) The impure metal is made to act as anode. A strip of the same metal in pure form is used as cathode. They are put in a suitable electrolytic bath containing soluble salt of the same metal. The more basic metal remains in the solution and the less basic ones go to the anode mud. [1] Q. 2. Write down the reactions taking place in different zones in the blast furnace during the extraction of iron. How is pig iron different from cast iron ? U [CBSE Comptt. Delhi 2015]  Ans.

3Fe2O3 + CO → 2Fe3O4 + CO2 (Iron ore) Fe3O4 + CO → 3FeO + CO2 CaCO3 → CaO + CO2 (Limestone) CaO + SiO2 → CaSiO3 (Slag) FeO + CO → Fe + CO2 C + CO2 → 2CO Coke C + O2 → CO2 FeO + C → Fe + CO [½ × 4 = 2] (Any four correct equations) Cast iron has lower carbon content (about 3%) than pig iron / cast iron is hard & brittle whereas pig iron is soft. [1] [CBSE Marking Scheme 2015]

Commonly Made Error  Students often write the process involved instead of mentioning the principle behind the process.

Answering Tip  The extraction of metals should be studied in detail. All the steps must be shown in proper order with balanced chemical equation.

Q. 3. (i) Write the principle of method used for the refining of germanium. (ii) Out of PbS and PbCO3 (ores of lead), which one is concentrated by froth flotation process preferably? (iii) What is the significance of leaching in the extraction of aluminium ?



R + U [CBSE Delhi Set-1, 3 2017]

[ 169

(3 marks each)

Ans. (i) Zone refining : The impurities are more soluble in the molten state (melt) than in the solid state of the metal. [1] (ii) PbS [1] (iii) Impurities like SiO2 etc, are removed by using NaOH solution and pure alumina is obtained.[1] [CBSE Marking Scheme 2017]

Q. 4. Write the role of (i) NaCN in the extraction of gold from its ore. (ii) Cryolite in the extraction of aluminium from pure alumina. (iii) CO in the purification of Nickel. R [CBSE Comptt. Delhi/OD 2018] Ans. (a) Gold is leached out in the form of a complex with dil. solution of NaCN in the presence of air/ NaCN acts as leaching agent. 1 (b) It lowers the melting point of alumina and makes it a good conductor of electricity. 1 (c) CO forms a volatile complex with nickel which is further decomposed to give pure Ni metal. 1 [CBSE Marking Scheme 2018]



Q. 5. (i) Indicate the principle behind the method used for the refining of zinc. (ii) What is the role of silica in the extraction of copper ? (iii) Which form of the iron is the purest form of commercial iron ? R [CBSE Delhi 2015]

Ans. (i) Zinc is refined by electrolytic refining. In this method, the impure metal acts as anode. A strip of the same metal in pure form is used as cathode. These are put in suitable electrolytic bath containing soluble salt of the same metal. The more basic metal remains in the solution and the less basic ones go to the anode mud. [1] (ii) Roasting of copper pyrite (CuFeS2) gives FeO, Cu2O and SO2. 4CuFeS2(s) + 11O2(g) → 4FeO(s) + 2Cu2O(s) +  8SO2(g) To remove FeO, SiO2 acts as flux and is added to form slag. FeO(s) + SiO2(s) → FeSiO3(l) [1] (slag) (iii) Wrought iron. [1] Q. 6. (i) Write the role of ‘CO’ in the purification of nickel. (ii) What is the role of silica in the extraction of copper? (iii) W  hat type of metals are generally extracted by electrolytic method?  R [CBSE Delhi Set-2 2019] 3 Ans. (i) To produce a volatile complex, which decomposes on further heating to give pure nickel. 1 (ii) To remove impurities (FeO) by forming a slag. / acts as a flux. 1



170 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(iii) More reactive metals having large negative electrode potential. [CBSE Marking Scheme, 2019] 1



(ii) The conversion of Zinc blende to Zinc metal is done by fractional distillation. (iii) The conversion of [Ag(CN)2]– to Ag is done by Leaching. Q. 8. Write the name and principle of the method used for refining of (a) Zinc, (b) Germanium, (c) Titanium. R [CBSE OD Set-1 2019] 3

Detailed Answer : (i) In Mond’s process for refining nickel, nickel is heated with CO to form Ni(CO)4 which is decomposed at a higher temperature to obtain the pure metal. So CO acts like a catalyst in this Ans. (a) Distillation/ Electrolytic refining : The process. impure metal is evaporated to obtain the Heat Heat Ni (impure) + 4CO ¾¾¾®  Ni (CO)  ¾¾¾® Ni (pure) + 4CO pure metal as distillate / The more basic metal 4

Heat

remains in the solution and the less basic ones go to the anode mud. [½ + ½] (b) Zone refining : Impurities are more soluble in the melt than in the solid state of the metal. [½ + ½] (c) van Arkel method : The metal should form a volatile compound which decomposes at higher temperature to pure metal. [CBSE Marking Scheme, 2019] [½ + ½]

Heat

Ni (impure) + 4CO ¾¾¾®  Ni (CO)4  ¾¾¾® Ni (pure) + 4CO

(ii) The ore is heated in a furnace after mixing with silica. In the furnace, iron oxide reacts with silica to form a slag of iron silicate. The copper is produced in the form of copper matte. This contains Cu2S and FeS. The copper matte then charged into silica line convertor and it converts any remaining FeS, FeO into slag and Cu2S/CuO into metallic copper. The role of silica in copper extraction is to remove the iron oxide obtained during the process of roasting. FeO + SiO2  ® FeSiO3  slag ( )    The silica act as a flux. (iii)  The metals which are highly reactive in nature are generally extracted through electrolytic process. For example, Sodium, Aluminium, Calcium, etc. Q. 7. How will you convert the following? (i) Impure Nickel to pure Nickel.  (ii) Zinc blende to Zinc metal. (iii) [Ag(CN)2]– to Ag. R [CBSE Delhi Set-3 2019] Ans. (i) Nickel is heated in a stream of carbon monoxide forming a volatile complex named as nickel tetracarbonyl. This complex is decomposed at higher temperature to obtain pure metal. 1 (ii) ZnS is roasted to give ZnO which is heated with reducing agent coke to give Zn. 1 (iii) Complex is treated with zinc , displacement reaction occurs to give pure Ag. 1 [CBSE Marking Scheme, 2019] Detailed Answer : (i) The conversion of impure Nickel to pure Nickel is done by the Mond’s process.



Q. 9. Write the name and principle of the method used for refining of (a) Tin, (b) Copper, (c) Nickel. R [CBSE OD Set-2 2019] Ans. (a) Liquation : Metals having low melting points than impurities. [½ + ½] (b) Electrolytic refining : The more basic metal remains in the solution and the less basic ones go to the anode mud. [½ + ½] (c) Mond’s process : Ni should form a volatile compound with a suitable reagent which decomposes at higher temperature to pure Ni.  [CBSE Marking Scheme, 2019] [½ + ½]

Detailed Answer : (a) Liquation. Principle : This method is based on the lower melting point than the impurities and tendency of the molten metal to flow on the sloping surfaces. (b) Electrolytic refining. Principle : This method is based on the phenomenon of electrolysis where pure metal gets deposited on cathode when electricity is passed through the electrolyte. (c) Vapour phase refining (Mond’s process). Principle : The metal is converted into its volatile compound and collected else where which gives pure metal when decomposed at high temperature.

Visual Case-Based Questions Q. 1. Read the passage given below and answer the following questions: (1×4=4) Chromatography is based on the principle where molecules in mixture applied onto the surface or into the solid, and fluid stationary phase (stable phase) is separating from each other while moving with the

(4 marks each) aid of a mobile phase. The factors effective on this separation process include molecular characteristics related to adsorption (liquid-solid), partition (liquidsolid), and affinity or differences among their molecular weights. Because of these differences, some components of the mixture stay longer in

GENERAL PRINCIPLES AND PROCESSES OF ISOLATION OF ELEMENTS

the stationary phase, and they move slowly in the chromatography system, while others pass rapidly into mobile phase, and leave the system faster. Source: Separation techniques: Chromatography (nih.gov) In these questions a statement of Assertion followed by a statement of Reason is given. Choose the correct answer out of the following choices. (a)  Assertion and reason both are correct statements and reason is correct explanation for assertion. (b)  Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. (i) Assertion: Different components of the mixture are adsorbed at different levels on the chromatographic column.  Reason: The adsorbed components are eluted by using suitable solvents. (ii) Assertion : Column chromatography is used for purification of substances in large quantities.  Reason : It produces coloured bands on the column. (iii) Assertion : Chromatography is used in analysis, isolation and purification of substances.  Reason: It requires a mobile and a stationary phase.

(iv) Assertion : Some components of the mixture stay longer in the stationary phase, and they move slowly in the chromatography system. 

Reason: Stationary phase is always composed of phase is always composed of “liquid” or a “gaseous component. Ans. (i) Correct option : (b)







Explanation : Different components of the mixture are adsorbed at different levels on the chromatographic column depending on mobile medium, the adsorbent material etc. The adsorbed components are later eluted using suitable solvents. (ii) Correct option : (d) Explanation: Column chromatography is used for purification of substances available in minute quantities.   (iii) Correct option : (b) Explanation: Chromatography is used in analysis, isolation and purification of substances where components are separated based on their distribution in two phases- a mobile and a stationery phase. (iv) Correct option : (c) Explanation: Stationary phase is always composed of a “solid” phase or “a layer of a liquid adsorbed on the surface a solid support”.

[ 171

Q. 2. Read the passage given below and answer the following questions: (1×4=4) In general, all blast furnace feed materials require some form of processing before being deemed suitable for blast furnace needs. Fine and ultra-fine ferrous ores must first be agglomerated to produce sinter or pellets respectively. Lump ore may be charged directly to a blast furnace but only after it has been suitably sized and screened to remove over and undersize material. Lump ore, however, usually comprises a minor portion of the total ferrous feed. Coal cannot be directly charged via the furnace top, it must first be transformed to coke. Raw materials are charged to the furnace in alternating layers of coke and ore. This alternating layer structure inside the furnace has a profound impact on the operation of the furnace and on the required quality of coke. Source: Blast Furnace Operation - an overview | ScienceDirect Topics (i) The main reactions occurring in blast furnace during extraction of iron from haematite are ________________. (a) Fe2O3 + 3CO → 2Fe + 3CO2 (b) FeO + SiO2 → FeSiO3 (c) Fe2O3 + 3C → 2Fe + 3CO (d) All of the above (ii) Choose the correct statement from the following: (a) In extraction of silver, silver is extracted as cationic complex. (b) Nickel is purified by zone refining. (c) Cast iron is obtained by remelting pig iron with scrap iron and coke using hot air blast. (d) None of the above (iii) For the metallurgical process of which of the ores, calcined ore cannot be reduced by carbon? (a) haematite (b) calamine (c) iron pyrites (d) All of the above (iv) Iron obtained from blast furnace with about 4% carbon and impurities like S,P,Mn,Si in smaller amounts is known as (a) Wrought iron (b) Pig iron (c) Cast iron (d) Commercial iron Ans. (i) Correct option : (a) Explanation : The main reactions occurring in blast furnace during extraction of iron from haematite is __________. Fe2O3 + 3CO → 2Fe + 3CO2   (ii) Correct option : (c) Explanation: Cast iron is obtained by remelting pig iron with scrap iron and coke using hot air blast.

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

 (iii) Correct option : (c) Explanation: Haematite (Fe2O3) and Calamine (ZnCO3) can be reduced by carbon during the metallurgical process as these are the oxides of iron and zinc (Calamine ore on calcination dissociates to give oxide ZnCO3 → ZnO + CO2) respectively. Iron pyrites (FeS2) is sulphide ore of iron. (iv) Correct option : (b) Explanation: Pig iron is obtained from the blast furnace which has 4% carbon and impurities like S,P,Mn,Si in smaller amounts. Q. 3. Read the passage given below and answer the following questions:  (1×4=4) Pfann’s technique of zone-refining has been applied to the purification of many metals, particularly silicon and germanium. The high purity silicon required for transistors for the electronic industry is produced by the concept of purification based on zone-refining. The concept is based on the nonequilibrium behaviour during solidification of the solid solution alloys. Purification of a metal by zonemelting (i.e., melting in zones, certain thickness at a time) technique is called zone-refining. The solidifying phase, in a solid solution type of liquid alloy (in diagrams having separate liquid and solids), is purer than the liquid.



The following questions are multiple choice questions. Choose the most appropriate answer :



(i)  Zone refining is based on the principle that ______________.



(a)  Impurities of low boiling metals can be separated by distillation



(b) Impurities are more soluble in molten metal than in solid metal.



(c)  Different components of a mixture are different absorbed on an adsorbent.



(d)  Vapour of volatile compound decomposed in pure metal.

can

be

(ii) Zone refining is useful for producing metals used in

(a) Pendulum

(b) Coinage alloy

(c) Wires

(d) Semiconductors

(iii) Zone refining is used in the purification of which of the following metals?



(a) Zn (c) In

(b) Tl (d) Se

(iv)



172 ]

In the given image, identify the zones X and Z.



(a) X-Impure Germanium, Z- Pure Germanium



(b)  X- Recrystallised pure Impure Germanium



(c) X- Molten Germanium, Z- Solid Germanium

Germanium,

Z-

(d) X- Solid Germanium, Z- Molten Germanium Ans. (i) Correct option : (b) Explanation: Zone refining is based on the principle that impurities are more soluble in molten metal than in solid metal.   (ii) Correct option : (d) Explanation: Zone refining is useful for producing metals used in semiconductors and other metals of very high purity e.g Ge, Si.  (iii) Correct option : (c) Explanation: Zone refining is used for the purification of Indium(In) metal.  (iv) Correct option : (b) Explanation: X- Recrystallised pure Germanium, Z- Impure Germanium As the heater moves along an impure solid rod of Germanium, the impurities move along the length in the adjacent molten zone. 

[ 173

SELF-ASSESSMENT TEST

Self Assessment Test - 6 Time : 1 Hour

Max. Marks : 25

Q. 1.  Read the passage given below and answer the following questions : (1 × 4 = 4) Note : Answer the questions on the basis of following figure : –200 eO 2F D A

–400

E

O2 e+ 2F C + O2

CO2

2C

0

+

–500

2

O

B

2C

G°/kJ mol–1 of O2

–300

+

O 2

–600

–700

2C

O

0

400

800 1200 1600 Temperature (°C)

2000



(i) Choose the correct option of temperature at which carbon reduces FeO to iron and produces CO. Below temperature at point A. (a) (b) Approximately  at the temperature corresponding to point A. (c) Above temperature at point A but below  temperature at point D. (d)  Above temperature at point A. A&E (ii) Below point ‘ A ’ FeO can ______________. (a) be reduced by carbon monoxide only. (b)  be reduced by both carbon monoxide and carbon. (c) be reduced by carbon only. (d)  not be reduced by both carbon and carbon monoxide. A&E (iii) For the reduction of FeO at the temperature corresponding to point D, which of the following statements is correct? (a)  ∆G value for the overall reduction reaction with carbon monoxide is zero. (b) ∆G value for the overall reduction reaction with a mixture of 1 mol carbon and 1 mol oxygen is positive. (c) ∆G value for the overall reduction reaction with a mixture of 2 mol carbon and 1 mol oxygen will be positive. (d) ∆G value for the overall reduction reaction with carbon monoxide is negative. A&E (iv) At the temperature corresponding to which of the points in Fig. FeO will be reduced to Fe by coupling the reaction 2FeO → 2Fe + O2 with all of the following reactions?

(A) C + O2 → CO2 (B) 2C + O2 → 2CO and (C) 2CO + O2 → 2CO (a) Point A and D (b) Point B and D (c) Point A and E (d) Point B and E Q. 2. Read the passage given below and answer the following questions :  (1 × 4 = 4) Froth flotation is a process for selectively separating hydrophobic materials from hydrophilic. This is used in mineral processing, paper recycling and waste-water treatment industries. Historically this was first used in the mining industry, where it was one of the great enabling technologies of the 20th century. It has been described as “the single most important operation used for the recovery and upgrading of sulfide ores”. The development of froth flotation has improved the recovery of valuable minerals, such as copper- and leadbearing minerals. Along with mechanized mining, it has allowed the economic recovery of valuable metals from much lower grade ore than previously. Choose the correct answer out of the following choices. (a)  Assertion and reason both are correct statements and reason is correct explanation for assertion. (b)  Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. (i) Assertion : Sulphide ores are concentrated by froth flotation method Reason : Cresols stabilize the froth in froth flotation method. (ii) Assertion : Pine oil is used to concentrate sulphide ores. Reason : It helps in the rise of impurities to the surface. (iii) Assertion : NaCN acts as a depressant in case of an ore containing ZnS and PbS. Reason : The role of the depressants is to separate two sulphide ores by selectively preventing one ore from forming froth. (iv) Assertion : Collectors enhance the non-wettability of the mineral particles. Reason : A suspension of the powdered ore is made with oil.

174 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q.3. In the metallurgy of aluminium _____________.

3+

(i) Al

is oxidised to Al(s).



 (ii) g  raphide anode is oxidised to carbon monoxide and carbon dioxide.



(iii) oxidation state of oxugen changes in the reaction at anode.



(iv) oxidation state of oxygen changes in the overall reaction involved in the process.

Q.4. In the following question a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.



(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.



(c)  Assertion is correct statement but reason is wrong statement.





Q. 7. (i) Name the method of refining of metals such as Germanium.

 (ii) In the extraction of Al, impure Al2O3 is dissolved in conc. NaOH to form sodium aluminate and leaving impurities behind. What is the name of this process?



(iii) What is the role of coke in the extraction of iron from its oxides?

Q. 8. (i) Indicate the principle behind the method used for the refining of zinc.  (ii) What is the role of silica in the extraction of copper ?

(iii)  Which form of the iron is the purest form of commercial iron ? R Q. 9. Explain the following :

(a) C  O2 is a better reducing agent below 710 K whereas CO is a better reducing agent above 710 K.

(d)  Assertion is wrong statement but reason is correct statement.



 (b) Generally, sulphide ores are converted into oxides before reduction.

Assertion : Hydrometallurgy involves dissolving the ore in a suitable reagent followed by precipitation by a more electropositive metal.



  (c) Silica is added to the sulphide ore of copper in the reverberatory furnace.



 (d) Carbon and hydrogen are not used as reducing agents at high temperatures.



(e) Vapour phase refining method is used for the purification of Ti.  A&E

Reason : Copper is extracted by hydrometallurgy. 

Q.5. Name the method that is used in the refining of nickel and copper. R Q.6.  Write the 2 reactions involved to obtain metallic copper from its sulphide ore by roasting/smelting.

 

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p-BLOCK ELEMENTS

[ 175

7

CHAPTER

p-BLOCK ELEMENTS

Syllabus ¾¾ Group–15 Elements : General introduction, electronic configuration, occurrence, oxidation states, trends in physical and chemical properties; Nitrogen preparation properties and uses; compounds of Nitrogen: preparation and properties of Ammonia and Nitric Acid, Oxides of Nitrogen (Structure only); Phosphorus - allotropic forms, compounds of Phosphorus: Preparation and properties of Phosphine, Halides and Oxoacids (elementary idea only). ¾¾ Group–16 Elements : General introduction, electronic configuration, oxidation states, occurrence, trends in physical and chemical properties, dioxygen: preparation, properties and uses, classification of Oxides, Ozone, Sulphur -allotropic forms; compounds of Sulphur: preparation properties and uses of Sulphurdioxide, Sulphuric Acid: industrial process of manufacture, properties and uses; Oxoacids of Sulphur (Structures only). ¾¾ Group–17 Elements : General introduction, electronic configuration, oxidation states, occurrence, trends in physical and chemical properties; compounds of halogens, Preparation, properties and uses of Chlorine and Hydrochloric acid, interhalogen compounds, Oxoacids of halogens (structures only). ¾¾ Group–18 Elements : General introduction, electronic configuration, occurrence, trends in physical and chemical properties, uses.

Trend Analysis List of Concepts

2018

2019

D/OD 2Q (2 marks) 1Q (2 marks)

2Q (1 mark)

Give reason

1Q (3 marks)

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1Q (5 marks)

Complete the reactions

1Q (1 mark)

1Q (2 marks)

1Q (2 marks)

Arrange the following

1Q (1 mark)/ 1Q (2 marks) 1Q (3 marks)

1Q (2 marks)

Draw the structure What happens when

Miscellaneous

D

2020

1Q (2 marks)

OD 2Q (1 mark)

D 3Q (2 marks)

OD

1Q (1 mark)

[ 177

p-BLOCK ELEMENTS

TOPIC-1

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Group-15 Elements, Properties and Some Important Compounds Revision Notes

p-block elements-I

 Elements of p-Block : Group 13 Boron family Group 14 Carbon family Group 15 Nitrogen family Group 16 Oxygen family Group 17 Halogen family Group 18 Noble gases

B, Al, Ga, In, Tl C, Si, Ge, Sn, Pb N, P, As, Sb, Bi O, S, Se, Te, Po F, Cl, Br, I, At He, Ne, Ar, Kr, Xe, Rn

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 General electronic configuration of p-Block elements : ns2 np1–6

p-block

elements- II  Group 15 elements : (i) Nitrogen family : 7N, 15P, 33As, 51Sb, 83Bi (ii) Atomic radii : Smaller than the corresponding elements of group 14. TOPIC - 1 Down the group, they increase due to addition of new shells. Group-15 Elements, Properties and Some Important Compounds (iii) Ionisation enthalpy : Higher than the corresponding elements of ... P. 177 group 14. Down the group, it decreases due to increase in atomic size. (iv) Electronegativity : Decreases down the group with increasing atomic TOPIC - 2 size. Group-16 Elements, Properties and  Physical properties of Group 15 elements : Some Important Compounds (i) Except dinitrogen, all are solids. .... P. 185 (ii) Metallic character increases down the group due to decrease in TOPIC - 3 ionisation enthalpy and increase in the atomic size. Group-17 Elements, Properties and (iii) Boiling point increases from top the bottom. Some Important Compounds (iv) Melting point increases upto arsenic and decreases upto bismuth. .... P. 197 (v) Oxidation states : –3 to +5.  Chemical properties of Group 15 elements : TOPIC - 4 Group-18 Elements, Properties and (i) Towards hydrogen : All the elements form hydrides of the type Some Important Compounds EH3 where E = N, P, As, Sb, Bi. The stability decreases from NH3 to .... P. 205 BiH3. Reducing character increases down the group. Basic character decreases down the group. Boiling point of NH3 is greater than PH3 because of intermolecular hydrogen bonding. Boiling point increases from PH3 onwards. (ii) Towards oxygen : Form two types of oxides E2O3 and E2O5. The acidic character decreases down the group. (iii) Towards halogens : Directly combine with halogens to form trihalides (EX3) and pentahalides (EX5). (iv) Towards metals : All the elements react with metals to form their binary compounds exhibiting –3 oxidation state.  Anomalous properties of nitrogen : N does not form pentahalides because of non-availability of d-orbitals in its valence shell. It has ability to form pp-pp multiple bonds with itself and other elements having high electronegativity. N differs from the rest of the members of group due to small size, high electronegativity, high ionisation enthalpy and non-availability of d-orbitals.  Oxides of Nitrogen: Preparation and Properties:

S. No.

Formula

Name

Preparation

Properties

O.N.

Colourless gas, rather unreactive.

+1

Colourless gas, paramagnetic, + 2NO + 4H2O neutral.

+2

Brown gas, reactive, paramag2Pb (NO3)2 673K → 2PbO + 4NO2 + O2 netic, neutral.

+4

1.

N2O

Dinitrogen monoxide NH4NO3 heat  → N2O + 2H2O

2.

NO

Nitrogen monoxide

3.

NO2

Nitrogen dioxide

3Cu + 8HNO3 → 3Cu(NO3)2

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

4.

N2O3

Dinitrogen trioxide

5.

N2O4

Dinitrogen tetroxide

6.

N2O5

Dinitrogen pentoxide

Dark blue in liquid or solid state, unstable in the gas phase.

+3

Colourless, exists in equilibrium with NO2 both in the gaseous and liquid state.

+4

2HNO3 + P2O5 → 2HPO3 + N2O5 Unstable as gas; in the solid state Metaphosphoric acid exists as [NO2]+ [NO3]–.

+5

250 K

→ 2N2O3 2NO + N2O4  273 K   2NO2    N2O4 373 K

 Structures of Oxides of Nitrogen : N N

(N2O)

(NO)

113 pm N 119 pm O

N 115 pm O

N

O

N

O

N

O

O

11

4p

(N2O3)

N 186 pm

N

130° m

N O

O

O

N

p 21

N

m 1p 15

N

(N2O5)

112°

9 11

N

O

pm

O N

134°

O

O N

O

O O

O

N

O

O O

O O

1

O

O

O

m

175 pm

O

N

O

O

O

N

N

N

N

m

O

(N2O4)

O

0p

134°

O

O

O

O

12

N (NO2)

N

12

1p

N

O

O

105°

m

.

O .

N

O

O O

O

O

N

N O

O

N

O N

O

O

Fig. 1 : The molecular and Lewis dot resonance structure of oxides of nitrogen  Dinitrogen (N2) : Preparation : (i) In laboratory :

heat

NH4Cl(aq) + NaNO2(aq)  → N2(g) + 2H2O(l) + NaCl(aq)

(ii) By thermal decomposition : Ba(N3)2 → Ba + 3N2 Properties : (i) N2 has very little reactivity at ordinary temperature. (ii) Forms nitrides with highly electropositive metals like

 Ammonia (NH3) : Preparation : (i) In laboratory : (ii) By Haber’s process :

3Mg + N2 → Mg3N2 6Li + N2 → 2Li3N

heat

2NH4Cl + Ca(OH)2  → CaCl2 + 2NH3 + 2H2O



–1 Fe/Mo N2(g) + 3H2(g)      2NH3(g), ΔH = – 46.1 kJ mol 773 K

Properties : (i) Extremely soluble in water. (ii) Acts as Lewis base

 Ag+ + 2NH3    [Ag(NH3)2]+

Scan to know more about this topic

Oxides of nitrogen

[ 179



p-BLOCK ELEMENTS

(iii) Forms salts with acids

 Cu2+ + 4NH3    [Cu(NH3)4]2+ Deep blue 2+    Cd + 4NH3  [Cd(NH3)4]2+

ZnSO4(aq) + 2NH4OH(aq) → Zn(OH)2(s) + (NH4)2SO4(aq) 2FeCl3(aq) + 3NH4OH(aq) → Fe2O3.xH2O(s) + 3NH4Cl(aq) (iv) Reaction with Nessler’s reagent :   

2K2[HgI4] + NH3 + 3KOH → [OHg2.NH2]I + 7KI + 2H2O Nitric Acid (HNO3) : Preparation : (i) In laboratory : NaNO3 + H2SO4 → NaHSO4 + HNO3 (conc.) (Brown ppt.) (ii) By Ostwald’s process : 4NH3 + 5O2 → 4NO + 6H2O  2NO + O2    2NO2 3NO2 + H2O → 2HNO3 + NO Properties : (i) Colourless liquid. (ii) Concentrated nitric acid is a strong oxidising agent. (iii) Reactions : – HNO3(aq) + H2O(l) → H3O+ (aq) + NO 3(aq) 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O (dil) Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O (conc.) 4Zn + 10HNO3 → 4Zn(NO3)2 + 5H2O + N2O (dilute) Zn + 4HNO3 → Zn(NO3)2 + 2H2O + 2NO2 (conc.) I2 + 10HNO3 → 2HIO3 + 10NO2 + 4H2O S8 + 48HNO3 → 8H2SO4 + 48NO2 + 16H2O P4 + 20HNO3 → 4H3PO4 + 20NO2 + 4H2O Uses : (i) In the manufacturing of nitrates used in explosives (nitroglycerine, trinitrotoluene etc), fertilisers (ammonium nitrate) etc. (ii) As a reagent in laboratory. (iii) In preparing aqua-regia. Allotropes of Phosphorus : (i) White or yellow phosphorus (ii) Black phosphorus (iii) Red phosphorus Differences between White and Red Phosphorus : Property

White Phosphorus

Red Phosphorus

State

Translucent white waxy solid

Iron grey lustrous powder

Odour

Garlic

Odourless

Physiological action

Poisonous

Non-poisonous

Stability

Less stable

More stable than white P

Solubility

Insoluble in water but soluble in CS2.

Insoluble in water as well as CS2.

Action of air

Readily catches fire to give dense white fumes of P4H10, Chemiluminescent with Does not glow in dark. greenish glow.

180 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Effect of heat

Changes to a-black P when heated at 473 K Changes to a-black P when heated at under high pressure and changes to red P 803 K in a solid form. when heated at 573 K.

Structure

P4 (tetrahedral)

Tetrahedral units of P4 joined together through covalent bond.







Phosphine (PH3) Preparation : In laboratory :

 

P4 + 3NaOH + 3H2O (white) (conc.)

∆



inert atmosphere of CO2

PH3 + 3NaH2PO2 Sodium hypophosphite

Properties : (i) Colourless gas with rotten fish smell (ii) Highly poisonous (iii) Pure sample is not spontaneously inflammable (iv) Burns in air or oxygen when heated at 150°C 2PH3 + 4O2 → P2O5 + 3H2O 3CuSO4 + 2PH3 → Cu3P2 + 3H2SO4 3HgCl2 + 2PH3 → Hg3P2 + 6HCl PH3 + HBr → PH4Br Phosphorus trichloride (PCl3) Preparation : (i) By passing dry chlorine over heated white phosphorus P4 + 6Cl2 → 4PCl3 (ii) By action of thionyl chloride with white phosphorus P4 + 8SOCl2 → 4PCl3 + 4SO2 + 2S2Cl2 Properties : (i) Colourless oily liquid (ii) Hydrolyses in the presence of moisture PCl3 + 3H2O → H3PO3 + 3HCl (iii) 3CH3COOH + PCl3 → 3CH3COCl + H3PO3 Shape : Pyramidal in which phosphorus is sp3 hybridised. Phosphorus pentachloride (PCl5) Preparation : (i) Reaction of white phosphorus with excess of dry chlorine. P4 + 10Cl2 → 4PCl5 (ii) By the action of SO2Cl2 on phosphorus. P4 + 10SO2Cl2 → 4PCl5 + 10SO2 Properties : (i) Yellowish white powder. (ii) In moist air, it hydrolyses to POCl3 and finally converts to phosphoric acid. PCl5 + H2O → POCl3 + 2HCl POCl3 + 3H2O → H3PO4 + 3HCl (iii) Decomposes on stronger heating heat

→ PCl3 + Cl2 PCl5  (iv) Gives corresponding halides with transition metals 2Ag + PCl5 → 2AgCl + PCl3 Sn + 2PCl5 → SnCl4 + 2PCl3

[ 181



p-BLOCK ELEMENTS

Shape : Trigonal bipyramidal

Cl

240 pm

Cl

202

P

Cl

pm Cl

Cl  Oxoacids of phosphorus with structure : (i) H3PO2 (ii) H4P2O5



H

(iii) H3PO3

(iv) H4P2O6

O

O

O

O

O

O

P

P

P

P

P

P

H

OH





HO

H

O

OH

H

OH OH

H

(v) H3PO4

HO

OH OH OH

HO

(vi) H4P2O7

O

O

O

P

P

P

OH O

OH OH



OH OH

HO



(vii) (HPO3)3



O

O P

O P OH

HO O

O P

O OH Trimer or polymer of HPO3  The acids which contain P-H bond have reducing characteristics and act as good reducing agent. Example : H3PO2 reduces AgNO3 to Ag. 4AgNO3 + 2H2O + H3PO2 → 4Ag + 4HNO3 + H3PO4  The P-H bonds are not responsible for basicity as they do not ionize to give H+. Only H attached with oxygen are responsible for basicity and are ionisable.

Know the Terms  Pnicogens : The nitrogen group is group-15 of the periodic table and is also collectively named the pnicogens or pnictogens. The word pnicogens is derived from the Greek word pnigein which means ‘to choke or stifle’ which is a property of nitrogen.  Inert pair effect : The reluctance of the valence s-electrons to be available for bonding as compared to the valence p-electrons due to their greater penetration in the nucleus.  Fuming nitric acid : Nitric acid containing dissolved NO2 is known as fuming nitric acid. It can be obtained by distilling concentrated HNO3 with a little of starch.  Phosphazenes : These are the cyclic compounds which contain both nitrogen and phosphorus atoms in the alternate position along with two substitutes on each phosphorus atom. These are cyclic trimers, tetramers or polymers in nature.  Fuming nitric acid : Nitric acid containing dissolved nitrogen dioxide.  Aqua-regia : Mixture of nitric acid and hydrochloric acid in ratio of 1 : 3.

182 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Mnemonics • Concept: Increasing order of atomic radii and atomic volume (Top to bottom-Group) • Mnemonic: Nice Phone Assigned to Sub Bidder. • Interpretation: Nitrogen(N), Phosphorus(P), Arsenic(As), Antimony(Sb), Bismuth(Bi)

Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS : Q. 1. In a cyclotrimetaphosphoric acid molecule, how many single and double bonds are present? (a) 3 double bonds; 9 single bonds (b) 6 double bonds; 6 single bonds (c) 3 double bonds; 12 single bonds (d) Zero double bonds; 12 single bonds A&E Ans. Correct option : (a) Explanation : Cyclotrimetaphosphoric acid molecule

Q. 2. Strong reducing behaviour of H3PO2 is due to : (a) low oxidation state of phosphorus (b) presence of two –OH groups and one P–H bond (c) presence of one –OH group and two P–H bonds (d) high electron gain enthalpy of phosphorus U Ans. Correct option : (c) Explanation : Strong reducing behaviour of H3PO2 is due to presence of one –OH group and two P–H bonds.



H3PO2 (Hypophosphoric acid) Q. 3. Which of the following statements is wrong? (a) Single N–N bond is stronger than the single P–P bond. (b) PH3 can act as a ligand in the formation of coordination compound with transition elements. (c) NO2 is paramagnetic in nature. (d) Covalency of nitrogen in N2O5 is four. U Ans. Correct option : (a) Explanation : N−N single bond is weaker than P−P bond due to smaller size of N as compared to P. Smaller size of N leads to smaller N−N bond

(1 mark each) length. Because of larger size of P atom, P−P bond length is more and lone pair-lone pair repulsion between P atoms is less which makes the P−P bond stronger than N−N bond. [B] ASSERTIONS AND REASONS: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Q. 1. Assertion (A) : N2 is less reactive than P4. Reason (R) : Nitrogen has more electron gain enthalpy than phosphorus. U Ans. Correct option : (c) Explanation : Due to high bond dissociation energy of triple bond between the two N atoms, nitrogen (N) is less reactive than P4 and its electron gain enthalpy is less than phosphorus. Q. 2. Assertion (A) : HNO3 makes iron passive. Reason (R) : HNO3 forms a protective layer of ferric nitrate on the surface of iron. R Ans. Correct option : (c) Explanation : HNO3 makes iron passive and its passivity is attained by formation of a thin film of oxide on iron.  Q. 3. Assertion (A) : Bismuth forms only one well characterised compound in +5 oxidation state. Reason (R) : Elements of group-15 form compounds in +5 oxidation state. Ans. Correct option : (b) Explanation : Elements of group-15 form compounds in +5 oxidation state. Bismuth forms only one well characterised compound in +5 oxidation state which is BiF5. Due to inert pair effect bismuth exhibit +3 oxidation state only and forms trihalides. But due to samll size and high electronegativity of fluorine, Bismuth forms BiF5. [C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1. Write the formula of the compound of phosphorus which is obtained when conc. HNO3 oxidises P4.  A [CBSE OD Set-1 2017]

[ 183

p-BLOCK ELEMENTS

[CBSE Marking Scheme 2017]

Ans.





Ans. H3PO4.

Q. 2. On heating Pb(NO3)2 a brown gas is evolved which undergoes dimerization on cooling. Identify the gas. A [CBSE OD Set-2/Comptt. OD 2016] 



Ans.

[Topper’s Answer 2015] Detailed Answer: O P OH OH The basicity of H3PO4 is three due to availability of three hydrogen atoms which can be released in aqueous solutions. HO

[Topper’s Answer 2016]



Detailed Answer:

Q. 4. On adding NaOH to ammonium sulphate, a colourless gas with pungent odour is evolved which forms a blue coloured complex with Cu2+ ion. Identify the gas.  A [CBSE Delhi 2016]

∆ → 2PbO + O2 + 4NO2↑ 2Pb(NO3)2 



(Brown gas)

Nitrogen dioxide (NO2) is evolved.

Ans. Ammonia (NH3).

Q. 3. What is the basicity of H3PO4 ?



(NH4)2SO4 + 2NaOH ® 2NH3 + 2H2O + Na2SO4 2+ 2+ Cu (aq) + 4NH3 (aq)  [Cu(NH3)4] (aq)







U [CBSE Delhi 2015]

Deep blue

Short Answer Type Questions-I



Q. 1. Explain the following :

(2 marks each)

(ii)

(ii) NF3 is an exothermic compound but NCl3 is an endothermic compound. 

F

F

(i) Nitrogen is much less reactive than phosphorus.

Xe

A [CBSE Comptt. OD 2015]

[CBSE Marking Scheme 2015]







[CBSE Marking Scheme, 2017]



Q. 3. Draw the structures of the following : (i) H4P2O7 (ii) XeOF4  Ans. (i)

A [CBSE Delhi Set-3 2017]





(ii) Due to low bond dissociation enthalpy of F2 than Cl2 and strong bond formation between N and F.  [1]

O

Q. 2. Draw the structures of the following :

(i) H3PO2

(ii) XeF4

O

P

A [CBSE Delhi Set-2 2017]

P

HO

O

OH OH





[1]

F

F

Ans. (i)  Due to high bond dissociation enthalpy of N ≡ N[1]



OH

Ans. (i)

(ii)

O

[1]

O



P



F

OH

Xe

[1] 

H

F

F 

H

F

[1]

[CBSE Marking Scheme, 2017]

184 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q. 4. “Orthophosphoric acid (H3PO4) is not a reducing agent whereas hypophosphorus acid (H3PO2) is a strong reducing agent.” Explain and justify the above statement with the help of a suitable example.  A [CBSE Comptt. OD Set-1, 2, 3 2017] Ans. Hypophosphorous acid is a good reducing agent as it contains two P-H bonds. There is no P-H bond in orthophosphoric acid, so it is not a reducing agent. [1] Example : It reduces AgNO3 to metallic silver / chemical equation [1] [CBSE Marking Scheme 2017] Q. 5. (i) What is the covalence of nitrogen in N2O5? (ii) BiH3 is a stronger reducing agent than SbH3, why?

Detailed Answer: (i) 4 (ii) Because BiH3 has longer bond length due to larger size of Bi it has lowest bond dissociation energy which makes it less stable than SbH3. Q. 6. Among the hydrides of Group-15 elements, NH3, PH3, AsH3, SbH3 and BiH3 which have the

(a) lowest boiling point ?

(b) maximum basic character ?

(c) highest bond angle ?

(d) maximum reducing character ? R [CBSE Delhi & OD 2018]  Ans.

A [CBSE Comptt. Delhi Set-1, 2, 3 2017]

Ans. (i) 4 [1] (ii) Due to lower bond dissociation enthalpy of BiH3 as compared to SbH3. [1] 

[CBSE Marking Scheme 2017]

[Topper’s Answer 2018]

Short Answer Type Questions-II Q. 1. Give reasons for the following : (a) Red phosphorus is less reactive than white phosphorus. (b) Electron gain enthalpies of halogens are largely negative. (c) N2O5 is more acidic than N2O3. [3] Ans. (a) Red phosphorus being polymeric is less reactive than white phosphorus which has discrete tetrahedral structure. [1] (b) They readily accept an electron to attain noble gas configuration. [1] (c) Because of higher oxidation state (+5) of nitrogen in N2O5. [1] [CBSE Marking Scheme, 2017] Detailed Answer : (a) In red phosphorus, P4 molecules are linked in an extended chain structure. So, it is much less reactive. P P

P P



P P

P P

P

P

P P

While white phosphorus contains discrete P4 molecules in which each phosphorus atom is tetrahedrally bonded to other phosphorus atoms. So, white phosphorus is highly reactive. [1]

(3 marks each)

(b) Electron gain enthalpies of halogens are largely negative due to smaller size and higher nuclear charge. Both these factors increase the force of attraction between the nucleus and the electron being added and hence the atom has a greater tendency to attract the electron towards it self.  [1] (c) N2O5 dissolves in H2O with a hissing noise to form nitric acid (HNO3) which is strong acid while N2O3 dissolves in H2O to form nitrous acid (HNO2). So, N2O5 is more acidic than N2O3. [1] N2O5 + H2O → 2HNO3 Nitric acid (strong acid) 0°C ® 2HNO2 N2O3 + H2O ¾¾¾ Nitrous acid (weak acid) Q. 2. (i) Arrange the following in decreasing order of bond dissociation enthalpy F2, Cl2 , Br2 , I2 (ii) Bi does not form pπ-pπ bonds. Give reason for the observation.  (iii) Electron gain enthalpy of oxygen is less negative than sulphur. Justify. A [CBSE SQP 2021] Ans. (i) Decreasing order of bond dissociation enthalpy:[1] Cl2 > Br2 > F2 > I2 (ii) Bi does not form pπ-pπ bonds as its atomic orbitals are large and diffuse so effective [1] overlapping is not possible. (iii)  Due to small size of oxygen, it has greater electron electron repulsions. [1]

[ 185

p-BLOCK ELEMENTS

Q. 3. Name three oxoacids of nitrogen. Write the disproportionation reaction of that oxoacid of nitrogen in which nitrogen is in +3 oxidation state. A Ans. Three oxoacids of nitrogen are :

(i) HNO2, Nitrous acid (ii) HNO3, Nitric acid (iii) H2N2O2, Hyponitrous acid

Disproportionation 3HNO2  → HNO3 + H 2O + 2NO



[1.5]

Long Answer Type Questions Q.1. On heating lead (II) nitrate gives a brown gas “A”. The gas “A” on cooling changes to colourless solid “B”. Solid “B” on heating with NO changes to a blue solid ‘C’. Identify ‘A’, ‘B’ and ‘C’ and also write reactions involved and draw the structures A&E of ‘B’ and ‘C’.  (Source- Exemplar) D Ans. 2 PbO + 4 NO2 Pb( NO3 )2 673 K (A) [½] ( Brown colour ) on cooling





  2 NO2    N 2O4 Heating ( B) (Colourless solid)

[½]

2 NO + N 2O4  → 2 N 2 O3 (C ) ( Blue solid)

[½]

D 250 K

[½] [½] [½]

‘A’ is NO2. ‘B’ is N2O4. ‘C’ is N2O3.

[½] [½] [½]

(5 marks each)

Ans. A = NH4NO2

B = N2 C = NH3

D = HNO3

(i) NH4NO2 ® N2 + 2H2O

(ii) N2 + 3H2 ® 2NH3 (iii) 4NH3 + 5O2 ® 4NO + 6H2O

4NO + O2 ® 2NO2



3NO2 + H2O ® 2HNO3 + NO

 [1] mark for each equation. Q.3. Colorless, oily liquid ‘B’ is obtained by reaction of Thionyl chloride with compound ‘A’. Compound ‘B’ hydrolyses in presence of moisture to give compound ‘C’. Compound ‘B’ also reacts with organic compounds containing -OH groups to give compound ‘C’. Identify compounds ‘A’,’B’,’C’ and write the equations involved in the given reactions. Draw the structure of compound ‘B’ and identify its hybridization.

A&E

(New) Ans: P4 + 8SOCl2 → 4PCl3 + 4SO2 + 2S2Cl2

(Structure of N2O4)

[1]





(A)



(White Phosphorus)



PCl3 + 3H2O → H3PO3 + 3HCl





(C)





(Phosphoric acid)



‘A’ is P4- White phosphorus

[½]



‘B’ is PCl3- Phosphorus trichloride

[½]



‘C’ is H3PO3- Phosphoric acid

[½]







PCl3 is Pyramidal in shape.

(B) (Colorless, oily liquid) [1]



(Structure of N2O3) Q.2. On heating compound (A) gives a gas (B) which is a constituent of air. This gas when treated with 3 mol of hydrogen (H2) in the presence of a catalyst gives another gas (C) which is basic in nature. Gas C on further oxidation in moist condition gives a compound (D) which is a part of acid rain. Identify compounds (A) to (D) and also give necessary A&E equations of all the steps involved.  (Source- Exemplar)

TOPIC-2



 Oxygen family : 8O, 16S, 34Se, 52Te, 84Po  Physical properties :

3

Phosphorus is sp hybridised.

Group-16 Elements, Properties and Some Important Compounds Revision Notes

[½]



[1]

[1]

[½] [½]



186 ]



Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Scan to know (i) Atomic and ionic radii : Smaller than the corresponding elements of group 15 due to more about increase in nuclear charge. Down the group they increase due to addition of a new shell. this topic (ii) Ionisation enthalpy : I.E.1 is lower than those of corresponding elements of group 15 due to increase in nuclear charge. I.E.2 is higher than those of group 15 due to smaller size of the ions and greater effective nuclear charge. Down the group I.E. decreases. (iii) Electron gain enthalpy : Oxygen has less electron gain enthalpy than sulphur. From Atomic and sulphur, the value again becomes less negative upto polonium. physical properties (iv) Electronegativity : More electronegative than group 15 elements. It decreases down the of Group 16 group due to increase in atomic size. elements (v) Oxygen and sulphur are non-metals, selenium and tellurium are metalloids and polonium is a metal. (vi) Melting and boiling points : Increase regularly from O to Te due to increase in size and hence greater van der Waals forces. Po has lesser melting and boiling point than Te due to maximum inert pair effect. (vii) Density : Increases down the group. (viii) Oxidation state : Oxygen shows an oxidation state of –2 only (except OF2 and H2O2). All other elements show +2, +4 and +6 oxidation states. Scan to know Chemical properties : more about this topic (1) With hydrogen : Form hydrides of type H2E (E = O, S, Se, Te and Po). Properties of hydrides : (i) Thermal stability : Decreases down the group. (ii) Acidic character : Increases down the group. Chemical (iii) Reducing nature : All are reducing agents except H2O. properties of Group 16 (iv) Boiling point : From H2O to H2S there is a sudden drop which increases from H2S to H2Te. elements (2) With halogens : Form halides of type EX2, EX4 and EX6 where E is an element of group and X is a halogen. Properties of halides : (i) Only hexafluorides are the only stable halides. (ii) All elements except selenium form dichlorides and dibromides. These dihalides are sp3 hybridised and have tetrahedral structure. 2Se2Cl2 → SeCl4 + 3Se (a) Preparation of SF4 and SF6 : 3SCl2 + 4NaF → SF4 + S2Cl2 + 4NaCl

Burn

S + 3F2 → SF6 (b) Properties of SF4 : SF4 is readily hydrolysed. SF4 + 2H2O → 4HF +SO2 (c) SF6 is sp3d2 hybridised and octahedral whereas SF4 is sp3d hybridised and is trigonal bipyramidal with one position occupied by a lone pair of electrons. (3) With oxygen : Form oxides of EO2 and EO3.  Dioxygen (O2) Preparation :





In laboratory,

2 2KClO3 MnO  → 2KCl + 3O2

Properties : (i) Slightly soluble in water and appreciably soluble in alkaline pyrogallol solution. (ii) Paramagnetic. Forms oxides with metals and non-metals. (iii) 2Mg(s) + O2(g) → 2MgO(s) C(s) + O2(g) → CO2(g) 4Al + 3O2 → 2Al2O3 P4 + 5O2 → P4O10 2ZnS + 3O2 → 2ZnO + 2SO2 CH4 + 2O2 → CO2 + 2H2O 4HCl + O2 → 2Cl2 + 2H2O

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



Classification of oxides : On the basis of chemical nature, (i) Basic : Na2O, CaO, etc. basic oxides. (ii) Acidic : CO2, SO2, etc. acidic oxides. (iii) Amphoteric : SnO2, Al2O3, etc. are amphoteric oxides. (iv) Neutral : H2O, CO, etc. are neutral oxides. (v) Poly-oxide : Oxides having oxygen more than required oxygen such as peroxide (Na2O2), super oxide (KO2), dioxide (PbO2), higher oxide (Mn2O7). (vi) Sub-oxide : Oxides having oxygen less than required, e.g., C3O2.  Ozone (O3) Preparation : By subjecting pure and dry oxygen to silent electric discharge.

3O2(g) Discharge  → 2O3(g), DH = + 284 kJ



Properties :



(i) Poisonous in nature



(ii) Powerful oxidising agent. It is a stronger oxidising agent than O2.

PbS + 4O3 → PbSO4 + 4O2



2I– + H2O + O3 → 2OH– + I2 + O2

Uses :

(i) As bleaching agent



(ii) As disinfectant and germicide



(iii) or purification of air in hospitals, railway tunnels, cinema house F



(iv) For bleaching oils, ivory flour, starch, etc











Structure :

Allotropes of Sulphur :

(a) Rhombic Sulphur (a-Sulphur) :



(i) Yellow in colour, melting point 385.8 K and specific gravity 2.06.



(ii) Formed by evaporating the solution of roll sulphur in CS2.



(iii) Insoluble in water but dissolves to some extent in benzene, alcohol and ether.



(iv) Readily soluble in CS2.



(b) Monoclinic Sulphur (b-Sulphur) :



(i) Melting point is 393 K and specific gravity 1.98.



(ii) Soluble in CS2.



(iii) Prepared by melting rhombic sulphur in a dish and cooling.



(iv) Stable above 369 K and transforms into a-sulphur below it. Both rhombic and monoclinic sulphur have S8 molecules. Shape of S8 and S6 molecules : 20

S

S

m

4p

S S S

107° S

S

20 S S

10 S

5 .7

S 2 .2 °

pm

S S S







188 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Sulphur dioxide (SO2) :

Preparation : S + 2H2SO4 → 3SO2 + 2H2O



(conc.) Cu + 2H2SO4 → CuSO4 +SO2 + 2H2O



(conc.) Na2SO3 + 2HCl → 2NaCl + SO2 + H2O



(dil.)



Properties :



(i) Colourless gas with pungent, suffocating odour



(ii) Highly soluble in water



(iii)

SO2 + H2O  H2SO3



(Sulphurous acid) 2NaOH + SO2 → Na2SO3 + H2O



Na2SO3 + H2O + SO2 → 2NaHSO3



Cl2 + SO2 → SO2Cl2



O2 + 2SO2 → 2SO3



SO2 acts as reducing agent. (iv)

2Fe3+ + SO2 + 2H2O → 2Fe2+ + SO42– + 4H+



5SO2 + 2MnO4– + 2H2O → 5SO42– + 4H+ + 2Mn2+

Uses :

(i) Used in refining petroleum and sugar.



(ii) As bleaching agent for wool and silk.



As anti-chlor, disinfectant and preservative. (iii)

 Sulphuric acid (H2SO4)

Preparation : By contact process which involves 3 steps :

(i) Burning of sulphur or sulphide ores in air :

S + O2 → SO2

(ii) Conversion of SO2 to SO3 with oxygen in the presence of a catalyst V2O5.

2 O5 2SO2 + O2 V → 2SO3

(iii)  Absorption of SO3 in H2SO4 to give oleum. Oleum is diluted with water to get H2SO4 of desired concentration. SO3 + H2SO4 → H2S2O7 (Oleum) H2S2O7 + H2O → 2H2SO4 Properties : Colourless, dense, oily liquid with a specific gravity of 1.84 at 298 K (i) (ii) Low volatility (iii) Strong acidic character Strong affinity for water (iv) (v) Acts as an oxidising agent Cu + 2H2SO4 → CuSO4 + SO2 + 2H2O (conc.) S + 2H2SO4 → 3SO2 + 2H2O (conc.)

C + 2H2SO4 → CO2 + 2SO2 + 2H2O (conc.)

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(vi) Acts as a dehydrating agent. 2 SO4 C12H22O11 H  → 12C + 11H2O



Uses : (i) In petroleum refining (ii) Manufacture of pigments, paints and dye stuff (iii) Detergent industry (iv) In storage batteries Oxoacids of Sulphur : O S

S

O

OH OH (H2SO4) Sulphuric acid

O

OH OH (H2SO3) Sulphurous acid

O

O

O

O

S

S

S

O OH O OH O (H2S2O8) Marshall’s acid Peroxodisulphuric acid

O

OH O — O — H (H2SO5) Caro’s acid Peroxomonosulphuric acid

S

O

O

O

O

S

S

S

S

S

HO OH O (H2S2O3) Thiosulphuric acid

O OH O OH

O

(H2S2O7) Pyrosulphuric acid (Oleum)

O OH

OH O

(H2S2O6) Dithionic acid

Know the Terms  Chalcogens : Group-16 elements are also known as chalcogens which means ore forming elements.  Oil of vitriol : Pure sulphuric acid is highly viscous due to the presence of intermolecular hydrogen bonding. It is known as oil of vitriol.  Oxide : A binary compound of oxygen with another element.

Mnemonics • Concept: Increasing order of atomic radii and atomic volume (Top to bottom-Group) • Mnemonic: Old Sultanpur Seems Terribly Polluted. • Interpretation: Oxygen(O), Sulphur(S), Selenium(Se), Tellurium(Te), Polonium(Po)

190 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q. Why is SO2 considered to be an air pollutant? Solution: Sulphur dioxide causes harm to the environment in many ways : STEP-1: When it combines with water vapour, it forms sulphuric acid (H2SO4). This causes acid rain. Acid rain damages soil, plants, and buildings, especially those made of marble.

STEP-2: Even in very low concentrations, SO2 causes irritation in the respiratory tract. It causes throat and eye irritation and can also affect the larynx to cause breathlessness. STEP-3: It is extremely harmful to plants. Plants exposed to sulphur dioxide for a long time lose colour from their leaves. This condition is known as chlorosis. This happens because the formation of chlorophyll is affected by the presence of sulphur dioxide.

Objective Type Questions

(1 mark each) O

[A] MULTIPLE CHOICE QUESTIONS :

HO − S − O − OH O

HO − S − O − O − S − OH

Q.1. Hot conc. H2SO4 acts as moderately strong oxidising agent. It oxidises both metals and nonmetals. Which of the following element is oxidised by conc. H2SO4 into two gaseous products? R (a) Cu (b) S (c) C (d) Zn Ans. Correct option : (c) Explanation : C + 2H2SO4 → CO2+ 2SO2 + 2H2O Reaction Condition Hot concentrated sulfuric acid should be used to oxidise carbon to carbon dioxide. Q.2. Which of the following are peroxoacids of sulphur? (a) H2SO5 and H2S2O8 (b) H2SO5 and H2S2O7 (c) H2S2O7 and H2S2O8 (d) H2S2O6 and H2S2O7 R Ans. Correct option : (a) Explanation : H2SO5 and H2S2O8 Peroxymonosulfuric acid and Peroxydisulfuric acid are peroxoacids of sulphur. O

O

O

O

Q. 3. Which of the following statements are correct for SO2 gas? (a) It acts as bleaching agent in moist conditions. (b) Its molecule has linear geometry. (c) It can be prepared by the reaction of dilute H2SO4 with metal sulphide. (d) All of the above Ans. Correct option : (a) Explanation : SO2 acts as a bleaching agent under moist conditions. SO2(g)+2H2O → H2SO4+ 2[H] SO2 is oxidized to sulphuric acid and releases nascent hydrogen which bleaches the material. But this is a temporary as atmospheric oxygen re-oxides the bleached matter after some time. [B] ASSERTIONS AND REASONS: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct state­ ments and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

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[C] VERY SHORT ANSWER TYPE QUESTIONS : Q.1. Write the formula of the compound of sulphur which is obtained when conc. HNO3 oxidises S8.  R [CBSE OD Set-2 2017] [CBSE Marking Scheme, 2017] OR



ns. H2SO4 A



[Topper’s Answer, 2017]

Q.2. Write the formulae of any two oxoacids of sulphur.  R [CBSE Comptt. OD 2015]

Ans. H2SO3 (Sulphurous acid) and H2SO4(Sulphuric acid). [CBSE Marking Scheme, 2017] Q.3. Arrange the following hydrides of Group-16 elements in the increasing order of their thermal stability : H2O, H2S, H2Se, H2Te.  A [CBSE Foreign Set-1,2,3 2017]

Ans. H2T < H2Se < H2S HBr > HI Q. 4. Write balanced chemical equations for the following processes : (a) Cl2 is passed through slaked lime. (b) SO2 gas is passed through an aqueous solution of Fe(III) salt. Ans. (a) 2Ca(OH)2 + 2Cl2 → CaCl2 + Ca(OCl)2 + 2H2O [1] (b) SO2 + 2Fe3+ + 2H2O → 2Fe2+ + SO42– + 4H+ [CBSE Marking Scheme, 2019] [1] Q. 5. (a) Write two poisonous gases prepared from chlorine gas. (b) Why does Cu2+ solution give blue colour on reaction with ammonia? A&E [CBSE OD Set-2 2019] Ans. (a) Mustard gas, tear gas, phosgene. (Any two ) [½ + ½] (b) Because it forms blue coloured complex [Cu(NH3)4]2+ (aq) or Equation. [CBSE Marking Scheme, 2019] [1]

Detailed Answer : (a) Phosgene (COCl2), tear gas (CCl3NO2), mustard gas (Any two) (C4 H8Cl 2S) . (b) Cu2+ ion reacts with ammonia solution to form a deep blue coloured complex because of the property of ammonia to form complex compounds. Ammonia acts as a lewis base due to presence of lone pair of electrons on the nitrogen atom. Hence, it is able to form coordinate bond with electron deficient molecules or a number of transition metal cations. Cu2+(aq) + 4NH3(aq) → [Cu(NH3)4]2+ Tetraammine copper (II) (blue colour) Q. 6. Account for the following: (i) The two oxygen-oxygen bond lengths in ozone molecule are identical. (ii) Most of the reactions of fluorine are exothermic. A&E [CBSE Comptt. Delhi Set-1 2017] Ans. (i) Due to resonance the two O-O bond lengths are identical. [1] (ii) Due to strong bond formed by it with other elements. [1] [CBSE Marking Scheme 2017] Detailed Answer: (i) In ozone, the central oxygen atom forms one single bond with a terminal oxygen atom and double bond with other terminal oxygen atom. The p electrons of double bond are delocalised over the three oxygen atoms. This results into resonance hybrids with the average bond distance of the single and double bond.





Short Answer Type Questions-II Q. 1. Explain the following giving appropriate reasons: (i) (CH3)3 P = O exists but (CH3)3 N = O does not. (ii) Oxygen has less electron gain enthalpy with negative sign than sulphur. (iii) H3PO2 is a stronger reducing agent than H3PO3. A&E

Ans. (i) Because as N can’t form 5 covalent bonds as its maximum covalency is three. The octet cannot be extended as it doesn’t have d orbital, while P can extend its octet as it has empty d orbital. [1]



Ans. (i) H2O < H2S < H2Se < H2Te [1] (ii) HF > HCl > HBr > HI [CBSE Marking Scheme, 2019] [1]



192 ]

[1]

(3 marks each)

(ii) This is due to very small size of oxygen atom, repulsion between electrons is large in relatively small 2p sub-shell. [1] (iii) In H3PO2 there are 2P – H bonds, whereas in H3PO3 there is 1 P-H bond. [1] Q. 2. Account for the following: (a) Moist SO2 decolourises KMnO4 solution. (b) In general interhalogen compounds are more reactive than halogens (except fluorine). (c) Ozone acts as a powerful oxidizing agent. U [CBSE SQP 2020]

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Detailed Answer : (a) The moist SO2 behaves as a reducing agent and decolourises KMnO4 solution.



Ans. (a) Moist sulphur dioxide behaves as a reducing agent, reduces MnO–4 to Mn2+.[1] (b) X − X’ bond in interhalogens is weaker than X –X bond in halogens except F-F bond. [1] (c) Due to the ease with which it liberates atoms of nascent oxygen. [1]  [CBSE SQP Marking Scheme 2020]

5SO2 + 2MnO-4 + 2H 2 O ® 5SO24 - + 4H + + 2Mn 2 +

Long Answer Type Questions Q.1. (a) Give reasons for the following : (i) Sulphur in vapour state shows paramagnetic behaviour.

(ii) N-N bond is weaker than P-P bond.



(iii)  Ozone is thermodynamically less stable than oxygen.



(b) Write the name of gas released when Cu is added to



(i) dilute HNO3 and

(ii) conc. HNO3 OR



Ans.





(5 marks each)

Ans. (a) (i) I n vapour state sulphur partly exists as S2 molecule which has two unpaired electrons like O2. [1] (ii) Due to greater interelectronic repulsion. [1] (iii) Because decomposition of ozone into oxygen results in the liberation of heat (DH is negative) and an increase in entropy (DS is positive), resulting in large negative Gibbs energy change (DG) for its conversion into oxygen. [1] (b) (i) NO gas/ Nitric oxide [1] (ii) NO2 gas / Nitrogen dioxide [CBSE Marking Scheme, 2019] [1]



194 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII





[5] [Topper’s Answer 2019]



OR

Detailed Answer : (a) (i) In gas phase, sulphur exists mostly as diatomic molecule S2. The bonding in this molecule is similar to that of dioxygen O2. According to molecular orbital theory it has two unpaired electrons in anti-bonding molecular orbitals. (ii) The N-N sigma bond length is less than the P-P sigma bond length. The shorter separation means the non-bonding electrons (lone pairs) on the nitrogen atoms repel each other strongly, making its bond weaker than the P-P bond. (iii) The enthalpy change for decomposition of ozone has a negative value, it results in liberation of heat :







2O3 → 3O2    …∆   Ho = −142 kJ / mol

The increase in number of moles of gas means the entropy change for the reaction is positive. This reinforces the effect of negative enthalpy change and results in a large negative value of DGo for the decomposition reaction, making it less stable than oxygen. (b) (i) Dilute HNO3: Nitric oxide, NO. (ii) Conc. HNO3: Nitrogen dioxide, NO2. Q. 2. (a) (i)  Write the disproportionation reaction of H3PO3.

   (ii) Draw the structure of XeF4. (b) Account for the following :

(i) Although fluorine has less negative electron gain enthalpy yet F2 is strong oxidizing agent. (ii) Acidic character decreases from N2O3 to Bi2O3 in group 15.



(c)  Write a chemical reaction to test sulphur dioxide gas. Write chemical equation involved. R [CBSE Delhi Set - 1, 2019]

Ans. (a) (i) 4H3PO3 → 3H3PO4 + PH3 (ii) F F

[1]

Xe F



F

[1] (b) (i) Due to small size and low bond dissociation enthalpy [1] (ii) As the size increases, electronegativity decreases / non-metallic character decreases . [1] (c) 5SO2 + 2MnO–4 + 2H2O → 5SO42– + 4H+ + 2Mn2+ [CBSE Marking Scheme, 2019] [1]

Detailed Answer : (b) (i)  The bond dissociation energy of F2 is quite low, and the small size of the fluoride ion means that its hydration energy is high. These two compensate for the less negative value of electron gain energy, making F2 a strong oxidizing agent. (ii) As we move down a group, the atomic size and metallic character increases while electronegativity decreases. This leads to the acidic character of the oxide decreasing from top to bottom, N2O3 to Bi2O3. (c) Sulphur dioxide decolourises acidified potassium permanganate solution. 5SO2 + 2KMnO4 + 2H2O → 2MnSO4 + K2SO4 + 2H2SO4

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Ans. (a) (i) Due to increase in size and metallic character. [1] (ii) Due to decrease in bond dissociation enthalpy. [1]

(iii) Due to low bond dissociation enthalpy of F-F bond than Cl-F bond whereas Cl-

Cl bond has higher bond dissociation [1] enthalpy than Cl-F bond.



(b) (i)



F

Xe

F



(ii)

O

O P HO





Q. 3. (a) Account for the following : (i) Tendency to show –3 oxidation state decreases from N to Bi in group 15. (ii) Acidic character increases from H2O to H2Te. (iii) F2 is more reactive than CIF3, whereas CIF3 is more reactive than Cl2. (b) Draw the structure of (i) XeF2, (ii) H4P2O7. [CBSE OD Set - 2 2019]

P O

OH

OH

OH

[CBSE Marking Scheme, 2019] [1 + 1] Detailed Answer : (a) (i) Due to increase in the size of the atom and the metallic character. (ii) Due to increase in size of the central atom as O < S < Se < Te, the distance between central atom and hydrogen atom also increases. This results in increase in bond length, decreasing bond dissociation enthalpy. The bond cleavage becomes easier, increasing the acidic character from H2O to H2Te. (iii)  Interhalogen compounds are more reactive than halogen compounds. But in case of fluorine, due to small size of fluorine it has high electronegativity and low bond energy so it is more reactive than ClF3. Therefore, ClF3 is more reactive than Cl2.

Q. 4. (a) Give one example to show the anomalous reaction of fluorine. (b) What is the structural difference between white phosphorus and red phosphorus? (c) What happens when XeF6 reacts with NaF? (d) Why is H2S a better reducing agent than H2O? (e) Arrange the following acids in the increasing order of their acidic character : HBr, Hl, HCl, HF R [CBSE OD Set - 2 2019] Ans. (a) 2F2(g) + 2H2O(l) → 4H+(aq) + 4F– (aq) + O2 (g)

(b) White phosphorus is discrete tetrahedral whereas red phosphorus is polymeric / or structures drawn.

(c) It forms Na+ [XeF7]– / XeF6 + NaF → Na+ [XeF7]–

(d) Due to lower bond dissociation enthalpy of H-S bond than H-O bond.

(e) HF < HCl < HBr< HI



[CBSE Marking Scheme, 2019] [1 × 5]



Detailed Answer : (a) It forms only one oxoacid as compared to the rest of the halogens that form a number of oxoacids. (b) White phosphorus : It exits as discrete to tetrahedral P4 unit.

Red phosphorus : It exits in the form of polymeric chain.

P P

P

P



P (c) XeF6 + NaF ¾¾ ® Na+ éë XeF7 ùû .

(d) D  ue to larger size of sulphur and weak S-H bond in comparison to size of oxygen and O-H bond. Therefore, H2S easily donates electron in comparison to H2O.

P

P

P P

P

P

P P

P

P P

(e) The strength of acidic character in halogen acids increases from HF to HI down the group, due to decrease in bond energy down the group. So, the increasing order of acidic character is: HF < HCl < HBr < HI.

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

A&E [CBSE Delhi/OD, 2018]



Ans. (a) (i) In H3PO3, phosphorus exhibits an oxidation state of +3 hence it tends to undergo disproportionation reaction. However, in H3PO4, phosphorus exhibits +5 oxidation state and cannot be oxidised further hence it does not show the disproportionation reaction. (ii) In interhalogen compounds, the central atom is of larger size and more electropositive than the surrounding atoms. As fluorine is more electronegative and smaller in size than chlorine, ClF3 is formed. (iii) Due to its small size and high electronegativity, oxygen forms pp-pp bonds and exists as a diatomic molecule. The intermolecular forces are weak van der Waals forces which result in

dioxygen to exist as gas at room temperature whereas, sulphur exists as S8 molecules held together by strong covalent forces and forms puckered ring structures packed in solid crystals. .. F H .. .. .. .. .. . — Cl — O (i) F.. —..Xe — F.. (ii) .O ..

(b)

..



..

.. ..



Q. 5. (a) Give reasons : (i) H3PO3 undergoes disproportionation reaction but H3PO4 does not. (ii) When Cl2 reacts with excess of F2, ClF3 is formed and not FCl3. (iii) Dioxygen is a gas while Sulphur is a solid at room temperature. (b) Draw the structures of the following : (i) XeF4 (ii) HClO3

.O . ..

.. ..

F ..

Q. 6. (a) When concentrated sulphuric acid was added to an unknown salt present in a test tube a brown gas (A) was evolved. This gas intensified when copper turnings were added to this test tube. On cooling, the gas (A) changed into a colourless solid (B). (i) Identify (A) and (B). (ii) Write the structures of (A) and (B). (iii) Why does gas (A) change to solid on cooling ? A&E [CBSE Delhi/OD, 2018] (b) Arrange the following in the decreasing order of their reducing character : HF, HCl, HBr, HI (c) Complete the following reaction :

XeF4 + SbF5



196 ]

[CBSE Delhi/OD 2018]

Ans.





[Topper’s Answer 2018]

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Detailed Answer : (a) (i) An addition of conc. H2SO4 to salt evolves brown gas, which further intensifies brown colour and shows that the salt is a nitrate and the gas evolved is NO2(A). On cooling, NO2 dimerizes to N2O4 which is colourless solid (B) is N2O4.

..

.. .O .

..

.O .

..

N ..

.. O

..

.. O

..

.. (ii) .O .

..



(iii) Due to dimerization of NO2. (b) HI > HBr > HCl > HF Decrease in bond dissociation enthalpy down the group makes it easier for the halogen to give away proton, thus the reducing property increases. (c) XeF4 + SbF5 → [XeF3]+ [SbF6]– Adduct is formed by XeF4 with lewis acid SbF5.

N–N

.. .O .

TOPIC-3 Group-17 Elements, Properties and Some Important Compounds Revision Notes  Halogen family : 9F, 17Cl, 35Br, 53I, 85At. Scan to know  Physical Properties : more about this topic (i) Atomic and ionic radii : They are the smallest in their respective periods due to increase in nuclear charge. Down the group, they increase. (ii) Oxidation state : Fluorine shows –1 oxidation state only. Other elements show oxidation states +1, +3, +5 and +7. Halogen family (iii) Ionisation enthalpy : Higher than the corresponding members of group 16. Down the group, it decreases. (iv) Electron gain enthalpy : Have maximum negative electron gain enthalpy in the corresponding periods. (v) Metallic character : Due to very high ionisation enthalpies they are non-metals. The last element I, shows some metallic character as it can form I+ by loss of electrons. (vi) Electronegativity : These are the most electronegative elements in their respective periods. Down the group, electronegativity decreases. (vii) F2 and Cl2 are gases, Br2 is a liquid while I2 is solid. (viii) All halogens are coloured. (ix) Melting point and boiling point : Increase down the group due to increase in size and nuclear charge causing greater van der Waals forces of attraction. (x) Bond dissociation enthalpy : Bond dissociation enthalpy decreases from Cl2 to I2 i.e., Cl2 > Br2 > F2 > I2.  Chemical Properties : (a) Highly reactive : The reactivity decreases down the group. Fluorine is the strongest oxidising halogen. A halogen oxidises halide ions of higher atomic number. F2 + 2X– → 2F– + X2 (X = Cl, Br or I) Cl2 + 2X– → 2Cl– + X2 (X = Br or I) Br2 + 2I– → 2Br– + I2 Fluorine oxidises water to oxygen whereas chlorine and bromine react with water to form corresponding hypohalic and hypohalous acids. 2F2(g) + 2H2O(l) → 4H+(aq) + 4F–(aq) + O2(g)

X2(g) + H2O(l) → HX(aq) + HOX(aq)

(X = Cl or Br)

(b) Towards hydrogen : All elements form hydrides of the type HX (X = F, Cl, Br, I). (i) Physical state : Except HF which is a liquid because of H-bonding, all are gases. (ii) Thermal stability : It decreases down the group due to increase in bond length. (iii) Reducing character : It increases from HF to HI due to decrease of stability. (iv) Acidic strength : HF < HCl < HBr < HI. (c) Towards halogens : They react with all elements except He, Ne and Ar to form binary halides. (d) Towards metal : (i) With particular metal, ionic character is M-F > M-Cl > M-Br > M–I. (ii) With metals having low I.E., halides are ionic.

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(iii) With metals having high I.E., halides are covalent. (iv) With metals showing more than one oxidation states, halides with higher oxidation states are more covalent. (v) Metals show highest oxidation state in fluorides. (e) Towards oxygen : Halogens form many binary compounds with oxygen, but most of them are unstable. Cl, Br and I form oxides in their oxidation state +1 to +7. Their stability decreases in the order I > Cl > Br.

O 170 pm

Cl 118°

112° Cl

O

Cl

141 pm O

(f) Oxoacids : Because of high electronegativity and small size, fluorine forms only one oxoacid, HOF (hypofluorous acid). The other halogens form acid of the type HOX—hypohalous acid, HOXO–halous acid, HOXO2 – halic acid and HOXO3 – perhalic acid. Acidic strength : HClO > HBrO > HIO Acidic strength of oxoacids containing the same halogen : HOCl > HClO2 < HClO3 < HClO4 (g) Structure of oxoacids of Chlorine : H O

O

O

O O

H Cl H Cl Hypochlorous Chlorous acid acid

Cl

O Chloric acid

H O

O

Cl O

O Perchloric acid

 Chlorine (Cl2) Preparation : In laboratory, ∆

MnO2 + 4HCl  → MnCl2 + Cl2 + 2H2O



∆

→ 2KCl + 2MnCl2 + 8H2O + 5Cl2 2KMnO4 + 16HCl 



(i)

CuCl

2 4HCl + O2  → 2Cl2 + 2H2O (Deacon’s process)

(ii) 4NaCl + MnO2 + 4H2SO4 → MnCl2 + 4NaHSO4 + 2H2O + Cl2 (iii) Electrolytic process : By electrolysis of brine. At cathode : 2H2O + 2e– → H2 + 2OH–; Na+­ + OH– → NaOH At anode : Cl– → Cl + e–; Cl + Cl → Cl2 Properties : (i) It is a greenish yellow gas with a pungent suffocating smell (ii) It is soluble in water (iii) About 2.5 times heavier than air (iv) With metals and non-metals form chlorides 2Al + 3Cl2 → 2AlCl3; 2Na + Cl2 → 2NaCl 2Fe + 3Cl2 → 2FeCl3; S8 + 4Cl2 → 4S2Cl2 P4 + 6Cl2 → 4PCl3; H2 + Cl2 → 2HCl (v) Reacts with compounds containing hydrogen to form HCl H2S + Cl2 → 2HCl + S; C10H16 + 8Cl2 → 16HCl + 10C NH3 + 3Cl2 → NCl3 + 3HCl; 8NH3 + 3Cl2 → 6NH4Cl + N2 (excess) (explosive) (excess) (vi) With cold and dilute alkalies 2NaOH + Cl2 → NaCl + NaOCl + H2O (cold and dilute) (vii) With hot and concentrated alkalies 6NaOH + 3Cl2 → 5NaCl + NaClO3 + 3H2O (hot and conc.) (viii) With dry slaked lime, it gives bleaching powder. 2Ca(OH)2 + 2Cl2 → Ca(OCl)2 + CaCl2 + 2H2O

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Electrolysis of Brine

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(ix) Chlorine water on standing loses its yellow colour. HOCl formed gives nascent oxygen. Cl2 + H2O → HCl + HOCl HOCl → HCl + O (x) It oxidises ferrous to ferric, sulphite to sulphate, sulphur dioxide to sulphuric acid and iodine to iodic acid. 2FeSO4 + H2SO4 + Cl2 → Fe2(SO4)3 + 2HCl Na2SO3 + Cl2 + H2O → Na2SO4 + 2HCl SO2 + 2H2O + Cl2 → H2SO4 + 2HCl I2 + 6H2O + 5Cl2 → 2HIO3 + 10HCl 

Uses : (i) For bleaching cotton and textiles. (ii) In sterilising drinking water. (iii) In the extraction of gold and platinum. (iv) Manufacture of dyes, drugs, refrigerant and other organic compounds like CHCl3, DDT, CCl4, etc. Hydrogen chloride (HCl) : Preparation : In laboratory 420 K

NaCl + H2SO4 → NaHSO4 + HCl



823 K

NaHSO4 + NaCl → Na2SO4 + HCl



HCl gas can be dried by passing conc. H2SO4 through it. Properties : (i) Colourless and pungent smelling gas. (ii) Extremely soluble in water. (iii) When three parts of conc. HCl and one part of conc. HNO3 are mixed, aqua-regia is formed which is used for dissolving noble metals e.g., gold, platinum. Au + 4H+ + NO3– + 4Cl– → AuCl4– + NO + 2H2O 3Pt + 16H+ + 4NO3– + 18Cl– → 3PtCl62– + 4NO + 8H2O



(iv) Reacts with NH3 giving white fumes of NH4Cl NH3 + HCl → NH4Cl

(v) Decomposes salts of weaker acids. Na2CO3 + 2HCl → 2NaCl + H2O + CO2 NaHCO3 + HCl → NaCl + H2O + CO2 Na2SO3 + 2HCl → 2NaCl + H2O + SO2 Uses : (i) In the manufacture of chlorine, glucose and NH4Cl. (ii) For extracting glue from bones and purifying bone black. Interhalogen Compounds : Halogen combines amongst themselves to form a number of compounds known as interhalogen compounds. Their general formulae are XX', XX3', XX5' and XX7' where X is halogen of larger size and higher electropositivity and X' of smaller size. Preparation : 

437 K

Cl2 + F2 → 2ClF; (Equal volume)





573 K



Properties : (i) Covalent compounds. (ii) Diamagnetic in nature.

I2 + 3Cl2 → 2ICl3 (excess)

Cl2 + 3F2 → 2ClF3; Br2 + 3F2 → 2BrF3 (excess) (Diluted with water) I2 + Cl2 → 2ICl; Br2 + 5F2 → 2BrF5 (Equimolar) (excess)

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(iii) More reactive than halogens. (iv) Undergo hydrolysis.

XX' + H2O → HX' + HOX

 Structure : On the basis of VSEPR theory, XX3' compounds have bent ‘T’ shape, XX5' compounds have square pyramidal and IF7 has pentagonal bipyramidal shape. Uses : (i) As non-aqueous solvents. (ii) As fluorinating agents. (iii) ClF3 and BrF3 are used for the production of UF6 in the enrichment of 235U.

Know the Terms  Pseudohalide ions : CN–, SCN– and OCN– ions are called pseudohalide ions while (CN)2, (SCN)2 and (OCN)2 are known as pseudohalogens.  Bleaching powder : Ca(OH)2 + Cl2 ® CaOCl2 + H2O Bleaching Powder

Mnemonics • Concept: Increasing order of atomic radii and atomic volume (Top to bottom-Group) • Mnemonic: First Class Biryani In Australia. • Interpretation: Fluorine(F), Chlorine(Cl), Bromine(Br), Iodine(I), Astatine(At)

Q. (a) Account for the following : (i) Tendency to show –2 oxidation state decreases from oxygen to tellurium. (ii) Acidic character increases from HF to HI. (iii) Moist SO2 gas acts as a reducing agent. (b) Draw the structure of an oxoacid of sulphur containing S–O–S linkage. (c) Complete the following equation : XeF2 + H2O → 

and bond dissociation enthalpy. HF also has intermolecular hydrogen bonding so hydrogen atoms get trapped in the hydrogen bond. STEP-III: In the presence of moisture SO2 is oxidised to H2SO4 liberating hydrogen. SO2 + 2H2O → H2SO4 + 2 [H]

Coloured substance + 2[H] → colourless Solution: (b) H2S2O7 STEP-1: (a) (i) –2 oxidation state decreases O O from oxygen to tellurium because of the S O S presence and filling of d-orbitals. As we O O move from O to Te, nuclear size increases OH OH and electronegativity decreases. (c) 2XeF2 + 2H2O → 2Xe + 4HF + O2 STEP-2: Acidic strength increases from HF  [3 + 1 + 1 = 5] to HI because HF has maximum polarity

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Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS : Q.1. Affinity for hydrogen decreases in the group from fluorine to iodine. Which of the halogen acids should have highest bond dissociation enthalpy? (a) H-F (b) HCl (c) HBr (d) HI

(1 mark each) [B] ASSERTION AND REASON TYPE QUESTIONS : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

(a) Assertion and reason both are correct statements and reason is correct explanation for assertion.



(b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.



(c) Assertion is correct statement but reason is wrong statement.



(d) Assertion is wrong statement but reason is correct statement.

U

Ion

ClO4-

IO4-

BrO4-

Reduction EQ=1.19V EQ=1.65V EQ=1.74V potential EQ/V (a) ClO–4 > IO–4 > BrO–4 

(c) BrO–4> IO–4 > ClO–4

(b) IO–4 > BrO–4 > ClO–4

(d) BrO–4 > ClO–4 > IO–4

A

Ans. Correct option : (c)

Explanation : The higher the reduction potential, the higher is its tendency to get reduced. Hence, the order of oxidising power is : BrO–4> IO–4 > ClO–4 Q. 3. Which of the following is iso-electronic pair? (a) ICl2, ClO2 (b) BrO–2, BrF+ 2 (c) ClO2, BrF (d) CN–, O3 A Ans. Correct option : (b) Explanation : (a) ICl2 = 53 + 2 × 17 = 87 ClO2 = 17 + 16 = 33 (b) BrO2− = 35 + 2 × 8 + 1 = 52 BrF2+ = 35 +9 × 2 − 1 = 52 (c) ClO2 = 17 + 16 = 33 BrF = 35 + 9 = 44 (d) CN− = 6 + 7 + 1 = 14 O3 = 8 × 3 = 24 Q. 4. A black compound of manganese reacts with a halogen acid to give greenish yellow gas. When excess of this gas reacts with NH3 an unstable trihalide is formed. In this process the oxidation state of nitrogen changes from : (a) – 3 to +3. (b) – 3 to 0. (c) – 3 to +5. (d) 0 to – 3. A&E Ans. Correct option : (a) Explanation : MnO2 + 4HCl ® MnCl2 + 2H2O + Cl2 (Greenish yellow gas) NH3 + 3Cl2 ® NCl3 + 3HCl When excess of chlorine reacts with ammonia then NCl3 and HCl will form. In this reaction on lefthand side chlorine has (−3) oxidation state and on the right-hand chlorine has (+3) oxidation state.

Q.1. Assertion (A) : HI cannot be prepared by the reaction of KI with concentrated H2SO4. Reason (R) : HI has lowest H–X bond strength among halogen acids. R Ans. Correct option : (b) Explanation : Both statements are correct but are independent of each other. HI cannot be prepared by the reaction of KI with concentrated H2SO4 as it results in the formation of HI which further oxidizes to I2 as H2SO4 is a strong oxidizing agent. Q.2. Assertion (A) : F2 is a strong oxidizing agent. Reason (R) : Electron gain enthalpy of fluorine is less negative. [CBSE OD Set-1 2020] Ans. Correct option : (b) Explanation : Fluorine is the best oxidising agent because it has more reduction potential (more ability to lose the electrons) which is attributed to its high electronegativity. Q.3. Assertion (A) : F2 has lower bond dissociation energy than Cl2. Reason (R) : Flourine is more electronegative than chlorine. [CBSE OD Set-2 2020] Ans. Correct option : (d) Explanation : F2 has higher bond dissociation enthalpy than Cl2. Q.4. Assertion (A) : F2 has lower reactivity. Reason (R) : F-F bond has low Δbond Ho. Ans. Correct option : (d) Explanation : Fluorine is the maximum reactive because of low bond dissociation enthalpy. [C] VERY SHORT ANSWER TYPE QUESTIONS : Q.1. Write the formula of the compound of iodine which is obtained when conc. HNO3 oxidises I2. R [CBSE OD Set-3 2017]

Ans. Correct option : (a) Explanation : F being smallest has the shortest H-F bond and therefore HF has the highest bond dissociation energy. Q. 2. Reduction potentials of some ions are given below. Arrange them in decreasing order of oxidising power.



Ans. HIO3.

[CBSE Marking Scheme 2017]

Q.2. Give reason for the following: Electron gain enthalpies of halogens are largely negative. A&E [CBSE O.D. Set-2, 2017]

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Ans.



[Topper’s Answer 2017]

Detailed Answer: Halogens are smallest in size in their respective periods resulting in high effective nuclear charge. They accept one electron to attain noble gas electronic configuration. Hence, they have negative electron gain enthalpy. Q. 3. Give reason: Fluoride ion has higher hydration enthalpy than chloride ion. 

A&E [CBSE Delhi Set-1, 2, 3, 2017]

Ans. Due to small size of fluoride ion/high charge density of fluoride ion/high charge size ratio of fluoride ion. [1] [CBSE Marking Scheme 2017] Detailed Answer : The size of fluoride ion is smaller to that of chloride ion. So, on dissolving in water, the hydration energy released in the case of fluoride ion is higher than chloride ion because of stronger interactions between the water and fluoride ion. Q. 4. Assign reason for the following: Reducing character increases from HF to HI.

Ans. As we move from HF to HI, the thermal stability of these hydrides decreases as bond dissociation enthalpy of HX bond decreases. So, on moving from HF to HI, the hydrogen is available for reduction. Therefore, reducing character increases. [1]

Commonly Made Error  Students often state the facts but do not explain the reason properly.

Answering Tip  While stating the reason, write the cause and the consequence. Q. 5. Give reason: ICl is more reactive than I2.  A&E [CBSE OD (Central) 2016] Ans. Inter halogen compound ICl is more reactive than Cl2 because I–Cl bond in inter halogen is weaker than Cl–Cl bond in halogens except F–F bond. [1]

A&E [CBSE O.D. Set-2, 2016]

Short Answer Type Questions Q. 1. Draw the structures of the following: (i) H2SO3 (ii) HClO3 R [CBSE OD Set-1 2017] (ii) 

Ans. (i)

S

H

S O



S O

O

OH

  (ii)

OH

[1]

F

Cl O Cl

O

[1+1] [CBSE Marking Scheme 2017]

Q. 2. Draw the structures of the following: (i) H2S2O8 (ii) ClF3 R [CBSE OD Set-2 2017] Ans.

O

F

F

[1] [CBSE Marking Scheme 2017]

HO

O

O

O HO

O



 

Ans. (i) 

(2 marks each)

OR



  

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[Topper’s Answer 2017] [2]

Commonly Made Error

[1]



 Students often do not follow the convention and draw inaccurate structures.

Ans. (a) I2 < F2< Br2 < Cl2   (b) H2O < H2S < H2Se < H2Te OR (i) CIF

Answering Tip

(ii)

O

 Draw the structures with lone pair of electrons (if present). Avoid over-writing. Q. 3. Arrange the following in order of property indicated for each set : (i) F2, Cl2, Br2, I2 – increasing bond dissociation enthalpy. (ii) PH3, AsH3, BiH3, SbH3, NH3 – increasing basic strength. U [CBSE SQP 2016; DDE] Ans. (i) I2 < F2 < Br2 < Cl2 [1] (ii) BiH3 HBr > HI

[2]

[Topper’s Answer 2019]

Short Answer Type Questions-II Q. 1. What is the reason behind the anomalous behaviour of fluorine in Group 16? R Ans: Fluorine’s anomalous behaviour is due to the following (a) Its small size (b) Its high electronegativity (c) Its low F-F bond dissociation enthalpy (d) Non-availability of d-orbitals in valence shell

(Any three correct reasons)

[3]

Q. 2. A greenish yellow gas ‘Y’ is obtained by oxidation of ‘X’ in the presence of CuCl2 as catalyst. ‘Y’ reacts with gas ‘Z’ to give HCl. Identify ‘X’, ‘Y’, ‘Z’ and write the equations involved. A&E

(3 marks each)

Ans: 4HCl + O2----CuCl2--- 2Cl2 + 2H2O (X ) (Y) (Greenish yellow gas) H2 + Cl2 --- 2HCl (Z) [3] Q. 3. Give any three characteristics of Interhalogen compounds. R Ans: Characteristics of Interhalogen compounds (a) All are covalent compounds. (b) Diamagnetic in nature. (c) Volatile solids or liquids except ClF which is a gas. (d) Physical properties are intermediate between those of constituent halogens but m.p and b.p are a little higher than expected. (Any three correct properties) [3]

Long Answer Type Questions Q. 1. (i) Answer the following questions :  (2+3) (a)  Write the balanced chemical reaction for reaction of Cu with dilute HNO3.

(b) F [1] CL

(b) Draw the shape of ClF3.

(ii) ‘X’ has a boiling point of 4.2K, lowest for any known substance. It is used as a diluent for oxygen in modern diving apparatus. Identify the gas ‘X’. Which property of this gas makes it usable as diluent? Why is the boiling point of the A [CBSE SQP 2021] gas ‘X’ so low? Ans. (i) (a) 3Cu + 8HNO3(dilute) ⟶ 3Cu(NO3)2 + 2NO + 4H2O [1]

(5 marks each)

F

F



(ii) ‘X’ is Helium [1] It is used as a diluent for oxygen in modern diving apparatus because of its very low solubility in blood. [1] It is monoatomic having no interatomic forces except weak dispersion forces and has second lowest mass therefore boiling point is lowest.[1]

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Ans. (a) (i) Endothermic compound / decomposition of ozone is exothermic in nature and ΔG is negative / decomposition of ozone is spontaneous. [1] (ii) Exists as [PCl4]+ and [PCl6]–.  [1] (iii) Shows only -1 oxidation state / most electronegative element/ absence of d-orbitals.  [1]

F

(b) (i)

F

F

— Br

F

F

Q. 2. (a) Account for the following : (i) Ozone is thermodynamically unstable. (ii) Solid PCl5 is ionic in nature. (iii) Fluorine forms only one oxoacid HOF. (b) Draw the structure of (i) BrF5 (ii) XeF4 [5] OR (i) Compare the oxidising action of F2 and Cl2 by considering parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy. (ii) Write the conditions to maximize the yield of H2SO4 by contact process. (iii) Arrange the following in the increasing order of property mentioned : (a) H3PO3, H3PO4, H3PO2 (Reducing character). (b) NH3, PH3, AsH3, SbH3, BiH3 (Base strength). A [CBSE Delhi Set-1 2016]

(ii)

F

F

Xe

F

F



[1 + 1] OR (i) F2 is the stronger oxidising agent than chlorine

(a) low enthalpy of dissociation of F-F bond



(b) less negative electron gain enthalpy of F



(c) high hydration enthalpy of F- ion



[½ × 4 = 2]



(ii) low temperature, high pressure and presence of catalyst  [1]

(iii) (a) H3PO4< H3PO3< H3PO2 

(b) BiH3< SbH3< AsH3< PH3< NH3



[1] [1]

[CBSE Marking Scheme, 2016]

TOPIC-4

Group-18 Elements, Properties and Some Important Compounds Revision Notes  Noble gases : 2He, 10Ne, 18Ar, 36Kr, 54Xe, 86Ra. Scan to know  Also known as rare gases as they are present in very small amounts in the air or as inert gases more about as they were considered chemically unreactive. this topic  Electronic configuration : ns2np6 (except He which has 1s2)  Physical Properties : (i) Gases (ii) Atomic radii increase down the group and highest in their respective periods. Noble Gases (iii) Highest ionisation enthalpy in their respective periods. Down the group, it decreases due to increase in atomic size. (iv) Large positive values of electron gain enthalpy as noble gases have stable electronic configuration and no tendency to accept electron. (v) Low melting points and boiling points due to weak dispersion forces. Down the group they increase because of increase in van der Waals forces. (vi) Liquefaction : They are difficult to liquify. Down the group, the ease of liquefaction increases.  In 1962, Bartlett studied the given reaction : O2 + PtF6 → O2+[PtF6]– Dioxygenyl hexafluoro platinate As ionisation enthalpy of molecular oxygen was almost similar to xenon, following reaction was also found to occur :

+ – 289 K Xe + PtF6 → Xe [PtF6]

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

 Preparation of Xenon fluorides : Xenon forms three binary fluorides, XeF2, XeF4 and XeF6. 673 K, 1 bar



Xe(g) + F2(g) → XeF2(s)



(excess) 873 K, 7 bar



Xe(g) + 2F2(g) → XeF4(s)



→ XeF6(s) Xe(g) + 3F2 (g)  (1 : 20 ratio)

573 K, 60-70 bar

143 K

XeF4 + O2F2 → XeF6 + O2



 Preparation of Xenon trioxide (XeO3) : 6XeF4 + 12H2O → 2XeO3 + 4Xe + 3O2 + 24HF XeF6 + 3H2O → XeO3 + 6HF  Preparation of Xenon oxyfluorides : XeF4 + H2O → XeOF2 + 2HF



Xenon oxydifluoride Partial

→ XeOF4 + 2HF XeF6 + H2O Hydrolysis Xenon oxytetrafluoride



Complete



→ XeO2F2 + 4HF XeF6 + 2H2O Hydrolysis Xenon dioxydifluoride  Structures of Xenon-fluorine compounds : F

F

Xe F

XeF2(sp3d) [Linear]

Xe F

F

F

F

XeF4 (sp3d2) [Square Planar]

 Uses : Helium

Xe

F F

(i) To lift weather For advertising. balloons and air ships.

F

O

F

Xe

Xe F O

O O

3

XeOF2 (sp d) [T-Shaped]

Argon

O

F

F

3 2

XeOF4 (sp d ) [Square Pyramidal]

Krypton

XeO3 (sp3) [Trigonal Pyramidal]

Xenon

To create an inert For runway and In electric flash atmosphere. approach lights in bulbs for high speed airports. photography.

(ii) As breathing For filling sodium In geiger counters. vapour lamps. mixture. (iii) For inflating In beacon light. the tyres of aeroplanes.

F

Xe

F

XeF6 (sp3d3) [Distorted Octahedral]

Neon

F

F

In high efficiency In gas filled lamps. miner’s cap lamps.

To date the age of rocks.

Mnemonics • Concept: Increasing order of atomic radii and atomic volume (Top to bottom-Group) • Mnemonic: Help Needy And Krippled (Crippled) on X-mas Religiously. • Interpretation: Helium(He), Neon(Ne), Argon(Ar), Krypton(Kr), Xenon(Xe), Radon(Rn)

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Q. (a) Compare the oxidizing action of F2 and Cl2 by considering parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy. (b) Write the conditions to maximize the yield of H2SO4 by contact process. (c) F2 is the stronger oxidising agent than chlorine : Solutions: STEP-1: Low enthalpy of dissociation of F-F bond.

STEP-2: Less negative electron gain enthalpy of F. STEP-3: High hydration enthalpy of F– ion. STEP-1: The main reaction in contact process is 2SO2 + O2 → 2SO3 ↑ STEP-2: This reaction is reversible and conditions to increase the yield of SO3 are : (a) High pressure (b) Low temperature (optimum temperture 720 K) should be maintained and (c) V2O5 as catalyst.

Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS : Q.1. Which one of the following does not exist? (a) XeOF4 (b) NeF2 (c) XeF2 (d) XeF6 R Ans. Correct option : (b) Explanation : Xe has least ionisation energy among the noble gases and hence it forms chemical compounds with oxygen and fluorine, however, Ne cannot form compounds with oxygen and fluorine so NeF2 does not exist. Q.2. In the preparation of compounds of Xe, Bartlett had taken O2+PtF6– as a base compound. This is because (a) both O2 and Xe have same size. (b) both O2 and Xe have same electron gain enthalpy. (c) both O2 and Xe have same ionisation enthalpy. (d) both Xe and O2 are gases. Ans. Correct option : (c) Explanation : In the preparation of compounds of Xe, Bartelett had taken O2+PtF6– as a base compound. This is because both O2 and Xe have almost same ionisation enthalpy. Q.3. Which of the following statements are true? (a) Only types of interactions between particles of noble gases are due to weak dispersion forces. (b) Hydrolysis of XeF6 is a redox reaction. (c) Xenon fluorides are not reactive. (d) None of the above

(1 mark each) Ans. Correct option : (a) Explanation : Only types of interactions between particles of noble gases are due to weak dispersion forces. [B] ASSERTIONS AND REASONS: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct state­ ments and reason is correct explanation for assertion. (b) Assertion and reason both are correct state­ ments but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Q. 1. Assertion(A) : Group 18 gases exhibit very high ionisation enthalpy.  Reason (R) : They have a stable electronic configuration. U Ans. Correct option : (a) Explanation : Group 18 gases exhibit very high ionisation enthalpy because they have a stable electronic configuration. Q. 2. Assertion(A) : The noble gases are inactive. Reason(R) : These gases have a closed shell structure. U

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Ans. Correct option : (a) Explanation : The noble gases are inactive as they have a closed shell structure. Q. 3. Assertion(A) : Helium diffuses through most commonly used laboratory materials. Reason(R) : This gas has a very low melting point. U Ans. Correct option : (c) Explanation : Helium diffuses through most commonly used laboratory materials which is an unusual property of this gas.

[C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1. Why do noble gases have large positive electron gain enthalpies? U Ans. Since noble gases have filled shells and stable electronic configurations, they do not accept electrons. Hence they have high positive electron gain enthalpies. [1] Q.2. What is the structure of XeOF4? R Ans. XeOF4 is square pyramidal. Q. 3. Complete the following hydrolysis equation and balance it. A XeF6 + H2O → HF + ________ Ans. XeF6 + 3H2O → 6HF + XeO3

Short Answer Type Questions - I Q.1. Draw the structures of the following : (i) H2S2O8 (ii) XeF6  [CBSE Delhi Set - I, 2020] ns. (i) H2S2O8 A O O O–O OH

O

(ii) XeF4

OH

(ii) XeF6 F

Q. 4. Write the structures of the following : (i) BrF3 (ii) XeF4 Ans. (i) BrF3

F

F

F





(ii) BrF5

OH

F



Br F F Q.3. Draw the structures of the following : [1+1] (i) HClO4 (ii) XeOF4 [CBSE Delhi Set - 3, 2020] Ans. (i) HClO4

Commonly Made Error  Students often draw rough structural diagrams without showing the lone pair.

. 5. What happens when : Q (i) SO2 gas is passed through an aqueous solution of Fe3+ salt ? (ii) XeF4 reacts with SbF5 ?  R [CBSE OD 2016] Ans. (i) When SO2 gas is passed through an aqueous solution of Fe3+ salt, SO2 acts as a reducing agent and reduces Fe3+ to Fe2+. The brown colour of iron (III) solution turns into green. [1]

O





O = Cl = O — H



O



O F

+ 2Fe3+ + SO2 + 2H2O → 2Fe2++ SO2– 4 + 4H

(ii) When XeF4 reacts with strong Lewis acids like SbF5, it forms complexes (addition compounds).

F



XeF4 + SbF5 → [XeF3]+[SbF6]–

[1]

[CBSE Marking Scheme 2016]

Xe F

[1]

 Draw the structures with lone pair of electrons (if present). Avoid over-writing.

F

(ii) XeOF4



Answering Tip O

OH

F

F





S O

Square planar

Xe

F F Q.2. Draw the structures of the following : (i) H2S2O7 (ii) BrF5 [CBSE Delhi Set - 2, 2020] Ans. (i) H2S2O7 O O O

[1]

F

F — Xe — F

S

R



S

S O

(2 marks each)

F

Q. 6. Draw the structures of the following : [1+1]



(i) H2SO4

(ii) XeF2

R [CBSE Delhi 2015]

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p-BLOCK ELEMENTS

Ans. (i) F2 + 2Cl- → 2F- + Cl2 [1] (ii) 2XeF2 + 2H2O → 2Xe + 4HF + O2 [1] [CBSE Marking Scheme 2017]





Ans. (i)

(ii)

S HO



F

. 11. What happens when Q (i) HCl is added to MnO2 ? (ii) PCl5 is heated ? Write the equations involved.

Xe O

HO

F

[1 + 1] [CBSE Marking Scheme 2015]

Q. 7. Complete the following reactions: (ii) XeF6 + 2H2O →

R [CBSE Delhi Set-1 2017]

Ans. (i) NH3 + 3Cl2 (excess) → NCl3 + 3HCl

[1]

(ii) XeF6 + 2H2O → XeO2F2

[1]

[CBSE Marking Scheme 2017] . 8. Q

What happens when (i) (NH4)2Cr2O7 is heated ? (ii) H3PO3 is heated ? Write the equation.

(ii) 4H3PO3 → 3H3PO4 +PH3 [CBSE Marking Scheme, 2017] Detailed Answer : (i) (NH4)2Cr2O7 decomposes on heating to produce Cr2O3, H2O and N2. The reaction that takes place is : (NH4)2 Cr2O7 ¾D¾ ® Cr2O3 + 4H2O + N2

[1]

(ii) H3PO3 disproportionates on heating to produce orthophosphoric acid or phosphoric acid and phosphine. The reaction that takes place is :

4H3PO3 ¾D¾ ® 3H3PO4 + PH3

[1]

Q. 9. Complete the following reactions: (i) Cl2 + H2O → (ii) XeF6 + 3H2O →

(ii) PCl5 ¾¾ ® PCl3 + Cl2



[1] [1]

Detailed Answer : (i) When HCl is added to MnO2, chlorine gas is formed along with other products. MnO2 + 4HCl → MnCl2 + 2H2O + Cl2 [1] (ii) When heated, PCl5 sublimes but decomposes on strong heating.

Ans. (i) (NH4)2Cr2 O7 → N2 + 4H2O + Cr2O3



D

[CBSE Marking Scheme, 2017]

(i) NH3 + 3Cl2 (excess) →



(i) MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O



O

R [CBSE Delhi Set-2 2017]

Ans. (i) Cl2 + H2O → 2HCl + [O] / HCl +HOCl [1] (ii) XeF6 + 3H2O → XeO3 + 6HF [1] [CBSE Marking Scheme 2017]

Commonly Made Error  Students often write wrong products, do not write all the correct products, write only the main product thus leading to error in balancing or do not balance the equations at all.

Answering Tip  Write correct, complete and balanced equations for all reactions mentioning all reactants and products. . 10. Complete the following chemical equation : Q (i) F2 + 2Cl- → (ii) 2XeF2 + 2H2O → R [CBSE Delhi Set-3 2017]

PCl5 ¾D¾ ® PCl3 + Cl2

[1]

Q. 12. Account for the following : (i) Acidic character increases from HF to HI. (ii) There is a large difference between the melting and boiling points of oxygen and sulphur. A&E [CBSE Delhi 2015] Ans. (i) Acidic character increases from HF to HI due to increase in size as a result attraction force decreases and acidity increases. HF < HCl < HBr < HI  → Increasing order of acidity size increasse

(ii) There is large difference between the melting point and boiling point of oxygen and sulphur due to small size and high electronegativity of oxygen. 2 Q. 13. Account for the following : (a) Cl2 acts as a bleaching agent. (b) Noble gases have very low boiling points. A&E Ans. (a) Because Cl2 in presence of moisture liberates nascent oxygen. [1] (b) Interatomic interactions are weak. [1]

Commonly Made Error  Students often lose time in writing long answers to 1 mark questions.

Answering Tip  While stating the reason, be specific. Avoid unnecessary explanations. . 14. (a) What happens when F2 reacts with water ? Q (b) Write the formula of a noble gas species which is isostructural with IBr2. U [CBSE Foreign Set-1 2017] Ans. (a) 2F2+2H2O → 4HF + O2 HF and O2 are produced. [1] (b) XeF2 [1] [CBSE Marking Scheme 2017]

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Short Answer Type Questions-II

Ans. (i) H2O < H2S < H2Se < H2Te, because of decrease in bond dissociation enthalpy.[1+1] O F

F Xe

F

F





Square pyramidal

[1]

[CBSE Marking Scheme 2017]

Q. 2. (i) F2 has lower bond dissociation enthalpy than Cl2. Why ? (ii) Which noble gas is used in filling balloons for meteorological observations ? (ii) Complete the following equation : XeF2 + PF5 → A&E [CBSE Delhi 2015] Ans. (i) F2 has lower bond dissociation energy than Cl2 because the size of F2 is much smaller than Cl2 as a result interelectronic repulsion works and make F2 weak. [1] (ii) Helium is used in filling balloons for meteorological observations.  [1] (iii) XeF2 + PF5 → [XeF2]+ [PF6]– [1] Q. 3. (i) Draw the structure of a noble gas species which is isostructural with BrO3–. (ii) Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F2 and Cl2. (iii) Why is Ka2 HCl > HF [1] (iv) Why fluorine is a stronger oxidizing agent than chlorine? Ans. Because fluorine has greater E° value (2.87V) than chlorine (1.36V). [1] (v) What are the sizes of X and X’ in the interhalogen compounds? Ans. Size of X is greater than X’. 

[ 213

SELF-ASSESSMENT TEST

Self Assessment PaperTest - 7 Time : 1 Hour [A] OBJECTIVE TYPE QUESTIONS :

Max. Marks : 25 (1 mark each)

1.  Read the passage given below and answer the following questions : (1 × 4 = 4) Interhalogen compounds are in the form of XYn where X and Y are halogens and n is one, three, five, or seven. Interhalogen compounds contain at most two different halogens. Large interhalogens, such as ClF3 can be produced by a reaction of a pure halogen with a smaller interhalogen such as ClF. All interhalogens except IF7 can be produced by directly combining pure halogens in various conditions.  Interhalogens are typically more reactive than all diatomic halogen molecules except F2 because interhalogen bonds are weaker. However, the chemical properties of interhalogens are still roughly the same as those of diatomic halogens. Many interhalogens consist of one or more atoms of fluorine bonding to a heavier halogen. Chlorine can bond with up to 3 fluorine atoms, bromine can bond with up to five fluorine atoms, and iodine can bond with up to seven fluorine atoms. Most interhalogen compounds are covalent gases. However, some interhalogens are liquids, such as BrF3, and many iodine-containing interhalogens are solids. (i)  Reduction potentials of some ions are given below: Arrange them in decreasing order of oxidising power. Ion ClO4– IO4– BrO4– o o o Reduction E =1.19V E =1.65V E =1.74V potential Eo/V (a) ClO4– > IO4– > BrO4– (b) IO4– > BrO4– > ClO4– (c) BrO4– > IO4– > ClO4– (d) BrO4– > ClO4– > IO4– (ii) Which of the following is isoelectronic pair? (a) ICl2, ClO2 (b) BrO2–, BrF2+ (c) ClO2, BrF (d) CN–, O3 (iii) Which of the following options are not in accordance with the property mentioned against them. (a) F2 > Cl2 > Br2 > I2        Oxidising power (b) Ml > MBr > MCl > MF  Ionic character of metal halide (c) F2 > Cl2 > Br2 > I2          Bond dissociation enthalpy. (d) Hl < HBr < HCl < HF    Hydrogen-halogen bond strength. (iv) Which of the following statements are correct? (a) Among halogens, radius ratio between iodine and fluorine is maximum. (b) Leaving F – F bond, all halogens have weaker X – X bond than X – X’ bond in inter-halogens.



(c) Among inter-halogen compounds maximum number of atoms ate present in iodine fluoride. (d) Inter-halogen compounds are more reactive than halogen compounds. OR Chlorine water on standing loses its yellow colour due to _________ (a) Formation of chlorine gas (b) Formation of HCl (c) Formation of HCl and HOCl. (d) Formation of HOCl. Following questions (No. 2 to 5) are Multiple Choice Questions carrying 1 mark each. 2.  Which of the following statements are correct? (a) All the three N−O bond lengths in HNO3 are equal. (b) All P−Cl bond lengths in PCl5 molecule in gaseous state are equal. (c) P4 molecules in white phosphorus have angular strain, therefore white phosphorus is very reactive. (d) PCl5 is ionic in solid state in which cation is tetrahedral and anion is octahedral. U 3. The oxidation state of central atom in the anion of compound NaH2PO2 will be __________ (a) +3 (b) +5 (c) +1 (d) –3 [A&E] 4.  Which of the following elements does not show allotropy? (a) Nitrogen (b) Bisumth (c) Antimony (d) Arsenic R [Exemplar page 92]  5. In solid state PCl5 is a ________ (a) Covalent solid. (b) Octahedral structure (c) Ionic solid with [PCl6]+ octahedral and [PCl4]– tetrahedral (d) Ionic solid with [PCl4]+ tetrahedral and [PCl6]– octahedra R The following questions (Q. no. 6 & 7), a statement of assertion followed by a statement of reason given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not the correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. 6. Assertion (A) : XeF6 has distorted octahedral structure.  Reason (R) : The orbitals of Xenon are sp3d3 hybridised. U 7.  Assertion (A) : The sp3d2 hybrid orbitals are responsible for square pyramidal structure of XeOF4.

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Reason (R) : The hybridization in Xenon is sp3d2 because there is a migration of two electrons of p to d orbital which results in the formation of sigma bond with F. U The following questions (Q.No. 8 & 9), are Short Answer Type-I and carry 2 marks each. 8. Mention two conditions which maximise the yield of H2SO4 during contact process. U 9. Complete the following equations : (i) Ag + PCl5 ® (ii) CaF2 + H2SO4 ® A 

Q.No. 10 & 11 are Short Answer Type-II carrying 3 marks each. 10. Give reasons for the following : (i) (CH3)3 P = O exists but (CH3)3 N = O does not. (ii) Oxygen has less electron gain enthalpy with negative sign than sulphur. (iii) H3PO2 is a stronger reducing agent than H3PO3. A&E

11. Draw the structures of the following compounds. (i) ClF3 (ii) HClO3 (iii) H2S2O7 A Q.No 12 is Long Answer Type Carrying 5 marks. 12. (i) Draw the structure of a noble gas species which is isostructural with BrO3–. (ii) Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F2 and Cl2. (iii) Why is Ka2 Mo, Tc, Ru, Rh, Pd, Ag, Cd

[ 221

‘d’ AND ‘f‘ BLOCK ELEMENTS

5-d series • Concept: Third Row Transition Elements-5-d series • Mnemonic: Late Harry Took Walk, Reached Office In Pajamas After an Hour. • Interpretation: La......, Hf, Ta, W, Re, OS, Ir, Pt, Au, Hg

Know the Terms  Oxidation state : The measure of the electronic state of an atom in a particular compound and is equal to the number of electron; it has in the valence shell, more than or less than the number of electrons in free atom.  Ferromagnetic substances : Substances which are attracted very strongly by the applied magnetic field. e.g., Fe, Co, Ni etc.



Q. (i) For M2+/M and M3+/M2+ systems, E° values for some metals are as follows : Cr2+/Cr = – 0.9 V  Cr3+/Cr2+ = – 0.4 V Mn2+/Mn = – 1.2 V Mn3+/Mn2+ = +1.5 V Fe2+/Fe = – 0.4 V    Fe3+/Fe2+ = +0.8 V Use this data to comment upon (a) the stability of Fe3+ in acid solution as compared to that of Cr3+ and Mn3+. (b) the ease with which iron can be oxidised as compared to the similar process for either Cr or Mn metals. (ii) What can be inferred from the magnetic moment of the complex K4[Mn(CN)6] ? (Magnetic moment : 2.2 BM) Solution:

3+ 2+ STEP - I : (i) (a) Cr /Cr has a negative reduction potential. Hence, Cr3+ cannot

be reduced to Cr2+. Cr3+ is most stable. Mn3+/Mn2+ have large positive E° values. Hence, Mn3+ can be easily reduced to Mn2+. Thus Mn3+ is least stable. Fe3+/ Fe2+ couple has a positive E° value but is small. Thus, the stability of Fe3+ is more than Mn3+ but less stable than Cr3+. [1] (b) If we compare the reduction potential values, Mn2+/Mn has the most negative value i.e., its oxidation potential value is most positive. Thus, it is most easily oxidised. Therefore, the decreasing order for their ease of oxidation is Mn > Cr > Fe. [1] STEP - I : (ii) (ii) K4[Mn(CN)6] Mn is in +2 oxidation state. Magnetic moment 2.2 indicates that it has one unpaired electron and hence forms inner orbital or low spin complex. In presence of CN– which is a strong ligand, hybridisation involved is d2sp3 (octahedral complex). [1]

Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS : Q. 1. Electronic configuration of a transition element X in +3 oxidation state is [Ar]3d5. What is its atomic number?

(1 mark each)

(a) 25 (c) 27

(b) 26 (d) 24 A [NCERT Exemp. Q. 1, Page 105] Ans. Correct option : (b)

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Explanation : It is formed by the loss of 3 electrons, the configuration of element X is [Ar] 3d64s2. Therefore, Atomic number = 26. Q. 2. The electronic configuration of Cu(II) is 3d9 whereas that of Cu(I) is 3d10. Which of the following is correct? (a) Cu(II) is more stable (b) Cu(II) is less stable (c) Cu(I) and Cu(II) are equally stable (d) Stability of Cu(I) and Cu(II) depends on nature of copper salts  U [NCERT Exemp. Q. 2, Page 105] Ans. Correct option : (a) Explanation : Cu(II) is more stable due to nuclear charge of Cu. Q. 3. When KMnO4 solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time because



(a) CO2 is formed as the product. (b) Reaction is exothermic. (c) MnO4– catalyses the reaction. (d) Mn2+ acts as auto-catalyst. A&E [NCERT Exemp. Q. 9, Page 107] Ans. Correct option  : (d)

Explanation  : When KMnO4 solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after sometime because Mn2+ acts as an auto-catalyst.



Reduction half-reaction :



[MnO4− + 8 H+ + 5e − → Mn 2 + + 4 H 2O] × 2



Oxidation half-reaction :



2 − [C 2O4 → 2CO2 + 2e ] × 5 Overall equation :



2− 2+ − + 2 MnO4 + 16 H + 5C 2O4 → 2 Mn + 10CO2 + 8 H 2O End point of this reaction  : Colourless to light pink.

−

Q. 4. When acidified K2Cr2O7 solution is added to Sn salts then Sn2+ changes to (a) Sn (b) Sn3+ 4+ (c) Sn (d) Sn+ A [NCERT Exemp. Q. 18, Page 108]  Ans. Correct option : (c) Explanation  : When acidified K2Cr2O7 solution is added to Sn2+ salt, Sn2+ changes to Sn4+. The reaction is given here  :

Oxidation Cr2 O72–+ 14H+ + 3Sn2+

Q.5. 

Reduction

2Cr3+ + 3Sn4+ + 7H2O

The oxidation state of Ni in [Ni(CO)4] is (a) 0 (b) 2 (c) 3 (d) 4 U [CBSE Delhi Set 1, 2020]

Ans. Correct option : (a)  0 Explanation : Let oxidation state of Ni = x x + 0 = 0 (CO is neutral ligand) x = 0 Q.6. Which of the following is the reason for Zinc not exhibiting variable oxidation state ? (a) inert pair effect (b) completely filled 3d subshell (c) completely filled 4s subshell (d) common ion effect Ans. Correct option : (b) [CBSE SQP 2021] Explanation : Zinc does not exhibit a variable oxidation state as it has completely filled 3d subshell. Zn — [Ar] 3d10 4s2 Atomic number 30 Q. 7. Which of the following is a diamagnetic ion ? (Atomic numbers of Sc, V, Mn and Cu are 21, 23, 25 and 29 respectively) (a) V2+ (b) Sc3+ 2+ (c) Cu (d) Mn3+ Anw. Correct option : (b) Explanation : Sc3+ is a diamagnetic ion. Atomic number is 21. [Ar]3d0 is Sc3+ electronic configuration. Since its d subshell is empty it is diamagnetic as it has noble gas configuration. .8. Which set of ions exhibit specific colours? (Atomic Q number of Sc = 21, Ti = 22, V=23, Mn = 25, Fe = 26, Ni = 28 Cu = 29 and Zn =30) (a) Sc3+, Ti4+, Mn3+ (b) Sc3+, Zn2+, Ni2+ (c) V 3+, V2+, Fe3+ (d) Ti3+, Ti4+, Ni2+ A&E [CBSE SQP 2021] Ans. Correct option : (c) Explanation : V3+, V2+, Fe3+ ions exhibit specific colours. Atomic number of V = 23, Electronic configuration of V - [Ar]3d3 4s2 Electronic configuration of V2+ - [Ar]3d3 Electronic configuration of V3+ - [Ar] 3d2 Atomic number of Fe = 26 Electronic configuration of Fe - [Ar]3d6 4s2 Electronic configuration of Fe3+ - [Ar]3d5 Since these ions have partially filled d- subshells, they exhibit colour. Most transition-metal ions have a partially filled d subshell. As for other ions, Atomic number of Sc = 21 Electronic configuration of Sc - [Ar]3d1 4s2 Electronic configuration of Sc3+ - [Ar]3d0 Since d subshell is empty, it shows no colour. Atomic number of Ti = 22 Electronic configuration of Ti- [Ar]3d2 4s2 Electronic configuration of Ti4+ - [Ar]3d0 Since d subshell is empty, it shows no colour. Atomic number of Mn =25 Electronic configuration of Mn- [Ar]3d5 4s2 Electronic configuration of Mn2+ [Ar]3d4

[ 223

‘d’ AND ‘f‘ BLOCK ELEMENTS

Since d subshell is partially filled, it shows colour. Atomic number of Ni = 28 Electronic configuration of Ni- [Ar]3d8 4s2 Electronic configuration of Ni4+ [Ar]3d8 Since d subshell is partially filled, it shows colour. Atomic number of Zn =30 Electronic configuration of Zn- [Ar]3d10 4s2 Electronic configuration of Zn2+- [Ar]3d10 Since d subshell is full, it shows no colour. [B] ASSERTION & REASON TYPE QUESTIONS : (a) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion. (b) Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion. (c) Assertion is correct, but reason is wrong statement. (d) Assertion is wrong, but reason is correct statement. Q. 1. Assertion (A) : Cu2+ iodide is not known. Reason (R) : Cu2+ oxidises I– to iodine. R [NCERT Exemplar] Ans. Correct option : (a) Explanation : Cu2+ oxidises iodide to iodine hence cupric iodide is converted to cuprous iodide. Q. 2. Assertion (A) : The highest oxidation state of osmium is +8. Reason (R) : Osmium is a 5d-block element. R Ans. Correct option : (b) Explanation : The highest oxidation state of osmium (Os) is +8. It is due to its ability to expand octet by using its all 8 electrons. Q. 3. Assertion (A) : Separation of Zr and Hf is difficult. Reason (R) : Because Zr and Hf lie in the same group of the periodic table. R [NCERT Exemplar] Ans. Correct option : (b) Explanation : Separation of Zr and Hf is difficult as both have same size. Q. 4. Assertion (A) : Cu cannot liberate hydrogen from acids. Reason (R) : Because it has positive electrode potential. R [NCERT Exemplar]

Ans. Correct option : (a) Explanation : Copper (Cu) does not liberate hydrogen from acids due to the presence of positive electrode potential. Q. 5. Assertion (A) : Transition metals have low melting points. Reason (R) : The involvement of greater number of (n – 1)d and ns electrons in the interatomic metallic bonding. U [CBSE OD Set-1, 2020] Ans. Correct option : (d) Explanation : Transition metals have high melting points because of the involvement of greater number of (n – 1)d and ns electrons in the interatomic metallic bonding. Q. 6. Assertion (A) : Transition metals have high melting point. Reason (R) : Transition metals have completely filled d-orbitals. U [CBSE OD Set-2, 2020] Ans. Correct option : (c) Explanation : Transition metals have high melting points because of the involvement of greater number of (n – 1)d and ns electrons in the interatomic metallic bonding. [C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1. Write the formula of an oxo-anion of Manganese (Mn) in which it shows the oxidation state equal to its group number. A [CBSE Delhi Set-1, 3 2017] Ans. MnO4–/KMnO4 [CBSE Marking Scheme 2017] Q. 2. Write the formula of an oxo-anion of Chromium (Cr) in which it shows the oxidation state equal to its group number. A [CBSE Delhi Set-2 2017] Ans. Cr2O72–/CrO42–/K2Cr2O7/K2CrO4 [CBSE Marking Scheme 2017] Q. 3. Give the structure of dichromate ion. Ans.

R



Short Answer Type Questions-I Q. 1. What are the transition elements ? Write two characteristics of the transition elements. R [CBSE Delhi 2015]

Ans. These atoms whose d-orbitals are incomplete in ground state or in one of the most common oxidation state are called transition elements or d-block elements. The valence shell electronic configuration of transition elements is (n-1)d1-10ns1-2 . [1]

(2 marks each)

Two characteristics of transition elements : (i) Transition metals show variable oxidation states. [1] (ii) All transition metals act as catalyst.

Commonly Made Error  Students tend to write irrelevant information on characteristics thus losing time.

Answering Tip  Use the precise definition and mention the main characteristics.

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q. 2. What is meant by ‘disproportionation’ ? Give an example of a disproportionation reaction in R aqueous solution. OR Suggest reasons for the following features of transition metal chemistry : (i) The transition metals and their compounds are usually paramagnetic. (ii) The transition metals exhibit variable oxidation states. A&E [CBSE Comptt. Delhi 2015]



Q. 3. Complete and balance the following chemical equations : (a) Fe2+ + MnO4– + H+ ® (b) MnO4– + H2O + I– ®  A [CBSE Delhi/OD 2018] Ans. (a) 5Fe2+ + MnO4– + 8H+ ® Mn2+ + 4H2O + 5Fe3+  [1] (b) 2MnO4– + H2O + I– ® 2MnO2 + 2OH– + IO3– [1] (Half mark to be deducted in each equation for not balancing) [CBSE Marking Scheme 2018] Detailed Answer :



Ans. Disproportionation is the reaction in which an element undergoes self-oxidation and selfreduction simultaneously. For example – [1] 2Cu+ (aq) → Cu2+ (aq) + Cu(s) [1] (Or any other correct equation) OR (i) Due to presence of unpaired electrons in d-orbitals. [1]

(ii) Due to incomplete filling of d-orbitals. Due to very small energy difference in between (n – 1) d and n s-orbitals. [1] [CBSE Marking Scheme 2015]

[Topper’s Answer 2018] [2]

Commonly Made Error  Many students write unbalanced equations. Some students fail to write the correct products.

Answering Tip



 Write all the reactants and products clearly. Balance the equation.

Q. 5. Explain the following observations : (i) Copper atom has completely filled d orbitals (3d10) in its ground state, yet it is regarded as a transition element. (ii) Cr2+ is a stronger reducing agent than Fe2+ in aqueous solution. A&E [CBSE Comptt. OD Set-1 2017]

Q. 4. In the following ions : Mn3+, V3+, Cr3+, Ti4+ (Atomic no : Mn = 25, V = 23, Cr = 24, Ti = 22)

(a) Which ion is most stable in an aqueous solution?



(b) Which ion is the strongest oxidizing agent?



(c) Which ion is colourless?



(d) Which ion has the highest number of unpaired electrons?



U [CBSE Foreign Set-1, 2, 3 2017]



Ans. (a) Cr

3+

4+

(c) Ti

[½] [½]

3+

[½]

3+

[½]

(b) Mn (d) Mn

[CBSE Marking Scheme 2017]

Ans. (i) Because it has incompletely filled d orbitals in one of its oxidation state (Cu2+). [1]

(ii) Cr2+(d4) changes to Cr3+(d3) while Fe2+(d6) changes to Fe3+ (d5). In aqueous medium d3 is more stable than d5. [1]

[CBSE Marking Scheme 2017]

Q. 6. Name the following : (i) A transition metal which does not exhibit variation in oxidation state in its compounds.

(ii) A compound where the transition metal is in the +7 oxidation state.

(iii) A member of the lanthanoid series which is well known to exhibit +4 oxidation state. (iv) Ore used in the preparation of Potassium R [CBSE SQP 2016] dichromate.

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‘d’ AND ‘f‘ BLOCK ELEMENTS

[½] [½] [½] [½]



Ans. (i) Scandium (Sc). (ii) KMnO4 or any other suitable example. (iii) Cerium (Ce) or any other example. (iv) Chromite ore. Q. 7. Explain the following observation (i) Zn2+ salt are colourless. o (ii) Copper has exceptionally positive E M

2+

/M

value.

Ans. (i) Due to absence of unpaired electrons. [1] (ii) Due to high DaH° and low DhydH°. [1] [CBSE Marking Scheme 2017] Detailed Answer : (i) Zinc has no unpaired electrons in its d orbital and has a stable fully filled d orbital state. Thus, due to absence of unpaired electrons, Zn2+ salts are colourless. [1] (ii) As copper has high energy of atomisation DaH° and low hydration energy DhydH°, due to which E° value is positive. [1]

Commonly Made Error  Students lose explanation.

time

in

giving



(B) on reaction with KCl forms a orange coloured crystalline compound (C). (i) Write the formulae of the compounds (A), (B) and (C). U (ii) Write one use of compound (C).

Ans. (i) 4FeCr2O4 + 16NaOH + 7O2 → 8Na2CrO4 [A] +

A&E [CBSE Comptt. OD Set-3 2017]





unnecessary

Answering Tip  Write the cause and consequence of the condition. Q. 8. Give reasons : (i) Zn is not regarded as a transition element. (ii) Cr2+ is a strong reducing agent. A&E [CBSE Comptt. Delhi 2016] Ans. (i) In both, Zn and Zn2+ ions absence of incompletely filled d-orbital therefore, Zn is not regarded as a transition element. [1] (ii) Cr2+ has d4 configuration while Cr3+ has more stable d3 (t23g ) configuration. Thus, Cr has a tendency to acquire Cr3+ due to greater stability of +3 oxidation state. Therefore, Cr2+ acts as a strong a reducing agent. [CBSE Marking Scheme, 2016] [1] Q. 9. Explain the following observation : (i) Silver atom has completely filled d-orbitals (4d10) in its ground state, yet it is regarded as a transition element. (ii) E° value for Mn3+/Mn2+ couple is much more positive than Cr3+/Cr2+. A&E [CBSE Comptt. OD Set-2 2017] Ans. (i) Silver can exhibit +2 oxidation state, wherein it will have incompletely filled d-orbital. [1] (ii) Much higher third ionisation energy of Mn where the required change is from d5 to d4. [1] [CBSE Marking Scheme 2017]

Q. 10. When chromite ore FeCr2O4 is fused with NaOH in presence of air, a yellow coloured compound (A) is obtained which on acidification with dilute sulphuric acid gives a compound (B). Compound



2Fe2O3 + 8H2O 2Na2CrO4 [A] + H2SO4 → Na2Cr2O7 [B] + Na2SO4 + H 2O Na2Cr2O7 [B] + 2KCl → K2Cr2O7 [C] + 2NaCl A : Na2CrO4 B : Na2Cr2O7 C : K2Cr2O7 [1] (ii) Use of K2Cr2O7 (C) : It is used as a strong oxidizing agent in industries. [1] Q. 11. Complete the following chemical equations : (i) KMnO4– + 3S2O32– + H2O ® ? (ii) Cr2O72– + 3Sn2+ + 14H+ ® ? R [CBSE Delhi 2016] – Ans. (i) MnO4– + 3S2O2– + H O → 2OH + 6SO2– 3 2 4 + MnO2 [1] 2– 2+ + 3+ (ii) Cr2O7 + 3Sn + 14H → 2Cr + 7H2O + 3Sn4+ [1] Q. 12. Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with oxalic acid ? Write the ionic equations for the reactions. Ans. Potassium permanganate is prepared by fusion of MnO2 with an alkali metal hydroxide and an oxidising agent like KNO3. This produces the dark green K2MnO4 which disproportionate in a neutral or acidic solution to give permanganate. 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O 3MnO42– + 4H+ → 2MnO4 + MnO2 + 2H2O [1] Oxalate ion or oxalic acid is oxidised at 333 K : 5C2O42– + 2MnO4– + 16H+ → 2Mn2+ + 8H2O + 10CO2 [1] [CBSE Marking Scheme 2015]

Commonly Made Error  Many students write unbalanced equations. Some students do not write the correct products.

Answering Tip  Write the balanced chemical equations involved in the preparation. Q. 13. Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with (i) an iodide (ii) H2S. R [CBSE Comptt. OD 2015] ns. (i) Cr2O72– + 6I– + 14H+ → 2Cr3+ + 3I2 + 7H2O [1] A (ii) Cr2O72– + 3H2S + 8H+ → 2Cr3+ +3S + 7H2O [1] [CBSE Marking Scheme 2015] Q. 14. When MnO2 is fused with KOH in presence of KNO3 as an oxidizing agent, it gives a dark green compound (A). Compound (A) disproportionate in acidic solution to give purple compound (B). An

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

alkaline solution of compound (B) oxidises KI to compound (C) whereas an acidified solution of compound (B) oxidises KI to (D). Identify (A), (B), (C) and (D).

compound (A). Compound (A) on acidification gives compound (B). Compound (B) on reaction with KCl forms an orange coloured compound (C). An acidified solution of compound (C) oxidises Na2SO3 to (D). Identify (A), (B), (C) and (D).

A&E [CBSE Delhi Set-I, 2019]



Ans. A = K2MnO4 / MnO24–, B = KMnO4 / MnO4– , C = KIO3 / IO3 , D = I2

Ans. A = Na2CrO4, B=Na2Cr2O7, C= K2Cr2O7, D= Na2SO4 [CBSE Marking Scheme, 2019] [½ × 4 = 2]

[CBSE Marking Scheme, 2019] [½ × 4]

Detailed Answer :

Detailed Answer :

4FeCr2O4 + 8Na2CO3 + 7O2 ¾¾ ® 8Na2CrO4

(A) K2MnO4 (green)

+ 2Fe2O3 + 8CO2



2 MnO2 + 4 KOH + O2 → 2 K 2 MnO4 + 2 H 2O Compound A



[A]

(B) KMnO4 (purple) 3MnO24− + 4 H+

→ 2 MnO− 4 + MnO2 Compound B



2Na2CrO4 + 2H+ ¾¾ ® Na2Cr2O7 + 2Na+ + H 2O [B]



Na2Cr2O7 + 2KCl ¾¾ ® K 2Cr2O7 + 2NaCl

+ 2 H 2O

(C) IO3

[C]

− − 2 MnO− 4 + H 2O + I → 2 MnO2 + 2OH +

IO− 3 Compound C

(D) I2 + 2+ 10 I− + 2 MnO− + 8 H 2O + 4 + 16 H → 2 Mn

K 2Cr2O7 + 3Na2SO3 + 4H 2SO4 ¾¾ ® 3Na2SO4 [D] + K 2SO4 + Cr2 (SO4 ) + 4H 2O

5I 2

Compound D

[2]

3



[A] Na2CrO4 : Sodium chromate, [B] Na2Cr2O7 : Sodium dichromate,



[C] K2Cr2O7 : Potassium dichromate, [D] Na2SO4 : Sodium sulphate. [2]

Q. 16. Write the balanced chemical equations involved in the preparation of KMnO4 from pyrolusite ore (MnO2).  → 2K 2 MnO4 + 2H 2O Ans. 2MnO2 + 4KOH + O2  3K 2 MnO4 + 2CO2 ¾¾ ® 2KMnO4 + MnO2 + 2K 2CO3 (ppt)





 [Topper’s Answer 2019] [2]

Q. 15. When FeCr2O4 is fused with Na2CO3 in the presence of air it gives a yellow solution of

Q. 17. Write the balanced ionic equations showing the oxidising action of acidified dichromate (Cr2O27). solution with(i) Iron (II) ion and (ii) tin (II) ion. A 2− + 2+ → 2Cr 3 + + 6Fe3 + + 7H 2O Ans. (i) Cr2O7 + 14H + 6Fe 

® 2Cr 3 + + 3Sn 4 + + 7H 2O (ii) Cr2O72 - + 3Sn 2 + + 14H + ¾¾

Short Answer Type Questions-II Q. 1. The magnetic moment of few transition metal ions are given below : Metal ion Magnetic moment (BM) Sc3+

0.00

Cr2+

4.90

Ni2+

2.84



3+ Ti

1.73



(Atomic no. Sc = 21, Ti = 22, Cr = 24, Ni = 28)



(3 marks each)

Which of the given metal ions : (i) has the maximum number of unpaired electrons? (ii) force colourless aqueous solution?

(iii) exhibits the most stable +3 oxidation state?

Ans. (i)Cr2+

[1]

(ii) Sc3+

[1]

(iii) Sc3+

[1]

Q. 2. Consider the standard electrode potential values (M2+/M) of the elements of the first transition series.

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‘d’ AND ‘f‘ BLOCK ELEMENTS



Ti

V

Cr

Mn

Fe

–1.63

–1.18

–0.90

–1.18

–0.44

Co

Ni

Cu

Zn

–0.28

–0.25

+0.34

–0.76

Explain :



(i) E° value for copper is positive. (ii) E° value of Mn is more negative as expected from the trend. (iii) Cr3+ is a stronger reducing agent than Fe2+. A [CBSE SQP 2017] Ans. (i) The high energy to transform Cu(s) to Cu2+(aq) is not balanced by its hydration enthalpy. [1]

(ii) Mn2+ has d5 configuration (stable half-filled configuration) [1] (ii) d5 to d3 occurs in case of Cr2+ to Cr3+.(More stable t32g) while it changes from d6 to d5 in case of Fe2+ to Fe3+. [1] [CBSE Marking Scheme 2017]

Q. 3. (i) Complete the following equations :

(a) 2MnO4– + 5SO32– + 6H+ ®

(b) Cr2O72– + 6Fe2+ + 14H+ ® (ii) Based on the data, arrange Fe2+, Mn2+ and Cr2+ in the increasing order of stability of +2 oxidation state. 3+

2+



E°Cr /Cr = – 0.4 V 3+ 2+ E°Mn /Mn = + 1.5 V



E°Fe

3+/Fe2+

R+U

= + 0.8 V –

2–

+

2+

Ans. (i) (a) 2MnO4 + 5SO3 + 6H ® 2Mn + 3H2O + 5SO42– [1] (b) Cr2O72– + 6Fe2+ + 14H+ ® 2Cr3+ + 6Fe3+ + 7H2O [1]

(ii) Cr2+< Fe2+ < Mn2+ Q. 4. Write the preparation of following :

[1]

(i) KMnO4 from K2MnO4

(ii) Na2CrO4 from FeCr2O4 (iii) Cr2O72– from CrO42– R [CBSE Comptt. Delhi/OD 2018] Ans. (i) 3MnO42– + 4H+ ® 2MnO4– + MnO2 + 2H2O [1] (or any other correct equation) (ii) 4FeCr2O4 + 8Na2CO3 + 7O2 ® 8Na2CrO4 + 2Fe2O3 + 8CO2 [1] (iii) 2CrO42– + 2H+ ® Cr2O72– + H2O

[1]

[CBSE Marking Scheme 2018] Q. 5. Account for the following : (i) CuCl2 is more stable than Cu2Cl2.

(ii) Atomic radii of 4d and 5d series elements are nearly same.

(iii) Hydrochloric acid is not used in permanganate titration.

A&E [CBSE Foreign Set-1, 2 2017]

Ans. (i) In CuCl2, Cu is in +2 oxidation state which is more stable due to high hydration enthalpy as compared to Cu2Cl2 in which Cu is in +1 oxidation state. [1] (ii) Due to lanthanoid contraction. [1] (ii) Because HCl is oxidised to chlorine. [1] [CBSE Marking Scheme 2017] Q. 6. (i) Give reasons for the following : (a) Compounds of transition elements are generally coloured. (b) MnO is basic while Mn2O7 is acidic. (ii) Calculate the magnetic moment of a divalent ion in aqueous medium if its atomic number is 26. A&E + A [CBSE Comptt. OD Set-1, 2, 3 2017]



Ans. (i) (a) Due to d-d transition. [1] (b) Due to higher oxidation state of Mn2O7 / Due to high polarizing power of Mn(VII). [1]

(ii) µ = 4( 4 + 2 ) = 4.90 B.M.  [1] [CBSE Marking Scheme 2017]

Q. 7. Give reasons : (i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4. (ii) Transition metals show variable oxidation states. (iii) Actinoids show irregularities in their electronic configurations. A&E [CBSE Delhi Set-1, 2016] Ans. (i) Mn can form pp -dp bond with oxygen by using 2p orbital of oxygen and 3d-orbital of Mn because of which it shows highest oxidation state of +7. With fluorine, Mn cannot form pp – dp bond thus shows the highest oxidation state of +4.  [1] (ii) Transition metal show variable oxidation state due to comparable energies of ns and (n – 1)d orbitals and partially filled d orbitals. So, both these orbitals take part in the reactions. [1] (iii) Due to comparable energies of 5f, 6d and 7s orbitals and the relative stabilities of f 0, f 7 and f14 occupancies of the 5f orbitals. [1]

Commonly Made Error  Students donot read the question properly or write long explanations to the questions asked.

Answering Tip  Be specific while writing reason. Avoid unnecessary explanations. Q. 8. (i) Account for the following : (a) Cu+ is unstable in an aqueous solution. (b) Transition metals form complex compounds. (ii) Complete the following equation :

Cr2O72- + 8H+ + 3NO2− → [CBSE OD 2015]

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



Ans. (i) (a) Because Cu+ undergoes disproportionation as 2Cu+ → Cu + Cu2+. 2+







(b) Because of small size of metal, high ionic charge and availability of vacant d-orbital. [1]

Hydration enthalpy of Cu is higher than that of Cu+ which compensates the I.E.2 of Cu involved in the formation of Cu2+ ions.  [1]



(ii) Cr2 O72− + 8H+ + 3NO–2 → 2Cr3+ + 3NO3− +



[CBSE Marking Scheme 2015]

4H2O (Balanced equation only)

[1]

Q. 9. Give reasons : (a) E° value for Mn3+/Mn2+ couple is much more positive than that for Fe3+/Fe2+.





(b) Iron has higher enthalpy of atomization than that of copper.







(c) Sc3+ is colourless in aqueous solution whereas Ti3+ is coloured.

A&E [CBSE Delhi/OD 2018]

2+

Ans. (a) Because Mn is more stable than Mn3+ due to half-filled d5 configuration whereas Fe2+ becomes unstable after loosing an electron from half-filled orbital. [1] (b) Due to presence of higher number of unpaired electrons in iron, they have stronger metallic bonding. Hence, the enthalpy of atomization is more of iron than that of copper. [1] (c) Sc3+ is colourless as it does not contain unpaired electrons to undergo d-d transition while Ti3+ is coloured as it contains unpaired electrons to undergo d-d transition by absorbing light from visible region and radiate complementary colour.  [1] [CBSE Marking Scheme 2018]

[ 229



‘d’ AND ‘f‘ BLOCK ELEMENTS

[Topper’s Answer 2018] Q. 10. A mixed oxide of iron and chromium is fused with sodium carbonate in free access of air to form a yellow coloured compound (A). On acidification the compound (A) forms an orange coloured compound (B), which is a strong oxidizing agent. Identify compound (A) and (B). Write chemical reactions involved.  A [CBSE Comptt. OD Set-1, 2, 3 2017] Ans. A : Na2CrO4 B : Na2Cr2O7

½+½

4FeCr2O4 + 8Na2CO3 + 7O2 → 8Na2CrO4 +



2Fe2O3 + 8CO2

+

1 +

2Na2CrO4 + 2H → Na2Cr2O7 + 2Na + H2O 1



[CBSE Marking Scheme 2017]

Commonly Made Error  Most of the students write either incorrect or incomplete equations. In many cases, the equations are unbalanced.

Answering Tip

(c) Mn can form multiple bonds with oxygen by using 2p orbital of oxygen and 3d orbital of Mn because of which it shows highest oxidation state of +7 with fluorine, Mn cannot form multiple bonds thus shows an oxidation state of +4. [3] Q. 12. Give reasons for the following : (i) Transition elements act as catalysts (ii) It is difficult to obtain oxidation state greater than two for Copper. (iii) Cr2O72– is a strong oxidising agent in acidic medium whereas WO3 and MoO3 are not. Ans. (i) Due to large surface area and ability to show variable oxidation states.[1] (ii) Due to high value of third ionisation enthalpy. [1] (iii) Mo(VI) and W(VI) are more stable than Cr(VI). [1] Q. 13. Observed and calculated values for the standard electrode potentials of elements from Ti to Zn in the first reactivity series are depicted in figure (1) :

 Write complete and balanced chemical equations. [3] Q. 11. Give reasons for the following : (a) Transition metals show variable oxidation states. (b) E° value for (Zn2+ /Zn) is negative while that of (Cu2+ /Cu) is positive. (c) Higher oxidation state of Mn with fluorine is +4 whereas with oxygen is +7. R & U [CBSE OD Set-2 2019]

Ans. (i) Because of comparable energies of (n – 1)d and ns orbitals / Incomplete filling of d-orbital. (ii) Because of stable 3d10 configuration of Zn2+ whereas due to low hydration enthalpy and high enthalpy of atomization of Cu2+. (iii) Due to the ability of oxygen to form multiple bonds with metal. [CBSE Marking Scheme, 2019] [1 × 3 = 3] Detailed Answer : (a) Because of availability of partially filled orbitals and comparable energies of ns and (n – 1)d orbitals. (b) E° value for (Zn2+/Zn) is negative due to stable completely filled d10 configuration in Zn2+. The positive value of (Cu2+/Cu) accounts for its ability to liberate H2 from acids due to its high enthalpy of atomization and low hydration energy.

FIGURE 1

Explain the following observations :

(i) The general trend towards less negative E° values across the series (ii) The unique behaviour of Copper (iii) More negative E° values of Mn and Zn  A&E [CBSE SQP 2021] Ans. (i) The general trend towards less negative E° V values across the series is related to the general increase in the sum of the first and second ionisation enthalpies.[1]

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



(ii) The high energy to transform Cu(s) to Cu2+ (aq) is not balanced by its hydration enthalpy.[1] (iii) The stability of the half-filled d sub-shell in Mn2+ and the completely filled d10 configuration in Zn2+ are related to their more negative E° V values[1] Q.14. Following ions are given : Cr2+, Cu2+, Cu+, Fe2+, Fe3+, Mn3+ Identify the ion which is (i) a strong reducing agent. (ii) unstable in aqueous solution. (iii) a strong oxidising agent. Give suitable reason in each.  A [CBSE OD Set-1 2020]

Ans. (i) Cr2+, because its configuration changes from d4 to d3 and having a half-filled t2g level. (ii) Cu+ in an aqueous medium energy is required to remove one electron from Cu+ to Cu2+, high hydration energy of Cu2+ compensates for it. Therefore Cu+ ion in an aqueous solution is unstable. 2Cu +  → Cu 2 + (aq) + Cu(s) (iii) Mn3+, because its configuration changes from Mn3+ to Mn2+ results in the half filled d5 configuration, which has extra stability.

Long Answer Type Questions





heat

(b) KMnO4  → [CBSE OD Set-1, 2, 3 2017]



3+





3

Ans. (i) (a) Cr , half filled t 3+

2g 5

[½+½]

(b) Mn , due to stable d configuration in Mn2+ [½+½] (c) Ti4+, No unpaired electrons [½+½] (ii) (a) 2MnO4- + 16H+ + 5S2- → 5S + 2Mn2+ + 8H2O [1] ∆ → K2MnO4 + MnO2 +O2 (b) 2KMnO4 

[1]

[CBSE Marking Scheme 2017]

Commonly Made Error  Students do not read and understand the question in a hurry.  Many students write unbalanced equations. Some students do not write the correct products.

Answering Tips  Read the question carefully. Do not forget to answer the sub-parts.  Write complete and balanced chemical equation. Q. 2. (i) Complete the following equations :

(a) Cr2O72– + 2OH– ® (b) MnO4– + 4H+ + 3e– ® (ii) Account for the following : (a) Zn is not considered as a transition element.

(b) Transition metals form a large number of complexes. (c) The E° value for the Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+ couple. R + A&E [CBSE OD 2014]



Ans. (i) (a) Cr2O72– + 2OH– → 2 CrO2− 4 + H2O

[1]







(b) MnO−4 + 4H+ + 3e– → MnO2 + 2H2O [1] (ii) (a) Because Zn/Zn2+ has fully filled d-orbitals. [1] (b) This is due to smaller ionic sizes, higher ionic charge and availability of d-orbitals.[1] (c) Because Mn2+ is more stable (3d5) than Mn3+(3d4), while Cr+3 is more stable due to t2g3/d3 configuration. [1] [CBSE Marking Scheme 2014] Q. 3. When chromite ore is fused with sodium carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (A) is obtained. On acidifying the yellow solution with sulphuric acid, compound (B) is crystallized out. When compound (B) is treated with KCl, orange crystals of compound (C) crystallize out. Identify (A), (B) and (C) and write the reactions involved. A [CBSE Comptt. OD Set-1, 2, 3 2017] Ans. A : Na2CrO4 B : Na2Cr2O7 C : K2Cr2O7 [½+½+1] 4FeCr2O4 + 8Na2CO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2[1] 2Na2CrO4 + 2H+ → Na2Cr2O7 + 2Na+ + H2O [1] Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl [1] [CBSE Marking Scheme 2017]



Q. 1. (i) Following are the transition metal ions of 3d series : Ti4+, V2+, Mn3+, Cr3+ (Atomic number : Ti = 22, V = 23, Mn = 25, Cr = 24) Answer the following : (a) Which ion is most stable in an aqueous solution and why? (b) Which ion is a strong oxidizing agent and why? (c) Which ion is colourless and why? (ii) Complete the following equations : (a) 2MnO4- + 16H+ + 5S2- ®

(5 marks each)

Commonly Made Error  Most of the students write either incorrect or incomplete equations. In many cases, the equations are unbalanced.

Answering Tip  Write complete and balanced chemical equations.

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‘d’ AND ‘f‘ BLOCK ELEMENTS

TOPIC-2

f-Block Elements : Lanthanoids and Actinoids Revision Notes

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 f-block elements : The elements in which filling of electrons takes place in (n–2) f-subshell which belongs to anti-penultimate (third to the outermost) energy shell. This block consists of two series of elements known as Lanthanoids and Actinoids. These elements are also known as inner transition elements. The general electronic configuration of the f - block elements is d and f Block (n–2)f1–14 (n–1)d0–1 ns2 elements For Lanthanoids, n is 6 while its value is 7 for Actinoids. There are many exceptions in the electronic configuration.  Lanthanoids : The series involving the filling of 4f-orbitals following lanthanum La (Z = 57) is called the lanthanoid series. There are 14 elements in this series starting with Ce (Z = 58) to Lu (Z = 71). l Electronic configuration : [Xe] 4f1-14 5d0-1 6s2 l Physical properties : (i) Highly dense metals, soft, malleable and ductile. (ii) High melting point. (iii) Forms alloys easily with other metals. (iv) Magnetic Properties : Among lanthanoids, La3+ and Lu3+ which have 4f 0 or 4f14 electronic configurations are diamagnetic and all other trivalent lanthanoid ions are paramagnetic due to presence of unpaired electrons. (v) Atomic and ionic sizes : With increasing atomic number, the atomic and ionic radii decreases from one element to the other but the decrease is very small. A steady decrease in the size of lanthanoids with increase in atomic number is known as lanthanoid contraction. Consequences of Lanthanoid contraction : (a) It leads to similar physical and chemical properties among lanthanoids. (b) Zr and Hf have same properties, due to similar atomic radii. (c) Chemical separation of lanthanoids become difficult. (vi) Oxidation state : They mainly give +3 oxidation state. Some elements show +2 and +4 oxidation states. (vii) Colour : Some of the trivalent ions are coloured. This happens due to the absorption in visible region of the spectrum resulting in f-f transitions. Chemical properties : All lanthanoids are highly electropositive metals and have almost similar chemical reactivity.

Ln



+C, 2773 K Ln C, LnC , Ln C 3 2 2 3 +O2 Ln2O3 +S Ln2S3 +N2 LnN +H2 O Ln(OH)3 + HX +X 2 LnX3 + acid H2

Uses :

(i) Misch metal is the alloy of cerium (about 55%) and various other Lanthanoid elements (40-43%). It also contains iron upto 5% and traces of sulphur, carbon, silicon calcium and aluminium. It is a pyrophoric material, hence it is used in lighter flints.

(ii) Lanthanoid oxides are used for polishing glass.

(iii) Cerium salts are used in dyeing cotton and also as catalysts. (iv) Lanthanoid compounds are used as catalyst for hydrogenated dehydrogenation and petroleum cracking. (v) Pyrophoric alloys are used for making tracer bullets and shells.  Actinoids : The series involving the filling of 5f orbitals from actinium, Ac (Z = 89) upto lawrencium, Lr (Z = 103) comprises of actinoids. l Electronic configuration : [Rn] 5f1-14 6d0-1 7s2

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

l Physical properties :

(i) Highly dense metals and form alloys with other metals.



(ii) Silvery white metals.



(iii) Highly electropositive.



(iv) High melting point.



(v) Ionic and atomic radii : The atomic and ionic size decreases with an increase in atomic number due to actinoid contraction. The electrons are added to 5f shell resulting in an increase in the nuclear charge causing the shell to shrink inwards. This is known as actinoid contraction.



(vi) Colour : Many actinoid ions are coloured.



(vii) Magnetic properties : Many actinoid ions are paramagnetic.

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(viii) Oxidation State : The common oxidation state exhibited is +3. They also exhibit oxidation state of +4, +5, +6 and +7. (ix) Many elements are radioactive. l Chemical reactivity : Less reactive towards acids. l Uses : (i) Thorium is used in the treatment of cancer and in incandescent gas mantles. (ii) Uranium is used in the glass industry, in medicines and as nuclear fuel. (iii) Plutonium is used in atomic reactors and in atomic bombs.

Difference between Lanthanoids and Actinoids

 Difference between Lanthanoids and Actinoids : S. No.

Lanthanoids

Actinoids

(i)

4f orbital is progressively filled.

(ii)

+3 oxidation state is most common along with +3 oxidation state is most common, but exhibit +2 and +4. higher oxidation state of +4, +5, +6, +7.

(iii)

Except promethium, all are non-radioactive.

All are radioactive.

(iv)

Less tendency of complex formation.

Strong tendency of complex formation.

(v)

Chemically less reactive than actinoids.

More reactive than lanthanoids.

5f orbital is progressively filled.

Mnemonics Lanthanides • Concept Name: Lanthanides Series • Mnemonic: Late CEO Promoted Nadia Palmer Smart to Europe so she said Goodbye to Toby a Day before Hoarding Eroding Timber in Yard of Lu. • Interpretation: La, Ce, Pr, Nd, Pm, Sm, Eu, Gd, Tb, Dy, Ho, Er, Tm, Yb, Lu Actinides • Concept Name: Actinide Series • Mnemonic : Active Thor Paid Ur Nephew for Pumpkins. • Amy Came Back from California. • Einstein and Fermi Made No. of Laws. • Interpretation: Ac, Th, Pa, U, Np, Pu, Am, Cm, Bk, Cf, Es, Fm, Md, No, Lr

Know the Terms  Transuranic elements : All the elements beyond uranium are known as transuranic or man-made elements. These elements do not occur in nature because their half-life periods are so short.

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‘d’ AND ‘f‘ BLOCK ELEMENTS



due to more effective shielding by inQ. (i) How would you account for the tervening 5f-electrons. following : (a) Actinoid contraction is greater STEP - II : (b) Transition elements genthan lanthanoid contraction. erally forms coloured compounds on (b) Transition metals form coloured account of d-d transition. When the compounds. visible light falls on the compounds, (ii) Complete the following equation : they absorb certain radiations and re– – + 2MnO4 + 6H + 5NO2 ® flect others. The colour observed corresponds to absorbed light. Solution: STEP - I : (i) (a) Actinoid contraction (ii) 2MnO4– + 6H+ + 5NO2– ® 5NO3– + 2Mn2+ + 3H2O. is greater than lanthanoid contraction

Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS : Q. 1. Which of the following statements is not correct? (a) La is actually transition element. (b) In Lanthanide series, Ionic radii decrease from La3+ to Lu3+. (c) La(OH)3 is less basic than Lu(OH)3 (d) Ionic radii of Zr and Hf are almost similar due to Lanthanoid contraction. U Ans. Correct option : (c) Explanation : La(OH)3 is more basic than Lu(OH)3. It is because of the fact that Due to lanthanoid contraction the size of lanthanoid ion decreases regularly with increase in atomic size. Thus covalent character between lanthanoid ion and OH− increases from La3+ to Lu3+. Thus the basic character of hydroxides decreases from La(OH)3 to Lu(OH)3. Q. 2. Lanthanoid contraction is caused due to (a) Atomic number (b) Size of 4f orbitals (c) Effective nuclear charge (d) Poor shielding effect of 4f electrons R Ans. Correct option : (d) Explanation : The lanthanoid contraction is due to poor shielding effect of 4f electrons. Q. 3. In which of the following elements, 5f orbitals are progressively filled? (a) Alkaline earth metals

(1 mark each) (b) Actinoids (c) Lanthanoids (d) Transition elements R Ans. Correct option : (b) Explanation : Actinoids are 5f block elements so in actinoids, 5f orbitals are progressively filled. [B] ASSERTION & REASON TYPE QUESTIONS : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

(a) Assertion and reason both are correct statements and reason is the correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not the correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Q. 1. Assertion (A) : Chromium is an actinoid. Reason (R) : In chromium, 3d orbitals are filled.  Ans. Correct option : (d)

U

Explanation : Chromium is a transition element and it belongs to 3d series because in chromium, 3d orbitals are filled.

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Q. 2. Assertion (A) : Chemistry of Actinoids is more complicated than Lanthanoids. Reason (R) : Actinoid elements are radioactive in nature. U Ans. Correct option : (a)

Explanation : Chemistry of actinoids is more complicated than Lanthanoids because actinoids are radioactive elements having relatively short half-lives.

Q. 3. Assertion (A) : Cerium (Ce) exhibits +4 oxidation state. Reason (R) : Ce4+ has 4f4 electronic configuration which is less stable. U Ans. Correct option : (c)

Explanation : Cerium exhibits +4 oxidation state because Ce4+ has 4f0 electronic configuration which is most stable.

[C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1. Why Lanthanoids cannot be easily separated? A Ans. Due to lanthanoid contraction, the chnge in the atomic or ionic radii of these elements is very small. So, their chemical properties are similar. Hence, Lanthanoids cannot be easily separated Q.2. Why does Zr (Z=40) and Hf (Z=72) exhibit almost identical atomic radii?  A&E Ans. Lanthanoid contraction is responsible for almost same atomic radii of 4d and 5d transition series elements. Q.3. Actinoid series members exhibit a large number of oxidation states compared to their corresponding lanthanoids. Why? A&E Ans. Due to comparable energies of 5f, 6d and 7s levels members of actinoid series exhibit a large number of oxidation states.

Short Answer Type Questions-I Q. 1. Write one similarity and one difference between the chemistry of lanthanoids and actinoids ? U [CBSE OD 2015] Ans. Similarity : (i) Both show contraction in size. (ii) Both show irregularity in their electronic configuration. (iii) Both are stable in +3 oxidation state. [1] Difference : (i) Actinoids are mainly radioactive but lanthanoids are not. (ii) Actinoids show wide range of oxidation states but lanthanoids do not. (iii) Actinoid contraction is greater than lanthanoid contraction. [1] (Write any one of these or any other one similarity and one difference)

Commonly Made Error  Students lose time in description for 1 mark questions.

Q. 2. Identify the following : (i) Oxo anion of chromium which is stable in acidic medium. (ii) The lanthanoid element that exhibits +4 oxidation state. 

U [CBSE SQP 2017] 2-

Ans. (i) Cr2O7 (ii) Cerium 

 Students write irrelevant content. Be specific. Read question carefully and write only what is asked.



(i) Transition metal of 3d series that exhibits the maximum number of oxidation states. (ii) An alloy consisting of approximately 95% lanthanoid metal used to produce bullet, shell and lighter flint. 

U [CBSE Comptt. Delhi/OD 2018]

Ans. (i) Mn (ii) Mischmetal

Short Answer Type Questions-II Q. 1. Account for the following : (i) Eu2+ is a strong reducing agent. (ii) Orange colour of dichromate ion changes to yellow in alkaline medium. (iii) E°(M2+/M) values for transition metals show irregular variation. A&E [CBSE Foreign Set-2 2017] 2+

Ans. (i) Eu is a strong reducing agent because Eu3+ is more stable than Eu2+.[1]

[1] [1]

Q. 3. Identify the following :



Answering Tip

(2 marks each)

[1] [1] [CBSE Marking Scheme 2018]

(3 marks each)

(ii) Dichromate ion changes to chromate ion/OH[1] Cr2O72- (orange) → CrO42- (yellow) (iii) Due to the irregular variation in ionisation enthalpies (sum of [1]st and 2nd ionisation enthalpies), heat of sublimation and enthalpy of hydration/due to irregular electronic configurations from left to right in a period, which changes the ionisation potential. [CBSE Marking Scheme 2017] [1]

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‘d’ AND ‘f‘ BLOCK ELEMENTS

Detailed Answer : (i) Electronic configuration of Eu2+ = 4f76s2. On oxidation, the evolution of the electrons takes place. Hence, after the removal of 2 electrons it achieves stable half filled electronic configuration acting as a strong reducing agent. 2− +  (ii) Cr2O72− + H 2O    2Cr2O4 + 2 H (Orange) (yellow) When an alkali is added to an orange solution of dichromate, a yellow solution is obtained due to the formation of chromate ions.



Q. 2. Explain the following : (a) Out of Sc3+, Co2+ and Cr3+ ions, only Sc3+ is colourless in aqueous solutions. (Atomic no. Co = 27; Sc = 21 and Cr = 24) 2+ (b) The E°Cu /Cu for copper metal is positive (+0.34), unlike the remaining members of the first transition series. (c) La(OH)3 is more basic than Lu(OH)3. A&E [CBSE SQP 2018-2019] Ans. (a) Co2+ : [Ar]3d7 Sc3+ : [Ar]3d0 Cr3+ : [Ar]3d3 Co2+ and Cr3+ have unpaired electrons. Thus, they are coloured in aqueous solution. Sc3+ has no unpaired electron. Thus it is colourless. [1] (b) Metal copper has high enthalpy of atomisation and enthalpy of ionisation. Therefore the high energy required to convert Cu(s) to Cu2+(aq) is not balanced by its hydration enthalpy. [1] (c) Due to lanthanoid contraction the size of lanthanoid ion decreases regularly with increase in atomic size. Thus covalent character between lanthanoid ion and OH– increases from La3+ to Lu3+. Thus the basic character of hydroxides decreases from La(OH)3 to Lu(OH)3. [1] [CBSE Marking Scheme 2018]

R [CBSE Delhi Set 2 2019]

Ans. (i) Due to comparable radii / comparable size.[1] (ii) In Mn2O3, Mn is in +3 (lower) oxidation state while in Mn2O7, Mn is in higher oxidation state (+7) [1]  (iii) Because its stable oxidation state is +3. [CBSE Marking Scheme, 2019] [1]

Q. 3. Give the reasons for following : (i) Transition elements and their compounds acts as catalysts. (ii) E° value for (Mn+2|Mn) is negative whereas for (Cu+2|Cu) is positive. (iii) Actinoids show irregularities in their electronic configuration. R [CBSE Delhi Set 1 2019] Ans. (i) Due to variable oxidation state. [1] (ii) Mn2+ is stable due to exactly half filled 3d5 configuration/ Due to high ΔaH0 and low ΔhydH0 for Cu2+/Cu is positive. [1] (iii) Due to comparable energies of 5f , 6d and 7s orbitals. [CBSE Marking Scheme, 2019] [1]

Detailed Answer : (i) Transition metals have the ability to adsorb many other substances on their surface and activate them as a result of chemisorption. Transition metals also exhibit a variety of oxidation states, and can change oxidation states relatively easily. This makes transition metals and their compounds good catalysts. (ii) Eo values are indicative of the stability of the oxidized form of the element. The lower the Eo value, more stable the oxidized form of the element. Mn2+ with a half filled d-subshell (d5) is stable, so Mn is easily oxidized to Mn2+, making the Eo value negative. Cu2+ with a partially filled d subshell (d9) is not stable, and is relatively easily reduced to element form. This makes its Eo value positive. (iii)  The electronic configurations of actinoids show irregularities because the energies of their 5f, 6d, and 7s orbitals are close to each other. Electrons can easily move between these subshells. Q. 4. Give reasons for the following : (i) Transition metals form alloys. (ii) Mn2O3 is basic whereas Mn2O7 is acidic. (iii) Eu2+ is a strong reducing agent.

Detailed Answer : (i) Transition metals easily form alloys with other transition metals because they have almost similar size. So they can easily replace each other in the crystal lattice. (ii) The transition metal oxides in the lower oxidation state of metals are basic in nature and in higher oxidation state they are acidic in nature. The oxidation state of Mn in Mn2O3 is +3 and Mn2O7 has +7. Therefore, Mn2O3 is basic and Mn2O7 is acidic. (iii) The common oxidation state of lanthanide metals is +3. Eu2+ is formed by losing the two s electrons acquires half filled (4f7) configuration. But still, they oxidize to their common +3 state. So the Eu2+ loses one electron and is oxidized to Eu3+. So, Eu2+ acts as a strong reducing agent.



OR

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[Topper’s Answer 2019] [3] Q. 5. Give reasons for the following : (a) Transition metals have high enthalpies of atomization. (b) Manganese has lower melting point even though it has a higher number of unpaired electrons for bonding. (c) Ce4+ is a strong oxidizing agent. R [CBSE OD Set 3 2019]

Ans. (i) Because of strong interatomic interactions / Strong metallic bonding between atoms. [1] (ii) Due to stable 3d5 configuration, interatomic interaction is poor between unpaired electrons. [1] (iii) Because Ce is more stable in +3 oxidation state.  [CBSE Marking Scheme, 2019] [1] Detailed Answer : (a) The transition elements have high enthalpies of atomization because they have large number of unpaired electrons in their atoms. This results in stronger interatomic interaction and stronger bonding between atoms. (b) Mn has low melting point because of 3d54s2 configuration, which is highly stable and

delocalised, so are not available for bonding, as a result interatomic forces becomes weaker. Hence Mn has low melting point bounded half-filled a orbital electrons with nucleus results in strong interatomic interaction. (c) The formation of Ce4+ is promoted by its noble gas configuration reverting to the common +3 state. E° value for Ce4+/Ce3+ is + 1.74 V thus readily gains an electron and acts as a strong oxidizing agent. Q. 6. (a) When a chromite ore (A) is fused with an aqueous solution of sodium carbonate in free excess of air, a yellow solution of compound (B) is obtained. This solution is filtered and acidified with sulphuric acid to form compound (C). Compound (C) on treatment with solution of KCl gives orange crystals of compound (D). Write the chemical formulae of compounds A to D. (b) Describe the cause of the following variations with respect to lanthanoids and actinoids : (i) Greater range of oxidation states of actinoids as compared to lanthanoids.  (ii) Greater actinoid contraction as compared to lanthanoid contraction. (iii) Lower ionisation enthalpy of early actinoids as compared to the early lanthanoids.

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‘d’ AND ‘f‘ BLOCK ELEMENTS

Ans. (a) A = FeCr2O4 B=Na2CrO4 C=Na2Cr2O7 D=K2Cr2O7 [½ × 4] (b) (i)  5f, 6d and 7s levels in actinoids are of comparable energies.  [1] (ii) This is due to poorer shielding by 5f electrons in actinoids as compared to shielding by 4f electrons in lanthanoids. [1] (iii) In actinoids, 5f electrons are more effectively shielded from the nuclear charge than the 4f electrons of the corresponding lanthanoids. Since the outer electrons are less firmly held, they are available for bonding in the actinoids.[1] Q. 7. (a) What happens when (i) Manganate ions (MnO2– 4 ) undergoes disproportionation reaction in acidic medium? (ii) Lanthanum is heated with sulphur? (b) Explain the following trends in the properties of the members of the First series of transition elements : (i) E0 (M2+/M) value for copper is positive (+0.34 V) in contrast to the other members of the series. (ii) Cr2+ is reducing while Mn3+ is oxidising, though both have d4 configuration. (iii) The oxidising power in the series increases in the

is required to transform Cu(s) to Cu2+(aq) which is not balanced by hydration enthalpy, therefore Eo(M2+ / M) value for copper is positive (+0.34 V). [1]



(iii) This is due to the increasing stability of the species of lower oxidation state to which they are reduced. [1] Q. 8. Give reasons for the following : (a) Transition metals form complex compounds. (b) Eo values for (Zn2+/Zn) and (Mn2+/Mn) are more negative than expected. (c) Actinoids show wide range of oxidation states.  R [CBSE OD Set 1 2019] Ans. (i)  Due to small size, high ionic charge and availability of d-orbital. [1] (ii) Due to stable 3d10 configuration in Zn2+ and 3d5 configuration in Mn2+. [1] (iii) D  ue to comparable energies of 5f, 6d and 7s orbitals / levels. [1] [CBSE Marking Scheme, 2019]

2− − order VO+ 2 < Cr2O7 < MnO 4 .

A [CBSE SQP 2020] 2− 4

Ans. (a) (i) MnO ions disproportionate in acidic medium to give permanganate ions and manganese(IV) oxide. [½] 3MnO24 − + 4H + → 2MnO−4 + MnO2 + 2H 2 O (ii) Lanthanum sulphide if formed.



heat 2La + 3S  → La 2 S 3 

[½] [½] [½]

(Deduct overall ½ mark if equation not balanced/  statements not written) (b) (i) Copper has high enthalpy of atomisation and low enthalpy of hydration, thus the high energy

(ii) Cr2+ is reducing as its configuration changes from d4 to d3, the latter having more stable half filled t2g level. On the other hand, the change from Mn3+ to Mn2+ results an extra stable d5 configuration. [1]



Detailed Answer : (a) Transition metals are able to form complex compounds due to the small size of metal, high ionic charge and availability of vacant d-orbital. (b) E° values of (Mn2+/Mn) and (Zn2+/Zn) are more negative than expected due to the greater stability of half filled d-subshell of Mn2+(3d5) and completely filled d-subshell of Zn2+(3d10). (c) Due to a very small energy gap between 5f, 6d and 7s subshells resulting in easier excitation of the outermost electrons to higher energy levels.

Long Answer Type Questions Q. 1. (i) Account for the following : (a) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4. (b) Zirconium and Hafnium exhibit similar properties.



(5 marks each)

(c) Transition metals act as a catalysts. (Atomic nos. : Mn = 25, Cr = 24) (ii) Complete the following equations : (a) 2MnO2 + 4KOH + O2 ∆ →

(b) Cr2O72− + 14H+ + 6I−  →



Ans.



Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII





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[Topper’s Answer 2016] [5]

Commonly Made Error  Students miss out on sub questions. Read and answer all sub questions carefully.

Answering Tips  Comprehend what is being asked before answering by reading the question carefully.  Don’t forget to answer further sub-parts of the questions. Q. 3. (i) Account for the following : (a)  Transition metals form large number of complex compounds. (b) The lowest oxide of transition metal is basic whereas the highest oxide is amphoteric or acidic. (c) E° value for the Mn3+/Mn2+ couple is highly positive (+1.57 V) as compared to Cr3+/Cr2+. (ii) Write one similarity and one difference between the chemistry of lanthanoid and actinoid elements.

U + A&E

Ans. (i) (a) Due to small size and high ionic charge/ availability of d orbitals. [1] (b) Higher is the oxidation state higher is the acidic character/as the oxidation state of a metal increases, ionic character decreases [1]

Q. 2. The elements of 3d transition series are given as : Sc Ti V Cr Mn Fe Co Ni Cu Zn Answer the following : (i) Write the element which is not regarded as a transition element. Give reason. (ii) Which element has the highest m.p? (iii) Write the element which can show an oxidation state of +1. (iv) Which element is a strong oxidizing agent in +3 oxidation state and why?  A [CBSE OD Set-2 2016] Ans. (i) Zn , because it does not have partially filled d-orbital in its ground state or ionic state. [1½] (ii) Cr has the highest melting point. As the number of unpaired electrons increases upto d5 configuration, it results in the increase in the strength of metallic bonds. To break the metallic bond, significant energy is required thus Cr with highest number of unpaired electrons i.e., 6 has the highest melting point. [1] (iii) Cu can show +1 oxidation state as it can loose one electron present in 4s orbital. [1] (iv) Mn is a strong oxidising agent in +3 oxidation state because change of Mn3+ to Mn2+ give stable half filled (d5) configuration.  [1½]

[ 239

‘d’ AND ‘f‘ BLOCK ELEMENTS

(c) Because Mn2+ has d5 as a stable, configuration whereas Cr3+ is more stable due to stable t32g [1] (ii) Similarity-both are stable in +3 oxidation state/ both show contradiction/irregular electronic configuration (or any other suitable similarity) [1] Difference–actinoids are radioactive and lanthanoids are not/actinoids show wide range of oxidation states but lanthanoids don’t (or any other correct difference) [1]

Q. 4. (i) (a) How is the variability in oxidation states of transition metals different from that of the p-block elements? (b) Out of Cu+ and Cu2+, which ion is unstable in aqueous solution and why? (c)  Orange colour of Cr2O72- ion changes to yellow when treated with an alkali. Why? (ii) Chemistry of actinoids is complicated as compared to lanthanoids. Give two reasons. A&E [CBSE Delhi Set-1, 2, 3 2017] Ans.





Ans. (i) (a) In p-block elements the difference in oxidation state is 2 and in transition metals the difference is 1. [1] (b) Cu+, due to disproportionation reaction and low hydration enthalpy [½+½]

(c) Due to formation of chromate ion/CrO42- ion, which is yellow in colour [1]



(ii) Actinoids are radioactive, actinoids show wide range of oxidation states [1+1] [CBSE Marking Scheme 2017] Q. 3. (i) Account for the following :



(a)  Transition metals show variable oxidation states.



(b) Zn, Cd and Hg are soft metals.



(c) E° value for the Mn3+/Mn2+ couple is highly positive (+1.57 V) as compare to Cr3+/Cr2+.

(ii) Write one similarity and one difference between the chemistry of lanthanoid and actinoid elements.

[CBSE OD Set-1, 2, 3 2017]

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

[Topper’s Answer 2017] [5] Cd and Hg are soft and have low Detailed Answer : (iii) Zn, melting point. (i) (a) The valence electrons of transition metals are (b) Write the preparation of the following : in (n-1)d and ns orbitals. As there is almost little energy difference between these orbitals, both the (i) Na2Cr2O7 from Na2CrO4 energy levels can be used for bond formation. Thus, (ii) K2MnO4 from MnO2 they exhibit variable oxidation states. (a) (i) The catalytic activities of transition metals Ans. (b) Because they contain fully filled d-orbitals, no and their compounds is due to the ability unpaired d electrons are present resulting in weak of adopt variable oxidation states and to metallic bonding. form complexes. It can also provide a large surface area for the reactants to be adsorbed. Q. 6. (i) (a) Which transition element in 3d series has 0 (Any one) positive E M value and why? /M (ii) Separation of lanthanoid elements is (b) Name a member of lanthanoid series which is difficult because all lanthanoid elements well known to exhibit +4 oxidation state and have almost similar physical as well as why? chemical properties. Due to the lanthanoid (ii) Account for the following contraction the change in the atomic or (a) The highest oxidation state is exhibited in ionic radii is very small. oxoanions of transition metals. (b) HCl is not used to acidify KMnO4 solution. (iii) Zn, Cd and Mg are soft and have low (c)  Transition metals have high enthalpy of melting point because no d-orbitals are atomisation. available for metallic bond formation and A&E [CBSE Comptt. Delhi Set-1, 2, 3 2017] bonds formed are very weak.  2+

Ans. (i) (a) Copper; Due to high DgH– and low DhydH–

[½+½] (b) Cerium; Due to stable 4f 0 configuration/Tb; Due to stable 4f 7 configuration [½+½] (ii) (a) Due to ability of oxygen to form multiple bonds to metal [1]



(b) HCl is oxidized to chlorine (c) Due to strong interatomic bonding

[1] [1]

[CBSE Marking Scheme 2017] Detailed Answer : (ii) (a) Due to high electronegativity and small size, oxygen acts as a strong oxidising agent. This results in oxygen’s ability to oxidise the metal to attain highest oxidation state. (b) As KMnO4 is a very strong oxidising agent, it oxidizes HCl resulting in evolution of chlorine gas. Therefore, HCl is not used to acidify KMnO4 solution. Q. 7. (a) Give reasons : (i) Transition metals and their compounds show catalytic activities. (ii) Separation of a mixture of Lanthanoid elements is difficult.



(b) (i) 2Na2CrO4 + H 2SO4  → Na2Cr2O7 + Na 2SO4 + H 2O

(ii) 2MnO2 + 4KOH + O2  → 2K 2 MnO4 + 2H 2O

Q. 8. (a) Account for the following : (i) Ti3+ is coloured whereas Sc3+ is colourless in aqueous solution. (ii) Cr2+ is a strong reducing agent. (b) Write two similarities between chemistry of lanthanoids and actinoids. (c) Complete the following ionic equation : 3MnO42 + 4H+ →  U [CBSE Delhi Set-1 2020] Ans. (a) (i) Ti3+ has incomplete d (3d1) orbital whereas Sc3+ has empty (3d°) d-orbital. (ii) Cr2+ ion can lose electron to form Cr3+, so acts as a strong reducing agent. (b) Similarities between chemistry of lanthanoids and actinoids : (i) Both show +3 oxidation state. (ii) Both are strong reducing agents.

(c) 3MnO24 − + 4H +  → 2MnO4− + MnO2 + 2H 2O

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‘d’ AND ‘f‘ BLOCK ELEMENTS

Visual Case based Questions Q.1. Read the passage given below and answer the following questions : Within the 3d series, manganese exhibits oxidation states in aqueous solution from +2 to +7, ranging from Mn2+(aq) to MnO−4 (aq). Likewise, iron forms both Fe 2+(aq) and Fe 3+(aq) as well as the FeO 2−4 ion. Cr and Mn form oxyions CrO2−4, MnO −4, owing to their willingness to form multiple bonds. The pattern with the early transition metals—in the 3d series up to Mn, and for the 4d, 5d metals up to Ru and Os—is that the maximum oxidation state corresponds to the number of ‘‘outer shell’’ electrons. The highest oxidation states of the 3d metals may depend upon complex formation (e.g., the stabilization of Co 3+ by ammonia) or upon the pH (thus MnO42− (aq) is prone to disproportionation in acidic solution). Within the 3d series, there is considerable variation in relative stability of oxidation states, sometimes on moving from one metal to a neighbour; thus, for iron, Fe3+ is more stable than Fe2+, especially in alkaline conditions, while the reverse is true for cobalt. The ability of transition metals to exhibit a wide range of oxidation states is marked with metals such as vanadium, where the standard potentials can be rather small, making a switch between states relatively easy. (CBSE QB 2021)  (Cotton, S. A. (2011). Lanthanides: Comparison to 3d metals. Encyclopedia of inorganic and Bio­ inorganic Chemistry.) In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices on the basis of the above passage. (a)  Assertion and reason both are correct statements and reason is correct explanation for assertion. (b)  Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c)  Assertion is correct statement but reason is wrong statement. (d)  Assertion is wrong statement but reason is correct statement. (i)  Assertion: Highest oxidation state is exhibited by transition metal lying in the middle of the series. Reason: The highest oxidation state exhibited corresponds to number of (n−1)d electrons. (ii)  Assertion: Fe3+ is more stable than Fe2+ Reason: Fe3+ has 3d5 configuration while Fe2+ has 3d6 configuration. (iii)  Assertion: Vanadium had the ability to exhibit a wide range of oxidation states. Reason: The standard potentials Vanadium are rather small, making a switch between oxidation states relatively easy. (iv) Assertion: Transition metals like Fe, Cr and Mn form oxyions



(4 marks each)

Reason: Oxygen is highly electronegative and has a tendency to form multiple bonds. (v)  Assertion: The highest oxidation states of the 3d metals depends only on electronic configuration of the metal. Reason: The number of electrons in the (n-1)d and ns subshells determine the oxidation states exhibited by the metal. Ans. (i) Correct option : (c) (ii) Correct option : (a) (iii) Correct option : (a) (iv) Correct option : (b) (v) Correct option : (d) Q.2. Read the passage given below and answer the following questions : The transition metals when exposed to oxygen at low and intermediate temperatures form thin, protective oxide films of up to some thousands of Angstroms in thickness. Transition metal oxides lie between the extremes of ionic and covalent binary compounds formed by elements from the left or right side of the periodic table. They range from metallic to semiconducting and deviate by both large and small degrees from stoichiometry. Since d-electron bonding levels are involved, the cations-exist in various valence states and hence give rise to a large number of oxides. The crystal structures are often classified by considering a cubic or hexagonal close-packed lattice of one set of ions with the other set of ions filling the octahedral or tetrahedral interstices. The actual oxide structures, however, generally show departures from such regular arrays due in part to distortions caused by packing of ions of different size and to ligand field effects. These distortions depend not only on the number of d-electrons but also on the valence and the position of the transition metal in a period or group. (source: Smeltzer, W. W., & Young, D. J. (1975). Oxidation properties of transition metals. Progress in Solid State Chemistry, 10, 17-54.) (CBSE QB 2021) In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices on the basis of the above passage. (a)  Assertion and reason both are correct statements and reason is correct explanation for assertion. (b)  Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c)  Assertion is correct statement but reason is wrong statement. (d)  Assertion is wrong statement but reason is correct statement. (i) Assertion: Cations of transition elements occur in various valence states Reason: Large number of oxides of transition elements are possible. (ii) Assertion: Crystal structure of oxides of transition metals often show defects.

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Reason: Ligand field effect cause distortions in crystal structures. (iii) Assertion : Transition metals form protective oxide films. Reason: Oxides of transition metals are always stoichiometric. (iv) Assertion: CrO crystallises in a hexagonal closepacked array of oxide ions with two out of every three octahedral holes occupied by chromium ions. Reason: Transition metal oxide may be hexagonal close-packed lattice of oxide ions with metal ions filling the octahedral voids. Ans. (i) Correct option : (b) (ii) Correct option : (a) (iii) Correct option : (c) (iv) Correct option : (d) Q.3. Read the passage given below and answer the following questions : The d block elements are the 40 elements contained in the four rows of ten columns (3-12) in the periodic table. As all the d block elements are metallic, the term d-block metals is synonymous. This set of d-block elements is also often identified as the transition metals, but sometimes the group 12 elements (zinc, cadmium, mercury) are excluded from the transition metals as the transition elements are defined as those with partly filled d or f shells in their compounds. Inclusion of the elements zinc, cadmium and mercury is necessary as some properties of the group 12 elements are appropriate logically to include with a discussion of transition metal chemistry. The term transition element or transition metal appeared to derive from early studies of periodicity such as the Mendeleev periodic table of the elements. His horizontal table of the elements was an attempt to group the elements together so that the chemistry of elements might be explained and predicted. In this table there are eight groups labeled I-VIII with each subdivided into A and B subgroups. Mendeleev recognized that certain properties of elements in Group VIII are related to those of some of the elements in Group VII and those at the start of the next row Group I. In that sense, these elements might be described as possessing properties transitional from one row of the table to the next. (source: Winter, M. J. (2015). D-block Chemistry (Vol. 27). Oxford University Press, USA.) In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices on the basis of the above passage. (a)  Assertion and reason both are correct statements and reason is correct explanation for assertion. (b)  Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c)  Assertion is correct statement but reason is wrong statement. (d)  Assertion is wrong statement but reason is correct statement. (CBSE QB 2021) (i) Assertion: Group 12 elements are not considered as transition metals. Reason: Transition metals are those which have incompletely filled d shell in their compounds.

(ii) Assertion: All d block elements are metallic in nature. Reason: The d –block elements belong to Group3 -12 of the periodic table. (iii) Assertion : Group VII elements of Mendeleev periodic table are transition elements. Reason: Group I –VIII in Mendeleev periodic table is divided into two subgroups, A and B. (iv) Assertion: Nickel is a transition element that belongs to group 10 and period 4 of the modern periodic table. Reason: Electronic configuration of Nickel is [Ar]183d84s2 Ans. (i) Correct option : (a) (ii) Correct option : (b) (iii) Correct option : (d) (iv) Correct option : (a) Q.4. Read the passage given below and answer the following questions :  (1 × 4 = 4) In transition elements, generally, ions of the same charge in a given series show progressive decrease in radius with increasing atomic number. This is because the new electron enters a d orbital each time the nuclear charge increases by unity. But the radii of the third (5d) series are virtually the same as those of the corresponding members of the second series. This phenomenon is associated with the intervention of the 4f orbitals which must be filled before the 5d series of elements begin. The filling of 4f before 5d orbital results in a regular decrease in atomic radii called Lanthanoid contraction. In these questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a)  Assertion and reason both are correct statements and reason is correct explanation for assertion. (b)  Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c)  Assertion is correct statement but reason is wrong statement. (d)  Assertion is wrong statement but reason is correct statement. Following are the transition metal ions of 3d series : Ti4+, V2+, Mn3+, Cr3+ (Atomic number : Ti = 22, V = 23, Mn = 25, Cr = 24) (i) Assertion : Among the given ions, Cr3+ is the most stable in an aqueous environment. Reason : Cr3+ has half filled t32g. (ii) Assertion : Among the given ions, Mn3+ is the most strong oxidizing agent. Reason : Mn3+ has an unstable configuration. (iii) Assertion : Ti4+ ion is colourless.  Reason : All valence electrons are unpaired in Ti4+ ion. (iv) Assertion : Atomic radii of third series of transition elements is similar to that of the second series.  Reason : According to Aufbau’s principle, 4f orbitals needs to be filled before the 5d series begin. Ans. (i) Correct option : (d) Cr3+, half filled t32g[½+½]

‘d’ AND ‘f‘ BLOCK ELEMENTS



(ii) Correct option : (c) Mn3+ is the strong oxidising agent because it has 4 electrons in its valence shell and when it gains one electron than it forms Mn2+, it results in the half-filled (d5) configuration that provides extra stability. [½+½] (iii) Correct option : (a) Ti4+, No unpaired electrons [½+½] (iv) Correct option : (b) Lanthanoid contraction [1] Q.5. Read the passage given below and answer the following questions :  (1 × 4 = 4) Although actinoids are similar to lanthanoids in that their electrons fill the 5f orbitals in order, their chemical properties are not uniform and each element has characteristic properties. Promotion of 5f - 6d electrons does not require a large amount of energy and examples of compounds with ππacid ligands are known in which all the 5f, 6d, 7s, and 7p orbitals participate in bonding. Trivalent compounds are the most common, but other oxidation states are not uncommon. Especially thorium, protactinium, uranium, and neptunium tend to assume the +4 or higher oxidation state. The following questions are multiple choice questions. Choose the most appropriate answer : (i) Which of the following oxidation state is common for all lanthanoids? (a) +2 (b) +3 (c) +4 (d) +5 (ii) There are 14 elements in actinoid series. Which of the following element does not belong to this series? (a) U (b) Np (c) Tm (d) Fm (iii) General electronic configuration of actinoids is (n – 2)f1–14 (n – 1)d0–2 ns2. Which of the following actinoids have one electron in 6d orbital? (a) U (Atomic no. 92) (b) Np (Atomic no.93) (c) Pu (Atomic no. 94) (d) Bk (Atomic no. 97) (iv) Gadolinium belongs to 4f series. Its atomic number is 64. Which of the following is the correct electronic configuration of gadolinium? (a) [Xe] 4f 75d16s2 (b) [Xe] 4f 65d26s2 8 2 (c) [Xe] 4f 6d (d) [Xe] 4f 95s1 Ans. (i) Correct option : (b) Explanation : All of the lanthanide elements are commonly known to have the +3 oxidation state. [1] (ii) Correct option : (c) Explanation : Tm is Thulium which belongs to Lanthanoids. Uranium(U), Neptunium(Np), Fermium(Fm) belong to Actinoid series. [1] (iii) Correct option : (a) Explanation : Uranium has an electronic configuration of 5f3 6d1 7s2. [1] (iv) Correct option : (a) Explanation : Gadolinium has an electronic configuration of [Xe] 4f 75d16s2 [1] Q.6. Read the passage given below and answer the following questions :  (1 × 4 = 4) Buffered aqueous solutions of potassium

[ 243 permanganate and sodium or potassium dichromate are used to remove completely traces of hydrogen sulfide from industrial gases. Processes employing such solutions are nonregenerative and, because of the high cost of the chemicals used, are only economical when very small amounts of hydrogen sulfide are present in the gas. Permanganate solutions are used quite extensively for the final purification of carbon dioxide in the manufacture of dry ice. The solution, which in the case of the permanganate process contains about 4.0% potassium permanganate and 1.0% sodium carbonate, is circulated until approximately 75% of the permanganate in either tower is converted to manganese dioxide. The following questions are multiple choice questions. Choose the most appropriate answer : (i) KMnO4 acts as an oxidizing agent in the acidic medium. The number of moles of KMnO4 that will be needed to react with one mole of sulphide ion in acidic solution is : (a) 2/5 (b) 3/5 (c) 4/5 (d) 1/5 (ii) KMnO4 acts as an oxidising agent in alkaline medium. When alkaline KMnO4 is treated with KI, iodide ion is oxidised to : (a) I2 (b) IO– – (c) IO3 (d) IO4– (iii) Generally, transition elements and their salts are coloured due to the presence of unpaired electrons in metal ions. Which of the following compounds are coloured? (a) KMnO4 (b) TiCl4 (c) Cu2Cl2 (d) All of the above (iv) Why is HCl not used to make the medium acidic in oxidation reactions of KMnO4 in acidic medium? (a) Both HCl and KMnO4 act as oxidising agents. (b) KMnO4 oxidises HCl into Cl2 which is also an oxidising agent. (c) KMnO4 is a weaker oxidising agent than HCl. (d) KMnO4 acts as a reducing agent in the presence of HCl. Ans. (i) Correct option : (a) Explanation : 2MnO–4 + 5S2– + 16H+ → 2Mn2+  + 5S + 8H2O  For 5 moles of S the number of moles of KMnO4 = 2  For 1 mole of S the number of moles of KMnO4 = 2/5 [1] (ii) Correct option : (c) Explanation : 2KMnO4 + KI + H2O → 2KOH + 2MnO2 + KIO3 [1] (iii) Correct option : (a) Explanation : KMnO4 is dark purple crystalline compound. [1] (iv) Correct option : (a) Explanation : KMnO4 oxidises HCl into Cl2 which is also an oxidising agent. Hence HCl is not used in the oxidation reactions of KMnO4 to make medium acidic. [1] 

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Self Assessment Test - 8 Time : 1 Hour

Max. Marks : 25

1. 

Read the passage given below and answer the following questions : (1 × 4 = 4)



Trends in atomic radii of transition metals.



2. 

The Lanthanide ions that have unpaired electrons are coloured and are paramagnetic. In several aspects, the magnetic and spectral behaviour of lanthanides is fundamentally different from that of the difference lies in the fact that the electrons responsible for the magnetic and spectral properties of lanthanide ions are 4f electrons , and the 4f orbitals are very effectively shielded from interaction with external forces by the over lying 6s2 and 6p6 shells. Hence, there are essentially only very weak effects of ligand fields. Hence the states arising from the various 4fn configurations are only slightly affected by the surroundings of the ions and remain practically invariant for a given ion in all of its compounds.



Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not the correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. (i) Assertion : Cerium also shows +4 oxidation state. Reason : Characteristic oxidation state of lanthanoids is +3. (ii) Assertion : The lanthanoid contraction leads to overall decrease of atomic and ionic radii from lanthanum to lutetium. Reason : The filling of 4f before 5d orbitals result in decrease of atomic radii. (iii) Assertion : The magnetic moment of Cr3+ is 3.87 B.M. and that of Co2+ is 4.87 B.M. Reason : Cr3+ and Co2+ ions have same number of unpaired electrons. (iv) Assertion : Ionisation enthalpies of Ce, Pr and Nd are higher than Th, Pa and U.

(i) Metallic radii of some transition elements are given below. Which of these elements will have highest density? Element Metallic radii/pm



Fe 126

Co 125

Ni 125

Cu 128



(a) Fe (b) Ni (c) Co (d) Cu (ii) Although Zirconium belongs to 4d transition series and Hafnium to 5d transition series even then they show similar physical and chemical properties because : (a) both belong to d-block. (b) both have same number of electrons. (c) both have similar atomic radius.



(d) both belong to the same group of the periodic table.



(iii)  The second and third rows of transition elements resemble each other much more than they resemble the first row.



(a) The statement is true.



(b) Initial elements of second and third rows resemble each other.



(c) Initial elements of first and second rows resemble each other.



(d) The statement is false. (iv)  Which of the following statements is not correct? (a) Copper liberates hydrogen from acids.

solution. Read the passage given below and answer the following questions :  (1 × 4 = 4)



The following questions are multiple choice questions. Choose the most appropriate answer.

(b)  In its higher oxidation states, manganese forms stable compounds with oxygen and fluorine. (c) Mn3+ and Co3+ are oxidising agents in aqueous solution. (d) Ti2+ and Cr2+ are reducing agents in aqueous







[ 245

SELF-ASSESSMENT TEST



Reason : Outer electrons are tightly held in actinoids as compared to lanthanoids.

Q.3. Transition elements form binary compounds with halogens. Which of the following elements will form MF3 type compounds? (i) Co (ii) Cu (iii) Ni In the following question (Q. No. 4) a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Q.4. Assertion : Actinoids form relatively less stable complexes as compared to lanthanoids.  Reason : Actinoids can utilise their 5f orbitals along with 6d orbitals in bonding but lanthanoids do not use their 4f orbitals for bonding. Q.5. Give reasons for the following : (i)  Transition metals have high enthalpies of atomization. (ii)  Manganese has lower melting point even though it has a higher number of unpaired electrons for bonding. Q.6. Complete and balance the following chemical equations :

(a) SO32– + MnO4– + H+ → Mn2+ + H2O + SO2– 4



(b) MnO4– + H2O + I– ®

Q.7. Give reason for the following :

(a) Compounds of transition elements are generally coloured.



(b) MnO is basic while Mn2O7 is acidic.



(c) Many of the transition elements are known to form interstitial compounds.

Q.8. The sum of first and second ionization energies and sum of third and fourth ionization energies of Nickel and Platinum are given as below: IE1 + IE2 (kJ mol−1) IE3 + IE4 (kJ mol−1)



Ni

2.49

8.80

Pt

2.66

6.70

By considering these values, state the following : (i) Most stable oxidation states of Ni and Pt and its cause. (ii)  Name the metal which easily forms com­ pounds in +4 oxidation state and why? (iii) Cr3+ is a stronger reducing agent than Fe2+.



Q.9. (i) Account for the following : (a) Transition metals show variable oxidation states. (b) Zn, Cd and Hg are soft metals. (c) E° value for the Mn3+/Mn2+ couple is highly positive (+1.57 V) as compared to Cr3+/Cr2+.





(ii) (a) Complete the following equation : MnO4– + 8H+ + 5e– →

(b) Out of Mn3+ and Cr3+, which is more paramagnetic and why ?

(Atomic nos. : Mn = 25, Cr = 24)

 

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

9

CHAPTER

COORDINATION COMPOUNDS

Syllabus 

Coordination compounds - Introduction, ligands, coordination number, colour, magnetic properties and shapes, IUPAC nomenclature of mononuclear coordination compounds. Bonding, Werner’s theory, VBT, and CFT; structure and stereoisomerism, importance of coordination compounds (in qualitative analysis, extraction of metals and biological system).

Trend Analysis 2018 D/OD

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Hybridization, Magnetic Character, Spin, Number of Unpaired Electrons and Isomers of Complex Write the formula for IUPAC name Properties

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Coordination Compounds : Properties and IUPAC Name Revision Notes  Coordination Compound : A coordination compound contains a central metal atom or ion surrounded by number of oppositely charged ions or neutral molecules. There is a coordinate bond between metal atom and these ions or molecules, e.g., [Cu(NH3)4]2+.  Double Salt : When two or more salts are added to form a stable solid together and break into constituent ions when dissolved in water or any solvent, e.g., FeSO4(NH4)2SO4.6H2O (Mohr’s salt).  Properties of double salts : (i) They give simple ions in aqueous solution because they are ionic compounds.

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Coordination compounds

TOPIC - 1 Coordination Compounds : Properties and IUPAC Name .... P. 247 TOPIC - 2 Werner's Theory, Bonding, VBT, CFT .... P. 259

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII





(ii) They do not contain coordinate bonds. (iii) They exist only in solid state as double salt. (iv) They are soluble in water.  Coordinate bonds : A type of covalent bond in which one of the atoms supplies both the electrons. It can be considered as a combination of transfer and sharing of electrons. Coordinate bonds are also called semipolar bonds.  Central metal atom or ion : The metal atom or ion surrounded by fixed number of ions or molecules is called central metal atom or ion, e.g., in K4[Fe (CN)6], Fe2+ is central metal ion.  Ligand : The neutral molecules or ions (usually anions) which are attached with the central metal atom or ion in complex compound. e.g., Cl–, OH–, CN–, CO, NH3, H2O, etc. A ligand may be neutral or charged species. It is always act as a Lewis base. l Types of ligands : (a) On the basis of number of donor sites :

..

..



(i) Unidentate ligands : Contain one donor atom. e.g., N H 3 ,H 2 O:



(ii) Bidentate ligands : Contain two donor atoms. e.g., (COO–)2, CH 2 — N H 2 | .. CH 2 — N H 2

(iii) Polydentate ligands : Contain several donor atoms. e.g., EDTA. (b) On the basis of charge : (i) Cationic ligands : Carry positive charge. e.g., NO2+, N2H5+. (ii) Anionic ligands : Carry negative charge. e.g., X–(halo), CN–(cyano).

..

..

..

(iii) Neutral ligands : Do not carry any charge. e.g., N H 3 (ammine), H 2 O : (aqua).

(c) On the basis of nature of ligand: (i) Chelate ligands : A bidentate or polydentate uses its two or more donor atoms to bind a single metal ion, then a ring like structure is obtained. It is called chelate and the ligand is known as chelate ligand. .. e.g., CH — N H 2

2

| .. CH 2 — N H 2



M

(ii) Ambidentate ligand : A ligand which contains two donor atoms but only one of them forms a coordinate bond at a time with central metal/ion is called ambidentate ligand. e.g., M

O N

M

SCN

M

NCS

O Nitrito— N (N donor atom)

Thiocyanato (S donor atom)

Isothiocynate (N donor atom)

 Coordination number : Number of monodentate ligands attached to central metal ion in a complex is called coordination number. It may also be defined as total number of chemical bonds formed between central metal ion and donor atom of ligand e.g., in [Ni(NH3)6]2+, the coordination number of Ni is 6.  Coordination polyhedron : The spatial arrangement of the ligand atoms which are directly attached to the central atoms or ions define a coordination polyhedron about the central atom e.g., [Pt Cl4]2– is square planar.  Charge on the complex ion : The charge on the complex ion is equal to the algebric sum of the charges on all the ligands coordinated to the central metal ion.  Donor atom : An atom in the Lewis base that forms the bond with the central atom/ion is called donor atom because it donates the pair of electrons.  Denticity : The number of ligating groups or coordinating atoms in a ligand is called denticity e.g., unidentate, bidentate, etc.  Applications of chelates : (i) In the softening of hard water. (ii) In the separation of lanthanoids and actinoids. (iii) In the detection as well as estimation of some metal ions such as nickel (II) ion.  Coordination sphere : The central atom/ion and the ligands attached to it are enclosed in square bracket and is collectively termed as coordination sphere e.g., in the complex K4[Fe(CN)6], the coordination sphere is [Fe(CN)6]4–.  Flexidentate character of ligands : Certain polydentate ligands have flexible character and are called flexidentates. e.g., EDTA is hexadentate in nature but in some cases, it may act as pentadentate or tetradentate ligand.  Oxidation number of central atom : It is defined as the charge it would carry if all the ligands are removed along with the electron pairs that are shared with the central atom. It is represented by Roman numerical.

[ 249



COORDINATION COMPOUNDS

 Homoleptic and Heteroleptic complexes : Complexes in which the metal atom or ion is linked to only one type of ligands are called homoleptic complexes, e.g., [Co(NH3)6]3+ and [Fe(CN)6]4–. The complexes in which the metal atom or ion is linked to more than one kind of ligands are called heteroleptic complexes. e.g., [Co(NH3)4Cl2]+ and [Cr(en)2Cl2]+.

 Homonuclear and Polynuclear complexes : Complexes in which only one metal atom is present are known as homonuclear complexes. e.g., [Co(NH3)6]Cl3 and [Cu(NH3)4]SO4. Complexes in which more than one metal atom is present are known as polynuclear complexes.  Counter ions : The ions which are not included in the primary coordination sphere are known as counter ions. e.g., in K4[Fe(CN)6], K+ ions are counter ions.  Coordination ions : The coordination entity with charge is called as coordination ion.  Nomenclature of Coordination compounds : (i) The cation whether simple or complex is named first followed by anion. (ii) Ligands are named in alphabetical order. Scan to know (iii) For indicating the number of each kind of ligand within the coordination entity, two more about this topic kinds of numerical prefixes are used (di, tri, tetra, etc.). For ligands containing any of these prefixes in their names, their numbers are indicated by prefixes bis, tris, tetrakis, etc. Anionic ligands end in –o. Neutral retain their names while cationic end in -ium. (iv) The coordination sphere is written in square bracket. Nomenclature (v) In naming, ligands are named first in alphabetical order followed by metal atom and then of coordination compounds the oxidation state of metal by a Roman numeral in parenthesis. (vi) Name of coordination compounds starts with a small letter and the complex part is written as one word. (vii) Oxidation number of central atom is indicated in Roman numerals. No space is left between the number and the rest of the name.  Isomers: Two or more coordination compounds which have the same molecular formula but differ in the arrangement of ligands around the central metal atom or ion are called isomers. This phenomenon is called isomerism.  Types of isomerism: Two types of isomerism are observed in coordination compounds.    •  Structural Isomerism is of the following types: (i)  Ionization isomerism: In this type of isomerism, isomers have same molecular formula but gives different ions in solution. e.g., [Co(NH3)5Cl]SO4 and [Co(NH3)5(SO4)]Cl. (ii)  Coordination isomerism: This type of isomerism is shown by those complexes in which both the cation and the anion are complex ions and they differ in the coordination of ligands. e.g., [Co(NH3)6][Cr(C2O4)3] and [Cr(NH3)6][Co(C2O4)3]. (iii) Solvate isomerism: Solvate isomers differ by whether or not a solvent molecule is directly bonded to the metal ion or merely present as free solvent molecules in the crystal lattice. e.g.,[Cr(H2O)5Cl]Cl2.H2O,[Cr(H2O)6]Cl3. (iv) Linkage isomerism: Isomers having the same molecular formula but different linking atom, this is due to the presences of ambident ligands. e.g.,[Co(NH3)5(NO2)]2+ and [Co(NH3)5(ONO)]2+.    •  Stereo isomerism is of following types: (i) Geometrical isomerism or cis-trans isomerism: In tetra coordinated square planar complexes, the cis-isomer has the same groups on the side whereas trans-isomer has same groups on opposite sides. e.g.

cis-diamminedichloro­ platinum (II)

trans-diamminedichloro­ platinum (II)

(ii)  Optical isomerism: Optical isomers are mirror images that cannot be superimposed on one another. These are called as enantiomers. The molecules or ions that cannot be superimposed are called chiral. The two forms are called dextro (d) or laevo (l) depending upon the direction in which they rotate the plane of polarized light in a polarimeter (d rotates to the right, l to the left). Optical isomerism is common in octahedral complexes involoving bidentate ligands

250 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Optical isomers (d and l) of [Co(en)3]3+

Mnemonics Spectrochemical Series • Concept: Ligands causing Crystal Field Splitting are arranged in order of increasing field strength-Spectrochemical Series • Mnemonic: I Brought SCaNned Classnotes to Study Fundamentals of Chemistry. • He Nurtured Excellence in NHew (New) CoordinatioN COmpounds chapter. • Interpretation: I– < Br– < SCN– < Cl– < S2– < OH– < C2O42– < H2O < NCS– < EDTA4– < NH2 < CN– < CO I– = I Br– = Brought SCN- = SCaNned Cl– = Classnotes S– = Study F– = Fundamentals – OH– = Of C2O42 = Chemistry (Chromate ion) H2O = He NCS– = Nurtured – EDTA4 =Excellence NH3 = NHew CN– = CoordinatioN CO = Compounds (Carbonyl group)

Know the Terms  Coordination chemistry : The study of the coordination compounds is known as coordination chemistry.  Labile complex : A complex in which ligand substitution is fast.  Inert complex : A complex in which ligand substitution is slow.  Synergic bonding : A ligand donates a pair of electrons to the metal atom or ion and then accepts a pair of electrons back in its vacant orbital also from d-orbitals of the metal or ion. This is called synergic bonding and the ligands involved are known as p-donor ligands.  Effective Atomic Number (EAN) : It can be calculated for the metal atom or ion in the coordination complex by using following relation : EAN = Atomic no. (Z) of metal atom – Oxidation number + 2 C.N. where, C.N. is coordination number.  Facial or fac isomer : When three ligands with donor atoms are on the same triangular face of the octahedron, the geometrical isomer is known as facial or fac isomer.  Meridional or mer isomer : When three ligands with donor atoms are on the same equatorial plane of the octahedron or around the meridian of the octahedron, the isomer is called meridional or mer isomer.  Perfect or penetrating complexes : These are the complexes in which complex ion is fairly stable and either completely or feebly dissociates in solution.  Imperfect or abnormal complexes : These are the complexes in which the complex ion is less stable and is dissociated reversibly to give enough simple ions.

COORDINATION COMPOUNDS

[ 251

Q. For the complex ion [Fe(en)2Cl2]+ write STEP-2: Magnetic character: Paramagnetic the hybridization type and magnetic STEP-3: behaviour. Draw one of the geometrical isomer of the complex ion which is optically active. [Atomic No.: Fe = 26] Solutions: STEP-1: Hybridisation: d2sp3

Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS : Q. 1. The oxidation of Ni in [Ni(CO4)] is (a) 0  (b) 2  (c) 3  (d) 4  [CBSE Delhi Set-1 2020] Ans:  Correct option: (a) Explanation :  CO is a neutral ligand and its oxidation state is zero. Since the overall charge on the complex is zero too, hence oxidation state of Ni is 0. Q. 2. Which of the following will give a white precipitate upon reacting with AgNO3? (a) K2 [Pt(en)2Cl2]  (b) [Co[NH3)3Cl3] (c) [Cr(H2O)6]Cl3  (d) [Fe(H2O)3Cl3]  R [CBSE Delhi Set-2 2020] Ans:  Correct option: (c) Explanation : AgNO3 + [Cr(H2O)6]Cl3 → AgCl ppt + NO3− Since Cl is outside the coordination sphere, it can react with AgNO3 forming the white AgCl precipitate. Q. 3.    The formula of the complex triamminetri(nitrito-O) Cobalt (III) is (a) [Co(ONO)3(NH3)3] (b) [Co(NO2)3(NH3)3] (c) [Co(ONO2)3(NH3)3] (d) [Co(NO2)(NH3)3]  U [CBSE Delhi Set-3 2020] Ans:  Correct option: (a) Explanation : [Co(ONO)3 (NH3)3] Q. 4.    How many ions are produced from the complex [Co(NH3)5Cl]Cl2 in solution? (a) 4  (b) 2  (c) 3  (d) 5  [CBSE OD Set-1 2020]

(1 mark each) Ans: Correct option: (c) Explanation : [Co(NH3)5Cl]Cl2 → [CO(NH3)5Cl](aq) + 2Cl−(aq) Q. 5.    The pair [Co(NH3)4Cl2]Br2 and [Co(NH3)4Br2]Cl2 will show (a)  Linkage isomerism (b)  Hydrate isomerism (c)  Ionization isomerism (d)  Coordinate isomerism  U [CBSE OD Set-1 2020] Ans: Correct option: (c) Explanation : Ionization isomers have identical central ion and the other ligands except for a ligand that has exchanged places with an anion or neutral molecule that was originally outside the coordination complex. [Co(NH3)4Cl2]Br2 → [Co(NH3)4Cl2]+ + 2Br− [Co(NH3)4Br2]Cl2 → [Co(NH3)4Cl2]+ + 2Cl−

Q. 6.    The coordination number of ‘Co’ in the complex [Co(en)3]3+ is (a)  3   (b)  6 (c)  4   (d)  5  [CBSE OD Set-2 2020] Ans: Correct option: (b) Explanation : Coordination number is the number of ligands joined to the central metal ion or atom. Since ethylenediamine is a bidentate ligand, Co has coordination number of 6. Q. 7.    Which of the following is the most stable complex? (a) [Fe(CO)5]   (b)  [Fe(H2O)6]3+ (c) [Fe(C2O4)3]3–  (d) [Fe(CN)6]3–  [CBSE OD Set-3 2020] Ans: Correct option: (c) Explanation : [Fe(C2O4)3]3– acts as the chelating ligands. because C2O42– is a bidentate ligand.

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q. 8.    What type of isomerism is shown by the pair [Cr(H2O)6]Cl3 and [Cr(H2O)5Cl] Cl2 . H2O? (a)  Ionisation isomerism (b)  Coordination isomerism (c)  Solvate isomerism (d)  Linkage isomerism  [CBSE OD Set-3 2020] Ans: Correct option: (c) Explanation : Solvate / Hydrate isomerism Solvate or hydrate isomers have the same composition but differ with respect to the number of solvent ligand molecules as well as the counter ion in the crystal lattice. Q. 9.    Ambidentate ligands like NO–2 and SCN– are : (a) unidentate (b) didentate (c) polydentate (d)  has variable denticity OR The formula of the coordination compound Tetraam mineaquachloridocobalt(III) chloride is (a) [Co(NH3)4(H2O)Cl]Cl2 (b) [Co(NH3)4(H2O)Cl]Cl3 (c) [Co(NH3)2(H2O)Cl]Cl2 (d) [Co(NH3)4(H2O)Cl]Cl R [CBSE SQP 2021] Ans: Correct option : (a) Explanation : Tetraammineaquachloridocobalt (III) chloride is [CO(NH3)4(H2O)Cl]Cl2. Central atom – Cobalt(III) Coordination sphere ligandsTetraammine - 4 NH3 groups neutral ligand Aqua – 1 H2O groups neutral ligand Chlorido -1 Cl group, negatively charged ligand, one negative charge Counter ion- Chloride ions, 2 Since Cobalt is 3+, one valency is satisfied with Cl in coordination sphere and 2 by chlorine counter ions. Hence the formula of the coordination compound is [CO(NH3)4(H2O)Cl]Cl2. [B] ASSERTIONS AND REASONS: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b)  Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c)  Assertion is correct statement but reason is wrong statement. (d)  Assertion is correct statement but reason is correct statement. Q. 1.  Assertion (A) : Actinoids form relatively less stable complexes as compared to lanthanoids. Reason (R) : Actinoids can utilize their 5d orbitals along with 6d orbitals in bonding but lanthanoids do not use their 4f orbital for bonding.

Ans. Correct option: (d) Explanation : Actinoids are more reactive and form relatively more stable complexes in comparison to lanthanoids as Actinoids can utilize their 5d orbitals along with 6d orbitals in bonding but lanthanoids do not use their 4f orbitals for bonding. Q. 2.  Assertion (A) : Toxic metal ions are removed by the chelating ligands. Reason (R) : Chelate complexes tend to be more stable. Ans. Correct option: (a) Explanation : When a solution of chelating ligand is added to solution containing toxic metal ligands chelates the metal ions by formation of stable complex. Q. 3. Assertion (A) : Linkage isomerism arises in coordi­ nation compounds containing ambidentate ligand. Reason (R) : Ambidentate ligand has two different donor atoms. Ans. Correct option: (a) Explanation : Linkage isomerism arises due to two different donor atoms in ambidentate ligand. [C] VERY SHORT ANSWER TYPE QUESTIONS: Q. 1. What is the IUPAC name of the complex [Ni(NH3)6]Cl2? [CBSE Comptt. Delhi 2015] Ans. Hexaamminenickel (II) Chloride. [1] [CBSE Marking Scheme, 2015] Q. 2. What is meant by chelate effect?  R [CBSE Comptt. OD Delhi 2015] Ans. Formation of stable complex with a polydentate ligand due to stronger bonding is known as chelate effect. [1] [CBSE Marking Scheme, 2015] OR Formation of stable complex with a polydentate ligand due to stronger bonding than the non chelate complexes is known as chelate effect. Q. 3. Which of the following is more stable complex and why? [Co(NH3)6]3+ and [Co(en)3]3+.  U [CBSE Comptt. OD Delhi 2015] Ans. [Co(en)3]3+: Because (en) is a chelating ligand/ bidentate ligand.[½ + ½] [CBSE Marking Scheme, 2015] Q. 4. How many ions are produce from the complex, [Co(NH3)6]Cl2 in solution? [CBSE SQP 2016] Ans. Three ions [CO(NH3)6]2+, 2Cl– [1] [CBSE Marking Scheme, 2015] Q. 5. Write the IUPAC name of the following coordination compound [NiCl4]2–.  [CBSE Comptt. Delhi 2016] Ans. Tetrachloridonickelate (II) ion. [1] [CBSE Marking Scheme, 2016]

[ 253

COORDINATION COMPOUNDS

Q. 6. Write IUPAC name of the complex: [CoCl2(en)2]2+.  [CBSE Comptt. OD Set-1, 3 2017]

Q. 8. Write the coordination isomer of [Cu(NH3)4] [PtCl4].  [CBSE Comptt. Delhi/OD 2018]

Ans. Dichloridobis (ethane-1, 2-diamine) cobalt (III) ion. [1] [CBSE Marking Scheme, 2017]

Ans. [Pt(NH3)4][CuCl4] [1] [CBSE Marking Scheme, 2018]

Commonly Made Error

Q. 9. Write the coordination number and oxidation state of Platinum in the complex [Pt(en)2Cl2].  [CBSE Comptt. Delhi/OD 2018]

 There is confusion in oxidation state of central metal ion. Answering Tip

Ans. Coordination Number = 6, Oxidation State = +2[½ + ½] [CBSE Marking Scheme, 2018]

 Practice and understand writing nomenclature and oxidation state of ligands. Q. 7. Write IUPAC name of the complex: [Co(NH3)4Cl(NO2)]+.  [CBSE Comptt. OD Set-2 2017]

Q. 10. Out of Cis-[Pt(en)2Cl2]2+ and Trans-[Pt(en)2Cl2]2+, which one is optically active? Ans. cis [Pt(en)2Cl2]2+.[½ + ½]

Ans. Tetraamminechloridonitrocobalt (III) ion. [1] [CBSE Marking Scheme, 2017]

Short Answer Type Questions-I

Q. 4. (i) Write the IUPAC name of the isomer of the following complex:

Q. 1. Using IUPAC norms write the formulae for the following:

(ii) Tetraamminechloridonitrito-N-platinum (IV) sulphate A [CBSE OD Set-1 2017]





Ans. (i) Na[Au(CN)2] 1 (ii)  [Pt(NH3)4Cl(NO2)]SO4 1 [CBSE Marking Scheme 2017] Q. 2. Using IUPAC norms write the formulae for the following: (i) Tris(ethane-1,2-diamine)chromium(III) chloride (ii) Potassium tetrahydroxozincate(II) A [CBSE OD Set-2 2017] Ans. (i) [Cr(en)3]Cl3 (ii)  K2[Zn(OH)4]

1 1 [CBSE Marking Scheme 2017] OR

[Topper’s Answer 2017]  2

Ans. (i) cis/trans-diamminedichloridoplatinum (II) 1 (ii)  [Co(NH3)4(H2O)Cl] (NO3)2 1 [CBSE Marking Scheme 2017] Q. 5. (i) Write the IUPAC name of the isomer of the following complex: [Co(NH3)5Cl]SO4 (ii) Write the formula for the following: Diamminechloridonitrito-N-platinum(II)  A [CBSE Foreign Set-2 2017]



(i) Potassium trioxalatoaluminate (III)

(ii) Dichloridobis(ethane-1,2-diamine)cobalt(III)







A [CBSE OD Set-3 2017]

Ans. (i) K3[Al(C2O4)3] 1 (ii)  [CoCl2(en)2]+ 1 [CBSE Marking Scheme 2017]

Ans. (i) Pentaamminechlorocobalt (III) sulphate 1 (ii)  [Pt(NH3)2Cl(NO2)] 1 [CBSE Marking Scheme 2017]

Q. 6. (i) Write the IUPAC name of the following complex : [Co(NH3)4Cl(NO2)]Cl (ii) Write the formula for the following: Dichloridobis(ethane-1,2-diamine)cobalt(III) chloride  A [CBSE Foreign Set-3 2017]

Q. 3. Using IUPAC norms write the formulae for the following:

[Pt(NH3)2Cl2] (ii) Write the formula for the following: Tetraammineaquachloridocobalt (III) nitrate A [CBSE Foreign Set-1 2017]

(i) Sodium dicyanidoaurate (I)





(2 marks each)

Ans. (i) Tetraamminechloridonitrito-N-cobalt (III) chloride.1 (ii)  [CoCl2(en)2]Cl 1 [CBSE Marking Scheme 2017]

Q. 7. (i) Write down the IUPAC name of the following complex : [Cr(NH3)2Cl2(en)2]Cl (en = ethylenediamine)





A [CBSE Delhi 2015]



(ii) Write the formula for the following complex : Potassium tetrachloridonickelate (II) A [CBSE Delhi 2015] Ans. (i) Pentaamminechloridocobalt (III) ion 1 (ii) K2[NiCl4] 1 Q. 9. When a coordination compound CoCl3.6NH3 is mixed with AgNO3, 3 moles of AgCl are precipitated per mole of the compound. Write (i) Structural formula of the complex, (ii) IUPAC name of the complex. A [CBSE OD 2016] Ans. (i) [Co(NH3)6]Cl3 Cl

Ans. Bonding in [Co(NH3)6]3+

d2sp3 hybridisation Atomic orbits of Co (III) ion Atomic orbits of Co (III) ion

Ans. (i) Diamminedichloridobisethylenediamine­chro­ mium (III) chloride (ii)  [Co(NH3)5(ONO)]2+ [CBSE Marking Scheme 2015] 1+1

Q. 8. (i) Write down the IUPAC name of the following complex : [Co(NH3)5Cl]2+





(ii) Write the formula for the following complex : Pentaamminenitrito-O-Cobalt (III).





Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

d 2sp 3 hybridised orbitals of Co (III) ion d 2sp 3 hybridised orbitals of Co Formation of (III) ion [Co(NH3)6] Formation of [Co(NH3)6]3+

4p

3d

4s

4p

3d

d2sp3 hybrid

½

3d XX XX 3d

XX XX

3d

d2sp3 hybrid XX XX XX XX Six pairs of electrons XX sixXX XXmolecules XX from NH 3 Six pairs of electrons from six NH3 molecules

½



Geometry : Octahedral

½

Diamagnetic

½



[CBSE SQP Marking Scheme 2020]

Q. 12. Using IUPAC norms write the formulae for the following : (i) Hexaamminecobalt(III) sulphate A [CBSE Delhi Set-1 2019]

NH3

Ans. (i) [Co(NH3)6]2(SO4)3

Cl

(ii) K3[Cr(C2O4)3]

Commonly Made Error  Students often make errors in naming the central atom and ligand in the coordinate compounds.

Answering Tip  Understand and practice the naming of coordinate compounds. Q. 10. When a coordination compound CrCl3.6H2O is mixed with AgNO3, 2 moles of AgCl are precipitated per mole of the compound. Write (i) Structural formula of the complex. (ii) IUPAC name of the complex. A [CBSE Delhi 2016] Ans. (i) [Cr(H2O)5Cl]Cl2.H2O 1 (ii)  Pentaaquachloridochromium (III) chloride monohydrate. 1 [CBSE Marking Scheme 2016]

Q. 11. Discuss the bonding in the coordination entity [Co(NH3)6]3+ on the basis of valence bond theory. Also, comment on the geometry and spin of the given entity. (Atomic no. of Co = 27)

1

[CBSE Marking Scheme, 2019] 1

Q. 13. Define the following terms with a suitable example of each : (a) Polydentate ligand (b) Homoleptic complex Ans. (i) A ligand having several donor atoms. ExampleEDTA 1 (ii) A complex in which a metal is bound to only one kind of donor groups / ligands. Example[Co(NH3)6]3+ 1 (or any other correct example) [CBSE Marking Scheme, 2019]

NH3

NH3 1 (ii) IUPAC name : Hexamminecobalt (III) chloride. 1 [CBSE Marking Scheme 2016]



4s

(ii) Potassium trioxalatochromate(III) NH3

Co

Cl

3d

3+

H3N

H3 N

½





254 ]

Detailed Answer : (a) A ligand which contains several donor atoms in a single ligand is known as polydentate ligand. Example, N(CH2 CH2 NH2)3. (b) Complexes in which a metal is bound to only one kind of donor groups are known as homoleptic complexes. Example, [Co(NH3)6]3+. Q. 14. Using IUPAC norms, write the formulae for the following complexes: (a) Potassium tri(oxalate)chromate(III) (b) Hexaaquamanganese(II) sulphate A [CBSE Delhi Set-3 2019] (i) K3[Cr(C2O4)3] 1 (ii) [Mn(H2O)6]SO4 [CBSE Marking Scheme, 2019] 1 Q. 15. Write IUPAC name of the complex [Pt(en)2Cl2]. Draw structures of geometrical isomers for this complex.[2]

[ 255

COORDINATION COMPOUNDS



1

Cl Cl



Pt



en



en

Ans. (i) [Co(NH3)6]2(SO4)3

en

(ii) K3[Cr(C2O4)3]

Cl

en

Trans-isomer

Cis-isomer  

Pt



Cl

½+½ [CBSE Marking Scheme, 2019]

1

[CBSE Marking Scheme, 2019] 1

Q. 17. Write IUPAC name of the complex [Cr(NH3)4Cl2]+. Draw structures of geometrical isomers for this complex. Ans. Tetraamminedichloridochromium(III) ion 

1



(II) 

Q. 16. Using IUPAC norms write the formulae for the following : (i) Hexaamminecobalt(III) sulphate (ii) Potassium trioxalatochromate(III)

Ans. Bis(ethan-1,2-diamine)dichloridoplatinum

Cl

Detailed Answer: IUPAC names of the complex [Pt(en)2Cl2] is Dichlorobis(ethylenediamine) platinum (II). The geometrical isomers of [Pt(en)2Cl2] is

Cl

H 3N

Cl

+ H 3N

Cr

NH3 cis



NH3 Cr NH3

NH3 H N 3

H 3N

+

Cl trans

 ½+½ [CBSE Marking Scheme, 2019]

[Topper’s Answer 2019]

Q. 18. Using IUPAC norms write the formulae for the following : (i) Pentaammine nitrito-o-cobalt (III) Chloride (ii) Potassium tetracyanonickelate (II) [2]

Ans.

(ii) K2[Ni(CN4)]





Ans. (i) [Co(NH3)5ONO]Cl2

Q. 21. Draw one of the geometrical isomers of the complex [Pt(en)2Cl]2+ which is optically inactive. Also write the name of this entity according to the IUPAC nomenclature.

1

[CBSE Marking Scheme, 2019] 1

Q. 19. Write IUPAC name of the complex [Co(en)2(NO2) Cl]+. What type of structural isomerism is shown by this complex?

Q. 20. Using IUPAC norms, write the formulae for the following complexes : (a)  Hexaaquachromium(III) chloride (b)  Sodium trioxalatoferrate(III) Ans.  (a) Cr(H2O)6]Cl3 (b) Na3[Fe(ox)3] 1+1 [CBSE Marking Scheme, 2019]



Ans.  Chloridobis(ethane-1,2-diamine)nitrito-N-cobalt(III) ion Linkage isomerism 1+1 IUPAC Name of the entity :

1

Dichloridobis (ethane-1,2-diamine) platinum (IV) ion 1 Q. 22. Discuss the bonding in the coordination entity [Co(NH3)6]3+ on the basis of valence bond theory. Also, comment on the geometry and spin of the given entity. (Atomic no. of Co = 27)

256 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q. 25. Write IUPAC name and hybridization of the following complexes :

Ans. Bonding in [Co(NH3)6]3+



(i) [Ni(CO)4] ,

(ii) [CoF6]3– (Atomic number Ni = 28,

Co = 27)

½

A [CBSE OD Set-2 2020]

Ans. (i) [Ni(CO)4] – Tetracarbonylnickel Hybridization - sp3



8

2

3d

4s

4p

3d

4s

4p

Z = 28 – 3d 4s

Hybridization –

Geometry : Octahedral ½ Diamagnetic ½  [CBSE SQP Marking Scheme 2020] Q. 23. (a) Write the IUPAC name and hybridisation of the complex [Fe(CN)6]3–. (Given : Atomic number of Fe = 26) (b) What is the difference between an ambidentate ligand and a chelating ligand?  A [CBSE Delhi Set-1 2020] Ans. (a) [Fe(CN)6]3 – hexacyanoferrate (III) ion hybridization – d2sp3 3d 4s 4p Z = 26 – Fe (III) – d2sp3 six pairs of e– from CN– ions (b) Ambidentate ligand can bond through different atoms to form different coordination compounds. e.g. NO2– can bind to the central atom or ion at either the nitrogen atom or one of the oxygen atom. Chelating ligand : If the ligands with two or more electron donor groups positioned in such a way that they form five or six membered ring with central metal ion are called chelating ligands. e.g. ethane - 1, 2-diamine (en)  1+1 = 2

Q. 24. Write the IUPAC name and hybridization of the following complexes : (i) [Ni(CN)4]2–. (ii) [Fe(H2O)6]2+. (Given : Atomic number of Ni = 28, Fe = 26)  A [CBSE OD Set-1 2020] Ans. (i) [Ni(CN)4]2– – tetracyanonickelate (II)

hybridization – dsp2

(ii) [Fe(H2O)6]2+ – hexaquairon (II)

hybridization – sp3d2

1+1=2

sp3 hybridization Ni(CO)4 four CO molecules

(ii)  [CoF6]3– – Hexafluorocobaltate (III) Hybridization - sp3d2 7

2

3d

4s

4p

4d

3d

4s

4p

4d

Z = 27 – 3d 4s

Hybridisation – sp3d2 – six pairs of e from six fluoride ions

 1+1=2 Q. 26.  Write IUPAC name and hybridization of the following complexes : (i)  [Co(NH3)6]3+     (ii)  [NiCl4]2– (Atomic number Ni = 28, Co = 27) Ans. (i) [Co(NH3)6]3+ : Hexaamminecobalt (III) Hybridization - d2sp3 Co

3d

4s

4p

Z = 27 7

3d 4s

2

2



d sp3 six pairs of electrons from six NH3

(ii) [NiCl4]2– : Tetrachloronickelate (II) Hybridization - sp3



Ni Z = 28 – 8

3d

4s

4p

2

3d 4s Ni(II) – sp3

four pairs of electron from four Cl– ions 1 + 1 = 2

[ 257

COORDINATION COMPOUNDS

Short Answer Type Questions-II Q. 1. Write the IUPAC name of the following : (i) [Co(NH3)6]Cl3 (ii) [NiCl4]2– (iii) K3[Fe(CN)6] A [CBSE Comptt. OD 2015]

Ans. (i) Hexaamminecobalt (III) chloride. 1 (ii) Tetrachloridonickelate (II) ion. 1 (iii) Potassiumhexacyanoferrate (III) 1 [CBSE Marking Scheme 2015]



[CBSE Marking Scheme. 2016] Q. 5. Write the hybridization, shape and magnetic character of [Fe(CN)6]4−. 

A [CBSE Comptt. Delhi 2016]

Ans.  Hybridization :    d2sp3 isomerism. Shape :      Octahedral Magnetic character : Diamagnetic 1 + 1+ 1 [CBSE Marking Scheme, 2016]



(b) Draw one of the geometrical isomers of the complex [Co(en)2Cl2]+ which is optically active.  A [CBSE OD Set-2 2016]

Ans. (a)  Hybridization : d2sp3 Magnetic character : Diamagnetic Spin type : Low spin complex. (b)  Cis-[Co(en)2Cl2]+ is optically active.

Q. 2. Indicate the types of isomerism exhibited by the following complexes : (i) [Co(NH3)5(NO2)]2+ (ii) [Co(en)3]Cl3 (en = ethylenediamine) (iii) [Pt(NH3)2Cl2] A [CBSE Comptt. Delhi 2015] Ans. (i) Linkage isomerism. 1 (ii) Optical isomerism. 1 (iii) Cis - trans / Geometrical isomerism 1 [CBSE Marking Scheme 2015]

(3 marks each)

 Some students write ‘ligand isomerism’ instead of ‘linkage isomerism’.

Answering Tip  Learn and understand different types of isomerism with examples. Q. 3. (i) For the complex [Fe (CN)6]3–, write the hybridization, magnetic character and spin nature of the complex. (At. number : Fe = 26). (ii) Draw one of the geometrical isomers of the complex [Pt(en)2Cl2]2+ which is optically active.  A [CBSE Delhi 2016]

Ans. (i)  Hybridization : d2sp3 Magnetic character : Paramagnetic Spin nature of the complex : Low spin.  1+½+½ (ii)

2+ Cl Cl en







Pt en



1 cis-isomer [CBSE Marking Scheme, 2016] 4−

Q. 4. (a) For the complex [Fe(CN)6] , write the hybridization, magnetic character and spin type of the complex. (At. number : Fe = 26).

Q. 6. (i) What type of isomerism is shown by the complex [Co(NH3)6] [Cr(CN)6] ? (ii) Why a solution of [Ni(H2O)6]2+ is green while a solution of [Ni(CN)4]2– is colourless ? (At. no. of Ni = 28). [KVS] (iii) Write the IUPAC name of the following complex : [Co(NH3)5(CO3)] Cl.  A [CBSE Delhi Set-1, 2 2017] Ans. (i) Co-ordination isomerism. 1 (ii) Unpaired electrons in [Ni(H2O)6]2+/d-d transition. 1 (iii) Pentaamminecarbonatocobalt (III) chloride. 1 [CBSE Marking Scheme, 2017]

Commonly Made Error

Detailed Answer: (ii) [Ni(CN)4]2− has no unpaired electron in its d-subshell therefore d-d transition is not possible whereas [Ni(H2O)6]2+ has unpaired electron in its d-subshell resulting in d-d transition imparting colour. Q. 7. (i) W  hat type of isomerism is shown by the complex [Co(en)3] Cl3 ?

(ii) Write the hybridisation and magnetic character

of [Co(C2O4)3]3–. (At. no. of Co = 27) (iii) Write IUPAC name of the following Complex [Cr(NH3)3Cl3].

258 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



Ans. (i) Optical isomerism1 ½+½ (ii) d2sp3, diamagnetic (iii) Triamminetrichloridochromiun (III)1 [CBSE Marking Scheme, 2017]







Q. 8. (a) What type of isomerism is shown by the complex [Co(NH3)5 (SCN)]2+ ? (b) Why is [NiCl4]2– paramagnetic while [Ni(CN)4]2– is diamagnetic ? (Atomic number of Ni = 28) (c) Why are low spin tetrahedral complexes rarely observed ? A [CBSE OD Set-1, 2, 3 2017]

ns. (a) Linkage isomerism. A 1 (b) In [NiCl4]2–, due to the presence of Cl–, a weak field ligand no pairing occurs whereas in [Ni(CN)4]2–, CN– is a strong field ligand and pairing takes place / diagrammatic representation. 1 (c) Because of very low CFSE which is not able to pair up the electrons. 1 [CBSE Marking Scheme, 2017]

[Topper’s Answer, 2017]

[ 259

COORDINATION COMPOUNDS

Ans. Hybridisation : sp3d2

1



Magnetic character : Paramagnetic

1



Spin nature : High spin

1

[CBSE Marking Scheme 2017] Q. 10. For the complex ion [Ni(CN)4]2– write the hybridization type, magnetic character and spin nature. [Atomic number: Ni = 28] A [CBSE Comptt. Delhi Set-3 2017]

Ans. Hybridisation : dsp2 1 Magnetic character : Diamagnetic 1 Spin nature : Low spin 1 [CBSE Marking Scheme 2017] Q. 11. When a coordination compound CrCl3.6H2O is mixed with AgNO3 solution, 3 moles of AgCl are precipitated per mole of the compound. Write: (i) Structural formula of the complex (ii) IUPAC name of the complex (iii) Magnetic and spin behaviour of the complex  A [CBSE Comptt. OD Set-3 2017]

Ans. (i)  Hexaamminenickel (II) chloride

1

(ii)  Potassium hexacyanidoferrate (III)1 (iii)  Tris(ethane-1,2-diamine)cobalt (III) ion1  [CBSE Marking Scheme 2018] Q. 14.  (a) Write the formula of the following coordination compound : Iron(III) hexacyanoferrate(II) (b) What type of isomerism is exhibited by the complex [Co(NH3)5Cl]SO4 ? (c) Write the hybridisation and number of unpaired electrons in the complex [CoF6]3 . (Atomic No. of Co = 27) Ans. (a) Fe4[Fe (CN)6]3 1 (b) Ionisation isomerism 1 (c) sp3d2, octahedral complex with 4 unpaired electrons 1  [CBSE Marking Scheme, 2018]





Ans. (i) [Cr(H2O)6]Cl3 1 (ii) Hexaaquachromium(III) chloride 1 (iii) Paramagnetic and high spin ½+½ [CBSE Marking Scheme 2017] Q.  12.  A metal complex having composition Cr(NH3)4Cl2Br has been isolated in two forms A and B. The form A reacts with AgNO3 to give a white precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia.

Q. 13.  Write IUPAC name for each of the following complexes: (i)  [Ni(NH3)6]Cl2 (ii)  K3[Fe(CN)6] (iii) [Co(en)3]3+ A [CBSE Comptt. Delhi/OD 2018] 



Q. 9. For the complex ion [CoF6]3– write the hybridization type, magnetic character and spin nature. [Atomic number: Co = 27] A [CBSE Comptt. Delhi Set-1, 2 2017]

Ans. (i)  Isomer A: [Cr(NH3)4BrCl]Cl

½

Isomer B: [Cr(NH3)4Cl2]Br ½



(i) State the hybridisation of chromium in each of them.

(ii) Calculate the magnetic moment (spin only value) of the isomer A. A [CBSE SQP 2018-2019] Ans. (i) Hybridisation of Cr in isomer A and B is d2sp3. 1½ 3+ 3 (ii) Number of unpaired electrons in Cr (3d ) is 3

(ii)  Hybridisation of Cr in isomer A and B is d2sp3.1

Q. 15. A metal complex having composition Cr(NH3)4Cl2Br has been isolated in two forms A and B. The form A reacts with AgNO3 to give a white precipitate readily soluble in dilute aqueous ammonia whereas B gives a pale yellow precipitate soluble in concentrated ammonia.

3+

3

(iii)  Number of unpaired electrons in Cr (3d ) is 3 Magnetic moment =

n( n + 2 )

n (n + 2 )



Magnetic moment =

= 3.87 BM1



=

(deduct half mark for wrong unit/unit not written)



= 3.87 BM 1½ (deduct half mark for wrong unit/unit not written) [CBSE Marking Scheme 2018]

=

3( 3 + 2 )



[CBSE Marking Scheme 2018]

3 (3 + 2 )

TOPIC-2

Werner's Theory, Bonding, VBT, CFT Revision Notes  Werner’s Theory of Coordination compounds : Different postulates of Werner’s coordinatron theory are given below: (i) Metal ions possess two types of valency (a) primary or ionisable valency and (b) Secondary or non ionisable valency.

260 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(ii) Every metal ion has a fixed number of secondary valency and this is known as coordination number.

Scan to know more about this topic

(iii) Primary valencies are satisfied by anions while secondary valancies are satisfied by negative group or neutral molecules with lone pair of electrons.

(ii) Secondary valencies are directed in space towards internal positions.  Limitations of Werner's theory : This theory does not explain the following queries—

Werner’s theory

(i) Why is the complex forming tendency limited to a few elements only ?

(ii) Why bonds in the coordination complexes are of directional nature ?

(iii) Why are certain complexes of magnetic nature and show geometrical and optical isomerism ?  Valence Bond theory : It was developed by Pauling. The brief points are : (i) A suitable number of vacant orbitals must be present in the central metal atom or ion for the formation of coordinate bond with the ligands. (ii) Central metal ion can use appropriate number of s, p or d-orbitals for hybridisation depending upon total number of ligands. (iii) The hybridised orbitals are allowed to overlap with those ligand orbitals that can donate an electron pair for bonding. (iv) The outer orbitals (high spin) or inner orbitals (low spin) complexes are formed depending upon whether outer d-orbitals or inner d-orbitals are used.  Limitation of Valence bond theory : (i) It cannot explain the detailed magnetic properties of complex compounds. Scan to know (ii) It cannot explain the optical absorption spectra of coordination compounds. more about this topic (iii) It cannot predict property whether a particular 4 coordinate complex is square planar or tetrahedral in nature. (iv) It fails to make distinction between strong and weak ligands. (v) It does not explain thermodynamic or kinetic stabilities of coordination compounds. Crystal field  Crystal field theory (CFT) : theory (i) The ligand is considered as point charge or point dipole. (ii) Interaction between metal ion and ligand is considered as electrostatic in nature. (iii) Metal ion is supposed to be present at the origin of the axis. Ligands approach to metal ion along the axis of octahedral complex between the axis of tetrahedral complex and in the case of square planar complex four ligand approach to metal ion along x, y plane. (iv) Due to the approach of the ligand hence due to the electrostatic interaction between ligands electrons and metal d-orbital electron, degeneracy of d-orbital is lost and spliting of d-orbitals occurs. (v) Some ligands are able to produce strong fields in which case, the splitting will be large whereas others produce weak fields and consequently result in small splitting of d-orbitals. In general, ligands can be arranged in a series in the order of increasing field strength as given below and called as spectrochemical series : I– < Br– < SCN– < Cl– < S2– < F– < OH– < C2O42– < H2O < NCS– < EDTA4– < NH3 < en < CN– < CO L L M

M

M

L

L L

Energy

[ Energy

[

d xy, d xz , d yz

[

2 ∆ 5 t

[ t2g

[

d orbitals free ion

[

[d

x2 – y2

d x 2 – y 2, d z 2

z y x

∆t

3 ∆t 5 , dz 2

Metal d orbitals

[ eg

Average energy of the Splitting of d orbitals d orbitals in spherical in tetrahedral crystal crystal field field

Fig. 2 : d orbital splitting in a tetrahedral crystal field

L

d x 2 – y 2, d z 2

dxy dxz dyz

Free metal ion

Bary centre dxy dxz dyz

Average energy of the d orbitals in spherical crystal field

eg 3/5 ∆o 2/5 ∆o

∆o t2g

Splitting of d orbitals in octahedral crystal field

Fig. 1: d orbital splitting in an octahedral crystal field

d-orbital splitting in a tetrahedral crystal field d-orbital splitting in an octahedral crystal field (vi) Explanation of colour and magnetic behaviour in complexes is possible by crystal field theory.

[ 261

COORDINATION COMPOUNDS

 Metal carbonyls : Homoleptic carbonyls are formed by d-block elements and contain carbonyl ligands only. e.g., V(CO)6, Cr(CO)6, [Mo(CO)6], [W(CO)6], [Mn2(CO)10], [Fe(CO)5], [Fe2(CO)9], [Co2(CO8)], [Co4(CO)12], [Ni(CO)4], etc. Metal carbonyls of outside the central part of d-block are unstable. Properties of metal carbonyls : (i) Metal carbonyls are mostly solids at room temperature and pressure. Exceptions being iron and nickel carbonyls which are liquids. (ii) The mononuclear carbonyls are volatile and toxic. (iii) Most of metal carbonyls are soluble in hydrocarbon solvents except [Fe2(CO)9]. (iv) Mononuclear carbonyls are either colourless or light coloured. (v) They are highly reactive due to metal centre and the CO ligands. (vi) Metal carbonyls are used as industrial catalyst and as precursor in organic synthesis. CO

CO OC

Ni

Ni CO

OC

CO

OC

CO

CO

Ni(CO)4 Tetrahedral

Fe(CO)5 Trigonal bipyramidal

OC

OC CO CO

CO Cr

CO

OC CO

CO Cr(CO)6 Octahedral

CO

CO CO

CO CO

Mn

Mn

CO CO

CO

CO

CO

Co

Co

CO

CO CO OC [Co2(CO)8]

CO

[Mn2(CO)10]

 Bonding in metal carbonyls : It also involves both s- and p-bond. s-bond is formed by overlapping of lone pair on CO to the vacant d-orbitals of metal whereas p-bond is formed by back donation of pair of d-electrons to vacant anitbonding orbital of carbonyl.  Factors affecting the stability of ‘coordination’ complexes : (i) Nature of the central ion : Greater the charge density on the central metal ion, greater is the stability of the complex. (ii) Nature of the ligand : More basic ligands have a tendency to donate the electron pairs to central metal ion more easily resulting in a stable complex. (iii) Chelate effect : Entropy increases when chelation occurs and so the formation of the complex becomes more favourable.  Applications of Complex compounds : (i) They are used in photography, i.e., AgBr forms soluble complex with sodium thiosulphate in photography. (ii) K[Ag(CN)2] is used for electroplating of silver, K[Au(CN)2] is used for gold plating. (iii) Some of ligands oxidise Co2+ to Co3+ ion. (iv) EDTA is used for estimation of Ca2+ and Mg2+ in hard water. (v) Silver and gold are extracted by treating Zn with their cyanide complexes. (vi) Ni2+ is tested and estimated by DMG (dimethyl glyoxime). (vii) cis-platin [Pt(NH3)2Cl2] is used as antitumor agent in the treatment of cancer. (viii) EDTA is used to remove Pb by forming Pb-EDTA complex which is eliminated in urine. (ix) Haemoglobin contains Fe, chlorophyll contains (Mg) and vitamin B12 contain Co2+ ‘which are all coordination compounds’. (x) Bauxite is purified by forming complex with NaOH. (xi) Coordination compounds are used as catalysts for many industrial processes.

Know the Terms  Homogeneous Catalysis : Organometallic compounds or intermediates derived from soluble transition metal complexes catalyse a variety of reaction in solutions. This is known as homogeneous catalysis.  Macrocyclic effect : Multidentate ligands happen to be cyclic in nature without causing any steric hindrance, the stability of the complexes is further increased. This is known as macrocyclic effect.  Stability constants (K) : The relative stabilities of coordination complexes can be compared in terms of stability constant (K) also denoted by b (Beta).  Metal carbonyl : Organometallic compounds in which carbon monoxide acts as the ligand.

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q. (i) Describe the type of hybridisation for the complex ion [Fe(H2O)6]2+. (ii) Write the IUPAC name of the ionisation isomer of the coordination compound [Co(NH3)5Br]SO4. Give one chemical test to distinguish between the two compounds. Solutions: STEP-1: (i) Fe exists as Fe2+. Fe (II) in [Fe(H2O)6]2+ = 3d6 4s0 4p0 4d0 STEP-2: As water is a weak ligand, pairing does not occur and the 6 lone pairs available from each water molecule moves

to one 4s, three 4p and two 4d orbitals. Thus, the hybridisation involved is sp3d2. (marks to be granted if hybridisation is depicted diagrammatically) STEP-3: (ii) The ionisation isomer is [Co(NH3)5SO4]Br. The IUPAC name is Pentaamminesulphatocobalt (III) bromide. STEP-4: The isomer [Co(NH3)5Br]SO4 gives a white precipitate of BaSO4 with BaCl2 solution whereas the isomer [Co(NH3)5SO4] Br does not form this precipitate. (or any other relevant test)

Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS : Q . 1. The crystal field splitting energy for octahedral (Do) and tetrahedral (Dt) complexes is related as (a) Dt =

2 Do 9

(c) Dt =

4 Do 9

(b) Dt =

5 Do 9

(d) Dt = 2 D o

 Ans. Correct option : (c) Explanation : Dt =

R [CBSE OD Set-2 2020]

4 Do 9

Q . 2. Which set of ions exhibit specific colours?(Atomic number of Sc=21, Ti=22, V=23, Mn=25, Fe=26, Ni=28, Cu=29 and Zn=30) (a) Sc3+, Ti4+, Mn3+ (b) Sc3+, Zn2+, Ni2+ (c) V3+, V2+, Fe3+

Atomic number of V = 23, Electronic configuration of V − [Ar]3d34s2 Electronic configuration of V2+ − [Ar]3d3 Electronic configuration of V3+ − [Ar]3d2 Atomic number of Fe = 26 Electronic configuration of Fe − [Ar]3d64s2 Electronic configuration of Fe3+ − [Ar]3d5 Since these ions have partially filled d-subshells , they exhibit colour. Most transtition metal ions have a partially filled d subshell. As for other ions, Atomic number of Sc = 21 Electronic configuration of Sc − [Ar]3d14s2 Electronic configuration of Sc3+ − [Ar]3d0 Q. 3. Atomic number of Mn, Fe and Co are 25, 26, 27 respectively. Which of the following inner orbital octahedral complex ions are diamagnetic? (a) [Co(NH3)6]3+  (b) [Mn(CN)6]3−

(d) Ti3+, Ti4+, Ni2+

[CBSE SQP 2021]

Ans. Correct option : (c) 3+

(1 mark each)

2+

Explanation : V , V , Fe colours.

3+

ions exhibit specific

(c) [Fe(CN)6]3−  

(d)  None of the above

Ans. Correct option : (a) Explanation : Molecular orbital electronic configuration of Co3+ in [Co(NH3)6]3+ is

[ 263

COORDINATION COMPOUNDS



(a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c)  Assertion is correct statements but reason is wrong statement. (d)  Assertion is wrong statements but reason is correct statement. Q. 1. Assertion : [Fe(CN)6]3− ion shows magnetic moment corresponding to two unpaired electrons. Reason : Because it has d2sp3 type hybridisation. R [NCERT Exemplar]  Ans. Correct option : (d) 1 Explanation : [Fe(CN)6]3− ion shows magnetic moment corresponding to two unpaired electrons. Q. 2. Assertion : [Cr(H2O)6]Cl2 and [Fe(H2O)6]Cl2 are reducing in nature. Reason : Unpaired electrons are present in their d-orbitals. R [NCERT Exemplar]  Ans. Correct option : (b) 1 Explanation : In the complexes, Co exists as Co2+ and Fe as Fe2+. Both of the complexes become satble by oxidation of metal ion to Co3+ and Fe3+. Q. 3. Assertion : Complexes of MX6 and MX5L type (X and L are unidentate) do not show geometrical isomerism. Reason : Geometrical isomerism is not shown by complexes of coordination number 6. U [NCERT Exemplar]  Ans. Correct option : (b) 1

[C] VERY SHORT ANSWER TYPE QUESTIONS :



Q. 1. Why are low spin tetrahedral complexes not formed? [CBSE Comptt. Delhi Set-1, 2, 3 2017]

Ans. Orbital splitting energies are not sufficiently large for forcing pairing. 1  [CBSE Marking Scheme 2017]

In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices

Q. 2. Low spin configuration are rarely observed in tetrahedral coordination entity formation. Explain.  [CBSE SQP 2018-2019] Ans. The Orbital splitting energies, ∆t, are not sufficiently large for forcing pairing of electrons in the tetrahedral coordination entity formation. 1  [CBSE Marking Scheme 2018]

[B]  ASSERTIONS AND REASONS :

Q. 3. A coordination compound with molecular formula CrCl3.4H2O precipitates one mole of AgCl with AgNO3 solution. Its molar conductivity is found to be equivalent to two ions. What is the structura; formula and name of the compound? [A][CBSE SQP 2017] Ans. [Cr(H2O)4Cl2]Cl Tetraaquadichloridochromium (III) chloride. 1  [CBSE Marking Scheme 2017]

Number of unpaired electron = 0 Magnetic property = Diamagnetic

Explanation : For complexes of MX6 and MX5L type, different geometric arrangements of the ligands are not possible due to presence of plane of symmetry. Q. 4. Assertion : Low spin tetrahedral complexes are rarely observed. Reason : Crystal field splitting is less than pairing energy for tetrahedral complexes. R [CBSE Delhi Set-1 2020]  Ans. Correct option : (a) 1 Explanation : In tetrahedral complexes, the splitting of the d-orbitals is inverted and is smaller in comparison to octahedral complexes.The Crystal field splitting energy is not large enough to force pairing and hence, low spin complexes are rarely observed.

Q. 4. On the basis of crystal field theort, write the electronic configuration of d6 in terms of t2g and eg in an octahedral field when ∆0 < P. A [CBSE SQP 2018-2019]  Ans. t32ge3g1

Q. 1. Out of

[CoF6 ]3−

and Co ( en )  3 

3+

, which one

complex is, (i) Paramagnetic, (ii) More stable, (iii) Inner orbital complex and (iv) High spin complex (Atomic number of Co = 27)

[CBSE Delhi Set 1 2019]



Short Answer Type Questions-I

(2 marks each)

Ans. (i) [CoF6]3–

(ii) [Co(en)3]3+ (iii) [Co(en)3]3+ (iv) [CoF6]3–

[CBSE Marking Scheme, 2019] ½ × 4

264 ]



Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Ans. (a) A complex formed by bi or polydentate ligands with metal. Example- [Co(en)3]3+. ½+½ (b) A ligand which can ligate through two different donor atoms. Example-SCN– . ½+½ [CBSE Marking Scheme, 2019] ½ +½

Q. 2. Out of [CoF6 ] and ëéCo (C 2O 4 )3 ûù which one complex



3-

3-

is : (i) diamagnetic, (ii) more stable, (iii) outer orbital complex and (iv) low spin complex. (Atomic number of Co = 27) [CBSE Delhi Set 2 2019]

Detailed Answer : (a) Chelate complex is the complex formed when a di – or polydentate ligand uses its two or more donor atoms to bind a single metal atom. (b) Ligand which can ligate through two different atoms is called ambidentate ligand. Q. 5. Using IUPAC norms, write the formula for the following complexes : (a) Tetraamminediaquacobalt(III) chloride (b) Dibromidobis(ethane-1,2-diamine)platinum(IV) nitrate A [CBSE OD Set-2 2019]



Ans. (i) [Co(C2O4)3]3– (ii) [Co(C2O4)3]3– (iii) [CoF6]3– (iv) [Co(C2O4)3]3– [CBSE Marking Scheme, 2019] ½ × 4



Q. 3. Write the hybridization and magnetic character of the following complexes:      (i) [Fe(H2O)6]2+    (ii) [Fe(CO)5] U [CBSE Delhi Set 2 2019]

The outermost electronic configuration of Fe is 3d6 4s2 4p0. H 2O is weak field ligand and do not cause pairing up of electrons.



(i) D0 < P and



(ii) D0 > P

A [CBSE OD Set-2 2019]  Ans. (a) d2sp3, diamagnetic ½+½ (b) (i) t2g4eg2 (ii) t2g6eg0 ½+½ [CBSE Marking Scheme, 2019]

xx xx x x xx xx

Detailed Answer : 4 (a) Oxidation number of Fe is + 2 in éë Fe (CN)6 ùû

Fe2+ – 3d6

xx xx xx

Q. 4. Define the following terms with a suitable example of each :

3d



4p

In this complex, there is no unpaired electron. Therefore, it involves dsp3 hybridisation and is diamagnetic. 2

(a) Chelate complex (b) Ambidentate ligand



3d 4s 4p 4d In this complex, there are 4 unpaired electrons. Therefore, it involves sp3d 2 hybridisation and is paramagnetic. (ii)  In [Fe(CO)5] the oxidation state of Fe is zero. The outermost electronic configuration of Fe is 3d6 4s2 4p0. CO is strong field ligand, causes pairing up of 4s electrons into the 3d orbitals.

4s

[CBSE Marking Scheme, 2019] 1



(i) In [Fe(H2O)6]2+the oxidation state of Fe is +2.

3d

1

Q. 6. (a)  Using valence bond theory, write the hybridization and magnetic character of the complex [Fe(CN)6]4- . (Atomic no. of Fe=26) (b) Write the electronic configuration of d6 on the basis of crystal field theory when :

Detailed Answer :

xx

(i) [Co(NH3)4(H2O)2]Cl3

(ii) [Pt Br2(en)2](NO3)2

Ans. (i) sp3d2, Paramagnetic ½+½ 3 2 (ii) sp d , Paramagnetic [CBSE Marking Scheme, 2019] ½ +½

xx

[Topper’s Answer 2019]



4s

4p

4d

–

CN being strong ligand pairs with 3d orbital of Fe2+



3d



–

4s

4p

Six CN ions give their six pairs of electron to the empty orbitals. The orbitals utilized for hybridization is d2 sp3. Magnetic character – Diamagnetic

[ 265

COORDINATION COMPOUNDS

Q. 7. (a) Although both [NiCl4]2– and [Ni(CO)4] have sp3 hybridisation yet [NiCl4]2– is paramagnetic and [Ni(CO)4] is diamagnetic. Give reason. (Atomic no. of Ni = 28) (b) Write the electronic configuration of d5 on the basis of crystal field theory when.

(i) Do < P and (ii) Do > P 





U [CBSE OD Set-3 2019] Ans. (a) In [NiCl4]2–, Cl– is a weak field ligand due to which there are two unpaired electrons in 3d orbital whereas in [Ni(CN)4]2–, CN- is a strong field ligand due to which pairing leads to no unpaired electron in 3d- orbital/ or structural representation. ½ + ½

(b) (i)

3

t2g eg2

(ii)

t2g5eg0

½+½

[CBSE Marking Scheme, 2019] Detailed Answer : (a) In [NiCl4]2– Ni is in +2 oxidation state and each Cl– donates a pair of electron. So, Cl– acts as a weak ligand and does not cause any forced pairing. Thus, electrons remains unpaired making it paramagnetic.



In [Ni(CO)4], Ni is in zero oxidation state and CO acts as a strong ligand causing forced pairing. Thus, no electron remains unpaired making it diamagnetic. Q. 8. (i) Using crystal field theory, write the electronic configuration of iron ion in the following complex ion. Also predict its magnetic behaviour: [Fe(H2O)6]2+ (ii)  Write the IUPAC name of the coordination complex: [CoCl2(en)2]NO3 Ans. (i) t2g4eg2 Paramagnetic

(ii) D  ichloridobis (ethane-1,2-diamine)cobalt (III) nitrate [1] Q. 9. (i) Predict the geometry of [Ni(CN)4]2− (ii) Calculate the spin only magnetic moment of [Cu(NH3)4]2+ ion. [A&E][CBSE SQP 2021] Ans. (i) Square planar (ii) Cu

2+

[1]

9

= 3d unpaired electron so

( 3) = 1.73 BM 

Short Answer Type Questions-II

[1]

(3 marks each)

(iii) Ni(CO)4 : The outermost electronic configuration will be 3d8 4s2 4po. CO is the strong ligand, causes pairing up of the 4s electrons into the 3d electrons

Q. 1. (i)  Draw the geometrical isomers of complex [Pt(NH3)2Cl2]. (ii)   On the basis of crystal field theory, write the electronic configuration for d4 ion if ∆o < P. (iii) Write the hybridization and magnetic behaviour of the complex [Ni(CO)4]. (At. no. of Ni = 28) A [CBSE Delhi 2015] 

[½ + ½]

Ans. (i) Thus, the hybridization will be sp3 (tetrahedral) Ni(CO)4 will be diamagnetic in nature.

eg

3

Electronic configuration : t



Cis-diamminedichloro Trans-diamminedichloro platinum (II) platinum (II) (ii)  When ∆o < P, it is weak field and high spin situation. As a result, one electron entered in eg orbital and 3 electrons in t2g.1

Q. 2. (i)  W  hat type of isomerism is shown by the complex [Cr(H2O)6]Cl3? (ii)   On the basis of crystal field theory, write the electronic configuration for d4 ion if ∆o > P. (iii) Write the hybridization and shape of [CoF6]3−. (Atomic number of Co = 27) A [CBSE OD 2015]  Ans. (i) Hydration isomerism 1 (ii) Electronic configuration is t42g or by diagram. 1 (iii) H  ybridization is sp3d2 and shape is octahedral. ½+½

t2g

2g

e1g 1

Q. 3. (i) For the complex [Fe(CN)6]3−, write the hybridization, magnetic character and spin nature of the complex. (At. number : Fe = 26). (ii) Draw one of the geometrical isomers of the complex [Pt(en)2Cl2]2+ which is optically active. A [CBSE Delhi 2016] 



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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Ans. (i) [Fe(CN)6]3−

(ii)



1

Commonly Made Error  Some students write configuration of ions.

incorrect

electronic

Answering Tip  Do practice for writing the electronic configuration of metal atom in coordination compounds.



Q. 4. (i)  For the complex [Fe(H2O)6]3+, write the hybridization, magnetic character and spin of the complex. (At. number : Fe = 26). (ii) Draw one of the geometrical isomers of the complex [Pt(en)2Cl2]2+ which is optically A [CBSE OD 2016] inactive. Ans. (i) [Fe(H2O)6]3+

Since H2O is a weak field ligand, it cannot cause pairing of electrons. Therefore, the number of unpaired electrons is 5. m = n( n + 2 ) = 5( 5 + 2 ) = 35 = 5.92 BM Thus, it is strongly paramagnetic (due to presence of unpaired electrons). In [Fe(H2O)6]3+ outer d-orbitals are used in hybridization to form high spin complex. (ii)  Geometrical isomers of [Pt(en)2Cl2]2+

trans-[PtCl2(en)2]2+ is optically inactive

Q. 5. (i) Define crystal field splitting energy. On the basis of crystal field theory, write the electronic configuration for d4 ion if Do < P. (ii) [Ni(CN)4]2– is colourless whereas [Ni(H2O)6]2+ is green. Why? (At. no. of Ni = 28) [CBSE Foreign Set-1 2017] Ans. (i) It is the magnitude of difference in energy between the two sets of d orbital i.e. t2g and eg 1 t32ge1g 1 (ii) In [Ni(H2O)6]2+, Ni+2(3d8) has two unpaired electrons which do not pair up in the presence of weak field ligand H2O. 1 [CBSE Marking Scheme 2017] Detailed Answer: (i) The difference in energy between the two sets of d-orbital (t2g and eg) caused by splitting of the degenerate levels due to the presence of ligands in a definite geometry. Electronic configuration: t32ge1g (ii) In [Ni(H2O)6]2+, Ni is present in +2 state with the configuration 3d8. It has two unpaired electrons which do not pair up in the presence of the weak H2O ligand. Therefore, it is green in colour. While undergoing d–d transition, red light is absorbed and complementary light emitted is green. In case of [Ni(CN)4]2–, Ni is in +2 state with the configuration 3d8 but in presence of the strong CN– ligand, the two unpaired electrons in the 3d orbitals undergoes pairing. As there is no unpaired electron present, it is colourless. Q. 6. (i) Define crystal field splitting energy. On the basis of crystal field theory, write the electronic configuration for d4 ion if Do > P. (ii) [Ni(CN)4]2– is diamagnetic whereas [NiCl4]2– is paramagnetic. Give reason. (At. no. of Ni = 28) R + A + A&E [CBSE Foreign Set-2 2017] Ans. (i) It is the magnitude of difference in energy between the two sets of d orbital i.e. t2g and eg 1 t42geg0 1

[ 267

COORDINATION COMPOUNDS

(ii) In [Ni(CN)4]2-, CN- is a strong field ligand and pairing takes place whereas in [NiCl4]2-, due to the presence of Cl-, a weak field ligand no pairing occurs/diagrammatic representation. 1 [CBSE Marking Scheme 2017] Q. 7. A metal ion Mn+ having d4 valence electronic configuration combines with three bidentate ligands to form a complex compound. Assuming Do > P. (i) Write the electronic configuration of d4 ion.. (ii) What type of hybridisation will Mn+ ion has? (iii) Name the type of isomerism exhibited by this complex. A [CBSE SQP 2017] Ans. (i) t24 g e 0g

1

(ii) sp3d2 (iii) optical isomerism

1 1 [CBSE Marking Scheme 2017]

Q. 8. (a) Write the formula of the following coordi­nation compound : Iron(III) hexacyanoferrate(II) (b) What type of isomerism is exhibited by the complex [Co(NH3)5Cl]SO4? (c) Write the hybridization and number of unpaired electrons in the complex [CoF6]3 . (Atomic No. of Co = 27) A [CBSE 2018] 3 Ans. (a) Fe4[Fe(CN)6]3 1 (b) ionisation isomerism 1 3 2 (c) sp d , 4  [CBSE Marking Scheme 2018] ½, ½



.





Visual Case based Questions Q. 1. Read the passage given below and answer the following questions: (1×4=4) The crystal field theory (CFT) is an electrostatic model which considers the metal-ligand bond to be ionic arising purely from electrostatic interactions between the metal ion and the ligand. Ligands are treated as point charges in case of anions or dipoles in case of neutral molecules. The five d orbitals in an isolated gaseous metal atom/ion have same energy, i.e., they are degenerate. This degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the metal atom/ ion. However, when this negative field is due to ligands (either anions or the negative ends of dipolar molecules like NH3 and H2O) in a complex, it becomes asymmetrical and the degeneracy of the d orbitals is lifted. It results in splitting of the d orbitals. The following questions are multiple choice questions. Choose the most appropriate answer :

[Topper’s Answer 2018]

(4 marks each)

(i) The colour of the coordination compounds depends on the crystal field splitting. What will be the correct order of absorption of wavelength of light in the visible region, for the complexes, [Co(NH3)6]3+, [Co(CN)6]3−, [Co(H2O)6]3+ (a)  [Co(CN)6]3− > [Co(NH3)6]3+ > [Co(H2O)6]3+ (b)  [Co(NH3)6]3+ > [Co(H2O)6]3+ > [Co(CN)6]3− (c)  [Co(H2O)6]3+ > [Co(NH3)6]3+> [Co(CN)6]3− (d)  [Co(CN)6]3−> [Co(NH3)6]3+> [Co(H2O)6]3+ Ans: Correct option: (c) Explanation : [Co(H2O)6]3+ > [Co(NH3)6]3+> [Co( CN)6]3− [1] (ii) The CFSE for octahedral [CoCl6]4− is 18,000 cm−1. The CFSE for tetrahedral [CoCl4]2− will be (a)  18,000 cm−1 (b)  16,000 cm−1 (c)  8,000 cm−1 (d)  20,000 cm−1 Ans: Correct option: (c) Explanation : 8,000 cm−1 [1]

268 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(iii)  An aqueous pink solution of cobalt(II) chloride changes to deep blue on addition of excess of HCl. This is because _____________. (a)  [Co(H2O)6]2+ is transformed into [CoCl6]4− (b)  [Co(H2O)6]2+ is transformed into [CoCl4]2− (c)  tetrahedral complexes have larger crystal field splitting than octahedral complex. (d)  None of the above Ans: Correct option: (b) Explanation : [Co(H2O)6]2+ is transformed into [CoCl4]2−. [1] (iv) A chelating agent has two or more than two donor atoms to bind to a single metal ion. Which of the following is not a chelating agent?. (a)  thiosulphato (b) oxalato (c)  glycinato (d)  ethane-1,2-diamine [1] Ans: Correct option: (a) Explanation : Thiosulphato is a monodentate ligand whereas Oxalato, glycinato and ethylene diamine are bidentate ligands and can form rings with the central metal ion. So, they all are also chelating ligands. Thiosulphato is a monodentate ligand and hence, cannot form chelate rings. Hence, it is not a chelating ligand Q. 2. According to Valence Bond Theory, the metal atom or ion under the influence of ligands can use its (n−1)d, ns, np or ns, np, nd orbitals for hybridisation to yield a set of equivalent orbitals of definite geometry such as octahedral, tetrahedral, square planar and so on. These hybridised orbitals are allowed to overlap with ligand orbitals that can donate electron pairs for bonding. In these questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement (d) Assertion is wrong statement but reason is correct statement. (i) Assertion : In the diamagnetic octahedral complex, [Co(NH3)6]3+, the cobalt ion is in +3 oxidation state. Reason : Six pairs of electrons, one from each NH3 molecule, occupy the six hybrid orbitals. Ans : Correct option: (b) Explanation : In the diamagnetic octahedral complex, [Co(NH3)6]3+, the cobalt ion is in +3 oxidation state and has the electronic configuration 3d6. (ii)  Assertion : [NiCl4]2− is an inner orbital complex. Reason : An inner orbital or low spin or spin paired complex uses inner d orbitals of the metal ion for hybridisation. Ans : Correct option: (d) Explanation : [NiCl4]2− is high spin complex. (iii) Assertion : In the square planar complexes, the hybridisation involved is dsp2. Reason : In [Ni(CN)4]2−. Here nickel is in +2 oxidation state and has the electronic configuration 3d8. Ans : Correct option: (b)

  (iv) Assertion : The paramagnetic octahedral complex, [CoF6]3− uses outer orbital (4d) in hybridisation (sp3d2). Reason : It is a high spin complex. Ans : Correct option: (c) Explanation : The paramagnetic octahedral complex, [CoF6]3− uses outer orbital (4d) in hybridisation (sp3d2 ). It is a low spin complex. ead the passage given below and answer the Q. 3. R following questions: (1×4=4) The existence of coordination compounds with the same formula but different arrangements of the ligands was crucial in the development of coordination chemistry. Two or more compounds with the same formula but different arrangements of the atoms are called isomers. Isomers are compounds with the same molecular formula but different structural formulas and do not necessarily share similar properties. There are many different classes of isomers, like stereoisomers, enantiomers, and geometrical isomers. There are two main forms of isomerism: structural isomerism and stereoisomerism. The different chemical formulas in structural isomers are caused either by a difference in what ligands are bonded to the central atoms or how the individual ligands are bonded to the central atoms. The following questions are multiple choice questions. Choose the most appropriate answer :   (i) Indicate the complex ion which shows geometrical isomerism. +     (a) [Cr(H (b) [Pt(NH 2O)4Cl2] 3)3Cl] 3+ 3−     (c) [Co(NH ) ] (d) [Co(CN) 3 6 5(NC)] Ans: Correct option: (a) Explanation : and (1)

(c) doc b

(2)

[1] (ii)  What kind of isomerism exists between [Cr(H2O)6]Cl3 (violet) and [Cr(H2O)5Cl]Cl2.H2O (greyish-green)? (a)  linkage isomerism (b)  solvate isomerism (c)  ionisation isomerism (d)  coordination isomerism Ans: Correct option: (b) Explanation : solvate isomerism [1] (iii) Which of the following complexes show linkage isomerism? (a)  [Co(H2O)5CO]3+ (b) [Cr(NH3)5SCN]2+ + (c)  [Fe(en)2Cl2] (d)  All of the above Ans: Correct option: (b) Explanation : Linkage isomerism is the existence of coordination compounds that have the same composition differing with the connectivity of the metal to a ligand. Typical ligands that give rise to linkage isomers are: thiocyanate, SCN −, isothiocyanate, NCS − [1] (iv)  Due to the presence of ambidentate ligands coordination compounds show isomerism. Palladium complexes of the type [Pd(C6H5)2(SCN)2] and [Pd(C6H5)2(NCS)2] are (a)  linkage isomers (b)  coordination isomers (c)  ionisation isomers (d)  geometrical isomers Ans: Correct option: (a) [1] Explanation : Same as above



[ 269

SELF-ASSESSMENT TEST

Self Assessment Test-9 Time : 1 Hour

Max. Marks : 25

1.  Read the passage given below and answer the following questions : (1 × 4 = 4) In an octahedral coordination entity with six ligands surrounding the metal atom/ion, there will be repulsion between the electrons in metal d orbitals and the electrons (or negative charges) of the ligands. Such a repulsion is more when the metal d orbital is directed towards the ligand than when it is away from the ligand. Thus, the dx2y2 and dz2 orbitals which point towards the axes along the direction of the ligand will experience more repulsion and will be raised in energy; and the dxy, dyz and dxz orbitals which are directed between the axes will be lowered in energy relative to the average energy in the spherical crystal field. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement (d) Assertion is wrong statement but reason is correct statement. (i) 

Assertion: The degeneracy of the d orbitals is removed in the octahedral complexes. Reason: The degeneracy is removed due to ligand electron-metal electron repulsions. (ii)  Assertion: The crystal field splitting, ∆o , depends upon the field produced by the ligand and charge on the metal ion. Reason: The energy of the two eg orbitals will increase by (3/5) ∆o and that of the three t2g will decrease by (2/5)∆o. (iii) Assertion: Spectrochemical series is theoretically determined series based on the absorption of light by complexes with different ligands. Reason: Ligands can be arranged in a series in the order of increasing field strength called the spectrochemical series. (iv) Assertion: In tetrahedral coordination entity formation, the d orbital splitting is inverted and is smaller as compared to the octahedral field splitting. Reason: The orbital splitting energies are not sufficiently large for forcing pairing and, therefore, low spin configurations are rarely observed in tetrahedral complexes. OR Assertion: [Fe(CN)6]3– complex is paramagnetic. Reason: Cyanide ion is strong field ligand hence favouring Spin pairing.

Following questions (No. 2 to 5) are Multiple Choice Questions Carrying 1 mark each. Q. 2  The correct IUPAC name of [Pt(NH3)2Cl2] is (a)  Diamminedichloridoplatinum (II) (b)  Diamminedichloridoplatinum (IV) (c)  Diamminedichloridoplatinum (I) (d)  Diamminedichloridoplatinum (IV)

A

Q 3.  Which of the following complexes are homoleptic? (a)  [Co(NH3)4Cl2]+ (b)  [Ni(CN)4]2− (c)  [NI(NH3)4Cl2] (d)  All of the above U Q 4. The stabilisation of coordination compounds due to chelation is called the chelate effect. Which of the following is the most stable complex species? (a)  [Fe(CO)5] (b)  [Fe(CN)6]3− (c)  [Fe(C2O4)3]3− (d)  [Fe(H2O)6]3+ U Q 5. Which of the following species is not expected to be a ligand? (a)  NO (b)  NH4+ (c)  NH2CH2CH2NH2 (d)  CO U The following questions (Q. no. 6 & 7), a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement (d) Assertion is wrong statement but reason is correct statement. Q 6. Assertion: [Fe(CN)6]3– involves d2sp3 hybridi­ sation with one unpaired electron and [Fe(H2O)6]3+ involves sp3d2 hybridisation with five unpaired electrons. Reason: This difference is due to the presence of strong ligand CN– and weak ligand H2O in these complexes. U Q 7. Assertion: [CoF6]3– and [Co(NH3)6]3+ , the former is paramagnetic and the latter is diamagnetic. Reason: The reason is presence of neutral ligands.  R Q. No. 8 & 9 are Short Answer Type-I Carrying 2 marks each. Q 8. Write the IUPAC name of the complex [Cr(NH3)4Cl2]+. What type of isomerism does it exhibit? A [2]

270 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q 9. Explain why [Fe(H2O)6]3+ has magnetic moment value of 5.92 BM whereas [Fe(CN)6]3− has a value of only 1.74 BM. Q. No. 10 & 11 are Short Answer Type-II Carrying 3 marks each. Q. 10. (i)   On the basis of crystal field theory, write the electronic configuration for d4 ion if ∆o < P. (ii)  Write the hybridization and magnetic behavi­ our of the complex [Ni(CO)4]. (At. no. of A [3] Ni = 28)  Q. 11.  (i)    Write the IUPAC name of the complex [Cr(NH3)4Cl2]Cl. (ii)  Why is [NiCl4]2− paramagnetic but [Ni(CO)4] is diamagnetic? (At. nos. : Cr = 24, Co = 27, Ni = 28)

Q.No 12 is Long Answer Type Carrying 5 marks each. Q 12.  Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex : (a)  K[Cr(H2O)2(C2O4)2].3H2O (b)  [Co(NH3)5Cl]Cl2 (c)  [CrCl3(py)3] (d)  Cs[FeCl4] A [NCERT] (e)  K4[Mn(CN)6] OR Specify the oxidation numbers of the metals in the  following coordination entities : (a)  [Co(H2O)(CN)(en)2]2+ (b)  [CoBr2(en)2]+ (c)  [PtCl4]2− (d)  K3[Fe(CN)6] A (e)  [Cr(NH3)3Cl3]

 

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HALOALKANES AND HALOARENES

[ 271

10 CHAPTER

HALOALKANES AND HALOARENES

Syllabus ¾¾ Haloalkanes : Nomenclature, nature of C–X bond, physical and chemical properties, optical rotation mechanism of substitution reactions. ¾¾ Haloarenes : Nature of C–X bond, substitution reactions (Directive influence of halogen in monosub­ stituted compounds only). Uses and environmental effects of - dichloromethane, trichloromethane, tetrachloromethane, iodoform, freons, DDT.

Trend Analysis List of concept names

2018 D/OD

IUPAC name Structure of Haloalkanes and Haloarenes SN1 Reaction, SN2 Reaction, Optical Activity and b-elimination Reaction Structure of Haloalkanes and Haloarenes Other Chemical Reactions

1Q (3 marks) 1Q (1 mark) 1Q (2 marks) 1Q (1 mark) 1Q (1 mark)

2019

2020

D 2Q (1 mark)

OD

D

2Q (1 mark) 1Q (3 marks) 1Q (3 marks) 1Q (3 marks)

2Q (1 mark) 1Q (3 marks) 1Q (3 marks) 1Q (3 marks)

2Q (1 mark)

OD

3Q (1 mark) 1Q (3 marks) 1Q (3 marks) 1Q (3 marks)

TOPIC-1

Haloalkanes and their Properties Revision Notes  Haloalkanes are aliphatic hydrocarbons where a hydrogen atom is replaced by halogen, while haloarenes are aromatic hydrocarbons where hydrogen in the benzene ring is replaced with halogen atom.

TOPIC - 1 Haloalkanes and their Properties .... P. 272 TOPIC - 2 Haloarenes and Polyhalogen Compounds .... P. 289

 Halogen atom is attached to sp3 hybridised carbon atom in haloalkanes while in haloarenes it is attached to sp2 hybridised carbon atom of the aryl group.

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 Classification : l On the basis of the number of halogen atom : These may be classified as mono, di or polyhalogen compounds depending on whether they contain one, two or more halogen atoms in their structures. For example,

Haloalkanes, haloarenesPhysical properties

[ 273

HALOALKANES AND HALOARENES

| 2 CH–X | CH2 –X Trihaloalkane

CH2 X | CH2 X Dihaloalkane

C2 H5X Monohaloalkane

X

X

X

X

X

Dihaloarene

Trihaloarene

X



Monohaloarene



Dihalogen compounds may be further classified as : (i) Geminal dihalides : Two halogen atoms are attached to the same carbon atom. For example,

H



H

Cl

H

C

C

C

H

Cl

H

H

H

H

H

C

C

H

Cl

Cl

Isopropylidene chloride Ethylidene chloride (2, 2–Dichloropropane) (1, 1–Dichloroethane) (ii) Vicinal dihalides : Two halogen atoms are attached to adjacent carbon atoms. For example, Cl | CH3 — CH — CH2 — Cl Propylene dichloride (1, 2-Dichloropropane)

Cl — CH2 — CH2 — Cl Ethylene dichloride (1, 2-Dichloroethane)

l On the basis of sp3 hybridisation : (i) Haloalkanes or alkyl halides (R—X) : General formula is CnH2n+1 X. They are further classified as primary, secondary and tertiary haloalkanes. H R

C

R R

X

H

R

X

H

Primary haloalkane



C

R C

X

R

Secondary haloalkane

Tertiary haloalkane

(ii) Allylic halides : Halogen is bonded to allylic carbon. X

CH2 =CH — CH —X



Allyl halide (3-Halo–1– propene)

3-Halocylohex–1–ene

(iii) Benzylic halides : Halogen atom is bonded to an sp3 hybridised carbon atom next to an aromatic ring. —

R1

C— X

—

CH2 X

Benzyl halide (1o )

R2

Here, R1=CH3, R2 =H (2°) or R1= R2 = CH 3 (3°)

2

l On the basis of sp hybridisation : (i) Vinylic halides : Halogen is bonded to one of the carbon atoms of a vinylic carbon.

CH2 =CH — X Vinyl halide



1–Halocylohex–1 –ene

(ii) Aryl halides : Halogen atom is directly bonded to sp2 hybridised carbon atom of an aromatic ring. X

Halobenzene







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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Common and IUPAC name of some halides : Structure

IUPAC Name

Common names

CH3Cl

Chloromethane

Methyl chloride

CH3CH2Cl

Chloroethane

Ethyl chloride

CH3 – CH2 – CH2Cl

1-Chloropropane

n-Propyl chloride

CH3 — CH — CH3

2-Chloropropane

Isopropyl chloride

CH3—CH2—CH2—CH2Cl

1-Chlorobutane

n-Butyl chloride

CH3 — CH — CH2Cl

1-Chloro-2-methyl propane

Isobutyl chloride

2-Chlorobutane

Sec. butyl chloride

2-Chloro-2-methyl propane

Tertiary butyl chloride

CH3F

Fluoromethane

Methyl fluoride

CH3Br

Bromomethane

Methyl bromide

CH3I

Iodomethane

Methyl iodide

CH2Cl2

Dichloromethane

Methylene dichloride

CHCl3

Trichloromethane

Chloroform

CCl4

Tetrachloromethane

Carbon tetrachloride

CHBr3

Tribromomethane

Bromoform

CHI3

Triiodomethane

Iodoform

ClCH2—CH2Cl

1, 2-Dichloroethane

Ethylene dichloride

CH3CHCl2

1, 1-Dichloroethane

Ethylidene chloride

CH2 = CHCl

Chloroethene

Vinyl chloride

CH2 = CH—CH2Br

3-Bromopropene

Allyl bromide

(CH3)3CCH2Br

1-Bromo-2, 2-dimethylpropane

neo-pentylbromide

CF3CF2CF3

Octafluoropropane

Perfluoropropane

CCl2F2

Dichlorodifluoromethane

Freon

CHCl2—CHCl2

1, 1, 2, 2-Tetrachloroethane

Acetylene tetrachloride

CHCl = CCl2

1, 1, 2-Trichloroethene

Acetylene trichloride

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—

IUPAC name and Preparation of Haloalkanes

—

Cl

CH3

CH3—CH2—CH—CH3 | Cl —

CH3

—

CH3 — C — CH3 Cl

F



Boiling Point

Benzene 80°CH C Cl 2

Cl

Fluorobenzene 85° CCHCl

2

Br

I

Chlorobenzene Bromobenzene 132° C CCl3 156° C

Iodobenzene Cl 189° C CH3

IUP AC nam e

1–Chloro –1– phenylmethane

Dichlorophenyl methane

Trichlorophenyl methane

1–Chloro–2–methyl benzene

Commo n nam e

(Benzyl chloride)

(Benzylidene chloride) (Benzal chloride)

(Benzotrichloride)

(2–Chlorotoluene)

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Nomenclature

[ 275



HALOALKANES AND HALOARENES

 Methods of preparation of Haloalkanes :



(a) From alcohols : Alkyl halides are prepared from alcohols, which are easily accessible. ZnCl



R – OH + HX 2 → R – X + H2O (X = Cl, Br, I)



R – OH + NaBr + H2SO4 ⎯→ R – Br + NaHSO4 + H2O



3R – OH + PX3 ⎯→ 3R – X + H3PO3 (X = Cl, Br)



R – OH + PCl5 ⎯→ R – Cl + POCl3 + HCl



Red P/X

2 R – OH X → R–X =Br ,I 2

2

2



R – OH + SOCl2 ⎯→ R – Cl +SO2 + HCl



The reactions of primary and secondary alcohols with HX require the presence of the anhydrous ZnCl2.

(b) From hydrocarbons : By free radical halogenation. Cl /UV Light

2 → CH3CH2CH2CH2Cl + CH3CH2CHClCH3 (Major Product) CH3CH2CH2CH3  Or heat





n-Butane

n-Butyl chloride

Sec.-Butyl chloride

(c) From alkenes : (i) Addition of hydrogen halide : C=C +H—X

C—C

+

—



—



X

H

(X = Cl, Br, I)

 – Markovnikov's Rule

R — CH — CH3

—

R — CH = CH2 + H — X

X Benzoyl peroxide R – CH = CH2 + H – Br  → R – CH2 – CH2 – Br Anti Markovnikov's addition



(ii) Addition of Halogens : H



H

H C=C

H

+ Br2

CCl4

BrCH2 — CH2Br Vic-Dibromide

(d) Halide Exchange : (i) By Finkelstein Reaction : Dry acetone

R – X + NaI → R – I + NaX (X = Cl, Br) (ii) By Swarts Reaction : AgF, Hg F ,CoF or SbF

2 2 2 3 → R–F R – X 

CH3 – Br + AgF ⎯→ CH3 – F + AgBr

 Nature of C-X bond in haloalkanes : The carbon-halogen bond is polarised. Carbon atom holds partial positive charge and halogen atom holds partial negative charge. This occurs due to difference in electronegativity. Halogens are more electronegative than carbon. Size of the halogen atoms increases down the group. Fluorine is the smallest and iodine is the largest. Carbon-halogen bond length also increases from C-F to C-I.  Physical properties of haloalkanes : Haloalkanes are colourless when pure but compounds of bromine are coloured. (i) Melting and boiling points : Haloalkanes, due to polar and strong dipole-interactions between their molecules, have high B.P. and M.P. This increase in B.P. and M.P. depends on increasing size, mass of halogens and magnitude of van der Waals forces of attractions. The increasing order is RF< RCl < RBr < RI. With respect to isomeric alkyl halides, B.P. decreases with increase in branching due to less surface area and weak inter-particle forces.

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



(ii) Density : It increases with increase in carbon atoms, halogen atoms and atomic mass of the halogen atoms. (iii) Solubility : Haloalkanes are insoluble in water but are soluble in organic solvents.  Chemical properties of haloalkanes : Reactivity of haloalkanes depends on the C-X bond cleavage. Higher the bond dissociation energy of C-X bond, lesser will be the reactivity. Dissociation energy of C-X bond decreases with increase in the halogen size, i.e., C–Cl > C–Br > C–I. Reactivity order of haloalkanes would be :

Chlorides < Bromides < Iodides There are four types of chemical reactions with haloalkanes :

(a) Nucleophilic substitution reactions : When an atom or group of atoms is replaced by a nucleophile, the reaction is called nucleophilic substitution reaction. e.g., – + – + : X C X + :Z C Z Nucleophile Product Leaving group KOH (aq) R — OH + KX Alcohol + – NaOR R — O — R' + NaX Ether

KCN(alc) AgCN

R — NC

Alkyl halide

+

+

KX AgX

Alkyl isocyanide

KNO2

R—X

R — CN

Alkyl cyanide

R—O—N=O Alkyl nitrite

AgNO2

R — NO2

Nitroalkane

O

O

R' — C — O — Ag

Ester

LiAlH4

R—H Alkane

NH3

R — NH2 + H — X

Amine

NaSH – Na C ≡ CH +

R — SH + NaX

Thioalcohol

R — C ≡ CH + NaX Higher alkyne



R — C — O — R' + AgX

Alkyl halides undergo nucleophilic substitution reactions.

..

.. –

CH3CH2Br +..OH CH3CH2OH + .. Br .. (b) Elimination reaction : Alkyl halides undergo b-elimination of hydrogen atom from b-carbon atom and halogen atom to form alkenes on being heated with KOH (alc.) or KNH2. e.g., CH3 CH3 —

—

CH3 — C — Cl + KOH







CH3

.. ..

–

ethanol

C = CH2 + KCl + H2O

heat

CH3

The reaction is called dehydrohalogenation. H Cl | H H |   H — C— C — H C=C + H2 O + Cl– | | H H H H –OH

[ 277

HALOALKANES AND HALOARENES

The following is order of reactivity : RCl < RBr < RI



RCH2X < R2CHX < R3CX





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(c) Reaction with metals :

(1) Reaction with magnesium : R–X



Chemical Reactions

Dry ether

Mg  → RMgX

+



Alkyl halide



CH3 – CH2 – Br + Mg  →



(2) Reaction with sodium (Wurtz reaction) :



Alkyl magnesium halide (Grignard reagent) Dry ether

CH3 – CH2 – MgBr

Ethyl magnesium bromide ether R – X + 2Na + X – R Dry  → R – R + 2NaX Dry ether

→ CH3 – CH2 – CH2 – CH3 + 2NaCl CH3 – CH2 – Cl + 2Na + Cl – CH2 – CH3  n-Butane

(d) Reduction : / HC l (conc.) R – X + 2(H) ⎯Zn ⎯⎯⎯⎯⎯ → R–H+H–X Zn / HC l (conc.)

C | R3 (Planar)

R3

CH3 — C — Cl

—

(slow)

..

–

CH3 — C + : Cl .. :

CH3

CH3

STEP II :

C | R3

+ Nu

CH3

R1 R2 R3

C—Nu

Or

+

CH3 — C + OH

—

–

CH3

–

CH3 — C — OH

CH3

—

R1

—

R2



Or

—



+ X–

—

slow

C—X

—

R2

—

CH3 – CH2 – Cl + 2(H) ⎯⎯⎯⎯⎯⎯→ CH3 – CH3 + HCl  Mechanism of Nucleophilic substitution reaction : Alkyl halides undergo two types of nucleophilic substitution reactions. (i) Unimolecular nucleophilic substitution reaction (SN1) : Those substitution reactions in which rate of reaction depends upon the concentration of only one of the reactants, i.e., alkyl halides are called SN1 reactions, e.g., hydrolysis of tertiary butyl chloride follows SN1 reaction. This reaction takes place in two steps: STEP I : R2 CH3 CH3 R1 R1

CH3

(Planar)

The slowest step is rate determining step which involves one species only. Therefore, rate of reaction depends only on the concentration of tertiary butyl chloride. Polar protic solvents like water, alcohol favour SN1 because they stabilize carbocation by solvation. Tertiary alkyl halides follow SN1 mechanism. (ii) Biomolecular nucleophilic substitution reaction (SN2) : The reaction whose rate depends on the concentration of two species, alkyl halide and nucleophile. They involve one step mechanism. Back side attack of nucleophile and departing of leaving group take place simultaneously.



Non-polar solvents favour SN2 mechanism. Primary alkyl halides follow SN2 mechanism.







278 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Reactivity of SN1 and SN2 mechanisms : S = Substitution, N = Nucleophilic, 2 = Bimolecular, 1 = Unimolecular. Nucleophilic substitution seldom occurs exclusively by one mechanism only. With a given halogen, primary, secondary and tertiary halides show opposite order of reactivity in the two mechanisms.

SN1 reactivity increases R'' CH3X , RCH2X , R — CH — X , R — C — X 1°

R' 2°

R' 3°







SN2 reactivity increases



  





Stereoisomerism : Isomerism exhibited by two or more compounds with the same molecular and structural formula, but different spatial arrangements of atoms or groups in space is called stereoisomerism. Plane-polarised light : The beam of light whose oscillations or vibrations are confined to one plane only is called plane-polarised light. It is obtained by passing a monochromatic light (light of single wavelength) through a nicol prism. Nicol prism : A nicol prism is a special type of prism made from calcite, a special crystalline form of calcium carbonate. It is a device for producing plane polarised light. Optical rotation : Property of rotating the plane of polarisation either towards left or right. Dextrorotatory : Those substances which rotate the plane of polarisation of light towards right, i.e., in clockwise direction are called dextrorotatory. It is conventionally given a positive sign. It is denoted by ‘d’ and a positive (+) sign is placed before the degree of rotation. Laevorotatory : Those substances which rotate the plane of polarisation of light towards the left, i.e., in anticlockwise direction are called laevorotatory. It is denoted by ‘l’ and a negative (–) sign is placed before the degree of rotation. Specific rotation : The extent of experimentally observed angle of rotation (optical rotation, represented by αobs) of a substance depends upon the following factors : (i) nature of substance, (ii) wavelength of the light used, (iii) the number of optically active molecules in the path of light beam (which depends upon concentration of sample), (iv) length of polarimeter tube, (v) solvent used. O



observed rotation (a

)

t C obs Specific rotation [α ]D = length of tube (d ) × concentration of solution (g mol -1 ) m

 Optically active substances : Those substances which rotate the plane of polarisation of plane-polarised light when it is passed through their solutions are called optically active substances. This phenomenon is called optical activity.  Polarimeter : The angle of rotation by which the plane-polarised light is rotated, can be measured by using an instrument called polarimeter. A schematic diagram of a polarimeter is shown in the figure below : α

Light Source a sodium lamp

Unpolarised Polariser light (Nicol pr ism)

Planepolarised light

Polarimeter tube

Rotated plane of polarised light

Analyser (Movable Nicol prism)

Viewer







Enantiomers : Those stereoisomers which are mirror images of each other but non-superimposable are called enantiomers, e.g., d(+) glucose and l(–) glucose are enantiomers.  Asymmetric molecule : If all the four substituents attached to carbon are different, the resulting molecule will lack symmetry. Such a molecule is called asymmetric molecule. Asymmetry of molecule is responsible for optical activity in such organic compounds.

[ 279

HALOALKANES AND HALOARENES





Symmetrical objects : Those objects whose projections are superimposable on their mirror images are symmetrical objects, e.g., a sphere, a cube, a cone, a tetrahedron are all identical to their mirror images and can thus be superimposed.  Chiral : An object which is non-superimposable on its mirror image is said to be chiral. The property of being chiral is known as chirality. A chiral object is also called dissymmetric. e.g.,



Non-superimposable hands  Achiral : Achiral objects are those objects which are superimposable on their mirror images.  Asymmetric carbon (Chiral carbon) : The carbon atom which is attached with four different groups of atoms is called asymmetric or chiral carbon atom.  Racemic mixture : A mixture containing equal amounts of enantiomers which does not show any optical activity. It is optically inactive due to external compensation.  Racemisation : The process of conversion of an enantiomer into racemic mixture is known as racemisation.  Absolute configuration : The three dimensional structure of a molecule that has one or more centres of chirality is referred to its absolute configuration.  Diastereoisomers : Those pairs of stereoisomers which are not mirror images of each other and are nonsuperimposable. (i) Diastereoisomers have different physical properties.



(ii) Diastereoisomers differ in magnitude of specific rotation. (iii) A compound with two chiral centres does not always have four stereoisomers.



 Example of racemisation in SN1 mechanism : When optically active alkyl halide undergoes SN1 mechanism, it is accompanied by racemisation because intermediate carbocation formed is sp2-hybridised and polar. Nucleophile (OH–) has equal probability of attacking it from either side leading to formation of equal amount of dextro and laevo-rotatory alcohols. As the products of SN1 mechanism has both inversion as well as rotation, the products formed by this reaction would be racemic mixture of alcohols.









Meso compounds : Those compounds which have two or more (even number) chiral carbon atoms and have an internal plane of symmetry are called meso compounds. They are optically inactive due to internal compensation.  Example of inversion in SN2 mechanism : When optically active d (+) alkyl halide is treated with OH–, we get optically active l (–) alcohol due to back side attack of nucleophile.

280 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



When d(+)-2-bromooctane is reacted with NaOH, l(–)-2-octanol is formed. Thus, SN2 mechanism leads to inversion of configuration. This inversion is called Walden’s inversion. l Important reactions of chloroethane :

Know the Terms  Synthetic tools : Alkyl halides are regarded as synthetic tools in the hands of chemistry due to their highly reactive nature.  Relative configuration : Arrangement of atoms in space of a stereo isomer of a compound relative to another compound chosen as arbitrary standard (like glyceraldehyde).

Q. What happens when : (i)  CH3–Cl is treated with aqueous KOH ? (ii) CH3–Cl is treated with KCN ? (iii) CH3–Br is treated with Mg in the presence of dry ether ? Solution STEP-1: (i) CH3 – Cl + KOH ® CH3OH + KCl (aq) Methanol

STEP-2: (ii) CH3 – Cl + KCN → CH3CN + KCl Methyl cyanide STEP-3: (iii) CH3Br + Mg ether ¾Dry ¾¾¾ ® CH3MgBr Methyl magnesium bromide

[ 281

HALOALKANES AND HALOARENES

Objective Type Questions



(i) CH3CH2 –CH2–OH



(ii) CH3CH 2  CH  OH | CH3 CH3 | (iii) CH3CH 2  C  OH | CH3





(a) (i) > (ii) > (iii) (b) (iii) > (ii) > (i) (c) (ii) > (i) > (iii) (d) (i) > (iii) > (ii) U [NCERT Exemp. Q. 1, Page 133] Ans. Correct option : (b) Explanation : The reactivity order of alcohols towards halogen acids is 3o>2o>1o as the stability of carbocations is of the order 3o>2o>1o. [1] Q. 2. Which of the following alcohols will yield the corresponding alkyl chloride on reaction with concentrated HCl at room temperature?

(a) CH3CH2 – CH2 – OH



(b) CH3CH 2  CH  OH | CH3 (c) CH3CH 2  CH  CH 2OH | CH3



CH3 |  C  OH CH CH (d) 3 2 | CH3



A [NCERT Exemp. Q. 2, Page 133]

Ans. Correct option : (d)

Explanation : As tertiary carbocation is more stable, so tertiary alcohols will yield the corresponding alkyl chloride on reaction with concentrated HCl at room temperature. While primary and secondary alcohols require the presence of a catalyst ZnCl2. [1]

Q. 3. Arrange the following compounds in increasing order of their boiling points: (i)

CH3 —

CH—CH2Br CH3 — (ii) CH3CH2CH2CH2Br

CH3 | (iii) CH3  C  CH3 | Br (a) (ii) < (i) < (iii)

(b) (i) < (ii) < (iii)



(d) (iii) < (ii) < (i)



[A] MULTIPLE CHOICE QUESTIONS : Q. 1. The order of reactivity of following alcohols with halogen acids is :

(1 mark each)

(c) (iii) < (i) < (ii)

U [NCERT Exemp. Q. 8, Page 135]

Ans. Correct option : (c) Explanation : Boiling points of isomeric haloalkanes decrease with increase in branching as with increase in branching surface area decreases which leads to decrease in intermolecular forces. [1] Q. 4. The conversion of an alkyl halide into an alcohol by aqueous NaOH is classified as

(a) a dehydrohalogenation reaction



(b) a substitution reaction



(c) an addition reaction



(d) a dehydration reaction R [CBSE Delhi Set 1 2020]

Ans. Correct option : (b) Explanation : R − X + NaOH ® R − OH + NaX [1] [B] ASSERTION & REASON TYPE QUESTIONS : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Q. 1. Assertion (A): Aryl halides undergo nucleophilic substitution reactions with ease. Reason(R) : The carbon halogen bond in aryl halides has partial double bond character. R [CBSE SQP 2020] Ans. Correct option: (d)

Explanation: Aryl halides are less reactive towards nucleophilic substitution reactions because of the carbon halogen bond in aryl halides has partial double bond character. [1]

Q. 2. Assertion (A): Hydrolysis of (–)-2-bromooctane proceeds with inversion of configuration.

Reason (R): This reaction proceeds through the formation of a carbocation. U [NCERT Exemplar]

282 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Explanation: Hydrolysis of (–)-2-bromooctane proceeds through the formation of a carbocation following SN1 reaction. [1]

Q. 3. Assertion (A): tert-Butyl bromide undergoes Wurtz reaction to give 2, 2, 3, 3-tetramethylbutane

Reason (R): In Wurtz reaction, alkyl halides react with sodium in dry ether to give hydrocarbon containing double the number of carbon atoms present in the halide. U [NCERT Exemplar] Ans. Correct option: (a) Explanation: In Wurtz reaction, alkyl halides react with sodium in dry ether to give hydrocarbon containing double the number of carbon atoms present in the halide so tert-Butyl bromide undergoes Wurtz reaction to give 2, 2, 3, 3-tetramethylbutane.[1]

Ans.

2-Bromo-2 methylpropane

 1 [CBSE Marking Scheme, 2016]

 



Q. 3. Write the structure of an isomer of compound C4H9Br which is most reactive towards SN1 reaction.  A [CBSE OD 2016]



Ans. Correct option: (d)



Commonly Made Error  Students often get confused between SN1 and SN2 reaction.

Answering Tip  Assign proper numbering to carbon atom before drawing the branch chains.

Q. 1. Which would undergo SN2 reaction faster in the following pair and Why?

—

Q. 4. Out of CH3 — CH — CH2 — Cl and

[C] VERY SHORT ANSWER TYPE QUESTIONS :

CH3

—

CH — CH — CH — Cl, which is more reactive 3 2



A&E [CBSE Delhi Set-1, 2, 3 2015]

Ans.  CH3 — CH2 — Br would undergo SN2 reaction faster it has less steric hindrance than tert - butyl bromide. 1 CH3 — CH2 — Br Because it is a primary halide / (1*) halide ½ + ½ [CBSE Marking Scheme, 2015]



Commonly Made Error  Sometimes, students can not identify correct alkyl halide for given reaction.

Answering Tip

CH3 towards SN1 reaction and why?

Ans. 

is more reactive as being secondary halide it forms more stable carbocation intermediate than primary halides.1 [CBSE Marking Scheme, 2016]



Q. 5. Amongst the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields a single monochloride. 

 Clearly understand the concept of SN2 reaction.

U

U [CBSE SQP 2016]



Ans. Neopentane or 2, 2-Dimethylpropane.



Q. 2. Which would undergo SN1 reactions faster in the following pair :



Ans. 1

Ans. BrCH2CH = CHCH2Cl 1  [CBSE Marking Scheme, 2017]





Q. 6. Write the structure of 1-Bromo-4-chlorobut-2ene. A [CBSE Delhi Set-1 2017]

 

U [CBSE OD 2015]

1

Answering Tip  Clearly understand the concept of SN1 reaction.

Q. 7. Write the structure of 3-Bromo-2-methylprop-1ene. A [CBSE Delhi Set-3 2017] Ans. BrCH2(CH3)C = CH2 1  [CBSE Marking Scheme, 2017]

Commonly Made Error  Students often confuse in reactivity of alkyl halidesprimary, secondary, teritary towards SN1 and SN2 reactions.

[ 283

HALOALKANES AND HALOARENES

Ans. Neopentane / C(CH3)4 1  [CBSE Marking Scheme, 2017]

Ans. 2-Methylprop-1-ene/isobutene/ structure 1  [CBSE Marking Scheme, 2018]

Q. 9. Identify the compound that on hydrogenation produces an optically active compound from the following compounds:



Q. 10. Predict the major product formed when sodium ethoxide reacts with tert. Butyl chloride.  A [CBSE Comptt. Delhi/O.D. 2018]



Q. 8. Among the isomers of pentane (C5H12), write the one which on photochemical chlorination yields a single monochloride. A [CBSE Foreign Set-1 2017]

Detailed Answer:

2-Methylprop-1-ene (Isobutene)



Ans. B. 

(B) [CBSE SQP 2018-19]

Ans. Benzyl chloride; ½ Due to resonance, stable benzyl carbocation is formed.½  [CBSE Marking Scheme, 2018]

(A)

Q. 11. Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous NaOH and why? A&E [CBSE Delhi/OD 2018]

1 [CBSE Marking Scheme, 2018]

[Topper’s Answer 2018]

Ans. Nucleophiles having two nucleophilic centres. CN−/SCN−/NO2−(any one) [½ + ½]  [CBSE Marking Scheme, 2019] Detailed Answer: An ambident nucleophile is an anionic nucleophile in which the negative charge is delocalized by resonance over two unlike atoms.





Q. 13. Write the IUPAC name of the given compound :

A [CBSE Delhi Set-2 2019]



Ans. 4-chlorobenzensulphonic acid 1  [CBSE Marking Scheme, 2019] Q. 14. Write on the sterochemical difference between SN1 and SN2 reactions.  R [CBSE Delhi Set-3 2019] Ans.

Detailed Answer: Benzyl chloride gets easily hydrolysed by aqueous NaOH as chlorobenzene possesses partial double bond character in the C−Cl bond. The lone pairs delocalized in the ring strethens C−Cl bond reducing its reactivity. Whereas benzyl chloride undergoes SN1 reaction to form stable benzyl carbocation. Q. 12. Define ambident nucleophile with an example.  R [CBSE Delhi Set-1 2019]



SN1

SN 2

Produces racemic mixture

Proceed with inversion

 1 [CBSE Marking Scheme, 2019]

284 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Detailed Answer: An SN1 reaction proceeds with racemisation and an SN2 reaction proceeds with complete stereochemical inversion. Q. 15. A hydrocarbon C5H12 gives only one monochloride on photochemical chlorination. Identify the compound. R [CBSE Delhi Set 3 2020] Ans.



(All the hydrogen atoms are equivalent and replacement of anyone hydrogen given monohalo derivative.) Q. 16. Out of CH3CH2CH2Cl and CH2 = CH − CH2 − Cl, which one is more reactive towards SN1 reaction?  R [CBSE OD Set 3 2020] Ans. CH2 = CH − CH2Cl

1



Short Answer Type Questions-I Q. 1. Which compound in each of the following pairs will react faster in SN2 reaction with –OH ? (i) CH3Br or CH3I (ii) (CH3)3CCl or CH3Cl Ans. (i) CH3I as I– ion is better leaving group than Br– ion.  1 (ii) CH3Cl as 1° alkyl halides are more reactive than tert-alkyl halides in SN2 reaction with OH–. 1 Q. 2. Write the equations for the preparation of 1-bromobutane from : (i) 1-butanol (ii) but-1-ene A [CBSE Comptt. Delhi 2016] HBr

Ans. (i) CH3CH2CH2 CH2 – OH → CH3CH2CH2CH2Br

1

HBr/Peroxide

(ii) CH3CH2CH = CH2 →

CH3CH2CH2 CH2Br 1 Q. 3. (i) Which alkyl halide from the following pair is chiral and undergoes faster SN2 reaction ? Br (a)

| Br (b)

A [CBSE Comptt. Delhi Set-I, 2 2017] (ii) Out of SN1 and SN2, which reaction occurs with (a) Inversion of configuration

(b) Racemisation

A [CBSE Delhi 2014]

Ans. (i) (b) is chiral ½ (a) undergoes faster SN2 ½ (ii) (a) SN2 ½ (b) SN1 ½ [CBSE Marking Scheme 2014] Q. 4. (i)  Allyl chloride can be distinguished from vinyl chloride by NaOH and silver nitrate test. Comment.



(ii) Alkyl halide reacts with lithium aluminium hydride to give alkane. Name the attacking reagent which will bring out this change. A&E + R [CBSE SQP 2017]

(2 marks each)

Ans. (i) Vinyl chloride does not respond to NaOH and silver nitrate test because of partial double bond character due to resonance. 1 (ii) Hydride ion / H– 1 [CBSE Marking Scheme 2017] Q. 5. Which alkyl halide from the following pair is (i) Chiral and (ii) undergoes SN1 reaction faster? (a) (CH3)3CBr (b) CH3CH2CHBrCH3 

A [CBSE Comptt. Set-2 2017]

Ans. (i) (b) is chiral. 1 (ii) (a)   [CBSE Marking Scheme 2017] 1 Q. 6. Which one of the following compounds is more reactive towards SN2 reaction and why? CH3CH(Cl)CH2CH3 or CH3CH2CH2Cl  A&E [CBSE Comptt. Delhi/OD 2018] Ans. CH3CH2CH2Cl , due to primary halide which has less steric hindrance. 1+1  [CBSE Marking Scheme 2018] Detailed Answer: In SN2 reaction, attack of nucleophile takes place from backward direction. Primary alkyl halide is the least sterically hindered among primary, secondary and tertiary alkyl halides. While in secondary halides presence of bulky –CH3 group cause steric hindrance for nucleophilic attack. Therefore, primary alkyl halides are more reactive towards SN2 reaction. Q. 7. The following haloalkanes are hydrolysed in presence of aq KOH. (i) 1-Chlorobutane (ii) 2-chloro-2-methylpropane Which of the above is most likely to give racemic mixture? Justify your answer.  A&E [CBSE SQP 2021] Ans. Racemic mixture will be given by 2-chloro2-methylpropane as it is an optically active compound. [1] When 2-chloro-2-methylpropane undergoes SN1 reaction, both front and rear attack are possible, resulting in a racemic mixture. [1]

[ 285

HALOALKANES AND HALOARENES

Short Answer Type Questions-II Q. 1. Give reasons : (i) C – Cl bond length in chlorobenzene is shorter than C – Cl bond length in CH3 – Cl. (ii) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride. (iii) SN1 reactions are accompanied by racemization in optically active alkyl halides.  A&E [CBSE Delhi 2016] Ans. (i) In chlorobenzene, each carbon atom is sp2 hybridised and due to resonance there is a partial double bond character, so bond length is short. 1 (ii) In chlorobenzene, carbon to which chlorine is attached to sp2 hybridised and is more electronegative than the corresponding carbon in cyclohexyl chloride which is sp3 hybridised. So the dipole moment is lower in chlorobenzene. In chlorobenzene, –I and +R effect oppose each other while in the other only –I effect is the only contributing factor resulting in lower dipole moment of cyclohexyl chloride. 1 (iii) In SN1 reaction, carbocation intermediate formed is a planar molecule which will lead to form dand l- products. Hence, racemization occurs. 1 [CBSE Marking Scheme 2016]

(3 marks each)

Q. 2. Following compounds are given to you : 2-Bromopentane, 2-Bromo-2-methylbutane, 1-Bromopentane (i) Write the compound which is most reactive towards SN2 reaction. (ii) Write the compound which is optically active. (iii) Write the compound which is most reactive towards β-elimination reaction. U [CBSE Delhi/OD, Set-1, 2, 3 2017]

Ans. (i) 1-Bromopentane 1 (ii) 2-Bromopentane 1 (iii) 2-Bromo-2-methylbutane 1 [CBSE Marking Scheme 2017]

Commonly Made Error  There is confusion in the order of reactivity of 1°, 2° and 3° towards SN1, optical activity and elimination reaction.

Answering Tip  Understand the variation in reactivity of 1°, 2° and 3° haloalkanes. OR



3 [Topper’s Answer 2017]

Detailed Answer:

286 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



(i) 1-Bromopentane as primary alkyl halides are most reactive towards SN2 reaction.





(iii) 2-Bromo-2-methylbutane as tertiary alkyl halide is most reactive towards b-elimination reaction.

(ii) 2-Bromopentane as it contains asymmetric carbon atom.

Q. 3. Give the IUPAC name of the product formed when : (i) 2-Methyl-1-bromopropane is treated with sodium is the presence of dry ether. (ii) 1-Methyl cyclohexene is treated with HI. (iii) Chloroethane is treated with silver nitrite. Ans. (i) 2,5-Dimethyl hexane. (ii) 1-Methyl-1-iodocyclohexane (iii) Nitroethane

A [CBSE SQP 2017]

1+1+1 [CBSE Marking Scheme 2017]

Detailed Answer:



(i) 2



CH3                             CH3 |   |                        ether H3C—CH—CH2 + 2Na Dry  → H3C—CH—CH2—CH2—CH—CH3 | |

Br 2-Methyl-1-bromopropane (ii) CH3

CH3 2,5-Dimethyl hexane CH3 I

+ HI

1-Methyl cyclohexene



+ 2NaBr

1-Methyl-1-iodocyclohexane

(iii) C2H5Cl + AgNO2 ⎯→ C2H5NO2 + AgCl



Q. 4. (a) Identify the chiral molecule in the following pair :



& OH

OH

(i)

(ii)



(b) Write the structure of the product when chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether.



(c) Write the structure of the alkene formed by dehydrohalogenation of 1-bromo-1-methylcyclohexane with alcoholic KOH. A [CBSE D/OD 2018] Ans.



[ 287

HALOALKANES AND HALOARENES

Ans.







Q. 5.  Among all the isomers of molecular formula C4H9Br, identify (a) the one isomer which is optically active. (b) the one isomer which is highly reactive towards SN2. (c) the two isomers which give same product on dehydrohalogenation with alcoholic KOH.  A&E [CBSE OD Set-2 2019]



[Topper’s Answer 2018] Ans.



Ans. (i) CH3CH2CH(Br)CH3 1 (ii) CH3CH2CH2CH2Br 1 (iii) (CH3)3CBr and (CH3)2CHCH2Br [CBSE Marking Scheme, 2019] ½ + ½





IUPAC-Name: (i) 2–Bromobutane (ii) 1– Bromobutane (iii) 2–Bromo–2–methylpropane and 1–Bromo–2– methylpropane. Q. 6.  Identify A, B, C, D, E and F in the following :

A&E [CBSE Delhi Set 1 2020]

[½ × 6 =3]

288 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII





Q. 7.  (i) Write the structure of major alkene formed by b-elimination of 2, 2, 3-trimethyl-3-bromopentane with sodium ethoxide in ethanol. (ii) Which one of the compounds in the following pairs in chiral?

(iii) Identify (A) and (B) in the following:

(A)

(B)

A&E [CBSE OD Set 1 2020]





(ii) (iii)

[1 +1 +1 = 3]









Ans. (i) 

Q. 8. How can you convert the following? (i)  But-1-ene to 1-iodobutane (ii)  Benzene to acetophenone (iii)  Ethanol to propanenitrile

A&E [CBSE OD Set 1 2020]



Ans. (i)

(ii) 

(iii)

Long Answer Type Questions Q. 1. Some alkyl halides undergo substitution reactions whereas some undergo elimination reactions on treatment with bases. Discuss the structural







[1 +1 +1 = 3]

(5 marks each) features of alkyl halides with the help of examples which are responsible for this difference. C [NCERT Exemp. Q. 95, Page 149]

CH3

CH3 aq.

C

H

H

HO

H

H

–

–

H 3C

CH 3



CH 3

C

TOPIC-2 H3C

C+

OH

or HO

CH

CH– OH 3 –H C+

HKOH CH3 alc. 3C

+

C — + Cl –

H3C

Elimination reaction CH3

CH3

— CH2+H2CH H3C—C — O 3 – — CH +H O 2-methylprop-1-ene — H C—C OH 3 2 2

—

+ OH

CH3

HC

C— Elimination H3CreactionCl CH3

—

C+

CH3

CH 3

CH 3

H3C

H 3C

Cl

CH3

+

C — + Cl –

—



CH 3

CH 3

—



Cl CH3

CH3

— —

H3C

+ Cl –

C+

KOH (Weak base)

C—

—

C

H3C



C

HO

CH3

3 alc. CH KOH

—

CH3 aq.

CH3

— —

CH3

CH 3

or OH

Here the reagent used i.e., aq. KOH It is a weak base, so substitution takes place.

—

Transition state tert-n-butyl halides prefer to undergo elimination reactions due to the formation of stable carbocations. 3o alkyl halides follow SN1 mechanism by forming tertiary carbocations. Now, if the reagent used is a weak base then substitution occurs while if it is a strong base then instead of substitution, elimination occurs.



—



+ Cl H

CH 3

C H 3C

3 3 –H 2-methylprop-1-ene H3C used CH3i.e., alc. Here the reagent KOH is a strong base, so elimination competes over substitution and alkene is formed. Secondary alkyl halide can undergo substitution or elimination depending on type of solvent and temperature conditions.  5

—

C

+ OH H3C

—

Cl

H

HO

(Weak base)

—

C ..... Cl

aq. KOH C

H3C

–

—

H

CH 3

CH 3 C+

[ CH 289 3

H 3C

—

H

CH3

+ Cl –

C+

—



Ans. Primary alkyl halides prefer to undergo substitution reactions by SN2 mechanism. A transition state is formed in which carbon is bonded to nucleophile and finally halogen atom is pushed out. H

Cl

H3C

HALOALKANES AND HALOARENES

KOH (Weak base)

C CH 3

CH 3

Haloarenes and Polyhalogen Compounds Revision Notes  Haloarenes : Haloarenes are the compounds formed by replacing one or more hydrogen atoms in an aromatic ring with halogen atoms. For example, monohalogen derivative

X |

(X = F, Cl, Br, or I).

 When two halogens are at 1, 2- positions of benzene ring, it is called ortho substituted derivative. e.g., Cl |

Cl

 When two halogens are at 1, 3- positions of benzene ring, it is called meta substituted derivative. e.g., Cl |

Cl







When two halogens are at 1, 4- positions of benzene ring, it is called para substituted derivative. e.g., Cl |

| Cl







Methods of preparation of Haloarenes : (i) Reacting benzene directly with halogen : + X2



Fe or FeX3 dark

—X +H—X

Scan to know more about this topic

(X = Cl, Br) Nucleophilic substituition reactions

290 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

+ I2

—I

HIO3

+ HI

or HNO3



Iodobenzene

(ii) Starting with diazonium salts : +

N ≡ NCl– |

NH2 | NaNO2/HCl 273 – 278K

Benzene diazonium chloride

≡

 Physical properties of Haloarenes : Isomeric haloarenes have similar boiling points. But para isomer has higher melting point than other two i.e., ortho and meta because of the symmetry in para isomer which occupies the crystal lattice better than ortho and meta isomers.

o-

mp-

K K

 Chemical properties of Haloarenes :

(a) Nucleophilic substitution reactions are very less possible with haloarenes as C—X bond attains partial double bond character because of resonance effect, difference in hybridization, instability of the formed phenyl cation. Due to these repulsions, it makes difficult for electron rich nucleophile to attack electron rich haloarenes. Under higher temperatures, some of the below mentioned nucleophilic reactions are possible.

(i) Substitution by —OH group (Dow’s process) :

Cl |

OH | NaOH 623 K, 300 atm



Chlorobenzene

Phenol

[ 291





HALOALKANES AND HALOARENES

(ii) Substitution by —CN group : Cl |

|

CuCN, DMF

CN

673 K



Benzonitrile

(iii) Substitution by —NH2 group :



Cl |

NH2 | NH 3 CuO, 475 K 60 atm

(iv) Reaction with metals :











Aniline



Nature of C—X bond in haloarene : The C—X bond of aryl halide is less reactive than that of haloalkanes. One reason is that in aryl halide, halogen atom is attached to sp2-hybrid carbon atom whereas in alkyl halides, it is attached to sp3-hybrid carbon atom. Due to more electronegativity of sp2-hybrid carbon in comparison to sp3carbon atom, there is less charge in separation of C—X bond in haloarene. Secondly in aryl halides like chlorobenzene, the lone pair of electrons present on chlorine atom migrate towards the aromatic ring by +R effect due to conjugation of lone pair of electrons with π-electrons of the aromatic ring. As a result of which, the halogen atom is attached with the aromatic ring by partial double bond. The phenyl cation formed is highly unstable. The replacement of such a halogen atom by other nucleophiles becomes difficult.



292 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(v) Electrophilic substitution reactions of haloarenes : It takes place at ortho and para positions :

Chlorobenzene

 Polyhalogen compounds : Carbon compounds containing more than one halogen atoms are usually referred to as polyhalogen compounds. Many of those compounds are useful in industry and agriculture. (i) Dichloromethane (Methylene chloride) : It is prepared industrially by the direct chlorination of methane. CH4 + 2Cl2 hν → CH2Cl2 + 2HCl The mixture so obtained is separated by fractional distillation. (ii) Chloroform : It is manufactured by chlorination of methane followed by separation by fractional distillation. hν CH4 + 3Cl2 Sunlight  → CHCl3 + 3HCl.

(iii) Iodoform (Triiodomethane) : It is prepared by heating ethanol or acetone with sodium hydroxide and iodine or Na2CO3 and I2 in water. It is insoluble in water, yellow precipitate of CHI3 is formed. This reaction is called iodoform reaction. CH3CH2OH + 6NaOH + 4I2 heat  → CHI3 + 5NaI + HCOONa + 5H2O CH3COCH3 + 4NaOH + 3I2 heat  → CHI3 + 3NaI + CH3COONa + 3H2O

[ 293

HALOALKANES AND HALOARENES

(iv) Carbon tetrachloride (Tetrachloromethane) : It is prepared by chlorination of methane or by action of chlorine on CS2 in the presence of AlCl3 as catalyst. AlCl

3 CS2 + 3Cl2  → CCl4 + SCl2 ∆

(Sulphur dichloride) CH4 + 4Cl2 hν → CCl4 + 4HCl  Some Important Conversions : (i)

Propene to propan-1-ol : aq. KOH, ∆

HBr/peroxide

CH3CH = CH2 (Anti-Markvonikov,  → CH3CH2CH2Br (Hydrolysis) → CH3CH2CH2OH addtion) Propene

(ii)

CH2 — CH2



KOH(alc), ∆ (Dehydrohalogenation)  →

CH3CH2OH  → CH3CH2I



Propan-1-ol

Ethanol to but-2-yne : P/I 2 , ∆



1-Bromopropane



Ethanol

Iodoethane

CH2 =

Br2 /CCl 4 CH2 ¾(Electrophilic ¾¾¾¾¾¾¾ ® addition)

Ethene

—





—





Br

Br



CH I(excess) KOH(alc), ∆ ,liq.NH  → CH3—C ≡ C—CH3 (Dehydrohalogenation)  → HC ≡ CH NaNH  → Na—C ≡ C—Na (Nucleophilic substitution) 196K 2



3

3



(iii)

But-2-yne

1-Bromopropane to 2-bromopropane :

Br | HBr KOH(alc), ∆ CH3CH2CH2Br  H3—CH = CH2 Markonikov,  → CH —CH—CH → C 3 3 addition



(Dehydrohalogenation)



1–Bromopropane (iv)

1-Propene

2–Bromopropane

Toluene to benzyl alcohol : CH2OH CH3



+ Cl2



CH2Cl h

Aq. KOH

heat

– KCl



Toluene

(v)

Benzene to 4-bromonitrobenzene : HNO3 + conc. H 2 SO4 — Br Conc.  → O2 N — (Nitration)

Br /FeBr

2 3 (Electrophilic  → substitution)



Benzene

(vi)

Benzyl alcohol to 2-phenylethanoic acid : 2

2

Benzyl alcohol



+



KCN(alc)/∆

 → — CH2Cl (Nucleophilic substitution) Benzyl chloride

H /H 2 O — CH2CN  → (Hydrolysis)

— CH2COOH 2-Phenylethanoic acid

Benzyl cyanide (vii)

Ethanol to propanenitrile : P/I , ∆

2 CH 3 CH 2 OH  → CH 3 CH 2 I



Ethanol

— Br

4-Bromonitrobenzene

SOCl — CH2OH  → ( - SO , - HCl)



Benzyl alcohol

1-Iodoethane

KCN ( alc ) / ∆ (Nucleophilic  → CH3CH2CN substitution)

Propanenitrile

294 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



(viii) Aniline to chlorobenzene : — NH2

Aniline

(ix)

CuCl/HCl — N  NCl  → (Sandmeyer reaction)

–

— Cl Chlorobenzene

Benzene diazonium chloride



+

NaNO2-HCl 273-278 K (Diazotisation)

2-Chlorobutane to 3, 4-dimethyl hexane :

Cl 2-chlorobutane

(x)



(xi)

CH3

CH3

3, 4-Dimethyl hexane

2-Methyl-1-propene to 2-chloro-2-methyl propane : CH3 CH3 HCl (Markovnikov  → CH3 — C — CH3 addition) CH3 — C = CH2 Cl 2-Methyl-1-propene 2-Chloro-2-methylpropane —

—

—



—



—

—

Dry ether 2CH3 — CH — CH2CH3 + 2Na (Wurtz's  → CH3CH2 — CH — CH — CH2CH3 + 2NaCl reaction)

Ethyl chloride to propanoic acid : KCN (alc)/∆ CH3CH2Cl (Nucleophilic  → CH3CH2CN substitution)



Ethyl chloride



H+ /H O

2 (Hydrolysis)  → CH3CH2COOH

Ethyl cyanide

Propanoic acid

(xii)

But-1-ene to n-butyl iodide :



NaI, acetone CH3CH2CH = CH2 (Anti-Markovnikov  → CH3CH2CH2CH2Br (Finkelstein  → CH3CH2CH2CH2I reaction) addition)

HBr/peroxide





But-1-ene

1-Bromobutane

—

(xiii) 2-Chloropropane to 1-propanol : CH3 — CH — CH3



Cl 2-Chloropropane



n-Butyl iodide

KOH (alc), ∆ (Dehydrohalogenation)  → CH3—CH = CH2 (Aniti-Markovnikov,  → CH3CH2CH2Br addition) HBr, peroxide

Propene

1-Bromopropane KOH(aq), ∆

Nucleophilic  → CH3CH2CH2OH substitution







(xiv) Isopropyl alcohol to iodoform :



Heat CH3 — CH — CH3 + 4I2 + 6NaOH (Iodoform  → CHI3 + CH3COONa + 5NaI + 5H2O reaction) Iodoform OH Isopropyl alcohol

—



(xv)

Chlorobenzene to p-nitrophenol :

— Cl

1-Propanol

Chlorobenzene

Conc. HNO +conc. H SO

3 2 4  → (Nitration)

15% NaOH, 433K (i)  → (ii) Dil. HCl

[ 295

HALOALKANES AND HALOARENES

(xvi) 2-Bromopropane to 1-bromopropane :



CH3 — CH — CH3

HBr/peroxide

KOH(alc), ∆  → CH3—CH2—CH2—Br (Dehydrohalogenation)  → CH3CH = CH2 (Peroxide effect)

—







Br 2-Bromopropane

Propene

1-Bromopropane



(xvii) Chloroethane to butane :



2CH3CH2—Cl + 2Na (Wurtz  → CH3CH2—CH2CH3 + 2NaCl reaction)

Dry ether, ∆

Chloroethane



Butane

(xviii) Benzene to diphenyl : Br /FeBr

2 3  →

2Na, dry ether,∆

 → Fittig reaction

(xix) tert-Butyl bromide to iso-butyl bromide : CH3

—

CH3 — C — CH3

CH3

—

—

CH3

KOH(alc), ∆ (Dehydrohalogenation)  → CH3 — C = CH2

HBr/peroxide

(Anti-Markovnikov,  → CH3 — CH — CH2Br addtion) Isobutyl bromide

2-Methyl-1-propene

Br

—



tert-Butyl bromide





(xx)



Aniline to phenyl isocyanide :

Warm + CHCl3 + 3KOH (alc.) (Carbylamine  → reaction)

 Important Name Reactions : (i) Sandmeyer’s Reaction :

+

–

N2 Cl

Cl CuCl, HCl

Benzene diazonium chloride – N2+Cl

+ N2

Chlorobenzene Br

CuBr, HBr

+

+ N2

Bromobenzene

–

N2 Cl

CN CuCN

+ N2

KCN



Cyanobenzene

(ii) Finkelstein Reaction :

Dry acetone CH3CH2Cl + NaI  → CH3CH2I + NaCl Ethyl iodide (iii) Wurtz Reaction :



Dry ether

® CH3CH3 + 2NaBr CH3Br + 2Na + BrCH3 ¾ ¾¾¾¾ Ethane

+ 3KCl + 3H2O

296 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII CH3 CH3 CH3 CH3 | | | | Dry ether CH3 — CH — Br + 2Na + Br— CH — CH3 CH3 — CH — CH — CH3 + 2NaBr



2, 3-Dimethyl butane

Isopropyl bromide

(iv) Wurtz-Fittig Reaction : Dry ether

Cl + 2Na + Cl — CH3

CH3 + 2NaCl

Dry ether



Cl + 2Na + Cl — C2H5

C2H5 + 2NaCl

(v) Fittig’s Reaction : Dry ether



Cl + 2Na + Cl —

+ 2NaCl

(vi) Friedel-Crafts Alkylation : CH3 + CH3Cl

Anhyd. AlCl3

Benzene

+ HCl Toluene

Toluene



(vii) Dow’s Process :

(viii) Hunsdiecker’s Reaction : CCl

4 R — COOAg + Br2 Reflux  → R — Br + AgBr + CO2

(ix) Gattermann’s Reaction : N2Cl

Cu/HCl

Cl + N2

[ 297

HALOALKANES AND HALOARENES

Q. (i) Write the structural formula of A, B, C and D in the following sequence of reaction : Cl | alc. HBr CH3—CH—CH A   3  KHO Perxide



B

Nal dry ether

Mg

C dry ether D

Solution STEP-1: (i) A: CH3–CH=CH2 STEP-2:

B: CH3–CH2–CH2Br

STEP-3: C: CH3–CH2–CH2l STEP-4: D: CH3–CH2–CH2Mgl (ii)

(ii) Illustrate Sandmeyer’s reaction with the help of a suitable example.

 

Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS : Q. 1. Identify the compound Y in the following reaction. NaNO2+HCI

Cu2Cl2

273–278 K





CH2–CH=CH2

Y+N2

Cl

(a)

Q. 2. What is ‘A’ in the following reaction?

+ – N2Cl

NH2



(1 mark each)







(b)

CH2–CH=CH2

Cl

(c)



(d)

(b)

(a)



Cl

Cl R [NCERT Exemp. Q. 3, Page 134] Ans. Correct option : (a) Explanation : When a primary aromatic amine is dissolved or suspended in cold aqueous mineral acid and treated with sodium nitrite, a diazonium salt is formed. When this freshly prepared diazonium salt is mixed with cuprous chloride, diazonium group is replaced by Cl. Then, chlorobenzene is formed which is Y in this reaction.

CH2– CH – CH3

Cl

NH2 NaNO2+HCl 273–278 K

CH2–CH2=CH2–Cl

Cl

Cl



A

+ HCl

+ – N2Cl

Cl Cu2Cl2 (Sandmeyer reaction)

+N2

(Y) [1]

CH – CH2 – CH3

Cl



(c)

(d) 

R [NCERT Exemp. Q. 15, Page 136]

Ans. Correct option : (c) Explanation : In this reaction, addition of HCl takes place on doubly bonded carbons in accordance with Markovnikov’s rule, that is, addition of negative addendum will take place on that carbon which has lesser number of hydrogen.

298 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

CH2 – CH = CH 2

Ans. Correct option: (b)

CH2 – CH – CH3 Cl



Explanation : Halogens are ortho-para directing due to (+M) or (+R) effect. Moreover, they are deactivating due to high electronegativity. [1]

+ HCl



 [1]



Q. 3. The IUPAC name of the compound shown below is:

Cl

Q. 3. Assertion (A) : It is difficult to replace chlorine by –OH in chlorobenzene in comparison to that in chloroethane.

Br

(a) 2-bromo-6-chlorocyclohex-1-ene (b) 6-bromo-2-chlorocyclohexene (c) 3-bromo-1-chlorocyclohexene (d) 1-bromo-3-chlorocyclohexene [CBSE SQP 2020] Ans. Correct option : (c) Explanation :

Reason (R) : Carbon-chlorine (C—Cl) bond in chlorobenzene has a partial double bond character due to resonance. U [NCERT Exemplar]

Ans. Correct option: (a)

Explanation : Chlorobenzene is very less reactive to nucleophilic substitution reaction by –OH group as Carbon-chlorine (C—Cl) bond in chlorobenzene has a partial double bond character due to resonance. [1]

[C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1.  Write the IUPAC name of the given compound.





A [CBSE OD 2016]

Ans. 2–phenylethanol.

IUPAC name: 3-bromo-1-chlorocyclohexene

Q. 1. Assertion (A): Presence of a nitro group at ortho or para position increases the reactivity of haloarenes towards nucleophilic substitution.



Q. 2. Write the IUPAC name of the given compound.



[B] ASSERTION & REASONS: In the following questions a statement of assertion followed by a statement of reasin is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion (c) Assertion is correct statement but reason is wrong statement (d) Assertion is wrong statement but reason is correct statement

Q. 3. Write the structure of 2,4-dinitrochlorobenzene. U [CBSE Delhi Set-2 2017]

 Ans.

Reason (R) : Nitro group, being an electron withdrawing group decreases the electron density over the benzene ring. R [NCERT Exemplar]

Reason (R) : Halogen atom is a ring deactivator. R [NCERT Exemplar]





Explanation: Nitro group being an electron withdrawing group, decreases the electron density of benzene ring thus increasing the reactivity of haloarenes towards nucleophilic substitution. [1]

Q. 2. Assertion (A) : In monohaloarenes, further electrophilic substitution occurs at ortho and para positions.

A [CBSE Delhi 2016]

Ans. 2, 4, 6-Tribromoaniline/2,4,6Tribromobenzenenamine. 1 [CBSE Marking Scheme 2016]

Ans. Correct option: (a)

1 [CBSE Marking Scheme 2016]

Q. 4. Out of

1 [CBSE Marking Scheme 2017]

which is an example

of allylic halide?



U [CBSE OD Set-1 2017]

[ 299

HALOALKANES AND HALOARENES

Q. 8. Out of Chlorobenzene and p-nitrochlorobenzene, which one is more reactive towards nucleophilic substitution reaction and why?

1 [CBSE Marking Scheme 2017]





Q. 5. Out of

and

which is an example

of vinylic halide?

U [CBSE OD Set-2 2017]

Ans. 





1 [CBSE Marking Scheme 2017]

Ans. p-Nitrochlorobenzene; Due to electron with­ drawing nature of −NO2 group. ½+½  [CBSE Marking Scheme 2019] Detailed Answer: The relative p-nitrochlorobenzene is more reactive towards nucleophilic substation reaction due to its electron –withdrawing inductive and resonance effects which resultd in the stabilized carbanion formed by the −NH2 group. Q. 9. Write the IUPAC name of the given compound : A [CBSE Delhi Set 1 2019]



OR



R [CBSE OD Set-2 2019]



Ans.

[Topper’s Answer 2017]

Q. 6. Out

of



which is an example of benzylic halide?

Ans.

A [CBSE OD Set-3 2017]





1 [CBSE Marking Scheme 2017] 



Ans.

[Topper’s Answer 2019]

Q. 7. Out of Chlorobenzene and Cyclohexyl chloride, which one is more reactive towards nucleophilic substitution reaction and why? R [CBSE OD Set-1 2019]



Q. 10. Out of

and

which will



Ans. Cyclohexyl chloride; Because of partial double bond character od C−Cl bond in Chlorobenzene /Resonance effect / sp3 hybridized carbon in cyclohexyl chloride whereas sp2 carbon in chlorobenzene. 1  [CBSE Marking Scheme 2019]

undergo SN1 reaction faster with OH−? A&E [CBSE Delhi Set 2 2020]

 Ans.

Detailed Answer: Cyclohexyl chloride is more reactive towards nucleophilic substitution reaction because in cyclohexyl chloride chlorine atom is attached to sp3 hybridized while in chlorobenze it is sp2 hybridized. Thus it has higher tendency to release electrons. In chlorobenzene, due to resonance, cleavage of C-Cl bond is difficult than cyclohexyl chloride which makes it less reactive towards nucleophilic substitution.

because conjunction between positive charge and double bond, resonance is possible.

300 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q. 11. Out of

and

which one is more reactive towards SN1 reaction? A&E [CBSE OD Set 1 2020]



Q. 12. Out of

and



Ans.

which will react faster in SN1 reaction with OH−? A&E [CBSE OD Set 2 2020]

 Ans.

Short Answer Type Questions-I

(2 marks each)

Q. 1. How will you carry out the following conversion : (i) 2-Bromopropane to 1-bromopropane (ii) Benzene to p-chloronitrobenzene Ans. (i) CH3 – CH(Br) – CH3

alc KOH

U [CBSE Comptt. OD Set-1, 2, 3 2017]

CH3 – CH=CH2 HBr, Peroxide CH3 – CH2 – CH2 – Br Cl



(ii)

+ Cl2

Cl

Fe

HNO3

dark

conc.H2SO4 NO2

Commonly Made Error

1+1 [CBSE Marking Scheme 2017] Cl

Cl

 A number of students cannot represent proper reagents used for a particular reaction.

H

E

E

Answering Tip Inductive effect destabilizes the intermediate carbocation

 Practice the organic reactions with reference to reagents used. Q. 2. Explain why in spite of being an electron withdrawing group, chlorine is ortho- and para- directing in electrophilic aromatic substitution reactions. A&E Ans. Although chlorine is an electron withdrawing group, yet it is ortho-para directing in nature in electrophilic aromatic substitution because when chlorine is present in benzene ring, it releases electron by resonance whereas it acts as withdrawing group only through inductive effect. By inductive effect, chlorine atom destabilizes the intermediate carbocation formation but by resonance, chlorine atom stabilises the intermediate carbocation and effect is more at o-and p-positions. Resonance effect opposes inductive effect. Inductive effect is stronger than resonance effect due to which reactivity is controlled by inductive effect and orientation by resonance.

H

E

(attack of ortho-position)

E

Cl

Cl E

(attack of para-position) E H Resonance effect stabilizes the intermediate carbocation





Cl

Cl

2

[ 301

HALOALKANES AND HALOARENES

Greater the stability of the carbocation, greater will be its ease of formation from the corresponding halide and faster will be the rate of reaction. The benzylic carbocation formed gets stabilised through resonance. ½



Commonly Made Error  Students write lengthy answers. Give point wise explanation, highlight the key points.

Answering Tip  Give resonance structures in support of answer. Q. 3.  Which one of the following compounds will undergo hydrolysis at a faster rate by SN1 mechanism? Justify.

CH2Cl

CH2

CH2

(i)

(ii)

CH2

(iii)

CH2 or

CH2

(iv)

CH2

CH3CH2CH2Cl U [CBSE SQP 2020]

Ans. The following compound will undergo SN1 faster:

(v)

CH2Cl

(vi)



½ CH3CH2CH2Cl forms a 1 carbocation, which is less stable than benzylic carbocation. ½ [CBSE SQP Marking Scheme 2020] 0

½





Short Answer Type Questions-II Q. 1. Give the IUPAC names of the following compounds : (i)

(iii) CH2 = CH – CH2 – Cl  A [CBSE Comptt. OD 2015] Ans. (i) 2-bromobutane 1 (ii) 1, 3-dibromobenzene 1 (iii) 3-chloropropene 1 [CBSE Marking Scheme 2015]

Br

—



—

(ii)



(3 marks each)

Br

Q. 3. How can the following conversion be carried out : (i) Aniline to bromobenzene (ii) Chlorobenzene to 2-chloroacetophenone (iii) Chloroethane to butane NH2

Ans. (i)

+ N2Cl

NaNO2 273 278 K

Aniline

Benzene diazonium chloride

N2Cl

Br

CuBr/HBr

2

273 278 K

(ii)





Bromobenzene

Cl

Cl

1

O

O + H3C–C–Cl Chlorobenzene

Anhy. AlCl3

CH3

2-Chloroacetophenone

1

302 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Na (iii) 2CH3CH2Cl dry  → CH3CH2CH2CH3 ether

Chloroethane

1

Butane

Q. 4. What happens when (i) Chlorobenzene is treated with Cl2/FeCl3, (ii) Ethyl chloride is treated with AgNO2, (iii) 2-bromopentane is treated with alcoholic KOH ? Write the chemical equations in support of your answer. Ans. (i)

A [CBSE OD 2015]

Cl

Cl

Cl Cl

+ Cl2

Anhy. FeCl3

1

+ Cl



(ii) CH3CH2Cl + AgNO2 → CH3CH2NO2 + AgCl

1

(iii) CH3CH2CH2CH(Br)CH3 + KOH (alc.) → CH3CH2CH=CH CH3

1

Q. 5. How do you convert : (i) Chlorobenzene to biphenyl, (ii) Propene to 1-iodopropane, (iii) 2-bromobutane to but-2-ene. Ans. (i) Chlorobenzene to biphenyl : When two molecules of chlorobenzene combine with sodium metal in the presence of dry ether,it forms biphenyl. —

Cl 2

+ 2Na

—

dry ether

Chlorobenzene (ii) Propene to 1–iodopropane :

+ 2NaCl



Biphenyl

1



HBr/Peroxide

NaI/acetone

→ CH3CH2CH2Br → CH3CH2CH2I CH3CH = CH2 



1

(iii) 2–bromobutane to but–2–ene Alc.KOH H3C—CH2—CH —CH3 → H3C—CH=CH—CH3 + CH3—CH2—CH=CH2 | Br But–2–ene But–1–ene 2–Bromobutane (80%) (20%)

(ii)

Q. 6. Write the major product (s) in the following : (i)



1 ?



AgCN (iii) CH3 — CH2 — Br → ? A [CBSE OD 2016]





OR

Br

—

CH2 — CH3

—

(i)

4 Br2, UV light 1

—

—

O2N

O2N

4 – Ethylnitrobenzene Na  → CH 3 — CH — CH 3 (ii) 2CH3— CH—CH3 dry ether | | CH 3 — CH — CH 3 Cl



CH — CH3

—



4 – (1 – Bromoethyl) nitrobenzene 1

2,3-dimethyl butane AgCN

(iii) CH3CH2Br → C2H5NC Carbylaminoethane (Ethyl isocyanide)

1 1 [CBSE Marking Scheme 2016]

[ 303

HALOALKANES AND HALOARENES

Commonly Made Error  Students sometimes do not draw the structure or not mention the name of the reagent or product.

Answering Tip  Write all the steps and reagents involved in the conversion. Q. 7. Give reasons : (i) n-Butyl bromide has higher boiling point than t-butyl bromide. (ii) Racemic mixture is optically active. (iii) The presence of nitro group (–NO2)at o/p positions increases the reactivity of haloarenes towards nucleophilic substitution reactions. A&E [CBSE Delhi 2015]

Ans. (i) Larger the surface area, higher the van der Waals’ forces, higher the boiling point. 1

Commonly Made Error  Students often do not answer in detail and the answer does not convey complete understanding of the concept.

Answering Tip  While attempting reason based question, write cause and consequence of the condition.

(ii) Rotation due to one enantiomer is cancelled by another enantiomer. 1 (iii) –NO2 acts as Electron withdrawing group or –I effect. 1 [CBSE Marking Scheme 2015]

Detailed Answer: (i) n-Butyl bromide is a straight chain molecule with strong intermolecular forces whereas t-butyl bromide is a branched chain molecule with weak intermolecular forces due to smaller surface area. Hence, n-Butyl bromide has higher boiling point than t-butyl bromide.

Q. 8. Write the major monohalo product(s) in each of the following reactions : (i)

 (iii)

(ii)

A [CBSE OD Set-2 2016] 

Ans. 

3 [Topper’s Answer 2016] Q. 9. Draw the structures of the major monohalo product for each of the following reactions :



(iii)







(i)

(ii)

A [CBSE Foreign Set-1, 2, 3 2017]

304 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



Br

Ans. (i)

(ii)

—CH3 — Br

1+1+1 [CBSE Marking Scheme 2018] 

CH—Me — Cl —

  OH

(a) Structure (i) contains chiral carbon and it is optically

(iii)

active so it is a chiral molecule. Cl.CH2

1+1+1 [CBSE Marking Scheme 2017]

(b) Wurtz-Fittig reaction Cl



Q. 10. (i) Account for the following : (a)  Electrophilic substitution reactions in haloarenes occur slowly. (b)  Haloalkanes, though polar, are insoluble in water. (ii) Arrange the following compounds in increasing order of reactivity towards SN2 displacement : 2–Bromo–2–Methylbutane, 1-Bromopentane, 2–Bromopentane  A&E + U [CBSE Comptt. OD Set-1, 2, 3 2017]

Ans. (i) (a) Due to-I effect of X, the ring set partially deactivated. (b)  They fail to form Hydrogen bonds with water / more energy is required to break hydrogen 1 (ii) 2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane [CBSE Marking Scheme 2017]1

+ CH3Cl +2Na

(c) (c)









Ans.

Br

Toluene CH2

CH3

CH3

+ 2NaCl

Dry ether

Alc KOH

+ (Major)

+ HBr (Minor)

Q. 12. A compound is formed by the substitution of two chlorine atoms for two hydrogen atoms in propane. Write the structures of the possible isomers. Give the IUPAC name of the isomer which can exhibit enantiomerism. A [CBSE SQP 2018, 2019] Ans.

Detailed Answer : (i) (a) It is due to the fact that non bonding pair of electrons on the halogen are in conjugation with the ring causing resonance stabilization of halo arenes by delocalisation of electrons. Q. 11. (a) Identify the chiral molecule in the following pair :

(b) Write the structure of the product when chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether. (c) Write the structure of the alkene formed by dehydrohalogenation of 1-bromo-1methylcyclohexane with alcoholic KOH. A [CBSE Delhi/OD 2018]

CH3



Cl | CH3 – CH – CH2Cl ½ CH3 – CH2 – CHCl2 ½ Cl | CH3 — C — CH3 ½ | Cl ClCH2 — CH2 — CH2Cl ½ The following isomer will exhibit enantiomerism: Cl | CH3 — CH — CH2Cl ½ IUPAC name: 1, 2-Dichloropropane. ½ [CBSE Marking Scheme 2018]

Q. 13. Write the product(s) formed when (i) 2-Bromopropane undergoes dehydrohalogenation reaction. (ii) Chlorobenzene undergoes nitration reaction. (iii) Methylbromide is treated with KCN. A [CBSE Comptt. Delhi/OD 2018]

Ans. (i) Propene 1 (ii) 4-nitrochlorobenzene and 2-nitrochlorobenzene / structures ½+½ (iii) Methylcyanide / Ethanenitrile / structure 1 [CBSE Marking Scheme 2018]



Detailed Answer:

[ 305

HALOALKANES AND HALOARENES

Detailed Answer: (i) CH3 – CHBr – CH3 + KOH ® CH3 – CH = CH2 + KBr + H2O Cl

(ii)

Cl + HNO3

Cl NO2

H2 SO4 50°C

+

NO2

(iii) CH3Br + KCN ® CH3CN + KBr

Q. 14. (i) Out of (CH3)3C-Br and (CH3)3C-I, which one is more reactive towards SN1 and why? (ii) Write the product formed when p-nitrochlorobenzene is heated with aqueous NaOH at 443 K followed by acidification. (iii) Why dextro and laevo – rotatory isomers of Butan-2-ol are difficult to separate by fractional distillation? A [CBSE Delhi Set 1 2019]

(iii) Butan-2-ol has four different groups attached to the second tetrahedral carbon. Therefore, it is chiral molecule and the dextro and laevo – rotatory isomers of Butan-2-ol are enantiomers. The enantiomers possess identical physical properties like melting point, boiling point, solubility, refractive index etc. They only differ with respect to the rotation of plane polarised light. Therefore, dextro and laevo – rotatory isomers of Butan-2-ol are difficult to separate by fractional distillation. Q. 15. Write the structures of main products when benzene diazonium chloride reacts with the following reagents :



(i) CuCN (ii) CH3CH2OH (iii) KI A&E [CBSE Delhi Set 2 2019]

CN

Cl



Ans. (i)

OH

—

—

(ii)

—

Ans. (i) (CH3)3C–I , Due to large size of iodine / better leaving group / Due to lower electronegativity. ½+½

(ii)

(i) NaOH, 443K (ii) H+

—

—

—

I

(iii)



NO2 NO2 1 (iii) Because enantiomers have same boiling points / same physical properties. [CBSE Marking Scheme, 2019] 1 Detailed Answer: (i) Within a family, larger atoms are better nucleophiles. I- > Br- > Cl- > F-, therefore I- is better leaving group than Br-.

Detailed Answer: N2+ Cl

OH

Cl

N2+ Cl

(i) NaOH, 443 K (ii) H+ NO2

p-nitrochlorobenzene

p-nitrophenol

T  he presence of an electron withdrawing group (-NO2) at ortho- and para- positions increases the reactivity of haloarenes.

N2+ Cl (iii)

Visual Case based Questions Q.1. Read the passage given below and answer the following questions: Nucleophilic substitution reaction of haloalkane can be conducted according to both SN1 and SN2 mechanisms. However, which mechanism it is based on is related to such factors as the structure

CH3CH2OH

(ii)



NO2

CN CuCN

(i)





(ii)

[CBSE Marking Scheme, 2019] 1 × 3



I KI



3

(4 marks each) of haloalkane, and properties of leaving group, nucleophilic reagent and solvent. Influences of halogen : No matter which mechanism the nucleophilic substitution reaction is based on, the leaving group always leave the central carbon atom with electron pair. This is

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII





1

Influences of solvent polarity: In SN reaction, the polarity of the system increases from the reactant to the transition state, because polar solvent has a greater stabilizing effect on the transition state than the reactant, thereby reduce activation energy and accelerate the reaction. In SN2 reaction, the polarity of the system generally does not change from the reactant to the transition state and only charge dispersion occurs. At this time, polar solvent has a great stabilizing effect on Nu than the transition state, thereby increasing activation energy and slow down the reaction rate. For example, the decomposition rate (SN1) of tertiary chlorobutane in 25° water (dielectric constant 79) is 300000 times faster than in ethanol (dielectric constant 24). The reaction rate (SN2) of 2-bromopropane and NaOH in ethanol containing 40% water is twice slower than in absolute ethanol. In a word, the level of solvent polarity has influence on both SN1 and SN2 reactions, but with different results. Generally speaking, weak polar solvent is favorable for SN2 reaction, while strong polar solvent is favorable for SN1 reaction, because only under the action of polar solvent can halogenated hydrocarbon dissociate into carbocation and halogen ion and solvents with a strong polarity is favorable for solvation of carbocation, increasing its stability. Generally speaking, the substitution reaction of tertiary haloalkane is based on SN1 mechanism in solvents with a strong polarity (for example, ethanol containing water). (CBSE QB 2021) The following questions are multiple choice questions. Choose the most appropriate answer : (i)  SN1 mechanism is favoured in which of the following solvents: (a) benzene (b) carbon tetrachloride (c) acetic acid (d) carbon disulphide (ii)   Nucleophilic substitution will be fastest in case of: (a) 1-Chloro-2,2-dimethyl propane (b) 1-Iodo-2,2-dimethyl propane (c) 1-Bromo-2,2-dimethyl propane (d) 1-Fluoro-2,2-dimethyl propane

(iii) SN1 reaction will be fastest in which of the following solvents? (a) Acetone (dielectric constant 21) (b) Ethanol (dielectric constant 24) (c) Methanol (dielectric constant 32) (d) Chloroform (dielectric constant 5) (iv) Polar solvents make the reaction faster as they: (a)  destabilize transition state and decrease the activation energy (b)  destabilize transition state and increase the activation energy (c) stabilize transition state and increase the activation energy (d)  stabilize transition state and decrease the activation energy (v) SN1 reaction will be fastest in case of: (a) 1-Chloro-2-methyl propane (b) 1-Iodo-2-methyl propane (c) 1-Chlorobutane (d) 1-Iodobutane Ans. (i) Correct option: (c) (ii) Correct option: (b) (iii) Correct option: (c) (iv) Correct option: (c) (v) Correct option: (b) Q. 2. Read the passage given below and answer the following questions: (1 x 4 = 4) Alkyl/Aryl halides may be classified as mono, di or polyhalogen compounds depending on one, two or more halogen atoms in their structures. Alkyl halides are prepared by free radical halogenation of alkanes, addition of halogen acids to alkenes and replacement of –OH group of alcohols with halogens using phosphorus halides, thionyl chloride or halogen acids. Aryl halides are prepared by electrophilic substitution to arenes.     The following questions are multiple choice questions. Choose the most appropriate answer : (i) Complete the reaction:     H3C-Br + AgF ® (a) H3C-Br + AgF ® H3C-F + AgBr (b) H3C-Br + AgF ® Br-CH2-F + AgH (c) H3C-Br + AgF ® [Ag(CH3)]F + Br (d) None of the above Ans.  Correct option: (a) Explanation: H3C-Br + AgF ® H3C-F + AgBr. [1] (ii)  Name the major monohalo product of the following reaction.



(a) 1-Iodo-1-methyl cyclohexane (b) 1-Iodomethyl cyclohexane (c) 1-Chloro cyclohexane (d) None of the above Ans. Correct option: (b)

just the opposite of the situation that nucleophilic reagent attacks the central carbon atom with electron pair. Therefore, the weaker the alkalinity of leaving group is , the more stable the anion formed is and it will be more easier for the leaving group to leave the central carbon atom; that is to say, the reactant is more easier to be substituted. The alkalinity order of halogen ion is I− < Br− < Cl− < F− and the order of their leaving tendency should be I− > Br− > Cl− > F−. Therefore, in four halides with the same alkyl and different halogens, the order of substitution reaction rate is RI > RBr > RCl > RF. In addition, if the leaving group is very easy to leave, many carbocation intermediates are generated in the reaction and the reaction is based on SN1 mechanism. If the leaving group is not easy to leave, the reaction is based on SN2 mechanism.



306 ]

[ 307

HALOALKANES AND HALOARENES



Explanation: According to Markovnikov’s rule, iodine will add to the carbon atom having less number of hydrogen atoms.  [1] (iii) 2-Bromopentane, 2-Bromo-2-methylbutane, 1-Bromopentane Write the compound which is most reactive towards β-elimination reaction. (a) 2-Bromopentane (b) 1- Bromopentane (c) 2-Bromo-2-methylbutane (d) None of the above Ans. Correct option: (c) Explanation :



Q. 3. Read the passage given below and answer the following questions: (1×4=4)













(iii) Which of the following reaction? (a)

is halogen exchange

(b)

(c) (d)

Ans. Correct option: (a) Explanation :



It is halogen exchange reaction as in this reaction both R and Na exchanges halogens. [1] OR Arrange the following compounds in increasing order of their boiling points.





(i) (ii) (iii)





(a) (ii) < (iii) < (i) (b) (i) < (ii) < (iii) (c) (iii) < (i) < (ii) (d) (iii) < (ii) < (i) Ans. Correct option: (c) Explanation : boiling point of (i) is 364 K, boiling point of (ii) is 375 K. boiling point of (iii) is 346 K

As the branching increases in the isomeric alkyl halides, the boiling point decreases. [1]

The objects which are non-superimposable on their mirror image (like a pair of hands) are said to be chiral and this property is known as chirality. Chiral molecules are optically active, while the objects, which are, superimposable on their mirror images are called achiral. These molecules are optically inactive. The above test of molecular chirality can be applied to organic molecules by constructing models and its mirror images or by drawing three dimensional structures and attempting to superimpose them in our minds. There are other aids, however, that can assist us in recognising chiral molecules. One such aid is the presence of a single asymmetric carbon atom. In these questions a statement of assertion followed by a statement of reason is given . Choose the correct answer out of the following choices.

(a) Assertion and Reason both are correct statements and Reason is correct explanation for Assertion. (b) Assertion and Reason both are correct statements but Reason is not correct explanation for Assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but Reason is correct statement. (i) Assertion: The stereoisomers related to each other as non-superimposable mirror images are called enantiomers Reason: Enantiomers possess identical physical properties. Ans. Correct option: (b) Explanation : The stereoisomers related to each other as non-superimposable mirror images are called enantiomers and they possess identical physical properties. [1] (ii) Assertion: A racemic mixture containing two enantiomers in equal proportions will have zero optical rotation. Reason: This is because the rotation due to one isomer will be cancelled by the rotation due to the other isomer. [1] Ans. Correct option: (a) (iii) Assertion: Butan-2-ol is a chiral molecule. Reason: It has 4 different groups attached to carbon atom. Explanation: A racemic mixture is an equimolar mixture of d and l forms. It is optically inactive due to external compensation as rotation of one form is cancelled by other form.[1] Ans. Correct option: (a) Explanation: Butan-2-ol is a chiral molecule as it has 4 different functional groups attached to the tetrahedral carbon atom. [1] (iv) Assertion: Propan-2-ol is an achiral molecule. Reason: Carbon is called asymmetric carbon or stereocentre.

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



Ans. Correct option: (c) Explanation: Propan-2-ol is an achiral molecule as it does not contain an asymmetric carbon, as all the four groups attached to the tetrahedral carbon are not different. [1] Q. 4. Read the passage given below and answer the following questions: (1×4=4)



308 ]

Aryl halides are extremely less reactive towards nucleophilic substitution reactions due to the following reasons: (i) In haloarenes, the electron pairs on halogen atom are in conjugation with π-electrons of the ring. (ii) In haloalkane, the carbon atom attached to halogen is sp3 hybridised while in case of haloarene, the carbon atom attached to halogen is sp2 -hybridised. (iii) In case of haloarenes, the phenyl cation formed as a result of self-ionisation will not be stabilised by resonance.

(ii)  

(iii)  



(iv)  

Ans. Correct option:(c) Explanation:

[1]   (d) Reaction of C6H5CH2Br with aqueous sodium hydroxide follows _______. (i) SN1 mechanism (ii) SN2 mechanism (iii) Any of the above two depending upon the temperature of reaction (iv) Saytzeff rule. Ans. Correct option: (iv) Explanation: C6H5-CH2 is stable cation so favours the progress of reaction by SN1 mechanism. [1] 

(a) A primary alkyl halide would prefer to undergo ________. (i)     SN1 reaction (ii)   SN2 reaction (iii)  α-Elimination (iv)  Racemisation Ans. Correct option: (ii) Explanation: A primary alkyl halide would prefer to undergo SN2 reaction. [1] (b) Which of the following alkyl halides will undergoes SN1 reaction most readily? (i) (CH3)3C—F (ii) (CH3)3C—Cl (iii) (CH3)3C—Br (iv) (CH3)3C—I Ans. Correct option: (d) Explanation: (CH)3C-I being a tertiary alkyl halide will most readily undergo SN1 reaction. [1]   (c) What is ‘A’ in the following reaction?



The following questions are Multiple Choice Questions. Choose the most appropriate answer :









(i)  

ll

[ 309

SELF-ASSESSMENT TEST

Self-Asessment Test-10 Time : 1 Hour

(a) Cl2 / UV light (b) NaCl + H2SO4 (c) Cl2 gas in dark (d) Cl2 gas in presence of iron in dark Q. 5.  Arrange the following compounds increasing order of their densities.

(i)   

in

the

(ii) 





(iii) 



(a) (b) (c) (d)

    

       

 The reaction in which a nucleophile replaces already existing nucleophile in a molecule is called nucleophilic substitution reaction. Haloalkanes are substrate in these reactions. In this type of reaction, a nucleophile reacts with haloalkane (the substrate) having a partial positive charge on the carbon atom bonded to halogen. A substitution reaction takes place and halogen atom, called leaving group departs as halide ion. Since the substitution reaction is initiated by a nucleophile, it is called nucleophilic substitution reaction. In these questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement (i) Assertion : Chlorobenzene is formed by reaction of chlorine with the benzene in presence of AlCl3. Reason : [AlCl4]– is the species that attacks the benzene ring in this reaction. (ii) Assertion : Electrophilic substitution reactions occurs slowly in haloarenes. Reason : The reason for this effect is –R effect of the benzene ring. (iii) Assertion : Haloalkanes, though polar are, insoluble in water. Reason : Haloalkanes do not form hydrogen bond with water. (iv) Assertion : In monohaloarenes , further electrophilic substitution occurs at the ortho and para positions. Reason : Halogen atom is a ring deactivator. OR Assertion : It is difficult to replace chlorine by –OH in chlorobenzene in comparison to that in chloroethane. Reason : Carbon-chlorine (C-Cl) bond in chlorobenzene has a partial double bond character due to resonance. Following questions (No. 2 to 5) are Multiple Choice Questions carrying 1 mark each. Q. 2.  Toluene reacts with a halogen in the presence of iron (III) chloride giving ortho and para halo compounds. The reaction is (i) Electrophilic elimination reaction (ii) Electrophilic substitution reaction (iii) Free radical addition reaction (iv) Nucleophilic substitution reaction U [NCERT exemplar page 134]





Q. 3. Which is the correct increasing order of boiling points of the following compounds? 1-Bromoethane, 1-Bromopropane, 1-Bromobutane, Bromobenzene (a) Bromobenzene < 1-Bromobutane < 1-Bromopropane < 1-Bromoethane (b) Bromobenzene < 1-Bromoethane < 1-Bromopropane < 1-Bromobutane (c) 1-Bromopropane < 1-Bromobutane < 1-Bromoethane < Bromobenzene (d) 1-Bromoethane < 1-Bromopropane < 1-Bromobutane < Bromobenzene [NCERT exemplar page 140] Q.4.  Which reagent will you use for the following reaction ? CH3CH2CH2CH3 → CH3CH2CH2CH2Cl + CH3CH2CHClCH3



Q.1. Read the passage given below and answer the following questions : (1 × 4 = 4)

Max. Marks : 25

(iv) 

(i) < (ii) < (iii) < (iv) (i) < (iii) < (iv) < (ii) (iv) < (iii) < (ii) < (i) (ii) < (iv) < (iii) < (i) [NCERT Exemplar page 135]

Q. 6. Assertion: Ethylidene chloride is a gem- dihalide.   Reason: 2 Halogen atoms are present on two carbon atoms.[R] Q.7. Assertion: 2-Bromopentane is a haloalkane. Reason: Haloalkanes contain halogen atom (s) attached to the sp3 hybridised carbon atom of an alkyl group.[R] Q.No. 8 & 9 are Short Answer Type-I carrying 2 marks each. Q.8. Draw the structure of major monohalo product in each of the following reactions:



 

310 ]



 

(b)  Aryl halides are extremely less reactive towards nucleophilic substitution. Predict and explain the order of reactivity of the following compounds towards nucleophilic substitution:

[II]

 

 

[I]

[III]



 

(c) (i)  Why iodoform has appreciable antiseptic R property? OR (a) (i) Allyl chloride can be distinguished from vinyl chloride by NaOH and silver nitrate test. Comment. (ii)  Alkyl halide reacts with lithium aluminium hydride to give alkane. Name the attacking reagent which will bring out this change. (b)  Identify the products A and B formed in the following reaction: CH3—CH2—CH=CH—CH3 + HCl → A + B   (c) Why is it necessary to avoid even traces of moisture during the use of a Grignard reagent?  





Q.9. Which halogen comound in each of the following pairs will react faster in SN2 reaction: (a) CH3Br or CH3I (b) (CH3)3 C—Cl or CH3—Cl Q.No. 10 & 11 are Short Answer Type-II carrying 3 marks each. Q.10. What happens when: (i) CH3—Cl is treated with aqueous KOH? (ii) CH3—Cl is treated with KCN? (iii) CH3—Br is treated with Mg in the presence of dry ether? U Q.11.  Write the major products in the following equations : PCl5 (i)  CH3  CH 2OH ?  R (ii) Write the structure of 1-bromo-4-chlorobut-2-ene.   (iii) Write down the structure and IUPAC name for neo-pentylbromide Q.No 12 is a Long Answer Type Carrying 5 marks each. Q.12. (a) Name the alkene which will yield 1-chloro-1methylcyclohexane by its reaction with HCl. Write the reactions involved.





Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

 

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ALCOHOLS, PHENOLS AND ETHERS

[ 311

11 CHAPTER

ALCOHOLS, PHENOLS AND ETHERS

Syllabus ¾¾ Alcohols : Nomenclature, methods of preparation, physical and chemical properties (of primary alcohols only), identification of primary, secondary and tertiary alcohols, mechanism of dehydration, uses with special reference to methanol and ethanol. ¾¾ Phenols : Nomenclature, methods of preparation, physical and chemical properties, acidic nature of phenol, electrophillic substitution reactions, uses of phenols. ¾¾ Ethers : Nomenclature, methods of preparation, physical and chemical properties, uses.

Trend Analysis List of concept names

2018 D/OD

2019

2020

D

OD

Reaction Mechanism

1Q (2 marks)

1Q (3 marks)

Conversion

1Q (2 marks)

Give reason

1Q (1 mark)

IUPAC Name

Write the products formed/reagents involved for a reaction

D

OD

1Q (1 mark)

1Q (3 marks)

Properties

1Q (3 marks) 1Q (3 marks)

1Q (1 mark) 1Q (1 mark)

TOPIC-1

Methods of Preparation and Properties of Alcohols and Phenols Revision Notes

1Q (1 mark)

1Q (1 mark)

TOPIC - 1 Methods of Preparation and Properties of Alcohols and Phenols .... P. 312 TOPIC - 2 Methods of Preparation and Properties of Ethers .... P. 339

 Alcohol : When one hydrogen atom of alkane is replaced by –OH group, the compounds obtained are called alcohols having general formula CnH2n+1OH.  Classification of Alcohols :

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(a) Mono, di, tri and polyhydric alcohols : (i) Those alcohols which contain one –OH group e.g., CH3OH, C2H5OH etc. are called monohydric alcohols.

Classification of alcohols

[ 313



ALCOHOLS, PHENOLS AND ETHERS

(ii) Those alcohols which contain two –OH groups, are called dihydric alcohols. Their general formula is CnH2n(OH)2. They are called diols. e.g., CH2OH Ethane-1, 2-diol. CH2OH

(iii) Those alcohols which contain three hydroxyl (–OH) groups, are called trihydric alcohols. They are also called triols. CH2OH

Their general formula is CnH2n–1 (OH)3. e.g., CHOH Propane-1, 2, 3-triol. CH2OH CH2OH (iii) Those alcohols which have more than one –OH groups are called polyhydric alcohols. e.g., Glycol. CH2OH

(b) 1°, 2° and 3° alcohols : (i) The alcohol in which –OH group is attached to primary (1°) carbon atom is called primary alcohol, e.g., CH3OH, C2H5OH etc. (ii) The alcohol in which –OH group is attached to secondary (2°) carbon atom is called secondary alcohol. e.g., CH3—CH—CH3 (2-propanol). OH



(iii) The alcohol in which —OH group is attached to tertiary (3°) carbon atom is called tertiary alcohol. CH3

e.g., CH3—C—OH or (CH3)3COH (2-methylpropan-2-ol). CH3



(c) Allylic and vinylic alcohols : (i) Those alcohols in which –OH group is attached to single bonded sp3-hybridised carbon next to carboncarbon double bond, that is to allylic carbon are called allylic alcohols. e.g., CH2 = CH—CH2—OH (ii) Those alcohols in which –OH group is attached to double bonded sp2-hybridised carbon atom are called vinylic alcohols. They are highly unstable and get tautomerised to form aldehydes. e.g., O

CH2=CHOH

CH3 – C – H

(Vinyl alcohol)

(Acetaldehyde)

(d) Benzyl alcohol : The alcohol in which –OH group is attached to single bonded sp3-hydridised carbon atom attached to aromatic ring is called benzyl alcohol. e.g., CH2OH

Benzyl alcohol (Phenyl methanol)

 Common and IUPAC Names of Some Alcohols Formula CH3OH

Common Name Methyl alcohol

IUPAC Name Methanol

CH3CH2OH

Ethyl alcohol

Ethanol

CH3CH2CH2OH

n-Propyl alcohol

Propan-1-ol

CH3CHOHCH3

iso-Propyl alcohol

Propan-2-ol

(CH3)2CH—CH2OH

iso-Butyl alcohol

2-Methyl propan-1-ol

CH3CH2CHOHCH3

sec- Butyl alcohol

Butan-2-ol

(CH3)3C—OH

tert- Butyl alcohol

2-Methyl propan-2-ol

CH2 — CH2 | | OH OH

Ethylene glycol

Ethane-1, 2-diol

CH2 — CH —CH2 | | | OH OH OH

Glycerol or Glycerine

Propane-1, 2, 3-triol

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Alcohol nomenclature

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

IUPAC names of some Phenols : OH OH





314 ]

OH

OH

OH



Common name Catechol IUPAC name Benzene-1, 2-diol

Resorcinol Benzene-1, 3-diol

OH

Hydroquinone or quinol Benzene-1, 4-diol

CH3

OH

CH3

CH3

OH OH

OH

m-Cresol 3-Methylphenol

o-Cresol 2-Methylphenol

p-Cresol 4-Methylphenol

Structure of alcohol : The oxygen of the —OH group is attached to sp3 hybridised carbon by a σ bond.





Phenol Phenol

Common name IUPAC name

H H–C

lone pairs O 108.50°

H 142 pm  Methods of preparation of Alcohols : (1) From haloalkanes : Haloalkanes are hydrolysed to the corresponding alcohols by treatment with aqueous alkali. CH3—Cl + KOH(aq) → CH3—OH + KCl

(2) From Aldehydes and Ketones : (i) Reduction : Aldehydes and ketones are reduced to primary and secondary alcohols respectively. The common reducing agents are lithium aluminium hydride (LiAlH4), sodium borohydride (NaBH4) or hydrogen gas in the presence of nickel or platinum as catalyst. O

(a)

CH3—C—H + H2



Ni or Pt

CH3—CH2—OH

Ethanal

Ethanol

O

(b)

OH

CH3—C—CH3 + H2



Ni

CH3—CH—CH3

2-Propanone

2-Propanol

(ii) Using Grignard reagent : This method is used to get all three types of alcohols. Formaldehyde (HCHO) reacts with Grignard reagent to give primary alcohol whereas other aldehydes give secondary alcohols. Ketones give tertiary alcohols. e.g., O

(a)

H—C—H + CH3MgBr Methanal Methyl magnesium (Formaldehyde) bromide





OMgBr H—C—H

H2O/H+

OH CH3CH2OH + Mg Ethanol

CH3

Hydroxy magnesium bromide

Adduct

(b)

(3) From Alkenes :



Adduct



H O

2 (i) CH2 = CH2 + H2SO4 → CH3CH2OSO3H  → CH3CH2OH + H2SO4

Ethene

Br

Ethanol

[ 315

ALCOHOLS, PHENOLS AND ETHERS

H2O

(ii) CH3 —CH = CH 2 + H2SO4  CH3 —CH – CH 3

OSO3H



CH3 +

CH3–C–CH3

2-methylpropene

OH





OH Isopropyl alcohol

CH3

(iii) CH3–C = CH2 + H2O H

CH3 — CH – CH3 + H2SO4

2-methylpropan-2-ol

(4) By hydroboration oxidation : As per anti-Markovnikov's rule H2O2 3R − CH=CH 2 +(BH3 )2  → (R − CH 2 − CH 2 )3 B   → R − CH 2 − CH 2 − OH+H3 BO3 OH− ,H2O (5) By oxymercuration-demercuration : As per Markovnikov's rule : CH3 – CH=CH2

(CH3COO)2Hg

CH3 –CH–CH2–HgOOCCH 3

H 2O

Propene

OH NaBH 4 OH – CH3 –CH–CH3 OH Propan–2–ol





(6) Reduction of carboxylic acids and esters : With the help of strong reducing agent, lithium aluminium hydride, carboxylic acids are reduced to primary alcohols. (i) LiAlH

4 RCOOH  → RCH2OH (ii) H O 2



Commercially, acids are reduced to alcohols by converting them to esters, followed by catalytic hydrogenation. H

R'OH 2 → RCOOR’ Catalyst  → RCH2OH + R’OH RCOOH  H+

 Preparation of Phenols : (i) From aryl halides



OH

O Na + NaOH



+

–

Cl 623 K 300 atm

HCl

Chlorobenzene

Phenol

Preparation of phenols

(ii) From benzene sulphonic acid SO3H

OH

(i) NaOH + (ii) H

Oleum



Phenol

(iii) From diazonium salts + –

NH2



Aniline

Scan to know more about this topic

N2Cl

OH

NaNO2/HCl

H2O, warm

273–278 K

–HCl – N2 Phenol

316 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

 Physical and Chemical properties of Alcohols and Phenols :

(1) Physical properties :



(i) Boiling points : The boiling points of alcohols and phenols increase with increase in the number of carbon atoms (increase in van der Waals forces). In alcohols, the boiling points decrease with increase of branching in carbon chain (because of decrease in van der Waals forces with decrease in surface area).



The –OH group in alcohols and phenols is involved in intermolecular hydrogen bonding resulting in high boiling point, which is lacking in ethers and hydrocarbons. R

R

H—O H—O H—O H—O— H

R—O

R

R H

H—O H—O H—O—



H—O H

O

H

O

(ii) Solubility : Solubility of alcohols and phenols in water is due to their ability to form hydrogen bonds with water molecules. The solubility decreases with increase in size of alkyl/aryl (hydrophobic) groups. R

R

H — O .... H — O .... H — O .... H — O .... H



R

(2) Chemical properties : Alcohols and phenols react both as nucleophiles and electrophiles. The bond between O–H is broken when alcohols react as nucleophiles. +

R — O — H+ + C —  R — O — C —  R — O — C — + H+ H



The bond between C–O is broken when they react as electrophiles.

+



Br– + CH2 — OH2  Br — CH2 + H2O



R The reactions of alcohols can be classified into : (a) Reactions involving the cleavage of O–H bond : (i) Reaction with metals :

R

– +

2R – O – H + 2Na



2R – O Na + H2

Sodium alkoxide

– + 1 CH3 – CH2 – ONa + H2 2 Sodium ethoxide

CH3 – CH2 – OH + Na

2 ((CH3) 3C – O(3 Al + 3H2

6(CH3)3 C – OH + 2Al 3° Butyl alcohol





Aluminium tert-butoxide

(ii) Esterification : H

+

R – COOH + R' – OH  R–COOR' + H2O Carboxylic acid

Alcohol

Ester H

+

(R – CO)2 O + R' – OH  R–COOR' + R – COOH Acid anhydride

Ester Pyridine

R – COCl + R' – OH



Acid chloride

R–COOR' + HCl Ester

[ 317



ALCOHOLS, PHENOLS AND ETHERS



(b) Reactions involving cleavage of C–O bond : Order of reactivity in such type of reaction is 3° Alcohol > 2° Alcohol > 1° Alcohol (i) Reaction with hydrogen halides : anhyd.ZnCl2

R – OH + H – X







(ii) Reaction with phosphorus halides :

(c) Reactions involving both the alkyl and hydroxyl group : (i) Dehydration : 3° Alcohol > 2° Alcohol > 1° Alcohol H2SO4

CH3CH2 – OH OH CH3 – CH– CH3

Ethene

85% H3PO4 440 K

Isopropyl alcohol

CH3

CH2 = CH2 + H2O

443 K

Ethyl alcohol



CH3 – CH = CH2+H2O Propylene

CH3

20% H3PO4

CH3 – C – CH3 OH

CH3 – C = CH2+H2O

358 K

Isobutylene

(ii) Oxidation : K2Cr2O7/H2SO4

R – CH2 – OH+ [O]



[O]

R – CHO

or alk.KMnO4 – H2O

1° Alcohol

R' CH – OH



R'

CrO3

C=O R

2° Alcohol

CH3



CH3 – C – CH3 OH

Hot conc. HNO3 [O], – H2 O

R – COOH Carboxylic acid

R



R – X + H 2O

Alkyl halide

Ketone

CH3 CH3 – C = CH2

– H2O, – CO2

CH3

[O]

CH3

[O]

C=O

– H2O, – CO2

Acetone

3° Butyl alcohol

CH3COOH Acetic acid

(iii) Dehydrogenation : R – CH2 – OH



Cu 573 K

1° Alcohol

Aldehyde

R CH – OH R'



Cu 573 K

2° Alcohol

CH3 CH3 – C – CH3 OH



R – CHO + H2 R' R

C = O + H2

Ketone

Hot conc. HNO3 – H2O

CH3 CH3 – C = CH2 + H2O Isobutylene

tert-Butyl alcohol

 Acidity of alcohols and phenols : The acidic character of alcohols is due to the polar nature of O–H bond. An electron releasing group (–CH3, – C2H5) increases electron density on oxygen tending to decrease the polarity of O–H bond. This decreases the acid strength. For this reason, the acid strength of alcohol decreases in the order : R R R

CH2OH >

CHOH R

Primary

> > R

C — OH R

Secondary

Tertiary

318 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



Alcohols can act as Bronsted acids as well as bases due to donation of proton and presence of unpaired electron on oxygen respectively.



Phenols are more acidic than alcohols and water. The hydroxyl group in phenol is directly attached to the sp2-hybridised carbon of benzene ring which acts as an electron withdrawing group. Due to this, the charge distribution in phenol molecule, as depicted in its resonance structures, causes the oxygen of –OH group to be positive.

O—H

+O—H

+O—H

+O — H

O—H



The ionisation of an alcohol and a phenol takes place producing alkoxide and phenoxide ions as shown in following equation : –

R—O—H

R—O +H

OH

O

+

–

+ H+

In alkoxide ion, the negative charge is localised on oxygen while in phenoxide ion, the charge is delocalised. The delocalisation of negative charge makes phenoxide ion more stable and favours the ionisation of phenol.







O



–

O

O

O

O

–

Distinction between Primary, Secondary and Tertiary Alcohols : (i) Lucas Test : Alcohol on treating with Lucas reagent forms a clear solution. Alkyl chlorides are formed on reaction which being insoluble results in the turbidity in the solution. Alcohol



HCl ZnCl2

Alkyl chloride + H2O

Tertiary alcohol is indicated, if turbidity appears immediately. Secondary alcohol is indicated, if turbidity appears within five minutes. Primary alcohol is indicated, if turbidity appears on heating. OH

(ii) Iodoform test : When ethanol or any alcohol containing the group CH3 – CH – is heated with iodine and aqueous NaOH or Na2CO3 solution at 333 – 343 K, a yellow precipitate of iodoform is obtained.



CH3OH + I2 + NaOH



CH3CH2CH2CHCH3 + 4I2 + 6NaOH OH





No reaction

CHI3+ CH3CH2CH2COONa+ 5NaI + 5H2O Iodoform

(iii) Ferric chloride test of phenols : Phenol gives a violet coloured water soluble complex with ferric chloride. 6C6H5OH+FeCl3







3–

+

[Fe (OC6H5)6] +3H + 3HCl Violet complex

[ 319







ALCOHOLS, PHENOLS AND ETHERS

Reactions of Phenol :

Na NaOH

C6 H5ONa+H2 C6H5ONa+H2O

Zn

+ZnO

Distillation NH 3

Alcohols, Phenols and Ethers

C6H5NH2+ H2O

ZnCl 2 PCl 5

C6 H5 C l

CH3 COCl

C6H5 COOCH3

Pyridine

C6 H5 COCl

C6H5OH

Scan to know more about this topic

NaOH

Na

C6H5COOC6H5

C6H5OCH3

CH3 l

OH Br

Br2 water

Br

Br

2,4,6- Tribromophenol

OH

OH

Br

Br2

+

CS 2

Br

o-and p-Bromophenol

OH

O2 N

Conc. HNO3

NO 2

Conc. H 2SO4

NO2

2,4,6-Trinitrophenol

OH

OH

Conc. H 2SO4

SO3 H

+ SO3 H

OH

NaOH+CO2 ,130 – 140 ° C

COOH

H+Kolbe- Schmidt reaction

o-Hydroxy benzoic acid

OH (i) CHCl 3 +NaOH (ii) H+

CHO

Reimer- Tiemann reaction

Salicylaldehyde

OH

OH NO2

Conc. HNO3

o-Nitrophenol

O

Na2Cr2 O7 H2SO4



O

Benzoquinone

+ NO2 p-Nitrophenol

320 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

 Reactions of Ethanol : (Lucas reagent)

 Differences between or identification of Methyl Alcohol and Ethyl Alcohol : S. No.

Test

Methyl alcohol

Ethyl alcohol

(i)

Boiling point

338K

351.2K

(ii)

On heating with I2 and NaOH.

No reaction.

Iodoform is obtained.

(iii) On heating with anhydrous sodium acetate and conc. sulphuric acid.

A specific odour of methyl acetate is produced.

Ethyl acetate is formed which has a sweet fruity odour.

(iv) On heating with salicylic acid and conc. H2SO4.

Methyl salicylate (oil of winter green) is formed which has a characteristic odour.

No specific smell.

(v)

On heating with bleaching powder. No reaction.

Chloroform with sweet smell is formed.

[ 321







ALCOHOLS, PHENOLS AND ETHERS

Methods of preparation of Glycol : CH2

H2O+[O]

CH2

Alkaline KMnO4 Ethane-1, 2-dioic acid 4

Ethane-1, 2-diol

Ethane-1, 2-diammine

 Methods of preparation of Glycerol : O CH2 — O — C — C17H35 + 3NaOH O Saponification CH — O — C — C17H35

CH2OH

CH2 — O — C — C17H35

CHOH

O Glyceryl stearate CH2

CH2Cl

CH + Cl2

273 K

CHCl

CH3

CH2

CH2OH

CH2Cl

CH CH2

HOCl (Cl2/H2O)

CHCl

CH2OH Aq.Na2CO3

NaOH

Glycerol or Propane1, 2, 3-triol

CH2

Know the Terms  Lucas reagent : An equimolar mixture of HCl and ZnCl2.  Wood spirit : Methanol is known as wood spirit as it is prepared by destructive distillation of wood.  Methylated spirit : Denatured ethyl alcohol which is unfit for drinking purposes is called methylated spirit.  Fusel oil : In the fermentation of starch, ethyl alcohol is prepared but in small amount. Some higher alcohols also form like isopentyl alcohol or isoamyl alcohol. This mixture is quite often called fusel oil.  Power Alcohol : It is the mixture of 20% alcohol and 80% petrol with ether, benzene or tetralin. It is used as a substitute for petrol for running internal combustion, engines in cars, scooters etc.  Lederer-Mannase reaction : Phenol condenses with formaldehyde in presence of acid or base to give bakelite (polymer). The reaction is known as Lederer-Mannase reaction.  Rectified spirit : It contains about 95.5% Ethyl alcohol +4.5% water.

322 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII





Q. Identify the product formed when propan-1-ol is treated with conc. H2SO4 at 413 K. Write the mechanism involved for the above reaction. Solutions: 1-Propoxypropane is formed.  Mechanism involved: STEP-1 : Formation of protonated alcohol



+ +

H2SO4 H2SO4 H2SO4



:: :

CH3CH2CH2OH + H CH3CH2CH2OH + H CH CH2CH2OH + H+ 3 Propan-1-ol Propan-1-ol Propan-1-ol attack Step 2 : Nucleophilic Step 2 : Nucleophilic attack STEP-2 : Nucleophilic attack Step 2 : Nucleophilic attack CH3CH2CH2 O : + CH3 – CH2 – CH2 CH3CH2CH2 O : + CH3 – CH2 – CH2 CH3CH2CH2 O : + CH3 – CH2 – CH2 H H H

++ +

Step 3 : Deprotonation Step 3 : Deprotonation Step 3 : Deprotonation + STEP-3 : Deprotonation + CH CH CH O CH2CH2CH3 CH33CH22CH22 O CH2CH2CH3 + CH3CH2CH2 O CH2CH2CH3 H H H

CH3CH2CH2 CH3CH2CH2 CH3CH2CH2 H H O H O O H H H

O++ O O+

H H H H H H

CH3CH2CH2 CH3CH2CH2 CH3CH2CH2

+ + O O + O H H + H + H+ O H22O H2O

CH CH CH CH22CH22CH33 CH2CH2CH3

CH CH CH CH CH CH + H+ O CH33CH22CH22 CH22CH22CH33 + H+ O CH3CH12CH CH CH2CH3 + H+ O - Propoxypropane 2 2 1 - Propoxypropane 1 - Propoxypropane

Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS : Q. 1. How many alcohols with molecular formula C4H10O are chiral in nature? (a) 1 (b) 2 (c) 3 (d) 4 A [NCERT Exemp. Q. 2, Page 154] Ans. Correct option : (a) Explanation: CH3 - CH 2 - C H - CH3 | OH Butan-2-ol is chiral in nature as it possesses chiral center. Q. 2. IUPAC name of m-cresol is ___________. (a) 3-methylphenol (b) 3-chlorophenol (c) 3-methoxyphenol (d) benzene-1,3-diol A [NCERT Exemp. Q. 8, Page 155] Ans. Correct option : (a) Explanation: CH3 3 2 1 OH

(1 mark each) (i) -OH is functional group and –CH3 is substituent. (ii) IUPAC name : 3-methylphenol. Q. 3. Phenol is less acidic than ______________. (a) ethanol (b) o-nitrophenol (c) o-methylphenol (d) o-methoxyphenol U [NCERT Exemp. Q. 12, Page 156] Ans. Correct option : (b) Explanation : In o-nitrophenol, nitro group is present at ortho position. Presence of electron withdrawing group at ortho position increases the acidic strength. On the other hand, in o-methylphenol and in o-methoxyphenol electron releasing group (–CH3 or –OCH3), at ortho or para positions of phenol decreases the acidic strength of phenols. So, phenol is less acidic than o-nitrophenol. [B] ASSERTION AND REASON TYPE QUESTIONS : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a)  Assertion and reason both are correct statements and reason is correct explanation for assertion.

[ 323

ALCOHOLS, PHENOLS AND ETHERS



 Remember IUPAC rules and write clearly as suffix -ol, -al look similar, you may lose marks for unclear writing. Q. 2.  Write the IUPAC name of the following compound: CH=CH–CH2–OH





A [CBSE OD Set-2 2017]

[Topper’s Answer 2017]





Ans. 3-phenyl-prop-2-en-l-ol. [CBSE Marking Scheme 2017]

Q. 3. Write the IUPAC name of the following compound : 

A [CBSE Foreign Set-1 2017]

C6H5 — CH2 — CH2 — OH Ans. 2-Phenylethanol. [CBSE Marking Scheme 2017]



Q. 4. Write the IUPAC name of the following compound : CH2 CH2 = CH — C — OH CH2



A [CBSE Foreign Set-2 2017]

Ans. 2-Methylbut-3-en-2-ol. [CBSE Marking Scheme 2017]



Q. 5. Write the IUPAC name of the following compound : CH3 C – OH CH3

C O

R

O

A [CBSE Foreign Set-3 2017]



R

Answering Tip

—

C O

 Students often make mistakes while using suffix.

—

C R

Commonly Made Error

—

 A&E [CBSE SQP 2021] Ans. Correct option: (b) Explanation: Carboxylic acids are more acidic than phenols as the carboxylate ion, the conjugate base of carboxylic acid is stabilized by two equivalent resonance structures. Thus, the negative charge is delocalized effectively. However, in phenols, negative charge is less effectively delocalized over oxygen atom and carbon atoms in phenoxide ion. O O O

Ans. 2-Bromo-3-methylbut-2-en-1-ol. [CBSE Marking Scheme 2017]



—

(b)  Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c)  Assertion is correct statement but reason is wrong statement. (d)  Assertion is wrong statement but reason is correct statement. Q. 1. Assertion (A): C2H5OH is a weaker base than phenol but is a stronger nucleophile than phenol. Reason (R): In phenol the lone pair of electrons on oxygen is withdrawn towards the ring due to resonance. U [CBSE SQP 2020] Ans. Correct option: (d) Explanation: C2H5OH is a weaker acid than phenol as in phenol the lone pair of electrons on oxygen is withdrawn towards the ring due to resonance, leaving oxygen with a positive charge thus increasing its acidity. Q. 2. Assertion (A): Ortho and para-nitrophenol can be separated by steam distillation. Reason (R): Ortho isomer associates through intermolecular hydrogen bonding while para isomer associates through intermolecular hydrogen bonding. U [CBSE Delhi Set 2 2020] Ans. Correct option: (c) Explanation: Ortho isomer has intermolecular H-bonding while para isomer has intermolecular hydrogen bonding. Q. 3. Assertion (A): (CH3)3C−O−CH3 gives (CH3)3C−I and CH3OH on treatment with HI. Reason (R): The reaction occurs by SN1 mechanism.  U [CBSE OD Set 1 2020] Ans. Correct option: (a) Explanation: (CH3)3C−O−CH3 gives (CH3)3C−I and CH3OH on treatment with HI. The reaction occurs by SN1 mechanism. Q. 4. Assertion (A): Carboxylic acids are more acidic than phenols. Reason (R): Phenols are ortho and para directing.

H3C — C = C — CH2 — OH | | CH3 Br 



[C] VERY SHORT ANSWER TYPE QUESTIONS Q. 1. Write the IUPAC name of the following compound :



Ans. Phenylethanol-2-ol. [CBSE Marking Scheme 2017]

Q. 6. Write the IUPAC name of the following compound :

CH3

CH3  — C  — CH  — CH3

A [CBSE OD Set-1 2017, 2013]

C2H5 

OH A [CBSE Delhi/OD 2018]

324 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q. 8. What happens when phenol is heated with zinc dust?  R [CBSE Comptt. OD Set-2 2017]

Ans. 3,3-Dimethylpentan-2-ol. [CBSE Marking Scheme 2017]  OR

Ans. OH

+ Zn

+ ZnO / Benzene is formed

[CBSE Marking Scheme 2017]





[Topper’s Answer 2018]

Q. 7. What happens when phenol is oxidized by Na2Cr2O7/H2SO4? R [CBSE Comptt. OD Set-1 2017] Ans.

OH

Q. 9. What happens when phenol is heated with bromine water?  R [CBSE Comptt. OD Set-3 2017] Ans.

OH

OH Br

O + 3Br2







+ ZnO



Na2Cr2O2 H2SO4

Br

O \ Benzoquinone is formed. [CBSE Marking Scheme 2017]



Br 2, 4, 6- Tribromophenol is formed. [CBSE Marking Scheme 2017]

Short Answer Type Questions-I

(2 marks each)

Q. 1. (i) Predict the major product of acid catalysed dehydration of 1-Methylcyclohexanol.

(ii) You are given benzene, conc. H2SO4, NaOH and dil.HCl. Write the preparation of phenol using these reagents.

Ans. (i) 1-Methylcyclohexene conc.H SO .∆

NaOH,fuse,575K

[1] dil.HCl

2 4 → C6 H 5SO3 H  → C6 H 5ONa  → C6 H 5OH (ii) C6 H 6 

Q. 2. Draw the structures of any two isomeric alcohols (other than 1° alcohols) having molecular formula C5H12O and give their IUPAC names. A [CBSE SQP 2015]

Ans. Any two isomers out of the following : (i) CH3—CH2—CH2—CH(OH)—CH3

entan-2-ol P (ii) CH3—CH2—CH(OH)—CH2—CH3 Pentan-3-ol (iii) CH —CH(CH )CH(OH)—CH 3-Methybutan-2-ol 3 3 3 (iv) CH3—CH2—C(CH3)(OH)—CH3 2-Methylbutan-2-ol

[1] [½ + ½] [½ + ½]

Commonly Made Error  Students often write incomplete or unbalanced equations.

Answering Tip  While writing reactions, do not forget to write the reagents and conditions involved. Ensure that the equations are balanced.

Q. 3. How do you convert: (i) Phenol to toluene (ii) Formaldehyde to Ethanol

A [CBSE OD Set-2 2016]

[ 325

ALCOHOLS, PHENOLS AND ETHERS

Ans.



2 [Topper's Answer 2016] Q. 4. (i) Arrange the following compounds in the increasing order of their acid strength:   p-cresol, p-nitrophenol, phenol (ii) Write the mechanism (using curved arrow notation) of the following reaction; +

CH2 = CH2 + H3O+ ® CH3 – C H2 + H2O



U+A



1



Ans. (i) p-cresol < Phenol < p-nitrophenol

O

(i)

(ii)





CH3 CH3

Cl H 3C

Answering Tip  Use arrows to show the transfer of electrons.

O

H 3C

OR

H 3C

Cl

1



 Students often get confused while showing the mechanism of reactions. Need to show the electron transfer, formation of intermediates, charges acquired by the reactants etc.

H 3C

CH3 CH3

1 [CBSE Marking Scheme 2017]



Commonly Made Error

Q. 5. Write the structures of the products when Butan-2-ol reacts with the following: (i) CrO3 (ii) SOCl2 A [CBSE OD Set-1, 2, 3 2017] O HC Ans. (i) 3 O CH 3 H 3C CH3



1

ClCH 3 CH3

[CBSE Marking Scheme 2017] 1





H 3C



H 3C



Cl

(ii)

[Topper's Answer 2017] 2



326 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q. 6. Write the mechanism of acid dehydration of ethanol to yield ethene. A [CBSE SQP 2018-2019]



Ans.

[CBSE Marking Scheme 2018]

Commonly Made Error  Students often write incomplete mechanism or do not show all the steps properly.

Answering Tip  Mention each step involved in the mechanism.

Short Answer Type Questions-II



—

1



Ethanoic acid

Br p-bromo anisole



[CBSE Marking Scheme 2015]



Commonly Made Error  Students often miss out on writing the reagents or conditions of the conversion reactions.

1

H+ – HCl

Answering Tip





CH2OH

—

+ NaOH

CH2ONa

—

CH2Cl

— Br

+

Anisole



OH —

(ii)



Br2 in

1

CH3 — CH — CH3

—

CH3CH = CH2 + H2O

OCH3

—

H+

OCH3

OCH3

—

Ans. (i)

(iii)

—

Q. 1. How are the following conversions carried out ? (i) Propene to propane-2-ol (ii) Benzyl chloride to Benzyl alcohol (iii) Anisole to p-Bromoanisole A [CBSE Comptt. Delhi 2015]

(3 marks each)

 Write the reagents involved in the conversions. Q. 2. How are the following conversions carried out ? (i) Benzyl chloride to Benzyl alcohol (ii) Ethyl magnesium chloride to Propan-1-ol (iii) Propene to Propan-2-ol Ans. (i) Benzyl chloride to Benzyl alcohol

A [CBSE Comptt. OD 2015]

CH2Cl

CH2OH

aq. NaOH

Benzyl chloride (ii) Ethyl magnesium chloride to propan-1-ol O

=



CH3–CH2MgCl + H – C – H

Benzyl alcohol 1

Dry ether

Ethyl magnesium chloride

CH3–CH2–CH2–OMgCl H3 O

+

CH3–CH2–CH2–OH Propan-1-ol

(iii) Propene to propan-2-ol







CH3CH = CH2 Propene

H2O H2SO4

OH CH3–CH–CH3 Propan-2-ol



1

1

[ 327



ALCOHOLS, PHENOLS AND ETHERS

Q. 3. Write the major products in the following equations : (i) CH3–CH2OH



PCl5

?

OH

(ii)

+ CH3Cl

anhyd.AlCl3

?

(iii) CH3–Cl + CH3CH2 – ONa → ?

A [CBSE Comptt. OD 2015]

PCl 5

Ans.(i) → CH3CH2Cl CH3—CH2OH 

1 OH

OH

OH —

Anhyd. AlCl3

+ CH3Cl

—

—

—

(ii)

CH3 +

—



1

CH3





(iii) CH3Cl + CH3CH2–ONa → CH3CH2–O–CH3

1 [CBSE Marking Scheme 2015]

Q. 4. How do you convert the following : (i) Phenol to anisole (ii) Propan-2-ol to 2-methylpropan-2-ol (iii) Aniline to phenol Ans. (i)





1



OH

O

CH3 – CH – CH3

K2Cr 2O7 / H2SO4

CH3 – C – CH3 Propanone CH3MgBr Dry ether

Propan-2-ol



OH CH3 – C – CH3

H2O / H+

CH3



1

OMgBr CH3 – C – CH3 CH3



(ii)

2-Methyl propan-2-ol (iii)

Q. 5. (i) Write the mechanism of the following reaction :



1

+

H 2CH3CH2OH → CH3CH2 – O – CH2CH3 (ii) Write the equations involved in the acetylation of Salicylic acid. 

A [CBSE Delhi 2015]

H+

½



..



→ CH3CH2 – O – CH2CH3 Ans. (i) 2CH3CH2OH  H .. .. + CH3–CH2–OH CH3–CH2–O– .. H .. + H + .. + H CH3–CH2–O +C H2–CH3–O CH3–CH2–O–CH2CH3 + H2O H H H

1

328 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII +

+

CH3CH2–O–CH2CH3 + H

CH3–CH2–O –CH2CH3 H

Ethoxy ethane

½



(ii)

1



Commonly Made Error  Students often write just the reactants and products without properly showing the electron transfer.

Answering Tip  Use arrows to show the transfer of electrons. Q. 6. Write the main product(s) in each of the following reactions : —

CH3

?

—

(i) CH3 — C — O — CH3 + HI

CH3



(i) B H

2 6  → ? (ii) CH3–CH = CH2 (ii) 3H O /OH -



2

2

(i) aq.NaOH

(iii) C6H5–OH  + → ?

A [CBSE Delhi 2016]

(ii) CO2 ,H

CH3

Ans.(i)

—

—

CH3 — C — O — CH3 + HI  CH3 — C — I + CH3OH





—

—

CH3

CH3

CH3

Methanol

1



(i) B 2 H 6 (ii) CH3 – CH = CH2 CH3CH2CH2OH (ii) 3H 2O2 / OH

1

Propanol

(iii)



2-Hydroxybenzoic acid (Salicylic acid)



1 [CBSE Marking Scheme 2016]

Commonly Made Error  Students often write incomplete equations or just write the name of the products thus losing marks.

Answering Tip  Draw the structure and the name of the product obtained.



—

CH3

—

(i) CH3 — C — O — CH3 + HI CH3

,



(ii) CH3 — CH2 — CH — CH3

Cu/573K

,

—

Q. 7. Write the final product(s) in each of the following reactions :

OH CHCl3 +aq.NaOH (iii) C6H5 — OH (i) → + (ii)

H

A [CBSE OD Central 2016]

[ 329

ALCOHOLS, PHENOLS AND ETHERS

Ans. (i)

CH 3 | CH3 ¾ C ¾ I +CH3I + H2O | CH 3 1

Q. 8. What happens when (i) (CH3)3C – OH is treated with Cu at 573 K, (ii) Anisole is treated with CH3Cl/anhydrous AlCl3, (iii) Phenol is treated with Zn dust? Write chemical equation in support of your answer. A [CBSE Foreign Set-1, 2, 3 2017]

CH 3 | CH3 ¾ C ¾ O—CH3 + 2HI | CH 3

Ans. (i) (CH3)3 C-OH undergoes dehydration. CH3

2-Iodo-2-methylpropane



Cu CH3 — C — OH 573 K CH3

/ 573 K (ii) CH3—CH2 ¾ CH ¾ CH3 Cu  → | OH

CH3—CH2 ¾ C ¾ CH3 + H2 || O

(iii) 

OCH3

CH3 — C = CH2

OCH3

OCH3

+CH3 Cl

OH —

C6H5 — OH

CH3

(ii) Methyl group is introduced at ortho and para positions. ½+½

1

Butanone-2

Anhyd. AlCl3

CH3 +

CS2

CH3

CHO

—

(i) CHCl3 + aq. NaOH

½+½

(ii) H+

Salicylaldehyde





1

(iii)

+ Zn

[CBSE Marking Scheme 2016]

+ ZnO

½+½ [CBSE Marking Scheme 2017]

Q. 9. (i) Why phenol is more acidic than ethanol? (ii) Write the mechanism of acid dehydration of ethanol to yield ether : H . .H+ .. + OCHCH + O— H 2CH OH CH CH CH — — — — — CH CH O H + H CH . . 3CH 2 3 2 2 3 .. 3 3 2 413K 2 AE + A [CBSE Comptt OD Set-I, III 2017]





H Ans. (i) Due to resonance, phenoxide ion is more stable .. . . than phenol whereas there is no resonance is alkoxide ion / + + O— H CH3— CHwith O H + H CH CH — — — — explained the help of resonating structures. 1 ..  3 2 .. 2 (ii) Mechanism of acid dehydration of ethanol : + .. + —H H ½ .—. CH CH3—CH2—O. . +CH3— +CH2— O CH —CH2— CH3+H2O (a) 3 2—O + CH3— CH2— O CH3H— CH2— O —H .. — H + H . . H H (b) + H .. + — 1 CH3—CH2—O —CH2— CH3+H2O CH3—CH2—O +CH3— CH2— O H H H

—

—

—

..

—

—

—

..

—

.. ——

—

+ H + (c) CH —CH —O. .—CH — CH — O + —CH + — CH CH2—3O —CH —CH CH33—CH22—O +CH 3—CH 2 3— CH 3 2— O 2CH 3 +2—HCH3+H2O 2—O H H H H



½ [CBSE Marking Scheme 2017]

+

—

Q. 9. (i) Write reaction. CH3Reimer-Tiemann —CH2—O—CH2— CH CH3—CH2—O— CH2CH3 + H+ 3— O (ii) Write the mechanism of acid dehydration of ethanol to yield ethene: H +





R + A [CBSE Comptt. OD Set-2 2017]

—

H CH3CH2OH+443  → CH2 = CH2 + H2O K CH3—CH2—O—CH2— CH3— O CH3—CH2—O— CH2CH3 + H+ H

1



Ans. (i)



330 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(ii) Step 1: Formation of protonated alcohol. H H

H H H —

Fast

—

—

—

—

—

—

+ H—C—C—O .. —H

—

—

.. + H—C—C—O—H+H .. H H

H H

Ethanol

Protonated alcohol [Ethyl oxonium ion]







Step 2: Formation of carbocation: It is the slowest step and hence, the rate determining step of the reaction. ½ H H H

—

—

—

—

—

Slow

H H

—

—

—

—

H—C—C++H2O H H







Step 3: Formation of ethene by elimination of a proton.

— H

H H

C=C —

+ H+

—



—

—



—

—

—

H—C—C

H

H

+

Ethene

H

[CBSE Marking Scheme 2017]

Q. 11. Explain the following behaviors: (i) Alcohols are more soluble in water than the hydrocarbons of comparable molecular masses.

(ii) Ortho-nitrophenol is more acidic than orthomethoxyphenol. (iii) Cumene is a better starting material for the preparation of phenol. A&E [CBSE SQP 2017] Ans. (i) Because of H-bond formation between alcohol and water molecule. (ii) Nitro being the electron withdrawing group stabilises the phenoxide ion. (iii) Side product formed in this reaction is acetone which is another important organic compound. 1+1+1 [CBSE Marking Scheme 2017]  Detailed Answer: (i) Alcohols are able to form H-bonds with water due to presence of (–OH) group. Hence, they are soluble in water. -----H—O-----H—O-----H—O-----H—O-----



R

—

H

—

R

—

—



1



H H



½

H H

+ H—C—C—O .. —H





H

Whereas hydrocarbons are not able to form H-bonds with water hence are insoluble in water. (ii) Nitro-group is an electron-withdrawing group. The presence of this group in the ortho position decreases the electron density in the O-H bond resulting in easier loss of proton. The o-nitrophenoxide ion formed after the loss of proton is stabilised by resonance. Hence, ortho-

nitrophenol is a stronger acid. Whereas, methoxy is an electron-releasing group which increases the electron density in the O-H bond, releasing proton difficult. Thus ortho-nitrophenol is more acidic that ortho-methoxyphenol. (iii) As oxidation of cumene results in the formation of phenol and acetone as by–product which is a commercial product used as chemical. Q. 12. (i) Complete the following reaction and suggest a suitable mechanism for the reaction : +

, 443 K CH3CH2OH H →



(ii) Why ortho-Nitrophenol is steam volatile while para-Nitrophenol is less volatile? A + A&E [CBSE Comptt. Delhi/OD 2018] Ans.

C2H5OH

H2SO4 443 K

CH2 = CH2 + H2O

½

[ 331



ALCOHOLS, PHENOLS AND ETHERS

(ii) o-Nitrophenol is steam volatile due to intramole­ cular hydrogen bonding while p-nitrophenol is less volatile due to intermolecular hydrogen bonding. 1

½





[CBSE Marking Scheme 2018]

. 13. Give reasons for the following: Q (a) Protonation of Phenols is difficult whereas ethanol easily undergoes protonation. (b) Boiling point of ethanol is higher than that of dimethyl ether. (c) Anisole on reaction with HI gives phenol and CH3– I as main products and not iodobenzene and CH3OH. A&E [CBSE OD Set-2 2016]



[Topper's answer 2016] 3 OR (i) In phenol, the lone pair of electrons on oxygen involves in delocalization which results in their nonavailability for the protonation. Whereas in ethanol, the electrons on oxygen atom are not delocalised which results in their availability for protonation. (ii) The variation can be established by intermolecular hydrogen bonding, where alcoholic hydrogen is bounded to a strongly electronegative oxygen atom to give rise to dipole of the form, Et–Oδ- – Hδ+ which arises the boiling point.



Whereas, C-O bonds in dimethyl ether are polar. Carbon bonded with hydrogen is moderately electronegative compared to oxygen. In the absence of intermolecular hydrogen bonding, ether becomes more volatile. (iii) Attack of I–ion on anisole is through SN2 mechanism, which attacks the lesser hindered CH3 group forming CH3I. Also, C-O bond in ether is difficult to break due to resonance.

332 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



Q. 14. Give the structures of final products expected from the following reactions : (i) Hydroboration of propene followed by oxidation with H2O2 in alkaline medium. (ii) Dehydration of (CH3)3C−OH by heating it with 20% H3PO4 at 358K. (iii) Heating of

with HI.

— CH2 —O—

Ans. (i) 6CH3CH  CH 2  ( BH3 )2

2(CH3CH 2CH 2 )3 B Alkaline medium +HO 2 2

6CH3CH 2CH 2OH Propanol-1



CH3

(ii)

CH3 – C – OH

20% H3PO4

CH3 – C = CH2 + H2O

358K

CH3 2 methylpropene (Isobutylene)

CH3 t-butyl alcohol (2 methyl-2-propanol)

(iii)

CH2 — O

+

ICH2

+ HI

OH

Q. 15. How can you convert the following? (i)  Phenol to o-hydroxybenzaldehyde (ii)  Methanal to ethanol (iii)  Phenol to phenyl ethanoate (i) By Reimer-Tiemann reaction : OH

A [CBSE Delhi Set 1 2020]

ONa CHCl2

NaOH

+ CHCl3

2NaOH

Chloroform

ONa

ONa CH(OH)2

ONa CHO

–H2O



CHO dil. HCl



(ii)

o-hydroxybenzaldehyde

H H

C=O

CHMgBr 3

H H

C

OMgBr

HO/H 2

H H

+

CH3

Ethanol

Methanal



OH CH3 + Mg(OH)Br

C

(iii) C 6 H 5OH  CH 3CO O 2

CH3COOC 6 H 5  CH3COOH Phenylethanoate

Acetic anhydride

OH

COOCH3 + CH3COCl

+ HCl

Acetyl chloride





Phenylethanoate



Q. 16. (a)  Write the mechanism of the following SN1 reaction : Aq. NaOH (CH3)3C−Br (CH3)3C−OH + NaBr (b)  Write the equation for the preparation of 2-methyl-2-methoxypropane by Williamson synthesis. A [CBSE OD Set 1 2020]  Ans. (a)

(CH3) 3 — C— B r

Aq. NaOH

(CH3 )3 — C — OH + NaBr CH3

Step-I :

CH3

CH3 – C – Br CH3 Step-II :

CH3

C CH3

+ Br CH3 CH3

–

(slow step)

(CH3) 3 — C— B r

Aq. NaOH

(CH3 )3 — C — OH + NaBr CH3

Step-I :

CH3

CH3 – C – Br

ALCOHOLS, PHENOLS AND ETHERS

Step-II :

+ OH–

(slow step)

[ 333

CH3

CH3 – C – OH + NaBr (fast step) CH3





–

CH3

CH3

H 3C

+ Br

CH3

CH3

CH3 C



C

(b) Williamson synthesis : CH3

CH3

CH3 – C – Br + NaOCH3

CH3 – C – O – CH3 + NaBr

CH3

CH3 OR

CH3

CH3

CH3 – C – ONa + CH3 – Br





Heat

CH3 – C – OCH3 + NaBr

CH3 Sodium-2-methyl-2-propoxide

CH3 2-methyl-2-methoxy propane

Long Answer Type Questions

(5 marks each)

—

Q. 1 (i) Write the formula of reagents used in the following reactions: (a) Bromination of phenol to 2,4,6-tribromophenol (b) Hydroboration of propene and then oxidation to propanol. (ii) Arrange the following compound groups in the increasing order of their property indicated: (a) p-nitrophenol, ethanol, phenol (acidic character) (b) Propanol, Propane, Propanal (boiling point) (iii) Write the mechanism (using curved arrow notation) of the following reaction: + + CH3CH2OH CH3 – CH2 – OH2 CH3 – CH2 – O – CH2 – CH3 + H2O A + U [CBSE Delhi Set-1,2,3 2017] H

1 1 1 1



Ans. (i) (a) Aq. Br2 (b) B2H6, H2O2 and OH- (ii) (a) ethanol (CH3)3N

 

[CBSE Marking Scheme 2019] Q.8. Arrange the following in decreasing order of the basic character : C6H5NH2, (CH3)3N, C2H5NH2



[CBSE, Outside Delhi Set - 1, 2019]





 Some students give the structure of primary or secondary amine.

Commonly Made Error



 Some students get confused and give the structure of secondary amine in place of primary amine.

[Topper’s Answer 2019]



Commonly Made Error



Ans. (CH3)3N > C2H5NH2 > C6H5NH2 [CBSE Marking Scheme 2019]

Detailed Answer :

C6H5NH2 is least basic due to presence of I-group i.e. – C6H5 group which withdraws unshared pair of electrons present on N-atom of –NH2 group due to which it does not undergo protonation easily; whereas (CH3)3N is most basic due to strong +I effect of –CH3 groups which release their bonding electrons towards N-atom and it undergoes protonation easily. Hence, the decreasing order of the basic character is (CH3)3N > C2H5NH2 > C6H5NH2

[ 413

AMINES

Q.9.  Arrange the following in increasing order of pK6 values : C6H5CH2NH2, C6H5NHCH3, C6H5NH2 U [CBSE, Outside Delhi Set - 2, 2019]



Q.11. Write IUPAC name of the following compound : (CH3CH2)2NCH3 U [CBSE, Delhi Set - 1, 2017]





Ans. C6H5CH2NH2 < C6H5NHCH3 < C6H5NH2 [1]  [CBSE Marking Scheme 2019]

part (hydrophobic) which retards the formation of hydrogen bonding. [1]



Ans. N-Ethyl–N–methylethanamine. [1] [CBSE Marking Scheme, 2017]





Ans. C2H5NH2 > (C2H5)2NH > C6H5NH2



[1]

[CBSE Marking Scheme 2019]

Detailed Answer: (C2H5)2NH and C2H5NH2 are aliphatic amines and C6H5NH2 is aromatic amine. Lower aliphatic amines can form hydrogen bonds with water molecules. Therefore, such amines are soluble in water. When number of alkyl groups (hydrophobic) increases then molar mass of amines increases. This usually results in a decrease in its solubility in water. Hence, C2H5NH2 is more soluble in water than (C2H5)2NH. Whereas, aromatic amines are insoluble in water because of large hydrocarbon

CH3NHCH(CH3)2 U [CBSE, Delhi Set - 2, 2017]



Ans. N-Methylpropan-2-amine.

[1]

[CBSE Marking Scheme, 2017]



Q.13. Write IUPAC name of the following compound : (CH3)2N — CH2CH3 U [CBSE, Delhi Set 3, 2017]



Ans. N, N-Dimethylethanamine [1] [CBSE Marking Scheme, 2017]

Q.10. Arrange the following in decreasing order of solubility in water : (C2H5)2NH, C2H5NH2, C6H5NH2 U [CBSE, Outside Delhi Set 3, 2019]

Q.12. Write IUPAC name of the following compound :

Q.14. Complete the following reaction equation : C6H5N2Cl + H3PO2 + H2O → ....................... 

R [CBSE Comptt. Delhi 2015] + –

Ans. ArN2Cl+ H3PO2 + H2O → ArH + N2 + H3PO3 Benzene + HCl (where Ar is C6H5) [1] [CBSE Marking Scheme 2015]

Detailed Answer: Greater the basic nature of amine, lesser will be the pK6 value of amine. Since, C6H5CH2NH2 is most basic, so, it possesses least pK6 value, while, C6H5NH2 is least basic, so, it possesses highest pK6 value. Hence, the increasing order of pK6 values is C6H5CH2NH2 < C6H5NHCH3 < C6H5NH2 [1]

Short Answer Type Questions-I

(2 marks each)

Q. 1. Write the chemical equations involved in the following reactions :

 (i) Hoffmann-bromamide degradation reaction, (ii) Carbylamine reaction.

R [CBSE Outside Delhi Set 1, 2 and 3, 2016]

O || Ans. (i) Ar/R — C — NH2 + Br2 + 4NaOH —→ Ar/R—NH2 + Na2CO3 + 2NaBr + 2H2O ∆

→ Ar/R—NC + 3KCl + 3H2O (ii) Ar/R—NH2 + CHCl3 + 3KOH 

(where R = alkyl group, Ar = aryl group)

[1] [1] [CBSE Marking Scheme, 2016]

Detailed Answer :

(i) Hoffmann – bromanide degradation reaction : When an amide is treated with bromine in aqueous or ethanolic solution fo sodium hydroxide, a primary amine with one carbon atom less than the origin amide is produced. This degradation is known as Hoffmann bromamide degradation reaction.

414 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

O || R — C — NH2 + Br2 + 4NaOH → R — NH2 + Na2 + Na2CO3 + 2NaBr + 2H2O



(Aqueous or alcoholic) 1° Amine

Example,



[1] (ii) Carbylamine reaction : It is used as a test for detection of primary amines. When aliphatic or aromatic primary amines are heated with chloroform and alcoholic potassium hydroxide, carbylamines or isocyanides having foul smell are formed. Secondary and tertiary amines do not respond to this test. 

[1]

Commonly Made Errors  Hoffmann’s degradation reaction : Some students are not able to write this equation correctly. On the product side, only alkyl amine was written in several cases; all the products formed were not mentioned by candidates.  A number of students do not mention alcoholic KOH. Some give incomplete equations and some do not mention by products.

Answering Tip  Write complete, balanced equations. Practice by writing all the named reaction equations. Q.2.   Give reasons : (i)   Electrophilic substitution in aromatic amines takes place more readily than benzene.

(ii)  Aniline does not undergo Friedel Crafts reaction. Ans. (i)

 U Ans. (i) —NH2 group of aromatic amines strongly activates the aromatic ring through delocalization of the lone pair of electrons of the N-atom over the aromatic ring. Due to the strong activating effect of the —NH2 group, aromatic amines undergo electrophilic substitution reactions readily than benzene. [1] (ii) Due to resonance, the lone pair of electrons on the nitrogen atom in CH3CONH2 is delocalized over the keto group.

+ – N2Cl

NH2

(ii)   CH3CONH2 is a weaker base than CH3CH2NH2.

OH H2O /H+

NaNO2 /HCl 273–278 K Aniline Aniline

Bemene Bazene diawnium diazonium chloride cloride (A) (A)

Phenol Phenol (B)

(B)

(ii) A Friedel–Crafts reaction is carried out in the presence of anhydrous AlCl3. But AlCl3, used as catalyst is acidic in nature i.e., Lewis acid whereas aniline is a strong Lewis base. Thus, aniline reacts with AlCl3 to form a salt.



As a result, electron density on the N-atom in CH3CONH2 decreases. On the other hand, in C2H5NH2, due to +I effect of the ethyl group, the electron density on the N-atom increases consequently, CH3CONH2 is a weaker base than CH3CH2NH2. [1]

Q.3.   Give reasons : (i)  Identify ‘A’ and ‘B’:





Due to the positive charge on the N–atom, electrophilic substitution in the benzene ring is deactivated. Hence, aniline does not undergo Friedel–Crafts reaction. [1] Q.4.   Why does acetylation of —NH2 group of aniline reduce its activating effect ? U Ans.  Direct nitration of aniline is not possible on account of –NH2 group present. However, nitration can be carried out after protecting the –NH2 group by





[ 415

AMINES



acetylation to give acetanilide which is then nitrated and finally hydrolysed to give o- and p-nitroanilines. The acetyl group being electron withdrawing attracts the lone pair of electrons of the N - atom towards carbonyl group. As a result, the activation effect of –NH2 group is reduced, that is, that lone pair of electrons on nitrogen is less available for donation to benzene ring by resonance. Therefore, activating effect of NHCOCH3 group is less than that of NH2 group. [2] Q.5.   Give reasons : (i) Aniline is a weaker base than cyclohexylamine. (ii) It is difficult to prepare pure amines by ammo­ nolysis of alkyl halides. A

Ans.  (i) Cyclohexylamine is more basic than aniline because aniline is a resonance hybrid of various resonance structures. As a result, in aniline the electron donating capacity of nitrogen for protonation is considerably decreased.[1] (ii)  Ammonolysis of alkyl halides does not give single amine but gives a mixture of primary, secondary and tertiary amines. NH / 343 K

3 → C2H 5NH 2 C2H 5I  - HI C2 H5 I

C H I

2 5  → - HI C H I

2 5 → (C H ) N ¾¾¾¾ ® C 2H 5) 2NH  - HI 2 53 –  [(C2H5)4N+]I [1] Q.6.    Give two chemical tests to identify primary, secondary and tertiary amines. R

Ans. Identification of primary, secondary and tertiary amines : S.No. Test (i) Reaction with nitrous acid

(ii)

Primary amine Gives alcohol with effervescence of N2 gas.

Secondary amine Gives oily nitrosoamine which gives Liebermann's nitrosoamine test. Reaction with benzene Gives N-alkyl benzene- Gives N, N-dialkyl sulphonyl chloride sulphonamide which is benzene sulphonamide (Hinsberg's reagent). soluble in alkali. which is insoluble in alkali.

Tertiary amine Forms nitrite in cold soluble in water and on heating gives nitrosoamine No reaction.

[1+1]

Short Answer Type Questions-II

(3 marks each)

Q.1.  Give reasons : (i)  Aniline does not undergo Friedal-Crafts reaction. (ii)  Aromatic primary amines cannot be prepared by Gabriel’s phthalimide synthesis. (iii) Aliphatic amines are stronger base than ammonia. U [CBSE, Delhi Set - 1, 2020] Ans. (i) Aniline does not undergo Friedal-Crafts reaction because aniline being a Lewis base forms a complex with AlCl3 which is a Lewis acid. The amino group is not in a position to activate the benzene ring towards electrophilic substitution. Therefore the reaction is not possible. 







C6 H 5 NH 2  AlCl 3   C6 H 5 NH 2  AlCl 3 



[1] (ii) Aromatic primary amines can not be prepared by Gabriel’s phthalimide synthesis because haloarenes have to react with potassium phthalimide and they are little reactive. So the bond cleavage does not take place. [1] (iii) Aliphatic amines are stronger bases than ammonia because the alkyl group in aliphatic amines has +I effect. So the alkyl group tends to increase the electron density on the nitrogen

atom whereas the electron releasing tendency of amines becomes more than that of ammonia.  [1] Q.2.  Arrange the following compounds as directed : (i) In increasing order of solubility in water : (CH3)2NH, CH3NH2, C6H5NH2 (ii) In decreasing order of basic strength in aqueous solution : (CH3)3N, (CH3)2NH, CH3NH2

(iii) In increasing order of boiling point :



(C2H5)2NH, (C2H5)3N, C2H5NH2 A [CBSE, Outside Delhi Set - 1, 2020]

Ans. (i) Increasing order of solubility in water : C6H5NH2 < (CH3)2 NH < CH3NH2  [1] (ii) Decreasing order of basic strength in aqueous solution : (CH3)2NH > (CH3)3 N > CH3NH2 [1] (iii) Increasing order of boiling point : (C2H5)3 N < (C2H5)2 NH < C2H5NH2 [1] Detailed Answer : (i) CH3NH2 and (CH3)2NH are aliphatic amines. Aliphatic amines are soluble in water because they can form hydrogen bonds with water. When number of alkyl group (hydrophobic) increases, molar mass laso increases which

416 ]



Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

tends to decrease in solubility in water. So, CH3NH2 is more soluble in water than (CH3)2NH. Whereas, C6H5NH2 is an aromatic amine which is insoluble in water due to more hydrphobic part which renders the formation of H-bond. Hence, increasing order of solubility in water: C6H5NH2 < (CH3)2NH < CH3NH2[1] (ii)  Greater the +I effect, greater will be the basic nature of amine in aqueous solution. However, tertiary amine also exhibits steric hindrance due to the presence of three bulky (–CH3) group due to which its basic strength is lower than secondary amine. Hence, decreasing order of basic strength in aqueous solution is: (CH3)2NH > (CH3)3N > CH3NH2 [1]



(iii)  Greater the extent of hydrogen bonding, greater will be the boiling point of amine. Primary amines have highest boiling point due to presence of two hydrogen atoms on N-atoms as more H-bonding takes place in primary amines. Hence, increasing order of boiling point is: (C2H5)3N < (C2H5)2 NH < C2H5NH2[1]

Commonly Made Error  Some students cannot write correct order of the properties of the compounds.

Answering Tip  Learn and understand the reason to explain the properties of the compounds.

Q.3.  An aromatic compound ‘A’ on heating with Br2 and KOH forms a compound ‘B’ of molecular formula C6H7N which on reacting with CHCl3 and alcoholic KOH produces a foul smelling compound ‘C’. Write the structure and IUPAC names of compound A, B and C. R [CBSE Delhi Set - 1, 2019]

Ans.



[CBSE Marking Scheme, 2019] [½ × 6]

Detailed Answer :



CN



Ans. (i)

(ii)



(i) CuCN



[3]



Q.4. Write the structures of main products when benzene diazonium chloride reacts with the following reagents.

(ii) CH3CH2OH



(iii) KI

R [CBSE Delhi Set 2, 2019]

I

[1 × 3] (iii)

[CBSE Marking Scheme, 2019]



Q.5.  Write equations of the following reactions:

(i) Acetylation of aniline (ii) Coupling reaction (iii) Carbylamine reaction



R [CBSE Delhi Set - 3, 2019]

Ans. (i)C6H5

N

H

H + CH3

C O

O

C O

CH3

C6 H5

N

C

H

O

CH3 +CH3COOH[1]

[ 417



AMINES

(ii)

N+

NCl + H

OH

OH

N

OH +Cl + H2O

N

p-hydroxyazobenzene (orange dye) N+

NCl + H



H+

NH2

N

N

NH2+Cl + H2O [1]

 (iii)

R - NH2 + CHCl3 + 3 KOH

(any one)



Heat

R - NC + 3 KCl + 3H2O

[1] [CBSE Marking Scheme 2019]



Q.6.  Complete the following reactions : CN H2 /Ni

(a) 

CH3

Br

(b) 

H3 PO2+ H2O

– N+ 2Cl

CH2 – NH2 + CHCl3

Ethanolic KOH

CH3



CH2NH2 Ans. (a) 

CH2 NC

Br

(b)

(c) 



(c) 

(or any other suitable method) [CBSE Marking Scheme 2019] Detailed Answer : CH2NH2

CN



(a) 



(b) 

H2 /Ni

CH3

Br

CH3 H3 PO2+ H2O

Br + N2 + H3 PO3 +HCl

– N+ 2Cl



(c)

CH2 – NH2 + CHCl3  

CH2 NC Ethanolic KOH

+ 3KCl + 3H2O

418 ]

How do you convert the following : (a)  N-phenylethanamide to p-bromoaniline (b)  Benzene diazonium chloride to nitrobenzene (c)  Benzoic acid to aniline

Ο || H– N – C – CH3

Ο || H– N – C – CH3

NH2

OH– or H+

Br

+ –

Br

+ –

N2Cl (b)

[1]





Br2 CH3COOH

Ans. (a) 

U [CBSE, Outside Delhi Set - 2, 2019]





Q. 7.

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

NO2

N2BF4 NaNO2

+ HBF4

[1]



Fluoro boricacid



NH2

COOH  

[1]



+ NH3



(c) 

CONH2 Br2+KOH

∆ −H2O

(or any other suitable method) [CBSE Marking Scheme 2019]

Detailed Answer :

(a)  N-pheny lethanamide to p-bromoaniline :

Ο || NH – C – CH3

NHCOCH3

Br2 CCl4 N-phenylethanamide

aq. NaOH ∆

Br

Br

(b)  Benzene diazonium chloride to nitrobenzene : +

–

+

N2Cl

–

N2BF4

NO2 NaNO2 Cu,

+ HBF4 Benzene Fluoroboric acid diozonium chloride

NH2

+ N2 + NaBF4

Nitrobenzene

(c)  Benzoic acid to aniline: COOH

COCl

SOCl2

Benzoic acid

CONH2 NH3

Benzoyl chloride

NH2

Br2 /NaOH

Benzamide

Aniline

[3]

[ 419

AMINES

Q.8.  Give reasons :

(i) Acetylation of aniline reduces its activation effect. (ii) CH3NH2 is more basic than C6H5NH2. (iii) Although –NH2 is o/p directing group, yet aniline on nitration gives a significant amount of m-nitroaniline. U [CBSE, Delhi Set - 1, 2017]



Ans. (i) Due to the resonance, the electron pair of nitrogen atom gets delocalised towards carbonyl group / [1 ] resonating structures. (ii) Because of +I effect in methylamine electron density at nitrogen increases whereas in aniline resonance takes place and electron density on nitrogen decreases / resonating structures. [1 ]

[CBSE Marking Scheme, 2017] [1]

(iii) Due to protonation of aniline / formation of anilinium ion.



—

—

Detailed Answer : (i) The lone pair of nitrogen will get involved in resonance with the carbonyl group. Hence it will reduce the activity of benzene ring in aniline. The resonance involved is as under : O O– —+ H— N — C — CH3 — N = C — CH3[1] — (ii) Aromatic amines are far less basic than aliphatic amines. This can be explained as follows : Resonance stabilization is there in aniline. It can be regarded as a resonance hybrid of these structures :

Hence, the lone pair of electrons on the nitrogen atom gets delocalized over benzene ring and thus is less available for protonation. The electron density on the nitrogen atom is increased by electron-donating inductive effect of the alkyl groups. As a result, aliphatic amines are much stronger bases than aniline. [1]

(iii) Nitration is usually carried out with a mixture of concentrated HNO3 and concentrated H2SO4. In the presence of these acids, most of aniline gets protonated to form anilinium ion. Therefore, in presence of acids, the reaction mixture consists of aniline and anilinium ion. Nitration of aniline due to stearic hindrance at ortho position, mainly gives para nitroaniline and the nitration of anilinium ion gives m-nitroaniline. In actual practice, approximately 1:1 mixture of p-nitroaniline and m-nitroaniline is obtained. NH2

— — NO2

o-Nitroaniline (2%)

p-Nitroaniline (51%) +

NH3

NH2

+

–H

NH4OH

+

—

Anilinium ion

—

—

—

+

NH3 NO2



NO2

+

–H+

+H+

—

—

NO+ 2

–H+

NH2

—

NH2

— NO2

NO2 m-Nitroaniline (47%)

Thus, nitration of aniline gives a substantial amount of m-nitroaniline due to protonation of the amino group.  [1]

420 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

OR



[3] [Topper's Answer, 2017] Q.9.  Write the structures of compounds A, B and C in the following reactions :

(b) (A)

3 /D ® (a) CH3 — COOH ¾NH ¾¾¾ Br2 /KOH (aq) CHCl3 + alc. KOH ® B ¾¾¾¾¾¾¾ ® C A ¾¾¾¾¾¾





NH2

NaNO2 /Cu ® C6H5N+2BF4– ¾¾¾¾¾ D CH3 COCl / pyridine ® C ¾¾¾ ® B ¾¾¾¾¾¾¾¾ A ¾Fe/HCl





(B)



(C)

[½] [½] [½]

[½]

R + U [CBSE, Outside Delhi Set 1, 2017]

ns. (a) (A) CH3CONH2 A (B) CH3NH2 (C) CH3NC

[½]

[½]



(b)

NO2

[CBSE Marking Scheme, 2017]

[ 421

AMINES

Detailed Answer : (a) CH3COOH



+ – NaNO2/Cu C6H5 N2 BF4 ∆

(b)

Br2/KOH(aq)

CH3 — CONH 2 Acetamide (A)

CH3 — NC Methyl isocyanide (C)

 



NH3/∆

 

CH3NH2 Methylamine (B)

CHCl3 + alc. KOH

[1½]

Fe/HCl

C6H5NO2 Nitrobenzene (A)

C6H5NH2

CH3COCl/Pyridine

Aniline (B)

C6H5NHCOCH3

N-phenyl ethanamide (C)

[1½]

Q.10.  Write the structures of A, B and C in the following :

Br / aq.KOH NaNO +HCl KI → A  → B → C . 0 − 5°C (a) C6H5 — CONH2  2

2

LiAIH CHCl + alc.KOH KCN ® B ¾¾¾¾¾¾¾ ®C . (b) CH3 — Cl ¾¾¾® A ¾¾¾¾  4

3

R + U [CBSE, Outside Delhi Set 1,2&3, 2016]

[½ + ½ + ½ ]

(ii) CH3CN, CH3CH2NH2, CH3 CH2NC

[½ + ½ + ½]



Ans. (i) C6H5NH2, C6H5N2+Cl–, C6H5I 



[CBSE Marking Scheme, 2016]

Detailed Answer :

Br /KOH

NaNO + HCl

Kl

2 2 → C6H5NH2[A]  → C6H5N2+Cl–[B] → C6H5I[C][½×3] (a) C6H5 – CONH2 

KCN

LiAlH

CHCl + KOH

4 3 → CH3CN[A]  → CH3CH2NH2[B]  → CH3CH2NC[C] (b) CH3Cl  ∆



Q.11.  Give reasons for the following : (i) Aniline does not undergo Friedal-Crafts reaction, (ii) (CH3)2 NH is more basic than (CH3)3 N in an aqueous solution, (iii) Primary amines have higher boiling point than tertiary amines. + U [CBSE, Outside Delhi Set - 1, 2 & 3 2016]



Ans. (i) Aniline is a Lewis base while AlCl3 is lewis acid. They combine to form a salt.  [1] (ii) Due to combined + I and solvation effects. [1] (iii) Due to presence of H-bonding in primary amines.  [1]  [CBSE Marking Scheme, 2016] Detailed Answer : (i) A Friedel Crafts reactions is carried out in the presence of AlCl3. But AICl3 is used as catalyst and is acidic in nature whereas aniline is a strong base. Thus, aniline reacts with AICl3 to form a salt. +

–

NH2 AlCl

NH2 + AlCl



Aniline

Salt

 [1]



Due to the positive charge on the N—atom, electrophilic substitution in the benzene ring is deactivated. Hence, aniline does not undergo Firedel—Crafts reaction. (ii) (CH3)2 NH is more basic than (CH3)3N in an aqueous solution. + I effect will increase in alkyl group that results in increasing the case of donation of lone pair electron. Amine accepts a proton and from cation which will be stabilised in water by solvation. Higher the solvation by hydrogen bonding, higher will be the basic strength.







[½×3]

[1]

Therefore, with increase in methyl group, hydrogen bonding and stabilisation by solvation decreases. This net effect results in decreases of basic strength from secondary to tertiary amine. (iii) In tertiary amines, there are no H-atoms whereas in primary amines, two H-atoms are present. Due to the presence of H-atoms, primary amines undergo extensive intermolecular H-bonding.

422 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

N—H

—

R R Primary amine

[1] (iii)  Butan-1-ol Alcohol forms stronger hydrogen bonds with water than formed by amine due to higher electronegativity of O in alcohol than N in amine. [1] [CBSE SQP Marking Scheme 2020]

     

R



N—H

—

R

—

N—R



As a result, extra energy is required to separate the molecules of primary amine. Therefore, primary amines have higher boiling point than tertiary amines. [1] Q.12. Arrange the following in increasing order of property specified: ethanamine, (i) Aniline, (solubility in water)

2-ethylethanamine

(ii) Ethanoic acid, ethanamine, ethanol (boiling point) (iii) Methanamine, N, N- dimethylmethanamine and N- methylmethanamine (basic strength in aqueous phase) Ans.   (i)  Aniline, N-ethylethanamine, Etanamine [1] [1] (ii)  Ethanamine, ethanol, ethanoic acid N, N dimethylmethanamine, methanamine, (iii)  N-methylmethanamine [1] Q.13.  (i) Give a chemical test to distinguish between N-methylethanamine and N,N-dimethyl ethanamine. (ii) Write the reaction for catalytic reduction of nitrobenzene followed by reaction of product so formed with bromine water.

Q.14.  How will you convert the following : (i)  Nitrobenzene into aniline, (ii)  Ethanoic acid into methanamine, (iii)  Aniline to N-phenylethanamide. (Write the chemical equations involved.) Q. 15 Give the structure of A, B and C in the following reactions : (i) CH3Br

U+

[CBSE, SQP, 2020–21]

A

LiAlH4

NH3 ∆

(ii) CH3COOH

B

HNO2 273 K

Br2+KOH

A

C B

CHCl3+NaOH

 Ans. (i) Nitrobenzene into aniline

NO2



C

U+ R

NH2 Sn/HCl



[1]

Ans. (ii) Ethanoic acid into methanamine Br2 CH3COOH NH3 CH3CONH2 +KOH CH3NH2 [1]



Ans. (iii) Aniline N-Phenylethanamide

O

(iii) Out of butan-1-ol and butan-1-amine, which will be more soluble in water and why? 

KCN

NH2

NH—C—CH3 (CH3CO)2O

(iii) 



R Tertiary amine



—

N—H

—

—

H

—

H

—

H

[1]

N-methylethanamine is a secondary Ans. (i)  amine. When it reacts with benzene sulphonyl chloride, it forms N- Ethyl -N methyl sulphonamide and N, N-dimethyl ethanamine is a tertiary amine it does not react with benzenesulphonyl chloride. [1] (ii) 

OR (i)  CH3Br



KCN

CH3CH2NH2 (B)

(ii)  CH3COOH

CH3NH2 (B)



CH3CN (A) HNO2 273K

NH3 ∆

LiAlH4

CH3CH2OH (C) CH3CONH2 (A)

CHCl3+NaOH

[½ + ½ + ½]

Br2+KOH

CH3NC (C) [½ + ½ + ½]

[ 423

AMINES

Long Answer Type Questions

(3 marks each) (b)  Give reasons : (i) (CH3)2NH is more basic than (CH3)3N in an aqueous solution. (ii) Aromatic diazonium salts are more stable than aliphatic diazonium salts. R + U [CBSE, Delhi/Outside Delhi, 2018]

Q.1.  (a)  Write the reactions involved in the following : (i)  Hoffmann bromamide degradation ­reaction (ii) Diazotisation (iii)  Gabriel phthalimide synthesis



Ans. (a) (i) Ar/ R-CONH2 + Br2 + 4 NaOH ® Ar/ R-NH2 + 2NaBr + Na2CO3 + 2 H2O

(ii) C H NH + NaNO + 2 HCI 6 5 2 2



273 - 278 K

¾¾¾¾® C6 H 5 N +2

[1]

-

Cl + NaCI + 2 H 2 O



[1]

(or any other correct equation)

(iii) [1]







(b) (i) Because of the combined factors of inductive effect and solvation or hydration effect

[1]



(ii) Due to resonance stabilisation or structural representation / resonating structures.

[1]



[CBSE Marking Scheme, 2018]

Detailed Answer :

(b) (i)

..

CH3 – N – H + H CH3 (2° amine)

H

⊕

CH3 – N – H CH3 (Salt) More stable

(acidic)

..

CH3 CH3 – N – H ⊕ CH3 (Salt) Less stable

⊕

(CH3)3 – N + H (acidic) (3° amine)



2° amine salt form are more stable than 3° amine due to inductive effect and higher degree of hydration. Therefore, higher the stability of salt, greater will be the reactivity of corresponding compound.



(ii) A  romatic diazonium salts are more stable than aliphatic diazonium salts due to dispersion of positive charge over the benzene ring caused by resonance. This is not found in aliphatic diazonium salts.



+ — N=N

+ — N— —N

+ — N=N

+ — N— —N

+

+



+ — N=N

+



[2]

424 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q.2.  (a) Write the structures of the main products of the following reactions : NH2 (CH3CO)2O

(i)

Pyridine







(iii)





(ii)



SO2Cl (CH3)2 NH +

N 2Cl

–

CH3CH2OH

(b) Give a simple chemical test to distinguish between aniline and N,N-dimethylaniline. (c) Arrange the following in the increasing order of their pKb values :



C6H5NH2, C2H5NH2, C6H5NHCH3 R [CBSE, Delhi/Outside Delhi, 2018]

 Ans. (a) (i) C6H5NHCOCH3

1



(ii) C6H5SO2N(CH3)2

1

(iii) C6H6

1



(b) Add chloroform in the presence of KOH and heat, Aniline gives a offensive smell while N, N dimethylaniline does not. (or any other correct test) 1



(c) C2H5NH2< C6H5NHCH3< C6H5NH2

Detailed Answer :

O

NH2

[CBSE Marking Scheme, 2018] 1

NH–C–CH3 O

(CH3CO)2O

(a) (i)

+ CH3–C–CH3

Pyridine Acetanilide SO2Cl

(ii)







O S–N–CH3

(CH3)2 NH

O CH3 N, N-dimethyl benzene sulphonamide

+ – N 2Cl

(iii)

CH3CH2OH

+ N2 + HCl + CH3CHO Benzene



(b) Carbylamine Test :



[3]

∆

C6H5NH2 + CHCl3 + 3KOH       C6H5NC + 3KCl + 3H2O Aniline Phenyl isocyanide In this reaction, aniline gives a offensive smell due to formation of phenyl isocyanide.

∆

C6H5N(CH3)2 + CHCl3 + 3KOH       No reaction

N, N-dimethylaniline does not give carbylamine test as no hydrogen is present on nitrogen atom. [1] (c) Greater the basic strength of amine, lesser will be the pK6 value. Alkyl amines are more basic than arylamines due to presence of +I effect of alkyl group. So, the increasing order of their pK6 values : C2H5NH2 < C6H5NHCH3 < C6H5NH2

[1]

[ 425

AMINES

Ans.





 [3] [Topper's Answer 2018]

Q. 3.  Write the structure of A, B, C, D and E in the following reactions : Sn / HCl

A

(CH3CO)2O

Pyridine

B

HNO3 + H2SO4 288 K

C

OH– or H+

D

H2SO4

E 

C6H5NO2

U+ R

426 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Ans.

Ans.



CONH2

NH2 Br2

NaNO2 , HCl 0°C

, KOH(aq)

(A)

N2Cl

(C)

(B)

CH3CH2OH

Br2 H2O

NH2

Br

Br

+ CH3CHO + N2 + HCl

Br (E)

(D)

(½ × 5 marks for structure and ½ × 5 for writing equations) Q.6. (i) Write the structures of main products when aniline reacts with the following reagents :

(a) Br2 water

Q. 4.  Predict the reagent or the product in the following reaction sequence : CH3

(CH3CO)2O

HNO3 H2SO4

Pyridine

NO2

NHCOCH3

NH2 CH3

CH3 5 NO2



(c) (CH3CO)2O / pyridine

CH3

CH3 1

(b) HCl

4

(ii) Arrange the following in the increasing order of their boiling point :

2

C2H5NH2, C2H5OH, (CH3)3N 3

(iii)  Give a simple chemical test to distinguish between the following pair of compounds :

NaNO2/HCl

NH2

(CH3)2NH and (CH3)3N

NO2

NH2

U+ R

Ans. (i) (a)

Ans.

A + U [CBSE Delhi 2015] NH2

Br

Br2 /H2O

Br

[1] Br 2,4,6-Tribromoaniline –

(b)



[1]

(c)





( CH3 )3 N < C2 H 5 NH 2 < C 2 H 5OH

(ii)

increasing order of b. p.

This order is due to H-bonding. [1] (iii) (CH3)2NH reacts with nitrous acid to form an oily layer of N-nitrosaemines which is insoluble in aqueous mineral acids.

formula C7H7NO reacts with Br2/aq. KOH to give compound B’, which upon reaction with NaNO2 & HCl at 0°C gives C’. Compound C’ on heating with CH3CH2OH gives a hydrocarbon D’. Compound B’ on further reaction with Br2 water gives white precipitate of compound E’. Identify the compound A, B, C, D & E; also justify your answer by giving relevant chemical equations.

[1]



Q. 5. An organic compound A’ with molecular

[ 427

AMINES

CH3

(a) H3PO2 + H2O (b) CuCN/KCN (c) H2O (ii)  Arrange the following in the increasing order of their basic character in an aqueous solution :

CH3 N H + HO NO

CH3

CH3

N – N = O + H2O

N-nitrosamine (yellow ppt)



Whereas (CH3)2NH reacts with nitrous acid to form soluble nitrite salts with no ppt.

CH3 CH3

N + HONO

CH3

⊕

C2H5NH2, (C2H5)2NH, (C2H5)3N Give a simple chemical test to distinguish (iii)   between the following pair of compounds :

–

(CH3)3 NH ONO

C6H5 – NH2 and C6H5 – NH – CH3

Trimethyl ammonium [1] nitrite



Commonly Made Error  Some students only mention the reagents but the observations are not given.

A + U [CBSE OD 2015]

Ans. (i) (a)

Q.7. (i) Write the structures of main products when benzenediazonium chloride reacts with the following reagents :

OH

Greater is the stability of the substituted ammonium cation, stronger should be the corresponding amine as a base. Thus, the order of basicity of aliphatic amines should be: primary > secondary > tertiary, which is opposite to the inductive effect based order. Secondly, when the alkyl group is small, like –CH3 group, there is no steric hindrance to H-bonding. In case the alkyl group is bigger than CH3 group, there will be steric hinderance to H-bonding. Therefore, the change of nature of the alkyl group, e.g., from –CH3 to –C2H5 results in change of the order of basic strength. In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement.

[1 + 1 + 1]





(c)

(ii) C2H5NH2 < (C2H5)3N < (C2H5)2NH

[1]

(iii) Add CHCl3 and alc. KOH, C6H5–NH2 gives foul smell of isocyanide whereas C6H5–NH–CH3 does not (or any other correct test). [1]

Visual Case-based Questions Q. 1. Read the passage given below and answer the following questions : (1 × 4 = 4)

CN



Answering Tip  When giving a test for differentiation, always write the key reagents, products, differentiable observations to give a complete answer.

(b)

(4 marks each) (i) Assertion: (CH3)2NH > CH3NH2 > (CH3)3N > NH3 is the order of basic strength in case of methyl substituted amines. Reason: The inductive effect, solvation effect and steric hindrance of the alkyl group decides the basic strength of alkyl amines in the aqueous state. (ii) Assertion: (C2H5) 2NH > (C2H5) 3N > C2H5NH2 > NH3 is the order of basic strength in case of ethyl substituted amines. Reason: The change of nature of the alkyl group, does not result in change of the order of basic strength. (iii) Assertion: Greater is the stability of the substituted ammonium cation, stronger is the corresponding amine as a base. Reason: The order of basicity of aliphatic amines is: primary > secondary > tertiary. (iv) Assertion: Amines behave as a Lewis base. Reason: Amines have an unshared pair of electrons on nitrogen atom. OR Assertion: Solubility of amines in water decreases with increase in molar mass. Reason: Intermolecular H bonds formed by the higher amines are weaker.

428 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Ans. (i) Correct option : (a) Explanation : (CH3)2NH > CH3NH2 > (CH3)3N > NH3 is the order of basic strength in case of methyl substituted amines as the inductive effect, solvation effect and steric hinderance of the alkyl group decides the basic strength of alkyl amines in the aqueous state.



(ii) Correct option : (c) Explanation : (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3 is the order of basic strength in case of ethyl substituted amines. The change of nature of the alkyl group, results in change of the order of basic strength.



(iii) Correct option : (c) Explanation : Greater is the stability of the substituted ammonium cation, stronger is the corresponding amine as a base but the inductive effect, solvation effect and steric hinderance of the alkyl group decides the basic strength of alkyl amines in the aqueous state.



(iv) Correct option : (a) Explanation : Amines behave as a Lewis base as they have an unshared pair of electrons on nitrogen atom.

OR

Correct option : (c) Explanation : Lower aliphatic amines are soluble in water because they can form hydrogen bonds with water molecules. However, solubility decreases with increase in molar mass of amines due to increase in size of the hydrophobic alkyl part.

Q. 2. Read the passage given below and answer the following questions : (1 × 4 = 4) Benzene ring in aninline is highly activated. This is due to the sharing of lone pair of nitrogen with the ring which results in increase in the electron density on the ring and hence facilitates the electrophilic attack. The substitution mainly takes place at ortho and para positions because electron density is more at ortho and para positions. On reaction with aqueous bromine all the ortho and para positions get substituted resulting in the formation of 2,4,6-tribromoaniline. To get a monobromo compound, the amino group is acetylated before bromination. After bromination, the bromoacetanilide is acid hydrolysed to give the desired halogenated amine. In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices : (a)  Assertion and reason both are correct statements and reason is correct explanation for assertion.

Assertion and reason both are correct statements (b) 

but reason is not correct explanation for assertion. Assertion is correct statement but reason is (c)  wrong statement. Assertion is wrong statement but reason is (d)  correct statement. (i) Assertion: Benzene ring is aniline is highly deactivated. Reason: In aniline, the sharing of lone pair of nitrogen with the ring increases the electron density on the ring. (ii) Assertion: In aniline –NH2 group facilitates the electrophilic attack. Reason: It is due to decrease in electron density on the ring. (iii) Assertion: In aniline, the substitution mainly takes place at ortho and para positions. Reason: The electron density is more at ortho and para positions. (iv) Assertion: The amino group of aniline is acetylated before bromination. Reason: It is due to the strong deactivating effect of –NH2 group. Ans. (i) Correct option: (d) Explanation : Benzene ring in aniline is highly activated. [1] (ii) Correct option: (c) Explanation : In aniline, –NH2 group facilitates the electrophilic attack because the sharing of lone pair of nitrogen with the ring increases the electron density on the ring. [1] (iii) Correct option: (a) Explanation : In aniline, the electron density is more at ortho and para positions than meta position, so, the substitution mainly takes place at ortho and para positions.

The above resonating structures of aniline show more electron density at the ortho and para positions.[1] (iv) Correct option: (c) Explanation: –NH2 group of aniline is acetylated before bromination due to the strong activating effect of –NH2 group.[1]

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Self Assessment Test - 13 Time : 1 Hour Q.1.  Read the passage given below and answer the following questions : (1 × 4 = 4) Lower aliphatic amines are gases, primary amine with more than two carbon atoms are liquid and higher amines are solids. Lower amines are readily soluble in water and solubility in water decreases and inorganic solvents like alcohol, ether, benzene etc increases with an increase in molecular weight. Amines are soluble in organic solvents. Boiling point of 1° amine is higher than 3° amine because of the presence of two H-atoms attached directly with N-atom in 1° amines resulting in hydrogen bonding. Boiling points of amines are lower than that of alcohols of almost similar molar mass. Amines show basic character, aliphatic amines are stronger bases than aromatic amines while less basic than ammonia. The following questions are Multiple Choice Questions. Choose the most appropriate answer: (i) Which of the following amines exist as gas ­under standard conditions? (a) methyl amine (b) butyl amine (c) ethyl amine (d) both a and c (ii) Which of the following amines possess highest boiling point? (a) Methyl amine (b) Dimethyl amine (c) Trimethyl amine (d) None of these (iii) Choose the most basic amine: (a) Aniline (b) Methyl amine (c) Ethyl amine (d) Diphenyl amine (iv) Which of the following amines is highly soluble in water? (a) Methyl amine (b) Ethyl amine (c) N-propyl amine (d) N-butyl amine The following questions (No. 2–5) are Multiple Choice Questions carrying 1 mark each. Q.2.  Amine that cannot be prepared by Gabrielphthalimide synthesis is (a)  ethyl amine (b)  methyl amine (c)  benzyl amine (d)  isobutyl amine Q.3. Tertiary amines have lowest boiling point amongst isomeric amines because : (a)  they have highest molecular mass (b)  they do not form hydrogen bond (c)  they are more polar in nature (d)  they are most basic in nature Q.4.  Primary and secondary amines are distinguished by (a)  Br2/ROH (b)  HClO (c)  HNO2 (d)  NH3

Max. Marks : 25 Q.5. The source of nitrogen in Gabriel synthesis of amines is _____________. (a)  Sodium azide, NaN3 (b)  Sodium nitrite, NaNO2 (c)  Potassium cyanide, KCN R (d)  Potassium phthalimide, C6H4(CO)2N–K+ In the following questions (Q No. 6–7), a Statement of Assertion followed by a statement of Reason is given. Choose the correct answer out of the following choices: (a) Both assertion and reason are correct statements, and reason is the correct explanation of the assertion. (b) Both assertion and reason are correct statements, but reason is not the correct explanation of the assertion. (c) Assertion is correct, but reason is wrong statement. (d)  Assertion is wrong, but reason is correct statement. Q.6. Assertion (A): Hoffmann’s bromamide reaction is given by primary amides. Reason (R): Primary amines are less basic than secondary amines. Q.7.  Assertion (A): Only a small amount of HCl is required in the reduction of nitro compounds with iron scrap and HCl in the presence of steam. Reason (R): FeCl2 formed gets hydrolysed to release HCl during the reaction. he following questions (Q No. 8–9) are short T answer Type – I and carry 2 marks each. Q.8. How will you convert: (i)  Aniline into Fluorobenzene. (ii)  Benzamide into Benzylamine.

R

Q.9.  Write the structures of A and B in the following: -

OH (i)  CH 3 CH 2 CN ¾¾¾¾¾¾ ®A Partial hydrolysis i)KCN (ii)  CH 3CH 2 Br  →A ii)LiAlH 4

NaOH+Br2 ¾¾¾¾® B

HNO2  →B  0°C

U

Questions (No. 10–11) are Short Answer Type – II and carry 3 marks each. Q.10.  An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br2 and KOH forms a compound ‘C’ of molecular formula C6H7N. Write the structures and IUPAC names of compounds A, B and C.

A

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q.12. (a)  Identify A-D CH2NO2

A

Sn/HCl

(i)  Aniline and Benzylamine. (ii)  Methylamine and Dimethylamine. (c)  Complete the following: A

OR (a)  Account for the following:

(i) Direct nitration of aniline yields significant amount of meta derivative.



(ii) Primary aromatic amines cannot be prepared by Gabriel phthalimide synthesis.

(b)  Carry out the following conversions:

—

—

CH2Cl

(b) Distinguish between the following pair of compounds:



Q.11.  An organic aromatic compound ‘A‘ with the molecular formula C6H7N is sparingly soluble in water. ‘A’ on treatment with dil HCl gives a water soluble compound ‘B’. ‘A’ also reacts with chloroform in presence of alcoholic KOH to form an obnoxious smelling compound ‘C’. ‘A’ reacts with benzene sulhponyl chloride to form and alkali soluble compound ‘D’. ‘A’ reacts with NaNO2 and HCl to form a compound ‘E’ which on reaction with phenol forms an orange dye ‘F’. Elucidate the structures of the organic compounds from ‘A’ to ‘F’. A No. 12 is a Long Answer Type carrying 5 marks.

NaOH(aq)/Br2

B

C

(i)  Ethanoic acid into methanamine. (ii)  Aniline to p-Bromoaniline. (c) Arrange the following in increasing order of basic strength:

D

CH2NC

—

Aniline, p-nitroaniline and p-toludine. A&E + A + U

 

Finished Solving the Paper ? Time to evaluate yourself !

OR SCAN THE CODE

SCAN

For elaborate Solutions

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BIOMOLECULES

Syllabus ¾¾ Carbohydrates : Classification (aldoses and ketoses), monosaccharides (glucose and fructose), D-L configuration, oligosaccharides (sucrose, lactose, maltose), polysaccharides (starch, cellulose, glycogen); Importance of carbohydrates. ¾¾ Proteins : Elementary idea of amino acids, peptide bond, polypeptides, proteins, structure of proteins—primary, secondary, tertiary and quaternary structures (qualitative idea only), denaturation of proteins; enzymes. ¾¾ Hormones : Elementary idea excluding structure. ¾¾ Vitamins : Classification and functions. ¾¾ Nucleic Acids : DNA and RNA.

Trend Analysis List of Concept Names

2018 D/OD

Carbohydrates and their types

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Structure of Glucose

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Chemical properties of Glucose, Lactose and Maltose Polysaccharides Amino acids Proteins

1Q (1 mark) 1Q (1 mark) 2Q (1 mark) 2Q (1 mark)

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Nucleic acids

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2019 D 2Q (1 mark)

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OD 1Q (1 mark) 1Q (2 marks)

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OD

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2Q (3 marks) 1Q (1 mark) 1Q (1 mark)

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Carbohydrates, their Classification and Importance TOPIC - 1 Revision Notes

Carbohydrates, their Classification and Importance .... P. 432

 Biomolecules : Biomolecules are the naturally occurring organic compounds present as essential constituents of living organism in different cells. e.g., polysaccharides, proteins, etc.

TOPIC - 2 Proteins, Hormones, Vitamins and Nucleic Acids .... P. 444

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



Carbohydrates : Carbohydrates may be defined as optically active polyhydroxy aldehydes or ketones or compounds which produce such units on hydrolysis. e.g., cellulose, glycogen, starch, etc.  Classification of carbohydrates : On the basis of the molecular size, carbohydrates have been classified into following four types : (i) Monosaccharides : Monosaccharides include non-hydrolysable carbohydrates. They are soluble in water. Those containing aldehydic group are called aldoses while other containing a ketonic group are called ketoses. Some monosaccharides with example are given in the table as shown below : Class Aldotriose Aldotetrose

Molecular Formula

Structural formula

Carbohydrates

Example

C3H6O3

Aldoses CH2OH.CHOH.CHO

Glyceraldehyde

C 4 H 8 O4

CH2OH.(CHOH)2.CHO

Erythrose, Threose

Aldopentose

C5H10O5

CH2OH.(CHOH)3 CHO

Arabinose, Ribose, Xylose, Lyxose

Aldohexose

C6H12O6

CH2OH.(CHOH)4CHO

Glucose, Mannose, Galactose, Talose, Idose, llose, Altrose

Ketotriose

C3H6O3

Ketoses CH2OH.CO.CH2OH

Dihydroxyacetone

Ketotetrose

C4H8O4

CH2OHCO CHOH CH2OH

Erythrulose

Ketopentose

C5H10O5

CH2OH.CO(CHOH)2CH2OH

Ribulose, Xylulose

Ketohexose

C6H12O6

CH2OHCO(CHOH)3.CH2OH

Fructose, Sorbose, Tagatose, etc.

(ii) Disaccharides : Those carbohydrates which on hydrolysis yield two molecules of monosaccharides Scan to know more about are called disaccharides. They are crystalline, soluble in water and sweet in taste. e.g., cane sugar, this topic maltose, etc. (iii) Oligosaccharides : Those carbohydrates which yield 2 to 10 monosaccharide molecules on hydrolysis are called oligosaccharides. e.g., raffinose on hydrolysis gives glucose, fructose and galactose. (iv) Polysaccharides : Those carbohydrates which produce large number of monosaccharide units Classification of Carbohydrates on hydrolysis are called polysaccharides. They are formed by linking together a large number of monosaccharide units through glycosidic linkages. e.g., starch, amylose and cellulose.  Sugar : In general, monosaccharides and oligosaccharides are crystalline solids, soluble in water, sweet in taste. They are collectively called sugars. e.g., glucose, fructose, sucrose etc. They are of two types : (i) Reducing sugars : Those carbohydrates which contain free aldehydic or ketonic group and reduce Fehling’s solution and Tollen’s reagent are known as reducing sugars. e.g., all monosaccharides, maltose. (ii) Non-reducing sugars : Those carbohydrates which do not have free aldehydic or ketonic group and do not reduce Fehling’s solution or Tollen’s reagent are known as non-reducing sugars. e.g., sucrose.  Glucose : Glucose occurs in nature in free as well as in combined form. It is present in sweet fruits and honey.  Methods of preparation of Glucose : (i) From sucrose :



+

C12H22O11 + H2O H → C6H12O6 + C6H12O6 Sucrose

Glucose

Fructose

(ii) From starch : Commercially, glucose is obtained by hydrolysis of starch by boiling it with dil H2SO4 at 393 K under pressure. +

(C6H10O5)n + nH2O H → nC6H12O6



393 K, 2-3 bar Glucose

 Structure of Glucose : It is a six carbon straight chain aldose which has one aldehydic group (–CHO), one primary hydroxyl group (–CH2OH) and four secondary hydroxyl groups (–CHOH). (a) Open chain structure : CHO CHO

H HO

OH H

H

H OH

H

OH

HO

H

H

OH

HO

H

CH2OH

HO

D–(+)–Glucose

CH2OH L–(–)–Glucose

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(b) Cyclic structure :

O

—



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H—C— OH

H—C

1

H

2

OH

HO

3

H

H

4

OH

H

5 6

HO—C— H

1

H

2

OH

HO

3

H

H

4

OH

H

H

5

OH

H

O

6

CH2OH

H

CH2OH β-D–(+)–Glucose

(c) Haworth structure : 6CH2OH O

H

HO

Pyran

6CH2OH

O

5

H

H

H

4

1

OH 3

H

2

HO

OH

OH

H

5

O

OH

H

H

4 3

—

CHO

—

(CHOH)4 HI → CH3 − CH 2 − CH 2 − CH 2 − CH 2 − CH3 ∆



n -Hexane

(ii) Reactions showing the presence of carbonyl group (> C = O) :

(iii) Acetylation of glucose :

(CH3CO)2O Acetic anhydride

(iv) Reaction showing the presence of alcoholic (– OH) group :

COOH Saccharic acid

—

(CHOH)4

COOH

Oxidation

(CHOH)4

—

CH2OH Glucose

—

(CHOH)4

COOH

Oxidation HNO3

—

—

CHO

—



2

H

OH

β-D –(+)–Glucopyranose

 Reactions of Glucose : (i) With HI :

CH2OH Glucose

OH 1

H

α-D –(+)–Glucopyranose



O

OH

CH2OH

α-D–(+)–Glucose

OH H

HO

CH2OH Gluconic acid

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 Fructose : It is a ketohexose obtained by hydrolysis of disaccharide.  Structure of Fructose : (a) Open chain structure :

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—

CH2OH

—

C=O Chemical properties of Glucose

—

HO—C—H

—

H—C—OH

—

H—C—OH CH2OH



(b) Cyclic structure : 1

CH2OH

HOH2C 3 4 5

Furan

6

Fructofuranose









Fructofuranose

(c) Haworth structure :

Sucrose : +

H C12H22O11 + H2O → C6H12O6 + C6H12O6 Sucrose D–(+)–Glucose D–(–)–Fructose  Structure :

All disaccharides and polysaccharides contain glycosidic linkages. How is it useful? It would be helpful if we can mark the carbon number in mnemonics for sucrose, maltose etc. e.g.,  Maltose molecule is composed of two α-D-glucose units in which C1 of one glucose (I) is linked to C4 of another glucose unit (II).











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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Lactose is a reducing sugar.

Polysaccharides : Carbohydrates which contain a large number of monosaccharide units joined together by glycosidic linkages.

(a) Starch is a polymer of α-glucose and consists of two components – amylose and amylopectin. Amylose is a long unbranched chain with 200 – 1000 α-D-(+)-glucose units held by Cl – C4 glycosidic linkage.







 in which the chain is formed by C – C glycosidic Amylopectin is a branched chain polymer of α-D-glucose units 1 4 linkage whereas branching occurs by C1 – C6 glycosidic linkage.





(b) Cellulose is a polysaccharide whose fundamental structural unit is β-D-glucose joined by glycosidic linkage between C1 of one glucose unit and C4 of the next glucose unit.

(c) Glycogen is an animal starch with structure similar to amylopectin and has more branches.

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



Importance of Carbohydrates : (i) They are essential for plants and animals as a source of energy. (ii) They form structural materials for cells. (iii) They provide raw materials for textile, paper and alcohol industry. (iv) Monosaccharides are also present in nucleic acids which control the transmission of hereditary effects from one generation to other and biosynthesis of proteins as well.  Distinction between Glucose (monosaccharide), Sucrose (disaccharide) and Starch (polysaccharide) :

S.

Test

Glucose (Monosaccharide)

No.

Sucrose (Disaccharide)

Starch (Polysaccharide)

1.

On heating with Fehling’s Red precipitate is obtained. No precipitate. solution

No precipitate.

2.

On heating with Tollens’ Silver mirror is formed. reagent (ammonical AgNO3)

No silver mirror is formed.

3.

On heating with phenyl Yellow coloured crystals No osazone is formed. hydrazine of osazone are formed.

No osazone is formed.

4.

On heating with resorcinol No colour. and HCl

Wine red colour.

No colour.

5.

On adding NaOH solution and 1-2 drops of cobalt nitrate

Violet colour.

No colour.

6.

On adding I 2 solution in No colour. aqueous solution

No colour.

Blue-violet colour.

7.

On heating in a dry test tube Melts into brown coloured Melts at 463 K, becomes Chars on heating substance and smells of brown at 473 K, chars at strongly. burnt sugar. Turns black high temperature. on heating further.

No colour.

No silver mirror is formed.

Know the Terms  Aldoses : Monosaccharides which contain an aldehyde (—CHO) group are called aldoses.  Ketoses : Monosaccharides which contain a keto (>C=O) group are called ketoses.  Invert Sugar : An equimolar mixture of glucose and fructose which is formed as a result of hydrolysis of sucrose is known as Invert sugar.  Anomers : Diastereomers of cyclic forms of sugar differing in configuration at the anomeric carbon, generally found in two forms a and b .  Glycosidic linkage : It is an oxide linkage between two or more monosaccharide units in polysaccharides.

Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS : Q. 1. Which of the following is a disaccharide? (a) Glucose (b) Starch (c) Cellulose (d) Lactose  R [CBSE, Delhi Set-3, 2020] Ans. Correct option : (d) Explanation : Glucose - monosaccharide Lactose - disaccharide Cellulose and Starch - polysaccharides Q. 2. Disaccharide that are reducing in nature are : (a) sucrose and lactose (b) sucrose and maltose (c) lactose and maltose (d) sucrose, lactose and maltose 

R [CBSE, SQP, 2020-2021]

(1 mark each) Ans. Correct option : (c) Explanation : Lactose and maltose are the disaccharides that are reducing in nature due to a free anomeric carbon that can convert to an aldehyde functional group. Q. 3. What is the structural difference between starch and cellulose ? (a) In starch, glucose monomers are in a- helix configuration, while in cellulose, glucose monomers are in b-sheet configuration. (b)  In starch, glucose monomers are in b-configuration, while in cellulose, glucose monomers are in a- configuration. (c)  In starch, glucose monomers are in aconfiguration, while in cellulose, glucose monomers are in b-configuration. (d) None of the above.

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Ans. Correct option : (c)

(i)

Explanation : In starch, glucose monomers are in a-configuration, while in cellulose, glucose monomers are in b-configuration.



Q. 4. Glycogen is a branched chain polymer of a-D glucose units in which chain is formed by C1–C4 glycosidic linkage whereas branching occurs by the formation of C1–C6 glycosidic linkage. Structure of glycogen is similar to _____. (a) Amylose

(b) Amylopectin

(c) Cellulose

(d) Glucose

(ii)

(iii)

U

Ans. Correct option : (b)



(b) H HO H H



CH2OH CHO OH H OH OH CH2OH

HO H HO HO

H H H H HOHO H H H H

CH2OH



(c)

  

OHOH HOHO OHOH H H O HO H O H HO OHOH H H H H

H H OHOH H O H O OHOH

CH2OH CH2OH

CH2OH CH2OH

(I)

   (II)

Structure (I) and (II) differ in structure at C-1.

Q.7. Which of the two components of starch are water soluble ? (a) Amylose (b) Amylopectin (c) Both of the above (d) None of the above  R





(a) (i) and (ii) (b) (ii) and (iii) (c) (i) and (iii) (d) (iii) is anomer of (i) and (ii) U Ans. Correct option : (a) Explanation : Cyclic structures of monosaccharides which differ in structure at carbon-1 are known as anomers.

CH2OH CHO H OH H H

(d)

U

Ans. Correct option : (c)

 



Explanation : Structure of glycogen is similar to amylopectin. It is a branched chain polymer of a D glucose units in which chain is formed by C1–C4 glycosidic linkage whereas branching occurs by the formation of C1–C6 glycosidic linkage. Q. 5. Which of the following pairs represents anomers? CHO CHO (a) HO H HO H HO HO H H H OH H OH H OH H OH





Explanation : The isomers, which differ only in the configuration of the hydroxyl group at C-1, are called anomers and are referred to as a and b-forms. Q. 6. Three cyclic structures of monosaccharides are given below which of these are anomers ?

Ans. Correct option : (a) Explanation : Starch generally consists of 20-25% of amylose. It is water soluble component. Q.8. Which of the following is an example of aldohexose? (a) Ribose (b) Fructose (c) Sucrose (d) Glucose R Ans. Correct option : (d) Explanation : Glucose is an example of aldohexose because it contains an aldehyde group. Q.9. The constituent units of sucrose are. (a) lactose and glucose (b) glucose and fructose (c) galactose and glucose (d) glucose and maltose R Ans. Correct option : (b)

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BIOMOLECULES



Explanation : The constituent units of sucrose are glucose and fructose because on hydrolysis, sucrose gives these products.

CH2OH O H H OH H OH H OH H

Q.10. Which disaccharide is present in milk? (b) Galactose

(c) Sucrose

(d) Lactose

R

Ans. Correct option: (d) Explanation : Lactose is a disaccharide which is present in milk. Q.11. Amylopectin is a polymer of: (a) b-D-glucose

(b) a-D-glucose

(c) b-D-fructose

(d) a-D-fructose

R

Ans. Correct option : (b) Explanation : Amylopectin is a branched chain polymer of a-D-glucose units in which the chain is formed by C1-C4 glycosidic linkage while branching occurs C1-C6 glycosidic linkage. Q.12. Glucose and fructose are : (a)  isomers of each other (b) homologuer of each other (c) anomers of each other (d) enantiomers of each other Ans. Correct option : (a) Explanation : Glucose and fructose differ structurally and stereochemically. They have same molecular formula i.e. C6H12O6. Hence, these are isomers of each other.



[B] ASSERTIONS AND REASONS In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion.

(c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement.

Q.1. Assertion : D(+)–Glucose is dextrorotatory in nature. Reason : ‘D’ represents its dextrorotatory nature.

Q.4. Assertion : Glucose reacts with hydroxylamine to form an oxime and also adds a molecule of hydrogen cyanide to give cyanohydrin. Reason : The carbonyl group is present in the openchain structure of glucose. Ans. Correct option : (a)  Explanation : Glucose reacts with hydroxylamine to form an oxime and also adds a molecule of hydrogen cyanide to give cyanohydrin which confirms the presence of the carbonyl group in the open chain structure of glucose.  [C] VERY SHORT ANSWER TYPE QUESTIONS : Q.1. What type of linkage is present in polysaccharides? R [CBSE, Delhi Set-2, 2020]  Ans. Glycosidic linkage.

Q.2. Write the name of component of starch which is water soluble.  R [CBSE,Outside Delhi Set-1, 2020] Ans. Amylose is water soluble component. Q.3. Write the name of linkage joining two monosac­ charides. R [CBSE,Outside Delhi Set-2, 2020]  Ans. Glycosidic linkage. Q.4.What is basic structural difference between glucose and fructose?  [CBSE Delhi Set-1, 2019]

Q.2. Assertion : b-glycosidic linkage is present in maltose.

Ans. Glucose has aldehydic group while fructose has ketonic group/ Glucose is aldose while fructose is ketose.  [CBSE Marking Scheme, 2019]

Explanation : ‘D’ corresponds to the position of –OH group on the right side on the farthest asymmetric C-atom.

OH

Q.3. Assertion : Deoxyribose, C5H10O4 is not a carbo­ hydrate. Reason : Carbohydrates are optically active polyhydroxy aldehyde or polyhydroxy ketone or substances which give aldehyde or ketone on hydrolysis. Ans. Correct option : (d)  Explanation : Deoxyribose, C5H10O4 is a carbohydrate and is the sugar moiety of DNA.

Ans. Correct option : (c)

H

Reason : Maltose is composed of two glucose units in which C–1 of one glucose unit is linked to C–4 of another glucose unit. Ans. Correct option : (d)  Explanation : a-glycosidic linkage is present in maltose as it is composed of two a-D-glucose units in which C1 of one glucose (I) is linked to C4 of another glucose unit (II). 

(a) Maltose

O

CH2OH O H H H OH H OH

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

OR



[Topper’s Answer 2019] [1] Q. 8. What is the difference between amylose and amylopectin? R

Commonly Made Error

Answering Tip  Understand the structure of glucose and fructose to remember the functional groups present in them. Q. 6.

  Write products obtained after hydrolysis of lactose. [CBSE, Delhi Set-1, 2019]

Amylose is a long unbranched chain of α-D-glucose units held by C1-C4 glycosidic linkage whereas



Ans. Glucose and galactose.  [CBSE Marking Scheme, 2019]

Ans. Amylose is a linear polymer of a- glucose whereas amylopectin is a branched polymer of a- glucose/ Amylose is water soluble whereas amylopectin is water insoluble.  (Or any other correct difference)  [CBSE Marking Scheme 2019] Detailed Answer :

 Some students do not write correct functional group for glucose and fructose.

amylopectin is a branched chain polymer of α-D-

Q. 7. What is the basic structural difference between starch and cellulose?  R [CBSE,Outside Delhi Set-2, 2019]

glucose units, units in which chain is formed by C1-C6 glycosidic linkage while branching occurs by C1-C6 glycosidic linkage.



Ans. Starch is a polymer of a- glucose whereas cellulose is a polymer of b- glucose.  [CBSE Marking Scheme 2019]

Short Answer Type Questions-I Q. 1.  Write the reactions showing the presence of following in the open structure of glucose : (i) a carbonyl group (ii) Straight chain with six carbon atoms U [CBSE,Outside Delhi Set-1, 2020]



Ans. (i) Reaction of glucose (a carbonyl group) CH 2OH CH 2OH | | Br2 / H 2 O (CHOH)4  → (CHOH)4 oxidation | | CHO COOH

Gluconic acid  (ii) Straight chain with six carbon atoms

[1]

CHO | HI (CHOH)4 → CH3 − CH 2 − CH 2 − CH 2 − CH 2 − CH3 ∆ n -hexane |

(2 marks each)

Detailed Answer : (i) Glucose reacts with bromine water to form gluconic acid in which an aldehyde group (–CHO) is oxidised into carboxylic acid group (–COOH). This reaction shows the presence of a carbonyl group in the open structure of glucose.

CH 2OH CH 2OH | | (CHOH)4 Br2 H2 O (CHOH)4  → | | ( Oxidetion ) CHO COOH Glucose Gluconic acid 

[1] (ii) Glucose on heating with HI to form n-hexane which shows that open structure of glucose contains straight chain with six carbon atoms.

CH 2OH

[1]

[CBSE Marking Scheme 2020] 



[1]

[ 441

BIOMOLECULES

Q.2. Write the reactions showing the presence of following in the open structure of glucose : (i) an aldehyde group (ii) a primary alcohol  U [CBSE,Outside Delhi Set-2, 2020]



When glucose reacts with HCN, it forms glucose cyanohydrin. It shows the presence of Carbonyl group (>C=O) in the open structure of glucose.

Ans.  (i) An aldehyde group : On reduction with sodium amalgam and water, the aldehydic group is reduced to primary alcohol.



[1]

Q.4. (i) What is an anomer ? [1]





[1]



(ii) A primary alcohol : (with nitric acid) On reaction with nitric acid, a primary alcohol group present in glucose is converted into carboxylic acid (-COOH) group.

[1]

Q.3. Write the reactions showing the presence of following in the open structure of glucose : (i) five –OH groups (ii) a carbonyl group  U [CBSE, Outside Delhi Set-3, 2020] Ans. (i) Presence of five –OH groups (with acetic anhydride) When glucose reacts with acetic anhydride, it forms pentacetyl glucose. It shows the presence of five –OH groups in the open structure of glucose.

(ii) Structure of a disaccharide formed by glucose and fructose is given below. Identify anomeric carbon atoms in monosaccharide units.

fCH OH 2 a e O O HOHC 2 H H H H a b d e OH H HO O H HO c CH 2OH b f c d H OH OH H R Ans. (i) ‘a’ carbon of glucose and ‘b’ carbon of fructose are the anomeric carbon atoms in the monosaccharide units.[1] (ii) The isomers which differ only in the configuration of the hydroxyl group at C-1 are called anomers and are referred to as a- and b-forms. [1] Q.5. Define the following terms : (i) Glycosidic linkage (ii)  Invert sugar

R

Ans. (i) Glycosidic linkage : The two monosaccharides units are joined together through an oxide linkage formed by loss of a molecule of H2O.[1] (ii) Invert sugar : Hydrolysis of sucrose brings about a change in sign of rotation from dextro (+) to laevo (–) and the product is known as invert sugar. [1]

Commonly Made Error  Students often write irrelevant information trying to explain the terms.

Answering Tip









(ii) Presence of a carbonyl group (with HCN)

 Write the precise definitions as given in the text.

Short Answer Type Questions-II Q. 1.  Define the following terms : (i) Glycosidic linkage (ii) Invert sugar (iii) Oligosaccharides

R



Ans. (i)  Glycosidic linkage : The two monosaccharides units are joined together through an oxide linkage formed by loss of a molecule of H2O.[1]

(3 marks each)

(ii)  Invert sugar : Hydrolysis of surcose brings about a change in sign of rotation from dextro (+) to laevo (−) and the product is known as invert sugar.[1] (iii)  Carbohydrates that yield teo to ten monosaccharides units on hydrolysis are called polysaccharides. Examples are starch, cellulose, glycogen etc.[1]

442 ]



Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(ii) H2N-OH (iii) (CH3CO)2O

Ans. (i)



U [CBSE, Delhi Set-3, 2019]

CHO (CHOH)4

COOH Br2 water

(CHOH)4

CH2OH



CH2OH

CHO (CHOH)4



Q.3. (i)  Glucose reacts with acetic anhydride in presence of pyridine to form pentaacetyl derivative of glucose. What does this reaction indicate?. (ii) How will you distinguish 1° and 2° hydroxyl groups in glucose? Explain with reaction.

Q. 2. Write the reactions involved when D-glucose is treated with the following reagents : (i) Br2 water

(ii) (iii)

CH NH2OH

N

OH

(CHOH)4

CH2OH

CH2OH CHO

CHO Acetic anhydride

[1] O

(CH

O

C

CH3)4

CH2

O

C

CH3

[1½]



(CHOH)4

 [1]

 R Ans. (i) Glucose on treatment with acetic anhydride in presence of pyridine or a few drops of conc. H2SO4. It forms penta-acetyl derivative indicating the presence of 5-OH groups. Out of which, one –OH group is primary (1o) alcoholic and four (C2, C3, C4 and C5) –OH groups are secondary (2o) alcoholic groups.

CH2OH





 [1]



[CBSE Marking Scheme 2019]

O

CHO (CHOH)4

(ii) Glucose (or gluconic acid) on oxidation with HNO3 gives saccharic acid (a dicarboxylic acid) indicating that one of the primary (1o) alcoholic group is oxidized to –COOH group but secondary (2o) hydroxyl groups undergo oxidation only in drastic conditions. [1½]

CH2OH Oxidation HNO3

CH2OH Glucose

COOH

(CHOH)4

Oxidation HNO3

(CHOH)4 COOH Saccharic acid

COOH Gluconic acid (Monocarboxylic acid)

(Dicarboxylic acid)

Long Answer Type Questions Ans. D-glucose reacts with hydroxylamine (NH2OH) to form an oxime because of the presence of aldehydic (−CHO) group or carbonyl carbon. This happens as the cyclic structure of glucose forms an open chain structure in an aqueous medium, which then reacts with NH2OH to give an oxime. CH=N

CHO (CHOH)4 CH2OH



(Glucose)

NH2OH





Q. 1. How do you explain the absence of aldehyde group in the pentaacetate of D-glucose ? R

(5 marks each) But pentaacetate of D-glucose does not react with NH2OH. This is because pentaacetate does not form an open chain structure. CH=O

HOH2 C HO

H

H

OH

H

OH CH2 OH

OH

Glucose (open chain form)

(CHOH)4 CH2OH Oxime

–H2 O



OH H

+5(CH3 CO)2 O

Pyridine Acetylation

O

(CH—O—C—CH3)4 O CH2 —O—C—CH3 Glucose pentaacetate NH2 OH No oxime formation

NH2 OH

CH=NOH H

CHO

OH

OH H OH



CH2 OH

NH2 OH

Glucose (open chain form) –H2 O

No oxime formation

NH2 OH

[ 443

BIOMOLECULES CH=NOH H OH





OH H

H

OH

H

OH

(a) When aqueous solution of glucose is treated with sodium amalgam (Na/Hg) or sodium borohydride, it is reduced to sorbitol (or glucitol) a hexahydric alcohol. Na/Hg

CH2 OH Glucose oxime

[5]

Q. 2. (i) Name any two reaction of glucose which cannot be explained by its chain structure.

(ii) What is the difference between reducing and non-reducing sugars?



(iii) Polysaccharides



[1]



(b) Prolonged heating with hydroiodic acid and

red phosphorus at 100°C gives a mixture of n-hexane and 2-iodohexane.

+

Ans. (i) (a) Glucose doesnot give 2,4-DNP test.  

(b) Glucose does not give Schiff’s test.

 

(c) The pentaacetate of glucose does not react with hydroxylamine.



(b) Non- reducing sugars : These are carbohydrates that cannot act as a reducing agents due to the absence of free aldehyde groups or free ketone groups. [1]

(iii) Carbohydrates that yield two to ten mono­ saccharides units on hydrolysis are called poly­ saccharides. Examples are starch, cellulose, glycogen, etc. [1]

Answering Tip  Be precise in your answer. Q. 3. On the basis of which evidences D-glucose was assigned the following structure?

CHO | (CHOH )4 | CH 2OH

 R Ans. This structure was assigned on the basis of the following evidences : (i) Molecular formula : C6H12O6 is molecular formula of glucose. (ii) Straight chain structure :





The formation of n-hexane suggests that all the six carbon atoms in glucose are arranged in a straight chain structure. [1]



(c) Presence of five –OH groups : On acetylation with acetic anhydride, glucose gives a pentaacetate. This confirms that glucose contains five –OH groups. We know that the presence of two or more –OH groups on the same carbon atom makes the molecules unstable.



Since glucose is a stable compound, therefore, the five –OH groups must be present on different carbon atoms. [1]

Commonly Made Error  Students often confused to give correct information.







[Any two of the above 1 + 1] (ii) (a) Reducing sugars : These are carbohydrates that can act as reducing agents due to the presence of free aldehyde groups of free ketone groups. [1]

(d) Presence of one primary alcoholic group : On oxidation with conc. HNO3, both glucose and gluconic acid give the same dicarboxylic acid and saccharic acid. The primary alcoholic group (–CH2OH) is always present at the end of the carbon chain. [1]

(e) Presence of an aldehyde (–CHO) group : Glucose reacts with hydroxylamine, NH2OH to form oxime. This suggests that glucose contains a carbonyl (>C=O) group.

On the basis of above observations, the following open CH2OH chain structure for glucose can be written as follows : CHO  ( CHOH)4  CH 2OH  [1]

444 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

TOPIC-2

Proteins, Hormones, Vitamins and Nucleic Acids Revision Notes  Proteins : Proteins are complex polyamides formed from amino acids. They are essential for O — —

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proper growth and maintenance of body. They have many peptide ( — C — NH — ) bonds.  Amino Acids : The compounds which contain carboxylic acid (–COOH) group and an amino group (–NH2) are called amino acids. Amino acids form proteins.  a-Amino Acids : Those amino acids in which —NH2 group and —COOH group are attached

Amino acids

to same carbon are called α-amino acids. These are obtained by hydrolysis of proteins. e.g., glycine.  Types of Amino Acids : l Acidic, Basic and Neutral Amino Acids : Amino acids are acidic, basic or neutral depending upon the relative number of amino and carboxyl groups in their molecule. If equal number of amino and carboxyl groups are present, it is neutral. More number of amino than carboxyl groups makes it basic and more carboxyl groups as compared to amino makes it acidic. l Essential and Non-essential Amino Acids : The amino acids which can be synthesised in the body are known as non-essential amino acids. e.g., glycine, alanine, glutamine, etc. On the other hand, those which cannot be synthesised in the body and must be obtained by diet are known as essential amino acids.  Classification of Proteins : Proteins are classified as follows : (a) Based on molecular shape : (i) Fibrous proteins : They have thread like molecules which tend to lie side by side to form fibres. e.g., keratin, collagen, etc. (ii) Globular proteins : They have molecules which are folded into compact units that often approach spheroidal shape. e.g., insulin, albumin, haemoglobin, etc. (b) Based on structure and shape : Scan to know (i) Primary structure : Each polypeptide in a protein has amino acids linked with each more about other in a specific sequence. This sequence of amino acids is called as primary structure this topic of proteins. (ii) Secondary structure : It refers to the shape in which a long polypeptide chain exists. They are of two types : 1. α-helix in which polypeptide chain forms intramolecular hydrogen bonds by Classification of twisting into a right handed helix with the — NH group of each amino acid residue proteins hydrogen bonded to the >C = O of an adjacent turn of the helix. e.g., keratin in hair, nails. 2. β-pleated sheet has all peptide chains stretched to nearly maximum extension and then arranged side by side held together with intermolecular hydrogen bonding. e.g., silk. (iii) Tertiary structure : It represents overall folding of polypeptide chains by H-bonds, disulphide linkages, van der Waals and electrostatic form of attraction. e.g., Fibrous and globular proteins. (iv) Quaternary structure : The spatial arrangement of two or more polypeptide chains with respect to each other is known as quaternary structure.  Name and structure of some naturally occurring α-Amino Acids : (H2N — CH — COOH) R

—

S. No.

Amino acids

Isoelectric point (Hydrophobic)

One alphabet code

Three alphabet code

—R (Side chain)

1.

Non-polar Glycine

5.97

G

Gly

—H

2.

Alanine

6.02

A

Ala

—CH3

3.

Valine*

5.97

V

Val

—CH(CH3)2

[ 445

BIOMOLECULES

4.

Leucine*

5.90

L

Leu

—CH2—CH(CH3)2

5.

Isoleucine*

6.02

I

Ile

—CH(CH3)—CH2—CH3

6.

Phenylalanine*

5.84

F

Phe

—CH2—Ph

7.

Methionine*

5.06

M

Met

—CH2—CH2—S—CH3

8.

Tryptophan*

5.88

W

Trp

—

—

H2C — CH2

6.30

P

H2C

Pro

CH—COOH

—

Proline

—

9.

+

N

— H

10.

Polar (Neutral) Serine

5.68

S

Ser

—CH2OH

11.

Cysteine

5.02

C

Cys

—CH2—SH

12.

Asparagine

5.41

N

Asn

13.

Glutamine

5.70

Q

Gln

14.

Threonine*

5.60

T

Thr

15.

Tyrosine

5.67

Y

Tyr

16.

Acidic Aspartic acid

2.98

D

Asp

—CH2—COOH

17.

Glutamic acid

3.22

E

Glu

—CH2CH2COOH

18.

Basic Lysine*

9.47

K

Lys

—CH2—(CH2)3—NH2

19.

Arginine*

10.76

R

Arg

20.

Histidine*

7.59

H

His

— —

O

— CH2CNH2

— —

O

— CH2 — CH2 — C — NH2

—CH(OH)—CH3

NH

— —

—CH2 — (CH2)2 — NH — C — NH 2



* These are essential amino acids.  Different Proteins and their functions : S. No.

Type of Protein

Example

Function

1.

Enzyme

Trypsin, Pepsin

As a catalyst in biochemical reactions.

2.

Structural

Collagen, Keratin

Structural and protective action in teeth, nails and hairs.

3.

Transport

Haemoglobin

Transport of oxygen from lungs by blood stream to different tissues.

446 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

4.

Motion

Myosin, Actin

For motion of muscles.

5.

Hormone

Insulin

Regulate body metabolism.

Storage

Ferritin, Casein

Store nutrients.

 Denaturation of Protein : When a protein in its native form is subjected to change, like change in temperature or pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix gets uncoiled and protein loses its biological activity. It is called denaturation of protein. e.g. coagulation of egg white on boiling, curdling of milk etc.  Enzymes : Enzymes are essential biological catalysts which are needed to catalyse biochemical reactions. e.g., maltase, lactase, invertase etc. Almost all enzymes are globular proteins.

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Nucleic Acids and Proteins

 Some specific Enzymes and the reactions catalysed by them : Enzyme Maltase Lactase Amylase or Ptyalin Invertase Urease Trypsin, Pepsin

Reaction catalysed

→ Glucose Maltose Lactose → Glucose + Galactose Starch → Glucose → Glucose + Fructose Sucrose → NH3 + CO2 Urea Protein → Amino acid

 Vitamins : Vitamins are group of organic compounds which are required in a very small amount for the healthy growth and functioning of higer organism. They cannot be made by organism and so have to be part of our diet.  Types of Vitamins : (i) Fat soluble vitamins : Vitamins A, D, E and K are fat soluble but insoluble in water. (ii) Water soluble vitamins : Vitamins belonging to group B (B1, B2, B6, B12 etc.) and vitamin C are soluble in water.  Different Vitamins : Vitamin Vitamin A

Name and Formula Retinol or Excerophytol C20H30O

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Vitamins and Classification

Sources

Function

Deficiency Diseases

Milk, butter, egg, fish, spinach, green vegetables, carrot, etc.

Essential for vision and growth. Develops resistance against diseases.

Night blindness, xerophthalmia, retarded growth and decreases the immunity of body towards various diseases.

Vitamin B1

Thiamine or Aneurin C12H17N4OS

Egg, fish, meat, rice, For proper functioning Beri-beri, anaemia, wheat, yeast, etc. of nervous system. weakness of muscles, etc.

Vitamin B2

Riboflavin or Lactoflavin C17H20N4O6

Milk, cheese, egg, meat, Essential for growth Cracking skin green vegetables, liver, of body. particularly at the etc. corners of mouth (Cheilosis), glossitis, dermatitis.

Vitamin B6

Pyridoxine C8H11O3N

Wheat, maize, husk of In blood formation. rice, meat, fish, egg, etc.

Convulsions, paralysis.

Vitamin B12

Cyanocobalamin C63H88CoN14O14P

Liver, egg, fish, meat, In blood formation. etc.

Macrocytic anaemia or pernicious anaemia.

Vitamin C

Ascorbic acid C 6H 8 O 6

Citrus fruits such as lemon, orange, tomatoes, amla, etc.

For bones, teeth and Scurvy, pyria, pain healing of wounds, in joints, loosening healthy skin. of teeth, mental depression, anaemia, bleeding of gums.

[ 447

BIOMOLECULES

Vitamin D

Calciferol or Ergocalciferol or Vitamin D2, C27H44O

Egg, meat, fish, liver oil, Control of metabolism Rickets, osteomalacia. butter, etc. of calcium and phosphorus in the formation of bones.

Vitamin E

α-Tocopherol C29H50O2

Milk, egg, meat, pulses, Anti sterility or green vegetables, seeds, reproduction. beans, etc.

Vitamin K (Vitamin K1 or K2)

Vitamin K1 or Phylloquinone C31H46O2, Vitamin K2, C41H56O2

Cabbage, spinach, green vegetables, egg, fish, etc.

Loss of reproductive ability or infertility.

Helps in clotting of Delay in blood clotting blood. (Haemophilia).

 Hormones : Hormones are the chemical substances produced by ductless glands called endocrine glands such as thyroid, adrenals etc. They find their way into the blood stream and influence and regulate the functions of the other organs of the body. Hormonal deficiency leads to specific biological disorders which can be cured by the administration of the specific hormones.  Steroid hormones : Those hormones which have structure similar to steroids. e.g., cortisone, testosterone, estrogen and progesterone.  Various hormones, gland of secretion and their functions : S. No. 1.

Hormones Steroid hormones : (i) Testosterone (androgens)

Gland Testes

(ii) Estrogen and progesterone Ovary 2.

3.

(iii) Cortisone Amine hormones : (i) Adrenalin

Adrenal cortex

(ii) Thyroxine

Thyroid

Peptide hormones : (i) Oxytocin

Posterior pituitary

Adrenal medulla

(ii) Vasopressin (ADH) (iii) Insulin (iv) Glucagon  

Function Responsible for development of male sex organs. Influences development of sex organs, maintains pregnancy. Regulates metabolism of water, mineral salts. Increases blood pressure and pulse rate. It also releases glucose from glycogen and fatty acids from fats. Stimulates rate of oxidative metabolism and regulates general growth and development.

Posterior pituitary

Causes constriction of some smooth muscles. It causes contraction of uterus during child birth. Controls the reabsorption of water in kidneys.

Pancreas Pancreas

Controls blood glucose level. Increases blood glucose level.

Nucleic acid : The polymers of nucleotides help in synthesis of protein and transfer genetic traits. Nucleic acids are of two types : (i) Deoxyribonucleic acid (DNA) Scan to know (ii) Ribonucleic acid (RNA) more about Constituents of nucleic acids : Pentose sugar, phosphoric acid and nitrogenous bases. this topic Nitrogen containing bases :  Pyrimidines : These are three bases derived from pyrimidines. These are cytosine (C), thymine (T) and uracil (U). In DNA, T is present but in RNA, U is present.  Purines : There are two bases derived from purine. These are adenine (A) and guanine (G). Hormones  Nucleoside : A unit formed by the attachment of a base to 1’-position of sugar is known as and Endocrine nucleoside. System 5’

1’

4’

3’

2’

448 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

 Nucleotide : When nucleoside is linked to phosphoric acid at 5’-position of sugar moiety, the unit obtained is called nucleotide.

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



Concept of Nucleic acids



Simple structure of nucleic acid chain : Base 

[

Base 

S T P S G C

Base 

]

 Sugar  Phosphate Sugar  Phosphate n Sugar   DNA : James Watson and Francis Crick gave double helix structure of DNA. Nucleic acid (DNA) hydrolysis Nucleotide hydrolysis Phosphoric acid + Nucleoside hydrolysis

S P

A T

S

P

Purine base (Adenine + Guanine)

A

P

S S

S

T

C G S C

G T

A G

S S

P

P

P S T A

S

G

C







P

P

P S

hydrolysis

S

A

S

Pyrimidine base (Cytosine + Thymine)

RNA : (i) m-RNA (Messenger RNA) (ii) r-RNA (Ribosomal RNA) (iii) t-RNA (Transfer RNA)  Nucleic acid (RNA)

P

T

C

Sugar (2-deoxy ribose)

S

A P

S

P

S

P

T P

C G S

P

P

Nucleotide hydrolysis Phosphoric acid

Sugar (Ribose)



Nucleoside hydrolysis Purine base (Adenine + Guanine)

Pyrimidine base (Cytosine + Uracil)

Properties of Nucleic Acids : (i) Nucleic acids are very important constituents (polynucleotide) found in nucleus of cell which help in biosynthesis of protein and act as carriers for transfer of hereditary characters. (ii) A molecule formed by the combination of one pentose sugar unit, a purine or pyrimidine base and a phosphate. Nucleotides combine among themselves to form polynucleotide (nucleic acid). (iii) A polynucleotide (DNA) which has a thymine base but not a uracil base. It contains deoxyribose sugar but not ribose sugar. It has double helix structure. (iv) A polynucleotide (RNA) which contains uracil base and ribose sugar but thymine base and deoxyribose sugar are absent. It has single stranded structure.

[ 449

BIOMOLECULES

Mnemonics • Mnemonic: PriVate PhIL Went For A TryM At Glyttery New Parlour. • Interpretation: Non-Polar Amino Acids-Proline, Valine, Phenylalanine, Isoleucine, Leucine, Alanine, Yryptophan, Methionine, Glycine. • Mnemonic: BaLy Likes His Art. • Interpretation: Basic Amino Acids-Lysine, Histidine, Arginine. • Mnemonic: ATGC • Interpretation: Nitrogen Bases Present In DNA = Adenine, Thymine, Guanine And Cytosine. • Mnemonic: Area Under The Growth Curve • Interpretation: Nitrogen Bases Present In RNA = Adenine, Uracil, Guanine And Cytosine

Know the Terms  N-Terminal end : There is a free amino group at one end of molecule of amino acid which is known as N-Terminal end.  C-Terminal end : There is a free carboxyl group at the end of amino acid molecule which is known as C-Terminal end.  Conjugated proteins : In this case, a protein part is linked to a non-protein part called prosthetic group which is mostly concerned with the special biological function of protein.  Derived proteins : These are the proteins formed by the partial hydrolysis of simple conjugated proteins such as proteases, peptones, peptides etc.  Enzyme inhibitors : These are the chemical substances which tend to reduce the activity of a particular enzyme instead of increasing it.  Transcription : Process of synthesis of RNA.  Replication : Process by which a single DNA molecule produces two identical copies of itself.  Native protein : Protein found in biological system with a unique 3-dimensional structure and biological activity.  Gene : Sequence of bases or nucleotides in the DNA molecule which regulates the synthesis of a specific protein.

Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS : Q.1. Amino acids are : (a) Acidic (b) Basic (c) Amphoteric (d) Neutral  R [CBSE, Delhi Set-1, 2020] Ans. Correct option : (c) Explanation : Amino acids are amphoteric because they contain both basic –NH2 group and acidic –COOH group. Q.2. An a-helix is a structural feature of : (a) Sucrose (b) Polypeptides (c) Nucleotides (d) Starch  R [CBSE, Outside Delhi Set-1, 2020] Ans. Correct option : (b) Explanation : An a-helix is a structural feature of proteins. i.e., polypeptides (Polymers of amino acids).

(1 mark each) Q.3.  Which one is the complementary base of cytosine in one strand to that in other strand of DNA? (a) Adenine (b) Guanine (c) Thymine (d) Uracil U [CBSE, Outside Delhi Set-3, 2020]  Ans. Correct option : (b) Explanation : Guanine (G) is the complementary base of cytosine (C) in one stand to that in other stand of DNA. C º G Q.4. Curdling of milk is an example of : (a) breaking of peptide linkage (b) hydrolysis of lactose (c) breaking of protein into amino acids (d) denaturation of protein R [CBSE, SQP, 2020-21] 

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Ans. Correct option : (d) Explanation : Curdling of milk is an example of denaturation of milk proteins. Q.5. Dinucleotide is obtained by joining two nucleotides together by phosphodiester linkage. Between which carbon atoms of pentose sugars of nucleotides are these linkages present ? (a) 5’ and 3’ (b) 1’ and 5’ (c) 5’ and 5’ (d) 3’ and 3’ A Ans. Correct option : (a) Explanation : These linkages are present between 5’ and 3’ of pentose sugars of nucleotides. Q.6. Proteins are found to have two different types of secondary structures, viz. a-helix and β-pleated sheet structure. a-helix structure of protein is stabilised by : (a) Peptide bonds (b) van der Waals forces (c) Hydrogen bonds (d) Dipole-dipole interactions U Ans. Correct option : (c) Explanation : a-helix structure of protein is stabilised by hydrogen bonds. A polypeptide chain forms all possible hydrogen bonds by twisting into right-handed helix with the –NH group of each amino acid residue hydrogen bonded to >C=O of an adjacent turn of helix. [B] ASSERTION AND REASON TYPE QUESTIONS : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.

(a) Assertion and reason both are correct state­ ments and reason is correct explanation for assertion. (b) Assertion and reason both are correct state­ ments but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Q.1. Assertion : The two strands of DNA are complementary to each other. Reason : The hydrogen bonds are formed between specific pairs of bases.  [CBSE, SQP, 2020-21]

Ans. Correct option : (a) Explanation : The two strands of DNA are ­complementary to each other and hydrogen bonds are formed between specific pairs of bases. Q.2. Assertion : All naturally occurring α amino acids except glycine are optically active. Reason : Most naturally occurring α amino acids have L-configuration. Ans. Correct option : (b) Explanation : Most of the naturally occurring amino acids have  L−configuration. All naturally occurring a−amino acids are optically active except glycine. Q.3. Assertion : Glycine must be taken through diet. Reason : It is non-essential amino acid. Ans. Correct option : (d) Explanation : Glycine must not be taken through diet because it can be synthesized in our body and a non-essential amino acid. [C] VERY SHORT ANSWER TYPE QUESTIONS : Q.1. What type of protein is present in keratin? R [CBSE, Outside Delhi Set-3, 2020]  Ans. Fibrous protein. Q.2. Write the products obtained after hydrolysis of DNA. R [CBSE, Outside Delhi Set-2, 2019] Ans. 2-deoxyribose + nitrogen containing heterocyclic base + phosphate  [CBSE Marking Scheme 2019]

450 ]

Detailed Answer : The products formed after hydrolysis of DNA are : (i) Pentose sugar, (ii) Phosphoric acid and (iii) Nitrogenous bases. Q.3. Name the linkage holding amino acids in a polypeptide. R Ans. Peptide bond. Q.4. Write the sequence of bases on mRNA molecule synthesised on the following stand of DNA. AGCGATTAC U Ans. UCGCUAAUG. Q.5. What is a nucleoside? R Ans.  It is a product formed by attachment of a nitrogenous base (purine or pyrimidine) to the sugar unit (ribose or deoxyribose) at C–1 by b-linkage.

Q. 1. (i)  Glucose on reaction with acetic acid gives glucose pentaacetate. What does it suggest about the structure of glucose ?

(ii) Write difference between fibrous protein and U+R globular protein. Give two points. Ans. (i) 5 −OH groups are present.

[1]



Short Answer Type Questions-I

(2 marks each)

(ii) Fibrous proteins : parallel polypeptide chain, insoluble in water,

Globular proteins : spherical shape, soluble in water, (or any 1 suitable difference)  [1] 

[CBSE Marking Scheme 2015]

[ 451

BIOMOLECULES

Detailed Answer : (ii) Fibrous protein

Globular protein

It contains linear thread like molecules which tend to lie side by side to form fibres. Example : keratin, collagen

It contains compact form of molecules which are folded together. Example : enzymes, hormones

Ans. (i) Nucleoside : It is formed when pentose sugar combines with nitrogen base. Nucleotide : When nucleoside bonds with phosphate group.[1] (ii) a-helix has intramolecular hydrogen bonding while b-pleated has intermolecular hydrogen bonding / a-helix results due to regular coiling of polypeptide chains while in a-pleated all polypeptide chains are stretched and arranged side by side.[1]



Q.2. (i) What is the difference between a nucleoside and nucleotide ? (ii) Write one difference between a-helix and b-pleated sheet structures of protein. R Q.3. Differentiate between peptide linkage and glycosidic linkage. Give two points. Ans.

Peptide linkage

R

Glycosidic linkage

Glycosidic linkage is formed between two Peptide linkage is formed between two amino acids. monosaccharides. Hydrolysis of glycosidic linkage forms two Hydrolysis of peptide linkage forms two amino acids. monosaccharides. It links carbon and nitrogen atoms. It links two carbon atoms through an oxygen atom. It is found in polypeptides. It is found in polysaccharides. It can be represented as –CONH–. It can be given as –C–O–C–. [2]  Q.4. Define the following as related to proteins : (b) Denaturation : When a protein is subjected to (a) Primary Structure physical change like change in temperature or (b) Denaturation chemical change like change in pH, protein loses its R Ans. (a)  Primary structure : Each polypeptide in a biological activity.[1] protein molecule having amino acids which are linked with each other in a specific sequence. [1]

Short Answer Type Questions-II Q.1. Differentiate between following : (i) Amylose and Amylopectin (ii) Globular protein and Fibrous protein (iii) Nucleotide and Nucleoside  R [CBSE, Delhi Set-1, 2020] Ans. (i) Amylose and Amylopectin : S. No. 1.

Amylose

Amylopectin

Amylose is a straight Amylopectin is a chain polymer of branched chain polymer D-glucose. of D-glucose.

2.

S. No. 1.

Globular protein

Fibrous protein

Polypeptide chains Polypeptide chains run are arranged as coils. parallel to each other.

They have spherical They have thread like shape. stucture.

3.

These are water These are insoluble in soluble. water. [1] (iii) Nucleotide and Nucleoside : S. No. 1.

2.

They are linked by 1, They are linked by a-1, 4- glycosidic linkage. 4 glycosidic and a–1, 6– glycosidic linkage. [1] (ii) Globular and Fibrous protein :

(3 marks each)

Nucleotide

Nucleoside

It consists of a nitrogenous base, sugar and one to three phosphate groups.

It consists of a nitrogenous base covalently bonded to a sugar without phosphate group.

2.

Example : 5’-uridine Example : Uridine monophosphate [1] Q.2. Define the following terms with a suitable example in each : (a) Polysaccharides (b) Denatured protein (c) Fibrous protein  R [CBSE, Delhi Set-2, 2020]

452 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



D-glucose

Q.5. (a)  What are the products of hydrolysis of maltose? (b)  What type of bonding provides stability to a-helix structure of protein ? (c) Name the vitamin whose deficiency causes pernicious anaemia. 

CH2OH

D-gluconic acid[1]



(iii) On denaturation secondary and tertiary proteins get converted into primary proteins. Denaturation disrupts the normal alpha-helix and beta sheets in a protein and uncoils it into a random shape. [1]

Q.4. Define the following with a suitable example in each: (i) Oligosaccharides (ii) Denaturation of protein (iii) Vitamins R [CBSE, Delhi Set-3, 2019] Ans. (i) Carbohydrates that yield two to ten monosaccharide units, on hydrolysis, are called oligosaccharides. Example: Sucrose or any other.[1]









Ans. (a) Glucose + Glucose [1] (b) Hydrogen bonding [1] (c) Vitamin –B12 [1] [CBSE Marking Scheme 2019] Q.6. Define the following with an example of each :

Br2 water (CHOH)4   (CHOH)4

CH2OH

(ii) W  hen a protein in its native form, is subjected to physical change like change in temperature or chemical change like change in pH, the hydrogen bonds are disturbed. Due to this, globules unfold and helix get uncoiled and protein loses its biological activity. Example: Boiling of egg.[1] (iii) O  rganic compounds required in the diet in small amounts to perform specific biological functions for normal maintenance of optimum growth and health of the organism. Example: Vitamin A[1] [CBSE Marking Scheme 2019]





Ans. (a) Polysaccharides : These are complex long chains of monosaccharides linked by glycosidic bonds e.g. starch, cellulose etc. [1] (b) Denatured protein : When native protein is subjected to physical change like change in temperature or change in pH, the hydrogen bonds are disturbed. Due to this, helix gets uncoiled and protein loses its biological activity. This is called denaturation and protein is called denatured protein. e.g. coagulation of egg. [1] (c) Fibrous protein : When polypeptide chains run parallel and are held together by hydrogen and disulphide bonds are called fibrous proteins. e.g. Keratin, myosin. [1] Q.3. (i) What are the hydrolysis products of DNA? (ii) What happens when D-glucose is treated with Bromine water? (iii) What is the effect of denaturation on the structure of proteins?  A [CBSE, Delhi Set-3, 2020] Ans. (i) On hydrolysis of DNA, the products are pentose sugar, phosphoric acid and bases. [1] (ii) When D-glucose is treated with bromine water D-gluconic acid is formed. CHO COOH



(a) Polysaccharides (b) Denatured protein (c) Essential amino acids

Ans. (a) Carbohydrates that give large number of monosaccharide units on hydrolysis / large number of monosaccharide units joined together by glycosidic linkage starch/ glycogen/ cellulose (or any other) ½+½ (b) Proteins that lose their biological activity / proteins in which secondary and tertiary structures are destroyed Curdling of milk (or any other) ½+½ (c) Amino acids which cannot be synthesised in the body. ½ Valine / Leucine (or any other) ½  [CBSE Marking Scheme, 2018]





OR

[ 453

BIOMOLECULES

[3] 



[Topper’s Answer 2018]





[1] (b) Due to the presence of carboxyl and amino group in the same molecule / due to formation of zwitter ion or dipolar ion. [1] (c) a-helix has intramolecular hydrogen bonding while b–pleated has intermolecular hydrogen bonding / a-helix results due to regular coiling of polypeptide chains while in b– pleated all polypeptide chains are stretched and arranged side by side. [1] [CBSE Marking Scheme, 2018]







Detailed Answer : (a) When D-glucose reacts with conc. HNO3, it forms saccharic acid. COOH CHO (CHOH)4 CH2OH



D-glucose

Conc. HNO3

(CHOH)4 COOH D-saccharic acid

OH

–

O

+

H

OH

– –

R–CH–C–O +NH 3

–

–

R–CH–C–OH +NH 3 –

NH2

O

+

H

– –



– –

R–CH–C–O

–

–

(a) Saccharic acid/COOH-(CHOH)4-COOH

(b) In aqueous solution, the carboxyl group present in amino acid can lose a proton and the amino group can accept a proton to form zwitter ion. This zwitter ion can act both as an acid and a base showing amphoteric behaviour. O

–

Q. 7. (a) Write the product when D-glucose reacts with conc. HNO3. (b) Amino acids show amphoteric behaviour. Why ? (c) Write one difference between a-helix and bpleated structures of proteins.  A + R [CBSE, Delhi/Outside Delhi, 2018]





Q. 8. (i)  Write the name of two monosaccharides obtained on hydrolysis of lactose sugar. (ii) Why vitamin C cannot be stored in our body? (iii) What is the difference between a nucleoside and nucleotide?  R [CBSE, Outside Delhi Set-1, 2, 3, 2016] Ans. (i) b-D glucose and b-D-galactose / glucose and galactose.  ½+½ (ii) water soluble, excreted out of the body. [1] (iii) In nucleotide, phosphoric acid/phosphate group attached to the nucleoside / structures of both nucleotide and nucleoside / nucleotide= base +sugar + phosphate group, nucleoside= base +sugar.  [1] Detailed Answer :

(i) beta-D-Glucose and b–D–Galactose. 



(ii) It is water soluble, hence it is excreted through urine. [1]

[1]

454 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(iii) Nucleoside : It is formed when pentose sugar combines with nitrogen base. Nucleotide : When Nucleoside combines with phosphate group. [1] Q. 9. (i) Write the structural difference between starch and cellulose. (ii) What type of linkage is present in nucleic acids ? (iii) Give one example each for fibrous protein and globular protein.  R [CBSE, Outside Delhi Set-1, 2, 3, 2016]

Q. 12. (i) Deficiency of which vitamin causes night blindness ? (ii) Name the base that is found in nucleotide of RNA only. (iii) Glucose on reaction with HI gives n-hexane. What does it suggest about the structure of glucose ?

Ans. (i) Starch – Polymer of a – D – glucose units / Polymer of a - glucose units. Cellulose – Polymer of b-D - glucose units / Polymer of b - glucose units.  [1] (ii) Phosphodiester linkage  [1] (iii) Fibrous protein – Keratin / myosin / collagen. Globular protein – Haemoglobin / insulin.  [½ + ½]  [CBSE Marking Scheme, 2016] Detailed Answer : (i) In starch, the glucose monomers are in alpha configuration while in cellulose, the glucose monomers are in beta configuration. Starch is a polymer consisting of amylose and amylopection while cellulose is a long chain composed only of b-D-glucose units. [1] (ii) Phosphodiester linkage between the 5’ and 3’ carbon atoms of sugar moieties is present in Nucleic acids. [1] (iii) Example of fibrous protein—Collagen, keratin, myosin. (Any one)  [1] Q. 10. How are vitamins classified ? Name the vitamin responsible for the coagulation of blood. R [CBSE Comptt. Delhi 2015] Ans. Vitamins are classified into : (i) Water soluble vitamins : Vitamins which are soluble in water but insoluble in fat and oils. e.g., vitamin B and C. (ii) Fat soluble vitamins : Vitamins are soluble in fat and oils but insoluble in water. e.g., vitamin A, D, E and K. Vitamin K is responsible for coagulation of blood. [2 + 1] Q. 11. (i) Which one of the following is a polysaccharide : starch, maltose, fructose, glucose. (ii) Write one difference between a-helix and b-pleated sheet structures of protein. (iii) Write the name of the disease caused by the deficiency of vitamin B12. R + U [CBSE OD 2015]  Ans. (i) Starch. [1] (ii) a  - Helix polypeptide chains are stabilized by intramolecular H-bonding whereas b-pleated sheet is stabilized by intermolecular H-bonding. (or any other difference) [1] (iii) Pernicious anaemia. [1]

Q. 13. (i) Deficiency of which vitamin causes scurvy ? (ii) What type of linkage is responsible for the formation of proteins ? (iii) Write the product formed when glucose is treated with HI. R Ans. (i) Vitamin C. [1] (ii) Peptide linkage. [1] (iii) n-hexane or its structure. [1] Q. 14. (i) A non-reducing disaccharide ‘A’ on hydrolysis with dilute acid gives an equimolar mixture of D–(+)–glucose and D-(-)-Fructose. HCI A+H O  C6H12O6+C6H12O6 2

R+U Ans. (i) Vitamin A [1] (ii) Uracil [1] (iii) I t suggests that six carbon atoms are in straight chain/CHO—(CHOH)4—CH2OH [1]

[a]D



= +66.50

+ 52.5°

– 92.4°



Identify A. What is the mixture of D–(+)–glucose and D-(-)-Fructose known as ? Name the linkage that holds the two units in the disaccharide. (ii) a-amino acids have relatively higher melting points than the corresponding halo acids. Explain. A + A&E Ans. (i) A–Sucrose (C12H22O11) [½] The mixture of D-(+)- glucose and D-(-)-Fructose is known as invert sugar. [½] The linkage which holds the two monosaccharide units through oxygen atom is called glycosidic linkage. [1] (ii) The amino acids exist as dipolar Zwitter ion. Due to their dipolar salt like character, they have strong dipole-dipole attractions. Thus, their melting points are higher than the corresponding halo acids which do not exist as Zwitter ions. [1] Q. 15. ( i) Which one of the following is a disaccharide : Starch, Maltose, Fructose, Glucose ? (ii) What is the difference between fibrous protein and globular protein ? (iii) Write the name of vitamin whose deficiency causes bone deformities in children. R + U [CBSE Delhi 2015]



Ans. (i) Maltose [1]

(ii) Fibrous proteins : parallel polypeptide chain, insoluble in water Globular proteins: spherical shape, soluble in water, (or any 1 suitable difference)1 (iii) Vitamin D [CBSE Marking Scheme 2015]  1

[ 455

BIOMOLECULES



Detailed Answer : (ii)

Fibrous protein

Globular protein

It contains linear thread like molecules which tend to lie side by side to form fibres.

It contains compact form of molecules which are folded together.

Example : keratin, Example : enzymes, collagen. hormones.







 1

Q. 16. (i)  Why water soluble vitamins must be supplied regularly in the diet? Give one example of it. (ii) Differentiate between the following: (a) Essential and non-essential amino acids. (b) Fibrous and globular proteins. A&E + U [CBSE Comptt OD Set-1, 2, 3 2017] Ans. (i) Because they are excreted in urine and cannot be stored in body; Vitamin C/B1/ B6 ½+½

(ii) (a) Essential amino acids are those which cannot be synthesized in the body and are supplied through diet whereas non-essential amino acids can be synthesized in the body. [1] (b) In fibrous proteins, the polypeptide chains run parallel and are held together by hydrogen or disulphide bonds while in globular, polypeptide chains coil around to give a spherical shape.[1] [CBSE Marking Scheme 2017]

Q 17. (i) What type of linkage is present in disaccharides? (ii) Write one source and deficiency disease of vitamin B12. (iii) Write the difference between DNA and RNA. R [CBSE Comptt. Delhi Set-1, 3 2017] Ans. (i) Glycosidic linkage.[1] (ii) Source: Meat, fish, egg, curd (any one); Pernicious anemia  ½+½ (iii) DNA is a double strand while RNA is a single strand molecule. (or any other correct difference)  [1]  [CBSE Marking Scheme 2017]

Long Answer Type Questions

(5 marks each)

Q. 1. Write the structures of fragments produced on complete hydrolysis of DNA. How are they linked in DNA molecule? Draw a diagram to show pairing of nucleotide bases in double helix of DNA.  R Ans. Complete hydrolysis of DNA yields a pentose sugar, phosphoric acid and nitrogen containing heterocyclic compounds called bases. Structures : (i) Sugar :

β-D-2-deoxyribose

(ii) Phosphoric acid :

[1]



|

|

OH | P HO || OH O (iii) Nitrogen base : DNA contains four bases : Adenine, Guanine, Thymine and Cytosine.

[1]

H 3C



[1] A unit formed by the attachment of a base to 1’-position of sugar is called nucleoside. When nucleoside links to phosphoric acid at 5’-position of sugar moiety, a nucleotide is formed. Nucleotides are joined H 3C together by phosphodiester linkage between 5’- and 3’- carbon atoms of the pentose sugar.





Base Base Base    Sugar  Phosphate  Sugaar  Phosphate n  Sugar





[1]

456 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

In DNA, two chains of nucleic acid coil around each other and held together by H-bonds between bases of two chains.

(ii) Classify the following into monosaccharides and disaccharides : ribose, 2-deoxyribose, maltose, galactose, fructose and lactose. Ans. (i) In aqueous solution, the carboxyl group of an amino acid can lose a proton and the amino group can accept a proton to give a dipolar ion known as zwitter ion. O O



Q.2. (i) How do you explain the amphoteric behaviour of amino acids ?



Double Helical Structure of DNA (Proposed by Watson and Crick)



R CH C O H



:NH2

R CH C O– +

NH3 Zwitter ion

[1]

Therefore, in zwitter ion form, the amino acid can act both as an acid and as a base. In the acidic medium, COO– ion of the zwitter ion accepts a proton to form the cation, and in the basic medium,+NH3 ion loses a proton to form the anion, as shown below :

O O O || || || + + H H R—CH—C—O– R—CH—C—O– R—CH—C—OH | OH– | | OH– + + NH2 NH3 NH3



Thus, amino acids show amphoteric behaviour.[3]

(ii) Monosaccharides : Ribose and fructose. 



Visual Case-Based Questions Q. 1. R  ead the passage given below and answer the following questions : EVIDENCE FOR THE FIBROUS NATURE OF DNA The basic chemical formula of DNA is now well established. As shown in Figure 1 it consists of a very long chain, the backbone of which is made up of alternate sugar and phosphate groups, joined together in regular 3’ 5’ phosphate di-ester linkages. To each sugar is attached a nitrogenous base, only four different kinds of which are commonly found in DNA. Two of these---adenine and guanine--- are purines, and the other two thymine and cytosineare pyrimidines. A fifth base, 5-methyl cytosine,



[1]

Disaccharides : Maltose and lactose.[1]

(4 marks each) occurs in smaller amounts in certain organisms, and a sixth, 5-hydroxy-methyl-cytosine, is found instead of cytosine in the T even phages. It should be noted that the chain is unbranched, a consequence of the regular internucleotide linkage. On the other hand the sequence of the different nucleotides is, as far as can be ascertained, completely irregular. Thus, DNA has some features which are regular, and some which are irregular. A similar conception of the DNA molecule as a long thin fiber is obtained from physicochemical analysis involving sedimentation, diffusion, light scattering, and viscosity measurements. These techniques indicate that DNA is a very asymmetrical structure approximately 20 A wide and many

BIOMOLECULES

thousands of angstroms long. Estimates of its molecular weight currently center between 5 X106 and X107 (approximately 3 X104 nucleotides). Surprisingly each of these measurements tend to suggest that the DNA is relatively rigid, a puzzling finding in view of the large number of single bonds (5 per nucleotide) in the phosphate-sugar back bone. Recently these indirect inferences have been confirmed by electron microscopy. [CBSE-QB 2021] (Source: Watson, J. D., & Crick, F. H. (1953, January). The structure of DNA. In Cold Spring Harbor symposia on quantitative biology (Vol. 18, pp. 123-131). Cold Spring Harbor Laboratory Press.) (i) Purines present in DNA are: (a)  adenine and thymine (b)  guanine and thymine (c)  cytosine and thymine (d)  adenine and guanine (ii) DNA molecule has ___________ inter­  nucleotide linkage and __________ sequence of the different nucleotides (a)  regular, regular (b)  regular, irregular (c)  irregular, regular (d)  irregular, irregular (iii) DNA has a ___________ backbone (a) phosphate-purine (b) pyrimidines-sugar (c) phosphate-sugar (d) purine-pyrimidine (iv) Out of the four different kinds of nitrogenous bases which are commonly found in DNA, ___________ has been replaced in some organisms. (a) adenine (b) guanine (c) cytosine (d) thymine Ans. (i) Correct option : (d) (ii) Correct option : (b) (iii) Correct option : (c) (iv) Correct option : (c) Q. 2. R  ead the passage given below and answer the following questions : Polysaccharides may be very large molecules. Starch, glycogen, cellulose, and chitin are examples of polysaccharides. Starch is the stored form of sugars in plants and is made up of amylose and amylopectin (both polymers of glucose). Amylose is soluble in water and can be hydrolyzed into glucose units breaking glycocidic bonds, by the enzymes a- amylase and b-amylase. It is straight chain polymer. Amylopectin is a branched chain polymer of several D-glucose molecules. 80% of amylopectin is present in starch. Plants are able to synthesize glucose, and the excess glucose is stored as starch in different plant parts, including roots and seeds. The starch that is consumed by animals is broken down into smaller molecules, such as glucose. The cells can then absorb the glucose.

[ 457 Glycogen is the storage form of glucose in humans and other vertebrates, and is made up of monomers of glucose. It is structurally quite similar to amylopectin . Glycogen is the animal equivalent of starch. It is stored in liver and skeletal muscles. Cellulose is one of the most abundant natural biopolymers. The cell walls of plants are mostly made of cellulose, which provides structural support to the cell. Wood and paper are mostly cellulosic in nature. Like amylose, cellulose is a linear polymer of glucose. Cellulose is made up of glucose monomers that are linked by bonds between particular carbon atoms in the glucose molecule. Every other glucose monomer in cellulose is flipped over and packed tightly as extended long chains. This gives cellulose its rigidity and high tensile strength—which is so important to plant cells. Cellulose passing through our digestive system is called dietary fiber. (Source: https://chem.libretexts.org) [CBSE-QB 2021] (i) In animals, Glycogen is stored in : (a) Liver (b) Spleen (c) Lungs (d)  Small Intestine (ii) Amylose is : (a)  straight chain, water insoluble compo­ nent of starch ,which constitutes 20 % of it (b) straight chain, water soluble component of starch, which constitutes 20 % of it (c) branched chain, water insoluble compo­nent of starch, which constitutes 80 % of it (d) branched chain, water soluble component of starch, which constitutes 80 % of it (iii)  hich biopolymer breaks down to release W glucose, whenever glucose levels drop in Our body : (a) starch (b) cellulose (c) chitin (d) glycogen (iv)  he linkages which join monosaccharides to T form long chain polysaccharides : (a)  Peptide linkage (b)  Disulphide bonds (c)  Hydrogen bonds (d)  Glycosidic linkage (v) Cellulose on complete hydrolysis yields : (a) amylose (b) amylopectin (c) glucose (d)  amylose and amylopectin Ans. (i) Correct option : (a) (ii) Correct option : (b) (iii) Correct option : (d) (iv) Correct option : (d) (v) Correct option : (c)  ead the passage given below and answer the Q. 4. R following questions : (1 × 4 = 4)  he two monosaccharides are joined together by T an oxide linkage formed by the loss of a water mol-

458 ]





Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

ecule. Such a linkage between two monosaccharide units through oxygen atom is called glycosidic linkage. In disaccharides, if the reducing groups of monosaccharides i.e., aldehydic or ketonic groups are bonded, these are non-reducing sugars, e.g., sucrose. On the other hand, sugars in which these functional groups are free, are called reducing sugars, for example, maltose and lactose. A non reducing disaccharide ‘A’ on hydrolysis with dilute acid gives an equimolar mixture of D– (+)–glucose and D-(-)-Fructose. HCl → C H O +C H O A + H O  2

6

12

6

6

12

6

[a]D = +66.50 + 52.5° – 92.4° (i) In the above reaction, reactant ‘A’ is : (a) Glucose (b) Sucrose (c) Maltose (d) Fructose (ii) What is the mixture of D-(+) glucose and D-(+) fructose known as ? (a) Anomers (b) Racemic mixture (c) Invert sugar (d) Optical mixture (iii) Name the linkage that holds the two units in the disaccharide ? (a) Nucleoside linkage (b) Glycosidic linkage (c) Peptide linkage (d) None of the above (iv)  Glucose on reaction with acetic acid gives glucose pentaacetate. What does it suggest about the structure of glucose ? (a) C-1 is anomeric carbon (b) C-5 is anomeric carbon (c) 3’-OH groups are present (d) 5’-OH groups are present OR Proteins are found to have two different types of secondary structures, viz. a-helix and b-pleated sheet structure. a-helix structure of protein is stabilised by (a) Peptide bonds (b) van der Waals forces (c) Hydrogen bonds (d) Dipole-dipole interactions Ans. (i) Correct option : (b) Explanation :



+

H C12 H 22O11 + H 2O ¾¾ ® C 6 H12O6 + C 6 H12O6 [1]

(A) (Sucrose)

Glu cose

Fructose

(ii) Correct option : (c) Explanation : The mixture of D-(+)- glucose and D-(-)-Fructose is known as invert sugar. [1] (iii) Correct option : (b) Explanation : The linkage which holds the two monosaccharide units through oxygen atom is called glycosidic linkage. [1] (iv) Correct option : (d) Explanation : It indicates that 5’-OH groups are present which react with acetic acid to give glucose pentaacetate. [1]

OR Correct option : (c) Explanation : a-helix structure of protein is stabilised by hydrogen bonds. A polypeptide chain forms all possible hydrogen bonds by twisting into right-handed helix with the –NH group of each amino acid residue hydrogen bonded to >C=O of an adjacent turn of helix. [1] Q.5.  Read the passage given below and answer the following questions : (1 × 4 = 4) The sequence of bases in m-RNA are read in a serial order in groups of three at a time. Each triplet of nucleotides (having a specific sequence of bases) is known as codon. Each codon specifies one amino acid. Many amino acids have more than one codons. The amino acids are brought to the mRNA by another type of RNA and called tRNA. Each amino acid has atleast one corresponding tRNA. At one end of the tRNA molecule is a trinucleotide base sequence that is complementary to some trinucleotide base sequence on mRNA. (i) Which of the following nitrogen bases is not present in RNA? (a) Thymine (b) Adenine (c) Guanine (d) Cytosine (ii) Each triplet of nucleotides is called : (a) Anticodon (b) Codon (c) mRNA (d) tRNA (iii) Each codon specifies : (a) 1 amino acid (b) 2 amino acids (c) 3 amino acids (d) None of these (iv) In mRNA, the complementary bases of AAT is : (a) CCG (b) UUA (c) AUU (d) UUU OR The amino acids are brought to the mRNA by: (a) rRNA (b) DNA (c) tRNA (d) None of these Ans. (i) Correct option : (a) Explanation : In RNA, thymine is not present. In place of thymine, uracil is present in RNA. [1] (ii) Correct option : (b) Explanation : Each triplet of nucleotides, which have a specific sequence of bases, is called codon. [1] (iii) Correct option : (a) Explanation : Each codon specifies one amino acid. Many amino acids possess more than one codons.  [1] (iv) Correct option : (b) Explanation : In mRNA A ::: U T ::: A So, the complementary bases of AAT in mRNA is UUA. [1] OR Correct option : (c) Explanation : The amino acids are brought to the mRNA by tRNA (transfer RNA).

BIOMOLECULES

Q. 3. R  ead the passage given below and answer the following questions : Adenosine triphosphate (ATP) is the energy-carrying molecule found in the cells of all living things. ATP captures chemical energy obtained from the breakdown of food molecules and releases it to fuel other cellular processes. ATP is a nucleotide that consists of three main structures: the nitrogenous base, adenine; the sugar, ribose; and a chain of three phosphate groups bound to ribose. The phosphate tail of ATP is the actual power source which the cell taps. Available energy is contained in the bonds between the phosphates and is released when they are broken, which occurs through the addition of a water molecule (a process called hydrolysis). Usually only the outer phosphate is removed from ATP to yield energy; when this occurs ATP is converted to adenosine diphosphate (ADP), the form of the nucleotide having only two phosphates. The importance of ATP (adenosine triphosphate) as the main source of chemical energy in living matter and its involvement in cellular processes has long been recognized. The primary mechanism whereby higher organisms, including humans, generate ATP is through mitochondrial oxidative phosphorylation. For the majority of organs, the main metabolic fuel is glucose, which in the presence of oxygen undergoes complete combustion to CO2 and H2O: C6H12O6 + 6O2 → 6O2 + 6H2O + energy The free energy (ΔG) liberated in this exergonic (ΔG is negative) reaction is partially trapped as ATP in two consecutive processes: glycolysis (cytosol) and oxidative phosphorylation (mitochondria). The first produces 2 mol of ATP per mol of glucose, and the second 36 mol of ATP per mol of glucose. Thus, oxidative phosphorylation yields 17-18 times as much useful energy in the form of ATP as can be obtained from the same amount of glucose by glycolysis alone. The efficiency of glucose metabolism is the ratio of amount of energy produced when 1 mol of glucose oxidised in cell to the enthalpy of combustion of glucose. The energy lost in the process is in the form of heat. This heat is responsible for keeping us warm. (Source: Ereciñska, M., & Silver, I. A. (1989). ATP and Brain Function. Journal of Cerebral Blood Flow & Metabolism, 9(1), 2–19. https://doi.org/10.1038/ jcbfm.1989.2 and https://www.britannica.com/science/ adenosine-triphosphate) [CBSE-QB 2021]

[ 459 The following questions are multiple choice questions. Choose the most appropriate answer : (i) Cellular oxidation of glucose is a: (a)  spontaneous and endothermic process (b)  non spontaneous and exothermic process (c) non spontaneous and endothermic process (d)  spontaneous and exothermic process (ii) What is the efficiency of glucose metabolism if 1 mole of glucose gives 38ATP energy? (Given: The enthalpy of combustion of glucose is 686 kcal, 1ATP = 7.3kcal) (a)  100% (b)  38% (c)  62% (d)  80% (iii) Which of the following statement is true? (a)   ATP is a nucleoside made up of nitrogenous base adenine and ribose sugar. (b) ATP consists the nitrogenous base, adenine and the sugar, deoxyribose. (c)   ATP is a nucleotide which contains a chain of three phosphate groups bound to ribose sugar. (d) The nitrogenous base of ATP is the actual power source. (iv) Nearly 95% of the energy released during cellular respiration is due to: (a)  glycolysis occurring in cytosol (b)  oxidative phosphorylation occurring in cytosol (c)  glycolysis in occurring mitochondria (d)   oxidative phosphorylation occurring in mitochondria (v) Which of the following statements is correct: (a) ATP is a nucleotide which has three phosphate groups while ADP is a nucleoside which three phosphate groups. (b)   ADP contains a nitrogenous bases adenine, ribose sugar and two phosphate groups bound to ribose. (c) ADP is the main source of chemical energy in living matter. (d) ATP and ADP are nucleosides which differ in number of phosphate groups. Ans. (i) Correct option : (d) (ii) Correct option : (b) Explanation : Glucose catabolism yields a Total of 38 ATP. 38 ATP x 7.3 kcal/mol ATP = 262 kcal. Glucose has 686 kcal. Thus the efficiency of glucose metabolism is 262/686 x 100 = 38%. (iii) Correct option : (c) (iv) Correct option : (d) (v) Correct option : (b) 

460 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

15 UNIT

Syllabus

POLYMERS

¾¾ Classification - natural and synthetic, methods of polymerization (addition and condensation), copolymerization, some important polymers: natural and synthetic like polythene, nylon, polyesters, bakelite, rubber. Biodegradable and non-biodegradable polymers.

Trend Analysis 2020

List of Concept names

D

OD

3Q (2 marks)

2Q (3 marks)

Classification of polymers and different polymers

1Q (1 mark)

–

Use of Polymers & mechanism of Polymerisation

2Q (1 mark)

–

–

–

Monomer units of (Name and structure)

Biogradable Polymers

Polymers

2019 D OD 1Q 1Q (1 mark) (1 mark) 1Q 1Q (3 marks) (3 marks) 3Q 2Q (1 mark) (1 mark) 1Q (3 marks) 1Q – (1 mark) 1Q – (1 mark)

Revision Notes  Polymers : Polymers are defined as the high molecular mass macromolecules, which consist of repeating structural units derived from the corresponding monomers.                Poly = many mers = parts or units  Polymerisation : The process by which monomers are converted into polymers is called polymerisation. 350 − 570 K, 100 − 200 atm

→ —CH nCH 2 = CH 2  ( )n 2 — CH 2 — Traces of O 2

Ethene Polythene (Monomer) (Polymer)  Natural Polymers : They are found in nature, mostly in plants and animals. e.g., proteins, natural rubber, etc.  Synthetic Polymers : These are man-made polymers prepared in the laboratory. e.g., polythene, teflon, nylon, etc.  Copolymerisation is a polymerisation reaction in which a mixture of more than one monomeric species is allowed to polymerise.  Copolymers : The polymers made by addition polymerisation from two different monomers.

2018 D/OD –

1Q (1 mark) – 2Q (1 mark)

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Polymers Scan to know more about this topic

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Types of polymerisation

462 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

—

—

CH = CH2 Polymerisation

CH2 + — nC CH6H=5CH 2= + CH 2==CH CH2n → nCH2 = CH —nCH 1, 3-Butadiene 1, 3-Butadiene Styrene Styrene

— CH2 — CH = CH — CH2 — CH — CH2 — n Butadiene - styrene copolymer [Buna-S]

 A ddition Polymers : These polymers are formed by intermolecular combination of repeating units without elimination of small molecule.  CH 2 − C H  Polymerisation nCH 2 = CH − Cl →  |   Peroxide Vinyl chloride Cl  n 

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PVC

Poly Vizylchloride

Polysters

 C ondensation Polymers : These polymers are formed by the repeated condensation reactions between different bifunctional or trifunctional monomer units usually with elimination of small molecules such as water, alcohol, hydrogen chloride etc. e.g., nylon 6, 6, nylon-6, terylene, etc. nH 2 N(CH 2)6NH 2 + nHOOC(CH 2)4 COOH —® — [ NH(CH 2)6 NHCO(CH 2)4CO—]n + 2nH 2O

Adipic acid Nylon 6, 6

Hexamethylene diamine

 Condensation Polymers : (a) Polyesters : These are polymers having large number of ester linkages. e.g., (i) Terylene : It is resistant to action of chemical and solvent. It has a low moisture absorbing power. It is also called dacron. It is used in dress materials like sarees. It is used as a blend with cotton and wool to give terrycot and terrywool respectively.

Ethylene glycol

— —

425 – 475 K

— O — CH2 — CH2 — O — C —

–(2n–1)H2O

—C— — n

Terylene (Dacron)

Terephthalic acid

Ethylene glycol

— —

heat –(2n–1)H2O

O

— O — CH2 — CH2 — O — C

Phthalic acid

C — + 2 n H 2O

—

C — OH

—

—

nHO — CH2 — CH2 — OH + nHO — C

O

— —

O

—

O

— —

(ii) Glyptal or alkyd resin : It is three dimensional cross-linked polymer. It is tough and flexible. It is used in adherent paints, lacquers and building materials like asbestos and cement. — —



— C — OH

— —

— —

nHO — CH2 — CH2 — OH + n HO — C —

O

O

O

— —

O

n

Glyptal

(b) Polyamides : Those polymers which have large number of amide linkages are called polyamides. e.g., (i) Nylon-6, 6 : It can be cast into sheets or fibres. Nylon fibres have high tensile strength. It is tough and resistant to abrasion. It is also elastic in nature. It is used to make bristles of toothbrush, climbing ropes, fishing nets and parachute fabrics. It is a condensation polymer of adipic acid and hexamethylene diamine.

–(2n–1)H2O Nylon-6, 6

(ii) Nylon-6 : It can be cast into sheets and fabrics. It is tough and strong.

H

—



O

— —

— —

Oxidation

N

NOH NH2OH

Beckmann Rearrangement

O2

Cyclohexane

H2SO4

Cyclohexanone

Cyclohexanoxime

H2C

C=O

H2C

CH2

H2C

CH2

Caprolactam

[ 463

POLYMERS



(iii) Nylon-6, 10 : It is a polymer of hexamethylenediamine and decanedioxyl chloride (sebacoyl chloride). O

O

nH2N — (CH2)6 — NH2 + nCl — C — (CH2)8 — C — Cl Hexamethylenediamine

Decanedioxyl chloride

H heat –(2n–1) HCl

O

O

— HN — (CH2)6 — N — C (CH2)8 — C — + 2nHCl Nylon-6,10

(c) Phenol-formaldehyde resin (Bakelite) : It is heat resistant thermosetting plastic.

Formaldehyde Phenol Scan to know more about this topic

Polysters and polymers

Bakelite









(d) Melamine-formaldehyde resin : It is thermosetting plastic which is unbreakable.

 

Low density polyethene : It is produced by free radical polymerisation at high temperature (200°C) and high pressure about 1000 – 2000 atm. It is a branched chain polymer. High density polyethene : It is produced by polymerisation of ethene in presence of Ziegler-Natta catalyst at temperature below 100°C and pressure 6-7 atmosphere. It is a linear polymer. •  Bake lite Pasta and Fries. •  Bakelite: Monomer units Phenol and Formaldehyde. Biodegradable polymers : Those polymers which are biodegradable, i.e., decomposed by micro-organisms and do not cause water pollution, e.g., PHBV, Poly (Glycolic acid) and Poly (Lactic acid), etc. (i) PHBV (Poly-β-hydroxybutyrate-co-β-hydroxy valerate) : It is a copolymer of 3-hydroxybutanoic acid and 3-hydroxypentanoic acid in which the monomers are connected by ester linkages. +

464 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII



The properties of PHBV vary according to ratio of both the acids. 3-hydroxybutyric acid provides stiffness and 3-hydroxypentanoic acid provides flexibility to copolymer. It is used in packaging, orthopaedic devices and even in controlled drug release. It is biodegradable. (ii) Poly (Glycolic acid) and Poly (Lactic acid) are biodegradable polymers and used in sutures. Dextron was the first bioabsorbable suture made from biodegradable polyesters for postoperative stitches.  Non-biodegradable polymers : Those polymers which do not degrade in environment and accumulate in the form of waste, e.g., polythene, polystyrene, etc. They consist of long chains of carbon and hydrogen atoms joined by strong interatomic bonding making it hard for microbes to break the bonds and digest them.  Natural, Synthetic and Condensation Polymers : Natural Polymers S. No. (i) (ii) (iii) (iv) (v) (vi) (vii)

Polymer Cellulose Starch Protein Nucleic acid Rayon (Artificial silk) Natural rubber Gutta percha

S. No. (i)

Polyethene

(ii)

Uses Occurs in cotton, cell wall Food material storage in plants Essential for growth Essential for life perpetuation Fabrics, surgical dressings Used for tyres after vulcanisation Rubber like material Uses

Ethene (CH2 = CH2)

Addition and Chain growth

Polypropene

Propene (CH3—CH=CH2)

(iii)

Polystyrene

Styrene (C6H5–CH=CH2)

(iv)

Polyvinyl chloride (PVC)

Addition and Chain growth Addition and Chain growth Addition and Chain growth Addition and Chain growth

Electrical insulator, packing materials, films, bottles etc. Storage battery tanks

(vi)

Monomer

Biodegradable Polymers

Class

(v)

Polymer

Monomer Class β-Glucose Biopolymer α-Glucose Biopolymer Amino acid Biopolymer Nucleotide Biopolymer β-Glucose Processed cellulose cis-Isoprene (cis-2Natural polymer methyl-1, 3-butadiene) Elastomer trans-Isoprene Elastomer Synthetic Polymers

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CH2 = CHCl Vinyl chloride Polytetrafluoroethene PTFE CF2 = CF2 (Teflon) Tetrafluoroethene Polymonochlorotrifluoroethene Cl F—C = CF2

(vii)

Polymethyl methacrylate (PMMA) (Perspex, Lucite or Acrylite)

Monochlorotrifluoroethene CH3—C—COOCH3 CH2

In combs, plastic handles, toys Pipes, raincoats, vinyl floorings Non-stick kitchenwares, electrical insulator

Addition and Chain growth

Non-stick kitchenwares

Addition and Chain growth

Substitute of glass and decorative material

Addition and Chain growth Addition and Chain growth

Lacquers, films, house piping Floor coverings and fibres

(ix)

Polyvinyl acetate

Methyl methacrylate CH2 = CH—COOC2H5 Ethyl-2-propenoate CH2 = CH—O—COCH3

(x)

Vinylite

Vinyl chloride and vinyl acetate

Addition and Chain growth

Vinyl floorings

(xi)

Polyacrylonitrile or Acrylic (orlon)

CH2 = CH—C ≡ N Vinyl cyanide (Acrylonitrile)

Addition and Chain growth

It closely resembles wool

(xii)

Buna-S

1, 3-Butadiene and styrene

Addition and Chain growth

Automobile tyres

(xiii) Buna-N

1, 3-Butadiene and acrylonitrile

Addition and Chain growth

Used for storing oil and solvents

(xiv) Neoprene

2-Chloro-1, 3-butadiene (Chloroprene)

Addition and Chain growth

Insulation, conveyor belt

(xv)

1, 2-Dichloroethene and sodium Condensation Rocket propellent polysulphide polymer

(viii) Polyethyl acrylate

Thiokol

[ 465

POLYMERS

S. No.

Polymer

Monomer

(xvi) Poly-β-hydroxybutyrate-co-βhydroxyvalerate (PHBV)

Class

OH CH3—CH—CH2—COOH OH

Uses

Condensation As packaging, orthopaedic and in controlled drug Polymer release.

CH3—CH2—CH—CH2—COOH

(xvii) Nylon-2-Nylon-6

S. Polymer No. 1. Polyesters (Terylene) Dacron 2. Glyptal (Alkyd resin) 3.

Nylon-6

4.

Nylon-6, 6

5.

Bakelite

6.

Melamine formaldehyde resin Urea formaldehyde resin

7.

Glycine (H2NCH2COOH) and aminocaproic acid H2N(CH2)5COOH

Condensation Biodegradable polymer polymer

Monomer

Class

Terephthalic acid and ethylene glycol Phthalic acid and ethylene glycol Caprolactam (cyclic amide)

Condensation and step growth Condensation and step growth Condensation and step growth Condensation and step growth Condensation and step growth Condensation and step growth Condensation and step growth

Adipic acid and hexamethylenediamine Phenol and formaldehyde Melamine and HCHO Urea and HCHO

Uses Ropes, safety belts, tyrecords. Binding material, paints and lacquers Fibre, plastic, tyre-cords and ropes Stockings, shirts, ropes Electric switches and switch-boards Crockery Crockery and laminated sheets

Know the Terms  Plasticizers : These are the substances which are added in the formation of polymers in order to alter their physical properties.  PMMA : It represents polymethylmethacrylate polymer.  PAN : It represents polyacrylonitrile polymer.  PTFE : It represents polytetrafluoroethylene polymer.  PCTFE : It represents polymonochlorotrifluoroethene polymer.

Q. Write the structure of the monomers used for getting the following polymers :

STEP-2:

(i) Dacron (ii) Melamine - formaldehyde polymer (iii) Buna-N Solutions: STEP-1:

STEP-3: CH2=CH—CH=CH2 + CH2=CHCN

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Objective Type Questions [A]  MULTIPLE CHOICE QUESTIONS : Q.1. Polymer are regarded as : (a) micromolecules (b) macromolecules (c) sub-micromolecules (d)  macrocyclic compounds R Ans. Correct option : (b) Explanation : Polymers possess high molecular weight (104 – 106). So, these are regarded as macro-molecules. Q.2. Which of the following is not a condensation polymer? (a)  Dacron (b)  Nylon-6, 6 (c)  Glyptal (d)  PTFE U Ans: Correct option: (d) Explanation:  PTFE is not a condensation polymer. It is an addition polymer obtained by the addition polymerisation of tetrafluoroethylene.    Q.3  Caprolactum is the starting material for : (a)  Nylon-6 (b)  Nylon-6,6 (c)  Nylon-6,10 (d)  Nylon-11 R Ans: Correct option: (a) Explanation:  Caprolactum is the starting material for Nylon-6.



Q.4  The S in Buna-S refers to : (a)  Sodium (b)  Sulphur (c)  Styrene (d)  Just a trade name R Ans: Correct option: (c) Explanation: Buna-S is a polymer of 1,3-butadiene and styrene. So, S refers to styrene. Q.5.  Bakelite is : (a)  addition polymer (b)  elastomer (c)  thermoplastic (d)  thermosetting R Ans: Correct option: (d) Explanation: Bakelite is a cross-linked polymer of phenol and formaldehyde. It does not regain original properties after heating. So, it is a thermosetting polymer.

(1 mark each) Q.6.  Which of the following is a natural fibre? (a)  Starch (b)  Cellulose (c)  Rubber (d)  Nylon-6 R Ans: Correct option: (b) Explanation: Cellulose is a natural fibre. Q.7.  The monomer unit of natural rubber is : (a)  butadiene (b)  chloroprene (c)  isoprene (d)  ethylene R Ans: Correct option: (c) Explanation: The monomer unit of natural rubber is isoprene.

[B]  ASSERTIONS AND REASONS In the following question, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statements but reason is wrong statement. (d) Assertion is wrong statements but reason is correct statement. Q.1.  Assertion : Bakelite is thermoplastic polymer. Reason : Bakelite possess high intermolecular forces of attraction due to cross-linking. Ans: correct option: (d) Explanation : Bakelite is thermosetting polymer. Q.2.  Assertion : PVC is a homopolymer. Reason : PVC has repeating monomeric units. Ans: Correct option: (a) Explanation : PVC is a homopolymer because it

contains same repeating units. i.e:



466 ]

Q.3. Assertion: Nylon-6,6 is an addition polymer. Reason: It is formed by polymerisation of adipic acid and hexamethylene diamine. Ans: Correct option: (d) Explanation: Nylon-6,6 is a condensation polymer as it is formed by condensation of adipic acid and hexamethylene diamine with the elimination of water molecules.

Q. 4. Assertion : Network polymers are thermo­setting.

Reason : Network polymers have high molecular mass.

Ans. Correct option : (a)

[ 467

POLYMERS

Q. 6. Assertion : Polytetrafluoroethene is used in making non-stick cookwares. Reason : Fluorine has highest electro­nega­tivity. Ans. Correct option: (b) Explanation: Teflon is used in making non-sticky cookwares, as it is chemically inert and thermally stable. Q. 7. Assertion : Buna-N is used for making oil seals, tank lining etc. Reason : Buna-N is reactive to petrol, lubricating oil and organic solvents. Ans. Correct option: (c) Explanation: Buna-N is used for making oil seals, tank lining etc. as it is resistant to petrol, lubricating oil and organic solvents.

Q.1. Name of the polymer which is used for making electric switches and combs.  Ans: Bakelite.

R (CBSE, Delhi Set-1, 2020)

Q.2. Name of the polymer which is used for making non-stick utensils.  Ans: Teflon. Q.3. Is

R (CBSE, Delhi Set 2, 2020)

—

[ CH2 – CH = CH – CH2 – CH2 – CH — ]n — CN

homopolymer or copolymer? U (CBSE, Delhi Set-3, 2020)  Ans: Copolymer (two different monomers are used) CH 2 = CH − CH = CH 2 + CH = CH 2



|

Buna-1-3-diene



Acrylonitrile

Q.4. What structural changes are produced during vulcanisation for natural rubber? R Ans: Sulphur cross-links are introduced between the polymeric chains. Q.5. Write the monomer units of Bakelite. Ans: Phenol and formaldehyde.

 

—



U [CBSE Delhi Set-1, 2020]

Ans. (i) Monomer–Ethylene glycol and phthalic acid HOOC COOH



n(CH2OHCH2OH) +

   

nCH2 = CHCN

[1]

Commonly Made Error

O O  U [CBSE Delhi Set-2, 2020] Ans. (i)  Monomer – o-hydroxymethylphenol OH CH2OH

[1]  (ii) Monomer – Hexamethylenediamine and adipic acid NH2(CH2)6NH2 and COOH(CH2)4COOH [1]

 Understand the synthesis of polymer form monomer units.

 

 Some students cannot write connect monomer units of a polymer.

Answering Tip

—n

[ NH – (CH2)6 –NH –C –(CH2)4 –C — ]n (ii)  —

 [1]

(ii) Monomer–vinyl cyanide or Acrylonitrile

[

—

—

—

]

Q. 2. Identify the monomers in the following polymers : OH OH CH2 CH2 (i)  — —

[

]

CO –n

CN

]

(2 marks each)



Q. 3. Identify the monomers in the following polymers : (i)

[ OCH2CH2 – O – C – —

—

 

O

– CH2 – CH–  (ii) n

R

Q.6. Give an example of a polymer which is polyamide.  R Ans: Nylon-6,6.

Q. 1.  Identify the monomers in the following polymers:

]



CN

Short Answer Type Questions-I

– O–CH2–CH2–O–C (i)

a

—

Ans. Correct option: (b) Explanation: Olefins like ethene undergo addition polymerisation by opening of double bond to form two new single bonds to give polymer like polyethene.

[C]  VERY SHORT ANSWER TYPE QUESTIONS :

O

– C— ]

—





Explanation : Network polymers are thermosetting and are cross linked and heavily branched molecules. Q. 5. Assertion : Olefinic monomers undergo addition polymerisation. Reason : Polymerisation of vinyl chloride is initiated by peroxides/ persulphates.

O

n

468 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

(ii)

[

— HN—

N

NH–CH2 — N

N NH

(ii) Melamine and formaldehyde. N NH2 H2N

[

—



CH2OHCH2OH and O O OH – C

C – OH

NH2

Q. 4. What are elastomers? Give one example. R Ans. Elastomers are the polymers which possess elastic character. The polymer chains in such type of polymers are held together by weakest intermolecular forces. These forces allow the polymers to be stretched under stress but they regain their former shape when the stress is relived. e.g. Rubber (synthetic or natural). [2]

Short Answer Type Questions-II Q. 1. Write the name and structures of monomer(s) in the following polymers : (i) Nylon-6 (ii) PVC (iii)  Neoprene R [CBSE Outside Delhi Set-1, 2020] Ans. (i)  Nylon - 6 Monomer - Caprolactum O

(3 marks each) O



[1] (iii) PHBV : Monomer - 3 hydroxybutanoic acid and 3-hydroxypentanoic acid. Terephthalic acid



OH

OH

| | CH3  CH  CH 2COOH  CH3  CH 2  CH  CH 2COOH

|

Cl

[1] Q.3.  Write the structure of monomers used for getting the following polymers : (i)  Nylon-6,6 (ii)  Glyptal (iii)  Buna – S

[1] Q. 2. Write the name of monomers used for getting the following polymers : (i) Nylon-6, 6 (ii) Terylene (iii) PHBV  [CBSE Outside Delhi Set-2, 2020] Ans. (i) Nylon 6, 6 : Monomer - adipic acid (hexanedioic acid) and hexamethylenediamine. H 2 N(CH 2 )6 NH 2 and Hexamethylenediamine

(ii) HO — CH2 — CH2 — OH, HOOC— CH = CH2

(iii) CH2 = CH – CH = CH2,

COOH(CH 2 )4 COOH Adipic acid

  [1] (ii) Terylene : Monomers - Ethylene glycol and terephthalic acid. OR



Ans. (i) HOOC(CH2)4COOH, H2N (CH2)6 NH2 COOH





3-hydroxypentanoic acid

—

(iii) Neoprene : Chloroprene - CH 2 = C − CH = CH 2

C – OH

Ethylene glycol

—

[1] (ii) PVC : Vinylchloride [1] CH2= CHCl

O

HOCH2CH2OH and HO – C

3-hydroxybutanoic acid

NH

[1]

and HCHO

N

N

n

 U [CBSE Delhi Set-3, 2020] Ans. (i)  Ethylene glycol and terephthalic acid

 

[CBSE Marking Scheme, 2019] 1 × 3

[ 469

POLYMERS

(Topper Answer's 2019) [3] CH3

—

CN

. 4. (i) Is [ CH2 CH n a homopolymer or copolymer? Q Give reason. (ii) Write the monomers of the following polymer:

[

[CBSE Marking Scheme, 2019]

NH – CH2

N NH

H

—

(iii) What is the role of Sulphur in vulcanization of rubber? R + U [CBSE, Delhi Set-1, 2019]

Detailed Answers : (a) Caprolactam N— C=O H2 C —

H2N —

—

N

H2 C

, NH2

H 2C

(b) Ethylene glycol

+ HCHO

N

CH2

CH2

HO — CH2 — CH2 — OH



—



N

—



(ii)

—



—

Ans. (i) Homopolymers, single repeating unit [½ + ½]

—

N

(c) CH2 = CH—CH = CH2, CH2 = CH [1 + 1 + 1]

—

N

NH



NH2

(Or names of monomers) [1] (iii) Sulphur forms cross links at the reactive sites of double bonds and thus the rubber gets stiffened / To improve the physical properties of rubber by forming cross links. [1] [CBSE Marking Scheme, 2019]

—

—

—

Answering Tip



—

Commonly Made Error  Some student get confused to identify homo, polymer and copolymer form a given structure.

HOH2C—CHacid HOOC— —COOH Terephthalic 2OH, (c) Buta–1, 3– diene CH2 = CH — CH = CH2 CH2 = CH — C ≡ N Acrylonitrile  [1 + 1 + 1] Q. 6. (a) Is –[ CH2 – CH(C6H5) –] n a homopolymer or copolymer ? Give reason. (b)  Write the monomers of the following polymer : –[ O—CH—CH2—C—O—CH—CH2—C –] n CH3

O

C2H5

O

(c) Write the role of benzoyl peroxide in polymerisation of ethene.  [CBSE, Outside Delhi Set-1, 2019]

 Understand the difference between homopolymer and copolymers.

(a)  Homopolymer; As the same monomer is repeated.  [½ + ½]

Q. 5. Write the structures of monomers used for getting the following polymers : (a) Nylon-6 (b) Terylene (c) Buna-N

(b) 

—

—

—

CH2

—

—



—

— N— C=O H2 C —

H 2C

CH2—CH2—CH—CH2—COOH

HOOC—

[1]

OH

(c)  It acts as an initiator.



CH2

(b) HOH2C—CH2OH,



[1]

[CBSE Marking Scheme, 2019] Detailed Answer : (a) It is a homopolymer because single type of monomer unit i.e., C6H5 — CH = CH2 undergoes polymerization. (b)  3–hydroxy butanoic acid OH O

H

H2 C

—

Ans. (a)

OH

CH3—CH—CH2—COOH,

—COOH

OH 3–hydroxy pentanoic acid

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

OH

Detailed Answer : (c) Strong intermolecular forces like hydrogen bonding and linear structure lead to close packing of polymer

O

OH (c) Benzoyl peroxide is the reagent which generates free radical formed on decomposition and for polymerization of ethene, the source of free radical is required to initiate the chain reaction. These free radicals are produced by benzoyl chloride. [1 + 1 + 1]

chains that imparts crystalline character. [1] Q. 9. Write the structures of monomers used for getting the following polymers : (a) Novolac (b) Neoprene (c) Buna-S

Q. 7. Write the structures of monomers used for getting the following polymers : (a) Nylon-6,6 (b) Bakelite (c) Buna -S Ans. (a) HOOC(CH2)4COOH,

—

OH

Ans. (a)

H2N(CH2)6NH2

—

Cl

—

OH

(b) H2C = C—CH = CH2 CH = CH2

, CH2O

—

(b)

, HCHO

—

CH = CH2

(c) CH2 = CH—CH = CH2

(c) CH2 = CH – CH = CH2, [CBSE Marking Scheme, 2019] 1 × 3 = [3] Q. 8. (a) Write one example each of : (i) Thermoplastic polymer (ii) Elastomers (b) Arrange the following polymers in the increasing order of their intermolecular forces : Polythene, Nylon-6,6, Buna-S (c) Which factor provides crystalline nature to a polymer like Nylon?  R [CBSE, Outside Delhi Set 2, 2019] Ans. (a) (i)  Polythene[½ + ½] (ii) Buna-S

(Or any other correct example)

(b)  Buna-S < polythene < Nylon-6,6[1] (c)  Hydrogen bonding[1] 

[CBSE Marking Scheme, 2019]

,



[CBSE Marking Scheme, 2019] 1 × 3 = [3] Q. 10. (a) Write one example each of : (i) Cross- linked polymer (ii) Natural polymer (b) Arrange the following in the increasing order of their intermolecular forces: Terylene, Buna-N, Polystyrene



(c) Define biodegradable polymers with an example.  R [CBSE, Outside Delhi Set-3, 2019] Ans. (a) (i) Bakelite (ii) Protein

(Or any other suitable example) [½ + ½] (b) Buna-N < Polystyrene < terylene [1] (c) Polymers which are easily degraded by microorganisms Example- PHBV (Or any other suitable example) [CBSE Marking Scheme, 2019] [½ + ½]

Q. 11. Explain the term ‘copolymerization’ and give two examples of copolymerization.





R [CBSE Comptt. OD 2015]

Ans. Copolymerisation is a polymerisation reaction in which a mixture of more than one monomeric species is allowed to polymerise and form a copolymer.

nCH2 = CH — CH = CH2 + 1, 3-Butadiene



n Styrene

—

—

CH = CH2

— CH2 — CH = CH — CH2 — CH — CH2 — n Butadiene - styrene copolymer [Buna-S]

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POLYMERS

CN

CN

—

—

Copolymerisation

— CH2 — CH = CH — CH2 — CH2 — CH — n Buna-N 3

n CH2 = CH — CH = CH2 + nCH2 = CH Acrylonitrile



1, 3-Butadiene







(or any other correct example) [CBSE Marking Scheme 2015]



Commonly Made Error  Students get confused between a homopolymer and a copolymer.

Answering Tip  Understand each with the help of examples for better clarity.  W  hile giving examples, draw the structure of the compounds.

Long Answer Type Questions (i)  Nylon-6, 6 (iii) Neoprene

Polymer

Name of monomer

(ii)   PHBV

Ans.  (a) (i)    Protein        (ii) Elastomer

 

Q. 1. (a)(i) Which of the following is natural polymer. Buna-S, Protein, PVC ?

(ii) Based on molecular forces what type of polymer is neoprene ? (b) Write the names and structures of the monomers of the following polymers : (b) Names and structures of monomers

(5 marks each)

[1] [1]

Structure of monomer

(i)

Nylon 6, 6

Hexamethylene diamine and Adipic acid

NH2 – (CH2)6 – NH2 COOH – (CH2)4 – COOH

(ii)

PHBV

3-hydroxybutanoic acid 3-hydroxypentanoic acid

OH | CH 3  CH  CH 2  COOH

1 1

OH | CH 3 —CH 2 —CH—CH 2 —COOH (iii)

Neoprene

2-chloro – 1, 3 - butadiene

Q. 2. (a)(i)    Write the name of the biodegradable polymer used in orthopaedic devices. (ii)  What type of reaction occurs in the formation of Nylon 6,6 polymer? (b)(i)    What is the role of Sulphur in the vulcanization of rubber ? (ii)  Identify the monomers in the following polymer :

CH 2 =C—CH=CH 2 | Cl

1

(iii)  Arrange the following polymers in the increasing order of their intermolecular forces : Terylene, Polythene, Neoprene.

R+U

ns:  (a) (i)    Poly(3-hydroxybutyrate-co-3-hydroxyvalerate) (PHBV)[1] A   (ii)  Consideration polymerisation[1] (b) (i) The physical properties of rubber can be altered to suit the requirements by adding sulphur in vulcanization of rubber. The process of introducing —S—S— polysulphide crosslinks between the adjacent chains tend to limit the motion of chains relative to each other.1

472 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

1



(ii) 

(iii)  Neoprene < Polythene < Terylene1

[½] Uses: It is used in making wash and wear fabrics.  [½]

Use: It is used in making carry bags, toys and electrical insulations etc. [½] (iv) PVC (polyvinyl chloride)

Structure:

[½]



Use: It is used in making rain coats, electrical insulations etc. [½] (v) Neoprene Structure: [½]



Visual Case-Based Questions Q. 1. Read the passage given below and answer the following questions : Biopolymers are polymers that are generated from renewable natural sources, are often biodegradable and nontoxic. They can be produced by biological systems (i.e. microorganisms, plants and animals), or chemically synthesized from biological materials (e.g., sugars, starch, natural fats or oils, etc.). Two strategies are applied in converting these raw materials into biodegradable polymers: extraction of the native polymer from a plant or animal tissue, and a chemical or biotechnological route of monomer polymerization. Biodegradable biopolymers (BDP) are an alternative to petroleumbased polymers (traditional plastics). Some BDP degrade in only a few weeks, while the degradation of others takes several months. In principle the properties relevant for application as well as biodegradability are determined by the molecular structure. According to the American Society for Testing and Materials, biopolymers are degradable polymers in which degradation results from the action of naturally occurring microorganisms such as bacteria, fungi and algae. Polylactic acid (PLA) is an example of biopolymer. It is a thermoplastic polyester. Generally, there are two major routes to produce polylactic acid from the lactic acid (CH3CH(OH)-COOH) monomer. The first route involves condensation-water removal by the use of solvent under high vacuum and temperature. This approach produces a low to intermediate molar mass polymer. An alternative method is to remove









[½]





Ans. (i)  Teflon Structure: [½] Uses: It is used in coating utensils to make them non-sticking. [½] (ii) Dacron Structure:

(iii) Polyethene Structure:



Q. 3. Give the structure and one use of the following polymers : (i)  Teflon (ii)  Dacron (iii) Polyethene (iv) PVC (v)  Neoprene R

Use: It is used in making hoses, conveyor belts etc. [½]

(4 marks each)

water under milder conditions, without solvent, to produce a cyclic intermediate dimer, referred to as lactide. This intermediate is readily purified by vaccum distillation. Ring opening polymerization of the dimer is accomplished under heat, again without the need for solvent. By controlling the purity of the dimer it is possible to produce a wide range of molar masses. PLA is a good material for production of clothing, carpet tiles, interior and outdoor furnishing, geotextiles, bags, filtration systems, etc. The primary biodegradability of PLA was tested using hydrolysis tests at various composting temperatures and pH. It was demonstrated that composting is a useful method for PLA biodegradation. The degradation rate is very slow in ambient temperatures. A 2017 study found that at 25oC in sea water, PLA showed no degradation over a year. As a result, it is poorly degraded in landfills and household composts, but is effectively digested in hotter industrial composts. [CBSE QB 2021] In the following questions, a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices on the basis of the above passage. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement.

[ 473

POLYMERS

(d) Assertion is wrong statement but reason is correct statement. (i) Assertion: Biodegradable polymers degrade in few weeks. Reason: Microorganisms bring about degradation of biopolymers. (ii) Assertion: Lactic acid on polymerisation forms CH 3 O O Reason: PLA is used in producing geotextiles. (iii) Assertion: Lactic acid undergoes condensation polymerisation Reason: Lactic acid is a bifunctional monomeric unit. (iv) Assertion: The degradation of PLA is very slow in ambient temperature. Reason: PLA is a thermoplastic. (v) Assertion: PLA is poorly degraded in landfills. Reason: The degradation rate of PLA is very slow in ambient temperatures. Ans. (i) Correct option is (d) (ii) Correct option is (b) (iii) Correct option is (a) (iv) Correct option is (b)

Q. 2. Read the passage given below and answer the following questions :(1 × 4 = 4)





Thermoplastics are the polymers which can be easily moulded into desired shapes by heating and subsequent cooling to room temperature. The intermolecular forces in thermoplastic polymers are intermediates to these of elastomers and fibers. There is no cross-linking between the chains. Infact, thermoplastic polymers soften on heating and become fluids but on cooling they become hard. They are capable of undergoing such reversible changes on heating and cooling repeatedly. e.g., polyethene Thermosetting polymers are the polymers which become hard and infusible on heating. They are normally made from semi-fluid substances with low molecular masses, by heating in a mould. Heating results in excessive cross-linking between the chains forming three dimensional network of bond. e.g., bakelite. The following questions are multiple choice questions. Choose the most appropriate answer. (i) Which of the following polymers become hard and infusible on heating ? (a) Thermoplastic (c)  Linear polymers polymers (d) Isotactic (b) Thermosetting polymers polymers

(ii) A thermosetting polymer is : (a) Nylon-6,6 (c) Bakelite (b) Polyethene (d)  All of these

(iii) Mark incorrect statement for thermoplastic polymers. (a)  They can be easily moulded into desired shapes by heating. (b)  There is no cross-linking between the chains. (c)  They do not undergo recycling. (d)  They soften on heating and become fluids. (iv) Polyethene is (a) Branched-linear polymer (b) Thermoplastic polymer

(c) Thermosetting polymer (d) Cross-linked polymer

Ans. (i) correct option : (b) Explanation: Thermosetting polymers become hard and infusible on heating because of cross-linking.



[1]

(ii) Correct option: (c) Explanation: Bakelite is a thermosetting polymer which is formed by the condensation polymerisation of phenol and formaldehyde and it possesses cross linking. [1] (iii)  Correct option: (c) Explanation: Thermoplastic polymers can undergo recycling because no cross-linking is present between the chains. [1] (iv)  Correct option: (b) Explanation: Polythene is thermoplastic polymers as it can be easily moulded into desired shape by [1] heating. Q.3. Shyam went to a grocery shop to purchase some food items. The shopkeeper packed all the items in polythene bags and gave them to Shyam. But Shyam refused to accept the polythene bags and asked the shopkeeper to pack the items in paper bags. He informed the shopkeeper about the heavy penalty imposed by the government for using polythene bags. The shopkeeper promised that he would use paper bags in future in place of polythene bags.

Answer the following : (a) Write the values (at least two) shown by Shyam. (b) Write one structural difference between lowdensity polythene and high-density polythene. (c) Why did Shyam refuse to accept the items in polythene bags ? (d) What is a biodegradable polymer ? Give an example. [CBSE, Delhi/outside Delhi, 2018] Ans. (a)  Concerned about environment, caring, socially alert, law abifing citizen (or any other 2 values). ½, ½ (b) Low density polythene is highly branched while high density polythene is linear. 1 (c) As it is non-biodegradable. (d) Which can be degraded by microorganisms, e.g. PHBV (or any other correct example).  ½+½ ll

474 ]

Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Self Assessment Test - 14 Time : 1 Hour

Max. Marks : 25

Q.1. Read the passage given below and answer the following questions : Chain growth polymerisation is a process of successive addition of monomer units to the growing chain by a chain mechanism. The monomer unit gets converted to some active intermediate species by a small amount of initiator such as organic peroxide or an acid or a base. Depending upon the conditions, the intermediate species may be free radical or an ion, and it reacts with other monomer unit to form still bigger intermediate species. The polymers formed by chain growth polymerisation are called chain growth polymers. e.g polyethene, teflon, PVC etc. The following questions are multiple choice questions. Choose the most appropriate answer: (i)  An initiator for chain growth polymerisation is : (a)  Benzoyl peroxide (b)  Lewis acid (c)  Lewis base (d)  All of these (ii)  In chain growth polymerisation, elemental composition of polymer : (a) Changes (b)  Remains same (c)  Can not predicted (d) Increases (iii)  Addition polymer means : (a)  A step growth polymer (b)  A chain growth polymer (c)  A thermosetting polymer (d)  A copolymer (iv) Which of the following is not referred to as chain growth polymer? (a) Nylon-6,6 (b) PVC (c) Teflon (d) Polyisoprene (i)  Correct option: (d) Following questions (No. 2–5) are Multiple Choice Questions carrying 1 mark each. Q. 2. Which of the following polymers of glucose is stored by animals? (a) Cellulose (b) Amylose (c) Amylopectin (d) Glycogen R Q. 3. Which of the following is not a semisynthetic polymer? (a) cis-polyisoprene (b) Cellulose nitrate (c) Cellulose acetate (d) Vulcanised rubber R Q. 4. Which of the following statements is not true about low density polythene? (a) Tough

(b) Hard (c) Poor conductor of electricity (d) Highly branched structure Q. 5. The incorrect statement about LDP is (a) It is obtained through the free radical addition of ethene. (b) It consists of linear molecules. (c) It is obtained by the H-atom abstraction. (d) Peroxide is used as an initiator.  Q. 6. Assertion : For making rubber synthetically, isoprene molecules are polymerised. Reason : Neoprene (a polymer of chloroprene) is a synthetic rubber. Q. 7. Assertion : Rayon is a semi synthetic polymer and is considered as a better choice than cotton fabric. Reason : Mechanical and aesthetic properties of cellulose can be improved by acetylation. The following questions (No. 8-9) are Short Answer type-I and carry 2 marks each Q.8. What are polymers? Explain the term poly dispersity index (PDI). Q.9. Write chemical equation for the synthesis of (i) Teflon (ii) Nylon-6,6 Q.No. 10-11 are Short Answer Type-II carrying 3 marks each. Q.10.  Name the type of reaction involved in the formation of the following polymers from their respective monomers. (i) PVC (ii) Nylon-6 (iii) PHBV Q.11. Give the starting material of each of the following : (i) PMMA (ii) PAN (iii) PVC Q.12.  Explain the following terms with example : (i) Polyesters (ii) Polyamides (iii)  Natural polymers (iv)  Synthetic polymers (v)  Biodegradable polymers OR What are condensation polymers? Give two examples? Give the synthesis and uses of bakelite.

 

Finished Solving the Paper ? Time to evaluate yourself !

OR SCAN THE CODE

SCAN

For elaborate Solutions

ll

CHEMISTRY IN EVERYDAY LIFE

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16 UNIT

CHEMISTRY IN EVERYDAY LIFE

Syllabus ¾¾ Chemicals in medicines : analgesics, tranquilizers, antiseptics, disinfectants, antimicrobials, antifertility drugs, antibiotics, antacids, antihistamines. ¾¾ Chemicals in food : preservatives, artificial sweetening agents, elementary idea of antioxidants. ¾¾ Cleansing agents : soaps and detergents, cleansing action.

Trend Analysis List of Concept names Analgesic & Tranquilizers Antiseptics & Disinfectants Antibiotics & Antacid Food Additives/Food preservatives Soaps & Detergents

2018 D/OD 2Q (2 marks) 1Q (1 mark) -

2019 D OD 2Q (1 mark) 1Q 2Q (1 mark) (1 mark) 3Q 1Q (1 mark) (1 mark) 2Q (1 mark) 4Q 3Q (1 mark) (1 mark)

2020 D 1Q (2 marks) 1Q (1 mark) 3Q (1 mark) -

OD 3Q (1 mark) 2Q (1 mark) 2Q (1 mark) 1Q (1 mark) 1Q (1 mark)

Revision Notes  Drugs : Drugs are the chemical substances of low molecular masses, which interact with macromolecular targets and produce a biological response.  Medicines : Drugs which produce a therapeutic and useful response.  Chemotherapy : Use of chemicals to elicit a therapeutic response in biological systems is chemotherapy.



 Classification of Drugs : (i) On the basis of pharmacological effect : Most useful for doctors as it provides the whole range of drugs available for the treatment of a particular type of problem. e.g., analgesics have pain killing effect, antiseptics kill or arrest the growth of microorganisms. (ii) On the basis of drug action : It is based on the action of a drug on a particular biochemical process. e.g., all anti-histamines inhibit the action of the compound histamines, which causes inflammation of the body. (iii) On the basis of chemical structure : Drugs which have common structural features and often have similar pharmacological activity. e.g., sulphonamides have common structural feature.

(iv) On the basis of molecular targets : Drugs generally interact with biological macromolecules such as carbohydrates, proteins, lipids and nucleic acids called target molecules. This classification is based upon the type of the molecular target with which the drug interacts.







CHEMISTRY IN EVERYDAY LIFE

Enzymes are the proteins which perform the role of biological catalysts in the body. Carrier proteins carry polar molecules across the cell membrane.

 Catalytic action of enzymes : (i) The prime function of an enzyme is to hold the substrate for a chemical reaction. Active sites of enzymes hold the substrate molecule in a suitable position, such that it can be attacked by the reagents effectively. Substrates bind to the active site of the enzymes through many kinds of interactions like ionic bonding, hydrogen bonding, van der Waals or dipole-dipole interactions.

[ 477 Scan to know more about this topic

Classifaction of Drugs

(ii) The second function of an enzyme is to provide functional groups that will attack the substrate and carry out chemical reaction.

Fig. 1 : (a) Active site of an enzyme, (b) Substrate, (c) Enzyme holding the substrate  Drug-enzyme interactions : Drugs can block the binding site of the enzyme and prevent the binding of substrate, or can inhibit the catalytic activity of the enzyme. Such drugs are called enzyme inhibitors. Drugs inhibit the attachment of substrate on active site of enzymes in two ways : (i) Drugs compete with natural substrate for their attachment on the active sites of enzymes. Such drugs are called competitive inhibitors.

Drug and substrate competing Drug blocks the active for active site of enzyme site of enzyme Fig. 2 : Drug and substrate competing for active site (ii) Instead of joining to the enzyme’s active site, some drugs bind to a different site of enzyme which is called allosteric site. This can change the shape of the active site in such a way that the substrate can’t recognize it. If the bond formed between an enzyme and an inhibitor is a strong covalent bond and cannot be broken easily, then the enzyme is blocked permanently. The body then degrades the enzyme inhibitor complex and synthesises the new enzyme.







Fig. 3 : Non-competitive inhibitor changes the active site of enzyme after binding allosteric site  Receptors as drug targets : Proteins that are crucial to body’s communication process are called as receptors. They are embedded in the cell membrane in such a way that their small part possessing active site projects out of the surface of the membrane and opens on the outside region of the cell membrane.

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Rate of reaction

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII





Message between two neurons and that between neurons to muscles is communicated through chemical receptors. To accommodate a messenger, shape of the receptor site changes. This brings about the transfer of message into the cell. Thus, chemical messenger gives message to the cell without entering the cell. Antagonists : Drugs that bind to the receptor site and inhibit its natural function are called antagonists. These are  useful when blocking of message is needed. Agonists : Drugs that mimic the natural messenger by switching on the receptor are called agonists. These are  useful when there is lack of natural chemical messenger. Types of drugs on the basis of therapeutic action :        (i) Analgesics : These are the medicines which give relief from pain. They are of two types : (a) Non-narcotic (non-addictive) analgesics : Aspirin (2-acetoxy benzoic acid), paracetamol, phenylbutazone or butazolidine etc. are the common examples of this group. Aspirin is the most common analgesic with antipyretic properties. It also has anti blood clotting action. It also reduces body temperature in fever. (b) Narcotic analgesics : These are the drugs which produce sleep and unconsciousness e.g., opium, alkaloids













like morphine, codeine, heroine (morphine diacetate), etc. These are, however, addictive drugs, hence used in severe pain only.       (ii) Antiseptics : These are the chemicals which prevent the growth of microorganisms or kill them but are not harmful to human beings. These are applied externally to the living tissues such as wounds, cuts and diseased skin surfaces. Dettol (chloroxylenol + a-terpineol), bithional, furacin, dilute Scan to know solution of boric acid are common example of antiseptics. more about      (iii) Disinfectants : These are chemicals which kill microorganisms or prevent their growth this topic but are not safe for human beings. These are applied to inanimate objects such as floors, drainage systems. Some substances can act as an antiseptic as well as disinfectant by varying the concentration. For example, 0.2% solution of phenol is an antiseptic while its one percent solution is disinfectant. Antiseptics and   (iv) Tranquilizers : It is a group of chemical substances which is used in the treatment of Disinfectants stress and severe mental stress. These are essential components of sleeping pills and psychotherapeutic drugs. These are of two types : (a) Barbiturates : These are derivatives of barbituric acid. These are sleep inducing and hence also called hypnotics. e.g., veronal, amytal, nembutal, luminal and seconal.  (b) Non-hypnotic tranquilizers : Chlordiazepoxide and meprobamate are relatively mild tranquilizers

which are used for relieving tension. Equanil is another non hypnotic tranquilizer which is used for controlling depression and hypertension. Valium, serotonin, reserpine, etc. are some other tranquilizers. (v) Antimicrobials : These are drugs which are used to cure diseases caused by a variety of microbes such as bacteria, fungi, virus, etc. Antibiotics, antiseptics, disinfectants, etc. are all antimicrobials. (vi) Antibiotics : These are the chemical substances (prepared wholly or partially by chemical synthesis) which in low concentration, either kill or inhibit the growth of microorganisms. Penicillin is a narrow spectrum antibiotic whereas ampicillin and amoxicillin are wide spectrum antibiotics which exert antimicrobial activity on more than one type of microorganisms. (vii) Sulpha drugs : These are derivatives of sulphanilamide. These have antibacterial powers and are used as medicines for various diseases. These are also used as antibiotics. Sulpha drugs are used against diseases like pneumonia, tuberculosis, diphtheria, etc. e.g. sulphadiazine, sulphathiazole. (viii) Antifertility drugs : These are the chemical substances used to control the pregnancy in woman. e.g., norethindrone and ethynylestradiol, etc. (ix) Antihistamines : These drugs are also called as anti allergy drugs and are used to treat allergy. e.g., skin rashes, conjunctivitis, inflammation of conjunctiva of eye and rhinitis (inflammation of nasal mucosa). e.g., diphenhydramine, chlorpheniramine. Antioxidants : These are the other important and necessary food additives. These compounds retard the action of  oxygen on the food and thereby help in its preservation. They also reduce the rate of involvement of free radicals in the aging process. e.g., Butylated hydroxy toluene (BHT) and Butylated hydroxy anisole (BHA) are used as antioxidants.  Antacid : An antacid is a substance that removes the excess of acid and raises the pH of stomach Scan to know more about to appropriate level. The most commonly used antacids are magnesium hydroxide, magnesium this topic carbonate and sodium bicarbonate etc.  Chemicals in food : Chemicals are added to food for various purposes like, for preservation, for enhancing their appeal and for adding nutritive value etc. Some uses are discussed below : (i) Artificial Sweeteners : These are the chemical compounds which are non-nutritive in Antibiotics nature and used as substituent for sugar in foods and beverages specially soft drinks. Some common artificial sweeteners are : (a) Saccharin : It is useful as a sugar substitute for diabetic persons and those who need to control their calorie intake.

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CHEMISTRY IN EVERYDAY LIFE

(b) Aspartame : It is methyl ester of dipeptide formed from aspartic acid and phenylalanine. Aspartame is used only in cold foods and soft drinks as it is unstable at cooking temperature. (c) Alitame : It is a high potency sweetener. The control of sweetness of food is difficult while using alitame. (d) Sucralose : It is trichloro derivative of sucrose. It is stable at cooking temperature. (ii) Food preservatives : Food preservatives are the substances which are capable of inhibiting or arresting the process of fermentation, acidification of the food. e.g., sodium benzoate, sodium metabisulphite.  Soaps : Soaps are sodium or potassium salts of long chain fatty acids and are prepared by a process called saponification, in which fat reacts with alkali.  Synthetic detergent : These are soapless soap and are of three types : (a) Anionic detergents : These are sodium salts of sulphonated long chain alcohols or hydrocarbons. e.g., sodium dodecylbenzene sulphonate. These are used in toothpaste and household works. (b) Cationic detergents : These are quaternary ammonium salts of amines with acetates, chlorides or bromides e.g., cetyltrimethyl ammonium bromide and are expensive. These have germicidal property and are expensive.

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(c) Non-ionic detergents : They do not contain any ion in their constitution. e.g., liquid dishwash detergents.  Biodegradable and non-biodegradable detergents : Detergents which contain straight chain hydrocarbons are biodegradable. On the other hand, detergents which have branched chain hydrocarbons are non-biodegradable and this leads to environmental pollution.

Food Additives

 Saponification : The process of manufacturing soap by the hydrolysis of oils and fats with aqueous alkalis.

CH2OH

— —

C HO—C—R + 3NaOH O Alkali CH2O—C—R

— Hydrolysis

CHOH

—

— —

CH2O—C—R O

CH2OH Glycerol

O

+ 3R—C—ONa Soap

Know the Terms Neurologically active drugs : The drugs which are used to cure tension and anxiety. Hypnotics : These are sleep inducing medicines. Salting out of soap : This is the process by which soap is separated from glycerol. Fillers : The chemical substances which are added to laundry soaps. e.g., sodium silicate, borax, etc.  Hard soaps : Sodium salts of fatty acids.  Soft soaps : Potassium salts of fatty acids.

   

Objective Type Questions [A] MULTIPLE CHOICE QUESTIONS : Q. 1. The most useful classification of drugs for medicinal chemists is : (a) on the basis of chemical structure. (b) on the basis of drug action. (c) on the basis of molecular targets. (d) on the basis of pharmacological effect.  R Ans. Correct option : (c) Explanation : Biomolecules, such as carbohydrates, lipids, proteins and nucleic acids, are target molecules of drugs and usually interact with drugs. These drugs possess some common structural feature, may have the same mechanism of action on a specific drug target. Q. 2. The compound that causes general anti-depressant action on the central nervous system belongs to the class of :

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O

Soaps and Detergents Scan to know more about this topic

Cleansing Action of Soaps and Detergents

(1 mark each) (a) analgesics. (b) tranquilizers. (c) narcotic analgesics. (d) antihistamines.  R Ans. Correct option : (b) Explanation : Tranquilizers cause general antidepressant action on the central nervous system. Q. 3. Compound which is added to soap to impart antiseptic properties is __________. (a) sodium laurylsulphate (b) sodium dodecylbenzenesulphonate (c) rosin (d) bithional R Ans. Correct option : (d) Explanation : Bithional is added to soaps to reduce the odour produced by bacterial decomposition of organic matter on the skin due to its antiseptic properties.

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Q. 4. Equanil is ______________. (a) artificial sweetener. (b) tranquilizer. (c) anti-histamine. (d) anti-fertility drug.  R Ans. Correct option : (b) Explanation : Equanil is a tranquilizer which is used in controlling depression and hypertension. Q. 5. A narrow spectrum antibiotic is active against __________ . (a) Gram-positive or Gram-negative bacteria. (b) Gram-negative bacteria only. (c) Single organism or one disease. (d) Both Gram-positive and Gram-negative bacteria. R

Ans.

Correct option : (a) Explanation : A narrow spectrum antibiotic is active against Gram-positive or Gram-negative bacteria. Q. 6. Which of the following enhances lathering property of soap? (a) Sodium carbonate (b) Sodium rosinate (c) Sodium stearate (d) Trisodium phosphate  R Ans. Correct option : (b) Explanation : A gum called rosin is added in soaps (e.g., shaving soaps) which form sodium rosinate that enhances lathering property of soap. Q. 7. The drug used for treatment of typhoid is : (a) Novalgin (b) Quinine (c) Chloromycetin (d) Paracetamol R Ans. Correct option : (c) Explanation : Typhoid is a disease caused by bacteria, so, antibiotic i.e. chloromycetin is used for treatment of typhoid. Q. 8. Which of the following drug is used as analgesic? (a) Sulphapyridine (b) Phenacetin (c) Novalgin (d) Chloroxylenol R Ans. Correct option : (c) Explanation : Novalgin is used as analgesic which relieves pain. [B] ASSERTIONS AND REASONS: In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Both assertion and reason are correct statements and reason is correct explanation for assertion. (b) Both assertion and reason are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Q. 1. Assertion : Sodium chloride is added to precipitate soap after saponification. Reason : Hydrolysis of esters of long chain fatty acids by alkali produces soap in colloidal form.

Ans.

Correct option: (b) Explanation : Hydrolysis of esters of long chain fatty acids by alkali produces soap in colloidal form. Sodium chloride is added to this colloid to precipitate soap after saponification process. Q. 2. Assertion : Competitive inhibitors compete with natural substrate for their attachment on the active sites of enzymes. Reason : In competitive inhibition, inhibitor binds to the allosteric site of the enzyme. Correct option: (c) Ans. Explanation: Competitive inhibitors compete with natural substrate for their attachment on the active sites of enzymes.  Q. 3. Assertion : Receptor proteins show selectivity for one chemical messenger over the other. Reason : Chemical messenger binds to the receptor site and inhibits its natural function. Correct option : (c) Ans. Explanation: Receptor proteins show selectivity for one chemical messenger over the other. Chemical messenger binds to the receptor site and brings about the transfer of the message.  Q. 4. Assertion : Artificial sweeteners are added to the food to control the intake of calories. Reason : Most of the artificial sweeteners are inert and do not metabolise in the body. Correct option: (b) Ans. Explanation: Artificial sweeteners are added to the food to control the intake of calories. Most of the artificial sweeteners are inert and do not metabolise in the body.  [C] VERY SHORT ANSWER TYPE QUESTIONS : Q. 1. Name the sweetening agent used in the cooking of sweets for a diabetic patient.  R [CBSE, Delhi Set-1, 2020] Ans.  Sucrolose.  Q. 2. Name the artificial sweetener whose use is limited to cold drinks R [CBSE, Delhi Set-2, 2020] Ans. Aspartame.  Q. 3. Which of the following is an antidepressant drug? Chloramphenicol, Luminal, Bithional  R [CBSE, outside Delhi Set-1, 2020] Ans. Luminal.  Q. 4. Which one of the following is a narcotic analgesic/Penicillin, Codeine, Ranitidine  R [CBSE, outside Delhi Set-2, 2020] Ans. Codeine.  Q. 5. Name the compound which is added to soap to provide antiseptic properties.  R [CBSE, outside Delhi Set 3, 2020] Ans. Bithionol.  Q. 6.  What is the cause of a feeling of depression in human beings ? Name a drug which can be useful treating depression. AE + R Ans. Low level of noradrenaline is responsible for the feeling of depression in human beings.[½] Iproniazid is a drug, used to counteract the effect of depression.[½] (or any other correct example)

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Short Answer Type Questions-I Q. 1. How do antiseptics differ from disinfectants? Name a substance which can be used as a disinfectant as well as an antiseptic.  A [CBSE, Delhi Set 1, 2020] Ans. Antiseptics are the chemical substances which prevent the growth of microorganisms to the living tissues and may also kill them. e.g. Dettol. While disinfectants are the chemical substances which kill microorganisms only on non-living objects e.g. - Phenyle.[1] Phenol - its 0.2 percent solution acts as an antiseptic and 1% solution is as disinfectant.[1] Q. 2. Define the following terms with a suitable example in each : (i) Bacteriocidal antibiotics (ii) Food preservatives  R [CBSE, Delhi Set 3, 2020] Ans. (i) Bacteriocidal antibiotics : The chemical agents that inhibit the growth of bacteria or kill them are called bacteriocidal antibiotics.

E.g. Penicillin, Ofloxacin.[1]

(ii) Food preservatives : The substances which are capable of inhibiting the process of fermentation, acidification or any other decomposition of food are called food preservatives. E.g. potassium metabisulphite, sodium benzoate etc.[1] Q. 3. Define the following terms with a suitable example in each : (i) Antibiotics (ii) Antiseptics  R [CBSE, Outside Delhi Set 1, 2020] Ans. (i) Antibiotics : It is a chemical substance produced by micro-organisms that can inhibit the growth or even destroy other micro-organisms which cause disease. e.g. Penicillin.[1]

(ii) Antiseptics : The chemical substances which prevent the growth of micro-organisms and may also kill them. e.g. - Bithional, Terpineol. [1]

Commonly Made Error



Some students get confused between antibiotics and antiseptics

Answering Tip



Prepare well about medicines.

Q. 4. Define the following terms with a suitable example in each : (i) Tranquilizers (ii) Anionic detergent  R [CBSE, Outside Delhi Set-2, 2020] Ans. (i) Tranquilizers : The drugs given to the patients suffering from anxiety and mental tension. They are the common constituents of sleeping pills.  E.g. Nardil, luminal[1] (ii) Anionic detergent : These detergents are either the sulphates or sulphonates of long chain hydrocarbons. E.g. Sodium alkyl sulphate.[1] Q. 5. Define the following terms with a suitable example in each : (i) Antacids (ii) Artificial Sweetener R [CBSE, Outside Delhi Set 3, 2020]  Ans. (i) Antacids : The chemical substances which can reduce or neutralise the acidity in stomach and raise the pH to some appropriate level are called antacids. e.g. Ranitidine (Zantac), milk of magnesia, Omeprazole.[1] (ii) Artificial sweetener : These are food additives which act as the substitute for sugar. They do not add calories to our body. E.g. Saccharin, aspartame.[1]

Short Answer Type Questions-II



Ans. (i) Tranquilizers (ii) Anionic detergents (iii) It is difficult to control the sweetness.  [CBSE Marking Scheme, 2019] Detailed Answer : (i)  The drug used in sleeping pills is tranquilizers. (ii) Anionic detergents are used in toothpaste. Anionic detergents are sodium salts of a sulphonated long chain of alcohols.

(3 marks each)



(iii) It is high potency sweetener, Alitame is more stable than aspartame. It is 2000 times sweeter than sucrose. It is difficult to control the sweetness of food while using it. [3] Q. 2. Define the following terms with suitable example in each : (i) Broad-spectrum antibiotics (ii) Disinfectants (iii) Cationic detergents  R [CBSE, Delhi Set 1, 2019]

Q. 1. (i) What type of drug is used in sleeping pills? (ii) W  hat type of detergents are used in toothpastes? (iii) W  hy the use of alitame as artificial sweetener is not recommended?

(2 marks each)

(i) Antibiotics which kill or inhibit a wide range of gram-positive and gram-negative bacteria. Example- Chloramphenicol (or any other) [1]

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Answering Tip











Learn and understand about antiseptics and disinfectants.

Q. 3. (i) Why bithional is added in soap? (ii)  Why magnesium hydroxide is a better antacid than sodium bicarbonate? (iii)  Why soaps are biodegradable whereas detergents are non-biodegradable?

Ans. (i) To impart antiseptic properties. [1] (ii)  Magnesium hydroxide is better alternatives because of being insoluble, it does not increase the pH above neutrality.[1] (iii) Because in soaps hydrocarbon chains are not branched. [1]  [CBSE Marking Scheme, 2019]

Commonly Made Error







(ii) The chemicals which either kill or prevent the growth of micro-organisms when applied to inanimate objects such as floors, drainage system, instruments, etc. Example – 1% Phenol solution (or any other) [½ + ½] (iii) Cationic detergents are quarternary ammonium salts of amines with acetates, chlorides or bromides as anions where Cationic part is involved in cleansing action. Example – Cetyltrimethylammonium bromide (Or any other) [½ + ½] [CBSE Marking Scheme, 2019]

Sometimes, students write about antiseptics in place of disinfectants as they get confused in these terms. OR

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[3] [Topper’s Answer 2019] Detailed Answer : (i) Biothional is generally added to soaps to impart antiseptic properties and reduces the odours produced by the bacterial decomposition of organic matter on the skin. (ii) Antacids neutralize excess acid produced in the stomach. Magnesium hydroxide is a better antacid because of its insolubility. Being insoluble, even an excess will not allow the pH to increase above neutral. Sodium bicarbonate being soluble and its excess can make stomach alkaline and trigger more production of acid.



(iii) S  oaps are sodium and potassium salts of fatty acids. Straight chain hydrocarbons are easily degraded by microorganisms. The fatty acids used to make soaps contain straight chain hydrocarbon part, making them biodegradable. Detergents mostly contain branched chain hydrocarbon. The bacteria cannot digest branched chain compounds. Hence they are non-biodegradable. [3] Q. 4.  Define the following terms with a suitable example in each : (i) Antibiotics (ii) Artificial sweeteners (iii) Analgesics U + R [CBSE, Outside Delhi Set 3, 2020] 

Ans. (i) An antibiotic refers to a substance produced wholly or partly by chemical synthesis, which in low concentrations inhibits the growth or destroys microorganisms by intervening in their metabolic processes. E.g., Penicillin.  [½+ ½] (ii) Chemicals which are sweet in taste and with low calories, i.e., Saccharin.  [½+ ½] (iii) Analgesics reduce or abolish pain without causing impairment of consciousness, mental confusion, incoordination or paralysis or some other disturbances of nervous system. E.g., Aspirin.      (Or any other correct example)  [½+ ½]  [CBSE Marking Scheme, 2019] Detailed Answer : (i) Antibiotics are used as drugs to treat infections because of their low toxicity for humans and animals. Antibiotics are chemical substances which are produced by microbes and inhibit the

(ii)  Artificial sweeteners : These are the organic substances which have been synthesized in the lab and are known to be many times sweeter than sugar. For example, Aspartame, Saccharin, Sucralose.



(iii) Analgesics : These drugs which are used for painkilling effect on human body. For example, Narcotic analgesics are used to get relief from the pain in terminal cancer. [3] Q. 5. (a) Pick out the odd one from the following on the basis of their medicinal properties : Equanil, Seconal, Bithional, Luminal (b) What type of detergents are used in dish washing liquids? (c) Why is the use of aspartame limited to cold foods? R [CBSE, Outside Delhi Set 2, 2019] Ans. (i) Bithional [1] (ii) Non-ionic detergents [1] (iii) Because it is unstable at cooking temperature  [1]  [CBSE Marking Scheme 2019] Detailed Answer : (a) Bithional as it is an antiseptic while others are tranquilizers. (b) Non-ionic detergents. (c) Because it is unstable at cooking temperature.[3] Q. 6. (a) Why is bithional added to soap ? (b) What is tincture of iodine ? Write its one use. (c) Among the following, which one acts as a food preservative ? Aspartame, Aspirin, Sodium benzoate, Paracetamol  R [CBSE, Delhi/Outside Delhi, 2019]



growth or even destroy other microorganisms. Examples – penicillin, salvarsan.

Ans. (a) To impart antiseptic properties (b) 2-3% solution of iodine in alcohol – water mixture / iodine dissolved in alcohol, used as an antiseptic/ applied on wounds. (c) Sodium benzoate / Aspartame[1+½+½+1]  [CBSE Marking Scheme, 2018] Q. 7. Define the following : (i) Anionic detergents (ii) Broad spectrum antibiotics (iii) Antiseptic R [CBSE, Delhi Set-1, 2019]

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

Ans. (i) Anionic detergents are sodium salts of sulphonated long chain alcohols or hydrocarbons / alkylbenzene sulphonate or detergents whose anionic part is involved in cleansing action.[1] (ii) Broad spectrum antibiotics : Antibiotics which kill or inhibit a wide range of gram-positive and gram-negative bacteria.[1] (iii) Antiseptics are the chemicals which either kill or prevent growth of microbes on living tissues.[1] [CBSE Marking Scheme, 2017] Detailed Answer : (i) Anionic detergents are sodium salts of sulphonated long-chain alcohols or hydrocarbons. Anionic part is involved in the cleansing action. Some commonly used anionic detergents are Sodium cetylsulphate (C16H33OSO3–Na+) and Sodium stearyl sulphate (C17H35CH2OSO3– Na+). [1] (ii) Broad spectrum antibiotics are the antibiotics that are effective against a wide range of gram-positive and gram-negative bacteria. A commonly used broad spectrum antibiotic is chloramphenicol. —

NHCOCHCl2

— CH — CH — CH2OH

—

O2N—

OH Chloramphenicol [1] (iii) Antiseptics are the chemicals which either kill or prevent the growth of micro-organisms. Antiseptics are applied to living tissues such as wounds, cuts, ulcers and diseased skin surfaces. Dettol and soframycine are some commonly used antiseptics. [1] Q. 8. Define the following : (i) Cationic detergents (ii) Narrow spectrum antibiotics (iii) Disinfectants R [CBSE, Delhi Set-2, 2017]

Ans. (i) Cationic detergents are quaternary ammonium salts of amines with acetates, chlorides or bromides as anions, cationic part has long chain hydrocarbon / detergents whose cationic part is involved in cleansing action.[1] (ii) Narrow spectrum antibiotics are effective mainly against gram-positive or gram-negative bacteria.[1] (iii) Disinfectants kill or prevent growth of microbes and are applied on inanimate / non living objects.[1] [CBSE Marking Scheme, 2017] Detailed Answer : (i) Cationic detergents are defined as the quaternary ammonium salts of amines with acetates, chlorides or bromides as anions. Cationic part possesses a long hydrocarbon chain and a positive charge on nitrogen atom. Cetyltrimethyl ammonium bromide is a cationic detergent. [1] (ii) Narrow spectrum antibiotics are those which are mainly effective against gram positive or gram negative bacteria. Penicillin G is a narrow spectrum antibiotic. [1] (iii) Disinfectants are the chemicals which either kill or prevent the growth of micro-organisms. These are generally applied to inanimate objects like floors, sewage, drainage, instruments, etc. 1% solution of phenol is a common disinfectant used at various places. [1] Q. 9. Define the following : (i) Anionic detergents (ii) Limited spectrum antibiotics (iii) Tranquilizers R [CBSE, Delhi Set-3, 2017] Ans. (i) Anionic detergents are sodium salts of sulphonated long chain alcohols or hydrocarbons / detergents whose anionic part is involved in cleansing action.[1] (ii) Limited spectrum antibiotics are effective against a single organism or disease.[1] (iii) Tranquilizers are class of chemicals used for treatment of stress or mild or severe mental diseases.[1] [CBSE Marking Scheme, 2017]

OR

[Topper’s Answer 2017] [3]

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Long Answer Type Questions Q. 1.

(a)  What are biodegradable and nonbiodegradable detergents? Give one example of each class. R (b) (i)  Give two examples of macromolecules that are chosen as drug targets. (ii) What are antiseptics ? Give an example. (iii) Why is the use of aspartame limited to cold foods and soft drinks ? R + A&E Ans. (a)  Biodegradable detergents : Detergents which are decomposed by microorganisms present in the environment are called biodegradable detergents. These detergents have linear alkyl chains. Sodium lauryl sulphate and sodium dodecylbenzene sulphonate are examples of biodegradable detergents. [1]

Non-biodegradable detergents : Detergents which are not degraded by microorganisms. Such detergents cause environmental problem. It is observed that such detergents contain branched chain which is not attacked by bacteria. Example of non-biodegradable detergents is given below :



[½ + ½] (b) (i) Carbohydrates, lipids, proteins, enzymes, nucleic acids (Any two). [½ + ½] (ii) Antiseptics are the chemical substances which are used to kill or prevent the growth of microbes. e.g., Dettol/Iodoform/Boric acid/phenol (or any other correct example). [½ + ½] (iii) Because it is unstable at cooking temperature.[1] Q. 2. (a) Differentiate between disinfectants and antiseptics. Give one example of each group. R (b) Give reasons for the following : (i) With reference to which classification has the statement, “ranitidine is an antacid” been given? (ii)  Metal hydroxides are better alternatives than sodium hydrogen carbonate for treatment of acidity.





(iii)  Aspirin is used in prevention of heart attacks.  A&E Ans. (a) Disinfectants : These are chemicals which kill microorganisms or prevent their growth but are not safe for human beings. These are applied to inanimate objects such as floors, drainage systems. Some substances can act as an antiseptic as well as disinfectant by varying the concentration. For example, 0.2% solution of phenol is an antiseptic while its one percent solution is disinfectant. Antiseptics : These are the chemicals which prevent the growth of microorganisms or kill them but are not harmful to human beings. These are applied externally to the living tissues such as wounds, cuts and diseased skin surfaces. Dettol (chloroxylenol + a-terpineol), bithional, furacin, dilute solution of boric acid are common examples of antiseptics. [1] + 1 (b) (i) It is unstable at cooking temperature.[1] (ii) Excessive hydrogencarbonate can make the stomach alkaline and trigger the production of even more acid. Metal hydroxides being insoluble do not increase the pH above neutrality. [1]   (iii) Aspirin has anti blood clotting action. [1] Q. 3. With reference to which classification has the statement, “ranitidine is an antacid” been given?  U Ans. The above statement refers to the classification of pharmacological effects of the drug. This is because any drug that is used to counteract the effects of excess acid in the stomach is called an antacid. Antacids are the medicinal agents which decreases the excess level of hydrochloric acid in stomach. Ranitidine belongs to the class of Histamine receptors blockers, where they block the action of H2 receptors present in stomach, which are responsible for the excess secretion of gastric acid in stomach. [2]

Commonly Made Error Students often write lengthy answers for a 1 mark question thus losing time.

Answering Tip Write the answer followed by example. Avoid unnecessary explanations.

Visual Case-Based Questions Q. 1. Read the passage given below and answer the following questions : (1 × 4 = 4) Chemical substances which prevent the growth of micro-organisms or kill them but are not harmful to the living human tissues are called antiseptics.

(5 marks each)

(4 marks each) These are applied to wounds, ulcers and diseased skin surfaces. The antiseptics are usually incorporated in face powders, deodorants, breath purifiers, etc to reduce the odours which may result from the bacterial decomposition on body

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Oswaal CBSE Question Bank Chapterwise & Topicwise Solved Papers, CHEMISTRY, Class – XII

or in the mouth. Dettol, a mixture of chloroxylenol and terpenol, is a commonly used antiseptic for wounds, cuts, diseased skin surfaces etc. Bithional is added to soaps to impart them antiseptic properties. Such soaps are used to reduce odour due to bacterial action on skin surface. Iodine is also used as an antiseptic in the form of tincture of iodine i.e. 2–3% solution of iodine in alcoholwater. A number of dyes like gentian violet and methylene blue are also used as antiseptics. The following questions are Multiple Choice questions Choose the most appropriate answer.   (i)  Chemical substances which prevent the growth of micro-organisms but are not harmful to the living human tissues are called  (a) Antiseptics (b) Disinfectants  (c) Antibiotics (d) Analgesics  (ii) What is added to soaps to impart the antiseptic properties?  (a) Phenol (b) Bithional  (c) Aspirin (d) Barbituric acid (iii) A mixture of chloroxylenol and terpenol is called :  (a) Bithional (b) Dettol  (c) Aspirin (d) Methylene blue   (iv) Tincture of iodine contains  (a) 1–2% solution of iodine in alcohol-water.  (b) 2–3% solution of iodine in alcohol-water.  (c) 1–2% solution of iodine in ether-water.  (d) 2–3% solution of iodine in ether-water. OR Gentian violet is used as an  (a) Antiseptic (b) Analgesic  (c) Antipysetic (d) Antibiotic Ans. (i) Correct option : (a) Explanation : Antiseptics prevent the growth of micro-organisms or kill them without causing harm to living human tissues.[1] (ii) Correct option : (b) Explanation : Bithional is added to soaps to impart them antiseptic properties.[1] (iii) Correct option : (b) Explanation : Dettol is a mixture of choloroxylenol and terpenol.[1] (iv) Correct option : (b) Explanation : Iodine is used as an antiseptic in the form of tincture i.e. 2–3% solution of iodine in alcohol water.[1] OR Correct option : (a) Explanation : Gentian violet is a dye which is used as an antiseptic.[1] Q. 2. Read the passage given below and answers the following questions : (1 × 4 = 4) Chemical substances which are produced by micro-organisms and are capable of destroying other micro-organisms are called antibiotics. The first antibiotic discovered was penicillin which is an effective drug for pneumonia, bronchitis, sore

throat etc. Chloramphenicol another antibiotic has prolonged action and no side effects. It is effective against pneumonia, typhoid, dysentry etc. Streptomycin another antibiotic is used for the treatment of tuberculosis. Antibiotics which are effective against several different types of harmful micro-organisms are called broad spectrum antibiotics. E.g. Chloromycetin etc. The following queations are multiple choice questions. Choose the most appropriate answer:  (i) Which of the following is the first antibiotic discovered?  (a) Chloramphenicol (b) Chloromycetin  (c) Penicillin (d) Streptomycin  (ii)  Chemical substances that are produced by micro-organism and are capable of destroying other micro-organisms are called  (a) Antiseptics (b) Antibiotics  (c) Analgesics (d) Antipyretics (iii) The chemical substances which are effective against several different types of harmful micro-organisms are called  (a) Antiseptics  (b) Narrow spectrum antibiotics  (c) Broad spectrum antibiotics  (d) Disinfectants   (iv) Chloromycetin is a  (a) Disinfectant  (b) Narrow spectrum antibiotic  (c) Broad spectrum antibiotic  (d) None to these OR    The antibiotic used for the treatment of tuberculosis is  (a) Chloromphenicol  (b) Penicillin  (c) Sulpha drugs  (d) Streptomycin Ans. (i) Correct option : (c) Explanation : Penicillin is the first antibiotic discovered by Alexander Fleming (1928). [1] (ii) Correct option : (b) Explanation : Antibiotics are produced by microorganism and are capable of destroying other micro-organism.[1] (iii) Correct option: (c) Explanation : Broad spectrum antibiotics are effective against several different types of harmful micro-organisms.[1] (iv) Correct option : (c) Explanation : Chloromycetin is broad spectrum antibiotic as it is effective against several different types of harmful micro-organisms.[1] OR Correct option : (d) Explanation : Streptomycin is an aminoglycoside antibiotic derived from Streptomyces griseus that is used for treatment of tuberculosis.[1] ll

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SELF-ASSESSMENT TEST

Self Assessment Test-15 Time : 1 Hour

Max. Marks : 25

Q. 1. Read the passage given below and answer the following questions : (1 × 4 = 4) Tranquilizers are substances which are used for the cure of mental diseases. These effect the central nervous system and induce sleep to the patients. Such compounds usually constitute sleeping pills. The tranquilizers are also called psycho-therapeutic drugs. The administration of these drugs makes the patient passive and helps to control their emotional distress which otherwise is likely to interface with their normal functions. Barbituric acid and its derivatives serve as good tranquilizers. Sedatives act as depressant and suppress the activities of central nervous system. Hypnotics are used to reduce mental tension and anxiety and also reduce pain. The following questions are multiple choice questions. Choose the most appropriate answer:    (i) The substances used for the cure of mental diseases are called  (a) Antibiotics (b) Antiseptics  (c) Tranquilizers (d) Analgesics (ii) The tranquilizers are also called  (a) Therapeutic drugs  (b) Psycho-therapeutic drugs  (c) Sulpha drugs (d) Antibacterial drugs (iii) Which of the following is tranquilizer?  (a) Penicillin (b) Ampicillin  (c) Barbituric acid (d) Bithional (iv)  Which of the following drugs act as depressant?  (a) Antipyretics (b) Sedatives  (c) Antibiotics  (d) Analgesics Following questions (No. 2–5) are multiple choice questions carrying 1 mark each: Q. 2. Which of the following is an analgesic? (a) Chloramphenicol (b) Penicillin (c) Paracetamol (d) Streptomycin R Q. 3. Antipyretics are medicinal compounds which (a) lower body temperature (b) relieve pain (c) control malaria (d) can kill other micro-organisms R Q. 4. Aspirin is (a) Analgesic (b) Antipyretic (c) Antimalarial (d) Both analgesic and antipyretic R Q. 5. Veronal, a barbituric drug is used as (a) Anaesthetic (b) Sedative (c) Antiseptic (d) None of these R

The following questions (Q. no. 6 & 7), a statement of assertive followed by a statement of reason id given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not the correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. Q. 6. Assertion : Sulpha drug contains sulphonamide group. Reason : Salvarsan is a sulphur drug. Q. 7. Assertion : Chemical messangers are chemicals that enable communication of message between neurons and muscles. Reason : Chemicals enter the cell through receptor. The following Questions (No. 8–9) are Short Answer Type – I and carry 2 marks each. Q. 8. Define and write an example for the following : (a) Broad spectrum antibiotics. (b) Analgesics R Q. 9. Define the following with suitable example. (a) Analgesic (b) Disinfectant R Q No: 10 – 11 are Short Answer Type – II carrying 3 marks each. Q. 10. (i) Name a substance which can be used as an antiseptic as well as disinfectant. (ii) Name an artificial sweetener whose use is limited to cold food and drinks.  (iii) What are cationic detergents? R Q. 11.  Write the therapeutic action of following on human body and mention the class of drugs to which each of these belong : (i) Ranitidine (ii) Morphine   (iii) Aspirin  R Q No 12 is Long Answer Type Question carrying 5 marks Q. 12. (a) Give the use of following compounds: (i) Dettol (ii) Bithional  (b) Write the therapeutic action of following on human body and mention the class of drugs to which each of these belong : (i) Equanil (ii) Aspirin (iii) Chloramphenicol R OR (a) Differentiate between soaps and detergents (b) (i)  Name sweetening agent used in the preparation of sweets for a diabetic patient.   (ii) What are food preservatives? Give an example. (iii) Give two examples of antifertility drugs.

 

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IMPHAL KARWI

Jain Book Shop, 9856031157 Ramji & Sons, 9026303241

Salem Book House, 9443324584, (0427) 2411315

KOLKATA

Eureka book Emporium, (033) 25934001, 9433089132 Schoolwale & Company, 9731715655

MIRZAPUR AGRA

Pustak Bhawan, 9936500469

Dutta Book Stall, 9402477623, 8729948473

Vidyarthi Book Centre (S.S.D.), 9879014367 Royal Stationery, 9824207514

TAMIL NADU HARYANA

Raghubar Dass Sat Prakash, 8950055303

MADHYA PRADESH

Raghav Pustak Sadan (S.S.D.), 7772988088

ANDHRA PRADESH

Akshaya Books Corner, 9666155555

Sri Rajeshwari Book Link, (0891) 6661718, 9848036014

ASSAM

PORT BLAIR

Krishna Book Centre, 9474205570

PURNIA RAIPUR DURG DELHI

NOIDA

MANIPUR

WEST BENGAL

UTTAR PRADESH

Maheshwari Pustak Bhandar, 9760602503

OUR DISTRIBUTORS

Y Renuka Devi, (0863) 2252308, 9490450750 Sri Balaji Book Depot, (040) 27613300, 9866355473 Sri Kanaka Durga Book Stall, 9603491009, 9849144007

Ashok Publication, 7002846982, Book Emporium, 9675972993, 6000763186 Kayaan Enterprises, (0361) 2630443, BLJ Publications, 7086099332 C. R. Book House, 9957886562

MUZAFFARPUR JHANJHARPUR PATNA

KOLHAPUR PUNE

ASSAM

TINSUKIA

GUWAHATI

MAHARASHTRA

Ashish Book Depot, 7798420420, Jai Book Co., 9422046679 Sai Shubham, 9975687687, (020) 69498635, Goel Mini Market, 9890825884, 9028943031, Praveen Sales, 9890683475 Vardhaman Educational, 9834678521 Anil Paper Mart, 9422722522, (02482) 230733 Unique Traders, (07152) 243617, 8600044411, 9960644752

ANDHRA PRADESH

ANDAMAN BIHAR

Pustak Bhandar, 9097046555 Krishna Book Agency, 9801019292, 9304977755 Bokaro Student Friends, (0612) 2300600, 2618400, Gyan Ganga Ltd., 9304826651 Nova Publisher & Distributors, (0612) 2666404, Sharda Pustak Bhandar, 8877279953, Shri Durga Pustak Mandir, 9334477386 Vikas Book Depot, (0612) 2304753, 7004301983 Chaurasia Book Centre, 9006717044, 7004456102

MUMBAI

NAVI MUMBAI NANDED NASHIK NAGPUR

PUNE

YAVATMAL SOLAPUR DHULE BHUSAWAL CUTTACK BHUBANESHWAR

CHATTISGARH

BARIPADA

DELHI

JALANDHAR

Agarwal Traders & Pub., (0771) 4044423, 7489991679, 8878568055 (W. app) Gupta Pustak Mandir, 9329100851 Bhagwati Bhawani Book Depot, 7882327620, 9827473100 Mittal Books, (011) 23288887, 9899037390 Bokaro,Student Friends Pvt Ltd., 7004074900 R. D. Chawla & Sons, 9899445522, Zombozone, 9871274082 Sharda Publication, 9971603248 Prozo (Global Edu4 Share Pvt. Ltd), 9899037390, 8587837835, 9318395520 Global Islamic Publication, 9873690828

GUJARAT

LUDHIANA

PATIALA BARNALA BHATINDA SANGRUR

Shivam Books & Stationery, (022) 28230897, 9892935799 Student Book Depot, 9821550165 Krishna Book Store, (022) 27744962, 9819537938 Abhang Pustakalaya, 9823470756 Rahul Book Centre, (0253) 2599608, 9970849681 Laxmi Pustakalay Stationers, (0712) 2727354, Novelty Book Depot, 9657690200 Renuka Book Distributor, 9765986633. Karamveer Book Depot, 9923966466, 7172725726

Vijay Book Depot, 9860122094, (0712) 2520496, 2534217

Shree Mahalaxmi Pustakalaya, (0712) 2283580, 7507099360 Natraj Book Depot., (020) 24485054, 9890054092 Vikas Book House, 9860286472, (020) 244683737 Mahesh Book & General Store, 9421036171, Kirti Book Agnecies, 9881190907 Dilip Book Agency, (0823) 2245450, 9423131275 Jitesh Vastu Bhandar (0217) 2741061, 9960000009 Navjeevan Book Stall, 7020525561 Anil Book Depot, 9403942906

ODISHA

A. K. Mishra Agencies, 9437025991, 9437081319 M/s Pragnya, 8847888616, 9437943777, Bharati Book House, 9438420527 Trimurthi Book World, (0679) 2253336 9437034735

PUNJAB

Amit Book Depot., 9815807871, Ravi Book Shop, 9815200925 Bhatia Book Centre, 9815277131, 7901814043 Cheap Book Store, 9872223458, 9878258592, Gaurav Book, 9478821183 Subhash Book Depot, 9876453625, Cheap Book Store, 9872223458 City Book Shop, 9417440753 Adarsh Enterprises, 9814347613 Navchetan Book Depot, 9779050692 Bhagwati Book, 9463120564, Agarwal Book, 9417816439 Jindal Book Depot, 9876141679

AHMEDABAD

Patel Book, 9898184248, 9824386112, 9825900335

KOTA

RAJASTHAN

VADODARA

Pooja Book Shop, 7600817198, (0265) 2280464 Umakant Book Sellers & Stationery, 9824014209, 9624920709 Goutam Book Sellers, 9081790813 College Store, 8141913750

BHILWARA JAIPUR

(0744) 231958, Radhe Traders, 8769802001, Perfect Stationers & Gen. Shoppe, 9829863904

BOKARO DHANBAD RANCHI BENGALURU

BANGLORE GULBARGA HUBLI BELLARY

JHARKHAND

Bokaro Student Friends, (0654) 2233094, 7360021503 Sahu Pustak Bhandar, 9431378296, 7979845480 Bokaro Student Friends, (0326) 3590527 Bokaro Student Friends, 9234628152, Gyan Ganga Ltd., 9117889900

MORENA GWALIOR INDORE REWA JABALPUR

Sri Sai Ram Book traders, 9738881088 Sapna Book House, (080) 46551999, 9343366670

Hema Book Store, 8041485110

L.E. Bhavikatti, (08472) 261400, 9880737400 Renuka Book Distributor, (0836) 2244124 Chaitanya Exhibition, Bellary - 9886393971

KERALA

MADHYA PRADESH

Shri Ram Book Store, 9424603124 Agarwal Book Depot, 9425116210 Student Book Depot, (0731) 2503333, 2535892, 9425322330 Arun Prakashan, (0731) 2459448, 9424890785 Siddharth Enterprises, 9425185072 Sangam General Store, (0761) 2412592, Akash Book Distributor, 7974264828 New Radhika Book Palace, 9425411533-66

GOA

MARGO

Golden Heart Emporium, (0832) 2725208, 9370273479

ROHTAK GURGAON

Swami Kitab Ghar, 9255121948 Pahuja & Co., 9999563778

SOLAN IMPHAL JALGAON

CHENNAI

KARNATAKA

Asad Book Centre, (0484) 2370431, 9447314548, Academic Book House, (0484) 2376613 Surya Book House, (0484) 2363721, 9847124217, Surya Book Centre, (0484) 2365149 Orbit Book Centre, 9847770749 KADAVANTHRA H & C Stores, (0484) 2203683, 2204683, 9446411997 JOMES SRINILAYAM H & C Store, (0484) 2351233 KOLLAM H & C Store, (0474) 2765421, 9447574778, H & C Store-2, 9995214104, 9809002519 KOTTAYAM H & C Store, (0481) 2304351, Book Centre, (0481) 2566992 PALARIVATTOM H & C Store, (0484) 2344337 TRIVANDRUM Academic Book House, (0471) 2333349, 9447063349, H & C Store, (0471) 2572010, 9446411996 Engineers Book Centre, (0471) 2596959 KANNUR Athulya Books, (0497) 2709294 CALICUT Aman Book Stall, (0495) 2721282, ERNAKULAM

COIMBATORE

HARYANA

HIMACHAL PRADESH

Mangla Enterprises, 9882050720

MANIPUR

Jain Book Shop, 9856031157

MAHARASHTRA

Sharma Book Depot, 9421393040

PUDUCHERRY TRICHY OOTY AGARTALA AGRA

ALIGARH ALLAHABAD AZAMGARH BALIA FARRUKHABAD GORAKHPUR JAUNPUR LUCKNOW MORENA MEERUT VARANASI

KOLKATA

ASANSOL SILIGURI

Raj Traders, (0744) 2429090, 9309232829, 8005529594, Bhandari Stationers,

Nakoda Book Depot, (01482) 243653, 9214983594 J K Enterprises, 9460983939, 9414782130, Education Point, 9269664791 Saraswati Book House, (0141) 2610823, 9829811155 Ravi Entreprises (0141) 2602517, 9828944479 Manohar Book Distrubutor, 9414072321

TAMIL NADU

Majestic Book House, (0422) 2384333 CBSE Book Shop, (0422) 2303533, 8056655337 Arraba Book Traders, (044) 25387868, 9841459105, Mr Book Store (044) 25364596, Indian Book House, (044) 24327784, 9094129595, Kalaimagal Store, (044) 5544072, 9940619404, Vijaya Stores, 9381037417 Ruby Books, (044) 26425958, 9884519471, Kamal Store. 9840276067 Bookmark It-Books & Stat. Store, 7305151653, M.K. Store, 9840030099 Tiger Books Pvt. Ltd., 9710447000, M. K. Store, 9840030099 Sri Saraswathi Book Stall, (04132) 222283, 9092626287 Trichy Book House, (0431) 2764198, 2766815, 9489764192. 9443238419 Bharat & Co., 9095552155

TRIPURA

Book Corner, 9856358594, Book Emporium, (0381) 2391509, 9436460399,

UTTAR PRADESH

Ajay Book Depot, (0562) 2254621, 9411205449 K. S. A. Book Distributor, 9149081912 Om Pustak Mandir, (0562) 2464014, 9319117771, Panchsheel Books, 9412257961, 9412257962, Shaligram Agencies (0571) 2421887, 9412317800 Mehrotra, (0532) 2266865, 9415636890 Sasta Sahitya Sadan, 9450029674

Vidya Kendra, 9415281234

Anurag Book, (0569) 2226843, 9839933650 Central Book House, 9935454590 Thakur Pustak, 9795198974, 5453-222298

Rama Book Depot, (0522) 4080133, 9956922433 (Retail), Vyapar Sadan, 7607102462 Shri Ram Book Store, 9424603124

Ideal Book Depot, (0121) 4059252, 9837066307 Garg Book Depot, 9927052149 Bokaro Student Friends, (0542) 2401250, 8299344607 Gupta Books, 9918155500, 8707225564 Shri Krishna Book & Stationery, 9415020103

WEST BENGAL

Eureka Book, (033) 25934001, 9433089132, Oriental Publishers & Distributor (033) 40628367, Katha 'O' Kahani, (033) 22196313, 22419071, Saha Book House, (033) 22193671, 9333416484, New National Book Store, 8697601392, United Book House, 9231692641, (033) 22418105 Book House, 9434747506 Agarwal Book House, (0353) 2535274, 9832038727, Metro Books, 9832477841

0906

VAPI NAVSARI

Shalibhadra Stationers, 9904477855