Oswaal CBSE Sample Question Papers Class 12 Chemistry (For 2023 Exam) 9789356343566

Table of contents :
Cover
Title Cover
Copyright Page
Contents
Syllabus
Solved Paper, Term-II, 2022-22
Mind Maps
On Tips Notes
¤ Sample Question Papers (Solved)
● Sample Question Paper - 1 (Issued by Board dated 16th Sep. 2022)
● Solutions of Sample Paper - 1 (CBSE Marking Scheme 2022-23)
● Sample Question Paper - 2
● Sample Question Paper - 3
● Sample Question Paper - 4
● Sample Question Paper - 5
¤ Self Assessment Papers
● Self Assessment Paper - 1
● Self Assessment Paper - 2
● Self Assessment Paper - 3
● Self Assessment Paper - 4
● Self Assessment Paper - 5
¤ Soutions
● Solutions of Sample Question Paper - 2
● Solutions of Sample Question Paper - 3
● Solutions of Sample Question Paper - 4
● Solutions of Sample Question Paper - 5
¤ Hints
¤ Solutions of Self Assessment Papers
● Solutions of Self Assessment Paper - 1
● Solutions of Self Assessment Paper - 2
● Solutions of Self Assessment Paper - 3
● Solutions of Self Assessment Paper - 4
● Solutions of Self Assessment Paper - 5

Citation preview

CHEMISTRY

(1)

15th EDITION



“978-93-5634-356-6”

ISBN

YEAR 2022-23

SYLLABUS COVERED

CENTRAL BOARD OF SECONDARY EDUCATION DELHI

PUBLISHED BY OSWAAL BOOKS & LEARNING PVT. LTD.

COPYRIG HT

RESERVED

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BY THE PUBLISHERS

All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, without written permission from the publishers. The author and publisher will gladly receive information enabling them to rectify any error or omission in subsequent editions.

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D I S C L AIMER This book is published by Oswaal Books and Learning Pvt Ltd (“Publisher”) and is intended solely for educational use, to enable students to practice for examinations/tests and reference. The contents of this book primarily comprise a collection of questions that have been sourced from previous examination papers. Any practice questions and/or notes included by the Publisher are formulated by placing reliance on previous question papers and are in keeping with the format/pattern/guidelines applicable to such papers. The Publisher expressly disclaims any liability for the use of, or references to, any terms or terminology in the book, which may not be considered appropriate or may be considered offensive, in light of societal changes. Further, the contents of this book, including references to any persons, corporations, brands, political parties, incidents, historical events and/or terminology within the book, if any, are not intended to be offensive, and/or to hurt, insult or defame any person (whether living or dead), entity, gender, caste, religion, race, etc. and any interpretation to this effect is unintended and purely incidental. While we try to keep our publications as updated and accurate as possible, human error may creep in. We expressly disclaim liability for errors and/or omissions in the content, if any, and further disclaim any liability for any loss or damages in connection with the use of the book and reference to its contents”.

Kindle (Edition 2)

TABLE OF CONTENTS







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5 - 10 13 - 24



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Latest CBSE Circular released on 20th May for Academic Year 2022-2023 (CBSE Cir. No. Acad 57/2022) Latest CBSE Syllabus released on 21st April 2022 for Academic Year 2022-2023 (CBSE Cir. No. Acad 48/2022) CBSE Solved Board Papers 2022 Term-II Examination (Delhi & Outside Delhi Sets) (To download Solved paper for Term-I 2021-22 & Latest Topper’s Answers 2020, scan the QR Code





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See Self Assessment Paper Solution at Last this PDF

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112 120 128 136

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Hints





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Mind Maps On Tips Notes Sample Question Papers (Solved) l Sample Question Paper - 1 (Issued by Board dated 16th Sep. 2022) l Solutions of Sample Paper - 1 (CBSE Marking Scheme 2022-23) l Sample Question Paper - 2 l Sample Question Paper - 3 l Sample Question Paper - 4 l Sample Question Paper - 5  Self Assessment Papers* l Self Assessment Paper - 1 l Self Assessment Paper - 2 l Self Assessment Paper - 3 l Self Assessment Paper - 4 l Self Assessment Paper - 5  Solutions l Sample Question Paper - 2 l Sample Question Paper - 3 l Sample Question Paper - 4 l Sample Question Paper - 5   





given on Page 24)

CBSE CIRCULAR 2022-23 dsUæh; ek/;fed f'k{kk cksMZ CENTRAL BOARD OF SECONDARY EDUCATION Dated: 20.05.2022



CBSE/DIR(ACAD)/2022/

Circular No. ACAD-57/2022

All Heads of schools affiliated to CBSE







Subject : Assessment and Evaluation Practices of the Board for the Session 2022-23 National Education Policy 2020 has affirmed the need to move from rote to competency-based learning. This will equip the learners with key competencies to meet the challenges of the 21st century proactively. Accordingly, the Board has taken multiple steps towards the implementation of Competency Based Education (CBE) in schools. These range from aligning assessment to CBE, development of exemplar resources for teachers and students on CBE pedagogy and assessment and continued teacher capacity building. In this context the Board has released Circular No. Acad-05/2019 dated 18.01.2019; Circular No. Acad-11/2019 dated 06.03.2019; Circular No. Acad-18/2020 dated 16.03.2020; Circular No. Acad-32/2020 dated 14.05.2020 and Circular No. Acad-31/2020 dated 22.04.2021. In continuation to these circulars, the Board is initiating further corresponding changes in the Examination and Assessment practices for the year 2022-23 to align assessment to Competency Based Education. Therefore, in the forthcoming sessions a greater number of Competency Based Questions or questions that assess application of concepts in real-life/ unfamiliar situations will be part of the question paper. The changes for classes IX-XII (2022-23) internal year-end/Board Examination are as detailed: (Classes IX-X) Year End Examination/ Board Examination (Theory)

(2021-22) Existing (As per Special Scheme of Assessment for Board Examination – Circular No. Acad51/2021 dated 05.07.2021)

(2022-23) Modified (Annual Scheme)

Composition

• Term I – Multiple Choice Question including case based and assertion reasoning type MCQs – 100% (30% questions competency based) • Term II – Case based/ Situation based, Open Ended- short answer/long answer questions (30% questions competency based)

• Competency Based Questions would be minimum 40% These can be in the form of Multiple Choice Questions, Case based Questions, Source based Integrated Questions or any other types. • Objective Type Questions will be 20% • Remaining 40% short

Composition

• Term I – Multiple Choice Question including case based and assertion reasoning type MCQs – 100% (30% questions competency based) • Term II – Case based/ Situation based, Open Ended- short answer/long answer questions (30% questions competency based)

• Competency Based Questions would be minimum 40% These can be in the form of Multiple Choice Questions, Case based Questions, Source based Integrated Questions or any other types. • Objective Type Questions will be 20% • Remaining 40% short answer/long answer questions (as per existing pattern)

Internal Assessment : No change Internal Assessment: End of year examination = 20:80 Curriculum document released by the Board vide circular No.Acad-50/2022 dated 28th April, 2022 and the forthcoming Sample Question Papers may be referred for the details of changes in the QP design of individual subjects.









Year End Examination/ Board Examination (Theory)

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(Dr. Joseph Emmanuel) Director (Academics)

SYLLABUS Latest Syllabus issued by CBSE dated 21st April 2022 for Academic Year 2022-23 CHEMISTRY (Code No. 043) CLASS-XII (Theory)

Total Periods (Theory 160 + Practical 60) 70 Marks



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 Coordination 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.

Coordination compounds - Introduction, ligands, coordination number, colour, magnetic properties and shapes, IUPAC nomenclature of mononuclear coordination compounds. Bonding, (5)

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.

Carboxylic Acids : Nomenclature, acidic nature, methods of preparation, physical and chemical properties; uses. 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



14 Periods

18 Periods







Unit XIII : Amines



15 Periods

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.







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

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.

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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

PRACTICALS SYLLABUS

Total

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. G. Preparation of Organic Compounds : Preparation of any one of the following compounds (i) Acetanilide (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. (7)

SYLLABUS

I.













J.

Characteristic tests of carbohydrates, fats and proteins in pure samples and their detection in given foodstuffs. Determination of concentration/ molarity of KMnO4 solution by titrating it against a standard solution of : (a) Oxalic acid, (b) Ferrous Ammonium Sulphate (Students will be required to prepare standard solutions by weighing themselves). 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-, (C2O4)2-, CH3COO -,NO3(Note: Insoluble salts excluded)  

ℓ

ℓ











K.





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 soyabean milk and its comparison with the natural milk with respect to curd formation, effect of a 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). 4 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.

Time Allowed : Two hours



Practical Examination for Visually Impaired Students of Classes XI and XII Evaluation Scheme Topic

Max. Marks : 30

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.

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SYLLABUS 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. 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 List of Practicals The experiments have been divided into two sections: Section A and Section B. The experiments mentioned in Section B are mandatory.



•



•





• •











1.









2.

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)

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.























F.

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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

qq

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.

( 10 )

40% Short/Long Types Questions

20% Objective Types Including MCQs

Long Questions divided into sub-parts

Know Your 2023 Board Sample Paper Better !

Number of Sections Increased

Number of Case-based Questions Increased

Analytical Questions Included

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Solved Paper, 2021-22 CHEMISTRY Term-II, (Delhi Set-I) Question Paper Code No. 56/5/1

Series: AABB5/5

Max. Marks : 35



Time : 2 Hours

General Instructions :

(i) This question paper contains 12 question. All questions are compulsory. (ii) This question paper comprises of three sections-Section A, B and C.





Read the following instructions very carefully and strictly follow them:





(v) Section C – Q. No 12 is case based question carrying 5 marks.



(iv) Section B – Q. No. 4 to 11 are short-answer type questions carrying 3 marks each. (vi) Use of log tables and calculators is not allowed.





(iii) Section A – Q. No. 1 to 3 are very short-answer type questions carrying 2 marks each.

56/5/1



Delhi Set-I

































(iii) What is Electrophoresis?











































Section-B 4. Observe the graph shown in figure and answer the following questions: 1×3=3

(a) What is the order of the reaction? (b) What is the slope of the curve? (c) Write the relationship between k and t1/2 (half life period) 5. (a) (i) Write the IUPAC name of the following complex: 1×3=3 K2[PdCl4] (ii) Using crystal field theory, write the electronic configuration of d5 ion, if D0 > P. (iii) What are Homoleptic complexes? OR (b) (i) Why chelate complexes are more stable than complexes with unidentate ligands? 1 (ii) What is ‘‘spectrochemical series’’? What is the difference between a weak field ligand and a strong field ligand? 2 6. (a) (i) Define coagulation. 1×3=3 (ii) State Hardy-Schulze rule.

















1. Answer the following questions (Do any two): 1×2=2 (a) Identify the order of reaction from the following unit for its rate constant: Lmol – 1s – 1 (b) The conversion of molecules A to B follow second order kinetics. If concentration of A is increased to three times, how will it affect the rate of formation of B? (c) Write the expression of integrated rate equation for zero order reaction. 2. Arrange the following in the increasing order of their property indicated: 1×2=2 (a) Ethanal, Propanone, Propanal, Butanone (reactivity towards nucleophilic addition) (b) 4-Nitrobenzoic acid, benzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxy benzoic acid (Acid strength) 3. Explain the following reactions: 1×2=2 (a) Wolff Kishner reduction (b) Cannizzaro reaction













Section-A

OSWAAL CBSE Sample Question Papers, CHEMISTRY, Class-XII OR (b) Write three differences between Physisorption and Chemisorption. 1×3=3 (a) Write any two consequences of Lanthanoid Contraction. 2+1=3 (b) Name the element of 3d series which exhibits the largest number of oxidation states. Give reason. Give reasons for the following statements: 1 × 3 = 3 (a) Copper does not displace hydrogen from acids. (b) Transition metals and most of their compounds show paramagnetic behaviour. (c) Zn, Cd and Hg are soft metals. (a) Account for the following: 1×3=3 (i) pKb of aniline is more than that of methylamine. (ii) Aniline does not undergo Friedel-Crafts reaction. (iii) Primary amines have higher boiling points than tertiary amines. OR (b) (i) Arrange the following compounds in the increasing order of their basic strength in aqueous solution: 1×3=3 CH3NH2,(CH3)3N,(CH3)2NH (ii) What is Hinsberg’s reagent? (iii) What is the role of pyridine in the acylation reaction of amines? A compound ‘A’ on reduction with iron scrap and hydrochloric acid gives compound ‘B’ with molecular formula C6H7N. Compound ‘B’ on reaction with CHCl3 and alcoholic KOH produces an obnoxious smell of carbylamine due to the formation of ‘C’. Identify ‘A’, ‘B’ and ‘C’ and write the chemical reactions involved. 1×3=3 (a) Complete the following: 1×3=3 l, AlH(i -Bu)2 (i) CH3CN ‘A’ H 2 N-OH ‘B’ 2. H2O H+

12. Read the passage given below and answer the questions that follow: 1+1+1+2=5

















The conductance of material is the property of materials due to which a material allows the flow of ions through itself and thus conducts electricity. Conductivity is represented by k and it depends upon nature and concentration of electrolyte, temperature, etc. A more common term molar conductivity of a solution at a given concentration is conductance of the volume of solution containing one mole of electrolyte kept between two electrodes with the unit area of cross-section and distance of unit length. Limiting molar conductivity of weak electrolytes cannot be obtained graphically.





(ii) Write lUPAC name of the following compound:





(a) Is silver plate the anode or cathode? 1 (b) What will happen if the salt bridge is removed? 1 (c) When does electrochemical cell behaves like an electrolytic cell? 1 (d) (i) What will happen to the concentration of Zn2+ and Ag+ when Ecell = 0. 1×2=2 (ii) Why does conductvity of a solution decreases with dilution?











































(iii) Write chemical test to distinguish between the following compounds: Phenol and Benzoic acid OR (b) Convert the following: 1×2=2 (i) Benzoic acid to Benzaldehyde (ii) Propan-1-ol to 2-Bromopropanoic acid (iii) Acetaldehyde to But-2-enal







11.





















10.



Oxidation-reduction reactions are commonly known as redox reactions. They involve transfer of electrons from one species to another. In a spontaneous reaction, energy is released which can be used to do useful work. The reaction is split into two half reactions. Two different containers are used and a wire is used to drive the electrons from one side to the other and a Voltaic/Galvanic cell is created. It is an electrochemical cell that uses spontaneous redox reactions to generate electricity. A salt bridge also connects to the half cells. The reading of the voltmeter gives the cell voltage or cell potential or 0 is positive the reaction electromotive force. If Ecell is spontancous and if it is negative the reaction is non-spontaneous and is referred to as electrolytic cell. Electrolysis refers to the decomposition of a substance by an electric current. One mole of electric charge when passed through a cell will discharge half a mole of a divalent metal ion such as Cu2+. This was first formulated by Faraday in the form of laws of electrolysis.



























9.













8.























7.





Section-C





14

Solved Paper-2022



15 an electrolyte is found to be 138.9 S cm2 mol – 1. Calculate the conductivity of this solution. 2

OR (d) The molar conductivity of a 1.5 M solution of





Delhi Set-II

56/5/2



Note: Except these all other Questions are from Set-I.

Section-A

2. Explain the following reactions:





(a) Clemmensen reaction



the follwing: 2+1=3 (i) Pentaamminenitrito-N-Cobalt (III) (ii) Tetrahydroxidozincate (II) (ii) What is crystal field splitting energy? 8. Give reason for the following statements: 1 × 3 = 3 (a) Scandium (Z = 21) is a transition element but Zn (Z = 30) is not. (b) [Ti (H2O)]3+ is coloured while [Sc (H2O6]3+ is colourless. (c) Physical and chemical properties of the 4d and 5d series of the transition elements are quite similar than expected. 10. A primary amine ‘A’ C2H7N reacts with alkyl halide (C2H5l) to give secondary amine ‘B’. ‘B’ reacts with C6H5SO2Cl to give a solid ‘C’ which is insoluble in alkali. Identify ‘A’, ‘B’, ‘C’ and write all the chemical reaction involved. 3









(b) Stephan reaction





1×2=2



Section-B



















































5. (a) (i) Write the IUPAC name of the following complex: 1×3=3 [Pt(NH3)6]Cl4 (ii) On the basis of crystal field theory, write the electronic configuration of d4 ion, if Do < P. (iii) What are Heteroleptic complexes? OR (b) (i) Using IUPAC norms write the formulas for













Outside Delhi Set-I

56/3/1



Section-A



5. (a)

Calculate Dr G° and log Kc for the following cell: 3 Ni(s) + 2 Ag+(aq) → Ni2+(aq) + 2 Ag(s) Given that E° cell = 1.05 V, 1 F = 96,500 C mol–1 OR Calculate the e.m.f. of the following cell at 298 K: 3 Fe(s) | Fe2+ (0.001 M) || H+ (0.01 M) | H2(g) (1 bar) | Pt (s) Given that E° cell = 0.44 V [log 2 = 0.3010 log 3 = 0.4771 log 10 = 1] Using valence bond theory, predict the hybridization and magnetic character of following: [CoF6]3– [Atomic number of Co = 27] Write IUPAC name of the following complex: [CoBr2(en)2]+ How many ions are produced from the complex [Co(NH3)6]Cl2 in solution? 1 ×3 = 3 Differentiate between the following: (i) Adsorption and Absorption (ii) Lyophobic Sol and Lyophilic Sol (iii) Multimolecular Colloid and Macromolecular colloid. 1×3=3 OR (I) Define the following terms: 3 (i) Zeta Potential (ii) Coagulation





1. Predict the reagent for carrying out the following transformations: 1×2=2 (a) Benzoyl chloride to Benzaldehyde (b) Ethanal to 3-hydroxy butanal (c) Ethanoic acid to 2-chloroethanoic acid 2. (a) Why on dilution the ∧m of CH3COOH increases very fast, while that of CH3COONa increases gradually? (b) What happens if external potential applied becomes greater than E° cell of electrochemical cell? 1×2=2 3. An Organic compound (A) with molecular formula C3H7NO on heating with Br2 and KOH forms a compound (B). Compound (B) on heating with CHCl3 and alcoholic KOH produces a foul smelling compound (C) and on reacting with C6H5SO2Cl forms a compound (D) which is soluble in alkali. Write the structure of (A), (B), (C) and (D). 2



















































(b)















6. (a)

























(b)



(c)







Section-B 4. Account for the following: 1×3=3 (a) Cu2+ salts are coloured while Zn2+ salts are white.















7. (a)

















(b) E° value of the Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+. (c) Transition metals form alloys









(b)























(iii) Draw structure of the semicarbazone of Ethanal. 1×3=3

Section-C









12. Read the following passage and answer the question that follow: The rate of reaction is concerned with decrease in concentration of reactants or increase in the concentration of products per unit time. It can be expressed as instantaneous rate at a particular instant of time and average rate over a large interval of time. A number of factors such as temperature, concentration of reactants, catalyst affect the rate of reaction. Mathematical representation of rate of a reaction is given by rate law: Rate = k[A]x [B]y x and y indicate how sensitive the rate is to change in concentration of A and B. Sum of x + y gives the overall order of a reaction. When a sequence of elementary reaction gives us the products, the reaction are called complex reaction. Molecularity and order of an elementary reaction are same. Zero order reaction are relatively uncommon but they occur under special condition. All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics. 1 + 1 +1 + 2 (a) What is the effect of temperature on the rate constant of a reasons. (b) For a reaction A + B → Product, the rate given by, rate k[A]2 [B]½. What is the order of the reaction? (c) How order and molecularity are different for complex reactions? (d) A first order reaction has a rate constant 2 × 10–3 s–1. How long will 6 g of this reactant take to reduce OR The half life for radioactive decay of 14C is 6930 years. An archaeological artifact containing wood had only 75% of the 14C found in a living tree. Find the age of the sample. [log 4 = 0.6021 log 3 = 0.4771 log 2 = 0.3010 log 10 = 1]





CaO

→ COONa +NaOH  ∆



























































































10. Give reasons: (a) Ammonolysis of alkyl halides is not a good method to prepare pure primary amines. (b) Aniline does not give Friedel-Crafts reaction (c) Although —NH2 group is o/p directing is electrophilic substitution reaction, yet aniline on nitration gives good yield of m-nitroaniline. 1×3=3 11 (a) (i) Which acid of the following pair would you except to be stronger? F—CH2–COOH or CH3—COOH (ii) Arrange the following compounds in increasing order of their boiling points: CH3CH2OH, CH3—CHO. CH3—COOH (iii) Give simple chemical test to distinguish between Benzaldehyde and Acetophenone. 1×3=3 OR (b) (i) Which will undergo faster nucleophilic addition reaction? Acetaldehyde Propanone (ii) What is the composition of Fehling's reagent?









(iii)







Zn(Hg)/Conc HCl

→ (ii) CH3 CH2 CHO 











































(II) Why a negatively charged sol is obtained when AgNO3 solution is added to KI solution? 8. Define transition metals. Why Zn, Cd and Hg are not called transition metals? How is the variability in oxidation states of transition metals different from that of p-block elements 3 9. (a) What happens when 1 ×3 = 3 (i) Propanone is is treated with CH3MgBr and then hydrolysed? (ii) Ethanal is treated with excess ethanol and acid? (iii) Methanal undergoes Cannizzaro reaction? (b) Write the main product in the following reaction? 1×3=3 (i) 2CH3COCl + (CH3)2Cd →



OSWAAL CBSE Sample Question Papers, CHEMISTRY, Class-XII

16

rr

SOLUTIONS 56/5/1



Delhi Set-I

unit of second order reaction.





1. (a) The unit L mol – 1 sec – 1 for rate constant is the



(b) For reaction A → B,



Section-A

Rate of reaction (r)

Solved Paper-2022 ...1

react with concentrated alkali is known as the Cannizzaro reaction. In this reaction, two molecules of aldehydes react where one is reduced to alcohol and the other is oxidized to carboxylic acid.



If the concentration of reactant increased to three times. Rate of reaction (r’)

= k[3A]2

...2



O

Thus, on dividing eq. 1 and 2.









1 = 9

Therefore, rate of formation of B increases to nine times.



(c) Integrated rate equation for zero order reaction is





[ R 0 −R] k= t

Where,

























R0 = Initial concentration of reactant R = Final concentration of reactant

Hydrazine





(Any two)

t = time taken





4-methoxy benzoic acid < benzoic acid < 4-nitrobenzoic acid < 3,4-dinitrobenzoic acid. (Due to presence of electron withdrawing group).

— C=N—NH2

glycol+KOH 180°C

R—

— C=N—NH2

R'













R—

— CH2+N2(g) R' Hydrocarbon



0.693 k



(ii) If D0 > P, the electronic configuration of 5 e 0 as it is associated with d5 ion will be t2g g strong field and low spin situation. Thus, no electron will enter into eg orbital.

(iii) The complex compounds in which all the ligand which are connected with central atom are same or identical are called homoleptic complexes.For example:[Ni(CO)4], [Co(NH3)6]+ 3.

OR (b) (i) Chelate complexes are more stable than unidentate ligand because chelate ligand forms a ring with the central metal ion and are held by strong force of attraction and are less likely to dissociate. However, unidentate ligands are attached to central metal at one point which involves less R—force of attraction and are more likely to glycol+KOH CH2+N2(g) —dissociate. 180°C R' (ii) A series in which ligands are arranged in Hydrocarbon the order of increasing magnitude of crystal field splitting energy (CFSE), is called spectrochemical series.





(b) Cannizzaro Reaction - The self-oxidation reduction (disproportionation) reaction of aldehydes having no a-hydrogens when

k 2.303

5. (a) (i) IUPAC name of K2[PdCl4] is Potassium tetrachloroethylene palladate (II).



R'



t1 / 2 =



— C=O+H2N—NH2 R' Hydrazine Carbonyl compound

R' bonyl pound

Slope =



Butanone 2° alc > 3° alc . 1 (b) Reimer-Tiemann reaction 1 (c) (i) R – CH2OH > RR'CHOH >> RR'R"COH Primary alcohol > Secondary alcohol > Tertiary alcohol As electron releasing effect of alkyl group increases from primary to tertiary alcohol, electron density. IUPAC nomenclature rules to be followed. 1 (ii) 2-Phenylpropan-2-ol 1 OR The reagents used for the oxidation of alcohols are alkaline KMnO4 and K2Cr2O7. The reaction of oxidation of butan-2-ol is:





(c)



















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 2



OR



















(a) Primary structure: Each polypeptide in a protein molecule having amino acids which are linked with each other in a specific sequence. 1 (b) Denaturation: When a protein is subjected to physical change like change in temperature or chemical change like change in pH, protein loses its biological activity. 1







Section - E 33.





OSWAAL CBSE Sample Question Papers, CHEMISTRY, Class-XII

1+1+1







6

































In p-block elements, the difference in oxidation state is 2 and in transition metals the difference is 1. 1















(ii) Cu+, due to disproportionation reaction and low hydration enthalpy. ½+½















(iii) Due to formation of chromate ion/CrO42– ion, which is yellow in colour. 1 (b) Actinoids are radioactive, actinoids show wide range of oxidation states 1+1 



















(a) (i)















OR























= –nFE° ½ = – 2 × 96500 × E° ½ = 0.226 V ½ = E° – 0.059/2 log ( [H+]2 [Cl–]2 / [H2] ) ½ = 0.226 – 0.059/2 log[ (0.1)2 ×(0.1)2 ] / 1 = 0.226 – 0.059 /2 log 10–4 = 0.226 + 0.118 = 0.344 V 1 (Deduct half mark if unit is wrong or not written) (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) 1 [CBSE Marking Scheme 2018]



(a) ΔGo – 43600 E° E





Copper has high enthalpy of atomisation and low enthalpy of hydration, thus the high energy 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 (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 (iii) This is due to the increasing stability of the species of lower oxidation state to which they are reduced. 1





(b) (i)





34.

½ ½

2La + 3S  → La 2S 3 heat









DTb = [3 × 0.512 K kg mol-1 × 1000 × 10 g 1 [111 g mol-1× 200 g] = 0.69 K 1 [CBSE Marking Scheme 2017]







(a) (i) The solution which obeys Raoult’s law over the entire range of concentration. 1 (ii) It is the excess pressure that must be applied to a solution to prevent osmosis. 1 (b) DTb = iKbm Here, m = wB × 1000/MB × wA 1 3 × 0.512 × 10 × 1000 DTb = 11 × 200

(a) (i) MnO24 − ions disproportionate in acidic medium to give permanganate ions and manganese (IV) oxide. ½ 2− + − ½ 3MnO4 + 4H → 2MnO4 + MnO2 + 2H 2O (ii) Lanthanum sulphide if formed.











35.







(b) (i) 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. (ii) 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 OR





[Topper's answer 2018]

[CBSE Marking Scheme 2017]

nnn

SOLUTIONS

Self Assessment Paper-2 Chemistry























18.





Option (D) is correct. 1 Explanation: An ideal solution obeys Raoult's law. Option (A) is correct. 1 Explanation: The C–O–C bond angle in ethers is slightly less than tetrahedral angle due to repulsive interaction between the two alkyl groups in ethers. Option (D) is correct. 1 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. 



17.







16.



C = O group in benzaldehyde.









15.





14.

Section - B 19.

(a) 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 (b) As copper has high energy of atomisation aH° and low hydration energy hydH°, due to which E° value is positive. 1 D



D















8.







7.

13.







6.







5.





4.







 

​



12.







11.

Explanation: Actinoids are 5f block elements so in actinoids, 5f orbitals are progressively filled. Option (D) is correct. 1 Explanation: N, N–dimethylbutan-1-amine Option (C) is correct. 1 Explanation: The unit of rate constant depends upon the order of the reaction. Option (C) is correct. 1 Explanation: It is formed by the loss of 3 electrons, the configuration of element X is [Ar] 3d74s2 Therefore, Atomic number = 27. Option (A) is correct. 1 Explanation: Homoleptic complexes are the ones in which central metal is attached to all identical ligands. In [Co(NH3)6]3+ is attached to 6NH3. Option (A) is correct. 1 Explanation: Thiosulphate or S2O3− is not a chelating agent since it is a monodentate ligand. Option (B) is correct. 1 Explanation: The +R effect of phenyl group decreases the electron density on the carbon atom of 













10.







3.







2.

Option (B) is correct. 1 Explanation: Guanine Option (A) is correct. 1 Explanation: ∆G = –ve E° = +ve Option (A) is correct. 1 Explanation: Let x be the oxidation number of Ni in Ni(CO)4. Total charge on the complex is 0 as it is a neutral complex. The sum of oxidation states of all elements in the complex should equal to 0. Therefore, x+4(0)=0. Hence, x=0. Thus, the oxidation number of Ni in Ni(CO)4 is 0. Option (D) is correct. 1 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. Option (C) is correct. 1 Explanation: [Fe(C2O4)3]3– is most stable as C2O42– acts as the chelating ligands, because it is a bidentate ligand. Option (D) is correct. 1 Explanation: (CH3)3C—I will undergo SN1 reaction most readily as C—I bond is weakest, due to the large difference in the size of carbon and iodine. Option (B) is correct. 1 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. Option (B) is correct. 1 Explanation: 



1.





Section - A

1





Option (B) is correct.



9.







Commonly Made Error Students lose time in giving unnecessary explanation.

OSWAAL CBSE Sample Question Papers, CHEMISTRY, Class-XII CH3CH2OCH2CH3







307.6 K

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 results in decrease in boiling point of water. 2











(b) A:

C6H5N2+Cl–

B : C6H5OH

1 1

[CBSE Marking Scheme 2018]





(a) (CH3)3N < CH3NH2 < (CH3)2NH







25.









24.





390 K







(a) Due to the presence of large number of unpaired electrons, metal-metal interaction is strong whereas mercury does not have unpaired electrons and forms weak metallic bonds. 1 (b) As oxygen can form double bond unlike fluorine. Therefore, MnF4 is highest fluoride but Mn2O2 is highest oxide. 1 21. (a) Presence of five –OH groups (with acetic anhydride)

n-Butylalcohol

The large difference in boiling points of alcohols and ethers is due to the presence of hydrogen bonding in alcohols. 2



20.

Write the cause and consequence of the condition.

CH3CH2CH2CH2OH

Diethyl ether







Answering Tip

 

2

(b) Presence of carbonyl group (with HCN)



(a) 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.







(b) Acetone reacts with alcohol to form a hemiacetal. 1 OR

R











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 ∝ cA or PA = KcA Raoult's law and Henry's law are similar as both gives equation to find partial pressure of gases. PA = KcA PA = KHcA 2

(b) (CH3)2 NH is more basic than (CH3)3 N in an aqueous solution. +I effect will increase in alkyl group that results in increasing the ease of donation of lone pair of electrons. Amine accepts a proton and forms cation which will be stabilised in water by solvation. Higher the solvation by hydrogen bonding, higher will be the basic strength.

(Henry's law)

R—N—H

R >

R—N—R

Therefore, with increase in methyl group, hydrogen bonding and stabilisation by solvation decreases. This net effect results in decrease of basic strength from secondary to tertiary amine. 1









Ethers have low polarity and as a result do not show any association by intermolecular hydrogen bonding. Therefore, ethers have low boiling points and lower than that of isomeric alcohols and almost same as those of alkanes of comparable molecular masses.

—

(Raoult's law)

—

23.





(a) Reverse osmosis will take place and the level of solution will decrease. 1 





22.



1









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



1









OR

Solutions

26.



3

Section - C

(a)

1





1





(b)

(b)



1

1



(c)





+ HI

CH2—O

(c) C H OH+(CH CO) O 6 5 3 2

Acetic Anhydride

∆

CH3COOC6H5+CH3COOH Phenylethanote

+

ICH 2

OR

OH

OH



1



OR (a) By Reimer-Tiemann reaction:



COOCH3 +CH5COCl



+ HCl

Acetyl chloride Phenylethanote



27.





(a) C6H5−CH(OH)−CN

1



(b) 2 CH3COCH2C6H5 + CdCl2

(c) (CH3)2−C(Br)COOH





1



1



[CBSE Marking Scheme, 2019] 1



Detailed Answer: H

CHO

(a)

OH Benzaldehyde Cyanohydrin

Benzaldehyde

N

NaCN / HCl



(b)



O O + CdCl 2 2 CH 3 2 + CdCl2 CH3 1-phenylpropan-2-one

(C6H 5CH 2)2Cd + 2 CH 3COCl (C6H5CH2)2Cd + 2 CH3COCl

1-phenylpropan-2-one

(c)

CH3



H 3C

C H

O C

CH3 (i) Br2/Red P4 (ii) H2O

OH

2-methylpropanoic acid

H 3C

C Br



O C OH

2-bromo-2-methylpropanoic acid

3



OSWAAL CBSE Sample Question Papers, CHEMISTRY, Class-XII

4

..

C6H5NH2 + AlCl3 → [C6H5NH2]+[AlCl3]–











1

3 (b) Aromatic primary amines can not be prepared by Gabriel's pthalimide synthesis because haloarenes have to react with potassium phthalimide and they are little reactive. So, the bond cleasvage does not take place. 1 (c) 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

(a) Aniline does not undergo Friedal-Crafts reaction because aniline being a Lewis base forms a complex with AlCl3, which is Lewis acid. The amino group is not in a position to activate the benzene ring towards electrophilic substitution. Therefore the reaction is not possible





28.





A and D are position isomers.



OR



(a) [Fe(CN)6]3−

Magnetic character – Paramagnetic,





Hybridization –d2sp3 (octahedra),







29.

Spin nature – Low spin complex







(b)

3 [CBSE Marking Scheme, 2016]





Commonly Made Error Some students write incorrect electronic configuration of ions.





Answering Tip Do practice for writing the electronic configuration of metal atom in coordination compounds.

Solutions

30.

5 bond with guanine, while adenine forms hydrogen bond with thymine. As a result, the two strands are complementary to each other. 2

(a) The high energy to transform Cu (s) to Cu2+(aq) is not balanced by its hydration enthalpy. 1 (b) Mn2+ has d5 configuration (stable half-filled configuration). 1 (c) 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









Section - E







33.



(a)

m =ZIt 56 × 2 × t 2.8 = 2 × 96500







[CBSE Marking Scheme 2017]



31.





Section - D

½

t = 4825 s / 80.417 min



(a) A: C6H5COCH3 'A' gives positive DNP test. Therefore, it is an aldehyde or a ketone. Since it does not reduce Tollens’ reagent, 'A' must be a ketone. 'A' responds to iodoform test. Therefore, it should be a methyl ketone. The molecular formula of 'A' indicates high degree of unsaturation, yet it does not decolourise bromine water. This indicates the presence of unsaturation due to an aromatic ring. The molecular formula of 'A' indicates that it should be phenyl methyl ketone (acetophenone). 1 (b) 2-aminopropanal 1 (c) (i) CH3CH2CH2COCH3 will give iodoform test as it has a terminal ketomethyl group. 1 (ii) Tollen’s reagent is also known as Ammonical silver nitrate. 1 OR (i) If an aromatic compound does not decolourise bromine water, it indicates presence of unsaturation due to an aromatic ring. 1 (ii) Tollen's reagent acts as a mild oxidising agent and oxidises the aldehydes to corresponding carboxylic acids and itself gets reduced forming silver mirror. Thus, it shows positive test for aromatic aldehydes. 1 (a) Glucose catabolism yields a Total of 38 ATP. 38 ATP × 7.3 kcal/mol ATP = 277.4 kcal. Glucose has 686 kcal. Thus the efficiency of glucose metabolism is 277.4/686 × 100 = 40.43 % approx 38%. 1 (b) The H-bonds formed between the −NH group of each amino acid residue and the C =O group of the adjacent turns of the α-helix help in stabilising the helix. 1 (c) 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 OR In the helical structure of DNA, the two strands are held together by hydrogen bonds between specific pairs of bases. Cytosine forms hydrogen

½





m1 E 1 = m2 E2





½





2.8 56 2 = × mZn 2 65.3



½

mZn = 3.265 (b) (i) A- Strong electrolyte, B- weak electrolyte













1 1

o (ii) ∧ m for weak electrolytes cannot be obtained o by extrapolation while ∧ m for strong electrolytes can be obtained as intercept. 1 [CBSE Marking Scheme, 2019]



























Detailed Answer: (a) Charge required to deposit 2.8 g of Fe: mass 2.8 g = = 0.05 mol mol Fe =  molar mass 56 g.mol −1





















32.







The quantity of charge is related to current as Q = It Therefore, the time needed to deposit 2.8 g Fe is: Q 9650 C t= = = 4825 s I 2 A











2 F charge is required to discharge 1 mol of Fe2+ ions as Fe, therefore deposition of 0.05 mol Fe will need 96500 C 0.05  × 2 = 0.1F = 0.1F× = 9650 C F



 



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 massof  Zn / charge on zinc ion = molar mass of Fe / charge on iron ion 







mass of Zn = 2.8 g ×











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











65.3 g / 2 = 3.265 g ≈ 3.3g 56 g / 2

















OSWAAL CBSE Sample Question Papers, CHEMISTRY, Class-XII



value of molar conductivity ∧ 0m . The plot of molar







conductivity of weak electrolyte tends to infinity as its concentration approaches zero; it does not 1 intersect the molar conductivity axis. OR (a) Zn(s)| Zn2+ (0.1 M) || Ag+ (0.01M)| Ag(s)

= 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 on dilution complete dissociation of weak electrolyte takes place and thus a sharp increase in molar conductivity is observed, while in strong electrolyte it has already dissociated completely. So on dilution molar conductivity does not rises much. 







(ii) As concentration of strong electrolyte approaches zero, the molar conductivity of the plot intercepts the molar conductivity axis, giving the limiting



6

E°

= −0.76 V

weak electrolyte



Zn 2+ / Zn

E°

= + 0.80 V

emf = ?



Ag+ / Ag

2





0.0591 [ Anode] log n [Cathode]



Ecell = E°cell −

Commonly Made Error



E°cell = E°cathode − E°anode



Zn 2+ / Zn

Some students get confused to find Eocell correctly.



= E°Ag / Ag − E°

= 0.80 – (–0.76) = 1.56 V 0.0591 [Zn 2+] log 2 [Ag+]2

Answering Tip



Ecell = 1.56 −

Understand to identify the Eocathode and Eoanode from given standard reduction potentials.





0.0591 [0.1] log 2 [0.01]2



= 1.56 −



[1+1]





(ii)









alc. KOH HBr, Peroxide → CH3 – CH2 – CH2 – Br (i) CH3 – CH(Br) – CH3 → CH3 – CH – CH=CH2  1-bromopropane







34. (a)



p-nitrochlorobenzene





Commonly Made Error A number of students cannot represent proper reagents used for a particular reaction.

Practice the organic reactions with reference to reagents used. Cl + Cl2



(b) (i)



Cl

Cl Cl

Anhy. FeCl3

+

1 

1







(iii) CH3CH2CH2CH(Br)CH3 + KOH (alc.) → CH3CH2CH=CH CH3





Cl

(ii) CH3CH2Cl + AgNO2 → CH3CH2NO2 + AgCl







Answering Tip

1

Solutions

35.



7

(a) t =

[ A ]0 2.303 log k [ A]

2.303 100 40 = log 75 k





2.303 100 log 20 k

t=

where [R]o = initial concentration, [R] = conc. after time t When half of the reaction is completed, [R] = [R]o/2. Representing, the time taken for half of the reaction





…(i)



…(ii)



Dividing (i) by (ii) 40 2.303 100 = log 75 t k



to be completed, by t1/2, equation becomes: [ R]o 2.303 log k= [ R]o / 2 t1 / 2  





⇒

100 2.303 log k 20



4 40 3 = t 3 log 5

⇒

log



⇒





40 0.6021 − 04.771 = t 0.6991

t1/2 =

 



 

2.303 log 2 k

½ 

t1/2 =

2.303 × 0.3010 k

t1/2 =

0.693 k



 

½







½



The above equation shows that half-life of first order reaction is independent of the initial concentration of the reactant. ½





40 0.1250 = t 0.6991





t=

0.6691 × 40 = 223.712 min. 0.1250



(b) For a first order reaction

t=



k=



2.303 100 2.303 100 log = log 100 − 25 40 75 t

2.303 = (log 4 − log 3) 40

t99% =







=

2.303 = ( 0.6021 − 0.4771) 40





3



=

4.606 k



and





t90% =



=



½



½

2.303 log 100 k 2.303 × 2 k





(b) Order of reaction: The sum of the co-efficients of the reacting species that are involved in the rate equation for the reaction, is called order of reaction. 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 changes hardly. 2 OR (a) For a first order reaction [ R] 2.303 log o , k= t [ R]



2.303 100 log 1 k

=



2.303 × 0.125 = 0.007196 = 7.196 × 10 −3 min −1 40

2.303 a log k a-x

½



2.303 100 log 10 k

½ 

2.303 2.303 log 10 = k k



½





t99%











t90%

=2

t99% = 2 × t90%



½



nnn

SOLUTIONS

Self Assessment Paper-3 Chemistry

Option (A) is correct.

CH2 — CH3

—

1 —

O2 N

1

—



O2 N

4 -(1-Bromoethyl) nitrobenzene

 













13.





Methyl amine

14.

2 + HCl CH 3 NH 2 + HNO2 NaNO  → CH 3 − N 2+Cl −





Option (C) is correct. 1 Explanation: Methylamine reacts with HNO2 to form CH3OH.

Option (B) is correct. 1 Explanation: Asymmetric/Chiral carbon atom is that in which all of its four valencies lie with four different groups or atoms. In molecules (i), (ii) and (iii), all have asymmetric carbon as each carbon has satisfied all four valencies with four different groups or atoms. In molecule (iv) carbon satisfies two of its valencies



15.







9.









2O H → CH 3OH + N 2 + HCl





(dehydrohalogenation)

Option (D) is correct. 1 Explanation: Alcohols are less acidic than phenol. Further electron withdrawing groups (like- Cl) increases the acidity of phenol, therefore, m-chlorophenol is most acidic. Option (B) is correct. 1 2+ Explanation: Cu has one unpaired electron. Therefore, in CuF2 it is coloured in solid state. Option (B) is correct. 1 Explanation: Na metal is basic and alcohols are acidic in nature. Hence, reactivity of Na metal towards alcohols decreases as the acidic strength of alcohols decreases due to steric hinderance of alkyl groups in tertiary alcohol and increase in electron density on an oxygen atom in the hydroxyl bond. 







12.

Option (D) is correct. 1 Explanation: m- and p-Nitrophenols exist as associated molecules due to intermolecular hydrogen bonding whereas o-nitrophenol shows 

alc. KOH









Cl

8.

Trimethylamine





7.





Option (C) is correct. 1 Explanation: 3-Bromo-1-chlorocyclohexene Option (A) is correct. 1 Explanation: Cu(II) has a greater charge density than Cu(I) ion and therefore forms much stronger bonds releasing more energy. That’s why Cu(II) is more stable due to nuclear charge of Cu. Option (B) is correct. 1 Explanation:

Option (B) is correct. 1 Explanation: A 3° amine is one in which all three hydrogen bonds are replaced by organic substituents. Trimethylamine is a tertiary or 3° amine as it is ammonia in which each hydrogen atom is substituted by a methyl group. 





11.





6.







5.

CH— CH3

—

1

[Topper Answer, 2020]



4

Br2, UV light

aromatic halides do not undergo nucleophilic



—

be used for preparation of aromatic amines, as substitution by salt formed by phthalimide. Option (A) is correct. 1 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.

Br

4 - Ethylnitrobenzene

Explanation: Gabriel phthalimide synthesis cannot

4.

with two hydrogen atoms, i.e., similar atom. So, it is not an asymmetric carbon atom. Option (C) is correct. 1 Explanation: 





10.



is

1 correctly







3.







2.

Option (B) is correct. Explanation: Arrhenius equation represented by k = Ae–Ea/RT. Option (A) is correct. Explanation: MnO4-/KMnO4 



1.





Section - A

OSWAAL CBSE Sample Question Papers, CHEMISTRY, Class-XII



When carbon disulphide is added to the acetone, the dipolar interaction between molecule of solute solvent (A – B) than the respective interaction between solute-solute (A – B) and solvent – solvent (B – B) weakens, which leads to increase in vapour pressure. Hence, shows positive deviation in vapour pressure from Raoult’s Law. 1 Minimum boiling azeotrope at a specific composition is formed by this mixture. 1 









23.















(a) Amines behave as a lesius base as they have unshared pair of electrons on nitrogen atom. 1







273 − 278 K (b) C6H6NH2 + NaNO2 + 2HCl → C6H5N2+Cl–





½

½ ½









(a) Molarity is number of moles of solute dissolved per litre of solvent, while molality is number of moles of solute present in 1 kg of solvent. (b) Molarity increases with increase in temperature while molality is not affected by increase or decrease in the tempertsure. 2

26.



CH 3 | CH3 ¾ C ¾ O—CH3 + 2HI | CH 3 CH 3 | CH3 ¾ C ¾ I +CH3I + H2O | CH 3













(a)



Section - C









25.





½

[CBSE Marking Scheme 2018]









Detailed Answer: Raoult's Law for a solution containing volatile components - The Partial vapour pressure of each volatile component 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: (a) The enthalpy of mixing of the pure components to form the solution is zero, ∆mixH = 0. That means no heat is absorbed or released during the mixing of the two pure components. (b) The volume of the mixing also be zero, ∆mixV = 0 That means the total volume of the solution is equal to the sum of the volume of the two components. (c) In pure components, A and B, the intermolecular attractions between solute-solute interactions and solvent-solvent interactions are almost similar to the solute-solvent interaction. 21. (a) Cr2O–7 1 (b) Cerium 1

CH3 – CH2 – CH = CH – CHO







(a) 2-Methylbut-2-enal CH3 | CH3 – CH = C – CHO (b) Pent-2-enal



















24.



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 (a) ∆mixH = 0 (b) ∆mixV = 0 (c) The components have nearly same intermolecular force of attraction (any two) [CBSE Marking Scheme 2019]



20.



+ NaCl + 2H2O 1











Section - B 19.





Disproportionation is the reaction in which an element undergoes self-oxidation and selfreduction simultaneously. For example – 1 + 2+ 2Cu (aq) → Cu (aq) + Cu(s) 1 (Or any other correct equation) OR (a) Due to presence of unpaired electrons in d-orbitals. 1 (b) Due to incomplete filling of d-orbitals. Due to very small energy difference in between (n-1)d and ns-orbitals. 1 [CBSE Marking Scheme 2015]



Option (D) is correct. 1 Explanation: Dimethyl ether is more volatile than ethyl alcohol. Option (B) is correct. 1 Explanation: An elementary reaction is a chemical reaction in which one or more chemical species react directly to form products in a single reaction step and with a single transition state. Molecularity is the number of molecules that participate in the reaction and it is calculated as every single step in a reaction whether it is elementary or complex while order is an experimental quantity. Thus, in elementary reaction, both molecularity and order are same.

OR







18.







17.





Explanation: When a solution of chelating ligand is added to solution containing toxic metal ligands chelates the metal ions by formation of stable complex.



1



Option (A) is correct.







16.

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



22.

intramolecular hydrogen bonding.



2

2-Iodo-2-methylpropane

1

Solutions

3

/ 573 K (b) CH3—CH2 ¾ CH ¾ CH3 Cu  → | OH



(b)





H+

+NaOH



(c)

(c)

1

– HCl



1

Butanone-2

—

CH3—CH2 ¾ C ¾ CH3 + H2 || O

CH2OH

—

CH2ONa

—

CH2Cl







OH —

CHO

—

(i) CHCl3 + aq. NaOH

C6H5 — OH

1 [CBSE Marking Scheme 2015]

(ii) H+





Salicylaldehyde

1

Commonly Made Error

[CBSE Marking Scheme 2016] OR



H+

H22 +H (a) CH3CH = C CH + H22OO

CH3 — CH — CH3 —



OH



Answering Tip

1



27.



(a)

Cl

Cl



+ Cl2

Students often miss out on writing the reagents or conditions of the conversion reactions.



Write the reagents involved in the conversions.



Cl Cl

Anhy. FeCl3

+

1

Cl



(b) CH3CH2Cl + AgNO2 → CH3CH2NO2 + AgCl





1



(c) CH3CH2CH2CH(Br)CH3 + KOH (alc.) → CH3CH2CH=CH CH3



1





OR

Br

—

CH2 — CH3

4

—

Br2, UV light

(a)

O2N

1

—

O2N

4 – Ethylnitrobenzene



—

4 – (1 – Bromoethyl) nitrobenzene



(b)



CH — CH3

—

Na  → 2CH3— CH—CH3 dry ether

CH 3 — CH — CH 3 | CH 3 — CH — CH 3

| Cl

2,3-dimethyl butane



1



1



AgCN

(c) CH3CH2Br → C2H5NC







1



Carbylaminoethane (Ethyl isocyanide) Commonly Made Error





Students sometimes do not draw the structure or not mention the name of the reagent or product.



[CBSE Marking Scheme 2016]

Answering Tip



Write all the steps and reagents involved in the conversion.

OSWAAL CBSE Sample Question Papers, CHEMISTRY, Class-XII



Cathode reaction: Pb2+(aq) + 2e– ® Pb(s) (c) According to Nernst equation:



 Zn 2+    Pb2+   





Ecell = E°cell − 0.0591 log 





It exists as a single It has a double helix chain structure. structure.

3.

The base units are: * Adenine (A) and guanine (G) as purine bases. * Cytosine (C) and uracil (U) as pyrimidine bases.



Do more practice of cell representation and numerical based on Nernst equation.









OR 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. 2 32. (a) Glycosidic linkage is connecting monosaccharide units in polysaccharides. 1 (b) Reducing sugars are carbohydrates that reduce Fehling's solution and Tollen's reagent. All monosaccharides and disaccharides, excluding sucrose, are reducing sugars. 1 (c)





The base units are: * Adenine (A) and guanine (G) as purine bases. * Cytosine (C) and thymine (T) as pyrimidine bases.

(a) 2CH 3  CO  CH 3 Propanone

Ba(OH)

CH 3 | CH 3  C  CH 2 CO  CH 3 1 | OH (Ketol)



(b)



1



(c)

Glycosidic linkage Glycosidic linkage is formed between two monosaccharides. Hydrolysis of glycosidic linkage forms two monosaccharides. It links two carbon atoms through an oxygen atom. It is found in polysaccharides. It can be given as –C–O–C–. 2



OR Glucose on reaction with acetic acid gives glucose pentaacetate which implies that 5 -OH groups are present in it. 2 



(a) Cell representation: Zn(s)|Zn2+(aq)||Pb2+(aq)|Pb(s)







Section - D 31.

Peptide linkage Peptide linkage is formed between two amino acids. Hydrolysis of peptide linkage forms two amino acids. It links carbon and nitrogen atoms. It is found in polypeptides. It can be represented as –CONH–.





1 [CBSE Marking Scheme, 2019]















30.



(Any one) 1 1 1







Thymine is present in DNA. (b) Amylose.















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

2.



The sugar present in Sugar present in DNA is RNA is ribose. deoxy- ribose.





DNA

1.











RNA

2

Commonly Made Errors







= 0.63 – 0.02955 × log 5 = 0.63 – 0.02955 × 0.6990 = 0.63 – 0.0206 = 0.6094 V





S. No

0.1 0.0591 log 2 0.02







2

Ecell = [– 0.13 – (– 0.76)] -





DIBAL - H (b) CH3–CH = CH – CH2 – CN (i)  → (ii) H 2 O Pent-3-enenitrile CH3 – CH = CH – CH2 – CHO Pent-3-ene-1-al 1 (c) Propanal and propanone can be differentiated by Tollen’s reagent, i.e., propanal will give silver mirror but propanone will not. CH3 – CH2 – CHO + 2[Ag(NH3)2]+ → CH3 – CH2 – COO– + 2Ag↓+ H2O + 4NH3 (Silver mirror) (or any other correct test) 1 29. (a) Structural differences between DNA and RNA

1









(b) Anode reaction: Zn(s) ® Zn2+(aq) + 2e–



(a) In F–CH2–COOH, fluorine is more electron withdrawing and has stronger –I effect than chlorine in Cl–CH2– COOH. So, F–CH2COOH is more acidic than Cl–CH2COOH hence its pKa value is lesser than Cl–CH2COOH. 1 



28.



4

1

Solutions

5

34.

Section - E 33.





(a) For a first order reaction



[ R] 2.303 log 0 , where [R]0 = initial concentration, t [ R]

k=

(a)



Double Salt

Complex Compound

They dissociate into They do not dissociate simple ions completely into simple ions when when dissolved in water. dissolved in water. e.g., KCl.MgCl2.6H2O.

[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:



e.g., [Co(NH3)6]Cl3

k=





2



(b) In lead poisoning, EDTA (Ethylene diamine tetraacetate) is used as ligand which is hexadentate, i.e., the denticity of EDTA is 6. EDTA binds with metal in octahedral manner by two N-atoms and four acetate oxygen atoms. 1



t1/2 =

2.303 log 2 k

t1/2 =

2.303 × 0.3010 k

t1/2 =

0.693 k



⇒







⇒













(c) Coordination Number = 6, Oxidation State = +2 ½+½ [CBSE Marking Scheme 2018] (d) Formation of stable complex with a polydentate ligand due to stronger bonding is known as chelate effect. 1 [CBSE Marking Scheme 2015]



OR





⇒



(a) Diamminedichloridoplatinum(II)

1

(b) [Co(NH3)4(H2O)Cl] (NO3)2

1









(c) The magnitude of the (stability or formation) equilibrium constant for the association quantitatively expresses the stability. Thus, in a reaction of the type:





























(b)

2





























K1 = [ML]/[M][L]















K2 = [ML2]/[ML][L]





ML2 + L  ML3

A = pr = 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 = [1.27 × 10–2 S cm–1] × 1000/0.05 mol/cm3 = 254.77 S cm2 mol–1 ½ [CBSE Marking Scheme 2017]









½



Larger the stability constant, higher the proportion of ML4 that exists in solution. Presence of free metal ions in the solution is rare 50 M will usually be surrounded by solvent molecules which will complete with the ligand molecules, L, and be successively replaced by them. This can be represented as follows : ML + L  ML2

½











½

The above equation shows that half-life of first order reaction is independent of the initial concentration of the reactant. ½ (b) Due to high activation energy 1 (c) Rate = k[A2]o[B2]o 1 (d) The unit of rate constant for a zero order reaction is mol L–1s–1. 1 35. (a) ‘B’ is a strong electrolyte. 1 A strong electrolyte is already dissociated into ions, but on dilution interionic forces overcome, ions are free to move. So, there is slight increase in molar conductivity on dilution. 1

M+L  ML









M + 4L  ML4





[ R]0 2.303 log [ R]0 / 2 t1 / 2

K3 = [ML3]/[ML2][L]









ML3 + L  ML4



K4 = [ML4]/[ML3][L]



where K1, K2, etc., are referred to as stepwise stability constants.



Alternatively, the overall stability constant thus is depicted as :

OR

(i)









M + 4L  ML4

b4 = [ML4]/[M][L]4



The stepwise and overall stability constant are therefore related as follows:





b4 = K1.K2.K3.K4 or more



bn = K1.K2.K3.K4 ........... Kn

EQCd2+ / Cd = - 0.40 V





EQCr3+/Cr = - 0.74 V

The galvanic cell of the given reaction is depicted as : 3+ 2+ Cr(s)|Cr(aq) ||Cd(aq) |Cd(s) Now, the standard cell potential is EQ = EQR - EQL = -0.40 - (-0.74) = + 0.34 V Q Q rG = – nFE cell  









  





3





Δ



OSWAAL CBSE Sample Question Papers, CHEMISTRY, Class-XII

 





 

 

 

 

 





  

 



 

 



= – RT ln K



Δ

= 0.5073

K = anti-log (0.5073)

= 3.2 (approximately)





2½



= 3.169 × 1034







K = antilog (34.501)







= 34.501





ø rG



Δ

 



Again,

 



 

= – 196.86 kJ mol –1

Δ





 

= – 196860.48 J mol –1

The galvanic cell of the given reaction is depicted as : Fe2+ (aq)│Fe3+ (aq)││Ag+ (aq)│ Ag(s) Now, the standard cell potential is Eø = EøR – EøL = 0.80 – 0.77 = 0.03 Here, n = 1. ø ø Then, rG = -nFE cell = - 1 × 96500 C mol -1 × 0.03 V = - 2,895 mol -1 = - 2.89 kJ mol -1 ø Again, rG = - 2.303 RT ln K ∆ rG log K = − 2.303 RT −( −2894.61) = 2.303 × 8.314 × 298

                 

 



= – 196860.48 CV mol –1



× 0.34 V



 

 

 

 

= –6 × 96500 C mol–1





Δ



Q rG





Then,

⇒ ∆r G = − 2.303RT ln K ∆r G ⇒ log K = − 2.303RT (−196.86 × 10 3 ) = 2.303 × 8.314 × 298







 



EQcell = +0.34 V

θ



Eø Fe3+ / Fe2+ = 0.77 V Eø Ag+ / Ag = 0.80 V

(ii)





In the given equation, n =6 F = 96500 C mol –1









6

2½

nnn

SOLUTIONS

Self Assessment Paper-4 Chemistry

electrochemical cell behaves like an electrolytic cell. Eexternal > Ecell Option (B) is correct. 1 Explanation: KCl (K+ + Cl–) and NaCl (Na+ + Cl–) ionize to give two ions and K2SO4 (K+ + SO2– 4 ) ionize to form three ions. So, van’t Hoff factors for KCl, NaCl, and K2SO4 are 2, 2 and 3 respectively. Option (B) is correct. 1 Explanation: Higher the molar mass, lower is the freezing point. Option (A) is correct. 1 Explanation: Value of Henry's constant increases with increase in temperature. Option (D) is correct. 1 Explanation: Interstitial compounds are chemically inert. Option (A) is correct. 1 Explanation: Cu2+ oxidises iodide to iodine hence cupric iodide is converted to cuprous iodide. Option (A) is correct. 1 Explanation: The half life period in first order 2.303 log 2 and is reaction, is given by t1/2 = k 



































16.











15.







independent of initial concentration. Option (B) is correct. 1 Explanation: Phenol is stronger acid than ethanol as phenoxide ion is stabilised by resonance whereas no such stabilisation occurs in ethoxide ion. Sodium ethoxide can be prepared by reaction of ethanol with sodium. Option (B) is correct. 1 Explanation: F– is highly electronegative atom which will withdraw electrons from nitrogen and create deficiency to donate electrons whereas –CH is electron donating group that will increase electron density on nitrogen to donate. Thus, N(CH3)3 is strong field ligand than NF3.





17.





4 ∆o 9

Section - B



Option (C) is correct. 1 Explanation: [Co(NH3)6]2(SO4)3 Option (C) is correct. 1 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 1.1 V. 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



(a) Silver can exhibit +2 oxidation state, wherein it will have incompletely filled d-orbital. 1 (b) Much higher third ionisation energy of Mn where the required change is from d5 to d4. 1 [CBSE Marking Scheme 2017]









19.







9.

18.











Explanation: ∆t =





8.





14.











m = 0.0036 Option (C) is correct. 1 Explanation: It is formed by the loss of 2 electrons, the configuration of element X is [Ar] 3d5 4s2. Therefore, Atomic number = 27. Option (C) is correct.



7.



13.









6.

10.

12.







5.







4.







3.







2.

11.



Option (B) is correct. 1 Explanation: When a small amount of solute is added to its solution, it does not dissolve and get precipitated and this type of solution is called as supersaturated solution. Option (C) is correct. 1 Explanation: Instantaneous reaction occurs in the smallest time interval, i.e., instantaneous rate of reaction at 40 s is rate of reaction during a small interval of time close to 40 s. Volume change during a small time interval close of 40 s, i.e., Instantaneous rate of reaction = Change in volume/Time interval close to 40 s Option (C) is correct. 1 Explanation: Coordination no. of platinum is 6, Oxidation state is +2, Pt(II). Option (C) is correct. 1 Explanation: Bromine water is a mild oxidising agent. So, it oxidises glucose into gluconic acid. Option (C) is correct. 1 Explanation: [Co(NH3)5Cl]Cl2 → [CO(NH3)5Cl]+(aq) + 2Cl−(aq) So, total three ions are produced from the given complex in solution. Option (B) is correct. 1 0.052 1000 × Explanation: m= 180 80.2



1.







Section - A

OSWAAL CBSE Sample Question Papers, CHEMISTRY, Class-XII

2



1+1





(b)









alc. KOH HBr, Peroxide → CH3 – CH2 – CH2 – Br (a) CH3 – CH(Br) – CH3 → CH3 – CH – CH=CH2  2-bromopropane 1-bromopropane





20.





Commonly Made Error Students often forget to mention all the reagents along with the reactants.







(a) —NH2 group of aromatic amines strongly activates the aromatic ring through delocalisation 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 OR

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 







22.

(b) Due to resonance, the lone pair of electrons on the nitrogen atom in CH3CONH2 is delocalised over the keto group. – O O + CH2 — C — NH2 CH3 — C = NH2





(a) Because it has incompletely filled d orbitals in one of its oxidation state (Cu2+). 1 (b) 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] 



21.

Reagents involved in each step must be shown.







Answering Tip





1





O || (a) Ar/R — C — NH2 + Br2 + 4NaOH —→Ar/R—NH2 + Na2CO3 + 2NaBr + 2H2O ∆



1



(where R = alkyl group, Ar = aryl group)







→ Ar/R—NC + 3KCl + 3H2O (b) Ar/R—NH2 + CHCl3 + 3KOH 

[CBSE Marking Scheme, 2016]



O || R — C — NH2 + Br2 + 4NaOH → R — NH2 + Na2CO3 + 2NaBr + 2H2O (Aqueous or alcoholic)

1° Amine

Example,















Detailed Answer: (a) Hoffmann – bromamide degradation reaction: When an amide is treated with bromine in aqueous or ethanolic solution for 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.

1

Solutions

3

(b) 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





23.



(a) CH3CH2CH(Br)CH3 1 (b) CH3CH2CH2CH2Br 1 [CBSE Marking Scheme, 2019]

 

8

Electronic configuration of Ni = [Ar] 3d 4s Ni2+ = [Ar] 3d8





































2



OR (a) 2-Bromopentane 1 (b) 2-Bromo-2methylbutane 1 [CBSE Marking Scheme 2017]









Since CN– is a strong field ligand. So. back pairing of e–s will occur





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



24.

Understand the variation in reactivity of 1°, 2° and 3° haloalkanes.



(a) 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 (b) In non-ideal solution, negative deviation shows the formation of maximum boiling azeotropes. 1





(a) (i) Due to-I effect of X, the ring set partially deactivated. (ii) They fail to form Hydrogen bonds with water / more energy is required to break hydrogen. 1 (b) 2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane 1 [CBSE Marking Scheme 2017]

















Detailed Answer: (a) (i) It is due to the fact that non bonding pair of electrons on the halogen are in conjugation with the ring causing resonance stabilisation of halo arenes by delocalisation of electrons. 1





Section - C 26.









Hybridization: dsp2 1 Magnetic character: Diamagnetic 1 Spin nature: Low spin1 [CBSE Marking Scheme 2017]















Detailed Answer: (a) 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: t32ge2g (b) 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. 27. (a) A :C6H5MgBr B : C6H5COOH C : C6H5COCl ½×3 (b) A : CH3CHO B : CH3CH(OH)CH2CHO C : CH3CH=CHCHO ½×3

















25.

(a) It is the magnitude of difference in energy between the two sets of d orbital, i.e., t2g and eg 1 t32ge2g 1 (b) 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]







 









Hybridisation: dsp2 Since, there is no unpaired electron present, so the complex is diamagnetic in nature. As, inner-d-orbital are involved in complex formation, so the spin nature is low spin. OR











 



Detailed Answer: [Ni(CN)4]2– Ni is in +2 oxidation state.

















 



 



OSWAAL CBSE Sample Question Papers, CHEMISTRY, Class-XII

4



(CHOH)4+5(CH3CO)2O





Pyridine Acetylation

(CH O C CH3)4 O CH2 O C CH3 Glucose pentaacetate





O

1½ 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



29.









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





28.

CHO









CHO

(a) CH3CH2CH3 1 (b) C6H5COONa + CHI3 ½+ ½ (c) CH4 1 [CBSE Marking Scheme 2015]



OR



[CBSE Marking Scheme, 2019] ½ × 6



Detailed Answer:













(a) Those substances which dissociates into electrically conducting ions in molten state or in aqueous solution. 1 (b) The molar conductivity of an electrolyte at infinite dilution is equal to sum of the conductivities of the individual ions. 1 (c) L°CH3COOH = L°CH3COO- + L°H+ ½ = 40.9 + 349.6 = 390.5 S cm2/mol ½ Now, a = Lm/L0m ½ = 39.05/390.5 = 0.1 ½



























(a) Since, Ti /Ti3+ has lower reduction potential than Fe3+/Fe2+, it cannot be reduced in comparison with Fe3+/Fe2+ ions. ½ Hence, Ti4+ cannot oxidise Fe2+ to Fe3+. ½























32.

Section - D 4+





















31.

















Detailed Answer: (a) In starch, the glucose monomers are in a configuration while in cellulose, the glucose monomers are in b configuration. Starch is a polymer consisting of amylose and amylopectin while cellulose is a long chain composed only of b-D-glucose units. 1 (b) Phosphodiester linkage between the 5’ and 3’ carbon atoms of sugar moieties is present in Nucleic acids. 1 (c) Example of fibrous protein—Collagen, keratin, myosin. (Any one) 1

(b) As the value of reduction potential increases, the stability of +2 oxidation state increases. ½ Therefore, correct order of stability is: Cr3+/Cr2+ < Fe3+/Fe2+ < Mn3+/Mn2+ ½ (c) (i) Mercury cell is used in hearing aids. (ii) The products of electrolysis depend on standard electrode potentials of the different oxidising and reducing species present in the electrolytic cell. 2 OR Inert electrodes do not participate in the reactions occurring on their surface. They do not participate in electron transfer but provide an interface for reaction. 2 



(a) 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 (b) Phosphodiester linkage 1 (c) Fibrous protein – Keratin / myosin / collagen. Globular protein – Haemoglobin / insulin. ½ + ½ [CBSE Marking Scheme, 2016] 





30.





3

Solutions

5

Commonly Made Error



(b) Given: Initial pressure, P0 = 0.30 atm Pt = 0.50 atm t = 300 s







Students sometimes copy wrong data from the question in a hurry.













Answering Tip

Write the working formula in each step followed by value assigned for each entity. Give appropriate unit along with the answer.



Rate constant,



k=





=

2.303 0.30 log 300 2 × 0.30 − 0.50

=

2.303 0.30 log 300 0.60 − 0.50

=

2.303 0.30 log 300 0.10





−5

1000 cm L × 4.1 × 10 S cm 10 −3 mol L−1

∧m =

½



3 −1

−1



∧ ∧

α=

α=



–1

= 41 S cm mol







c m 0 m



41 = 0.105 390.5









½









1 1 [CBSE Marking Scheme 2017]





(i)

= 3.6 × 10–3 s–1 1



[CBSE Marking Scheme, 2016]



Commonly Made Error



Students tend to overlook or incorrectly copy the data.

Answering Tip



Read the concept, understand and practice the interpretation of graphs of various orders.

34. (a)

= 0.0036s–1

(deduct [½] mark if unit is not written)

Answering Tip

1.099 300 s



Students often misinterpret graph either in hurry or overlook the details.



=

Commonly Made Error



1 

2.303 × 0.4771 300

½ 

(a) (i) First order. (ii) s-1/time-1





2.303 log 3 300

=



Section - E 33.

=

½ 

2



1 



OR 1000κ ∧m = C

P0 2.303 log t 2 P0 − Pt

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.

o-Nitrophenol is steam volatile due to intramolecular hydrogen bonding while p-nitrophenol is less volatile due to intermolecular hydrogen bonding. 1





(ii) Due to the formation of stable intermediate tertiary carbocation / CH3O– being a strong base favours elimination reaction. 1







O– Na+

(b) (i)

—

OH

CHO H+

—

CHCl3 + aq NaOH

—

—

—

OH

CHO





(c)



1







(ii) (Award 1 mark if attempted in any way) 1 Add neutral FeCl3 to both the compounds, phenol will give violet colouration while ethanol does not. 1 [CBSE Marking Scheme, 2019]









OSWAAL CBSE Sample Question Papers, CHEMISTRY, Class-XII

6 Detailed Answer:





(a) (i) o-Nitrophenol forms intramolecular H bond whereas molecules of p-Nitrophenol get associated through intermolecular H bond. During boiling, the strong intermolecular H bonding increases the boiling point but intramolecular H bonding cannot do so.





1

ene 2-methylprop-1-eue

As the given alkyl halide is tertiary in nature, therefore on reaction with sodium methoxide, elimination reaction takes place in place of substitution and hence, alkene is formed as a major product. Therefore, in the given reaction, 2-methylpropene is formed as a major product.





(b) (i) The Reimer-Tiemann reaction is a chemical reaction used for the o-formylation of phenols.



3

heat

CI



2

NaOCH3

(ii)

(ii)



OH 1 2

OH

R

6 +

3

OH

R – Cl

5

AlCl3

+

–HCl

4 Phenol

o- isomer



Friedel crafts alkylation involves formation of carbocation intermediate derived from halide, since phenyl carbocation is highly unstable. (c) Phenol being more acidic than ethanol will turn the blue litmus paper red but ethanol doesn’t. Haloform test is also used to distinguish ethanol and phenol. Ethanol when treated with NaOH and iodine, iodoform is obtained which is yellow needle shaped. But phenol does not show positive haloform test.









1

(ii) HOOC CH2 CH2 CH2 CH2 COOH O





The mechanism of the reaction of HI with methoxymethane involves the following steps: Step 1: Protonation of methoxymethane:

NOH

(a) (i)

35.

OR











R p-isomer

1

C NH

1









(iii)

Step 2: Nucleophilic attack of I–:

C

O





(b) (i) Reaction: Hell-Volhard- Zelinsky reaction. IUPAC name: 2-Bromopropanoic acid. (ii) Reaction: Rosenmund reduction reaction.













½ ½ ½











Step 3: When HI is in excess and the reaction is carried out at a high temperature, the method formed in the second step reacts with another HI molecule and gets converted to methyliodide









IUPAC name: Benzaldehyde. OR

(a) (i) CH3 — CH2 — CH2 — CH2OH Butan-1-ol

½





O KMnO4









5

CH3CH2CH2CHO

O KMnO4

CH3CH2CH2COOH Butanoic acid



1

Solutions





7

(ii)

(ii) As the compound reduces Tollen’s reagent and undergoes Cannizzaro reaction, it is an aldehyde and not a ketone. (c) On vigorous oxidation, it gives 1, 2-benzenedicarboxylic acid. So, it must have an alkyl group at ortho position with respect to CHO group on the benzene ring. (d) Molecular formula suggests that it should be













1







( (

)

)





2-ethyl benzaldehyde.

(b) (i) It is an aldehyde or ketone as it forms 2, 4-DNP derivative.





Commonly Made Error

CHO

o





C2H5

Students get confused in aromatic reactions and are not able to analyse that the given molecular formula will form aromatic rings.

3

Answering Tip



Draw a rough structure of the formula to get a clear idea of the reactions.

nnn

SOLUTIONS

Self Assessment Paper-5 Chemistry

Molarity =







Option (B) is correct. 1 Explanation: Phenol and benzoyl chloride in the presence of pyridine. Option (B) is correct. 1 Explanation: 0.059 E Mg 2+ /Mg = E0 Mg 2+ / Mg + log  Mg 2 +  2





16.















17.



9.











8.

Option (C) is correct. 1 Explanation: In equilibrium state, the rate of dissolution of a solid solute in a volatile liquid solvent is equal to the rate of crystallisation. Option (A) is correct. 1 Explanation: [Co(ONO)3 (NH3)3] Option (A) is correct. 1 5 Explanation: Moles of NaOH = = 0.125 mol 40



7.





straight line with a positive slope and intercept E° Mg 2+ /Mg .

Option (B) is correct. 1 2+ Explanation: In the complexes, Co exists as Co and Fe as Fe2+. Both of the complexes become stable by oxidation of metal ion to Co3+ and Fe3+. 





Compare this equation with the equation of straight line y = mx + c. The graph of E 2+ vs. log [Mg2+] is a Mg /Mg











15.









6.







5.



14.

[Topper Answer, 2020]









13.









12.









4.

Option (A) is correct. 1 Explanation: Copper will not liberates hydrogen from acids. Cu + H2SO4 → Cu2SO4 + SO2 + 2H2O 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O Option (D) is correct. 1 Explanation: Aldehydes with no α-hydrogen undergo Canizzaro reaction. Option (B) is correct. 1 Explanation: R − X + NaOH ® R − OH + NaX

Option (C) is correct. 1 Explanation: 2KMnO4 + KI + H2O → 2KOH + 2MnO2 + KIO3 Option (B) is correct. 1 Explanation: Electromotive force (emf) is the difference between the electrode potentials of two electrodes when no current is drawn through the cell. Option (B) is correct. 1 Explanation: Nucleic acids are the polymers of nucleotides in which nucleic acids are linked together by phosphodiester linkage. Option (A) is correct. 1 Explanation: Phenol being more acidic reacts with sodium hydroxide solution in water to give sodium phenoxide which is resonance stabilised. Alcohols are very weak acids. Option (B) is correct. 1 Explanation: Conversion of alkyl halides into alcohols involves substitution reaction. Option (A) is correct. 1 Explanation: Nitro group being electron withdrawing group, decreases the electron density of ring hence increases the reactivity of haloarenes towards nucleophilic substitution. Option (A) is correct. 1 Explanation: Since, en (ethylene diamine) is a chelating ligand/bidentate ligand, [Co(en3)]3+ is a more stable complex as compared to the other one due to the formation of chelating ring. 





3.











2.



11.

2O H → CH 3OH + N 2 ↑ + HCl



methyl amine

0.125 × 1000 = 0.278 mol/L 450





10.

2 + HCl CH 3 NH 2 + HNO2 NaNO  → CH 3 − N +2 Cl −

450 1000



Option (B) is correct. 1 Explanation: Methylamine reacts with HNO2 to form CH3OH and nitrogen gas is evolved.

Volume of solution in litres=





1.





Section - A

OSWAAL CBSE Sample Question Papers, CHEMISTRY, Class-XII



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.

Student get confused in order of reactions.

Answering Tip

22.





 



























1 1 [CBSE Marking Scheme 2017]

[CBSE Marking Scheme, 2019] ½ × 4 

Dry Cell/Leclanche cell 1 Anode: Zn(s) → Zn2+ + 2e– ½ Cathode: MnO2 + NH4+ + e– → MnO(OH) + NH3 ½ [CBSE Marking Scheme, 2017] 











(b) Presence of a carbonyl group (with HCN) When glucose reacts with HCN, it forms glucose cyanohydrin. It shows the presence of carbonyl group (>C=O) in the open structure of glucose.





25. 1







(a) [CoF6]3– (b) [Co(en)3]3+ (c) [Co(en)3]3+ (d) [CoF6]3–



Acetic anhydride



(a) K3[Al(C2O4)3] (b) [CoCl2(en)2]+

24.







= 444.379 s 2 (a) 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. (CH3CO)2O

Detailed Answer: The cell which is used in the transistors is Dry cell. At anode: Zn(s) → Zn2+ + 2e– 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

1

















1

[CBSE Marking Scheme 2017] OR



2.303 × 0.2219 1.15 × 10 −3

(b) [Pt(NH3)4Cl(NO2)]SO4



=

1



2.303 5 log 1.15 × 10 −3 3

(a) Na[Au(CN)2]



=

(a) It is a zero order reaction, molecularity = 2 (Bimolecular reaction) (b) Unit of k = mol L–1 s–1. [1 × 2 = 2] [CBSE Marking Scheme 2016]

23.











OR Initial amount = 5 g Final concentration = 3 g Rate constant (k) = 1.15 × 10–3 s–1 We know that for a First order reaction, [R ] 2.303 t= log initial k [R final ]

Always write the working formula followed by the value substitution for each entity. Do not forget to mention units wherever required.





Learn the units of various order reactions and also try and understand the underlying concept.







Answering Tip

20.











Commonly Made Errors

Commonly Made Error





1 1 [CBSE Marking Scheme 2017]















(a) Zero order (b) Pseudo-first order



Section - B 19.

DrG° = -nFE°cell, n = 6 = -6 × 96500 C/mol × 0.30 V = -173700 J/mol = -173.7 kJ/mol 1 E°cell = 0.059/n × log Kc log Kc = 0.30 × 6/0.059 = 30.5 1 [CBSE Marking Scheme 2017]

21.





Option (C) is correct. 1 Explanation: 'D' corresponds to the position of –OH group on the right side on the farthest asymmetric C-atom.





18.



2

Solutions

3

Section - C 26.



=

(a) HCl: Graph 1, NaCl: graph 2, NH4OH : Graph 3





(b) NH4OH is a weak electrolyte, thus it show a curved decreasing graph.







Among HCl and NaCl , HCl is stronger electrolyte thus it is above the weaker electrolyte (NaCl). OR 2+



or,

2 mol

t=







1 mol



Cu 2 + ( aq .) + 2e −  → Cu( s) (2 g given) 2 mol

3039.37C = 1519.68 s 2

= 25 min. 33 s

1



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 mol





= 2A × t = 3039.37 C





1



Let 2 A of current be passed for time t, quantity of electricity used

−

Zn ( aq ) + 2e  → Zn(s)



96500 × 2 = 3039.37 C 63.5

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 g mol

Q 63.5 g of Cu is deposited by electric charge = 96500C ∴ 2 g of Cu is deposited by electric charge



Amount of Zn deposited:

65 = 2 × 2 = 2.0472 g 63.5 2

–1





1











[Topper’s Answer 2020]

27.



(a) Phosphodiester linkage between the 5’ and 3’ carbon atoms of sugar moieties is present in Nucleic acids. 1



(d)





(b)

CHO (CHOH)4 CH2OH

COOH Br2 water

(c) Example of fibrous protein—Collagen, keratin, myosin. (Any one) 1



CHO (CHOH)4

(CHOH)4 CH2OH

CH2OH

1 





CH NH2OH

N

OH

(CHOH)4 CH2OH

1

OSWAAL CBSE Sample Question Papers, CHEMISTRY, Class-XII





k2 = 0.693/20, k1 = 0.693/40







CH3 | CH3 – CH– CH2 – OH CH3 | B: CH3 – CH – COOH CH3 | C: CH3 – C = CH2



29.

A:

log



1











log 2 =  

 320 − 300  Ea   2.303 × 8.314  320 × 300  



















 



3 1 ½+½





1



½ ½













Ea = 27663.8 J/mol or 27.66 kJ/ mol (a) CH3CH2CH3 (b) 2CH3COCH2C6H5 + CdCl2 (c)

30.





CH3 | D: CH3 – C – CH3 | OH A and D are position isomers.



½











 





½

Ea  1 1  k2 =  −  k1 2.303R  T1 T2 

k2/k1 = 2

 







28.



4

Section - D 1





(a) H3C-Br + AgF → H3C-F + AgBr

(b)

CH3 + HI







31.

I

1-iodo-1 -methyl cyclohexane 

According to Markovnikov's rule, iodine will add to the carbon atom having less number of hydrogen atoms. 1



(c) (i)

1 [Topper’s Answer 2017]











1-methyl cyclohexene

Markovnikov’s rule

CH3



another between the reactants. 1 OR 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. Increasing order of boiling point of given compound (c) < (a) < (b) 2



(a)



















1

carbon and therefore, decrease the magnitude of positive charge on it and make it less reactive toward nucleophilic attack. 1 (ii) Steric effect In ketones, the bulk of two alkyl groups also hinders the approach of the nucleophile to the carbonyl carbon. 1 OR Oximes are a class of organic compounds having the general formula RR′CNOH, where R is an organic side chain and R′ is either hydrogen or an organic

This is called Etard reaction. (b) This is because polarity of the carbonyl group is reduced in benzaldehyde due to resonance. Hence, it is less reactive towards nucleophilic addition reaction. 1 (c) Ketones are less reactive than aldehydes due to following facts: (i) Electron releasing effect In ketones, the carbonyl carbon is attached to alkyl groups which are electron releasing in nature. These alkyl groups push electrons towards carbonyl









32.















Detailed Answer: 2-Bromo-2-methylbutane as tertiary alkyl halide is most reactive for elimination. (ii) The given reaction is halogen exchange reaction, also called the Finkelstein reaction, is an SN2 reaction (Substitution Nucleophilic Bimolecular reaction) that involves the exchange of one halogen atom for

Solutions

5

side chain. If R′ is H, then it is known as aldoxime and if R′ is an organic side chain, it is known as ketoxime. Ketals are gem−dialkoxyalkanes in which two alkoxy groups are present on the same carbon atom within the chain. The other two bonds of the carbon atom are connected to two alkyl groups. 2



34.



Section - E

(a) (i) Cr3+/Cr2+ has a negative reduction potential. Hence, Cr3+ cannot be reduced to Cr2+. Cr3+ is most stable. Mn3+/Mn2+ has large positive E° value. 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 (ii) 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 (b) (A) = MnO2 (B) = K2MnO4 (C) = KMnO4 (D) = KIO3 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O (A) (B) 3MnO42– + 4H+→ 2MnO4– + MnO2 + 2H2O (C) 2MnO4– + H2O + KI → 2MnO2 + 2OH– + KIO3 (A) (D) 3





33.

= Kf m ½ = w2 × 1000/M2 × W1 = Kf × 10 × 1000/342 × 90 1 = 12.3 K kg/mol ½ = 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 (b) (i) Number of moles of solute dissolved in per kilo gram of the solvent. 1 (ii) Abnormal molar mass: If the molar mass calculated by using any of the colligative properties to be different than theoretically expected molar mass. 1 [CBSE Marking Scheme 2017]





(a) Here,





DTf m 273.15 - 269.15 Kf DTf

























































































OR









OSWAAL CBSE Sample Question Papers, CHEMISTRY, Class-XII



6

[Topper’s Answer 2015] 5















Detailed Answer: (a) (i) The valence electrons of transition metals are in (n-1)d and ns orbitals. As there is almost little energy difference between these orbitals, both the energy levels can be used for bond formation. Thus, they exhibit variable oxidation states. (ii) Because they contain fully filled d-orbitals, no unpaired d electrons are present resulting in weak metallic bonding.

5







35.





(a) (i) Ar/ R-CONH2 + Br2 + 4 NaOH → Ar/ R-NH2 + 2NaBr + Na2CO3 + 2 H2O − 278 K (ii) C H NH + NaNO + 2 HCl 273   → C6 H 5 N +2 Cl − + NaCl + 2 H 2 O 6 5 2 2





1

1 (or any other correct equation)







OR



1







(iii)

Solutions







7

(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:





..

CH3 – N – H + H

(b) (i)

CH3 (2° amine)

⊕

 

(acidic)

..

⊕

(CH3)3 – N + H (acidic) (3° amine)

H CH3+ – N – H CH3 (Salt) More stable CH3 CH3 – N⊕– H CH3 (Salt) Less stable

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) Aromatic 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.







2



nnn

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