Inorganic Chemistry, 1Ed [1 ed.] 9781259062858

Table of contents :
Title
Contents
1 Structure of Atom
2 Nuclear Chemistry
3 Chemical Bonding
4 Molecular Symmetry
5 Redox Reactions
6 Non-aqueous Solvents
7 Extraction of Elements
8 Periodic Table and Periodic Properties
9 Hydrogen and its Compounds
10 Chemistry of Group 1 Elements
11 Chemistry of Group 2 Elements
12 Chemistry of Group 13 Elements
13 Chemistry of Group 14 Elements
14 Chemistry of Group 15 Elements
15 Chemistry of Group 16 Elements
16 Chemistry of Group 17 Elements
17 Chemistry of Group 18 Elements
18 Chemistry of d-block Elements
19 Chemistry of Elements of 3d Series
20 Chemistry of Elements of 4d Series
21 Chemistry of 5d Series
22 Chemistry of Lanthanides and Actinides
23 Coordination Compounds-I Basics Concepts: Nomenclature and Stereochemistry
24 Coordinate Compounds-II Theories of Bonding
25 Coordination compounds III: Quantitative Basis of Crystal Field Theory
25 Coordination Complexes IV: Spectroscopic and Magnetic Properties of Coordination Compounds
27 Coordination compunds - The Reaction Mechanisms of Transition - Metal Complexes
28 Complexes of Pi - Acceptor Ligands
29 Chemistry of Organaometallic compounds
30 Metak Clusters
31 Inorganic Nomenclature
32 Inorganic Polymers
33 Bioinorganic Chemistry
34 Pollution
35 Analytical Chemistry
Index

Citation preview

INORGANIC CHEMISTRY

ABOUT THE AUTHORS Rajni Garg is working as Head, Department of Applied Chemistry, RITM, Faridabad. She received her doctorate degree in Chemistry from Gurukula Kangri University, Haridwar, Uttar Pradesh, and postgraduate degree in Chemistry (Honours) from Panjab University, Chandigarh. She has more than 12 years of research and teaching experience and has authored several national and international publications.

Randhir Singh is working as Professor and Head in the Department of Chemistry, Gurukula Kangri University, Haridwar, Uttar Pradesh. He received his doctorate degree in Chemistry from University of Roorkee, India (now IIT) and postgraduate degree in Chemistry from University of Meerut, India. He has been Postdoctoral Research Associate in the Department of Biochemistry and Molecular Biology, University of Southern California, Los Angeles, USA, and visited University of Southern California, Los Angeles, as scientist exchange programme. He has more than 35 years of research and teaching experience and has guided many MSc, MPhil. and PhD students in their theses work. He has received many academic honours from Council of Scientific and Industrial Research, India, for his research contributions and has several national and international publications.

INORGANIC CHEMISTRY

Rajni Garg Associate Professor and Head Department of Applied Sciences Rattan Institute of Technology and Management Faridabad, Haryana

Randhir Singh Professor and Head Department of Chemistry Gurukul Kangri Vishwavidyalaya Haridwar, Uttar Pradesh

McGraw Hill Education (India) Private Limited NEW DELHI McGraw Hill Education Offices New Delhi New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited P-24, Green Park Extension, New Delhi 110016 Inorganic Chemistry Copyright © 2014 by McGraw Hill Education (India) Private Limited. No part of this publication can be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, McGraw Hill Education (India) Private Limited Print Edition: ISBN (13) : 978-1-25906285-8 ISBN (10) : 1-25-906285-6 Ebook Edition: ISBN (13) : 978-9-38328690-4 ISBN (10) : 9-38-328690-3 Managing Director: Kaushik Beuami Head—Higher Education Publishing and Marketing: Vibha Mahajan Sr Publishing Manager—SEM & Tech. Ed.: Shalini Jha Associate Sponsoring Editor: Smruti Snigdha Sr Editorial Researcher: Amiya Mahapatra Sr Development Editor: Renu Upadhyay Manager—Production Systems: Satinder S Baveja Asst Manager—Editorial Services: Sohini Mukherjee Production Executive: Anuj K Shriwastava Asst General Manager—Higher Education Marketing: Vijay Sarathi Senior Graphic Designer—Cover: Meenu Raghav General Manager—Production: Rajender P Ghansela Manager—Production: Reji Kumar Information contained in this work has been obtained by McGraw Hill Education (India), from sources believed to be reliable. However, neither McGraw Hill Education (India) nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought. Typeset at Print-O-World, 2579, Mandir Lane, Shadipur, New Delhi 110 008, and printed at Cover Printer :

CONTENTS Preface

ix

1. Structure of Atom 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 1.10 1.11 1.12 1.13 1.14

2. Nuclear Chemistry 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8 2.9 2.10 2.11 2.12

1.1–1.27

Introduction 1.1 Rutherford Scattering Experiment 1.1 Planck’s Quantum Theory of Radiation 1.2 Photoelectric Effect 1.3 Atomic Spectrum of Hydrogen 1.3 Bohr’s Model of the Atom 1.4 Sommerfeld’s Extension of Bohr’s Atomic Model 1.9 Dual Character of Matter 1.10 Heisenberg’s Uncertainty Principle 1.11 Compton Effect 1.12 Schrodinger Wave Equation 1.12 Quantum Numbers 1.17 Probability Distribution Curves 1.18 Rules for Filling of Orbitals and Electronic Configuration of Elements 1.20 Summary 1.23 Solved Examples 1.23 Exercises 1.27 Nucleus 2.1 Composition of the Nucleus 2.1 Nuclear Forces 2.4 Nuclear Stability 2.5 Nuclear Models 2.7 Nuclear Reactions 2.8 Radioactivity 2.14 Radioactive Disintegration 2.15 Law of Successive Disintegration: Radioactive Equilibrium 2.15 Soddy-Fajans and Russel Group Displacement Law 2.16 Artificial Radioactivity 2.17 Applications of Radioactive Isotopes 2.18 Summary 2.19 Solved Examples 2.20 Exercises 2.23

2.1–2.23

vi

Contents

3. Chemical Bonding

3.1–3.65

3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12 3.13 3.14 3.15

Introduction 3.1 Ionic Bond or Electrovalent Bond 3.1 Covalent Bond (Lewis–Langmuir Concept) 3.3 Dipole Moment 3.7 Coordinate Covalent Bond or Dative Bond 3.8 Van der Waals’ Forces or Intermolecular Forces 3.9 Hydrogen Bond 3.10 Orbital Overlap Theory 3.13 Molecular Orbital Theory 3.19 Metallic Bond 3.36 Hybridisation 3.39 Sidgwick – Powell Theory 3.49 Valence Shell Electron-pair Repulsion Theory (VSEPR theory) 3.49 Shapes of Some Common Molecules 3.50 Linnett Double Quartet Theory (LDQ Theory)— Modification of Lewis Longmuir Octet Theory 3.57 3.16 Resonance 3.60 Summary 3.62 Solved Examples 3.63 Exercises 3.64

4. Molecular Symmetry 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8

5. Redox Reactions 5.1 5.2 5.3 5.4 5.5

4.1–4.72

Introduction 4.1 Symmetry Element 4.1 Multiplication of Symmetry Operations 4.5 Mathematical Group 4.5 Matrix Representation of Symmetry Operations 4.10 Terms Symbols of Diatomic Molecules 4.20 Applications of Group Theory 4.22 Structure of Soilds 4.40 Summary 4.67 Solved Examples 4.68 Exercises 4.71 Introduction 5.1 Electrochemical Cell 5.2 Kinetics of Redox Reactions 5.6 Redox Reactions in Aqueous Systems 5.8 Diagrammatic Representation of Potential Data 5.12 Summary 5.17 Solved Examples 5.18 Exercises 5.20

5.1–5.21

Contents

6. Non-aqueous Solvents 6.1 6.2 6.3 6.4 6.5 6.6 6.7 6.8 6.9 6.10 6.11

8.1–8.23

Introduction 8.1 Mendeleef’s Periodic Table 8.1 Modern Periodic Law and Periodicity 8.2 Long form of Periodic Table 8.2 Periodic Properties 8.6 Shielding or Screening Effect 8.17 Summary 8.19 Solved Examples 8.20 Exercises 8.22

9. Hydrogen and its Compounds 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9

7.1–7.10

Introduction 7.1 Occurrence of Elements 7.1 Metallurgy 7.2 Purification of Impure Metals or Refining 7.5 Thermodynamics of the Metallurgy: Ellingham Diagram 7.6 Summary 7.9 Solved Examples 7.9 Exercises 7.10

8. Periodic Table and Periodic Properties 8.1 8.2 8.3 8.4 8.5 8.6

6.1–6.39

Introduction 6.1 Classification of Solvents 6.2 Liquid Ammonia 6.3 Liquid Sulphur Dioxide 6.10 Anhydrous Hydrogen Fluoride 6.14 Anhydrous Sulphuric Acid 6.16 Acetic Acid 6.18 Liquid Dinitrogen Tetroxide, N2O4 6.20 Molten Salts and Ionic Liquids 6.21 Concept of Acid – Base 6.24 Acid Strength Behaviour in the Periodic Table 6.33 Summary 6.35 Solved Examples 6.36 Exercises 6.38

7. Extraction of Elements 7.1 7.2 7.3 7.4 7.5

vii

Introduction 9.1 Position of Hydrogen in the Periodic Table 9.2 Occurrence and Production of Hydrogen 9.3 Physical Properties of Hydrogen 9.3 Chemical Properties of Hydrogen 9.3 Uses of Hydrogen 9.4 Different Forms of Hydrogen 9.5 Spin Isomers of Hydrogen 9.6 Isotopes of Hydrogen 9.7

9.1–9.16

viii

Contents

9.10 Compounds of Hydrogen 9.9 9.11 Water H2O 9.12 9.12 Heavy Water (D2O) 9.13 Summary 9.14 Solved Examples 9.15 Exercises 9.15

10. Chemistry of Group 1 Elements 10.1 10.2 10.3 10.4 10.5 10.6 10.7

11. Chemistry of Group 2 Elements 11.1 11.2 11.3 11.4 11.5 11.6 11.7 11.8 11.9 11.10

11.1–11.20

Introduction 11.1 General Characteristics of Group 2 Elements 11.1 Chemical Properties of Alkaline Earth Metals 11.3 Beryllium (Be) 11.6 Magnesium (Mg) 11.10 Calcium (Ca) 11.12 Strontium (Sr) 11.14 Barium (Ba) 11.14 Radium (Ra) 11.15 Portland Cement 11.16 Summary 11.18 Solved Examples 11.19 Exercises 11.19

12. Chemistry of Group 13 Elements 12.1 12.2 12.3 12.4 12.5 12.6 12.7 12.8 12.9 12.10

10.1–10.20

Introduction 10.1 General Characteristics of Group I Elements 10.1 Chemical Properties of Alkali Metals 10.3 Lithium (Li) 10.7 Sodium (Na) 10.11 Potassium (K) 10.16 Rubidium, Caesium and Francium 10.18 Summary 10.19 Solved Examples 10.19 Exercises 10.20

Introduction 12.1 Electronic Structure 12.1 General Physical Properties 12.1 Diagonal Relationship between Boron and Silicon 12.3 Chemical Properties of Group 13 Elements 12.4 Boron 12.5 Aluminimum (Al) 12.22 Gallium (Ga) 12.26 Indium and Thallium (Th) 12.27 Comparision of Compounds of Group 13 Elements 12.29

12.1–12.31

Contents

ix

Summary 12.30 Solved Examples 12.30 Exercises 12.31

13. Chemistry of Group 14 Elements 13.1 13.2 13.3 13.4 13.5 13.6 13.7 13.8 13.9 13.10

Introduction 13.1 General Properties of Group 14 Elements 13.1 Anomalous Behaviour of Carbon 13.3 Carbon and Silicon—Comparison of Properties 13.3 Carbon 13.4 Silicon (Si) 13.18 Germanium (Ge) 13.29 Tin (Sn) 13.31 Lead (Pb) 13.34 Comparative Account of Compounds of Group 14 Elements Summary 13.38 Solved Examples 13.38 Exercises 13.39

14. Chemistry of Group 15 Elements 14.1 14.2 14.3 14.4 14.5 14.6 14.7 14.8

13.37

14.1–14.42

Introduction 14.1 General Properties of Group 15 Elements 14.1 Chemical Properties of Group 15 Elements 14.3 Nitrogen (N) 14.5 Phosphorus (P) 14.23 Arsenic (As) 14.35 Antimony (Sb) 14.37 Bismuth (Bi) 14.39 Summary 14.40 Solved Examples 14.41 Exercises 14.42

15. Chemistry of Group 16 Elements 15.1 15.2 15.3 15.4 15.5 15.6 15.7 15.8 15.9

13.1–13.39

Introduction 15.1 General Properties of Group 16 Elements 15.1 Anomalous Behaviour of Oxygen 15.3 Oxygen (O2) 15.3 Sulphur (S2) 15.7 Selenium (Se) 15.22 Tellurium (Te) 15.24 Polonium (Po) 15.26 Comparative Account of Compounds of Group 16 Elements Summary 15.27 Solved Examples 15.27 Exercises 15.28

15.1–15.28

x

Contents

16. Chemistry of Group 17 Elements 16.1 16.2 16.3 16.4 16.5 16.6 16.7 16.8 16.9 16.10 16.11

17. Chemistry of Group 18 Elements 17.1 17.2 17.3 17.4 17.5 17.6 17.7 17.8 17.9

16.1–16.46

Introduction 16.1 General Characterisation 16.1 Chemical Properties 16.4 Fluorine (F) 16.6 Chlorine (Cl) 16.17 Bromine (Br) 16.23 Iodine (I) 16.25 Astatine (At) 16.31 Interhalogen Compounds 16.32 Polyhalides 16.38 Pseudohalogens and Pseudohalides 16.41 Summary 16.44 Solved Examples 16.44 Exercises 16.45

17.1–17.15

Introduction 17.1 History and Discovery 17.1 Occurrence and Isolation of Noble Gases 17.2 Uses of Noble Gases 17.4 Physical Properties 17.4 Chemical Properties 17.5 Chemistry of Xenon (Xe) 17.6 Compounds of Krypton (Krf2) 17.12 Compounds of Radon (Rn) 17.13 Summary 17.13 Solved Examples 17.14 Exercises 17.14

18. Chemistry of d-block Elements

18.1–18.10

18.1 Introduction 18.1 18.2 Classification of d-block Elements 18.2 18.3 General Characteristic of d-block Elements 18.3 Summary 18.8 Solved Examples 18.8 Exercises 18.9

19. Chemistry of Elements of 3d Series 19.1 19.2 19.3 19.4 19.5 19.6 19.7

Introduction 19.1 Scandium (Sc) 19.1 Titanium (Ti) 19.3 Vanadium (V2) Chromium (Cr) 19.13 Manganese (Mn) 19.20 Iron (Fe) 19.26

19.1–19.53

Contents

19.8 19.9 19.10 19.11

Cobalt (Co) 19.36 Nickel (Ni) 19.38 Copper (Cu) 19.42 Zinc (Zn) 19.46 Summary 19.49 Solved Examples 19.51 Exercises 19.51

20. Chemistry of Elements of 4d Series 20.1 20.2 20.3 20.4 20.5 20.6 20.7 20.8 20.9 20.10 20.11

21.1–21.19

Introduction 21.1 Hafnium (HF) 21.1 Tantalum (Ta) 21.2 Tungsten (W) 21.2 Rhenium (Re) 21.4 Osmium (Os) 21.5 Iridium (Ir) 21.6 Platinum (Pt) 21.8 Gold (Au) 21.10 Mercury (Hg) 21.14 Summary 21.17 Solved Examples 21.17 Exercises 21.18

22. Chemistry of Lanthanides and Actinides 22.1 22.2 22.3 22.4 22.5 22.6

20.1–20.21

Introduction 20.1 Yttrium (Y) 20.1 Zirconium (Zr) 20.2 Niobium (Nb) 20.3 Molybdenum (Mo) 20.6 Technetium (Tc) 20.10 Ruthenium (Ru) 20.11 Rhodium (Rh) 20.12 Palladium (Pd) 20.13 Silver (Ag) 20.15 Cadmium (Cd) 20.18 Summary 20.19 Solved Examples 20.19 Exercises 20.20

21. Chemistry of 5d Series 21.1 21.2 21.3 21.4 21.5 21.6 21.7 21.8 21.9 21.10

xi

Introduction 22.1 Lanthanides 22.1 Lanthanum (La) 22.11 Actinides 22.12 Thorium (Th) 22.15 Uranium (U) 22.17

22.1–22.26

xii

Contents

22.7 Plutonium (Pu) 22.22 Summary 22.24 Solved Examples 22.25 Exercises 22.25

23. Coordination Compounds-I Basics Concepts: Nomenclature and Stereochemistry 23.1 23.2 23.3 23.4 23.5 23.6

Introduction 23.1 Important Terms 23.2 Rules for Nomenclature of Coordination Compounds 23.5 Rules for Formula of the Coordination Compounds 23.6 Classification of Complexes 23.7 Isomerism 23.9 Summary 23.15 Solved Examples 23.16 Exercises 23.17

24. Coordination Compounds— II Theories of Bonding 24.1 24.2 24.3 24.4 24.5

25.1–25.35

Introduction 25.1 Determination of Octahedral Crystal Field Potential 25.2 Determination of Tetragonal Crystal Field Potential 25.11 Determination of Square Planar Crystal-Field Potential 25.15 Determination of Tetrahedral Crystal-Field Potential 25.16 Determination of Cubic Crystal-Field Potential 25.20 Structural and Thermodynamic Effects of Splitting of Orbitals 25.21 Jahn–Teller Effect (Distortion of Geometry) 25.28 Summary 25.33 Solved Examples 25.33 Exercises 25.34

26. Coordination Complexes IV: Spectroscopic and Magnetic Properties of Coordination Compounds 26.1 26.2 26.3 26.4

24.1–24.35

Introduction 24.1 Techniques for Study of Complexes 24.1 Theories of Coordination 24.3 Crystal Field Theory (CFT) 24.9 The Ligand Field Theory–Molecular Orbital Theory 24.17 Summary 24.32 Solved Examples 24.33 Exercises 24.34

25. Coordination Compounds III: Quantitative Basis of Crystal Field Theory 25.1 25.2 25.3 25.4 25.5 25.6 25.7 25.8

23.1–23.18

Introduction 26.1 Coupling Schemes 26.2 Energy Terms and Energy States 26.3 Electronic Spectra of Transition–Metal Compounds 26.11

26.1–26.58

Contents

26.5 26.6 26.7 26.8 26.9 26.10

xiii

Orgel Diagrams 26.16 Racah Parameters 26.24 Terms Correlation Diagrams under the Effect of Weak and Strong Field Effects 26.26 Tanabe-sugano Diagrams (T-S Diagram) 26.29 Charge-Transfer Transitions 26.34 Types of Magnetism 26.39 Summary 26.55 Solved Examples 26.56 Exercises 26.57

27. Coordination Compounds -V The Reaction Mechanisms of Transition-Metal Complexes

27.1–27.30

27.1 Introduction 27.1 27.2 Ligand-substitution Reactions 27.10 27.3 Oxidation–reduction Reactions in Coordination Compounds 27.22 Summary 27.27 Solved Examples 27.28 Exercises 27.29

28. Complexes of p-Acceptor Ligands 28.1 28.2 28.3 28.4 28.5

29. Chemistry of Organometallic Compounds 29.1 29.2 29.3 29.4 29.5 29.6 29.7

29.1–29.22

Introduction 29.1 Organometallic Compounds of Alkali Metals 29.3 Organometallic Compounds of Alkaline Earth Metals 29.4 Organometallics of Group 13 Elements 29.5 Organometallics of Group 14 Elements 29.6 Organometallics of Group 15 Elements 29.8 Organometallic Compounds of Transition Elements 29.9 Summary 29.19 Solved Examples 29.20 Exercises 29.21

30. Metal Clusters 30.1 30.2 30.3 30.4

28.1–28.12

Introduction 28.1 Complexes of Carbonyls 28.1 Complexes of Nitric Oxide 28.7 Complexes of Phosphines 28.8 Complexes of Cyanide and Isocyanide Ligands 28.10 Summary 28.10 Solved Examples 28.11 Exercises 28.11

Introduction 30.1 Polynuclear Compounds of Oxygen and other Chalcogens 30.1 Clusters of p-block Elements other than Chalcogens 30.2 Low-valent Metal Clusters 30.3

30.1–30.13

xiv

Contents

30.5 High-Valent Metal Clusters or Halide-type Clusters 30.7 Summary 30.11 Solved Examples 30.12 Exercises 30.12

31. Inorganic Nomenclature

31.1–31.10

31.1 Introduction 31.1 31.2 General Nomenclature and Formulae of Compounds 31.4 Solved Examples 31.9 Exercises 31.9

32. Inorganic Polymers 32.1 32.2 32.3 32.4

33. Bioinorganic Chemistry 33.1 33.2 33.3 33.4 33.5 33.6 33.7 33.8 33.9

34.1–34.24

Introduction 34.1 Air Pollution 34.3 Water Pollution 34.15 Soil Pollution 34.22 Summary 34.23 Exercises 34.24

35. Analytical Chemistry 35.1 35.2 35.3 35.4 35.5 35.6

33.1–33.14

Introduction 33.1 Metalloporphyrins 33.2 Cytochromes 33.6 Peroxidases (Molar Mass ~40,000) 33.7 Catalases 33.7 Ferredoxins 33.7 Metallo-enzymes 33.8 Biological Nitrogen Fixation 33.10 Na-K pump 33.11 Summary 33.12 Solved Examples 33.12 Exercises 33.13

34. Pollution 34.1 34.2 34.3 34.4

32.1–32.30

Introduction 32.1 Classification of Inorganic Polymers 32.1 General Characteristics of Inorganic Polymers 32.2 Important Inorganic Polymers 32.2 Exercises 32.29

Errors 35.1 Detection and Minimisation of Errors 35.3 Precision 35.4 Ways of Expressing Precision 35.4 Analysis of Data by Using Statistical Techniques 35.6 Detecting Outliers 35.7

35.1–35.34

Contents

35.7 35.8 35.9 35.10 35.11 35.12 35.13

Index

xv

Significance Tests 35.8 Significant Figures 35.8 Expressing Error or Accuracy of a Measurement 35.10 Error Propagation in Final Results 35.11 Volumetric Analysis 35.13 Preparation of Standard Solution 35.14 Volumetric Methods 35.14 Summary 35.27 Solved Examples 35.28 Exercises 35.33

I.1-I.10

PREFACE “Most of the fundamental ideas of science are essentially simple, and may, as a rule, be expressed in a language comprehensible to everyone.” Albert Einstein Inorganic chemistry is a dynamic and fascinating field of chemistry growing at a rapid pace in both research and theoretical aspects. The lusty impact of this field has introduced the subject as an essential part in the curricula of all universities. The first edition of Inorganic Chemistry aims to provide the essentials of the subject in an easy and understandable manner. The book is an outcome of the teaching and research experience of the authors so that students can learn concept formulation instead of just rote memorization. Target Audience The book is primarily aimed for students at undergraduate (BSc pass and honours) and postgraduate (MSc pass and honours) levels taking inorganic chemistry as a special subject for a one semester or a full-year course. Secondary Readers This book is designed to provide concise information about various aspects of inorganic chemistry that can also be used by students from various fields involving inorganic compounds, such as environmental science, polymer science, industrial chemistry, bioinorganic chemistry and metallurgy. However, they can skip the irrelevant topics as per their field. This book will also be a source of reference for the students doing BTech courses or taking inorganic chemistry as an ancillary subject. It will also be helpful for the challenging requirements of various competitive exams such as CSIR, SLET and GATE. About the Book The content of this book has been framed in an easy-to-understand language that would generate interest in the subject. All the chapters provide descriptive information and are enriched with illustrations, comprehensible articles, solved examples, both numerical as well as theoretical, to satisfy the needs of students. At the end of each chapter, a concise summary has been given for quick revision before examinations. The chapters have been enriched with exercises comprising theory-based general questions and objective-type questions to provide an insight about the examination pattern. The book introduces descriptive and illustrative information about structure of atoms and nuclei, radioactivity, chemical bonding, molecular symmetry, structure of solids, redox reactions, non-aqueous solvents, acids and bases, extraction of elements, Periodic Table, chemistry of known elements, coordination chemistry, organometallics, inorganic polymers, bioinorganic chemistry, environmental chemistry and analytical chemistry. Although it is very difficult to include such a vast subject in a single

xviii

Preface

book, a reasonable attempt has been made to cover a variety of important topics. We hope this book will prove very helpful in providing the complex concepts of inorganic chemistry in an easy way. Salient Features Comprehensive coverage of topics as per the latest syllabi of various universities—covers all important topics such as Atomic Structure, Periodic Table, Chemical Bonding, Group Theory, Coordination Chemistry, Organometallic Compounds Special emphasis given to theoretical aspects of the concepts Applications/Case Studies provided throughout for better understanding of the subject In-depth exploration through solved examples and exercises for self-assessment and skill evaluation Concise presentation of chemistry of elements and their compounds Special emphasis on coordination chemistry, including various theories and organometallics Chapter-end pedagogy designed as per the Indian university examinations Learning Objectives and Summary with each chapter More than 750 figures and tables for correlation of data 150 illustrative solved examples Over 950 unsolved exercises in the form of Review Questions and Objective-Type Questions Online Learning Center The text is accompanied by an Online Learning Center that can be accessed at https://www.mhhe.com/ sing/ic1 which contains supplementary material. Acknowledgements We feel immense pleasure to place on record our sincere thanks to McGraw Hill Education (India), New Delhi, for presenting our efforts into reality in the form of this book. Special thanks are due to Ms Smruti Snigdha, Mr Amiya Mahapatra and Ms Renu Upadhyay, Mr Satinder Baveja, Ms Sohini Mukherjee and Mr Anuj Shriwastava for their sustained interest in this project. We are also grateful to our students for their much valuable criticism and suggestions during the development of this book. Our most heartfelt acknowledgment must go to Prof. Rishav Garg for his assistance and encouragement in all ways. Above all, we would like to thank Ayush Garg and Anusha Garg who silently endured all the unpleasantness and sacrificed most. Note of Thanks to Reviewers We are highly indebted to all the expert reviewers for their generous comments and suggestions on the manuscript. Ramesh C Kambhoj Sandeep Kaur Nand Kishor Singh Ashu Chaudhary Debasis Das A Jeya Rajendran B B V Sailaja

Kuruskshetra University, Haryana University of Delhi, New Delhi Banaras Hindu University (BHU), Varanasi, Uttar Pradesh Kurukshetra University, Haryana University of Calcutta, Kolkata, West Bengal Loyola College, Chennai, Tamil Nadu Andhra University, Hyderabad, Andhra Pradesh

Preface

xix

Feedback Request from Author Some unintentional discrepency and errors might be encountered in spite of our best efforts. Constructive suggestions and criticisms from readers are highly welcome and will be gracefully acknowledged. Readers can approachus at [email protected]. Publisher’s Note Do you have any further request or a suggestion? We are always open to new ideas (the best ones come from you!). You may send your comments to [email protected].

chapter

Structure of Atom

1

After studying this chapter, the student will be able to discovery of the nucleus effect

principle

1.1

INTRODUCTION

Atom was considered as the smallest indivisible particle of matter till the discovery of fundamental particles, namely the following: 1 —A negatively charged particle having a mass of about the mass of a proton

1836

(9.1091 × 10–31 kg) and charge equal to 1.60206 × 10–19 C (unit negative charge) —A positively charged particle having a mass equal to 1.6727 × 10–27 kg and charge equal and opposite to that of electron (unit positive charge) —A neutral particle having a mass equal to 1.6748 × 10–27 kg.

1.2

RUTHERFORD SCATTERING EXPERIMENT

The location of the fundamental particles was first described by Rutherford’s a-particle scattering experiment, in which a-particles from a radioactive element were allowed to strike on a very thin gold foil. The direction of the deflected a-particles was detected with the help of a movable circular screen (Fig. 1.1). Rutherford’s observations and conclusions are as follows: 1. Most of the a-particles passed through the gold foil without any deflection. It was concluded that most of the space in the atom was empty.

1.2

Inorganic Chemistry

Radium

–

Lead block

a-particles Slit

Undeflected a-particles

a-particles Rebounded a-particles

a-particles

–

+

–

– –

Thin gold foil Circular ZnS screen

Largely deflected a-particles

Fig. 1.1 Rutherford experiment and simplified representation of three types of deflection 2. Some a-particles were deflected back through large angles. It was concluded that there is a positively . charged part in the centre of the atom, called as 3. A few a-particles completely rebounded to their original path. It was concluded that nucleus is very dense and the whole mass of the atom is concentrated in it. On the basis of these conclusions, Rutherford proposed nuclear model of atom.

1.2.1 Rutherford Model : Nuclear Model or Solar Model or Planetary Model of the Atom According to Rutherford’s model, the following postulates hold: 1. An atom consists of a positively charged body located at its centre, called nucleus. The nucleus is made up of protons and neutrons. Since, the mass of electron is negligible, the mass of the atom is almost concentrated in the nucleus. 2. Electrons revolve around the nucleus in different orbits to balance the positive charge on the nucleus. Therefore, atom is electrically neutral. 3. Most of the surrounding space around the nucleus is empty (volume of the atom is very large as compared to that of nucleus).

1.2.2 Drawbacks 1. According to classical electromagnetic theory, if a charged particle is subjected to acceleration around an oppositely charged particle, it emits radiations and hence its energy and speed go on decreasing. It means, electrons revolving around the nucleus should continuously emit radiations and lose energy. Due to this, the electron would gradually fall into the nucleus following a spiral path and atom should be unstable. However, it is not unstable. 2. Due to continuous emission of light radiations, the atomic spectrum should be continuous instead of a line spectrum. This means that Rutherford’s model of atom failed to explain the stability of atom and line spectrum of atom.

1.3

PLANCK’S QUANTUM THEORY OF RADIATION

In 1901, Max Planck studied the spectral distribution of energy emitted by a black body and put forward a theory known as Planck’s Quantum Theory of Radiation.

Structure of Atom

1.3

The main postulates of this theory are as follows: 1. The emission or absorption of energy by a body does not take place continuously, but discontinuously, in the form of small packets of energy called (singular quantum). 2. The energy of each quantum is directly proportional to the frequency, n, of radiation, i.e., n or = hn where, h = 6.626 × 10–34 Js (in S.I. units) and is called Planck’s constant. 3. A body can emit or absorb energy in whole number multiples of quantum, i.e. = n where, = 1, 2, 3 ............. etc. This is called quantisation of energy.

1.4

PHOTOELECTRIC EFFECT

Sir J. J. Thomson and P. Lenard observed in some of their experiments that when light beam of suitable frequency is allowed to strike on the surface of a metal, the electrons are ejected (emitted) from the surface of the metal. This phenomenon is called photoelectric effect, and the ejected electrons are called photoelectrons. It was further observed that only these radiations, which have a certain minimum frequency, called threshold frequency, can eject the electrons. Different metals have different magnitudes of threshold frequency. The metals with low ionisation energy mainly show photoelectric effect. An increase in the intensity of the incident light increases the number of electrons emitted per second from the surface; however, kinetic energy of photoelectrons remains the same. An increase in the magnitude of the frequency of the incident radiations increases the magnitude of the kinetic energy of photoelectrons.

Explanation of Photoelectric Effect In 1905, Einstein afforded an explanation of the photoelectric effect with the help of Planck’s quantum theory of radiation. He suggested that when a photon of light of frequency, , strikes on the surface of a metal, some of its associated energy (equal to binding energy called threshold Light beam ejected photoelectron energy or work function h 0) is consumed to separate the electron of energy ho – from the atom and the remaining energy is imparted to the ejected electron as kinetic energy (= h =h

0

1 mu2) (Fig. 1.2). That means, 2 +

1 mu2 2

(1.1) ho0

1 or mu2 = h ( – 0) (1.2) Fig. 1.2 Diagrammatic representation of photoelectric effect 2 where, 0 is the threshold frequency. Millikan used Eq. (1.2) to calculate the value of h equal to 6.570 × 10–34 Js, an excellent agreement with the experimental value equal to 6.626 × 10–34 Js.

1.5

ATOMIC SPECTRUM OF HYDROGEN

If a discharge is passed through hydrogen gas kept at low pressure and the emitted light is studied by a spectrometer, hydrogen spectrum is obtained, consisting of a number of lines in the visible, ultraviolet and infrared region. The lines observed in the spectrum are grouped as given in Table 1.1.

1.4

Inorganic Chemistry

In 1885, Balmer found that the wave number n of any line in the visible region of the atomic spectrum of hydrogen is given by

Table 1.1 Spectral series for atomic spectrum of hydrogen

1ˆ Ê 1 - 2˜ 2 ¯ 2

n=R Á Ë

where = 3,4,5 ...... and R is the Rydberg constant or Rydberg number with its value equal to 109,678 cm–1.

Name of series Lyman series Balmer series Paschen series Brackett series Pfund series Humphries series

Spectral region Ultraviolet region Visible region Infrared region Infrared region Infrared region Infrared region

Rydberg’s Formula In 1889, Rydberg gave a general relationship for the lines in the hydrogen spectrum as È1 1 1˘ n= = RÍ 2 - 2 ˙ l ÍÎ n1 n2 ˙˚ for Lyman series, 1 = 1; for Balmer series, 1 = 2; for Paschen series, 1 = 3; for Brackett series, 1 = 4; for Pfund series, 1 = 5; for Humphries series, 1 = 6;

= 2, 3, 4, 5 2 = 3, 4, 5, 6 2 = 4, 5, 6, 7 2 = 5, 6, 7, 8 2 = 6, 7, 8, 9 2 = 7, 8, 9, 10 2

(1.3)

................ ................ ................ ................ ................ ................

In order to explain the line spectrum of hydrogen and to remove the drawbacks of Rutherford’s model, Neil Bohr put forward an atomic model in 1913.

1.6

BOHR’S MODEL OF THE ATOM

The main postulates of Bohr’s model are as follows: 1. In an atom, electrons revolve around the nucleus in fixed concentric circular shells called energy shells or energy levels, numbered as = 1, 2, 3, ... and designated by K, L, M..., respectively. 2. Energy levels are associated with a definite amount of energy , which increases with the increase of distance from the nucleus, i.e.. 1 < 2 < 3 < ......... As long as an electron revolves in a particular orbit, it cannot absorb or emit energy. Thus, these orbits are also called stationary states or ground states. 3. Energy is emitted or absorbed, by an electron, when electron jumps from one energy level to another. This energy is equal to the difference in the energies associated with these two levels, i.e. D =

2

–

1

= hn

where n is the frequency of the energy emitted or absorbed. This means, energy of an electron cannot change continuously but changes abruptly by a fixed amount, i.e. energy of an electron is quantised. 4. The angular momentum of an electron in a particular orbit is also quantised, i.e. the angular momentum of an electron can have only definite or discrete values as mnr = n

h where = 1, 2, 3 ... 2p

1.5

Structure of Atom

1.6.1 Bohr’s Model for Hydrogen Atom and Hydrogen-like Species (One-electron Species) Consider a hydrogen-like species with atomic number equal to Z and an electron of charge e, revolving around the nucleus, in an orbit of radius r. The charge on nucleus must be +Ze. Let m be the mass of electron and n be its tangential velocity. The centripetal force acting on the electron tends to attract the electron towards the nucleus and is equal

Ze2

(in CGS system of units). While, the centrifugal force acting on the electron tends to take it away r2 mn 2 . from its orbit and is equal to r For an electron to remain in its orbit, these two forces must be equal, i.e. 2 Ze2 = mn (1.4) r r2 Ze2 or n2 = (1.5) mr According to Bohr’s postulates, to

h 2p

(1.6)

nh 2p mr

(1.7)

mnr = n or

n=

Squaring Eq. (1.7),

n2 h 2

n2 =

(1.8)

4p 2 m 2 r 2

From equations (1.5) and (1.8), we get

Ze2 n2 h 2 = mr 4p 2 m 2 r 2 or For radius of the

r= th

(1.9)

n2 h 2

(1.10)

4p 2 Zme2

orbit, r =

In the CGS system,

n2 h 2

(1.11)

4p 2 me2 Z

h = 6.626 × 10–27 erg s, p = 3.14, Thus,

r =

=

m = 9.108 × 10–28 g,

n2 ¥ (6.624 ¥ 10 -27 erg s)2 4 ¥ (3.14) ¥ (9.108 ¥ 10 -28 g ) ¥ (4.8 ¥ 10 -10 esuu)2 ¥ Z 2

n2 ¥ (6.624 ¥ 10 -27 )2 erg 2 s2 4 ¥ (3.14)2 ¥ (9.18 ¥ 10 -28 ) ¥ (4.8 ¥ 10 -10 )2 ¥ g esu2 ¥ Z

e = 4.8 × 10–10 esu

1.6

Inorganic Chemistry

=

Thus,

( (

)

g cm 2 s-2 s2 0.529 ¥ 10 -8 n2 ¥ Z g. g1/ 2 cm 3 / 2 s-1

)

2

Èerg = g cm 2 s-2 ˘ Í ˙ ÍÎesu = g1/ 2 cm 3 / 2 s-1 ˙˚

0.529 ¥ 10 -8 n2 0.529 ¥ n2 cm = = Å Z Z 2 0.529 ¥ n r = Å Z

(1.12)

For a hydrogen atom, Z = 1, the radius of ground state is designated as the Bohr’s radius (a0). r1 = a0 =

0.529(1)2 = 0.529 Å (1)

Dividing Eq. (1.5) by Eq. (1.7), we get

Ze2 2p mr n2 ¥ = mr nh n

(1.13)

Ze2 .2 nh Again putting all the values, we get or

n=

n=

(1.14)

Z (4.8 × 10 −10 )2 × 2 × 3.14 cm s−1 n × 6.626 × 10 −27

, n=

0.2183 × 109 Z cm s−1 n

0.2183 × 107 Z m s−1 (1.15) n 1 The total energy of the electron revolving is an orbit in equal to the sum of its kinetic energy  mν 2  2  2   and the potential energy  − Ze  , i.e.  r    1 Ze2 = mν 2 − (1.16) 2 r Ze2 From Eq. (1.5), mn2 = (1.17) r From Equation (1.16) and Eq. (1.17), we get or

n=

Ze2 Ze2 − Ze2 − = r 2r 2r From Eq. (1.11) and Eq. (1.18), we get =

=

− Ze2 4π 2 Zme2 × 2r n2 h 2

=

−2π 2 Z 2 me 4 n2 h 2

(1.18)

(1.19)

Structure of Atom

th

Hence, the energy of an electron in the 2

=

orbit

2

−2π Z me n2 h 2

1.7

is given by

4

(1.20)

Again, putting all the values, we get =

−2 × (3.14)2 × Z 2 × (9.108 × 10 −24 ) × (4.8 × 10 −10 )4 errg n2 × (6.624 × 10 −27 )2

−11 = −2.18 × 10

Z2 erg n2

Z2 erg atom–1 2 n Z2 = –(2.18 × 10–11) × (6.022 × 1023) erg mol–1 2 n Z2 =– 13.12 × 1012 2 erg mol–1 n 1 erg = 107 J Z2 = – 13.12 × 105 2 J mol–1 n Z2 = – 1312 2 kJ mol–1 n = −2.18 × 10 −11

More accurately,

Since

or

(1.21)

1 erg = 6.2419 × 1011 eV

Since

= –(2.18 × 10–11) × (6.022 × 1023) × (6.2419 × 1011)

Z2 eV mol–1 n2

Z2 eV mol–1 n2 If an electron jumps from the 1, level to the 2 level then = 13.6

(D ) Since

2– 1

1

= =

Equation (1.22) reduces to

Since

2

–

1

= hn

-2p 2 Z 2 me 4 n12 h 2

& En2 = -

(1.22)

2p 2 Z 2 me 4 n22 h 2

hn =

2p 2 Z 2 me 4 Ê 1 1ˆ Á 2 - 2˜ 2 h Ë n1 n2 ¯

(1.23)

n=

2p 2 Z 2 me 4 Ê 1 1ˆ Á 2 - 2˜ 3 h Ë n1 n2 ¯

(1.24)

n = nc

1.8

Inorganic Chemistry

from Eq. (1.24), we get nc =

2p 2 Z 2 me 4 Ê 1 1ˆ Á 2 - 2˜ 2 h Ë n1 n2 ¯

For H atom, Z = 1

n=

1 2p 2 me 4 Ê 1 1ˆ = - 2˜ Á 3 2 l h c Ë n1 n2 ¯

and

n=

 1 1 1  =R 2 − 2   n λ  1 n2  2p 2 me 4

where R is Rydberg’s constant and is equal to

h3c

The value of R can be calculated as R=

(1.25)

1ˆ Ê ÁË since v = l ˜¯

(1.26)

(1.27)

.

2 × (3.14)2 × (4.8 × 10 −10 )4 × (9.108 × 10 −28 ) (6.624 × 10 −27 )3 × (3 × 10 −10 )

= 109,678 cm–1

1.6.2 Explanation of Spectral Lines of Hydrogen Atom The emission spectrum of the hydrogen atom consists of a large number of lines. Bohr provided an explanation based on his postulates. Any given sample of hydrogen contains a very large number of atoms. When energy is supplied to this sample of the gas, different atoms will absorb different amounts of energy. The single electron present in different hydrogen atoms shifts to different energy levels depending upon the amount of energy absorbed by the atoms. The electrons in higher energy levels are unstable and drop back to the lower energy levels, emitting energy in the form of line spectrum containing various lines of particular frequency and wavelength. As depicted in Fig. 1.3, the wavelength of these spectrum lines can be calculated by using Rydberg’s formula. An excellent argument between the experimental and the calculated values strengthened Bohr’s model of the hydrogen atom. 8 7 6 5

Humphries series Pfund series

4

Brackett series

4101.7 Å

Hα

4340.5 Å

Balmer series

4861.3 Å

2

Paschen series

6562.8 Å

Energy level (not to scale)

3

Hb

Hγ

Hδ

Continuum

H3

1 Lyman series

Fig. 1.3 Energy-level diagram for hydrogen spectrum

Structure of Atom

1.9

1.6.2 Limitations of Bohr’s Model 1. Bohr’s model cannot explain the origin of the spectra given by multi-electron species. 2. Bohr’s model cannot explain the fine spectrum of hydrogen atom. 3. When an excited atom giving a line emission spectrum is put in a magnetic field, its spectral lines get split up into a number of closely-spaced lines. This phenomenon is called Zeeman effect. Similar splitting of the spectral lines is observed in the presence of an electric field. This phenomenon is called Stark effect. Bohr’s model fails to explain these effects. 4. Bohr’s model considers the electron as a material particle of small mass moving around the nucleus. But according to De broglie, the electron has a dual character. 5. Bohr’s model contradicts with Heisenberg’s uncertainty principle.

SOMMERFELD’S EXTENSION OF BOHR’S ATOMIC MODEL

1.7

Sommerfeld extended Bohr’s model in 1915 to account for the fine structure of the line spectrum of hydrogen atom by putting forward the idea of elliptical orbits. According to Sommerfeld’s modification, electrons revolving around the nucleus can set their motion in elliptical orbits under the influence of the nuclear charge. These elliptical orbits have a major axis and a minor axis with different lengths. As the size of the orbit increases, these two axes become equal in length and the orbit becomes circular. In Bohr’s model, orbits are numbered as = 1, 2, 3, 4 .... where is known as the principal quantum number. Sommerfeld used two quantum numbers, radial quantum number nr and azimuthal quantum number, nf. These quantum numbers are related to the geometry of the ellipse as =

nr + nφ

=

nφ

Length of the major axis Length of the minor axis

There can be the following possibilities: 1. If f < , length of the minor axis will be greater than that of the major axis and the orbit remains elliptical. 2. If f = , length of the major axis will become equal to that of the minor axis and the orbit becomes circular. This means, if

= 3; = 2; = 1;

f

and have maximum of values, i.e.

= 1, 2, 3; two elliptical orbits and one circular orbit = 1, 2; one elliptical and one circular orbit f = 1; one circular orbit (Fig. 1.4) f f

Applying the principle of quantisation of momentum to the electron revolving in an elliptical orbit, Sommerfeld deduced the energy of the electron in the hydrogen atom as =

-2p 2 me 4

(n

f

+ nr

2

)

h2

This means, the energy of an electron depends upon the two quantum numbers f and r. Hence, the electronic transition from one energy level 1 to another energy levels 2 will result into closely spaced lines in the hydrogen spectrum depending upon the various possible values of f. However, Sommerfeld’s theory could not predict the correct number of lines observed in a fine structure.

1.10

1.8

Inorganic Chemistry

DUAL CHARACTER OF MATTER

In 1905, Einstein suggested that light has a dual character, i.e. particle as well as wave character. In 1924, de Broglie postulated that all forms of matter like electrons, protons, atoms, molecules, etc., also have dual character. He also derived an expression showing the relationship between momentum and wavelength.

1.8.1 Derivation of de Broglie’s Equation According to Planck’s quantum theory, the energy of a photon is given by = hn (1.28) According to Einstein’s mass–energy relationship, = mc2 (1.29) Equating equation (1.28) and (1.29) hn = mc2 Since n = c/l, Eq. (1.30) becomes

h

c

or

nφ = 2 nφ = 1

(1.30)

= mc2

(1.31)

l mc

(1.32)

l=

nφ = 3

Fig. 1.4 Bohr–Sommerfeld orbits when n = 3

For an ordinary particle, c can be replaced by n, the velocity of the particle l=

h h = mν p

(1.33)

where p is the momentum of the particle. Equation (1.33) is known as de Broglie’s equation.

1.8.2 Experimental Verification of de Broglie’s Equation The wave nature of electrons was verified experimentally by Davison and Germer in 1927 (Fig. 1.5). They obtained diffraction pattern of electrons similar to that of X-ray diffraction. Since X-rays have wave character, electrons should also have wave character. Beam of electrons Heated tungsten filament Photographic plate Collimator Nickel plate Diffraction pattern

Fig. 1.5 Representation of Davisson and Germer experiment

1.8.3 Quantisation of Angular Momentum Consider an electron moving round the nucleus in a circular orbit. According to de Broglie, an electron behaves as a standing or stationary wave. Its motion can be described in two waves as shown in Fig. 1.6.

Structure of Atom

1.11

For an electron wave to be in phase, the circumference of the Bohr’s orbit (2pr) must be equal to the whole number multiple of the wavelength (l) of the electron wave, i.e. 2pr = l

(1.34)

λ

2 r (1.35) n h According to de Broglie l = (1.36) m From equations (1.35) and (1.36), we get Out of phase In phase 2 r h = (1.37) n m Fig. 1.6 (a) Electron wave out of phase (b) Electron wave in phase nh or mnr = (1.38) 2 It is quite evident that electrons can move in only those orbits for which the angular momentum is an h , otherwise the electron wave will be out of phase and will emit radiations. integral multiple of 2 Thus, de Broglie concept leads to Bohr’s postulate of quantisation of angular momentum. or

l=

1.8.4 Significance of de Broglie’s Concept The de Broglie concept is significant only for microscopic objects such as atomic and sub-atomic particles. However, in case of macroscopic bodies, this concept loses its significance. For example, the wavelength associated with the electron of mass, 9.11 × 10–31 kg with a velocity of 105 ms–1 can be determined as l=

h 6.626 × 10 −34 Js = mν 9.11 × 10 −31 kg × 106 ms−1

0.7 Å

whereas, the wavelength associated with a ball weighting 2 × 10–3 kg moving with the same velocity as that of electron will come out to be l=

h 6.626 × 10 −34 Js = mν 2 × 10 −3 kg × 105 ms−1

3.3 × 10–36 m

3.3 × 10–26 Å

This wavelength is too small to be measured and hence is insignificant. This proves that the de Broglie’s concept is insignificant for macroscopic bodies.

1.9

HEISENBERG’S UNCERTAINTY PRINCIPLE

According to this principle, it is impossible to measure simultaneously the exact position and momentum (or velocity) of a microscopic moving particle.

h 4 where, Dx and Dp are the uncertainties in position and momentum of the particle respectively. Since Dp = mD , where D is the uncertainty in velocity and m is the mass of the moving particle. Equation (1.39) can be rewritten as h Dx . mD 4 Mathematically,

Dx . Dp

(1.39)

(1.40)

1.12

Inorganic Chemistry

It means that Dx and Dp (or D ) are inversely proportional to each other. Thus, if Dx is very small, i.e. the position of the particle is determined with more precision, Dp (or D ) would be large, i.e. uncertainty in determination of the momentum (or velocity) would be more.

Significance of de Broglie Concept It is well known that all observations are done by the impact of light rediations. When a moving electron collides with a photon of light, its velocity and path changes due to transfer of energy from the photon to the electron. But the position or velocity of an object of reasonable size, will not be altered by the impact of light radiations. Its quite evident that this concept is significant only for microscopic particles.

1.10

COMPTON EFFECT

Arthur Compton suggested that the photon-electron interaction can be considered as a collision between two balls—one moving and the other at rest. He found that if monochromatic X-rays are allowed to fall on solid matter, an electron is ejected and X-rays associated with smaller energy are scattered from their original path. This decrease in energy or increase in wavelength of the received photon is known as Compton effect. According to Compton, if l is the wavelength of the incident X-rays, l is the wavelength of the scattered X-rays, m is the mass of the electron and q is the angle of scattering, then

Scattered photon

θ Photon

Electron at rest

Scattered electron

Fig. 1.7 Illustration of Compton effect

l (1 – cos q) mc Here, Dl is called Compton shift (Fig. 1.7). It is evident that Dl is independent of the nature of the substance and the wavelength of the incident X-rays. It depends only on the magnitude of the angle of scattering. He suggested the following: 1. If q = 0°, Dl = 0. It means the scattered photon is parallel to the incident radiation, i.e. there is no wavelength shift. h 6.626 × 10 −34 Js = 2. If q = 90°, Dl = mc (9.11 × 10 −31 kg) (3 × 108 ms−1 ) Dl = l – l =

= 0.2424 × 10–11 m = 0.0242 Å

l is called Compton wavelength. mc 2h 3. If q = 180°; Dl = = 2 × 0.0242 Å = 0.0484 Å. mc It is evident that the wavelength of the scattered photon is always longer than that of the incident photon. Here,

1.11

SCHRODINGER WAVE EQUATION

Erwin Schrodinger, in 1926, proposed that if an electron behaves as a wave, there must be a wave equation to describe its wave motion. This wave equation is called Schrodinger wave equation and for an electron wave propagating in three dimensions in space, it can be written as

Structure of Atom

∂ 2ψ ∂x 2

+

∂ 2ψ ∂y 2

+

∂ 2ψ

(E − V ) ψ = 0

(1.41)

 8π 2 m ψ + 2 ( E − V ) ψ = 0 h 

(1.42)

2 y + 8π m ( E − V ) ψ = 0 h2

(1.43)

∂z 2

 ∂2 ∂2 ∂2  2 + 2 + 2 ∂y ∂z  ∂x

or

8π 2 m

1.13

or

2

where,

2

+

h2

 ∂2 ∂2 ∂2  + 2 + 2  2 ∂y ∂z   ∂x

=  

m is the mass of the electron. is the total energy of the electron. V is the potential energy of the electron. 2 is known as the Laplacian mathematical operator. y is a mathematical function of the space coordinates x, y and z and should be written as y (x, y, z). For the sake of convenience, it is generally written as y only.

∂ 2ψ ∂ 2ψ ∂ 2ψ are the double differentials of y with respect to x, y and z respectively. , 2 and 2 ∂x ∂y ∂z 2

1.11.1 Derivation of Schrodinger Wave Equation Schrodinger assumed that the electron waves are similar to the stationary waves. For stationary waves of wavelength l and displacement in x direction, the amplitude y can be written as

2π x λ Differentiating Eq. (1.44), with respect to x, we get y = A sin

∂ψ  2π x   2π  = A cos ∂x  λ   λ  ∂ψ 2π A 2π x = or .cos ∂x λ λ Differentiating Eq. (1.46) again with respect to x, we get

or

(1.44)

(1.45) (1.46)

∂ 2ψ 2π A  2π x   2π  = − sin  2 λ  λ   λ  ∂x

(1.47)

∂ 2ψ

(1.48)

∂x 2

=

−4π 2  2π x  A sin 2  λ  λ 

From equations (1.44) and (1.48), we get

∂ 2ψ −4π 2 = 2 ψ ∂x 2 λ

(1.49)

1.14

Inorganic Chemistry

This is the equation for unidimensional stationary wave propagating along the x-axis. For wave motion extended to three dimensions, Eq. (1.49) can be written as

∂ 2ψ ∂ 2ψ ∂ 2ψ −4π 2ψ + + = ∂x 2 ∂y 2 ∂z 2 λ2

(1.50)

This equation is applicable to all microscopic particles like electrons. The total energy of an electron is

1 the sum of its kinetic energy  mν 2  and the potential energy, V 2

 1 = mν 2 + V 2 = ( – V)

i.e.

1 m 2

or

2

(1.51) (1.52)

2

or

1 ( mν ) = ( – V) 2 m

or

1 h2 = ( – V) 2 2m

(1.53) (1.54)

1 2m = (E −V ) λ 2 h2

or

(1.55)

From equations (1.50) and (1.55), we get

∂ 2ψ ∂ 2ψ ∂ 2ψ −8π 2 m ( E − V )ψ + + = h2 ∂x 2 ∂y 2 ∂z 2 or or

∂ 2ψ ∂x 2

+

∂ 2ψ

+

∂y 2

∇ 2ψ +

∂ 2ψ ∂z 2

+

8π 2 m h2

8π 2 m h2

( E − V )ψ = 0

( E − V )ψ = 0

from de Broglie equation    λ = h    mν   h  mν =  λ   (1.56)

(1.57) (1.58)

1.11.2 Other Forms of Schrodinger Wave Equation -h 2

Multiplying both sides of Eq. (1.58) by

-h 2 8p 2 m or

8p 2 m

2

, we get

y + (V – ) y = 0

-h 2

2

(1.59)

y + Vy = y

(1.60)

or

 −h 2 2   2 ∇ +V y = y  8π m 

(1.61)

or

Hˆ y = y

(1.62)

2

8p m

Structure of Atom

1.15

 −h 2 2  ∇ + V  is called Hamiltonian operator and consists of two parts, i.e. the kinetic 2  8π m 

where Hˆ = 

2   energy part  −h ∇ 2  and the potential energy part (V).  8π 2 m 





In terms of polar coordinates,

1 ∂ Ê r 2 ∂y (r , q , f ) ˆ 1 1 ∂2y (r , q , f ) ∂ Ê sin q ∂y (r , q , f ) ˆ + + Á ˜ Á ˜ ¯ r 2 sin 2 q ∂r ∂q r 2 ∂r Ë ∂f 2 ¯ r 2 sin q ∂q Ë

+

8p 2 m Ê e2 ˆ y (r , q , f ) = 0 E + Á ˜ h2 Ë r¯

Schrodinger wave equation has many solutions for y. However, only some significant values which give certain definite and acceptable values of the total energy are acceptable. The acceptable values of the wave function are called eigen (acceptable) wave functions and the acceptable values of total energy are called eigen values.

1.11.3 Conditions for an Eigen Wave Function 1. It should be finite. 2. It should be single valued. 3. It should be continuous 4.

∂ 2ψ ∂ 2ψ ∂ 2ψ must be continuous functions. , and ∂x 2 ∂y 2 ∂z 2

5. The function should be normalised, i.e.

Ú y dt = 1.

Fig. 1.8 Electron charge cloud

It should be noted that y has no physical significance. However, y2 gives the probability of finding an electron in a given region round the nucleus. Thus, there are regions of space around the nucleus in which there is higher probability of finding an electron. This three-dimensional region is called atomic orbital. An orbital can be represented by means of electron probability distribution in three-dimensional space around the nucleus, known as electron charge cloud (Fig. 1.8).

1.11.4 Schrodinger Wave Equation for Hydrogen and Hydrogen-like Species In terms of polar coordinates,

1 ∂ Ê r 2 ∂y (r , q , f ) ˆ 1 1 ∂2y (r , q , f ) ∂ Ê sin q ∂y (r , q , f ) ˆ + + Á ˜¯ Á ˜ ∂r ∂q r 2 ∂r Ë r 2 sin 2 q ∂f 2 ¯ r 2 sin q ∂q Ë

+

8p 2 m h2

( E - V ) y (r , q , f ) = 0

(1.63)

1.16

Inorganic Chemistry

e2 r

For a hydrogen atom, potential energy of the electron V = From equations from (1.63) and (1.64), we get

(1.64)

1 ∂ Ê r 2 ∂y (r , q , f ) ˆ 1 1 ∂2y (r , q , f ) ∂ Ê sin q ∂y (r , q , f ) ˆ + + 2 + 2 2 Á ˜ ˜ 2 ∂r Á ¯ r sin q ∂r ∂q r ∂f 2 Ë ¯ r sin q ∂q Ë 8p 2 m Ê e2 ˆ E + y (r , q , f ) = 0 Á r ˜¯ h2 Ë

The function y can be written in the form y (r, q, f) =

(1.65)

R(r)

Q (i) Φ (z)

radial wave function

angular wave function

(1.66)

where, R(r) is a function which depends only on r .Q (q ) is a function which depends on q only and F(f) is a function which depends only on f. It corresponds to

∂y ∂R ˘ =QF◊ ∂r ∂r ˙˙ ∂y ∂Q ˙ = R◊F◊ ∂q ∂q ˙ ˙ ∂y ∂F ˙ = R◊Q ∂f ∂f ˙˚

(1.67)

2 2 After substituting equations (1.66) and (1.67) in Eq. (1.65) and multiplying both sides by r sin q , we get

RQF sin 2 q ∂ Ê 2 ∂R ˆ sin q ∂ Ê e2 ˆ ∂Q ˆ 1 ∂2 F 8p 2 m Ê ◊ Ár sin q + ◊ + ◊ 2 + 2 Á E + ˜ r 2 sin 2 q = 0 ˜ Á ˜ R dr Ë ∂r ¯ r¯ ∂q ¯ f ∂f Q ∂q Ë h Ë

(1.68)

This equation can be split into three simpler equations as 2  2  1. 1 ∂  r 2 ∂R  − β R(r ) + 8π m  E + Ze  R = 0   2 2 2  

r ∂r 

∂r 

r

Its solution gives R ( r )

h

n,l



r 

3  2 Z   n - l - 1  - ρ / 2 l 2l +1 e =  ρ Ln + l ( ρ )   3  na0   2 n ( n + 1) 

 2Z  2 l +1  r ; Ln + l ( ρ ) is the associated Laguere polynomial; na  0 quantum numbers and a0 is Bohr’s radius. where ρ = 

2.

ml2 1 ∂ Ê ∂ˆ sin q +b =0 sin q ∂q ÁË ∂q ˜¯ sin 2 q

(1.69)

and are the two integral

(1.70)

Structure of Atom

Its solution gives F l,ml (q ) =

(2l+1) l - ml 2 l + ml

Pl

ml

(cos q )

1.17

(1.71)

where, Plml (cos q ) is the associated Legendre polynomial and m is the third quantum number. 3.

∂2 F ∂f

2

+ ml2 F = 0

Its solution gives F =

1.12

(1.72)

1

e ± i mlf

(1.73)

2p

QUANTUM NUMBERS

Solution of Schrodinger’s wave equation results into three integral quantum numbers, viz. , and m. The fourth quantum number has been included to discuss the spin angular momentum of the electron. These numbers are discussed below in detail.

1. Principal Quantum Number (n) As already discussed, the principal quantum number represents the number of the main energy level or energy shell (orbit) in which the electron revolves round the nucleus. It gives the following information: (a) = 1, 2, 3, 4 ............ represents 1st, 2nd, 3rd, 4th shell, etc. designated as K, L, M and N respectively. (b) It gives the radius of an orbit (distance of electron from the nucleus) by the equation

n2 h 2 (CGS units) 4 2 me2 Z (c) It gives the energy of an electron in an orbit by the equation -2p 2 Z 2 me 4 = (CGS units) n2 h 2 (d) It gives the maximum number of electrons that can be accommodated in a given shell as 2 2. r =

2. Azimuthal Quantum Number or Subsidiary Quantum Number or Orbital Angular Momentum Quantum Number (l) It represents the existence of energy subshells in the principal shell and explains the appearance of the group of closely spaced lines in the hydrogen spectrum. For a given value of , total number of values is equal to . The possible values of range from 0 to ( – 1). This means: For = 1, = 0; only one subshell for 1st shell = 2, = 0, 1; two subshells for 2nd shell = 3, = 0, 1, 2; three subshells for 3rd shell, and so on. The subshells are designated as = 0, 1, 2, 3, 4..... , and so on. It gives the following information: (a) Total number of subshells in a given shell = (b) Maximum number of electrons that can be accommodated in a given subshell = 2 (2 + 1). Thus subshells can have 2, 6, 10, 14 electrons respectively.

1.18

Inorganic Chemistry

(c) Orbital angular momentum of an electron =

h l (l + 1) 2π

3. Magnetic Quantum Number or Orientation Quantum Number (ml or m) Orbital angular momentum is a vector quantity and can have a number of orientations in space. As a result, the accompanying magnetic momentum can also have a number of orientations in space represented by magnetic quantum number, m. For a given subshell, m can have its value from – to + (through zero) and each represents a particular orbital, a three-dimensional space in which probability of finding an electron is maximum. It means, For = 0, m = 0; only one orbital in -subshell, known as -orbital

Z-axis

= 1, m = –1, 0, +1; three orbitals in p-subshell, known as p-orbitals = 2, m = –2, –1, 0, +1, +2; five orbitals in d-subshell, known as d orbitals It gives us the following information: (a) The maximum number of orbitals in a given subshell = 2 +1 (b) The maximum number of electrons in any orbital = 2 (c) It accounts for the Zeeman effect, i.e. splitting of spectral lines in presence of strong magnetic field. Orbital angular momentum can have (2 + 1) components along any chosen direction. For example, l=2 the d-orbital can have five orientations of its orbital angular Fig. 1.9 Orientation of orbital angular momentum momentum in the presence of external magnetic field as shown in Fig. 1.9. Thus, a spectral line splits into five lines in the presence of an external magnetic field corresponding to electronic transition from a higher to a lower orbital of an atom.

4. Spin Quantum Number (s) An electron spins about its own axis either in a clockwise direction or anticlockwise direction and behaves like a small magnet. Thus, the spinning of the electron generates a magnetic moment known as spin angular

h , where is known as spin quantum number of the 2π electron corresponding to the two possible spin directions. The spin quantum number can have two values, 1 1 i.e. (corresponding to clockwise direction) and (corresponding to anticlockwise direction). 2 2

momentum of the electron given by

1.13

s(s +1)

PROBABILITY DISTRIBUTION CURVES

Depending upon the probability of finding the electron in a particular region at a given radial distance and in a particular direction from the nucleus, two types of probability distributions are discussed below:

1. Radial Probability Distribution Curves of Electrons The curves obtained by plotting the radial distribution function (4pr2R2) as a function of radial distance, r give the probability of finding of the electron at different radial distance from the nucleus. The radial probability distribution curves for a hydrogen atom are shown in Fig. 1.10. As can be seen in these diagrams, the probability is zero at r = 0, i.e. at the nucleus for every orbital. Each plot shows one or more peak, i.e.

Structure of Atom

1.19

Fig. 1.10 Radial probability distribution for hydrogen atom region of maximum probability of electrons. The number of peaks for an orbital is equal to ( – ). The surface at which the probability of finding an electron comes out to be zero is called node. Any orbital can have a maximum of ( – – 1) nodes (neglecting the nodes at infinity) (Fig. 1.10)

2. Angular Probability Distribution Curves or Shapes of Orbitals The angular function depends only on the direction and is independent of the radial distance, r. The angular probability distribution curves are noted as polar diagrams and symmetry of the angular functions are shown by the sign (+) and (–) as shown in the Fig. 1.11.

Fig. 1.11 Angular probability distribution curves It is evident that there is a different shape for every orbital. The s-orbitals can have only one orientation and hence are spherically symmetrical. The p-orbitals can have three orientations (px, py and pz orbitals depending upon the direction) and are of dumb-bell shape. The two lobes of a p-orbital are separated by

1.20

Inorganic Chemistry

a nodal plane of zero electron density perpendicular to the corresponding axis. The d-orbitals can have five orientations, i.e. there are five d-orbitals. The three d-orbitals, namely dxy, dyz and dxz have their lobes lying in the planes, and the two d-orbitals, namely dx2–y2 and dz2, have their lobes lying along the axis.

1.14

RULES FOR FILLING OF ORBITALS AND ELECTRONIC CONFIGURATION OF ELEMENTS

The atomic orbitals are filled up according to the following rules:

1. Aufbau Rule According to this rule, the orbitals are filled in increasing order of energy, i.e. the orbital with the lowest energy is filled up first followed by the orbital with the higher energy. Energy of the orbital is determined using the ( + ) rule. According to this rule, the orbital with lowest ( + ) value has lowest energy. For two orbitals with same ( + ) values, the orbital with lower has lower energy. Using these rules, the sequence of filling of the various orbitals has been represented in Fig. 1.12.

2. Pauli’s Exclusion Principle Fig. 1.12 Sequence of filling of This principle was put forth by Wolfgang Pauli in 1925. According various atomic orbitals to this principle, it is impossible for any two electrons in the same atom to have all the four quantum numbers same. For example, for a K shell, two combinations of the quantum numbers can be =1 =0 m=0 =+½ =1 =0 m=0 =–½ It is evident that three quantum numbers are same but the fourth is different. In other words, two electrons in the same orbitals have opposite spins. Since, the third electron has to acquire either of these two combinations, which is not possible, it is excluded from the orbitals. Thus, an orbital can have maximum of two electrons, with opposite spin. This principle is used to determine the maximum number of electrons in a subshell and shell.

Table 1.2 Maximum number of electrons for subshells and shells Principal quantum number (n)

Azimuthal quantum number (l)

Magnetic quantum number (m)

Spin quantum number (s)

1

0 1 -subshell

0 1 -orbital

+½ –½

2

0 2 -subshell

0 2 -orbital –1

+½ –½ +½ –½ +½ –½ +½ –½

1 2p-subshell

0 +1 2p-orbital

Maximum number of electrons 2 electrons 2 electrons

8 electrons

6 electrons

Structure of Atom

Principal quantum number (n)

Azimuthal quantum number (l)

3

0 3 -subshell

Magnetic quantum number (m) 0 -orbital –1

1 3p-subshell

0 +1 3p-orbitals –2

2 3d-subshell

–1 0 +1 +2 3d-orbitals

Spin quantum number (s)

1.21

Maximum number of electrons

+½ –½

2 electrons

+½ –½ +½ –½ +½ –½

6 electrons

18 electrons

+½ –½ +½ –½ +½ –½ +½ –½ +½ –½

10 electrons

3. Hund’s Rule of Maximum Multiplicity Spin multiplicity is represented by 2S + 1, where S is the total spin of electrons. According to Hund’s rule, in case of degenerate orbitals (orbitals of same energy), orbitals are filled up to have maximum multiplicity. For example, in case of 2p orbitals with 3-electron arrangement I has maximum spin multiplicity and is more stable because of the following reasons:

I

2S + 1 = 2 × 3 +1 =4 2

II

2S + 1 = 2 × 1 +1 =2 2

III

2S + 1 = 2 × 1 +1 =2 2

(a) Symmetry An arrangement with symmetrical distribution of charges is more stable because of lesser electrostatic repulsion as in case of the arrangement I. (b) Exchange Energy Two electrons with parallel spins can exchange their positions and lead to decrease in energy known as exchange energy. It means that arrangement I has least energy because of more possible exchanges and hence is more stable. (c) Pairing Energy Due to considerable repulsion between two electrons in the same orbital, some energy for pairing of two electrons known as pairing energy is required. This means that pairing increases the energy of the system and destabilises it. In short, Hund’s rule can be stated as : in the ground state, pairing of degenerate orbitals is not done till each orbital is singly occupied.

1.22

Inorganic Chemistry

n=6

6p 5d

6s n=5

5p 4d

5s n=4

n=4

4p 3d

Energy

n=3

4p

4d

3s

3p

3d

2s

2p

4f

Energy

4s 4s

n=3

4f

3p

3s

n=2 n=2

2p

2s n=1

n=1 1s

1s For hydrogen atom

For multi-electron atoms

Fig. 1.13 Energy-level diagrams Table 1.3 Electronic configurations of first twelve elements Name

Symbol

Electronic configuration

Hydrogen

1H

1

1

Helium

2He

1

2

Lithium

3Li

1 22

1

Beryllium

4Be

1 22

2

Boron

5B

1 2 2 2 2p1

Carbon

6C

1 2 2 2 2p2

Nitrogen

7N

1 2 2 2 2p3

Oxygen

8O

1 2 2 2 2p4

Fluorine

9F

1 2 2 2 2p5

Neon

10Ne

1 2 2 2 2p6

Sodium

11Na

1 2 2 2 2p6 3

1

Magnesium

12 Mg

1 2 2 2 2p6 3

2

Representation

Structure of Atom

1.23

Anomalous configurations of chromium and copper are the result of more stability of exactly halffilled and completely filled orbitals, reasons being the decrease in energy due to more symmetry and more exchange energy in half-filled and completely filled orbitals (as discussed earlier).

¼ ¼ ¼ ¼ ¼

¼ ¼ ¼ ¼ ¼

¼ ¼ ¼ ¼ ¼

d5

d10

(The more symmetrical distribution and more the chances of exchanges between electrons of parallel spin)

¼ ¼ ¼ ¼

¼ ¼ ¼ ¼

¼ ¼ ¼ ¼

d4

¼

d9

(the less symmetrical distribution and less chances of exchanges between electrons of parallel spin.)

The first successful attempt to explain the stability of an atom was made by Rutherford during his scattering experiment. He proposed that an atom consists of a positively charged central nucleus surrounded by electrons revolving in their orbits. However, this model couldn’t justify the stability of an atom and its spectrum. To remove these drawbacks, Bohr formulated his theory of atomic structure. He postulated that electrons move in circular orbits with a definite amount of energy associated with the orbits. These orbits are known as stationary or ground states. An electron can absorb some fixed amount of energy and move to a higher energy state. When it comes back to its ground state, it releases energy equal to the difference in energies of the 0.529 n2 Å, velocity of an electron, two levels. According to Bohr, the radius of the th orbit is equal to r = Z 7 1312 Z 2 = kJ/mol. However, v = 0.2183 ¥ 10 Z ms–1 and energy of an electron in the th orbit, is n2 n according to de Broglie, every matter has a dual character, i.e. particle as well as wave character and the wavelength associated with a particle, λ = h . Heisenberg pointed out in his uncertainty principle that it

mν is impossible to determine both the position and momentum of a microscopic particle simultaneously with complete accuracy. The uncertainty in measurement is given by

h 4 Thus, the probability concept came into light and Schrodinger formulated the wave equation for the Dx.mDn

∂ 2ψ ∂ 2ψ ∂ 2ψ 8π 2 m + + + 2 ( E − V )ψ = 0 h ∂x 2 ∂y 2 ∂z 2 The accepted valued of these wave function y are known as eigen functions and the corresponding energy values are known as eigen values. behaviour of a small particle as

EXAMPLE 1 When light radiation of wavelength 300 mm strikes a metal surface, the ejected electrons possess the kinetic energy equal to 1.68 × 105 J mol–1. Calculate the work function of the metal and the threshold wavelength.

1.24

Inorganic Chemistry

Kinetic energy of 1 mol of electrons = 1.68 × 105 Jmol–1 Kinetic energy of one electron = Now, or

1.68 × 105 = 2.789 × 10 −19 J 6.022 × 1023

hn = hn0 + K.E. hn0 = hn – K.E. hn0 = =

hc

– K.E.

 6.626 × 10 −34 Js × 3 × 10 −8 ms−1  −19   − 2.789 × 10 J 300 × 10 −9 m  

= 6.626 × 10–19J – 2.789 × 10–19J = 3.837 × 10–19 J Work function of electron = 3.837 × 10–19 J i.e.

hc

= 3.837 × 10–19 J

0

l0 =

6.626 × 10 −34 Js × 3 × 10 −8 ms−1 = 5.181 × 10–7 m 3.837 × 10 −19 J

EXAMPLE 2 Using Bohr’s theory, calculate the energy of an electron in first shell of H atom and Li2+ ion.

According to Bohr’s theory, energy of an electron is given by the expression

1312 Z 2 kJ/mol n2 For electron in first shell of H-atom; Z = 1 and = 1 =

1

=

1312 (1)2 (1)2

= – 1312 kJ/mol

For 1s electron of Li2+ ion, Z = 3 and = 1 =

1312 (3)2 = – 11808 kJ/mol (1)2

EXAMPLE 3 Calculate the wavelength of the emitted photon, when an electron in the H-atom returns from n = 2 to n = 1.

 1 1 1  = 109678  2 − 2  cm −1   λ  1 2  1 1 1  = 109678  2 − 2  cm −1 λ 1 2  1 3 = 109678 × cm −1 4 λ

Structure of Atom

l=

1.25

4 cm–1 = 12.20 × 10–6 cm 109678 3

= 122 nm

EXAMPLE 4 Calculate the ionisation energy of the Be3+ ion. For ionisation energy of the Be3+ ion, D =

2–

1=

1 and

2

=

1

 −1312 Z 2   −1312 Z 2   kJ/mol  −  n22 n12    

=  

1  1 1  1  = −1312 (4)2  − 2  = 1312 × 16 −  n2 n2   ∞ (1)   2 1 

= −1312 Z 2 

= 20 992 kJ/mol.

EXAMPLE 5 Calculate the radii of the first and third orbits of the hydrogen atom. Radius of an orbit, r =

0.529 2 n Å Z

For hydrogen atom, Z = 1 and for first orbit, = 1 r1 = 0.529 Å For third orbit, = 3r3 = 0.529(3)2 = 4.761 Å

EXAMPLE 6 Calculate the velocity of an electron in second orbit of hydrogen atom. Velocity of an electron,

n = 0.2183 × 107

Z ms–1 n

For hydrogen atom, Z = 1 and for second orbit, = 2 n2 =

0.2183 ¥ 107 = 1.095 × 106 ms–1 2

EXAMPLE 7 Calculate the wavelength associated with a body of mass 5 mg moving with a velocity of 2 ms–1. According to de Broglie’s concept, l =

h m

h = 6.626 × 10–34 Js, l=

m = 5 mg = 5 × 10–6 kg, n = 2 ms–1

h 6.626 × 10 −34 = = 6.625 × 10 −29 m mν 5 × 10 −6 × 2

1.26

Inorganic Chemistry

EXAMPLE 8 Calculate the uncertainty in the velocity of a body of 0.02 kg mass whose position is known with an uncertainty of 0.5 × 10–5 m.

According to the uncertainty principle,

h h or Dx = 4 π × m∆x 4 –34 –5 h = 6.626 × 10 Js; m = 0.02 kg and Dx = 0.5 × 10 m Dx × m Dn =

D =

6.626 × 10 −34 ms–1 = 5.3 × 10–28 ms–1 −5 4 × 3.14 × 0.02 × 0.5 × 10

EXAMPLE 9 Which of the following orbitals are allowed: 2p, 5p, 6d, 5h 2p 5p

n=2 n=5

l =1 l =1

Allowed Allowed

6d 5h

n=6 n=5

l=2 l=5

Allowed Not allowed

Èl = n - 1 to 0 ˘ Í ˙ and Í ˙ ÍÎl π n ˙˚

EXAMPLE 10 What are the possible values for the quantum numbers for an electron in the 3p orbital?

For the 3p orbital, = 3, = 1, m = –1, 0, +1 and ms = +½ and -½ for each value of m.

EXAMPLE 11 How many subshells and orbitals are possible for the n = 5 energy level? For

= 5, = 0, 1, 2, 3, 4

Thus, there are five corresponding subshells, viz. 5 , 5p, 5 5 and 5 . The total number of orbitals for any shell =

2

For = 5, the total number of orbitals = (5)2 = 25

EXAMPLE 12 Calculate the wavelength of the first and limiting line of the Lyman series in the H atom.

For first line of Lyman series, = 1,

2

=2

4  1 1 1  3 = 109678  2 − 2  = 109678 × , λ = = 122 nm 109678 × 3 4 λ  (1) (2)  For limiting line of Lyman series,

1

= 1,

2

=0

1  1 1 1  –6 = 109678  2 − = 109678 , λ = 109678 cm = 9.117 × 10 cm = 9117.6 nm 2  λ  (1) (∞) 

Structure of Atom

1.27

QUESTIONS Q.1 Discuss Rutherford’s model and its drawbacks. Q.2 Enumerate the postulates of Bohr’s model. How does Bohr’s model explain the spectrum of hydrogen? Q.3 Derive an expression for the radius of an orbit, velocity, and energy of an electron in the hydrogen atom. Q.4 Calculate the radii of the second and fifth orbits in hydrogen. Q.5 Calculate the frequency of light emitted when an electron in an excited state undergoes transition from the third orbit to the first orbit. Q.6 Calculate the wavelength of the first and limiting line of the Balmer series. Q.7 Calculate the ionisation energy of the Li+ ion. Q.8 Calculate the de Broglie wavelengths of an electron moving at 20% the speed of light. Q.9 Calculate the uncertainty in the position of an electron if uncertainty in its velocity = 3 × 10–2 ms–1. Q.10 State Heisenberg’s uncertainty principle and discuss its significance. Q.11 What are the drawbacks in Bohr’s theory of atomic structure? Q.12 Write down the Schrodinger wave equation and define each of the terms in it. Q.13 What is eigen function and what is its significance? Q.14 Define quantum numbers and explain the significance of each of these numbers. Q.15 Explain the following: (a) Photoelectric effect (b) Compton effect (c) Pauli’s exclusion principle (d) Exchange energy Q.16 Describe Hund’s rule of maximum multiplicity. Q.17 Differentiate between an orbit and orbital. Q.18 Why are half-filled and completely filled orbitals more stable? Justify your answer with the help of suitable examples. Q.19 Discuss Schrodinger wave equation for hydrogen atom. Q.20 Draw probability distribution curves for 2s and 3p orbitals.

MULTIPLE-CHOICE QUESTIONS 1. The electronic level in hydrogen atom that can absorb a photon but not emit is (a) 1s (b) 2s (c) 2p (d) 3s 2. What will be the uncertainty in the momentum of an e–, if uncertainty in its position is zero (a) Zero (b) ½ (c) 2 (d) > h/2 3. The ratio of energy of two radiations with wavelengths 1000 Å and of 2000 Å respectively would be equal to (a) ¼ (b) ½ (c) 4 (d) 2 4. In the ground sate of chromium atom, the total no. of orbitals populated by the electrons are equal to (a) 14 (b) 20 (c) 16 (d) 15 5. In the third Bohr orbit of Li2+ ion, the velocity of the electron is equal to (a) 6.6 × 108 cm/s (b) 19.8 × 108 cm/s (c) 2.2 × 108 cm/s (d) 4.4 × 108 cm/s

chapter

Nuclear Chemistry

2

After studying this chapter, the student will be able to

disintegration

2.1

NUCLEUS

We have already discussed the structure of an atom in Chapter 1. The Rutherford scattering experiment revealed the existence of a positively charged central part of an atom termed nucleus. The experiment proved that size of the nucleus is incredibly small, of the order of 10–15 m. Further experiments have shown that the radius of the nucleus is given by r = R0A1/3 where R0 is a constant with a value equal to 1.5 × 10–15 m and A is the mass number of the element. Nuclear radii are measured in femtometres (1 fm = 10–15 m). For example, the nucleus of oxygen has a radius of 2.5 fm. Studies have proved that most of the mass of an atom is concentrated in the nucleus. Hence, the density of the nucleus is very

Nuclear mass g/cm 3 4p 3 ¥ (Nuclear radius) 3 Density of the nucleus is found of the order of 2.4 × 1014 g/cm3. high and is given by the relation, r =

2.2

COMPOSITION OF THE NUCLEUS

Early studies established that the nucleus is composed of protons and neutrons collectively called nucleons and were considered as fundamental particles. But after the experimental verification of the dual character of the electron, a British scientist, Dirac presented a quantum mechanical treatment of an electron in the form

2.2

Inorganic Chemistry

of antiparticle theory. He suggested that in the universe, there are some unstable fundamental particles along with the stable fundamental particles. He considered that to every particle in the universe, there is a corresponding antiparticle. Thus, there must exist anti-electrons, antiprotons and antineutrons corresponding to electrons, protons and neutrons respectively. The first antiparticle was discovered by Carl Anderson in 1932. He bombarded light elements with alpha particles and was successful to discover a new particle as anti-electron which was later called positron. 10 5B

+ 42He $ 137N + 10n

13 7N

$ 136C + +10 e anti-electron (positron)

An electron and positron mutually annihilate each other producing gamma ray photons. 0 +1e

+ (–1)0e $ hn + hn´

In 1930, Wolfgang Pauli put forth the theory of beta decay and indicated the existence of the neutrino to explain the loss of mass during the decay of free neutron into a proton and an electron. A neutrino has zero charge and variable mass, always less than the mass of an electron. The existence of other particles was demonstrated later in 1952, by Allen and Rodeback. In 1934, Enrico Fermi indicated the existence of the antineutrino, a particle identical to neutrino but with opposite spin. The existence of the antineutrino was proved by Reines and Cowan in 1953. A Japanese physicist, Hideki Yukawa in 1934, combined the theory of relativity and quantum theory and described the nuclear interactions by the exchange of new particles called mesons, between protons and neutrons. Mesons were actually detected later in certain radiation experiments. In 1949, Enrico Fermi and C.N. Yang suggested that a nucleon and an antinucleon exist as a composite particle called pyion. Nowadays, with the existence of the modern high-technology equipments, many more particles have been discovered. These particles can be divided into two categories, i.e. elementary particles and composite particles.

2.2.1 Elementary Particles The elementary particles are the fundamental particles and are not composed of other particles. These are classified according to spins of these particles into two types:

1. Fermions These elementary particles have half integer spin (½) and have their own distinct antiparticle. These are further divided its two types, viz. quarks and leptons.

(a) Quarks Quarks are the fundamental constituents of nucleons and are subjected to strong nuclear forces. The name was given by Murray Gell–Mann in 1964. These are in six forms and can be arranged into three families as given in the Table 2.1. Their respective antiparticles are known as antiquarks which carry the opposite electric charge. (b) Leptons Leptons are lighter particles and are not subjected to strong forces. They are not found in the nucleus, but may be produced in the nucleus and are quickly expelled. The leptons also come in six forms arranged in three families as given in Table 2.1. The neutral leptons are collectively called neutrinos.

2. Bosons Bosons are the particles which mediate the fundamental force of nature and carry integer spin. The elementary bosons are listed in Table 2.2.

Nuclear Chemistry

2.3

Table 2.1 Fermions (spin = ½) Family

Quarks Name

I

II

III

Leptons

Symbol

Electrical charge

Up Down

u d

+2/3 –1/3

Strange Charged

s c

–1/3 +2/3

Bottom Top

b t

–1/3 +2/3

Rest mass in MeV

Name

Symbol

Electrical charge

Rest mass in MeV

~5 ~7

Electron Electronneutrino

e

–1 0

0.511 liquid > gas. The van der Waals forces exist due to different types of long-range attractive forces between molecular interactions as discussed here:

1. Dipole-Dipole Interactions (Keesom forces) All polar molecules have permanent dipole moment. The electrostatic interactions between the positive end of dipole of the molecule and negative end of the dipole of another molecule result in dipole-dipole interactions. Greater the dipole moment of the molecules, greater is the dipole-dipole interactions, meaning stronger are the van der Waals forces. For example, NH3, SO2, HCl are polar molecules showing this type of interactions. Further, because of these interactions, these gases can be easily liquified. The electrostatic energy, U between two interactive dipoles 1 and 2 is given by the expression, Umm =

2 m12 m22

indicating that these interactions are inversely proportional to the sixth 3kT(4p e 0 )2 r 6 power of the intermolecular separation.

2. Induced Dipole-induced Dipole Interactions or London Forces (Dispersive Forces] Nonpolar molecules like O2, N2, I2, He, Ne, etc., have no permanent dipole moment (Fig. 3.2). However, the existence of van der Waals forces in such molecules has been established. –+– Fig 3.2 Average charge distribution

3.10

Inorganic Chemistry

The existence of these forces has been explained by London. According to London, electrons are in a constant motion in + + –– –– an atom, i.e. the electron density may be concentrated round the nucleus in one region at a given instant resulting in an Fig. 3.3 Included dipole-induced unsymmetrical distribution of charges. As a result, the nondipole interaction polar molecules gets instantaneously polarised and develops an induced dipole in another neighboruing non-polar molecule (Fig. 3.3). Such types of interactions are also called dispersive forces due to association of phenomenon of dispersion of light with these forces. However, the magnitude of these forces is very small and are the exclusive source of van der Waals forces in such molecules. The dispersion or London energy for two are inversely proportional to the sixth power of the intermolecular separation polar molecules of the same kind given by Udisp =

-3a 2 hn 0 4(4p e 0 )2 r 6

3. Dipole Induced Dipole Interactions (Debye Forces) A polar molecule having a permanent dipole may also induce dipole in a nonpolar molecule resulting in dipole induced dipole type of interactions. The average induction energy, or debye energy, between a nonpolar molecule and a -m 2a polar molecule is given by Uind = U(ma) = (4p e 0 )2 r 6 The interactions also vary as r–6 and are not temperature dependent. It should be noted that the total long-range intermolecular energy for interaction between two polar molecules is the sum of these three interactions, i.e. U = Uel + Uind + Udisp

Factors Affecting Van der Waals forces 1. Molecular Size Greater the size of the molecule, more are the interactions and hence stronger are the van der Waals forces. That is why, boiling point of noble gases increases in the order He < Ne < Ar < Kr < Xe < Rn

2. Surface Area More is the surface area of a molecule, more is the interaction and stronger are the van der Waals forces for a homologous series of compounds. That is why, boiling point of ethane is higher than that of methane. A special type of intermolecular force is the hydrogen bond discussed in the next article.

3.7

HYDROGEN BOND

As already discussed, van der Waals forces increase with decrease in size of the molecule resulting in an increase in the boiling point of the molecule, but it has been seen that boiling points of hydrides of groups 15, 16 and 17 show abnormal variation as shown in Table 3.6. This abnormal behaviour is justified on the basis of a unique force known as hydrogen bond. It is defined as the electrostatic force of attraction between a hydrogen atom bonded to a highly electronegative element (like F, O or N) and another highly electronegative element (like F, O or N) present in a molecule of the same or different substance and is represented by a dotted line. Consider the case of the HF molecule. Due to linkage of hydrogen atom with an electronegative atom, F, the hydrogen atom aquires partial positive charge which is represented as H + – F –. This hydrogen atom is now attracted towards the fluorine atom of

3.11

Chemistry Bonding

Table 3.6 Boiling point variation for some hydrides Hydrides of Group 15 elements PH3 AsH3 SbH3 219K 237K 255K Hydrides of Group 16 elements H2S H2Se H2Te 212K 232K 271K Hydrides of Group 17 elements HCl HBr HI 188K 206K 238K

NH3 240K H2O 373K HF 293K

BiH3 295K

other HF molecule and is linked by electrostatic force of attraction known as hydrogen bond. This can be represented as follows: H +—F – ------- H +—F – ------ H +—F – Thus, hydrogen bond is a type of dipole-dipole attraction but is more stronger than other dipole–dipole interactions not involving the cases discussed here. These bonds are very weak with bond energy lying between 4 to 45 kJ/mol and are yet quite important in various physiochemical systems.

1. Types of Hydrogen Bonding Depending upon the linkage formed between two different molecules of same or different compounds and linkage formed within a molecule, the hydrogen bond is of two types:

(a) Intermolecular Hydrogen Bonding If a hydrogen bond is formed between two or more molecules of same or different compounds, it is known as intermolecular hydrogen bonding. It results in the association of molecules as shown in Fig. 3.4. F H

O H

H

O

H Liquid H2O R

H

O

H

H F

R

O

H

O

H

H

O

O

H

H

O

O

H

H

H

O R

H

H F

O

C

C O

Alcohols

H

F H

F Solid HF

R

O

H

H

H

H

O

Carboxylic acid O

Cu

H

O

H

O

O

O

N O

S

H H Solid CuSO4.SH2O

O

O

N O

H

O

p-nitrophenol

Fig. 3.4 Association of molecules due to intermolecular H–bonding

(b) Intramolecular Hydrogen Bonding If a hydrogen bond is formed between hydrogen and an electronegative atom, both present in the same molecule, it is known as intramolecular hydrogen bonding. It leads to the formation of a five-membered or six-membered ringlike structure and hence is also known as chelation.

3.12

Inorganic Chemistry

2. Condition for the Formation of Hydrogen Bond

O H

The two main conditions to be satisfied for hydrogen bonding are

O

(a) Presence of highly electronegative atoms like F, O or N directly linked to H atom by covalent bond

C

H O

O

(b) Small size of the highly electronegative atom for strong Fig. 3.5 polarisation It should be noted that electronegativities of nitrogen and chlorine are exactly same = 3.0. But nitrogen can form hydrogen bond due to its small size (0.76 Å) and chlorine cannot (0.99 Å).

Intramolecular H–bonding in salicyclic acid

3. Effects of Hydrogen Bonding (a) Variation in the Boiling Points of Hydrides of Groups 15, 16 and 17 As discussed earlier, boiling point of a compound increases with the increase in magnitude of the van der Waal’s forces. However, the boiling points of hydrides of the first elements in these groups is exceptionally higher as compared to the corresponding hydrides of the next elements in the same group. This is due to the reason that nitrogen, oxygen and flourine form hydrogen bonds and result in the formation of (NH3)x, (H2O)x, (HF)x clusters. More heat energy is required to break hydrogen bonds in these clusters resulting in higher boiling point. The remaining elements of these groups cannot form hydrogen bonds due to large size and low electronegativity. Hence, their boiling points vary according to the variation in van der Waal’s forces. That is why water has a high boiling liquid due to strong intermolecular hydrogen bonding and H2S is a gas due to absence of hydrogen bonding. (b) Low Density of Ice than Water In the solid state, water shows extensive intermolecular hydrogen bonding to form a cagelike structure in which every water molecule is associated with four other molecules in tetrahedral fashion. Thus, ice has an open structure with large empty space. When temperature is increased up to zero degree centigrade (Fig. 3.6), a large number of hydrogen bonds are broken and molecules come closer to one another resulting in a sharp increase in the density. On further heating up to 4°C, more molecules come close together resulting in constriction of volume and increase of density. But on further heating, the effect of expansion predominates leading to increase in volume. That is why, water has higher density than ice with its maximum value at 4°C.

H O H

H

H

O

O H

H

H O

H

Vacant spaces

H O

H

H O

H

Fig. 3.6 Large structure of ice

(c) Comparison Between Ortho, Meta and Para-isomers of an Aromatic Compound The ortho-isomer of an aromatic compound shows intermolecular H-bonding while meta and para-isomers show intermolecular H-bonding and get associated to form clusters. As a result, ortho- isomers have lesser melting point and boiling point as compound to meta- and para-isomers and hence are more volatile. Due to intramolecular H–bonding in an ortho-isomer, its solubility in water decreases as it now cannot form intermolecular hydrogen bond with H2O molecules (Fig. 3.7).

Chemistry Bonding

O N

3.13

O H O

O HO

N

O O

H—O

N O

o-nitrophenol

p-nitrophenol

Fig. 3.7 Comparison between intermolecular and intermolecular H-bonding

3.8

ORBITAL OVERLAP THEORY

Heitler and London in 1927 gave a theoretical treatment for covalent-bond formation which was further modified by Pauling in the form of orbital overlap theory. According to this theory, (a) Covalent bond formation takes place by the overlapping between valence-shell orbitals of the two atoms each containing one unpaired electron, resulting in the formation of a bond orbital. The two electrons get paired in the bond orbital with opposite spin and are called bond pair of electrons. The combining orbitals have proper orientation. (b) The valence-shell orbitals with paired electrons don’t participate in the overlapping process and their electrons are called nonbonding electron pairs or lone pair of electrons. (c) Strength of the covalent bond depends upon the extent of overlapping between the atomic orbitals. The relative strength of the covalent bond obtained by overlapping between s–s, s–p and p–p overlap is s–s < s–p < p–p. (d) The atomic orbitals overlap together to form two types of covalent bonds, i.e. s and p-bond. (e) Covalent bond has directional characteristics, for example the bonds formed by overlapping between p-orbitals are directed towards the axis of the two orbitals.

3.8.1 Types of Covalent bonds 1. Sigma Bond (s) The sigma bond is formed by head-on overlapping of the two half-filled orbitals of the valence shells of the two combining atoms. This bond can be formed by overlapping of s–s, s–p and p–p atomic orbitals with proper orientation as shown below (Fig. 3.8). z s–s overlap

s–p overlap

pz–pz overlap

Fig. 3.8 Head-on overlapping of atomic orbitals

2. Pi-bond ( ) Pi-bond is formed by side wise overlapping of two half filled p orbitals of the valence shell of the two combining atoms as shown in Fig. 3.9. The already present -bond poses a hinderance between the two atoms for further overlapping for p-bond formation. Hence, the extent of overlapping decreases and sigma bond becomes stronger than pi bond.

3.14

Inorganic Chemistry

H—1s1 H 1s H

1s(H)

1s s–s overlap

Fig. 3.9

1s(H)

H2

H : H or H – H

Fig. 3.10 Formation of H2 molecule

Sidewise overlapping of p-orbitals

Examples

1. H2 Molecule The Hydrogen atom has only one electron in its 1s orbital. The 1s orbitals of the two

hydrogen atoms with unpaired electrons overlap with each other to form sigma bond as shown in Fig. 3.10.

2. F2 Molecule The fluorine atom has one unpaired electron in its p-orbital. The two half-filled orbitals

?

?

?

?
bp–bp 3. The lone pair and bond pair of electrons arrange themselves around the central atom in order to have minimum electrostatic forces of repulsion. 4. The geometry of the molecule is said to be symmetrical if the atom is surrounded by hybrid orbitals containing only bond pair of electrons and /or the surrounding atoms are same. However, if the central atom is surrounded by lone pair of electrons/or the surrounding atom are not same, the geometry of the molecule is said to be unsymmetrical. The shapes of molecules depending upon these points has been summarised in Table 3.18. Table 3.18 Shapes of molecules depending on VSEPR theory Total e´ pair

No. of bp

No. of lp

AB2

2

2

0

Linear (180°)

Linear

BeF2, BeCl2

AB3 AB2L

3

3 2

0 1

Triangular planar (120°)

Triangular planar V shape

BCl3, BH3 SnCl2, PbCl2

AB4 AB3L AB2L2

4

4 3 2

0 1 2

Tetrahedral (109.5°)

Tetrahedral pyramidal V-shape

CH4, NH+4, SiCl4 NH3, PH3 H2O, SCl2, SeCl2

AB5

5

5

0

Trigonal bipyramidal (120°, 90°)

Trigonal bipyramidal

PCl5, SbCl5

4 3 2

1 2 3

see-saw T-shape Linear

SF4, SeCl4 ClF3, BrF3 XeF2, ICl2–

Type of the species

AB4L AB3L2 AB2L3

Spatial arrangement of electron pairs

Geometry of the species

E.g.

AB6 AB5L AB4L2

6

6 5 4

0 1 2

Octahedral (90°, 90°)

Octahedral Square pyramidal Square plannar

SF6, TeF6 IF5, (SbF5)– XeF4, ICl4–

AB7

7

7

0

Pentagonal bipyramidal (72°, 90°)

Pentagonal bipyramidal

IF7

6

1

distorted Octahedral

XeF6

AB6L

3.14

SHAPES OF SOME COMMON MOLECULES

Geometry of some molecules has been discussed by using the two concepts—hybridisation and VSEPR theory, as discussed here:

3.51

Chemistry Bonding

1. AB4 Type Species This type of species have four bond pairs of electrons accommodated in sp3 hybrid orbitals of the central atom arranged in tetrahedral geometry. (a) CH4 Molecule This was discussed in the previous section. (b) SO42– Ion The valence shell electronic configuration of the central atom S, in ground state is 3s2 3px2 3py1 3p1z and in the excited state is 3s1 3px1 3py1 3p1z 3d1xy 3d1yz (in order to form four s bonds, four unpaired electrons are required and formation of two p-bonds require further two unpaired electrons). The 3s and three 3p orbitals undergo hybridisation to form four equivalent sp3 hybrid orbitals aligned along the corners of a tetrahedron. In case of SO42–, two oxygen atoms are considered to carry one negative charge each, so that these have only one half-filled 2pz orbital. These half-filled 2pz orbitals of the two O– ions and half-filled 2pz orbitals of the two O atoms overlap with the four sp3 hybrid orbitals of S atom to form four s covalent bonds. The remaining unhybridised half-filled 3dxy and 3dyz orbitals of S atom overlap with the remaining half-filled p-orbitals of each of the two O atoms to form pp-dp bond (fig. 3.64). Thus O–S–O bond angle is 104.9° and the molecule is tetrahedral. O– ion

O atom 2p

2s

2s s bond2p

s bond

– O

pp–dp bond

S in excited state

S 3s s bond

O–

3d

3p

O

pp–dp bond

s bond

O

O– ion

O atom 2s

2p

2s

2p

Fig. 3.64 Geometry of SO42– ion

2. AB3L Type Species This type of species are having three bond pairs of electrons and one lone pair of electrons accomodated in sp3 hybrid orbitals of the central atom arranged in tetrahedral geometry. However, due to lp–bp repulsion, the bond angle of the species is slightly less than 109.5°. NH3 Molecule The valence shell electronic configuration for N atom is 2s2 2px1 2py1 2p1z . It has three unpaired electron required for the formation of three s covalent bonds. Thus 2s and three 2p orbitals undergo hybridization to form four equivalent sp3 hybrid orbitals, out of which three orbitals overlap with half-filled 1s orbitals of each H atom to form three s covalent bonds. The H–N–H bond angle should be 109.5°, but studies reveal the bond angle as 107.3°. This is due to the reason that lp–bp repulsion is greater than bp–bp repulsion and the molecule gets distorted from the normal tetrahedral geometry to pyramidal and the bond angle decreases to 107.3° (Fig. 3.65).

N N in ground state

H 2s

2p

sp3 hybridisation

107.3° H

H H

Fig. 3.65 Geometry of NH3 molecule

H

H

3.52

Inorganic Chemistry

3. AB2L2 Type Species This type of species are having two bond pairs of electrons and two lone pairs of electrons accommodated in sp3 hybrid orbitals of the central atom arranged in tetrahedral geometry. However, due to lp–lp repulsion, the bond angle of the species is quite less than 109.5°. H2O Molecule The valence-shell electronic configuration for O atom is 2s2 2px2 2py1 2p1z with two unpaired electrons required for the formation of two s covalent bonds. The hybridization of 2s and three 2p orbitals results in the formation of four equivalent sp3 hybrid orbitals out of which two overlap with half-filled 1s orbitals of each H atom to form two s covalent bonds. As a result, the central atom is surrounded by two lone pairs of electrons and two bond pair of electrons. Since lp–lp > lp–bp > bp–bp repulsion, O in the H2O molecule is more distorted than ground 2p state 2s NH3 molecule and bond angle further 3 104.5° sp hybridisation decreases to 104.5°. Thus, the molecule H H has a V-shaped, angular or bent geometry Fig. 3.66 Geometry of H2O molecule (Fig. 3.66). Comparison of the Bond Angle of the Hydrides of Group 16 Elements The bond angle order for the hydrides of Group 16 elements is H2O > H2S > H2Se > H2Te 104.5o

92.5o

91.05o

89.5o

This is due to the reason that as the size of central atom increases, its electronegativity decreases. As a result, the bond pairs move away from the central atom. Hence, repulsion between bond pairs decreases and these come closer. Also the lone-lone pair repulsion between the orbitals of the largesized atom are higher, so that the bond angle decreases.

4. AB5 Type Species This type of species are having five bond pairs of electrons accommodated in sp3d hybrid orbitals of the central atom arranged in trigonal bipyramidal geometry. PF5 molecule The valence-shell electronic configuration of P atom in the ground state is 3s23p3. It requires promotion of one electron from 3s to 3dz2 orbital in order to form five sigma covalent bonds. Now the 3s, three 3p and one 3d orbital undergo hybridization to form five sp3d hybrid orbitals. Out of these, three orbitals are oriented towards the corners of an equilateral triangle and form equatorial bonds with bond angle 120o. The other two orbitals are oriented at right angles to the plane of the equatorial orbitals and form axial bonds with bond angle 180o. Thus, the bond angle between an axial and equatorial bond is 90o. As a result, the axial bonds feel greater repulsion and the axial bonds get slightly lengthened (Fig. 3.67)

Fig. 3.67 Geometry of PF5 molecule

Comparison of the Bond Angles of the Halides of Group 15 Elements PF3 < PCl3 < PBr3 < PI3

Chemistry Bonding

3.53

This is due to the reason that as the electronegativity of surrounding atoms decreases with increasing size, bond pairs get closer to the central atom. So the repulsion between the bond pairs increases and these are pushed farther apart so that the bond angle increases.

5. AB4L Type Species This type of species are having four bond pairs of electrons and one lone pair of electrons accommodated in sp3d hybrid orbitals of the central atom arranged in trigonal bipyramidal geometry. Due to lp-bp repulsion, the molecule is distorted and the geometry is seesaw shaped. SF4 Molecule The valence-shell electronic configuration of S atom in the ground state is 3s23p4. It requires promotion of one electron from 3p to 3dz2 orbital in order to form four sigma covalent bonds. Now the 3s, three 3p and one 3d orbital undergo hybridisation to form four sp3d hybrid orbitals. Out of these, four orbitals overlap with half-filled 2pz orbitals of each F atoms to form four covalent bonds, while the fifth orbital at equatorial position (lesser repulsion than axial position) contains a lone pair of electrons. Due to lp-bp repulsion, the molecule gets distorted and bond angles, 90o, 120o and 180o decrease to 89o, 118o and 177o respectively (Fig. 3.68).

Fig. 3.68 Geometry of SF4 molecule

6. AB3L2 Type Species This type of species are having three bond pairs of electrons and two lone pair of electrons accommodated in sp3d hybrid orbitals of the central atom arranged in trigonal bipyramidal geometry. Due to lp-lp repulsion, the molecule is distorted and the geometry is T- shaped. ClF3 Molecule The valence-shell electronic configuration of Cl atom in the ground state is 3s23p5. It requires promotion of one electron from 3p to 3dz2 orbital in order to form five sigma covalent bonds. Now the 3s, three 3p and one 3d orbital undergo hybridisation to form five sp3d hybrid orbitals. Out of these, three orbitals overlap with half-filled 2pz orbitals of each F atom to form three covalent bonds, while the remaining two orbitals at equatorial position (lesser repulsion than axial position) contain a lone pair of electrons each. Due to lp-bp repulsion, the molecule gets distorted and bond angle, decreases from 90° to 87.6o (Fig. 3.69).

Fig. 3.69 Geometry of ClF3 molecule

7. AB3L3 Type Species This type of species are having two bond pairs of electrons and three lone pair of electrons accommodated in sp3d hybrid orbitals of the central atom arranged in trigonal bipyramidal geometry. However, the lone pairs occupy the equatorial positions and the molecule is linear.

3.54

Inorganic Chemistry

XeF2 Molecule The valence shell electronic configuration of Xe atom in the ground state is 3s23p6. It requires promotion of one electron from 3p to 3dz2 orbital in order to form two sigma covalent bonds. Now the 3s, three 3p and one 3d orbital undergo hybridisation to form five sp3d hybrid orbitals. Out of these, two orbitals overlap with half-filled 2pz orbitals of each F atoms to form two covalent bonds, while the remaining three orbitals at equatorial position (lesser repulsion than axial position) contain a lone pair of electrons each. Thus, the molecule is linear (Fig. 3.70).

Fig. 3.70 Geometry of XeF2 molecule

8. AB6 Type Species This type of species are having six bond pairs of electrons accommodated in sp3d2 hybrid orbitals of the central atom arranged in octahedral geometry. SF6 Molecule The valence-shell electronic configuration of S atom in the ground state is 3s23p4. It requires promotion of one electron each from 3s and 3p orbitals to 3dz2 and 3dx2 – y2 orbitals in order to form six sigma covalent bonds. Now the 3s, three 3p and two 3d orbitals undergo hybridisation to form six sp3d2 hybrid orbitals oriented towards the corners of an octahedron with F–S–F bond angle 90o (Fig. 3.71). F

S in ground state

F 3s

3p

F

3d S

S in excited state

F sp3d hybridisation

F F

Fig. 3.71 Geometry of SF6 molecule

9. AB5L Type Species This type of species are having five bond pairs of electrons and one lone pair of electrons accommodated in sp3d2 hybrid orbitals of the central atom arranged in octahedral geometry. Due to lp-bp repulsion, the molecule is distorted and the geometry is square pyramidal. ClF5 Molecule The valence-shell electronic configuration of Cl atom in the ground state is 3s23p5. It requires promotion of one electron each from 3s and 3p orbitals to 3dz2 and 3dx2 – y2 orbitals in order to form six sigma covalent bonds. Now the 3s, three 3p and two 3d orbitals undergo hybridisation to form six sp3d2 hybrid orbitals oriented towards the corners of an octahedron. Out of these, five orbitals overlap with half-filled 2pz orbitals of each F atoms to form five covalent bonds, while the sixth orbital contains a lone pair of electrons. Due to lp-bp repulsion, the molecule gets distorted and the geometry is square pyramidal (Fig. 3.72).

Chemistry Bonding

3.55

F

Cl in ground state 3s

3p

3d

F

F Cl

Cl in excited state

F

F

sp3d 2 hybridisation

Fig. 3.72 Geometry of ClF5 molecule

10. AB4L2 Type Species This type of species are having four bond pairs of electrons and two lone pair of electrons accommodated in sp3d2 hybrid orbitals of the central atom arranged in octahedral geometry. The lone pairs are oriented opposite to each other at axial positions and the geometry is square planar. XeF4 Molecule The valence shell electronic configuration of Xe atom in the ground state is 3s23p6. It requires promotion of two electrons 3p orbitals to 3dz2 and 3dx2 – y2 orbitals in order to form four sigma covalent bonds. Now the 3s, three 3p and two 3d orbitals undergo hybridization to form six sp3d2 hybrid orbitals oriented towards the corners of an octahedron. Out of these, four orbitals overlap with half-filled 2pz orbitals of each F atoms to form four covalent bonds, while the remaining two orbitals each containing a lone pair of electrons are oriented opposite to each other at axial positions and the geometry is square planar (Fig. 3.73). Xe in ground state 3s

3p

3d

F

F Xe

Xe in excited state

F

F

sp3d 2 hybridisation

Fig. 3.73 Geometry of XeF4 molecule

11. AB2 Type Species This type of species are having two bond pairs of electrons accommodated in sp hybrid orbitals of the central atom arranged in linear geometry. CO2 Molecule The valence shell electronic configuration of the central atom, C in ground state is 2s2 2p2. In the excited state, one electron from 2s orbital is promoted to one of the empty p-orbitals. Now these singly occupied 2s and 2pz orbitals undergo hybridization to form two equivalent sp hybrid orbitals aligned in opposite directions (Fig. 3.74). These two half-filled sp-hybrid orbitals of carbon atom overlap with half-filled 2pz orbitals of two oxygen atoms to form 2s bonds. The remaining two unhybridised half-filled 2px and 2py orbitals of carbon atom undergo lateral overlap with each of the two half-filled 2p orbitals of two oxygen atoms to form pp–pp bond. Since p-bonds are not included in hybridisation and do not affect the shape of the molecule, hence CO2 molecule is linear in shape.

3.56

Inorganic Chemistry

O atom

p bond p bond

s bond

s bond

C in excited state s bond

s bond C

O

sp hybridisation

p bond

O

z

p bond

O atom

: :

: :

s s C O p p

O

Fig 3.74 Geometry of CO2 molecule

12. AB3 Type This type of species are having three bond pairs of electrons accommodated in sp2 hybrid orbitals of the central atom arranged in trigonal planar geometry. Carbonate ion CO32– In carbonate ion, two oxygen atoms carry uninegative charge and have only one half-filled p orbital. The third oxygen atom, has the normal two half-filled p-orbitals (Fig. 3.75). The 2s, 2px and 2pz orbitals of the carbon atom undergo hybridisation to form three equivalent sp2 hybrid orbitals aligned at the corners of the trigonal. These three sp2 hybrid orbitals overlap with single filled 2p orbitals of two oxygen atoms and one O– ion to forms three sigma bonds. Now the unhybridised 2py orbital of carbon atom overlaps with the 2p orbital of oxygen atom to form pi bond. Hence, the molecule is triangular planar in shape. –

–

–

O ion

O ion 2s

2p

s bond

2s

O–

O 120°

2p s-bond

C C in excited state sp2 hybridisation

2s

s bond

2p

p-bond O

O atom 2s

2p

Fig 3.75 Geometry of CO32– ion

13. Nitrate Ion (NO3-) In nitrate ion, one of the oxygen atom carries uninegative charge and has only half-filled 2p orbitals. While the other two oxygen atoms donot carry any charge and have two half-filled 2p orbitals. The 2s, 2px and 2pz orbitals to N atom hybridise to form three equivalent sp2 hybrid orbitals arranged in trigonal planar geometry. Now, one oxygen atom rearranges its electrons to have one empty 2p orbital which overlaps with one of the sp2 hybrid orbital having lone pair of electrons to form a sigma coordinate bond. While the other two oxygen atoms form 2s bonds by the overlapping of their 2p orbitals with the two sp2 hybrid orbitals. The unhybridised 2p orbital of N atom undergoes lateral overlapping with 2p orbital of third oxygen atom to form p-bond as shown in Fig. 3.76. This type of species are having three bond pair of

3.57

Chemistry Bonding

O:

:

:O

:

–

:

Rearrangement of electrons in – O ion O atom

:

electrons and one lone pair of electrons in sp2 hybrid orbitals of the central atom aligned at the corners of a triangle.

2p 2s Coordinate s bond

2p

2s

120° s bond N

N atom sp2 hybridisation 2s

s bond

2p

p-bond :O:

O atom 2p

2s

Fig 3.76 Geometry of NO3– ion

14. AB3 with one Lone Pair or AB2L In SnCl2 molecule, the valence-shell electronic configuration of the central atom, Sn, in ground state is 5s2 5px1 5py1. It has two unpaired electrons required for the formation of two covalent bonds. The 5s, 5px and 5py orbitals undergo hybridization to form three equivalent sp2 hybrid orbitals aligned at the corners of a trigonal. Thus, the bond angle Cl–Sn–Cl should be 120° after the pz orbitals of Cl atom overlap with sp2 hybrid orbital of Sn to form covalent bond. However, studies several that Cl–Sn–Cl bond angle is lesser than 120°. This is due to the reason that after bond formation, Sn atom is surrounded by two bond pairs and one lone pair of electrons. As bp-bp repulsion is lesser than lp-lp repulsion, the orbitals are pushed inwards and the bond angle decreases than 120° and the molecule is V-shaped (Fig. 3.77). Lone pair Sn

Sn in ground state 5s sp2

Bond pair

5p hybridisation

Cl Cl

Fig 3.77 Geometry of SnCl2

3.15

LINNETT DOUBLE QUARTED THEORY (LDQ THEORY): MODIFICATION OF LEWIS LONGMUIR OCTET THEORY

This theory was presented by J W Linnett in 1961 as a modification of octet theory put forward by Lewis. He proposed that pairing of electron and concept of hybridization is not required for prediction of the geometry of molecules. Rather he emphasized on the concept of double quartet. The main point of his theory are as follows: 1. In case of molecules of first short period, the eight valence electrons should not be considered as four pairs of electrons. These should be considered as two groups (double) of four electrons (quartet) each, hence the theory is named as Linnett double quartet theory.

3.58

Inorganic Chemistry

2. The four electrons in a particular quartet are considered to have the same spin (a), which is opposite to the spin of the four electrons in the other quartet (b). 3. The interelectronic repulsion and spin correlation in the four electrons in a quartet will be minimized by the arrangement of these electrons in a regular tetrahedral configuration around the nucleus of the atom. It means that the double quartets will arrange themselves in two tetrahedra at such a relative position that interelectronic repulsions Fig. 3.78 Two interconnected are minimum. This is achieved by an arrangement in which the tetrahedron with double vertices of one tetrahedron lie at the face centres of the other quartet of electrons tetrahedron as shown in Fig. 3.78. 4. The electrons participating in a double bond lie at the vertices of two tetrahedra with one edge common between the two nuclei (Fig. 3.79b). On the other hand the electrons participating in a triple bond lie at the vertices of two tetrahdra with one face common between the two nuclei (Fig. 3.79c)

(b)

(a)

(c)

Fig. 3.79 Representation of single, double and triple bonds

It is considered that electrons of the opposite spin set may or may not be close together. We can illustrate this theory with the help of following examples; The Lewis structure of F2 with fourteen valence electrons can be represented as

: : : :

(a) F2 Molecule follows:

: F : F: However, according to Linnett theory, in F2 molecule, each F is considered to be surrounded by a double quartet of electrons with two spin sets of four electrons each. The spin set of electrons around (x) (Fig. 3.80). x xF x

x x F x x

x or

F

F

x xF

x

x F x x

Fig. 3.80 Linnett representation for F2

The spin set of one quartet is opposite to that of other and the molecule is diamagnetic. The Lewis structure of O2 molecule with sixteen valence electrons can be represented

: : : :

(b) O2 Molecule as follows:

O ::O

Chemistry Bonding

3.59

According to this structure, the molecules should be paramagnetic, but actually it is diamagnetic. According to Linnett theory, a set of 7 electrons of one spin and another set of 5 electrons of another spin are considered to have a tetrahedral orientation around the two oxygen atoms as shown in Fig. 3.81. x

x

x

x

x

x

x

Fig. 3.81 Orientation of electrons in O2

The combination of these two representations yields the resulting representation as follows: x x x

or

x x x

x

or

x x

x x

The short line indicates an electron pair and overall there will be two unpaired electrons correlating to the paramagnetic nature of O2 molecule. X

X X

X

X X

X

Fig. 3.82 Linnett representation for O2 molecule

(c) N2 Molecule The Lewis structure of N2 molecule with ten valence electrons can be represented as follows: :N N : In LDQ approach, each N atom is surrounded by a set of five electrons. N

N

or

x xN x N x x

Thus, the two spin set cancel each other and the molecule is diamagnetic.

:

(d) NO Molecule The Lewis structure for NO molecule with 11 valence electrons can be represented as follows: : N O: According to LDQ approach, the molecular structure can be represented as a set of 5 electrons of one type of spin and another set of 6 electron of opposite spin, each with a tretrahedral orientation around N and O atom as shown follows: X

X X

or

N

O

or

X

X

X

x x x x N x O x

3.60

Inorganic Chemistry

These structure can be combined as follows: xNx x x

x O x

x N

or

x

Ox

x x x x

x

Fig. 3.83 Linnett representation of NO molecule

Thus Linnett representation correlates with the paramagnetic character of NO. Linnet representations of some simple molecules have been given in Fig 3.84. H x

H x

Ox

F x x

Hx

x

x H2O

HF

H x x C

xN H x

Hx

H x

Hx

H

H CH4

NH3

Fig. 3.84 Linnett representation for some simple molecules

However, their theory could not gain interest due to its empirical formation.

3.16

RESONANCE

According to valence bond theory, the molecules containing multiple bonds can be represented by several possible Lewis structures. These Lewis structures are known as conical forms or resonating structures. The true structure of the molecules is taken to a weighted average of these structures and is said to be the resonance hybrid of these structures and this phenomenon is known as resonane; introduced by Heisenberg (1920) and further developed by Pauling and Ingold. For example, O3 molecule can be represented by two Lewis structures I and II: O

O O

O

O I

or

O

O

O

O O

or

O II

O

Chemistry Bonding

3.61

According to these structures, O3 molecule should have two different oxygen-oxygen bond length i.e. bond length of O = O should be around 121 pm (double bond length) and that of O—O should be around 148 pm, (single bond length). However, experiments reveal the oxygen-oxygen bond length in O3 molecule as 126 pm, which is intermediate between that of O = O double bond length and O–O single bond length. It means that true structure for O3 molecule is not represented by either of these Lewis structures, but is rather an intermediate of these two structures. Double-headed arrows are used to represent resonance as shown below for O3 molecule and the resonance hybrid is represented by mean of the dotted lines representing the delocalisation of p-electron cloud. O O

O

O O I II Resonating structures

O O

O

O

Resonance hybrid

Resonance results in decrease in energy of the resonance hybrid than the energy of the resonating structures. As a result, resonance hybrid gets more stable. E 3 III This extra stability of the resonance hybrid is reflected in terms of resonance energy which is the difference in the energies of the most stable resonating structure and the E2 II true structure of the molecule. The resonance energy can be determined from E1 I the difference of the experimentally observed heat of Resonance energy formation for the molecule and the calculated heat of formation for the formula of the resonating structure. For E0 True structure example, the experimentally observed heat of formation for CO2 is 1464 kJ/mol and the calculated heat of formation for one resonating structure of the molecules is Fig. 3.85 Representation of resonance energy 1590 kJ/mol. This means that resonance energy of CO2 is equal to 126 kJ/mol (1590–1464) or the molecules is 128 kJ more stable than the resonating structure. We can represent this concept with the help of fig. 3.85. Here E1, E2 and E3 are the energies of the three different resonating structures of a molecule, where E3 > E2 > E1. It means that E1 is the energy of the most stable resonating structure. If E0 is the energy of the true structure of the molecule, then E1 – E0 gives the resonance energy. Although, the resonating structures a convenient mode of representation of the structure of a molecule, however these structure have no real existence.

Rules for Writing Resonating Structures 1. The resonating structures should differ only in the position of electrons and the atomic positions should remain the same. 2. There should be same number of paired and unpaired electrons for all the resonating structures of a molecule. 3. The resonating structures should have almost same energy. 4. The negative charge should reside on an electronegative atom and positive charge should reside on an electropositive atom. 5. Like charges should not reside on the adjacent atoms and unlike charges should reside on the adjacent atoms. We can illustrate these rules with the help of following examples.

3.62

Inorganic Chemistry

(a) CO C

– C

O I

+ C

+ O II

– O III

Experimental studies reveal that structure II is more predominant due to close agreement of carbonoxygen bond length with triple bond length.

(b) NO3–

_ O

O

O

N

N

N

O _

O

O

O _

O

O

(c) SO3

O

O

O

O

S

S

S

O

O

O

O

O

The important factors for the formation of an ionic bond are the low I.E. of the atom of form the cation, high E. A. of the atom to form the anion and high lattice energy to form the crystal lattice. Ionic bond is formed by complete transfer of electrons, while covalent bond is formed by mutual sharing of electrons between the combining atoms. However, if the electrons are shared by both the atoms, but donated by only one atom, the bond formed is termed as coordinate bond. According to Fajan’s rule, smaller the cation, larger the anion, more the charge on ions, greater is the covalent character. The atoms with pseudo-noble gas configuration have greater covalent character and lesser melting point than the other atoms. All the symmetrical molecules have zero dipole moment and are non-polar, while the unsymmetrical molecules have non zero dipole moment and are polar. According to orbital overlap concept, sigma bond is formed by head on overlapping, while pi bond is formed by sidewise overlapping of the orbitals. According to molecular orbital theory, the atomic orbitals of similar energy and same symmetry overlap to form bonding and antibonding molecular orbitals. The energy of MOs increase as: Upto N2 molecule s 1s < s*1s < s2s < s*2s < p2px = p2py < s2pz < p*2px = p*2py < s*2pz From O2 molecule s1s < s*1s < s2s < s*2s < s2pz < p2px = p2py < p*2px = p*2py < s*2pz According to the concept of hybridization, the atomic orbitals of similar energy hybridise together to form new orbitals of equivalent orbitals called hybrid orbitals. These hybrid orbitals overlap with the orbitals of the other atoms to form strong sigma bond. According to VSEPR theory, the geometry of a compound depends upon the arrangement of the electron pairs of the valence shell of the central atom and the molecules with lone pairs of electrons in the central atom have irregular geometry.

Chemistry Bonding

3.63

EXAMPLE 1 Arrange in increasing order of their melting points: LiCl, MgCl2, AlCl3, CCl4 According to Fajan’s rule, the compound with smaller cation has higher covalent character and lesser melting point. The size of cations of these compounds increases as Li+ > Mg2+ > Al3+ > C4+ Thus the order of decreasing covalent character and decreasing melting point is CCl4 < AlCl3 < MgCl2 < LiCl

EXAMPLE 2 Determine the percentage ionic character of a compound with dipole moment 1.98 D and internuclear distance 0.92 Å.

Calcuated dipole moment = q × d = 4.8 × 10–10 esu × 0.92 × 10–8 cm = 4.42 D Experimental dipole moment

% ionic character = _______________________ × 100 Calculated dipole moment

1.98 D = × 100 = 44.8% 4.42 D

EXAMPLE 3 Arrange the following in increasing order of their stability: O2, O2+, O2–, O22– The bond order of these species can be determined as follows: O2 : KK(s2s)2 (s*2s)2 (s2pz)2 (p2px)2 (p2py)2 (p*2px)1 (p*2py)1

1 (8 – 4) = 2 2 O2+ : KK(s2s)2 (s*2s)2 (s2pz)2 (p2px)2 (p2py)2 (p*2px)1 Bond order =

1 1 (8 – 3) = 2 2 2 O2– : KK(s2s)2 (s*2s)2 (s2pz)2 (p2px)2 (p2py)2 (p*2px)2 (p*2py)1 Bond order =

1 1 (8 – 5) = 1 2 2 : KK(s2s)2 (s*2s)2 (s2pz)2 (p2px)2 (p2py)2 (p*2px)2 (p*2py)2 Bond order =

O22–

1 (8 – 6) = 1 2 Higher the bond order, higher is the stability of the species. Thus, the order of increasing stability is: O22– < O2– < O2 < O2+ Bond order =

3.64

Inorganic Chemistry

EXAMPLE 4 Draw the resonance structures of chlorate ion. O

Cl O – O

O

Cl O

O–

– O

Cl

O

O

Resonance structure of ClO3–

QUESTIONS Q.1 What do you mean by an ionic bond? Discuss the formation of an ionic bond and the necessary conditions. Q.2 Discuss the formation of a covalent bond using the orbital overlap concept. Q.3 Differentiate between electrovalency and covalency with the help of suitable examples. Q.4 Discuss the formation of a sigma and a pi bond. Use suitable examples in support of your answer. Q.5 Give reasons for the following: (a) Tripositive or trinegative charged ions are rare. (b) Some elements show variable convalency. (c) CCl4 is non-polar but CHCl3 is polar (d) PCl5 exists but NCl5 does not exist. Q.6 A nonpolar molecule may have polar bonds. Justify the statement with the help of suitable example. Q.7 Differentiate between covalent and coordinate bond. Q.8 Comment upon the statement: A fully filled orbital can participate in bonding. Q.9 Discuss valence bond theory in context of H2 molecule. Q.10 Differentiate between bonding and antibonding molecular orbitals. Q.11 Draw molecular orbital energy-level diagram for N2 molecule. Q.12 Use MO diagram to predict the magnetic behaviour of the following molecules: (a) H2+ (b) B2 (c) O2+ (d) N22– Q.13 Draw MO diagrams for the following molecules: (a) CO (b) HF (c) F2 Q.14 Use MO diagram to compare bond length and bond order of following species : O2+, O2, O2–, O22–. Q.15 (a) Neon molecule is monoatomic but Hydrogen molecule is diatomic. (b) The loss of an electron from a diatomic molecule results in decrease of its bond energy. (c) Bond length of NO is higher than that of NO+ (d) Bond order of N2 is higher than that of N2+. Q.16 Write short notes on the following: (a) Electron sea model (b) Band theory (c) London forces (d) Term symbols Q.17 Use bond theory to differentiate between conductors, semiconductors and insulators. Q.18 Differentiate between intermolecular and intramolecular hydrogen bond. Q.19 Account for the following: (a) H2O has abnormally high boiling point. (b) O-nitrophenol is steam volatile.

Chemistry Bonding

3.65

Q.20 Discuss the concept of hybridization with the help of suitable examples. Q.21 Beryllium is divalent white carbon is tetravalent in their compounds. Justify the statement on the basis of hybridisation. Q.22 Discuss the characteristics of hybridisation. Discuss the stereochemistry of the molecules with the following hybridisation. (a) sp (b) sp2 (c) sp3 Q.23 Compare the bond angles in the following molecules: (a) CH4 (b) NH3 (c) H2O Q.24 Discuss VSEPR theory and the trend of repulsive interactions in lone pair-lone pair and lone-pairbond pair. Q.25 Discuss the geometry of the following molecules: (a) ClF3 (b) XeF2 (c) IF7 (d) PCl5 Q.26 Discuss the geometry of the following ions: (a) ClO3– (b) SO42– (c) NO3– Q.27 What do you mean by regular and irregular geometry? Use VSEPR theory to illustrate your answer. Q.28 Construct the wave functions for the sp and sp3 hybrid orbitals. Q.29 Give reasons for the following: (a) XeF2 and ClF3 have same hybridisation but different geometries (b) SnCl2 and H2O have same geometry but different hybridization Q.30 Discuss the concept of resonance. Draw resonance structure for the following molecules or ions: (a) NO (b) CO32– (c) CO

MULTIPLE-CHOICE QUESTIONS 1. Which of the following has highest melting point? (a) LiCl (b) BeCl2 (c) AlCl3 2. Which of the following has highest dipole moment? (a) CCl4 (b) CHCl3 (c) CO2 3. The molecule with least bond angle is (a) BeF2 (b) H2O (c) CH4 4. The molecule with linear geometry is (a) SnCl2 (b) NH2 (c) XeF2 5. The molecule with two lone pairs of electron on the central atom is (a) PbCl2 (b) SeCl2 (c) XeF2

(d) CCl4 (d) HF (d) ICl2– (d) None of these (d) SF4

chapter

Molecular Symmetry

4

After studying this chapter, the student will be able to

calculations

4.1

INTRODUCTION

Symmetry implies that each part of an object is well balanced and in the right measure. A geometrical object is said to possess symmetry if its figure is symmetrical about a point, plane or line. In chemistry, molecules and crystals are subjected to symmetry operations in order to study their structure, molecular spectra and associated properties. Symmetry operation is defined as a process or an operation which when carried out on a body brings it from its original form to an equivalent or indistinguishable form. After this operation, every point of the body before and after the operation are coincident with each other. H H rotation at 180° H H For example, the rotation of a homonuclear diatomic molecule at an angle of 180° about an axis perpendicular to its bond axis results Fig. 4.1 Rotation of a diatomic in an equivalent or indistinguishable orientation. Thus it is known as molecule at an angle of 180° a symmetry operation as shown in Fig. 4.1. A symmetry operation which when carried out on a body brings it back to its original position is known as an identity operation.

4.2

SYMMETRY ELEMENT

A symmetry operation is carried out about a point, axis or plane. This point, axis or plane is known as a symmetry element. Different symmetry elements has been described below.

4.2

Inorganic Chemistry

1. Centre of Symmetry and Inversion Centre A molecule is said to possess a centre of symmetry if an imaginary line drawn from each atom through the centre of the molecule encounters an equivalent atom in the same direction and these atoms are equidistant from the centre. This symmetry element is imaginary and as such inverts the coordinates of every atom through the origin lying within the geometry of the molecule. Hence, it is also known as an inversion centre and the associated symmetry operation is known as inversion operation denoted by i. Some examples of molecules possessing centre of symmetry are SF6, H2O2, C2H4, [PtCl4]2–, [Ni(CN)4]2–, N2O2, trans C2H2Cl2, benzene and homonuclear diatomic molecules as shown in Fig. 4.2. On the other hand, heteroatomic asymmetrical molecules do not possess centre of symmetry, The inversion operation on [PtCl4]2– has been shown in Fig. 4.3. This symmetry operation completely inverts the molecule to an equivalent position and another such operation brings the molecule back to its original position. Thus i2 is known as an identity operation.

F

F

F

Cl

Cl Pt

S F

Cl

F

F

H

Cl

H C

C

H

H O

H—H N

N N

N

O

Fig. 4.2

Some molecules with centre of symmetry Cl 1

Cl

i

Pt Cl

Cl

1 Cl Pt 2 Cl

Cl 2

Cl

Fig. 4.3 Inversion operation

2. Plane of Symmetry A molecule is said to possess a plane of symmetry, if reflection of each atom in one-half of the molecule through an imaginary plane (bisecting the molecule) produces an equivalent geometry of the molecule. This plane of symmetry is represented by s (sigma). A molecule can have any or all of the following planes of symmetry. (a) Vertical Plane of Symmetry (sn) It passes through the principal axis of symmetry and is perpendicular to the molecular plane as shown in Fig. 4.4. BF3 has three (sn) planes denoted by sn , sn , and sn each passing through B atom and one of F atoms. (b) Horizontal Plane of Symmetry (sh) It is perpendicular to the principal axis of symmetry as shown in Fig. 4.5. (c) Dihedral Plane of Symmetry (sd) Passing through the principal axis between two subsidiary axis as shown in Fig. 4.6 for allene molecule.

F1

s'n

F2

s"n' s'n

B

B

F3

F2

F3

F1

s'n'

Fig. 4.4 Vertical planes of symmetry in BF3 Cl

H C

C Cl

H

Fig. 4.5

Cl

H sh

C

C

Cl

H

Horizontal plane of symmetry in C2H2Cl2 C2

H3

H1 C

C

C

C2

Some Noteworthy Points (i) If the reflection operation is carried out an even number of times through a plane of symmetry, it gives an equivalent or original form. It means that s2n is an identity operation. Thus, s2 = E.

H2 H4

C2

Fig. 4.6 Dihedral plane of symmetry in allene

4.3

Molecular Symmetry

(ii) If the reflection operation is carried out an odd number of times through a plane of symmetry, it gives an equivalent form as obtained from a single reflection operation. It means that s2n+1 is same as s. Thus s3 = s2s = Es = s.

3. Axis of Symmetry A molecule is said to possess an axis of symmetry, if on rotation of the molecule with respect to an imaginary line passing through the molecule bring it to an equivalent or identical form. Since the two forms are super imposable on each other, this element is also known as proper axis of symmetry. The imaginary line is known as proper axis of rotation and the symmetry operation is known as proper rotation, denoted by Cn, here n is the number of times an equivalent form of the molecule is obtained in a complete rotation of 360° on the axis and is known as the order of the axis. It means that proper rotation is carried out at an angle of 360°/n. Depending upon the order of axis, the possible axis of symmetry are as follows: (a) One-Fold Axis of Symmetry The molecule is said to possess one-fold axis of symmetry if it requires a rotation of complete 360° to have the original form. Thus, n = 1 and this axis of symmetry is termed as C1 axis (b) Two-Fold Axis of Symmetry The molecule is said to possess two-fold axis of symmetry if it requires a rotation of 180° to have the original form. This axis of symmetry is termed as C2 axis. Thus n = 2, i.e. two times rotation at 180° will bring the molecule back to its original form as shown in Fig. 4.7 for H2O molecule. The complete operation is termed as C22.

Rotation at 180°

H2

H1

Rotation at 180°

H1

H2

H2

H1

Fig. 4.7 Two-fold axis of symmetry

(c) Three-Fold Axis of Symmetry The molecule is said to possess three-fold axis of symmetry, if it requires a rotation of 120° to have the original form. This axis of symmetry is termed as C3 Thus n = 3, i.e. three times rotation at 120° will bring back the molecule to original form as shown in Fig. 4.8 for NH3 molecule. The complete operation is termed as C33. Rotation at 120°

N H3 H1

H2

Rotation at 120°

N

Rotation at 120°

N

H1 H2

H3

N H3 H2

H2 H3

H1

H1

Fig. 4.8 Three-fold axis of symmetry

(d) Four-Fold Axis of Symmetry The molecule is said to possess four-fold axis of symmetry, if it requires a rotation of 90° to have the original form. This axis of Cl4 Cl3 Cl1 Cl4 symmetry is termed as C4. Thus, n = 4, i.e. four times rotation rotation at 90° will bring back the molecule to original form as shown Pt Pt at 90° in Fig. 4.9 for [PtCl4]2– molecule. The complete operation is Cl2 Cl1 Cl3 Cl2 termed as C44. (e) Five-Fold Axis of Symmetry The molecule is said to possess five-fold axis of symmetry, if it requires a rotation of 72° to have the original form. This axis of symmetry is termed as C5. Thus, n = 5, i.e. five times rotation at 72° will bring back the molecule to its original form, as in case of ferrocene (Fig. 4.10). The complete operation is termed a C55.

rotation at 90°

Cl1

rotation at 90° rotation at 90°

Pt Cl4

Cl2

Cl2

Cl3

Cl1

Cl3 Pt Cl4

Fig. 4.9 Four-fold axis of symmetry

4.4

Inorganic Chemistry

C6

C2

C2 Fe

Fig. 4.10 Ferrocene with five-fold axis of symmetry

Fig. 4.11 Benzene with six-fold axis of symmetry

Fig. 4.12 Possible axis of symmetry in benzene

(f) Six-Fold Axis of Symmetry The molecule is said to possess six – fold axis of symmetry, if it requires a rotation of 60° to have the original form. This axis of symmetry is termed as C6. Thus, n = 6, i.e. six times rotation at 60° will bring back the molecule to its original form, as in case of benzene (Fig. 4.11). The complete operation is termed as C66. Some molecules possess more than one axis of symmetry for example, in benzene, there are six 2-fold axis of symmetry, three 2-fold axis of symmetry passing through C – H bonds, three 2-fold axis of symmetry passing through C–C bonds and three 4-fold axis of symmetry (Fig. 4.12) However, the axis which passes through maximum number of atoms in the molecule is known as the principal axis. Thus C6 is the principal axis of symmetry for benzene. (g)

-Fold Axis of Symmetry The molecule is said to possess -fold axis of symmetry, if it can be rotated at any angle through a lengthwise axis passing through the atoms. It is termed as C and it is found in linear molecules such as HCl, CO2, etc. Some Noteworthy Points (i) The maximum possible number of rotation around proper axis, Cn is (n-1). Thus different possible successive operations around various axis of symmetry are as follows :C2 $ 1C2 (180°) $ one operation C3$ C13 (120°), C23 (240°) $ Two operations C4 $ C14 (90°), C24 (180°), C34 (270°) $ Three operations C5 $ C15 (72°), C25 (144°), C35 (216°), C46 (288°) $ Four operations (ii)

(iii)

C6 $ C16 (60°), C26 (120°), C36 (180°), C46 (240°), C56 (300°) $ Five operations Last operation after any successive operations around a proper axis would give the original form, hence it is known as an identity operation, denoted by E, It means that for any Cn axis, Cnn is always an identity operation, E is Cnn = E. Any rotation operation can be represented by small numbers. For example, C12, C24 and C36 are at an angle of 180°. This means that C42 and C63 be represented by C21. Similarly C31 and C62 are at angle of 120°. Thus C62 is equal to C31. S4

4. Axis of Improper Rotation or Rotation–Reflection Axis A molecule is said to possess an axis of improper rotation, if a rotation around an axis at an angle of 360°/n followed by reflection through a plane perpendicular to this axis or vice versa brings it to an equivalent form. The improper axis of rotation is represented by Sn , where n is the order of the axis. Any Sn axis can generate many symmetry operations out of which some may or may not be genuine

L

X

L

M L

L X

Fig. 4.13 S4 axis for ML4X2

4.5

Molecular Symmetry

improper rotation operations, proper rotation operations, reflection operations and identity operations. For example, for a distorted octahedral complex, ML4X2 (Fig. 4.13), an axis passing through X – M – X is S4 axis which can be simplified as follows: (i) S41 can be simplified as s C4 and is a genuine rotation reflection or improper rotation operation. (ii) S42 means s 2C42. It can be simplified as follows: S42 = s2C42 = EC42 = C42 = C2 . Thus it is a proper rotation operation. (iii) S43 = s3C43 = s2sC43 = EsC43 = sC43. Thus it is a genuine improper rotation operation. (iv) S44 will bring molecule to its original form as S44 = s4C44 = E × E = E. Thus it is an identity operation.

MULTIPLICATION OF SYMMETRY OPERATIONS

4.3

When two or more symmetry operations are carried out in succession, their combination gives another symmetry operation and this combination is called as multiplication of symmetry operations. Suppose a symmetry operation A is followed by another symmetry operation B, then their combination is written as A × B = C, where C is another symmetry operation. The order of operation is form right to left. In case of more than two symmetry operations, say ABCD = E, in the order D, C, B, A, the product A × B may or may not be equal to B × A. If A × B = B × A, these symmetry elements are known as equivalent symmetry elements and they are said to commute with each other, The associated symmetry operations are known as commutative. For example, in H2O molecule, C2 × sxz = sxz × C2 = syz as shown in Fig. 4.14. On the other hand, if AB BA, the symmetry elements do not commute with each other and the associated symmetry operations. For example, in BF3 sn × C3 C3 × sn as shown in Fig. 4.15. Here sn × C3 = C3 , while C3 × sn = C23. syz z C2

sn''

C2 H2

H1

x H2

H1

H1

H2

F3

F2

B

C3 F3

F1 C3'

sn sn''

H2

y

F1

C3

B

x H2

H1

H1

Commutative symmetry operations in H2O molecule

H2

F3

F2

sn''

s'n

C2

syz

4.4

sn''

F3 B F1

F2

z y s xz

Fig. 4.14

F2

s'n

B

sxz

z

x H1

F1

y

F3 B

F1

F1 sn'' F2

B F3

F2

sn

Fig. 4.15

Noncommutative symmetry operations in BF3 molecule

MATHEMATICAL GROUP

A complete set of symmetry operations is known as a mathematical group if following conditions are fulfilled: 1. Combination of symmetry elements A and B must give another symmetry element C, i.e. A × B = C. This means that the result of operation B followed by operation A is equivalent to the result of operation C.

4.6

Inorganic Chemistry

2. In each set, there is an identity element E for every symmetry element, so that EA = AE = A. 3. The elements of a group hold associative law of multiplication. This means that A(BC) = (AB)C. 4. The combination of an element and its reciprocal is always the identity element, so that AA–1 = A–1A = E. 5. The reciprocal of combination of two or more elements is same as that of combination of reciprocals of these elements in the reverse order, so that (ABC)–1 = C–1B–1A–1. 6. If B–1AB = C, then A and C are called the conjugate elements and C is called as the similarity transform of A by B. The Conjugate elements hold following characteristics: (a) In each group, atleast one element is the similarity transform of another element. This element is identity element E. It means that each element is conjugate of itself, so that E–1AE = A. (b) If in a group B–1AB = C, then there must be D–1CD = A in the same group. (c) If an element A is conjugate with two element B and C, which are conjugate with each other, then these elements A, B and C are conjugate with each other. The total number of elements in a finite group is called as the order of the group, whereas the complete set of conjugate elements for this group is called as the class. The number of elements in each class is called as its order and it is an integral factor of the order of the group. A group of symmetry operations which leave the positions of a point invariant under each operation is called as a point group denoted by Schoenflies notation using a capital letter and a subscript as given in Table 4.1. Table 4.1 Schoenflies notation for point group Capital letter C D S T O I

Symmetry element Simple rotation axis n-fold rotation axis to principal axis Rotation-reflection axis Symmetry based on tetrahedron Symmetry based on octahedron Symmetry based on icosahedron

Subscript S i n nn nh nd

Symmetry element Plane of symmetry Centre of symmmetry n-fold rotation axis vertical symmetry plane horizontal symmetry plane dihedral symmetry plane

4.4.1 Classification of Point Groups 1. C1 This group is for molecules which donot possess any element of symmetry except E. For example, CFCl, BrI. 2. Cs This group is for such molecules which possess only a plane of symmetry as in case of hypochlorous acid, quinoline and mono-substituted naphthalene. Identity element is invariably present for every molecule. 3. Ci This group is for such molecules which possess only a Cl H centre of symmetry along with identity element as in case of 1, C Cl H C 2 – dichloroethane (staggered form) and meso-tartaric acid. 4. Cn The molecules which possess n-fold axis of symmetry Cl H and identity element belong to this group. For example 1, 1, Fig. 4.16 1,1,1- trichloroethane 1- trichloroethane belongs to C3 group. (Fig. 4.16) 5. Cnn The molecules which possess n-fold axis of symmetry and n vertical planes of symmetry belong to this group. Some common examples have been given below: C2n $ H2O, SO2, HCHO, NO2 (C2 and 2Csv)

Molecular Symmetry

4.7

C3n $ NH3, [Cr(h6–C6H6)(CO)3] (C3 and 3Csv) Cun $ XeOF4, SF5Cl, [Co(NH3)4Cl2] (C4 and 4Csv) C

n$

CO, HCl (Heteronuclear diatomic & unsymmetrical linear triatomic)

6. Cnh The molecules which possess Cn axis and horizontal plane of symmetry belong to this group. For example: C2h $ trans-dichloroethylene, trans H2O2 (C2 and sn) C3h $ Boric acid (C3 and sh) 7. Sn The molecules which possess n-fold improper axis of rotation coincident with the principal axis Cn belong to this group. 8. Dn The molecules which possess Cn axis and nC2 axis all perpendicular to Cn axis belong to this group as in case of ethene. 9. Dnh The molecules which possess Cn axis, nC2 axes perpendicular to Cn and a horizontal plane of symmetry belong to this group. These molecules essentially possess n vertical planes of symmetry and an Sn improper rotation axis. Thus here 4n symmetry operations are possible. For example

D h — Homonuclear diatomic and linear symmetrical triatomic molecular (CO2, CS2). (C , C2, sn and sh planes) 10. Dnd The molecules which possess nC2 axes perpendicular to Cn and n vertical planes of symmetry passing through C2 axes, but no horizontal plane of symmetry, belong to this group. In these molecules a S2n improper rotation axis and 4n symmetry operations are possible. For example D3d — Ethane (C3, 3C2, 3sd) D2d — Allene (3C2, 2sd, S4) D5d — Ferrocene (C5, 5C2, 5sd) D

h

— Homonuclear diatomic and linear symmetrical triatomic molecular (CO2, CS2). (C3, C2, sn and sh planes)

11. Td The regular tetrahedral molecules AB4 (Fig. 4.17) with 24 symmetry operations belong to this group. The symmetry elements Fig. 4.17 Representation of a regular tetrahedron and the associated symmetry operations are as given below: (a) There are four C3 axes of symmetry, passing through A and one B atom and the centre of the opposite trigonal face. The associated symmetry operations are C31 and C32 giving in total 8C3 symmetry operations. (b) There are three C2 axes of symmetry, each passing through the centre of two opposite edges.

4.8

Inorganic Chemistry

Each C2 axes gives one symmetry operation, i.e. total 3C2 operations. (c) There are six planes of symmetry each passing through one edge and the centre of the opposite edge. These are 6s symmetry operations associated with these elements. (d) There are three S4 axes of improper rotation about X, Y and Z axes. The associated symmetry operations with each S4 axis are S41 = sC41 S43 = s3C43 = s2sC43 = sC43 S42 = s2C42 = EC2 = C2

S44 = s4C44 = E

It means that for each S4 axis, there are two genuine improper rotation operations giving in total 6S4 operations. Thus, there are in total 24 symmetry operations (1E, 8C3, 3C2, 6s, 6S4 ) possible for a molecule belonging to Td group. Some examples are CH4 , SiCl4 , ZnCN42–, etc. However, irregular tetrahedral molecules such as CHCl3 do not possess 24 symmetry operations and are not included in this group. 12 Oh The molecules with regular octahedral geometry (Fig. 4.18) and B1 48 symmetry operations belong to this group. The symmetry elements B6 B2 and the associated symmetry operations are given below: A (a) There are three C4 axis of symmetry along X, Y and Z axes passing through two B’s at the opposite corners. The associated symmetry B3 B5 operations for these C4 are: B 4

C41 , C42 = C2 , C43 , C44 = E Fig. 4.18

(b)

(c) (d) (e)

Representation of

This means that each C4 axis gives two genuine C4 symmetry octahedron operations giving in total 6C4 symmetry operations. There are four C3 axes passing through the centre of opposite triangular faces. The associated symmetry operations are: C31, C32 , C33 = E This means that each C3 axis gives two genuine C3 symmetry operations giving in total 8C3 operations. There are three C2 axes collinear with C4 axis giving in total 3C2 symmetry operations. There are six C2 axes passing through six opposite axes giving in total 6C21 operations. There are three S4 axis coincident with each C4 axis. Each S4 axis can generate following symmetry operations: S41 = sC41

S43 = s3C43 = s2sC43 = sC43

S42 = s2C42 = EC2 = C2

S44 = s4C44 = EE = E

Thus each S4 axis can generate only two genuine S4 improper rotation operation giving in total 6S4 operations. (f) There are four S6 axis collinear with C3 axes. The possible symmetry operation for each S6 axis are S61 = sC6

S64 = s4C64 = EC32 = C32

S62 = s2C62 = EC3 = C3

S65 = s5C65 = s4sC65 = EsC65 = sC65

S63 = s3C63 = s2sC2 = sC2 = i

S66 = s6C66 = E × E = E

This means that there are only two genuine improper rotation operations on each S6 axis giving in total 8C6 operations.

Molecular Symmetry

4.9

(g) There are six vertical planes of symmetry, each passing through C4 axes generating 6sn operations. (h) There are three horizontal planes of symmetry, passing through four vertices generating 3sh operations. (i) There is one centre of symmetry, i. Thus the total number of symmetry operations for a regular octahedral molecules are 48 (1E, i, 6C4, 3C2, 8C3, 6C21, 6S4, 8S6 , 6sn, 3sh), Such examples are SF6 , PtCl62–. 13. Ih The molecules with icosahedron geometry such as B12H12 belong to this group.

4.4.2 Steps for Determination of Point Group of a Molecule 1. Look for the special point groups, C n, D h, Td, Oh or Ih. 2. If special point groups cannot be assigned, look for Cn axis. If no Cn axis is present proceed as follows: (a) If element i is present, point group is Ci (b) If plane of symmetry is present, point group is Cs. (c) If no element of symmetry except E is present, point group is Cn. If Cn is present proceed to step 3. 3. Look for any Sn (n is even) axis. If only Sn coincident with the principal axis is present, point group is Sn. If Sn is absent or other elements of symmetry are also present, proceed to step 4. 4. Look for nC2 axis perpendicular to Cn axis, if these are present proceed to step 5 and if absent, proceed to step 6. 5. (i) If nC2 axis perpendicular to Cn axis are present point group is Dn. (ii) If in addition to (i), sh is present, point group is Dnh. (iii) If in addition to (i),sd is present, point group is Dnd. 6. (i) If more than one proper axes of rotation are present, choose the principal axis with highest order and the point group is Cn. (ii) If sh is also present, point group is Cnh. (iii) If sn is present, point group is Cnn. We can use these steps to determine point groups of some typical molecules as discussed ahead : 1. Acetic Acid, CH3 C O

This molecule has no element of symmetry except E. Thus it belongs to point group C1. This molecule has only a plane of symmetry, i.e. the plane of the molecule. Thus it belongs to point group Cs.

OH

2. Quinoline,

N

3. 1,2 – Dichloro, 1,2 – Difluoroethane (Staggered form) This molecule has only centre of symH metry along with E. Hence it belongs to point grup Ci. Cl

F H

Cl

F C

C

Cl

H F

or Cl

F H

4. Trans-dichlorodiamine Pt(II) Complex This molecule has one C2 axis, the principal axis and H3N Cl two C2 axis perpendicular to the principal axis. In addition, there is one sh plane. Thus, this molecule belongs to point group of D2h. Pt Cl

NH3

4.10

Inorganic Chemistry

This molecule has one C5 principal axis and five C2 axis perpendicular to the principal axis. In addition, there are five sd planes. Thus, this molecule belongs to the point group D5d.

5. Ferrocene Fe

6. HCl It is a linear heteroatomic molecule. It has a C axis collinear with the bond axis and infinite number of sn planes. Thus, this molecule belongs to C n. 7. CO2 O = C = O It is a linear symmetrical triatomic molecule with a C axis collinear with the bond axis. There are C2 axis perpendicular to the C axis. Similarly there are planes of symmetry passing through the bond axis as a plane of symmetry well as perpendicular to the bond axis. Thus, it belongs to D h group. 8. Ammonia, NH3 This molecule has one C3 axis and three sn passing through N atom and each of one hydrogen atom. Thus, this molecule belongs to C3n point group. 9. Trans–dichloroethylene H C

C

Cl

10. [CoF5Cl]3–

F

F

F

Co

H

This molecule is an irregular octahedral molecule and has C4 as principal axis, C2 axis, two sn and two sd. Thus, it belongs to the point groups C4n.

F

F

Table 4.2 Group multiplication table for H2O molecule

Cl

4.4.3 Group Multiplication Table We can obtain a table including the products of the symmetry elements of a mathematical group. This table is known as the group multiplication table. The symmetry elements are represented at the heads of each row and each column and their product is represented in a matrix form as shown below in table Table 4.2 for H2O molecule. In H2O molecule belonging to C2n point group, the symmetry elements are E, C2, sn and sn .

4.5

This molecule has a C2 axis and sh plane. Thus, this molecule belongs to the point group C2h.

Cl

E C2 sn sn

E EE = E C2E = C2 sn E = sn sn E = sn

sn Esn = sn C2 sn = sn sn sn = E sn sn = C2

C2 EC2 = C2 C2C2 = E sn C2 = sn sn C2 = sn

sn E sn = s n C2 sn = sn sn sn = C2 sn sn = E

Or E

E E

C2 C2

sn sn

sn sn

C2

C2

E

sn

sn

sn

sn

sn

E

C2

sn

sn

sn

C2

E

MATRIX REPRESENTATION OF SYMMETRY OPERATIONS

The symmetry operations can be represented in mathematical formulations by using Cartesian coordinate system. In this system, each atom, i of the molecule is specified by means of coordinates xi, yi and zi. which define the position vector from the origin to the atom. The position vector is represented as

È xi ˘ Í ˙ Í yi ˙ ÍÎ zi ˙˚

Molecular Symmetry

4.11

Suppose the molecule is subjected to a symmetry operations so that its position vector changes as

È xi¢ ˘ Í ˙ Í yi¢ ˙ ÍÎ zi¢ ˙˚ This transformation of coordinates is written as a set of linear equations x i = a11xi + a12yi + a13zi y i = a21xi + a22yi + a23zi z i = a31xi + a32yi + a33zi and in matrix form as È xi¢ ˘ Èa11 a12 a13 ˘ È xi ˘ Í ˙ Í ˙Í ˙ Í yi¢ ˙ = Ía21 a22 a23 ˙ Í yi ˙ ÍÎ zi¢ ˙˚ ÍÎa31 a32 a33 ˙˚ ÍÎ zi ˙˚ where the coefficients aijs form the transformation matrix. In case of identity operation, the coordinates are not affected, thus there is no change in sign of x, y or z. It gives the equation as x2 = (+1)x1 + (0)y1 + (0)z1 y2 = (0)x1 + (+1)y1 + (0)z1 z2 = (0)x1 + (0)y1 + (+1)z1 The matrix representation is

È x1 ˘ È x2 ˘ È+1 0 0 ˘ È x1 ˘ E = Í y1 ˙ = Í y2 ˙ = Í 0 +1 0 ˙ Í y1 ˙ Í ˙ Í ˙ Í ˙Í ˙ ÍÎ z1 ˙˚ ÍÎ z2 ˙˚ ÍÎ 0 0 +1˙˚ ÍÎ z1 ˙˚ It is clear that the transformation matrix can be written as

È1 0 0 ˘ Í0 1 0 ˙ Í ˙ ÍÎ0 0 1 ˙˚ Now, consider the reflection operation through the plane syz as shown in Fig. 4.19. For z-axis lying in the mirror plane the vectors can be represented as x2 = (–1)x1 + (0)y1 + (0)z1 y2 = (0)x1 + (+1)y1 + (0)z1

y r

z x

Fig. 4.19

z2 = (0)x1 + (0)y1 + (+1)z1 There is no change in y and z but only x changes its sign. The matrix representation is

È x1 ˘ È x2 ˘ È-1 0 0 ˘ È x1 ˘ Í ˙ Í ˙ Í ˙Í ˙ Í y1 ˙ = Í y2 ˙ = Í0 1 0 ˙ Í y1 ˙ ÍÎ z1 ˙˚ ÍÎ z2 ˙˚ ÍÎ0 0 1 ˙˚ ÍÎ z1 ˙˚

(x1, y1, z1)

(x2, y2, z2)

Reflection operation through plane syz

4.12

Inorganic Chemistry

This gives the transformation matrix as

È-1 0 0 ˘ Í0 1 0 ˙ Í ˙ ÍÎ0 0 1 ˙˚ For sn = sxz, there is change in position of y only Thus, transformation matrix for sxz is

È1 0 0 ˘ Í ˙ sxz = Í0 -1 0 ˙ ÍÎ0 0 1 ˙˚ Now, we will determine the transformation matrix for rotation operation about an axis Cn at an angle f as shown in Fig. 4.20. x1, y1 The vectors can be represented as È x1 = r sin b ˘ È x2 = r sin{p - (b + f )} = r sin(b + f ) ˘ Í ˙&Í ˙ Î y1 = r cos b ˚ Î- y2 = r cos{p - (b + f )} = -r sin(b + f )˚ b

x2 = r sin b cos f + r cos b sin f = x1 cos f + y1 sin f y2 = r sin b cos f – r cos b sin f = y1 cos f – y1 sin f

f

x

(p – q + f)

There is no change in z-coordinate. The transformation matrix is obtained as

(x2,–y2)

È cos f sin f 0 ˘ Í- sin f cos f 0 ˙ Í ˙ ÍÎ 0 0 1 ˙˚

Fig. 4.20

Rotation operation about Cn axis at an angle f

Now, we have to put the value of b and f to obtain the result for the symmetry operations for a molecule. For example, we can obtain the values for E, C2, sn and sn for C2v group as follows

È1 0 0 ˘ E = Í0 1 0 ˙ Í ˙ ÍÎ0 0 1 ˙˚ For C2 in C2n, rotation is at angle 180° or p. Thus.

È cos p C2 = Í- sin p Í ÍÎ 0

sin p cos p 0

0 ˘ È-1 0 0 ˘ 0 ˙˙ = ÍÍ 0 -1 0 ˙˙ 1 ˚˙ ÎÍ 0 0 1 ˙˚

4.5.1 Matrix Representation of Point Groups It is known that each atom has three degrees of freedom and a molecule with N atoms will hold dimensional basis of 3N. Thus, H2O molecule has a 9-dimensional basis (Fig. 4.21) and each symmetry operation can be represented by a 9 × 9 matrix.

Molecular Symmetry

4.13

z1

z1 x1 z3 H

y1 x1 z 2 y3 x3

H

x2

y1 z2

y2

H

C2 y2 x2

z3 x3 y3

H

Fig. 4.21 Representation of 9-dimensional basis for H2O molecule

We can represent the matrix representation for C2 as follows:

0 0 0 0 0 0 ˘ È x1 ˘ È x1 ˘ È- x1 ˘ È È-1 0 0 ˘ ˙ ÍÍ Í ˙ Í ˙ Í ˙ y y 0 0 0 0 0 0 ˙˙ Í y1 ˙ Í 1 ˙ Í 1 ˙ Í Í 0 -1 0 ˙ Í z1 ˙ Í+ z1 ˙ Í ÍÎ 0 0 +1˙˚ 0 0 0 0 0 0 ˙ Í z1 ˙ ˙ Í Í ˙ Í ˙ Í ˙ 0 0 0 Í x 2 ˙ Í - x2 ˙ Í 0 0 0 È-1 0 0 ˘ ˙ Í x2 ˙ Í 0 -1 0 ˙ ˙ Í y ˙ 0 0 0 C2 = Í y2 ˙ = Í- y2 ˙ = Í 0 0 0 ˙ Í Í ˙ Í Í ˙˙ Í 2 ˙ Í z2 ˙ Í+ z2 ˙ Í 0 0 0 ÍÎ 0 0 +1˙˚ ˙ Í z2 ˙ 0 0 0 ˙ Í Í ˙ Í ˙ Í ˙ 0 0 0 ˙ Í x3 ˙ È-1 0 0 ˘ Í x3 ˙ Í- x3 ˙ Í 0 0 0 Í 0 -1 0 ˙ Í y ˙ Í- y ˙ Í 0 0 0 0 0 0 ˙ Í y3 ˙ Í ˙ Í 3˙ Í 3˙ Í ˙ Í ˙ ÍÎ 0 0 +1˙˚ 0 0 0 ˙˚ ÎÍ z3 ˙˚ ÍÎ z3 ˙˚ ÍÎ+ z3 ˙˚ ÍÎ 0 0 0 Similarly, for sn = syz with change only in x-coordinate, the matrix representation can be given as follows: È x1 ˘ È- x1 ˘ È È-1 ˙ ÍÍ Í ˙ Í Í y1 ˙ Í+ y1 ˙ Í Í 0 Í z1 ˙ Í+ z1 ˙ Í ÍÎ 0 ˙ Í Í ˙ Í Í x 2 ˙ Í - x2 ˙ Í 0 syz = Í y2 ˙ = Í+ y2 ˙ = Í 0 ˙ Í Í ˙ Í Í z2 ˙ Í+ z2 ˙ Í 0 ˙ Í Í ˙ Í Í x3 ˙ Í- x3 ˙ Í 0 Í y ˙ Í+ y ˙ Í 0 Í 3˙ Í 3˙ Í ÍÎ z3 ˙˚ ÍÎ+ z3 ˙˚ ÍÎ 0

0˘ 0 ˙˙ 0 +1˙˚

˘ 0 ˙˙ 0 ˙ ˙ 0 ˙ 0 ˙ ˙ 0 ˙ ˙ 0 ˘˙ 0 ˙˙ ˙ ˙ +1˙˚ ˙˚

0 0 0

0 0 0

0 0 0 0 0 0

0 0 0 0

0 0 0 0 0 0

È-1 0 0 ˘ Í 0 +1 0 ˙ Í ˙ ÍÎ 0 0 +1˙˚

0 0

0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 1

0 0 0 0 È-1 0 Í 0 +1 Í ÍÎ 0 0

È x1 ˘ Í ˙ Í y1 ˙ Í z1 ˙ Í ˙ Í x2 ˙ Íy ˙ Í 2˙ Í z2 ˙ Í ˙ Í x3 ˙ Íy ˙ Í 3˙ ÍÎ z3 ˙˚

It is obvious that we have to write a large set of matrices. However we meet to specify only the trace of the matrix in the form of sum of the diagonal elements giving a number called character (c) of the matrix. For example, (c) (C2) = (–1 –1 +1) = –1 and (c) (syz) = (–1 +1 +1) = +1 Some Noteworthy Points 1. The position vectors of the atoms which remain in the same position and direction, contribute (+1) to the character of the matrix. 2. The position atoms which reverse their direction, contribute (–1) to the character of the matrix. 3. The atoms which change their position contribute zero to the character of the matrix. Thus, in E operation of C2n point group, all the position vectors remain the same and c(E) = 9. Similarly, for sxz operation, c(sxz) = 1

4.14

Inorganic Chemistry

On this basis, we can write a set of matrices for the complete set of symmetry operations of C2n point group called as its representation (G) as given below.

E C2 G

9

-1

s yz

s xz

3

1

4.5.2 Rules for Determination of Irreducible Representation (IR’s) 1. The total number of IR’s for a group is equal to the number of classes of symmetry operations of that group. 2. The sum of the squares of the characters of an IR gives the order of the group, i.e. where R is the symmetry operation of the group and h is the order of the group. 3. The sum of squares of the dimensions of IR’s also gives the order of the group.

2

 [ c i ( R )]

= h;

R

 l12 = l12 + l22 + l32 + ---- = h i

4. The characters of the IR’s of the same group are orthogonal to each other.

 ÈÎ ci ( R) c j ( R)˘˚

=0

R

5. The characters of the conjugate elements (elements of the same class) of an IR are same. Now, we can use these rules to find out IR’s of a C2n group. The symmetry operations for this group are E, C2 , sn (syz) , sn (sxz) , means order of the group is (1 + 1 + 1 + 1) = 4. 1. Since each of these symmetry operation is a distinct class, there are 4 IR’s possible, i.e. G1, G2, G3 and G4. (Rule 1). 2. According to Rule 2, cE2 + c2C2 + c2sn + c2sn = 4 It means that c for each operation in IR should be ±1. 3. According to Rule 3,

l12 + l22 + l33 + l42 = 4

It means that l12 + l22 + l33 + l42 = 1, i.e. each IR should be 1- dimensional 4. For any IR, at least one IR is symmetrical to all the operations i.e. its characters are ±1. Thus, G1 can be written as

G1

E C2 1 1

s v¢ 1

s v ¢¢ 1

The sum of the squares of these characters should be 4 as there are 1E, 1C2, sn and 1sn operations. 12 × 1 + 12 × 1 + 12 × 1 + 12 × 1 = 4 5. According to Rule 4, the characters of IR’s are orthogonal to each other. Since character of E is always positive, i.e. 1, the character of other operations should be such that c1E ciE + c1C2 ciC2 + c1sn cisn + c1sn cisn = 0 where i = 2, 3 and 4. It means that at least two of the characters should be –1, this gives the complete representations of C2n group as

G1 G2 G3 G4

E C2 s n ¢ s n ¢¢ 1 1 1 1 1 1 -1 -1 1 -1 1 -1 1 -1 -1 1

Molecular Symmetry

4.15

Similarly, we can obtain representations for C3n group with symmetry operations as 1E, 2C3 and 3sn, i.e. in total three classes of symmetry operations. 1. There are 3 IR’s for 3 classes of symmetry operations. 2. Order of the group is (1 + 2 + 3) = 6. It means that l12 + l22 + l32 = 6, i.e. l1 = 1 , l2 = 1 and l3 = 2 as 12 + 12 + 22 = 6. Thus, these should one two 1-dimensional representations and one 2-dimensional representation or G1 = 1, G2 = 1 and G3 = 2 It gives characters for E as 1,1 and 2 for G1, G2 and G3 respectively. 3. The symmetrical IR should have all characters as +1, i.e.

G1

E 2C3 1 1

3s v ¢ 1

The sum of the squares of the characters must be 6, cE2 + 2c2C + 3c2s = 6 i.e. 3

n

12 + 2(1)2 + 3(1)2 = 6 4. According to orthogonality of the characters of IR’s , for G2 1 × c1Ec2E + 2c1C3c2C3 + 3c1sn c2sn = 0 Since character of E for G2 is 1, characters for C3 and sn can each be either +1 or –1. This gives G2 as

G2

E 2C3 1 1

3s v ¢ -1

Thus, 1 × 1 × 1 + 2 × 1 × 1 + 3 × 1 × (–1) = 0 5. The character of E for G3 in 2. According to orthogonality of G1 and G3, 1 × c1Ec3E + 2c1C c3C + 3c1s c3s = 0 3

or or

3

1 × 1 × 2 + 2 × 1 × c3C3 + 3 × 1 × 3c1sn = 0

n

n

2c3C3 + 3c3sn = –2

Similarly, for orthogonality of G2 and G3. 1 × c2E c3E + 2c2C3c3C3 + 3c1sn c3sn = 0 or 1 × 1 × 2 + 2 × 1 × c3C3 + 3 × (–1) × 3c3sn = 0 or 2c3C – 3c3s = –2 3

n

Subtracting the two equations, we get 6c3sn = 0 and c3sn = 0 Using this value, we get 2c3C3 = –2 and c3C3 = –1 Thus, complete set of IR’s for C3n point group is

G1 G2 G3

E 2C3 1 1 1 1 2 -1

3s n ¢ 1 -1 0

4.5.3 Reduction of Reducible Representation into IR’s Any reducible representation G can be represented as a linear combination of IR’s, i.e.

4.16

Inorganic Chemistry

G = n1G1 + n2G2 + n3G3 + -----where G1, G2, G3 are IR’s and n1, n2, n3 are the number of times each IR is obtained in G. This means that c(R) = Sni ci (R)

 gR c j ( R) c ( R) =   ni gR c j ( R) ci ( R)

and

R

According to the rules, if i π j

R

  ni c j ( R) ci ( R) = 0

and if i π j

  ni c j ( R) ci ( R) = ni h R

R

1 Â gR c j ( R) c ( R) h R R where h is the order and gR is the number of elements in the class. This reduction formula can be used to reduce any reducible representation into its IR’s. We can use this formula for C2v group with h = 4, gR = 1 and reducible representation as given in the table below: It gives

 gR c j ( R) c ( R) = ni h

(

ni =

or

)

E C2 1 1 1 1

G1 G2 G3

s n ¢ s n ¢¢ 1 1 -1 -1 -1 1 -1 1 -1 -1 -1 1 3

1 1

G4 G red

9

Gred = n1G1 + n2G2 + n3G3 + n4G4

1 [1 × 1 × 9 + 1 × 1 × (–1) + 1 × 1 × 1 + 1 × 1 × 3] = 3 4 1 n2 = [1 × 1 × 9 + 1 × 1 × (–1) + 1 × (–1) × 1 + 1 × (–1) × 3] = 1 4 1 n3 = [1 × 1 × 9 + 1 × (–1) × (–1) + 1 × 1 × 1 + 1 × (–1) × 3] = 2 4 n1 =

n4 = It means that

1 [1 × 1 × 9 + 1 × (–1) × (–1) + 1 × (–1) × 1 + 1 × 1 × 3] = 3 4

Gred = 3G1 + G2 + 2G3 + 3G4

In general, Gred can be determined in a simple way without writing any complete matrix. In this method, the character of the matrix is equal to the number of vectors that remain unshifted by the symmetry operation. For example, in case of C2n point group, under E operation all the three atoms remain unshifted, i.e.

È x1 ˘ È x1 ˘ Í ˙ Í ˙ E Í y1 ˙ = Í y1 ˙ ÍÎ z1 ˙˚ ÍÎ z1 ˙˚

or

È1 0 0 ˘ E = ÍÍ0 1 0 ˙˙ and c E = 3 ÍÎ0 0 1 ˙˚

Thus, for coordinate x, y, z of all the atoms, yE = 3 × 3 = 9

Molecular Symmetry

4.17

Under C2 operation, only the O atom remains unshifted, i.e.

È x1 ˘ È- x1 ˘ Í ˙ Í ˙ C2 Í y1 ˙ = Í- y1 ˙ ÍÎ z1 ˙˚ ÍÎ z1 ˙˚

or

È-1 0 0 ˘ C2 = ÍÍ 0 -1 0 ˙˙ and cC2 = -1 ÍÎ 0 0 1 ˙˚

Under reflection through sxz operation, only the O atom remains unshifted.

È x1 ˘ È x1 ˘ Í ˙ Í ˙ s xz Í y1 ˙ = Í- y1 ˙ ÍÎ z1 ˙˚ ÍÎ z1 ˙˚

or

s xz

È1 0 0 ˘ = ÍÍ0 -1 0 ˙˙ and cs xz = 1 ÍÎ0 0 1 ˙˚

Similarly, under reflection through syz plane, all the three atoms remain unshifted, i.e.

È x1 ˘ È- x1 ˘ Í ˙ Í ˙ s yz Í y1 ˙ = Í y1 ˙ ÍÎ z1 ˙˚ ÍÎ z1 ˙˚

or

s yz

È-1 0 0 ˘ = ÍÍ 0 1 0 ˙˙ and cs yz = 1 ÍÎ 0 0 1 ˙˚

Total character of all the atoms csyz = 3 × 1 = 3

E C2

Thus, we get

G red

9

-1

s xy

s yz

1

3

4.5.4 Character Tables In order to study characters of IR’s for a group, a character table is used. The character table can be described as below: (a) A standard format of a character table consists of four parts as shown in Table 4.3. Table 4.3 Standard format of a character table Schoenflies symbols of group Mullikan’s symbols for IR’s

Symbol for each class in group and number of elements in the class Characters of IR’s of each class

Symmetry properties

Translations Rotations

Transformation characteristics of quadratic functions Product function of coordination

(b) Mullikan’s symbols are used to represent the dimension of an IR in place of G1, G2, G3, etc. 1-dimension IR – A or B; 2-dimension IR – E and 3-dimension IR – T (i) A 1-dimension IR which is symmetrical with respect to Cn is labelled as A and the antisymmetrical is labelled as B. This means that characters for symmetrical rotation are +1 while for unsymmetrical are –1, i.e. cCn = +1 $ A and cCn = –1 $ B (ii) Subscript 1 is used to indicate symmetric representation and 2 is used for antisymmetric representation with respect to rotation axis other than the principal axis (subsidiary axis). cC2 = +1 $ A1, B1, E1,T1, and cC2 = –1 $ A2, B2, E2, T2 If there is no subsidiary axis, 1 is used for symmetric representation and 2 for antisymmetric representation with respect to sn. csn = +1 $ A1, B1, El,T1, and csn = –1 $ A2, B2, E2, T2

4.18

Inorganic Chemistry

(iii) Single prime ( ) is used for symmetric representation and double prime ( ) for antisymmetric representation with respect to sh. i.e. csh = +1 $ A , B , E ,T , and csh = –1 $ A , B , E , T (iv) If a point group has centre of symmetry, subscript g is used for symmetric representation and u is used for antisymmetric representation with respect to inversion through centre of symmetry. i.e. ci = +1 $ g and ci = –1 $ u If the point group has no centre of symmetry, these subscripts are not used. (c) Symmetry properties of translations are indicated by x, y, z and rotations are indicated by Rx, Ry and Rz. Now we will use these points for the character table of C2v point group. (Table 4.4)

1. Mulliken’s Symbols

Table 4.4 Character table of C2v point group

G1 — cE = 1, 1-dimensional, cC2 = +1, csxz = +1 $ A1

C2v A1 A2 B1 B2

G2 — cE = 1, 1-dimensional, cC2 = +1, csxz = –1 $ A2 G3 — cE = 1, 1-dimensional, cC2 = –1, csxz = +1 $ B1 G4 — cE = 1, 1-dimensional, cC2 = –1, csxz = –1 $ B2

E 1 1 1 1

C2 1 1 –1 –1

sxz 1 –1 1 –1

syz 1 –1 –1 1

x2, y2, z2 xy xz yz

z Rz x, Ry y, Rx

2. Symmetry Properties For C2v point group, the principal axis C2 is collinear with z-axis. It means that there is no change in z-coordinate by E, C2, sxz and syz. Thus, z-transformation is A1. Rz remains unchanged under E and C2 but its direction changes under sxz and syz. Hence it is A2. x coordinate changes under C2 and sxz to –x. Similarly Ry changes under C2 and syz. Hence it is B. y coordinate changes under C2 and sxz to –y. Similarly Rx changes under C2 and sxz. Hence, it is B2.

4.5.5 Symmetry Properties of Wave Functions In order to determine the symmetry group of a molecule, the orbitals used by the central atoms need to be specified in terms of various symmetry operations. We can illustrate this concept as follows:

1. s-orbital The total wave function of s-orbital, is always symmetrical to every symmetry operations, as it is spherically z z z symmetrical.

2. p-Orbital

+

These orbitals (Fig. 4.22) can be represented in terms of following eigen functions: px = yr sinq cosf py = yr sinq sinf pz = yr cosq

y

y x

+

–

y – x

x +

– pz orbital

px orbital

py orbital

Fig. 4.22 Shapes of p-orbitals

where yr is a constant and angle independent. Now, we will consider the symmetry operations on these orbitals.

(a) Reflection Consider the case of px-orbital. If there is a change in the sign on performing a symmetry operation, the orbital is antisymmetric and otherwise it is symmetric. Under reflection by xz and xy plane, px-orbital is symmetric as there is no change in sign. But under reflection by yz plane, it is antisymmetric as there is a change in sign.

Molecular Symmetry

4.19

For example, under C2v point group, the result is as follow: (i) Reflection through sxz plane, f change to –f and q remains unchanged. sxy(px) = sxz(r sinq cosf) = r sinq cos(–f) = r sinq cosf = px, symmetric c = +1 sxz(py) = sxz(r sinq sinf) = r sinq sin(–f) = –r sinq sinf = –py, antisymmetric c = –1 sxz(pz) = sxz(r cosq) = r cosq = pz, symmetric c = +1 (ii) Reflection through syz plane, f changes to (f – p) and q remains unchanged. syy(px) = syz(r sinq cosf) = r sinq cos(f–p) = –r sinq cosf = –px, antisymmetry, c = –1 syz(py) = syz(r sinq cosf) = r sinq sin(f – p) = +r sinq sinf = +py, symmetric c = +1 syz(pz) = syz(r cosq) = r cosq = pz, symmetric c = +1

(b) C2 Operation Around z-axis This operation does not affect q but f changes to q + f. Thus C2(px) = C2(r sinq cosf) = r sinq cos(p + f) = r sinq (–cosf ) = –px, antisymmetric c = –1 C2(py) = C2(r sinq sinf) = r sinq sin(p + f) = –r sinq sinf = –py, antisymmetry, c = –1 C2(pz) = C2(r cosq) = r cosq = pz

(c) E-operation This operation does not result in change in size, hence the characters of p-orbitals for this operation are always +1. The character – table for C2v point group can be represented as follows: orbital

E C2

s xz

s yz

I . R.

s px

1 1

1 -1

1 1

1 -1

a1 b1

py

1

-1

-1

1

b2

pz

1

1

1

1

a1

Thus, applying all the symmetry operations, total characters can be obtained as follows:

È p x ˘ È1 0 0 ˘ È p x ˘ Í ˙ Í ˙ E Í py ˙ = ÍÍ0 1 0 ˙˙ Í py ˙ Í p ˙ ÍÎ0 0 1 ˙˚ Í p ˙ Î z˚ Î z˚

or

È1 0 0 ˘ E = ÍÍ0 1 0 ˙˙ and c E = 3 ÎÍ0 0 1 ˙˚

È px ˘ È-1 0 0 ˘ È px ˘ Í ˙ Í ˙ C2 Í py ˙ = ÍÍ 0 -1 0 ˙˙ Í py ˙ andd cC2 = -1 Í p ˙ ÍÎ 0 0 1 ˙˚ Í p ˙ Î z˚ Î z˚ È p x ˘ È1 0 0 ˘ È p x ˘ Í ˙ Í ˙ s xz Í py ˙ = ÍÍ0 -1 0 ˙˙ Í py ˙ and cs xz = 1 Í p ˙ ÍÎ0 0 1 ˙˚ Í p ˙ Î z˚ Î z˚

4.20

Inorganic Chemistry

È px ˘ È-1 0 0 ˘ È px ˘ Í ˙ Í ˙ s yz Í py ˙ = ÍÍ 0 1 0 ˙˙ Í py ˙ Í p ˙ ÍÎ 0 0 1 ˙˚ Í p ˙ Î z˚ Î z˚ Thus, and

and d cs yz = 1

E C2 sxz syz 3 –1 1 1 G = a1 + b1 + b2

4.5.6 Transformation of Atomic Orbitals Transformation properties of any p or d orbital of an atom can be determined from the subscript in the area III and IV of the character table. An s-orbital has no angular dependence, hence it always transforms according to the totally symmetric representation. The p and d orbitals transform as per their angular dependence as given by the subscripts in the area III and IV of the related character table of the point group to which the molecule belongs. For example, PCl3 belongs to the point group C3n. It is clear from the character table of the C3n paint group that, s-orbital transforms into A1 representation. px, py and pz orbials transform into E and A1 representation respectively. dxy, dx2–y2 , dxz and dyz transform into E representation or form the basis for E representation. The overall symmetry of the molecule is obtained by the direct product of the symmetry characters of the occupied orbitals as shown for C2v point group.

4.6

C2n

E

C2

sxz

syz

A1

1

1

1

1

A2

1

1

–1

–1

B1

1

–1

1

–1

B2

1

–1

–1

1

A1 × A2 = A2

1×1

1×1

1 × –1

1 × –1

A1 × AF = A1

1×1

1×1

1×1

1×1

A1 × B1 = B1

1×1

1 × –1

1×1

1 × –1

B1 × B1 = A1

1×1

–1 × –1

1×1

–1 × –1

A1 × B2 = B2

1×1

1 × –1

1 × –1

1×1

B2 × B2 = A1

1×1

–1 × –1

–1 × –1

1×1

A2 × B2 = B1

1×1

1 × –1

–1 × –1

–1 × 1

TERMS SYMBOLS OF DIATOMIC MOLECULES

Due to electrostatic repulsions between the electrons of a diatomic molecule, various molecular orbitals have different energies. These electronic states are expressed by means of molecular quantum numbers analogous to that of atomic quantum numbers used for atomic orbitals. In case of a diatomic molecule, the combining atomic orbitals with mL values give arise to the molecular orbitals with analogous quantum number, equal to the magnitude of ML , i.e. = |ML|, where, ML = ml1 + ml2 + ----- + mln Thus, the possible values of = 0, 1, 2, 3 ……… coded with a capital Greek letter S, p, f-respectively and represent the molecular or bitals designed with symbols s, p, d, f-respectively. The electron spin multiplicity is as usual indicated by 2S + 1 and the molecular term symbol is expressed as 2S+1 .

4.21

Molecular Symmetry

–

+ ss $ sg (+) +

–

+

–

+

+

–

ss* $ su (+)

– (+)

–

+

–

+

+

–

s p*z $ su (+)

spz $ sg +

+

+

–

–

–

pp* x (or pp*y ) $ pg (–)

ppx(or p py ) $ pu (–)

Fig. 4.23(a) Parity of molecular orbitals

s (g)

p (u)

d (g)

Fig. 4.23(b) Parity of atomic orbitals Table 4.5

Designation of molecular orbitals

Molecular orbital s p d f

0

State S P D F

In addition, subscripts and superscripts are added to express 1 the symmetry properties. The subscripts g or u is added to indicate 2 the parity. If the wave function changes sign on inversion through 3 the centre of symmetry, the state is ungerade (u) and if it remains unaffected, the state is gerade (g) (Fig. 4.23a and b). The possible molecular states for any two wave functions can be determined by using the direct products : g × g = g; u × u = g and g × u = u The superscript (+) or (–) is added to express the behaviour of the orbitals towards reflection in a plane containing the molecular axis. (+) is added for a symmetric state, while (–) is added for an unsymmetric state. Thus, for all the sMOs a(+) sign is added. The multiplication rule is as follows: (+) × (+) = (+); (+) × (–) = (–) × (+) = (–); (–) × (–) = (+). When two electrons reside in p or d MOs with parallel spin, S– state is arised. However the p and D states occur as degenerate pairs and are not classified in + or –.

Rules for the Determination of Terms Symbols for Diatomic Molecules 1. Only the HOMO are considered. 2. For completely filled MOs (s2 or p4), ML = 0, = 0, S = 0, 2S + 1 = 1 and parity is always gerade (u × u = g). Thus, the term symbol will be 1Sg+. 3. For MOs with one electron, following cases arise: 1 (a) If electron is present in bonding sMO, (s1g); = 0; S = , 2S + 1 = 2 and parity is g. 2 Thus, the term symbol will be 2Sg+. 1 (b) If electron is present in bonding PM.O., (P1u); = ±1; S = , 2S + 1 = 2 and parity is u. 2 Thus, the term symbol will be 2Pu. 4. For MO’s with two electrons (other than s2), following cases may arise: (a) If the electrons are present in two degenerate bonding PMO’s, (p1up1u) with parallel spins. = (+1) + (–1) = 0, state is S; S = 1 , 2S + 1 = 3, parity is g (u × u = g). Thus, the term symbol is 3Sg+. (b) If the electrons are present in two degenerate bonding PMO’s, (p1up1u) with antiparallel spin, the term symbol is 1Sg+. (c) If both electrons are present in the same bonding PMO’s, (p2) with antiparallel spins. = (+1) + (+1) = 2, state is D; S = 0, 2S + 1 = 1, parity is g (u × u = g). Thus, the term symbol is 1Dg.

4.22

Inorganic Chemistry

Table 4.6

Term symbols for ground states of some homodiatomic molecules

Table 4.7

Term symbols for ground states of some heterodiatomic molecules

Molecule

Term symbol

Molecule

Term symbol

H2

1 + Sg 1 + Sg 1 + Sg 3 – Sg 1 + Sg 1 + Sg 3 + Sg 1 + Sg

LiH

1 +

BeH

2 +

Li2 Be2 B2 C2 N2 O2 F2

S

S

BH

1 +

NH

3 –

CO

1 +

S

S

S

HF

1 +

LiF

3 +

HCl

1 +

S

S

S

5. In case of more than two electrons, such as three electrons occupying the bonding PMO’s, (pu3). (p)2(p)1 Here, = (+1) + (+1) + (–1) = +1, state is P, S = + 0, 2S + 1 = 2. Parity is u × u × u = u Thus, the term symbol is 2Pu We can determine the possible term symbols depending upon the possible molecular electronic configuration. In case of more than one possibilities, the ground state term is determined using Hund’s rule. Thus, the ground state term must have highest spin multiplicity. In case of two terms with same spin multiplicity, the term with higher value is lower in energy. We can consider the example of O2 molecule with the following molecular electronic configuration: (s1s)2 (s*1s)2 (s2s)2 (s*2s)2 (s2pz)2 (p2px)2 (p2py)2 (p*2px)1 (p*2py)1 The completely filled MO’s will not be considered. Hence, the possible configuration of the outer most two electrons are as follows: (a) (p*2px)1 (p*2py)1 = (+1) + (–1) = 0, S state. S = 1, 2S + 1 = 3 and if S = 0, 2S + 1 = 1 and Parity is g × g = g. Thus, the term symbol is 3Sg– for parallel spin and 1Sg+ for antiparallel spin. (b) (p*2px)2 = (+1) + (+1) = 2, D state. S = 0, 2S + 1 =1 and Parity is g × g = g state. Thus, the term symbol is 1Dg The ground state for O2 can be determined on the basis of Hund’s rule. Hence, 3Sg– must be the ground state due to its highest spin multiplicity.

4.7

APPLICATIONS OF GROUP THEORY

We will briefly discuss the applications of group theory as below:

1. Solution of Integrals We can determine the solution of integrals used in molecular quantum mechanics on the basis of following guidelines: (a) The direct product Gij of any representation Gi with Gj is simply Gi, if Gj is totally symmetric representation. (b) The direct product Gij of any representation Gi with itself is a totally symmetric representation.

Molecular Symmetry

4.23

Consider I1 , the integral of the product of two functions, i.e. I1 =

Ú y i y j dt

I1 will be equal to zero or vanish unless the integrand yiyj is invariant or at least some term in it is invariant, under all the symmetry operation of the point group to which the system (molecule) belongs. It means that the representation for yiyj or Gij is a totally symmetric representation or in other words, Gi and Gj should be one and the same (Gi = Gj). Thus, for non-zero overlap of two atomic orbitals, they must have the same symmetry.

2. Probability of Spectral Transition The probability of spectral transitions between two states ith and jth with respective wave functions yi and yj, in x, y and z coordinates can be determined as follows: If I is the intensity of spectral transition, then we can obtain three integrals corresponding to the three coordinates:

I x a Ú y i x y j dt (direct product as G i ¥ G x ¥ G j ) I y a Ú y i y y j dt (direct product as G i ¥ G y ¥ G j ) I z a Ú y i z y j dt (direct product as G i ¥ G z ¥ G j ) The spectral transition between two states with wave function yi and yj will be allowed, if the integrand is non-zero. This is possible only if yi × yj contain Gx, Gy or Gz (the representation to which x, y or z respectively belong). In other words, the direct product should contain A1, the totally symmetric representation. We can illustrate this with the help of the a molecule with C2n point group. Consider the electronic transition between dyz and dxy orbitals or the A2 and B2 states i.e. Gi = A2 and Gj = B2. From character table of C2n point group, we can check that x-vector corresponds to B1, so we have Gi × Gx × Gj = A2 × B1 × B2 = A1. Thus A2" B2 transition is allowed in x-direction or is said to be x-polarized. In the ground state, each yi and hence yv is totally symmetric (A1g). But in the excited state for one of the normal mode, the corresponding yi may belong to the representation other than the total symmetric. As an approximation, the complete wave function yi for the molecule can be represented as the product of yelec, yvib, yrot and ytrans wave functions of the molecule, all independent of one another. i.e. yi = yelec yvib yrot ytrans and the transition integral can be represented as m = yelec yvib yrot ytrans mˆ y elec ¢ y vib ¢ y rot ¢ y trans ¢ dt . To another approximation, if the sample is in the solid or liquid phase during spectral transition, yrot and ytrans are not likely to change during the excited electronic state, i.e. m = yelec yvib mˆ y ¢ y ¢ dt elec

vib

As the Laporte forbidden transitions get slightly allowed, yelec & yvib may get interdependent and combined state is represented as yvibronic, i.e. m = yvibronic mˆ y vibronic dt ¢ This is known as vibronic coupling and results in distortion of octahedral symmetry with loss in g-character. As discussed earlier, if this integral is non-zero, the transition is vibronically allowed. For example, consider the case of [Co(NH3)6]3+ with ground state 1A1g and excited states 1T1g and 1T2g. From the character table of Oh point group, x, y and z jointly form T1u representation. For the electronic transition, 1A1g $ T1g, the direct product representation is written as G[y elec (x, y, z) yelec] = T1g × T1u × A1g = T1g × T1u The direct product T1g × T1u can be reduced to A1u + Eu + T1u + T2u. Similarly, for the electronic transition 1 A1g $ 1T2g, the direct product A1g × T1u × T2g or T1u × T2g can be reduced to A2u + Eu + T1u + T2u. None of these reduced representations contain A1g. Hence, the integral is zero and these electronic transitions are Laporte forbidden.

4.24

Inorganic Chemistry

However, during these electronic transitions, there may be simultaneous excitation of a T1u or T2u vibrational mode. In the case of 1A1g $ 1T2g transition with excitation of simultaneous T1u vibrational mode, G[y vibronic (x, y, z) yvibronic] the direct product will be written as A1g × (A1g × T1u × T1g) × T1u obtained by multiplying the direct product of Gelectronic with Gvibronic and further reduced to (A1u + Eu + T1u + T2u) × T1u. The term T1u × T1u can be reduced to the term A1g. Thus the transition becomes slightly allowed.

3. Vibronic Polarisation

1E

g

1

1

Energy

A1g $ Eg 1T For an octahedral complex, the directions x, y and z are 2g 1B interchargeable by the symmetry operations and the 2g 1A $ 1B 1g 2g components of the electric dipole vector of the light 1A 2g are equivalent. Hence, the direction of the vibration 1A $ 1A 1T 1g 2g 1g of the oscillating electric vector of light has no effect 1E g on the extent of interaction. However, in case of less symmetrical complexes (x, y and z belong to different 1A g representations) polarization is encountered. In such Fig. 4.24 Electronic transitions for tetragonal cases, all the expected transitions may not take place Co3+ complex when electronic absorption spectra is recorded using plane polarized light source. Consider the tetragonal complex, trans [Co(en)2Cl2]+ with D4h symmetry and the ground state as A1g. Here the excited, T1g and 1T2g undergo splitting as follows: 1

T1g $ (1A2g + 1Eg) and 1T2g $ (1B2g + 1Eg)

As a result, the possible symmetry based transitions (Fig. 4.23) are as follows: For a tetragonal complex, z component of the electric dipole belongs to the A2u representation, while the x and y components belong to Eu representation of the D4h character table. We can obtain the following transition integrals for the possible electronic transitions: 1

A1g $ 1A2g

Integral

1

A1g $ 1Eg

1

A1g $ 1B2g

¢ z y elec dc Ú y elec

A1g × A2u × A2g = A2u A1g × A2u × Eg = Eu

A1g × A2u × B2g = B1u

¢ ( x, y) y elec dc Ú y elec

A1g × Eu × A2g = Eu

A1g × Eu × B2g = Eu

A1g × Eu × Eu = A1u+ A2u + B1u + B2u

All these transitions are Laporte forbidden, as none of the representation contains A1g. However, some low intensity absorption bands have been observed and can be assigned to the vibronic coupling due to the interdependence of the electronic and vibrational states of the compound. The first excited vibrational states belong to the following symmeties: 2A1g, B1g, B2g, Eg, 2A2u, B1u and 3Eu The probability of possible electronic transitions can be determined by multiplying the direct product of Gelec with Gvibronic (excited state) as the Gvibronic (ground state) is A1g and would not change the result. All the allowed transitions must relate to A1g representation. Thus, in case of A1g $ A2g with simultaneous excitation of a vibration of any of the first excited vibrational mode is forbidden with z-polarization as the direct product does not contain A1g. G[y vibronic (z) yvibronic] = A1g × (A1g × A2u × A2g) × Eu = A2u × Eu = Eg On the other hand, the A1g $ A2g with simultaneous oxcitation of Eu vibrational mode is allowed with (x, y) polarisation as the direct product contains A1g.

Molecular Symmetry

4.25

G[y vibronic (x, y) yvibronic] = A1g × (A1g × Eu × A2g) × Eu = Eu × Eu = A1g Similarly, the other allowed transitions have been predicted as given below: Transition

z

xy

A1g $ B2g

A1g × (A1g × A2u × B2g) × B1u = B1u × B1u = A1g Allowed by B1u vibrational mode

A1g × (A1g × Eu × B2g) × Eu = Eu × Eu = A1g Allowed by Eu vibrational mode

A1g $ B2g

A1g × (A1g × A2u × Eg) × Eu = Eu × Eu = A1g Allowed by Eu vibrational mode

A1g × (A1g × Eu × Eg) × B1u or A1u = (A1u + A2u + B1u + B2u) × B1u or A1u Allowed by B1u and A1u vibrational mode

We can use these predictions to analyse the 1A $ 1E ,1T 1A $ 1 E g 1g 2g g 1g experimental observations for electronic transition 1.5 1A $ 1 A + 1g 2g spectra of trans [Co(en)2Cl2] in plane polarised light. If the spectrum is carried out with light polarized 1.0 log ε perpendicular to the principal (z) axis of the complex, four transitions are expected. Experimentally, three 0.25 absorption bands have been observed first two bands assigned to 1A1g $ 1Eg (~16,000 cm–1), 0 1 15,000 20,000 25,000 30,000 A1g$1A2g (~22,000 cm–1) respectively and v (cm–1) third broad band is a composite band (~ 27,000 + cm–1) assigned to 1A1g$1T2g and 1A1g$ 1Eg Fig. 4.25 Dichroism of trans [Co(en)2Cl2] with solid line for light polarized parallel to the principal transitions due to smaller energy gap between Eg axis and dotted line for light polarized and B2g states (Fig. 4.25) The band at ~ 22,000cm–1 perpendicular to the principal axis of the is strongly polarized band which disappears when complex light is polarized parallel to the principal axis of the complex, as A1g $ A2g transitions is Transition Polarisation of incident radiation vibronically forbidden for z-polarization. Thus, in x, y z this case only two absorption bands are observed. A2 × A2 × A2 = A2 A2 $ A2 A2 × E × A2 = E (Fig. 4.25) (forbidden) (forbidden) In case of complexes with an approximate A2 × A2 × A2 = A2 A2 $ A1 A2 × E × A1 = E octahedral array of ligands, the true molecular (forbidden) (allowed) symmetry is lower than the octahedral symmetry. A $ E A × E × E = A + A + E A × A × E = E 2 1 2 2 2 2 For example, in case of tris(acetylacetonato) and (allowed) (forbidden) tris (oxalato) complexes of Cr(III) ion, the actual molecular symmetry is D3. In this symmetry, A2g, T1g and T2g states of Oh symmetry are reduced to A2, (A2 + E) and (A1+ E) states respectively. D3 symmetry has no centre of symmetry, hence the pure electronic selection rules are not followed. In this case, the z component of the electric dipole vector of the complex transforms as A2 while the x,y components transform as E representation of D3 group. Now, we can predict the possible transitions as follows: As usual, the transition related to the A1 representation will be allowed. Thus A2 $ A1 is z allowed but x, y forbidden, while A2 $ E is x, y allowed but z forbidden. It means that spectrum in light polarized perpendicular to the principal axis will show A2 $ E transition band, while spectrum in light polarized parallel to the principal axis will show A2 $ A1 transition band.

4.26

Inorganic Chemistry

4. Symmetry Adapted Linear Combinations of Atomic Orbitals (SALCs of AOs) Group theory helps to identify the combination of atomic wave functions of individual atoms of a molecule to form Molecular orbitals corresponding to the molecular symmetry. The steps involved are as follows: (a) Identification of the point group to which the molecule belongs (b) Formation of reducible representation on the basis of contribution to the character equal to the unshifted vectors under symmetry operation. (c) Reduction to reducible representation into irreducible representation using reduction formula. (d) Writing the expression for the SALCs corresponding to the component of the irreducible representation

(a) For a Tetrahedral Complex

(i)

For Sigma Bonding

A tetrahedral molecule AB4 belongs to Td point group with the following set of characters for the generated reducible representation.

Ts

E 8C3 4 1

3C2 0

6 S4 0

6s d 2

The reducible representation can be reduced to irreducible representation Ts by using the reduction formula Ni = gR (S4) = 6, gR(sd) = 6

1 Â c ( R) gR ci ( R) ; here h = 24, gR(E) = 1, gR(C3) = 8, gR(C2) = 3, h R

1 [4.1.1 + 1.8.1 + 0.3.1 + 0.6.1 + 2.6.1] = 1A1 24 1 n2(A2) = [4.1.1 + 1.8.1 + 0.3.1 + 0.6.(–1) + 2.6.(–1)] = 0A2 24 1 n2(E) = [4.1.2 + 1.8.(–1) + 0.3.(2) + 0.6.0 + 2.6.0] = 0E 24 1 n2(T1) = [4.1.3 + 1.8.0 + 0.3.(–1) + 0.6.1 + 2.6.(–1)] = 0T1 24 1 n2(T2) = [4.1.3 + 1.8.0 + 0.3.(–1) + 0.6.(–1) + 2.6.(1)] = 1T2 24 n1(A1) =

Thus, Gs = A1 + T2 The complete set of s bonds can be formed by the AOs on the central atom A and SALCs on the B atoms belonging to these representations (A1 and T2). It is clear from the character table that the AOs of A transform as follows: Table 4.8 Character table for Td point group Td

E

8C3

3C2

6S4

6sd

A1

1

1

1

1

1

A1

1

1

1

–1

–1

E

2

–1

2

0

0

T1

3

0

–1

1

–1

(Rx, Ry, Rz)

T1

3

0

–1

–1

1

(x, y, z)

(x2 + y2 + z2) (2z2 – x2 – y2), x2 – y2 (xy, xz, yz)

4.27

Molecular Symmetry

s-orbtial $ A1;

(px, py, pz) and (dxy, dxz, dyz) orbitals $ T2

Out of these s and p orbitals are suitable for s bonding while d orbitals are suitable for p bonding. SALCs of A1 symmetry must be unchanged by all symmetry operations and should be positive everywhere, i.e.

1

 (s) = 2 (s z

1

+ s z2 + s z3 + s z4

)

A1

Figure 4.26 shows vectors representing p-orbitals along the principal axis for s bonding and perpendicular to the principal axis for p bonding. We see that y vectors are all parallel to xy plane of the coordinate system. The coordinates of ligand atoms used for s and p bonding have been represented as (px1, py1, sz1), (px2, py2, sz2), (px3, py3, sz3) and (px4, py4, sz4). C3 axis of symmetry is collinear with pz orbitals involved in s bonding. z In case of SALCs matching the symmetries of p px4 orbitals (T2), the positive amplitude should match with positive amplitude and the negative amplitude py sz4 4 px1 should match with negative amplitude, i.e. py1

1 Â (px) = 2 (sz1 – sz2 + sz3 – sz4) T

sz1

2

y sz2

1

 (py) = 2 (sz1 + sz2 – sz3 – sz4)

px3

x py3

T2

1 Â (pz) = 2 (sz1 – sz2 – sz3 + sz4) T

px2 py2

sz3

Fig. 4.26 Representation of a tetrahedral molecule

2

Now the metal orbitals and LGO’s of appropriate symmery undergo overlap to form sigma bonding and antibonding MO’s

(b)

For p-Bonding The reducible represenation in this case will be Td Gp

E 8C3 8 -1

3C2 0

6 S4 0

6s d 0

This can be reduced into Gp = E + T1 + T2. This means that LGO’s of E, T1 and T2 symmetry are required. However, there are no metal orbitals of T1 symmetry and T2 set is involved in s-bonding, hence the metal orbitals of only E symmetry will combine with LGO’s for p-bonding. SALC’s for E

1 1 (px1 – px2 – px3 + px4) and (py1 – py2 – py3 + py4) 2 2 Now, metal orbitals and LGO’s of appropriate symmetry undergo overlap to form p-bonding and antibonding MO’s as explained further in chapter 24. symmetry can be obtained as

(b) For An Octahedral Complex

(i)

For Sigma Bonding An octahedral molecules AB6 belongs to Oh point group with the following set of characters for the generated reducible representation.

Oh Ts

E 8C3 6 0

6C2 0

6C4 2

3C2 2

i 6 S4 0 0

8S6 0

3s h 4

The reducible representaion can be reduced to Gs = A1g + Eg + T1u.

3s d 2

4.28

Inorganic Chemistry

Table 4.9 Character table for Oh point group. Oh

E

8C3

6C2

6C4

3C2

i

6S4

8S6

3sh

6sd

A1g

1

1

1

1

1

1

1

1

1

1

A2g

1

1

–1

–1

1

1

–1

1

1

–1

Eg

2

–1

0

0

2

2

0

–1

2

0

T1g

3

0

–1

1

–1

3

1

0

–1

–1

T2g

3

0

1

–1

–1

3

–1

0

–1

1

A1u

1

1

1

1

1

–1

–1

–1

–1

–1

A2u

1

1

–1

–1

1

–1

1

–1

–1

1

Eu

2

–1

0

0

2

–2

0

1

–2

0

T1u

3

0

–1

1

–1

–3

–1

0

1

1

T2u

3

0

1

–1

–1

–3

1

0

1

–1

(ii)

x2 + y2 + z2 (2z2 – x2 – y2, x2 – y2) (Rx, Ry, Rz) (xz, yz, xy)

(x, y, z)

For -Bonding The representation corresponding to 12 p bonds of AB6 molecules can be obtained as

Oh Gp

E 8C3 12 0

6C2 0

6C4 0

3C2 -4

i 6 S4 0 0

8S6 0

3s h 0

6s d 0

Gp = T1g + T2g + T1u + T2u This topic has been explained further in Chapter 24.

5. Hybridisation of Atomic Orbitals On the basis of group theory, suitable atomic orbitals for hybridisation (equivalent orbitals) can be recognised. For example, H2O belongs to C2v point group. As the character of the reducible representation is equal to the number of unshifted vectors by the symmetry operation, the reducible representaion can be generated as

C2n

R C2

G red

9

-1

s xz

s yz

1

3

Gred = 3A1 + A2 + 2B1 + 3B2 From the character table, it is clear that s orbital transforms as A1, while px, py and pz orbital transform as B1, B2 and A1 respectively. One the other hand dxy, dxz and dyz orbitals transform into A2, B1 and B2 respectively. However, the combination with d-orbitals are higher in energy. Hence, s, px, py and pz hybridise together to undergo sp3 hybridisation.

(a) For PF5, D3h Symmetry Point Group D3h G red

E 2C3 5 2

3C2 1

sh 3

2 S3 0

3s n 3

Gred = 2A 1 + A 2 + E The possible combination for sp3d hybridisation is s(A1 ) + dz2 (A1 ) + pz (A1 ) + px (E ) + py (E )

Molecular Symmetry

4.29

(b) For (PtCl4)2–, D4h Point Group

D4 h G red

E C4 4 0

C2 0

2C2¢ 2

2C2¢¢ i 2 S4 0 0 0

sh 4

2s n 2

2s d 0

Gred = A1g + B1g + Eu The possible combinations for dsp2(I) or sp2d(II) hybridisation are I $ s(A1g) + dx2 – y2 (B1g) + px (Eu) + py (Eu) and II $ dz2 (A1g) + dx2 – y2 (B1g) + px (Eu) + py (Eu) The first combination is energtically and symmetrically more favorable

(c) For BCl3, D3h Point Group D3h G red

E 2C3 3 0

3C2 1

sh 3

2 S3 0

3s v 1

Gred = A 1 + E

6. Energy Correlation Diagram: Walsh Diagram (a) For AH2 Molecule with Angular Geometry (C2n Symmetry) In C2n point group, for AH2 molecule, the reducible representation for the unshifted vectors can be obtained as follows:

C2n G(H )

E C2 2 0

sn 2

sn ¢ 0

a + b2

SALC’s for H-atoms can be obtained as follows: y1 = y1s(HA) + y1s(HB) (a1 symmetry) y2 = y1s(HA) – y1s(HB) (b2 symmetry) From the character table for C2n point group, A-atom has 2s and set of three p orbitals available for bonding. The symmetry of 2s, 2px, 2py and 2pz orbitals are respectively a1, b1, b2 and a1 respectively. Out of these 2s and 2pz orbitals are of appropriate symmetry for s bonding with H atoms. It means that y1 (a1 symmetry) of H atoms can combine with mixed 2spz orbital (a1 symmetry) of A atom to give bonding MO 1a1, antibonding MO 3a1 and non-bonding MO 2a1. There is no effective overlap between 2px orbital (b1 symmetry) of A atom with y2 (b2 symmetry) or y1 (a1 symmetry) of H-atoms as shown in Fig. 4.27. Hence, it forms nonbonding MO1b1 . 2py orbital (b2 symmetry) of A atom combines with y2 (b2 symmetry) of H atoms to form bonding MO 1b2 and antibonding MO 2b2. The complete MO diagram for AH2 molecule has been shown in Fig. 4.27. (b) For AH2 Molecule with Linear Geometry (D h Symmetry) From the character table for D h point gruop, A-atom has orbital of sg or a1g symmetry and 2px orbitals of a1u or su symmetry. On the other hand 2px and 2py orbitals are of e1u or pu symmetry. Linear combinations of 1s atomic orbitals of H atoms are [y1s(HA) + y1s(HB)] and [y1s(HA) – y1s (HB)] of a1g and a1u symmetry respectively. AO’s of A-atom and SALC’s of H atom of appropriate symmetry combine together to form s bonding and antibonding MO’s. However, the AO’s with pu symmetry of A-atom donot match with the symmetry of SALC’s of H atoms and remain non-bonding. Thus combination of sg of A-atom with that of H-atom gives 1sg (bonding MO) and 2sg (antibonding

4.30

Inorganic Chemistry

+ + + –

–

3a1 +

2b2

– + +

+

+

– – a1 b1

+

–

b2 + a1

1b1 a1, b1, b2 (2p)

+

a1, b2

–

+

+

b2

–

– –

+ +

1b2

a1

2a1 a1 (2s) –

+ –

– +

A 1a1

H, H

+

+

Fig. 4.27 MO diagram for angular geometry of AH2 molecule

MOs). Similarly, combination of su of A-atom with that of H-atoms gives 1su (bonding MO) and 2su (antibonding MO) as shown in the MO diagram (Fig. 4.28). –

2su

+

+

+

+

+

– su + u

+

– 1su +

–

sg + su

– –

1pu

+

+

–

2sg

– –

–

–

+

–

+

+

+

+

sg +

+

+

1sg

Fig. 4.28 MO diagram for linear geometry of AH2 molecule

Walsh Diagram Prediction of Linear or Bent Geometry for AH2 Molecules We generally use VSEPR theory to predict the geometry of molecules. However, for simple AH2 molecules (A belongs to second period) A.D. Walsh put forth an empirical and qualitative method to determine their geometry. In this method, the geometry of the molecules is determined on the basis of Walsh diagram, a correlation diagram between energies of C2n and D h orbtials. z z x x For linear molecules (D h), y axis is selected as the unique sigma y H A H A y bond direction. For angular molecules (C n), the y and z axis lie in the plane of paper and pass through A. Thus, the decrease of angle from H H linear to angular geometry makes no difference in the axes.

4.31

Molecular Symmetry

Comparison of Energies of HOMOs The reduction of symmetry from D h to C2n results in the mixing of the 2s and 2pz orbitals of the central A atom. As a result, the 2s and 2pz AOs of a1 symmetry mix to give 2spz orbitals of a1 symmetry. Some important points are the following: (i) 1sg is stabilised due to an increase in bonding overlap on going to 1a1 (ii) 1su is destabilised due to reduction in bonding interation but increase in antibonding interaction on going to 1b2 (iii) There is no change in the energy of non-bonding 1pu as it changes to 1b1. (iv) Due to mixing of 2s and 2pz orbital, on reduction of symmetry, 1pu is stabilised on changing to 2a1, but 2sg is highly destabilised on changing to 3a1. (v) 2su is stabilised as it changes to 2pz. According to Walsh’s rule, the structure in which HOMO is stablised, is adopted by the molecule. If HOMO remain unperturbed, the next occupied MO is Character table for D3h point group considered. Linear function, E ... quadratic rotations For BeH2 (Fig. 4.29.), HOMO is 1su which gets x2 + y 2, z 2 A1g = Σ +g 1 ... destabilised on reduction of symmetry, hence it is – ... linear. But for H2O, HOMO is 1b1 which remains A2g = Σ g 1 Rz unperturbed. Hence, 2a1 will be considered, it is E1g = Πg 2 ... (xz, yz) (Rx, Ry) stabilised on reduction of symmetry. As a result, ... 2 2 E2g = ∆ g (x – y 2, xy) H2O is bent. Similarly, the AH2 molecules with 5 – 8 ... ... ... electrons in valence shell adopt C2n structure. + (c) Molecular Orbital Energy Level Diagram of CO2 Molecule (D h symmetry) CO2 molecule belongs to D h point group with C axis taken as the internuclear axis (z-axis) for sigma bonding. From the character table, the symmetry of 2s-orbital of C atom is a1g. The totally symmetric 2s orbitals of C atoms are also of a1g symmetry, but due to higher electronegativity and lesser energy of oxygen atom, these orbitals do not participate in bonding and remain non-bonding. In order to match with 2s orbital of C atom with a1g symmetry, 2pz orbitals of O atoms form SALC’s as follows: y1 = [y2pz(O1) + y2pz (O2)] " a1g

z

E1u = Πu 2 ... ... ... ...

(x, y)

2su (b2)

3a1

2sg (a1)

2b2

1b1 1pu (a1 + b1)

y2 = [y2pz(O1) – y2pz (O2)] " a1u Now y1 (a1g symmetry) oxygen group orbital combines with 2s (a1g symmetry) carbon orbital to form 2a1g bonding MO and 2a*1g antibonding MO. The carbon 2px and 2py orbitals are of e1u symmetry and 2pz orbital is of a1u symmetry. y2 (a1u symmetry) oxygen group orbital combines with (a1u symmetry) carbon orbital to from 2a1u bonding MO and 2a*1u antibonding MO. The p-bonds are formed by 2px and 2py atomic

A1u = Σ u 1 ... A2u = Σ –u 1 ...

1b2

1su (b2) 1sg 180° (D3u)

2a1

1a1

(a1)

Fig. 4.29 Walsh diagram for AH2 molecule

90° (C2n)

4.32

Inorganic Chemistry

orbitals of C atom and O atom. In order to match the symmetry, 2px and 2py orbitals form SALC’s as a two pairs of group orbitals.

y 3 = ÈÎy 2px (O1 ) + y 2px (O2 )˘˚ ¸Ô ˝e y 4 = Èy 2py (O1 ) + y 2py (O2 )˘ Ô 1u Î ˚˛ y 5 = ÈÎy 2px (O1 ) + y 2px (O2 )˘˚ ¸Ô ˝e y 6 = Èy 2py (O1 ) + y 2py (O2 )˘ Ô 1g Î ˚˛ e1u oxygen group orbitals combine accordingly with e1u orbitals of carbon atom to form p-bonds. e1g orbitals remain non–bonding. The molecular orbital energy level diagram with orbital overlapping has been shown in Fig 4.30. The molecular electronic configuration of CO2 can be represented as 1a1g2 1a1u2 2a1g2 2a1u2 e1u4 e1g4 (d) Molecular Orbital Energy Level Diagram of NH3 Molecule (C3n Symmetry) NH3 molecule belongs to C3n symmetry. From the character table, 2s and 2pz orbitals of N atom are of a symmetry while 2px and 2py orbitals are of e symmetry. Due to difference in energy of 2s orbitals of three H atoms with that of N atom (each of a symmetry), SALC’s are formed to give one combination of a symmetry and two combinations of e symmetry. After combination of appropriate orbitals, the molecular electronic configuration can be represented as s21a s42e s23a –

3a1u 3a1g

–

+ + +

–

+

2px (e ) 2py 1u 2p z (a1u)

–

+

+ –

+

–

p e1u

–

+ –

– +

+ +

+

–

– +

– e1g (nb) 2s (a1g)

e1g

–

e1u a1u

p e1u

–

+

+ –

+

2p

+ –

a1g

– 2a1u

C

1a1u (nb) 1a1g (nb)

–

+

2a1g

–

+

+

– –

– + +

–

+

+

–

–

a1u a1g

–

– +

+

+

–

2s + O

+ MO's

Fig. 4.30 Molecular orbital energy-level diagram of CO2

+

Molecular Symmetry

7. Splitting of Levels and Terms in Ligand Field Environment

4.33

5a 4c

We know that an octahedral complex has Oh symmetry. However, Oh point group can be obtained from pure rotational subgroup O by adding the inversion, i. The O point group has only rotation symmetry operations and five irreducible representations (A1, A2, E, T1 and T2). On the other hand Oh point group has ten irreducible representations (A1g, A2g, Eg, T1g, T2g, A1u, A2u, Eu, T1u and T2u). To an assumption, all the d-orbitals can be consider in the general form as y = R(x).Q(q).F(f).ys

3a a+e (2p)

a (2s) e N a

Where the radial function R(r) depends only on the radial 2e H distance r from the nucleus, the angular functions Q(q) and F(f) depend only on the angles q and f and the spin function 1a ys which does not depend on r, q and f. Thus, R(r) is invariant to all symmetry operations in Fig. 4.31 Molecular orbital energy level diagram of NH3 a point group while Q(q) and F(f) will be affected by rotations. However, if the rotation is carried out about the z-axis, Q(q) will also be invariant and only F(f) will be altered. Thus, we can describe the symmetry operation A as follows: A[R(r) Q(q) F(f)] = R(r) Q(q) A F(f) The explicit form of F(f) function can be given as F(f) = eimf where m ranges from +l to –l for the five d-orbitals (l = 2) i.e. 2, 1, 0, –1, –2. Thus, the parts of the d- orbital wave functions variant with the symmetry operations of the O point group are ei2f, eif, eiof, e–i1f, e–i2f, e–i2f (the normalising constants have not been considered as these are invariant under all symmetry operations. The matrix required for identity operation, E can be given as

Èe2if ˘ È1 Í if ˙ Í0 Íe ˙ Í E Íe 0 ˙ = Í0 Í - if ˙ Í Íe ˙ Í0 ÍÎe -2if ˙˚ ÍÎ0

0 0 0 0 ˘ È e i 2f ˘ 1 0 0 0 ˙˙ ÍÍeif ˙˙ 0 1 0 0 ˙ Íe0 ˙ and c ( E ) = 5 ˙ ˙Í 0 0 1 0 ˙ Íe - if ˙ 0 0 0 1 ˙˚ ÍÎe -2if ˙˚

While, for rotation operation, C(a) with rotation by angle a can be given as

Èe2if ˘ Èe2ia Í if ˙ Í0 Íe ˙ Í C (a ) Íe0 ˙ = Í0 Í - if ˙ Í Íe ˙ Í0 ÍÎe -2if ˙˚ ÍÎ0

0 eia 0 0 0

0 0 e0 0 0

0 0 0 e - ia 0

˘ È e i 2f ˘ ˙ Í if ˙ ˙ ˙ Íe ˙ Íe 0 ˙ ˙ Í - if ˙ ˙ ˙ Íe - i 2a ˙ Í -2 if ˙ e ˚ ˚ Îe

0 0 0 0

4.34

Inorganic Chemistry

1ˆ Ê sin Á l + ˜ a Ë 2¯ ’ and c C (a ) = , where (a 0) and (l = 2) sin (a /2) sin 5a /2 i.e., c C(a ) = sin p /2 sin 5p /2 1 For two-fold rotation a = p and, c (C2 ) = = =1 sin p /2 1 sin 5p /3 - sin p /3 = = -1 For three-fold operation, a = 2p/3, c (C3 ) = sin p /3 sin p /3 sin 5p /4 -1 2 For four-fold rotation, a = 2p/2, c (C4 ) = = = -1 sin p /4 1 2 It means that Gred for the point group O can be written as

O G red

E 6C4 5

3C2 (= C42 ) 8C3

-1

1

-1

6C2 1

The reducible representation can be reduced by considering h = 24 and using character table as follows;

1 [1.1.5 + 6.1.(–1) + 3.1.1 + 8.1.(–1) + 6.1.1] = 0A1 24 1 n(A2) = [1.1.5 + 6.(–1).(–1) + 3.1.(–1) + 8.1.(–1) + 6.(–1).1] = 0A2 24 1 n(E) = [1.2.5 + 6.0.(–1) + 3.2.1 + 8.(–1).(–1) + 6.0.1] = 1E 24

n(A1) =

Table 4.10 The character table for the point group, O O

E

6C4

3C2(C24)

8C3

6C2

A1

1

1

1

1

1

A2

1

–1

1

1

–1

E

2

0

2

–1

0

T1

3

1

–1

0

–1

T2

3

–1

–1

0

1

x2 + y2 + z2 (2z2 – x2 – y2, x2 – y2) (Rx, Ry, Rz) (x, y, z) (xy + xz + yz)

1 [1.3.5 + 6.1.(–1) + 3.(–1).(1) + 8.0.(–1) + 6.(–1).1] = 0T1 24 1 n(T2) = [1.3.5 + 6.(–1).(–1) + 3.(–1).1 + 8.0.(–1) + 6.1.1] = 1T2 24 n(T1) =

Thus Gred = E + T2 Similarly, we can obtain Gred for the point group Oh as follows:

and

Oh

E 6C4

G red

5

Gred = Eg + T2g

-1

3C2 = (C42 ) 8C3 1

-1

6C2 1

iE 6iC4 5

-1

3iC2

8iC3

6iC2

1

-1

1

Molecular Symmetry

4.35

Similarly, the expression for rotation operation on a p orbital [m = –1, 0, + 1] wave function can be obtained as follows:

È eif ˘ Èeia Í ˙ C (a ) Í e0 ˙ = ÍÍ 0 ÎÍe - if ˙˚ ÍÎ 0 with c(E) = 3 and c C(a ) =

0 e0 0

0 ˘ È eif ˘ Í ˙ 0 ˙˙ Í e0 ˙ Í - if ˙ e - ia ˙˚ Îe ˚

sin 3a 2 sin a 2

sin 3p 2 where, c (C2 ) = = - 1;

c (C3 ) = sin p = 0 and c (C4 ) = sin 3p 4 = 1 sin p 2 sin p 3 sin p 4 Thus, Gred can be written for the point group O as O E 6C4 3C2 (= C42 ) 8C3 6C2 G red

3

1

-1

0

1

3C2 (= C42 ) 8C3

6C2

and Gred = T1 Similarly, for the point group Oh,

Oh

E 6C4

G red

3

1

-1

0

iE 6iC4

-1

3

-1

3iC2

8iC3

6iC2

+1

0

+1

and Gred = T1u Using the same treatment to electrons in other orbitals, the results can be summarised as given in Table 4.11. Table 4.11 Splitting of orbitals in an Octahedral ligand environment Orbital s p d f g h i

l 0 1 2 3 4 5 6

c(E) 1 3 5 7 9 11 13

c(C2) 1 –1 1 –1 1 –1 1

c(C3) 1 0 –1 1 0 –1 1

Thus, the totally symmetric s-orbital transforms into A1g while the p-orbitals transform into T1u representation for the character table of Oh point group. On the other hand, the orbitals with higher l values, split into two or more sets. It should be noted that the states of free ion are represented with capital letters while that of free atoms are represented with small letters as shown in Table 4.12. The use of subscript u and g is done as per the following rules:

c(C4) 1 1 –1 –1 1 1 –1

Irreducible representation/splitting A1g A1u Eg + T2g A2u + T1u + T2u A1g + Eg + T1g + T2g Eu + 2T1u + T2u A1g + A2g + Eg + T1g + 2T2g

Table 4.12 Orbital

Splittings of orbitals of free atom in various symmetries Oh

Td

s p d f g h

a1g t1g eg + t2g a2u + t1u + t2u a1g + eg + t1g + t2g eu + 2t1u + t2u

a1 t2 e + t2 a2 + t1 + t2 a1 + e + t1 + t2 e + t1 + 2t2

i

a1g + a2g + eg + t1g + 2t2g

a1 + a2 + e + t1 + 2t2

4.36

Inorganic Chemistry

Table 4.13 Splitting of spectroscopic terms under 1. For centrosymmetric environment such as Oh: ligand field environment (a) The atomic orbitals with even value of l (s, d, g) Free ion Oh Td are symmetric inversion and are of g character. term (b) The atomic orbitals with odd value of l (p, f, 1 1 S 1A1g A1 h) are antisymmetric to inversion and are of u 3 3 3 character. P T1g T1 1 1 1 2. For non-centrosymmetric environment such as Td, D Eg, 1T2g E, 1T2 3 3 3 no g or u subscripts are used. F A2g, 3T1g, 3T2g A2, 3T1, 3T2 1 Now, we will discuss the splitting of spectroscopic G 1A1g, 1T2g, 1Eg, 1T1g 1A1, 1T2, 1E, 1T1 terms under the effect of weak ligand field environment using the same concept as for a single d electron. This is because of the reason that, the only variable part of the wave function for a D term is also eimf due to its completely analogous five fold degeneracy and five values of Ml (+2, +1, 0, –1, –2) as for eimf (+2, +1, 0, –1, –2) of the single electron d-orbital wave function. Similar is the analogy of s-orbital and S state, p orbital and P states, f orbitals and F state and so on, as given in the table 4.13 for d2 ion in ligand field environment. However, if the electrostatic octaqhedral / tetrahedral ligand field environment is strong and greater than the interelectronic repulsions between the d electrons, there is an extremely large splitting of the d orbitals. As a result, the possible strong field configurations for d1 metal ion are t12g or eg1 depending upon the electron occupancy in t2g or eg level and the corresponding terms are 2T2g and 2Eg. In case of d2 metal ion, the possible strong field configurations (in the order of increasing energy) are t22g, 1 1 t 2geg and eg2. We can discuss their origin as follows: (a) The configuration t22g corresponds to the 2-electron occupancy in two t2g levels i.e. t12gt12g with the direct product t2g × t2g in Oh point group. This direct product can be reduced into A1g + Eg + T1g + T2g using the character table for Oh point group. Considering the spin multiplicity with the total spin either 1 or 0 ( for two R R electrons in t2g levels), the possible terms are 3A1g, 3Eg, 3T1g, 3T2g, 1A1g, 1Eg, 1T1g and R S 1 T2g. However, only some of terms actually exist, with the sum of total degeneracies R S equal to the total number of electronic arrangements. The possible fifteen electronic S S arrangements for t22g configuration can be shown: R 2 S R R This means that the total degeneracies for the set of the terms for t 2g configuration is fifteen and must remain so, even if the field is changed. It means that t2g × t2g $ R a SR A1g + bEg + cT1g + dT2g, where a, b, c and d are the spin multiplicities (either 1 or 3) S Thus, 1.a + 2.b + 3.c + 3.d = 15 This equation can have only three solutions as given below: RS S

I

a b c d 3 3 1 1

R R

II III

1 1 3 1 1 1 1 3

S S

S

R SR S R S

Out of these, the correct assignment of multiplicity will be decided by using the correlation diagrams (discussed in Chapter 26) and using two principles: (a) One to one correspondence between the states at the two extreme sides of the abscissa (b) Non-crossing of the states of the same symmetry and spin degeneracy with the increasing ligand field strength. It is clear from the correlation diagram for d2 ion that there are only A1g states on the left and 3 A1g states are not present. This rules out the possibility of solution I. Before moving to the other possibilities, we will discuss the other two possible strong field configurations e2g and t12g e1g.

Molecular Symmetry

4.37

(b) The configuration eg2 corresponds to the direct product eg × eg and can be reduced to A1g + Eg + A2g using the character table of Oh point group. The possible 6 electronic arrangements can be shown.

R

R

R

S S

R

S

S

S

S

S S

Thus, the total degeneracy must be 6 with the following possible equations : eg × eg $ aA1g + bA2g + cEg and 1.a + 1.b + 2.c = 6 with the possible solutions

I II

a b c 3 1 1 1 3 1

Ê 1.3 + 1.1 + 2.1 = 6ˆ ÁË 1.1 + 1.3 + 2.1 = 6˜¯

Again moving to the correlation diagram, these is only 1A1g state on the left and 3A1g state is not present. This rules out the possibility of solution I. Thus the correct strong field terms for eg2 configurations are 1 A1g, 3A2g and 1Eg. (c) The configurations t12g e1g corresponds to the electron occupancy in t2g and eg level with direct product t2g × eg which can be reduced to T1g + T2g using the character table of Oh point group. The possible 24 electronic arrangements can be shown. We notice that, all the spin arrangements are either paired or unpaired, i.e. the two terms may be both triplet or singlet. This means that the correct possible terms for t12g e1g configuration are 1T1g, 3T1g, 1T2g and 3 T2g. Again moving to the correlation diagram, we observe that there are two 3T1g states on the left. The higher one must correlate to the t12g e1g configuration and the lower one to the t22g configuration. This means that the correct solution for t22g configuration is II and the correct terms are 1A1g, 1Eg, 3T1g and 1 T2g. It should be noticed that the other terms have been correlated using the non-crossing rule and indicated by the solid lines in the correlation diagram.

Bethe’s Method of Descending Symmetry Instead of this somewhat oblique procedure of assigning the correct states, Bethe used the method of descending symmetry as described below: This method makes use of the correlation table which represents the decomposition of the representations of Oh point group on lowering of symmetry (Table 4.14). We have discussed that eg2 configuration can go over into 1A1g + 3A2g + 1Eg or 3A1g + 1A2g + 1Eg states. According to this method, the Oh symmetry of the complex is lowered to D4h (by moving the trans pair of ligands to a greater distance than the remaining four ligands). As a result, the degenerate eg orbitals of the Oh symmetry now gives two non-degenerate levels of symmetries i.e. a1g and b1g of D4h symmetry. Similarly, the states of eg2 configuration of Oh symmetry go over to the states of D4h symmetry as shown in the correlation table, i.e. Oh $ D4h A1g $ A1g A2g $ B1g Eg $ A1g B1g

a1g eg (Oh)

b1g (D4h)

4.38

Inorganic Chemistry

Table 4.14 Oh

O

Td

D4h

Correlation table

D2d

C4n

C2n

D3d

D3

C2h

A1g

A1

A1

A1g

A1

A1

A1

A1g

A1

Ag

A2g

A2

A2

B1g

B1

B1

A2

A2g

A2

Bg

Eg

E

E

A1g + B1g

A1 + B1

A1 + B1

A1 + A2

Eg

E

Ag + Bg

T1g

T1

T1

A2g + Eg

A2 + E

A2 + E

A2 + B1 + B2

A2g + Eg

A2 + E

Ag + 2Bg

T2g

T2

T2

B2g + Eg

B2 + E

B2 + E

A1 + B1 + B2

A1g + Eg

A1 + E

2Ag + Bg

A1u

A1

A1

A1u

B1

A2

A2

A1u

A1

Au

A2u

A2

A2

B1u

A1

B2

A1

A2u

A2

Bu

Eg

E

E

A1u + B1u

A1 + B1

A2 + B2

A1 + A2

Eu

E

Au + Bu

T1u

T1

T1

A2u + Eu

B2 + E

A1 + E

A1 + B1 + B2

A2u + Eu

A2 + E

Au + 2Bu

T2u

T2

T2

B2u + Eu

A2 + E

B1 + E

A2 + B1 + B2

A1u + Eu

A1 + E

2Au + Bu

Now, the electron occupancy can occur in the following possible ways:

Direct product

Spin multiplicity

a12g

a11g b11g

b12g

A1g ¥ A1g

A1g ¥ B1g

B1g ¥ B1g

= A1g

= B1g

= A1g

1

1

1

A1g

B1g , 3 B1g

A1g

The A1g state is singlet due to different spins of the two electrons in the same a1g level (according to the exclusion principle). On the other hand, B1g state can be singlet or triplet as the two electrons occupy the different levels, a1g and b1g . Accordingly, the spin multiplicity of the states in Oh and the corresponding states of D4h symmetry must remain the same. It means that A1g state of D4h symmetry corresponds to A1g state of Oh symmetry and so on. Thus the A1g state of Oh must be singlet. Similarly, we can obtain the following result using the correlation table: Oh $ D4h;

1

A1g $ 1A1g;

3

A2g $ 3B1g and

1

Eg $ 1A1g + 1B1g

Hence, the eg2 strong field configuration gives 1A1g, 3A2g and 1Eg states. Now, we can apply this method to determine the correct states arising from the strong field t22g configuration out of three possible sets (1A1g + 1Eg + 1T1g + 1T2g), (3A1g + 3Eg + 1T1g + 1T2g) and (1A1g + 1Eg + 1T1g + 3T2g). We require that subgroup of Oh in which, these possible state split into a different state or sum of different states. Thus, we can choose C2h symmetry in which the correlation is as shown below: Oh $ C2h A1g $ Ag Eg $ Ag + Bg T1g $ Ag + Bg + Bg T2g $ Ag + Ag + Bg Here, the t2g of Oh symmetry gives ag + ag + bg of C2h symmetry. The electron occupancy can occur in the following ways:

4.39

Molecular Symmetry

a2g

a12g

a 1a1g

a 1gb1g

a 1gb1g

b2g

Direct product

Ag × Ag = Ag

Ag × Ag = Ag

Ag × Ag = Ag

Ag × Bg = Bg

Ag × Bg = Bg

Bg × Bg = Ag

Spin multiplicity

1

1

1

1

1

1

Ag

Ag, 3A1g

Ag

Bg, 3B1g

Bg, 3B1g

Ag

For the sake of simplicity, the two different ag orbitals have been represented by ag and a g. The spin multiplicity of the states have been decided using the exclusion principle i.e. the electrons occupying the same orbitals give a singlet state and the electrons occupying the different orbitals may give a singlet or a triplet state. Thus we obtain in total the following states of C2h symmetry with the sum of degeneracy equal to 15: 41Ag + 3Ag + 21Bg + 23Bg [4(1 × 1) + 1(3 × 1) + 2(1 × 1) + 2 (3 × 1) = 15] From the correlation table, it is clear that the triplet 3Ag and 3Bg states of C2h symmetry must correlate with the 3T1g state of Oh symmetry. Similarly, the order states can be correlated as given below:

Oh $ C2h 1A $ 1A g 1g 1E $ 1 A g g 1B g 3T 1g

$ 3Ag 3B g 3B g 3T $ 1A 2g g 1A g 1B g

The same procedure can be applied for tetrahedral ligand field environment. The results can be summarised as follows: Electronic Configuration States e2

A1 + A2 + E

et2

T1 + T 2

t22

A1 + E + T1 + T2

8. Application to Predict IR Active Vibrations Group theory helps to predict the IR active vibrations out of the possible vibrations for simple molecules (NH3, CO2, H2O, etc.). For example, we have discussed that the reducible representation of C2n point group can be reduced to obtain the following result:

E C2 G red

9

-1

s xz

s yz

1

3

and

Gred = 3A1 + A2 + 2B1 + 3B2

For H2O molecule, these are total 9 modes of different symmetries, out of which 3 modes are translations, 3 rotations and 3 are vibrations. From the character table, the translational modes with x, y, z vectors have symmetry B1, B2 and A1 respectively. Similarly, the rotational modes with Rx, Ry and Rz vectors have symmetry A2, B1 and B2 respectively. Thus, we have the sum of translational and rotational modes as A1 + A2

4.40

Inorganic Chemistry

+ 2B1 + 2B2. We can obtain the vibrational modes by subtracting this sum from the total modes as 9 modes – 6 (translational + rotational modes) = 3 vibrational modes or (3A1 + A2 + 2B1 + 3B2) – (A1 + A2 + 2B1 + 2B2) = 2A1 + B2 It means that we have possible vibrations as

H

H

H

n1

H n2

Symmetric stretch

Asymmetric stretch

H

H n3

Bending

We can obtain the symmetries of these vibrations as follows:

E C2

s xz

s yz

n1 n2

1 1

-1 1

1 1

-1 1

B2 A1

n1

1

1

1

1

A1

A vibration is IR active if the excited normal mode belongs to the same representation as any one of the Cartesian coordinates. From the character table of C2n point group, it is clear that z coordinate and the normal modes (n2 and n3 ) belong to A1 while the y coordinate and the normal mode n1 belong to B2. Thus all these modes will be IR active.

4.8

STRUCTURE OF SOLIDS

The physical state of a substance depends upon the balance between two opposing effects; ordering effect of the cohesive forces of attraction and disordering effect of thermal agitation. The solid state is characterised by definite shape, definite size, rigidity, incompressibility, mechanical strength and negligible diffusibility. In solids, the constituent particles, i.e. atoms or ions, or molecules are held together by strong cohesive forces in rigid structural array. Solids are classified into two broad categories: crystalline solids and amorphous solids. A solid substance in which the constituent atoms, ions or molecules are packed in a definite geometrical configuration is called a crystalline solid. This regular arrangement in the three-dimensional network of crystals extends over a large distance. Thus, the crystalline solids have long-range order. Freshly cut crystalline solids (cut with a sharp edged tool) give two pieces with plane B surfaces. A crystalline solid has a sharp melting point, i.e. it undergoes a definite and abrupt change into liquid state at a fixed temperature. The most D important characteristic of a crystalline solid is anisotropy (Fig. 4.32), i.e. their physical properties such as thermal conductivity, electrical conductivity, refractive index and mechanical strength are different in different directions. Consider a simple two-dimensional arrangement of a crystal constituting only two different kinds of atoms. The arrangement of atoms along the vertical line AB is different from that along the slanting line CD. As a result, C A the properties measured along the directions of these lines will differ from Fig. 4.32 Anisotropy in a each other. crystalline solid

Molecular Symmetry

4.41

A solid substance in which the constituent atoms, ions or molecules are not packed in a definite geometrical configuration is called an amorphous solid. Here, the bonding arrangements determine short-range order. The distinction can be observed with the irregular surface of fractured glass. These forms are metastable and hence do not have a sharp melting point. For instance, when glass is heated, it does not change abruptly into the liquid state but softens and starts to flow. Due to random arrangement of constituent particles, in an amorphous solid, their physical properties are the same in all directions. This is known as isotropy. Nevertheless, only the crystalline solids can be considered as true solids. In this chapter, we will discuss only about crystalline solids. Most pure compounds can be obtained as crystalline solids. Every crystal has a characteristic property of taking the shape of a polyhedron with planar faces, sharp vertices and linear edges and is defined by the regularity of arrangement of its constituent particles. The size and shape of a crystal also depends upon the rate of crystallisation. Several naturally occurring solids can be well recognised due to their definite crystalline shapes. However, there are some solids with very small crystal sizes, so that they can be recognised only under a powerful microscope. Such solids are said to be microcrystalline solids. The branch of science which deals with the structure, geometry and properties of crystals and crystalline solids is known as crystallography.

4.8.1 Laws of Crystallography Three fundamental laws of Crystallography are:

1. Law of Constancy of Interfacial Angles (Steno, 1669) This law states that for a given crystalline solid, the angles between the corresponding faces or planes are equal irrespective of its size and shape. Thus, the size and shape of a particular crystal may vary according to the method of crystallisation, but the angle of intersection of any two corresponding faces, i.e. 120° 120° 120° the interfacial angles remains invariably the same. For example, the crystals seem very different in Fig. 4.33 but possess the same interfacial angles. Fig. 4.33 Section of an ideal crystal

2. Law of Constancy of Symmetry

and the possible shapes

This law states that all crystals of one and the same substance have the same symmetry. The elements of symmetry of the crystal include the total number of planes, lines and centres of the symmetries possessed by a crystal. These elements of symmetry have been described in the next article.

3. Law of Rational Indices This law states that the intercepts on a face of a crystal along the crystallographic axes are given in the ratio ma: mb: pc; where m, n and p are integral whole numbers. The crystallographic axes are the three noncoplanar coordinate axes selected arbitrarily and may coincide or be parallel to the edges. The intercepts are the distances of the points at which the standard phase cuts these axes from the origin and are denoted by a, b and c. Thus, if any other phase cuts these axes at intercepts x, y and z, then x: y: z = ma: mb: pc.

4.8.2 Elements of Crystal Symmetry The main elements of crystal symmetry are plane of symmetry, centre of symmetry and axes of symmetry.

1. Plane of Symmetry A crystal is said to have a plane of symmetry when an imaginary plane divides the crystal into two parts which are exact mirror images of each other.

4.42

Inorganic Chemistry

For a cubic crystal like NaCl, there are nine such planes of symmetry including three rectangular planes of symmetry at right angles to the planes and six diagonal plane of symmetry passing diagonally through the cube as shown in Fig. 4.34 and 4.35.

Fig. 4.35 Diagonal planes of symmetry

2. Centre of Symmetry

Fig. 4.34 Rectangular planes of symmetry

Fig. 4.36

Centre of symmetry in a crystal

It is an imaginary point within a crystal, through which if any straight line is drawn, intersects the crystal surface at equal distances from each side (Fig. 4.36). Every crystal has only one centre of symmetry.

3. Axes of Symmetry or Rotation Axis A crystal is said to possess an axis of symmetry if complete rotation about an imaginary line within the crystal results in the more than once same appearance. In general, a crysal asscumes the same appearance for every rotation by an angle of 360°/n, where n is the fold of the axis. This means that a crystal can have only 1-fold, 2-fold, 3-fold, 4-fold and 6-fold rotation axes corresponding to the angles of rotation as 360°, 180°, 120°, 90° and 60°. (a) 1-fold Rotation Axis A crystal is said to possess a 1-fold rotation axis if it requires a rotation of complete 360° to have the same appearance. (b) 2-fold Rotation Axis or Diad Axis A crystal is said to possess a 2-fold rotation axis if it requires a rotation of complete 180° to have the same appearance. For example, there can be six 2-fold rotation axies in a cube as shown in (Fig. 4.37). (c) 4-fold or Tetrad Rotation Axis In this case, the same appearance comes after the fourth rotation with every rotation of 90°. A cube can have three 4-fold rotation axis (Fig. 4.37).

Fig. 4.37 2-fold rotation axis in a cube

(d) 3-fold or Triad Rotation Axis Here, the same appearance comes after the rotation by an angle of 120°. A cube can have four triad rotation axis (Fig. 4.37). (e) 6-fold or Hexad Rotation Axis Here, the same appearance comes after the rotation by an angle of 60°. A hexagonal crystal can have one 6-fold rotation axis (Fig. 4.37).

4.8.3 Space Lattice and Lattice Points Crystals have their constituent particles arranged in a definite regular order. These positions of the constituent particles in the crystal are usually represented by points known as lattice points or lattice sites. The regular arrangement of the infinite set of these lattice points result in the formation of a space lattice or crystal lattice.

Molecular Symmetry

It gives an idea of the arrangement of the constituent particles in threedimensional space as shown in Fig. 4.38. It is clear from the figure that the crystal lattice is formed by repetition of an infinite number of adjacently placed small units known as the unit cell. Thus, a unit cell is defined as the smallest unit of the crystal lattice which on repeating again and again, results in the formation of entire crystal of the given substance. A unit cell is described in terms of the distances a, b and c equal to the length of the edges of the unit cell and the angle a, b and g equal to the angles between the three imaginary axes OX, OY and OZ as shown in Fig. 4.38.

Z

Unit cell

c b a O g a

4.43

Lattice point

b

X

Y

Fig. 4.38

Representation of space lattice

4.8.4 Bravais Lattice A Bravis deviced from the geometrical considerations that all the possible three-dimensional space lattices for the crystalline solid can be represented by means of 14 distinct types of lattices known as Bravais lattice. These are named as (i) simple cubic, (ii) body-centred cubic, (iii) face-centred cubic, (iv) hexagonal, (v) rhombohedral, (vi) simple tetragonal, (vii) body centred tetragonal, (viii) simple orthorhombic, (ix) end centred orthorhombic, (x) face-centred orthorhombic, (xi) body-centred orthorhombic, (xii) simple triclinic, (xiii) simple monoclinic, and side-centred monoclinic. These have been represented in Fig. 4.39. It can be visualised that the crystals have four types of arrangements in the different shapes as given below:

(a) Simple or Primitive Lattice (P) In this lattice, the points are present at all the eight corners of the unit cell as in the primitive cubic unit. This means that there is only one-eighth of the atom in one unit cell. 1 Hence the number of atoms per unit cell in a simple lattice is equal to 8 × =1. 8 (b) Body-Centred Lattice (I) In this lattice, the points are present at all the corners as well as at the centre of the unit cell as in the body-centered cubic unit cell. The atom present at the centre belongs to only one unit cell.

Simple cubic

Body-centred cubic

Simple tetragonal

Body-centred orthorhombic

Body-centred tetragonal

Simple triclinic

Face-centred cubic

Simple orthorhombic

Simple monoclinic

Fig. 4.39 Bravais lattices

Hexagonal

Base-centred orthorhombic

Rhombohedral

Face-centred orthorhombic

Base-centred monoclinic

4.44

Inorganic Chemistry

1 + 1 = 2. 8 3. Face-Centred Lattice (F) In this lattice, the points are present at the corners as well as the centre of each face as in a face centred cubic unit cell. The atom present at the centre of the face is shared by the two unit cells. Thus, the number of atoms 1 1 = 4. per unit cell is equal to 8 × + 6 × 8 2 4. End-Centred Lattice (C) In this lattice, the points are present at the corner and at the centres of the two end faces, as in an end centred orthorhombic unit cell. 1 1 +2× = 2. Thus, the number of atoms per unit cell is equal to 8 × 8 2 Thus, the number of atoms per unit cell in the body-centered lattice is equal to 8 ×

4.8.5 Crystal Systems The geometry and internal structure of a crystal is determined by 32 different combinations of elements of symmetry. These are known as 32 point groups or systems. However, some of these can be grouped together to form an overall seven different categories and are known as the seven basic crystal systems, namely cubic, orthorhombic, tetragonal, rhombohedral, hexagonal and monoclinic. These systems have been summarised along with their types, properties, parameters and number of space lattices in Table 4.15. The first column of the table shows the names of the seven crystal systems, the second column gives the axial distances, the third column gives the axial angles, the fourth columns indicates the type of the space lattice, the fifth column indicates the symmetry elements and the last column gives the examples. y

4.8.6 Methods of Designating Planes of Crystal

A

There are many methods of designating the various planes of crystals. These methods are based on the law of rationality of indices. The two important methods are Weiss indices and Miller indices as described here.

D

C E

1. Weiss Indices

B

F

x

H

G Weiss used integers to index the planes on the bases of the intercepts made z by the plane on the axes. Consider the unit cell ABCDEFGH with intercepts Fig. 4.40 Unit cell ABCDEFGH a, b and c along the x, y and z axis respectively (Fig. 4.10). Consider the plane AFH with intercepts on all the three axis. If the plane intercepts the three axis at unit distances, then the rational indices are given as 1a: 1b: 1c, where Weiss indices are 1, 1, 1. Similarly, for all the other crystal faces, Weiss indices can be given as

Plane ABCD Plane ABEF Plane ADHE

, 1, 1, 1, , 1, 1

Plane BCGF Plane DCHG Plane EFGH

1,

, ,

1,

,1 ,

However, the Weiss indices are generally in terms of infinity and fractions and hence are difficult to designate.

2. Miller Indices Miller indices have been given by W H Miller and are the most common method used for the designation of the planes in terms of the set of integers h, k and l. Miller indices are taken as the reciprocals of the intercepts made by the plane on the various axes, i.e. these can be obtained from Weiss indices. The steps for designation are as follows:

a=b

a=b

a

a

a

Tetragonal

Hexagonal

Orthorhombic

Monoclinic

Triclinic

b

b

b

a=b=c

Rhombohedral or trigonal

c

c

c

c

c

a=b=c

Axial distances

Cubic

Name of the crystal system

b

g

90°

Simple triclinic ]1

Simple monoclinic 2 Base-centered monoclinic

a = g = 90° b 90° a

Simple orthorhombic, Body centered orthorhombic end-centered orthorhombic Face centered orthorhombic

Simple hexagonal ]1

Simple tetragonal 2 Body centered tetragonal

Simple hexagonal ] 1

Simple cubic Body centered cubic 3 Face centered cubic

Types of space lattices = 14

a = b = g = 90°

a = b = 90° g = 120°

a = b = g = 90°

a = g = 90° = 90°

a = b = g = 90°

Axial angles

4

Table 4.15 Seven crystal systems

One 1-fold axis

One 2-fold axis

Three naturally perpendicular 2-fold axis

One 6-fold axis

One 4-fold axis

One 3-fold axis

Four 3-fold axes

Minimum symmetry elements

CuSO4 . 5H2O, H3BO3, K2Cr2O7

Na2SO4 . 10H2O, FeSO4, CaSO4 . 2H2O, Na2B4O7.10H2O

KNO3, K2SO4, BaSO4, MgSO4, PbCO3, Mg2SiO4, a-sulphur.

AgI, ZnO, CdS, HgS, PbI2, graphite, quartz, ice, beryl, Zn, Mg, Cd.

SnO2, TiO2, NiSO4, ZrSiO4, KH2PO4, PbWO4, Sn

NaNO3, ICl, Sb, Bi, magnetic calcite.

NaCl, KCl, ZnS, Cu, Pb, Ag, Au, Hg, diamond.

Examples

Molecular Symmetry

4.45

4.46

Inorganic Chemistry

Step 1. Designation of Weiss indices, i.e. the intercepts of the plane with axes. Step 2. Inversion to take reciprocals of the intercepts. Step 3. Clearing of fraction by multiplication with the LCM of the integers. These steps can be illustrated with the help of the plane XHE and plane CBFG in Fig. 4.41. 1

1

½

Step 2. Reciprocals

1

1

2

Step 3. Clear fractions (Miller indices)

1

1

2

Step 1.

Fractional intercepts

1 0

1 1 1

1

A x

D

B C

E

z

F

y

G

H

Fig. 4.41 Plane XHE and CBFG

0

4.8.7 Close Packing of Constituent Particles in Crystals The packing of constituent particles (atoms, ions or molecules) in a crystal can be illustrated by consideration of packing of hard spheres of identical sizes in a layer (two dimensions) and extending the arrangement by putting these layers one above the other (three dimensions). Two common ways of packing have been shown in Fig. 4.42. It is quite evident that the packing utilising maximum available space will be more economical and is termed close packing. In Fig. 4.42 around 60.4% of the available space is occupied by the spheres in hexagonal I II array (I) in comparison to the occupation of 52.4% of Fig. 4.42 Close packing of spheres in hexagonal the space in square array (II). Hence, the arrangement I and square array represents close packing. A second layer of spheres can be arranged on the top of the first layer so as to fill the empty space (holes or voids) of the first layer. As a result each sphere in the first layer is in contact with the three spheres of the second layer. Now if we try to add a third layer of spheres, two alternative arrangements are possible as shown in Fig. 4.43 and 4.44. If the first layer is labelled ‘A’ and the second layer is labelled ‘B’, one way is to put each spheres of the third layer on the voids of the second layer so that these sphere lie directly above the spheres present in the first layer. The continuation of this arrangement in the same sequence is represented as ABABABA… arrangement with hexagonal symmetry and hence is known as hexagonal close packed arrangement or hcp. (Fig. 4.43). Alternatively, the spheres in the third layer can be placed over the unoccupied voids of the first layer (marked by c), as a result the spheres in the third layer will form a new layer labelled C. This pattern can Tetrahedral void

Octahedral void A B

C

Fig. 4.43

Hexagonal close packing of spheres (AB ABABA)

Fig. 4.44

First layer Second layer Third layer

Cubic close packing of spheres (ABC ABCA..)

Molecular Symmetry

4.47

also be continued to form ABCABCABCA…arrangement with cubic symmetry and hence is termed cubic closed packed arrangement or face centered cubic arrangement or ccp and fcp respectively (Fig. 4.44). It can be seen that in hcp and ccp arrangement, each sphere is in contact with twelve other spheres in the above layer and three spheres in the below layer. The number of nearest neighbours in contact with the given sphere is known as its coordination number. Thus, the coordination number of each sphere in hcp and ccp (fcc) structure is 12 (Fig. 4.45). The third common close packed arrangement is known as body-centered cubic arrangement. It is formed by using arrangement II as shown in Fig. (4.46). The second layer of spheres is placed above the holes of the first layer while the third layer is placed over the holes of the second layer, i.e. immediately above the first layer. However, this arrangement is less efficient (coordination number 8) and only 68% of the total available space is occupied in comparison to the 74% of space occupation in hcp and ccp arrangement as calculated in the next article.

A A

B

B

A

C

Fig. 4.45 Coordination number in hcp and ccp arrangements

Fig. 4.46 Body centred cubic close packing

4.8.8 Calculation of Percentage Occupied Space 1. Calculation of Percentage Occupied Space in Simple or Primitive Arrangement The simple or primitive cubic arrangement can be represented as shown in Fig 4.47, i.e. 8 spheres at the corner of the cube. The number of spheres per unit cell = 8 ×

1 =1 8

From Fig. 4.47, AD = 2r = a or r = a/2 3

3 Total volume occupied by spheres = 4 p ÊÁ a ˆ˜ = p a

3 Ë 2¯

6

Percentage of the volume occupied by the spheres Fig. 4.47

=

p a 3 /6 a3

× 100 = 52%

Representation of simple arrangement

4.48

Inorganic Chemistry

2. Calculation of Percentage of the Total Volume Occupied in fcp Arrangement The cubic close arrangement can be represented as shown in Fig. 4.48, i.e. 8 spheres at the corners of the cube and six spheres at the face centres. Let the length of each side of the cube = a Volume of the cube = a3 In ABC, AC2 = AB2 + BC2 AC2 = a2 + a2 = 2a2

or

AC = 2a (face diagonal)

&

If ‘r’ is the radius of the sphere, AC = 4r ‘;’ 2a = 4r or

r=

2a 4

Fig. 4.48

a 2 2

Number of spheres per unit cell = 8 × Volume of the sphere =

Representation of ccp arrangement and one face

1 1 +6× =4 8 2

4 3 r 3

Volume occupied by four spheres = 4 ×

3

4 3 16 Ê a ˆ p a3 pr = p Á = ˜ 3 3 Ë2 2¯ 3 2

Percentage of the volume occupied by the spheres =

p a3 3 2 a3

¥ 100 = 74.06%

3. Calculation of Percentage of the Total Volume Occupied in bcc Arrangement The body-centred cubic arrangement can be represented as shown in Fig. 4.49, i.e. 8 spheres at the corners of the cube and one sphere at the body centre. In

ABC, AC = 2a as calculated earlier

In

ACD,

AD2 = AC2 + CD2 = ( 2a)2 + a2 = 3a2

or

AD = 3a

AD is the body diagonal of the cube and is equal to 4r. Thus, or

Fig. 4.49

AD = 4r = 3a r = 3 a/4

The number of spheres per unit cell = 8 ×

Representation of bcc arrangement and body diagonal

1 +1×1=2 8 3

4 4 Ê 3a ˆ 3 Volume occupied by two spheres = 2 × p r 2 = 2 ¥ p Á ˜¯ = 3 8 p a Ë 3 3 4 Percentage of the volume occupied by the spheres =

3 8 p a3 a3

¥ 100 = 68.05%

Molecular Symmetry

4.49

4. Calculation of Percentage of Total Volume Occupied in hcp Arrangement Hexagonal close packing can be represented as shown in Fig. 4.50, i.e. 12 spheres are present one at each corners of the top and bottom face, 2 spheres are present, one at each centre of the top and bottom face and three spheres are present inside the unit cell forming an equilateral triangle. As the atoms touch each other along the edges of the hexagon, so a = 2r. The number of atoms in the top layer = 7. The corner atom is shared by 6 surrounding hexagonal cells. The atoms at the centre are shared by the two surrounding hexagonal cells. This means that three atoms are contributing fully to the hexagonal cells. 3 3 Thus, the total number of atoms in a hexagonal unit cell = + + 3 = 6 2 2 Let the edge of the unit cell be ‘a’ and its height be ‘c’. The three atoms present in the body are lying in a horizontal plane at a distance of c/2 and at the corners of the triangle.

Fig. 4.50

Representation of hcp arrangement

Consider the bottom layer of the hexagonal unit cell. In ABY,

a 3 2 2 2 a 3 a AX = AY = ¥ = 3 3 2 3 AY = AB cos 30° =

And In

AZX,

(4.1) (4.2)

AZ2 = AX2 + ZX2 From Eqs (4.2) and (4.3), 2

Ê a ˆ Ê cˆ ˜ +Á ˜ 3 ¯ Ë 2¯

(4.3) 2

a2 = Á Ë

1/ 2

a2 c2 c2 2a2 c  3 + or = or =  3 4 4 3 a  8 Area of the base = 6 × Area of AOB 1 = 6 × (BO) × (AY) 2 1 a 3 = 6× ×a× 2 2 3 2 = 3a (4.5) 2 3 Volume of the unit cell = 3a2c (4.6) 2 4 3 Total volume of all the atoms in the unit cell = 6 × pr 3 3 = 6 × 4 a ÊÁ a ˆ˜ = p a 3 3 Ë 2¯ p a3 Percentage occupied space = ¥ 100 3 3a 2 c 2 or

a2 =

(4.4)

Fig. 4.51 Bottom layer of hexagonal unit cell

(4.7)

4.50

Inorganic Chemistry

=

2p a ¥ 100 3 3c

= 2p ÊÁ 3 ˆ˜ Ë ¯

1/ 2

3 3 8

¥ 100 =

p 3 2

¥ 100 = 74%

4.8.9 Interstitial Voids in Close Packing of Spheres We have seen in the close packing of the constituent particles in the crystal that some vacant space is left in between the spheres. This vacant space is known as the interstitial voids or interstitial site or interstitial hole. Consider the first layer of the spheres closely packed in a single planes with the centres of three adjacent spheres lying at the vertices of an equilateral triangle. There is some vacant space between these three spheres, known Fig. 4.52 Triangular site as triangular or trigonal site (Fig 4.52). Consider again the hexagonal close packing of spheres. These are marked as ‘a’, ‘b’ and ‘c’ as shown in Fig. 4.54. The holes ‘a’ and ‘b’ are the vacant spaces between the centres of the group of three spheres which are in contact with one sphere at their bottom or top. The only difference is that the hole ‘b’ is hole ‘a’ in its inverted position. In this arrangement, these four spheres are lying at the vertices of a regular tetrahedron (Fig. 4.53) and hence, the holes a and b are known as tetrahedral holes. c The hole ‘c’ is in contact with three spheres in the bottom layer and three spheres in the top layer. These six spheres are lying at the vertices of a regular octahedron (Fig. 4.54) and hence the hole ‘c’ is known as the octahedral hole. Fig. 4.53 Tetrahedral holes Fig. 4.54 Octahedral hole

4.8.10 Classification of crystals Crystals can be divided into four types depending upon the type of bonding and the nature of the constituents particles occupying the lattice points, i.e. ionic crystals, covalent crystals, metallic crystals and molecular crystals. These are described briefly here as follows:

1. Ionic Crystals The constituent particles occupying the lattice points of the ionic crystals are positive and negative ions (cations and anions). These ions are held together by strong electrostatic forces of attraction (ionic bonds), and it requires a large amount of energy to separate these ions from one another. As a result, ionic solids have very high melting and boiling points and very low vapour pressures at ordinary temperatures due to very high heat of vaporisation. Ionic solids are hard, brittle and insulating in the solid state but in molten state or aqueous solutions become good conductors of electricity. This is due to the reason that in the solid state, the ions are arranged in a definite order in the crystal lattice and not free to move; whereas in the molten state or in the aqueous solutions, the well-ordered arrangements of the crystal are disturbed and free to move. These solids are soluble in aqueous and polar solvents. The ionic bonds are nondirectional and extend equally in all directions. As a result ionic, solids donot have any geometry or exhibit space isomerism. On the other hand, the anions are arranged in either cubicclose packed or hexagonal close-packed arrangement with some interstitial sites left behind. These sites can

Molecular Symmetry

4.51

be trigonal, tetrahedral, octahedral or cubic and some or all of these are occupied by the cations. For any close-packed arrangement, the number of octahedral sites is equal to the number of anions and the number of tetrahedral sites is equal to double the number of anions. The coordination number of a cation and the geometric arrangement of anions is determined according to the limiting radius ratio (to be discussed in the next article).

Effect of Pressure and Temperature on the Structure of Crystals In general, the increase of pressure results in increase of coordination number of the cation. For example, the chlorides, bromides and iodides of Li, Na, K, Rb and Ag have NaCl type structure with coordination number of cations and anions equal to 6 (6:6 ionic crystal) at ordinary temperatures and pressures. However, when pressure is applied on these crystals, their coordination increases up to 8 (8:8 ionic crystals) and the crystal structure changes to CsCl type. On the other hand, increase in temperature results in the decrease of coordination number. For example, ammonium chloride, ammonium bromide, ammonium iodide have cesium chloride type of crystal structure at ordinary temperature and pressure with coordination number of cations and anions equal to 8 (8:8 ionic crystal). However, when these crystals are heated to about 760 K, the CsCl structure converts to NaCl structure. This can be summarised as follows: Pressure NaCl CsCl (6:6) 760 K (8:8)

2. Covalent Crystals The constituent particles occupying the lattice points of covalent crystals are the neutral atoms of either same or different kind. These atoms are held together by covalent bonds (electron pairs). A covalent crystal is also known as a covalent network crystal due to presence of a large network of covalently linked atoms. Some particular examples are boron nitride, diamond, graphite, silicon carbide, silicon dioxide, quartz and rhombic sulphur. Covalent bond is quite strong so that the covalent solids do not evaporate or melt easily. In general, these solids are bad conductors of electricity as all the electrons are engaged in bonding and hence are not available for conduction. These compounds are insoluble in water and dissolve in nonpolar solvents. These solids have generally low coefficients of expansions and high heats of fusion.

3. Metallic Crystals The constituent particles occupying the lattice points of the metallic crystals are the fixed metallic cations (kernels) surrounded by mobile electrons (free to move throughout the crystal). The mobile electrons are present in the interstices of the lattice constituted by the metallic cations. As a result, strong electrostatic forces of attraction known as the metallic bond holds the compact solid structure of metals. This accounts for the quite high melting and boiling points of metals (except alkali metals, Hg). Metals are good conductors of heat and electricity due to presence of mobile electrons which also imparts metallic lusture. Metals are malleable, ductile and quite hard and tough. Metals can resist stretching without breaking and have high tensile strength.

4. Molecular Crystals The constituent particles occupying the lattice points of a molecular crystal are the discrete or independent covalent molecules and atoms in case of noble gases. The molecules may be polar (held together by weak dipole-dipole forces) or nonpolar (held together by van der Waals forces). Some examples of polar molecules are ice, PCl3, etc., and some nonpolar molecules are CO2, sugar etc. Since dipole-dipole forces and van der Waal’s forces are much weaker than the electrostatic forces of attraction, hence these crystals have comparatively low heats of vaporisation and low melting and boiling

4.52

Inorganic Chemistry

points. Further, the forces in polar molecules are comparatively much stronger than that of nonpolar molecules; hence polar molecules have comparatively higher heats of vaporisation and high melting and boiling points than the nonpolar molecules. Since the molecules are neutral, these are bad conductors of electricity.

4.8.11 Limiting Radius Ratios for Ionic Crystals In case of an ionic crystal, the crystal lattice is generally constituted by anions, whereas the cations are present in interstitial sites. This means that size of a cation is a measure of the size of the interstitial site. The size of a cation or radius of a cation (r+) is determined in terms of the ratio of radius of the cation (r+) to that of anion (r_). This ratio (r+/ r_) is known as limiting radius ratio as it gives the minimum value of the (r+/ r_) for the cation which can occupy a given site.

1. Determination of the Limiting Radius Ratio During close packing of ions in an ionic crystal, a certain particular arrangement of cations and anions is preferred which leads to minimum energy and maximum stability. The energy of an ionic crystal, depends upon some factors, i.e. interionic repulsions and the attraction between the cations and the anions. The greater the coordination number of a cation (more number of anions surrounding the cation), greater is the attraction and hence lesser is the energy. On the other hand, interionic repulsions increase with increase in the number of anions of the same charge and size squeezed together. This means that for an arrangement of minimum energy and maximum stability, these two factors should be optimised. These factors are also affected by the internuclear distance (r) taken as the sum of radii of the cations and anions. All these points are taken in consideration in calculation of limiting ratio as discussed here.

(a) Limiting Ratio for Trigonal Site (Coordination Number of Cation = 3) As already discussed, trigonal site is formed by the vacant space left between the three adjacent spheres, i.e. anions lying at the vertices of an equilateral triangle and lying close packed in a plane. This site is occupied by a cation in an ionic crystal with triangular or trigonal structure. Let r+ and r– be the radii of the cation and anion respectively. Consider EBC in Fig. 4.55. In an equilateral triangle EBC, CE = BC = BE = 2r– Since A is the centre of the triangle as well of the cation, AB = r– + r+ and BD = r– In right-angled

ADB,

BD = cos 30˚ AB or

r3 = r- + r+ 2

or r+ = 0.155 r–

or

r+ 2 = - 1 = 0.155 r3 Fig. 4.55 Trigonal arrangement of anions touching the cation at the centre of the equilateral triangle

Thus, the limiting r+/r– ratio for a trigonal site is 0.155. This means that for a cation to occupy the trigonal site, the lowest value of the r+/r– ratio is 0.155. In other words, the lowest size for a cation to occupy the trigonal site is 0.155 times the radius of the anion.

4.53

Molecular Symmetry

(b) Limiting Radius Ratio for Tetrahedral Site (Coordination Number = 4) The empty space left by placing one sphere over three adjacent spheres lying at the vertices of a triangle in a plane is known as the tetrahedral site. Consider an ionic crystal such as ZnS, where the cations occupy the tetrahedral sites. This site is represented by placing the cation at the centre of the tetrahedron whose vertices are occupied by the anions (the alternate corners of the cube). Thus, all the four anions are in touch with each other as well as with the cation present at the centre as shown in Fig. 4.56 (for the sake of simplicity, the cations and anion have been shown by small separate spheres). If a is the length of each side of the cube then the face diagonal, AB = 2 a = 2r–

B

(4.8)

And the body diagonal, AD = 3 a = 2r– + 2r+

(4.9)

A

Dividing Eq. (4.9) by (4.8), E

2r- + 2r+ = 2r-

3

r− + r+ 3 = r− 2

or

2

or

r 3 1.732 r 3 or + = −1 = − 1 = 0.225 1+ + = 1.414 r− 2 r− 2

or

r+ = 0.225 r–

Thus, the limiting r+ /r– ratio for a tetrahedral site is 0.225.

+ D – C

–

Fig. 4.56 Representation of a tetrahedral site occupied by a cation which is touching all the four anions which in turn are touching each other

This means that for a cation to occupy the tetrahedral site, the lowest value of the r+ /r– ratio is 0.225. In other words, the lowest size for a cation to occupy the tetrahedral site is 0.225 times the radius of the anion.

(d) Limiting Radius Ratio for an Octahedral site (Coordination Number Cation = 6) As discussed earlier, an octahedral site is formed by three spheres lying at the vertices of a triangle in one layer superimposed by three spheres lying at the vertices of an inverted triangle in the other layer. This means that the six spheres are lying at the six vertices of a regular octahedron constituted by the two triangles with their vertices in the opposite directions (Fig. 4.57). Consider the unit cell of an ionic crystal with octahedral sites, say NaCl. Here, six anions are present in the centre of the six faces of the cube and are touching each other to form an octahedral site which is occupied by a cation. It is clear from Fig. 4.57 that the octahedral site is constituted at the centre of a square plane at the corners of A

A

B

– – y

– w – – –

+

D

+

B

– D C

x –

–

Fig. 4.57

Representation of an octahedral site in a unit cell

Fig. 4.58

z –

C

A cross section through the octahedral site

4.54

Inorganic Chemistry

which are present the four atoms A,B,C and D, as shown by cross sections through the site in Fig. 4.58. (4.10) Consider the square plane WXYZ, with sides XZ =YZ = a = 2r– In the right angled XYZ, XY = 2 a = 2r+ + 2r– (4.11) From Eq (4.10) and (4.11), 2 (2r–) = 2r+ + 2r– or or

2 r– = r+ + r– or

r+ = 0.414 r–

r+ = 2 – 1 = 1.414 – 1 = 0.414 r−

r+ ratio for an octahedral site is 0.414. This means that for a cation to occupy the r− r octahedral site, the lowest value of the + ratio is 0.414. In other words, the lowest size for a cation to r− occupy the octahedral site is 0.414 times the radius of the anion. Thus, the limiting

(d) Limiting Radius Ratio for Cubic Site (Coordination Number of Cation = 8) A cubic site is formed when eight spheres are placed at the corners of the cube in such a way that all the spheres are touching each other, as in case of an ionic crystal with CsCl type structure. The cubic site is occupied by the cation which is in touch with all the eight spheres (Fig. 4.59). Let a be the length of the each side of the cube. B Since the spheres touch each other, AD = a = 2r– (4.12) A D And the body diagonal, AC = 3 a = 2r+ + 2r–

(4.13)

Dividing Eq. (4.13) by (4.12),

C

r +r 3= + − r− r+ Fig. 4.59 Representation of a cube site or = 3 – 1 = 1.732 – 1 = 0.732 r− r Thus, the limiting + ratio for a cubic site is 0.732. This means that for a cation to occupy the r− r cubic site, the lowest value of the + ratio is 0.732. In other words, the lowest size for a cation to r− occupy the cubic site is 0.732 times the radius of the anion.

2. Applications of Limiting Radius Ratio The limiting radius ratio is useful in the following ways:

r+ ) or r− the relative size of the ions changes the coordination number and the structure of the crystals. Consider an ionic crystal with the cations occupying the octahedral sites each surrounded by six closely packed anions touching each other, as well as touching the cation. Now if the radius of the cation is decreased, i.e. r+/r– ratio is decreased, the anions still touch each other but are not able to touch the cation. This means the force of attraction between the cation and anion decrease leading to instability of the structure. Further, if the anions try to come closer to the cation these would feel extra-interionic repulsions. To overcome this repulsion and increase the force of attraction, one or more anions have to

(a) Structures of Ionic Crystals

It is clear from the above discussion that as the radius ratio (

Molecular Symmetry

4.55

be pushed away. This results in decrease of coordination number and a change in structure of the ionic crystal so that finally a coordination number of four is achieved with tetrahedral structure (Fig. 4.60). On the other hand, if size of the cation is increased, i.e. r+/r_ ratio is increased, the anions would be pushed further apart and close packing of anions would be disturbed. In order to restore the close packing, more anions have to be accommodated leading to an increase in coordination number. Thus, finally a coordination number of eight is achieved with cubic close packing (Fig. 4.60). –

– –

–

–

CN = 3 (Trigonal)

– +

–

Decrease of r+/r–

–

–

CN = 4 (Tetrahedral)

– +

Increase of r+/r–

–

– –

CN = 6 (Octahedral)

– +

–

CN = 8 (Cubic)

Fig. 4.60 Change in structure of ionic crystals with change in coordination number

Similarly, if the radius of the cation is decreased further, a coordination number of 3 is achieved with trigonal structure.

(b) Range of Radius Ratio Values for Coordination Number The limiting radius ratio for coordination number 6 (octahedral) is 0.414. As the radius ratio is increased, the coordination number 8 (cubic) is attained at the limiting radius ratio 0.732. This means that the range of radius ratio for coordination number 6 is 0.414–0.732, while above 0.732, a cubic arrangement with coordination number 8 is favoured. Table 4.16 lists the ranges of radius ratios for the coordination number of cations and the structure of ionic crystals with the examples. Table 4.16 Range of radius ratio for ionic crystals Range of limiting radius ratio (r+/r–)

Coordination number of cation

Structural arrangement

Examples

0.732–1.000

8

Cubic

CsCl, CsI, TlCl, TlBr

0.414–0.732

6

Octahedral

NaCl, AgCl, AgBr

0.225–0.414

4

Tetrahedral

ZnS, ZnO, CuCl, SiO44–

0.155–0.225

3

Trigonal

B2O3, BN, BO3–

(c) Determination of Geometry of Ionic Crystal From the radii of cations and anions, the radius ratio for a particular set of cations and anions can be determined and hence from table 4.16, the coordination number of a cation and its structural arrangement can be predicted. For example, the radius ratio for B3+ (r+ = 0.23 Å) and O2– (r– = 1.40 Å) is equal to 0.23/1.40 = 0.164. It lies in the range of 0.155–0.225. Thus, boron oxide would prefer a coordination number of 3 with trigonal arrangement. Some other examples are given in Table 4.17. Table 4.17 Examples of ionic crystals Cation 2+

Zn (0.74 Å) +

Na (0.95 Å) +

Cs (1.69 Å)

Anion

Radius ratio

S (1.84 Å)

0.74/1.84 = 0.40

4

Tetrahedral

0.95/1.81 = 0.524

6

Octahedral

1.69/1.81 = 0.93

8

Cubic

2–

_

Cl (1.81 Å) _

Cl (1.81 Å)

Coordination number

Geometry

4.56

Inorganic Chemistry

4.8.12 Structure of Ionic Crystals In general, the structure of an ionic crystal depends on the relative sizes of the cations and the anions. In most of the cases, the anions are larger form the close-packed arrangement while the cations occupy the interstitial sites. Depending upon the general formula, the structures of the ionic crystals can be described as follows:

1. Ionic Crystals of the Type AX In this type, we will discuss the structures of NaCl, ZnS and CsCl with three common structural arrangements.

(a) Rock Salt or Sodium Chloride Type Structure The radius ratio for NaCl is equal to (95 pm/ 181pm) = 0.524. This value corresponds to the range of octahedral arrangement (0.414–0.732) with coordination number equal to 6. Thus, in NaCl, Cl– ions are arranged in cubic close-packed arrangement in which Cl– ions are present at all the eight corners as well as the centre of the six faces of the cube. The Na+ ions are present in the octahedral voids, i.e. 12 Na+ ions are present at the centre of the twelve edges and one Na+ ion is present at the centre of the cube. 1 1 ×8+ ×6=4 Therefore, the total number of Cl– ions in one unit cell = 8 2 1 × 12 + 1 = 4 And the total number of Na+ ions in one unit cell = 4 It means that one unit cell of NaCl crystal is constituted by 4 Cl– ions and 4 Na+ ions. Since the number of octahedral voids in a unit cell is equal to the number of anions, in this case for 4 Cl– ions, there are 4 octahedral voids, each occupied by the Na+ ions. Thus, there are 4 Na+ Cl– ion ions for 4 Cl– ions resulting in the Cl– ion is Na+ ion formula of the crystal as Na4Cl4 or octahedrally + Na ion is surrounded by NaCl, i.e. the stochiometry as 1:1. + octahedrally Some examples of ionic crystals 6 Na ions surrounded by with NaCl type crystals are halides of 6 Cl– ions alkali metals and silver (except AgI), NH4Cl, NH4Br and oxides of alkaline Fig. 4.61 Structural representation of NACl crystal earth metals, etc. (Fig. 4.61). (b) Zinc Blende Type Structure The radius ratio for ZnS is equal to 0.40 (74 pm/184 pm). This value corresponds to the range of tetrahedral arrangement (0.414–0.225) with coordination number equal to 4. Thus, in ZnS, S2– ions are arranged in cubic close-packed arrangement in which S2– ions are present at all the eight corners as well as the centre of the six faces of the cube. Thus, the total number of S2– ions in one unit cell = 4. The Zn2+ ions are present in half of the tetrahedral voids. Since the number of tetrahedral Zn2+ ion is voids in a unit cell is equal to double the number of anions, tetrahedrally in this case for 4 S2– ions there are 8 tetrahedral voids, out surrounded by 2– of which four are occupied by the Zn2+ ions. This means four S ions. that in each unit cell, there are 4 S2– ions and 4 Zn2+ ions S2– resulting in the formula of the crystal as Zn4S4 or ZnS, i.e. Zn2+ the stereochemistry of the crystal is 1:1. Some examples of crystals with zinc blende type Fig. 4.62 Structural representation of structure are halides (except fluoride) of Cu, sulphides of zinc blende crystal Be, Cd and Hg (II), AlP, SiC, ZnO, etc. (Fig 4.62).

Molecular Symmetry

(c) Wurtzite-type Structure ZnS also exists in wurtzite type structure with coordination number of 4 due to its radius ratio (0.40) falling in the range of 0.414–0.225 for tetrahedral arrangement. In this structure, S2– ions are arranged in hexagonal close-packed arrangement in which each S2– ion has two tetrahedral sites out of which, only one is filled by Zn2+ ion. This means for every one S2– ion, one Zn2+ ion is present and hence the stereochemistry is 1:1. Further, each Zn2+ ion is tetrahedrally surrounded by four S2– ions and each S2– ion is tetrahedrally surrounded by four Zn2+ ions. Thus, the coordination number is 4:4.

4.57

Zn2+ S2–

Fig. 4.63

Structural representation

(d) CsCl Type Structure The radius ratio for CsCl is equal to of Wurtzite structure 160 pm/181 pm = 0.884. This value corresponds to the range of cubic arrangement (0.732–1). Thus, in CsCl, Cl– ions are arranged in simple cubic arrangement while the Cs+ ions are present at the cubic interstitial sites. Therefore, the total number of Cl– ions in one unit cell is equal to 1/8 × 8 = 1 and the total Cl– number of Cs+ ion in one unit cell is equal to 1. The unit cell Cs+ – can be extended to show that each Cl ion is also surrounded by eight Cs+ ions as each Cs+ ion is surrounded by eight Cl– ions forming the coordination number 8:8 and the formula of the crystal as CsCl with stereochemistry 1:1. It should be noted Fig. 4.64 Structural representation that CsCl structure is not a body-centred cubic arrangement of CsCl type structure because in that arrangement, the same ions are present at the corners as well as at the body centres. Thus, it is clear from the above discussion that as the radius ratio changes, there is a change in the coordination number and the structure of the ionic crystal. Hence, greater the radius ratio, greater is the coordination number and vice versa (Fig. 4.63 and 4.64).

2. Ionic Crystals of the Type AX2 In this type we will discuss the structures of CaF2, TiO2, CaC2 and CdI2 with their common structural arrangement.

(a) CaF2, or Calcium Fluorite Type Structure The radius ratio for CaF2 is equal to 99 pm/136 pm = 0.73. This value corresponds to the range of cubic arrangement 0.732–1.000 with a coordination number equal to 8. Thus, in CaF2, Ca2+ ions are arranged in cubic close-packed arrangement, i.e. present at all the eight corners as well as the centre of the six faces of the cube. On the other hand, the F- ions are present in all the tetrahedral voids. The total number 1 1 of Ca2+ ions in one unit cell is equal to × 8 + × 6 = 4. 8 2 2+ Since for every Ca ion, these are two tetrahedral voids and F– ions are present in all tetrahedral voids, it means that for the 4 Ca2+ ions, there are 8 F– ions so that this stereochemistry of the compound is 1:2. Further, each Ca2+ ion is surrounded by F– ions, while each F– ion is surrounded by 4 Ca2+ ions present at the 4 corners of the tetrahedral resulting in 8:4 ionic crystal as shown in Fig. 4.65. Some examples of CaF2 type structures are fluorides Fig. 4.65 Structural representation of CaF2 crystal of Ba, Sr, Pb, Hg, Cu, etc.

4.58

Inorganic Chemistry

(b) TiO2, Titanium Dioxide or Rutile Structure TiO2 exists in a distorted structure as one of the axes of the cube is shorter than the other by about 30%. In this structure, Ti4+ ions are arranged in distorted body centered cubic arrangement, i.e. are present at all the eight corners as well as at the centre of the body of the distorted cube. On the other hand, O2– ions are present in octahedral voids and occupy positions of three-four coordination. This means that each O2– ion is surrounded by 3 Ti4+ ions distorted towards the corners of an equilateral triangle whereas each Ti4+ ion is octahedrally surrounded by 6O2– ions resulting in the coordination numbers as 6:3 as shown in Fig. 4.66. (c) CaC2, Calcium Carbide Structure The radius ratio for CaC2 crystal (0.52) is similar to that of the NaCl crystal (0.524). Thus, the structure of CaC2 is very similar to that of NaCl, i.e. a 6 × 6 crystal. Here, Ca2+ ions are present at the site of Na+ ions, while C22– ions are present at the place of Cl– ions. This means that the carbon atoms are associated to form C2 groups which are aligned in parallel resulting in a slight distortion in the crystal (Fig. 4.67). However, in case of FeS2 crystal with similar structure to that of CaC2, the S2 units are not aligned in parallel. (d) CdI2, Cadmium Iodide In this structure, I– ions are arranged in hexagonal closed-packed arrangement, whereas the Cd2+ ions occupy the octahedral sites between every two layers of the iodide ions. It results in the surrounding of each Cd2+ ion by 6 I– ions in an octahedral arrangement, whereas the surrounding of each I– ion by 3 Cd2+ ions forming the base of a triangular pyramid with an I– ion at its apex as shown in Fig. 4.68. This results in the formula of the compounds as CdI2 and its structure as a layered lattice.

Ti4+ O2–

Fig. 4.66

Structural representation of TiO2 crystal

8

8 8

8

8

8 C2 group

8 8

Ca2+

8 8 8

8 8 8

Fig. 4.67

8

Structural representation of CaC2 crystal

Fig. 4.68

Structural representation of CdI2 crystal

4.8.13 Lattice-energy Calculations for an Ionic Crystal During formation of an ionic crystal, a large number of cations and anions in their gaseous state come closer and arrange themselves in the close-packed regular pattern known as the ionic lattice under the effect of electrostatic forces. As a result, a large amount of energy is released resulting in the formation and stabilisation of the crystal. This energy is known as lattice energy of the ionic crystal and is defined as the amount of energy released when one mole of the cations and one mole to the anions in their gaseous state come close to each other from infinity and combine together to form one mole of the ionic crystal. For example, when one mole of A+ (g) and one mole of B– (g) combine together, the lattice energy of the ionic + – U crystal is given by A+(g) + B– æ-æ Æ A B (s). where U is the lattice energy and the negative sign signify the release of energy.

1. Born-Lande Equation Lattice energy of an ionic crystal can be determined in terms of interionic coulombic intereractions as discussed here:

Molecular Symmetry

Consider the two ions Az+ and Bz– as the two point charges at a distance r. Using the simple electrostatic model, the electrostatic energy of attraction is expressed as: 2

z+ z – e (4.14) 4pe o r where z+ and z– are the magnitudes of the charges on the cation and anion respectively, e is the electronic charge and eo is the permittivity of free space (8.854 × 10–12 F m–1). Since the ions are not the point charges but have their electron clouds, hence as the two ions approach each other, short-range interelectronic repulsions operate as given by Born as B Erep = (4.15) rn where B is a constant, Born coefficient or repulsion coefficient and n is the Born exponent whose value can be obtained from the compressibility (K) as Eatt =

4.59

+ Erep U 0

r Eatt

Uo

– ro

Fig. 4.69

Plot of energy terms verus r, the interionic distance

Table 4.18 Values of Born exponent Type of ions 2 10 18 36 54

n

Examples

5 Li+, Be2+ 7 Na+, Mg2+, O2–, F– 9 K+, Ca2+, Cu+, S2–, Cl– 10 Rb+, Ag+, Br– 12 Ca2+, Au+, I–

18r 4 Ae2 (n 1) The value of n varies, with the electronic configuration of the ion, as shown in Table 4.18. For an ionic crystal, the value of n is taken as the average of the n values of the two constituent ions. 7+9 =8 Thus, for the NaCl crystal, n =    2  The total energy i.e. the lattice energy of the ion pair, can be written as the sum of these two terms as z+ z- e 2 B + U = Eatt + Erep = (4.16) 4pe o r r n K=

Since the two terms are with opposite signs, the plot of U versus r show a minima at ro, the equilibrium internuclear distance (Fig. 4.69). r = ro, dU = 0

Thus, at

dr

2r o

Hence, differentiating equation (4.16) w.r.t. r, we obtain

- z+ z- e2 ron -1

2

dU = - z+ z- e - nB = 0 or B = 4pe o n 4pe o ro2 ron +1 dr Substituting Eq. (4.17) in Eq. (4.16), we obtain at r = ro 2 n -1 2 Uo = z+ z- e - z+ z- e ro n

6ro

3ro

(4.17)

5ro

ro 2ro

Fig. 4.70

Representation of interionic distances in NaCl crystal

z+ z- e 2 Ê 1 ˆ (4.18) Á1 - ˜ 4pe o ro 4pe o ro Ë n ¯ 4p ne o ro However, in case of an ionic crystal, the coulombic interactions of more ions need to be considered. For example in case of NaCl crystals, for each Na+ ion, there are and so on as shown in Figure 4.70. (i) 6 Cl– ions at a distance ro (iv) 6 Na+ ions _________ 2ro + (ii) 12 Na ions at a distance 2ro (v) 24 Cl– ions _________ 5ro – (iii) 8 Cl ions _________ 3r (vi) 24 Na+ ions _________ 6r or

o

Uo =

o

4.60

Inorganic Chemistry

Hence, the Eatt for the ion pair in the crystal lattice can be expressed as Eatt = =

6 z+ z- e2 12 z+ z- e2 8 z+ z- e2 6 z+ z- e2 24 z+ z- e2 24 z+ z- e2 + + + ....... 4pe o ro 4pe o 2ro 4pe o 3ro 4pe o ro 4pe o 5ro 4pe o 6ro z+ z- e 2 4pe o ro

12 8 6 24 24 z+ z- e 2 Ê ˆ = + + + 6 .... A ÁË ˜¯ 4pe o ro 2 3 2 5 6

(4.19)

where A is known as the Madelung’s constant and is the summation of the infinite series of the interactions due to the geometrical arrangement of the ions and is independent of the ionic charges but depends on the geometry of the crystal. Comparing Eq. (4.18) with (4.19), the lattice energy of the ion pair in a lattice can be expressed as

z+ z- e 2 Ê 1 ˆ A Á1 - ˜ 4pe o ro Ë n ¯ And for one mole of ion pairs with N pairs (N = 6.022 × 1023) Uo =

Uo = NUo

z+ z- e2 NA Ê 1 ˆ Á 1 - ˜ (in SI units) 4pe o ro Ë n ¯ Table 4.19 Value of Madelung’s constant for some common crystals:

This is known as the Born-Lande equation for the lattice energy of the Type of structure Coordination A M ionic crystals. By using the value of number Madelung’s constant and knowing the Rocksalt (NaCl) 6:6 1.74756 1.74756 value of ro, the lattice energy for various Zinc blende (ZnS) 4:4 1.63806 1.63806 crystals can be interpreted. Wurtzite (ZnS) 4:4 1.64132 1.64132 For some common crystal structures, 8:4 2.51939 5.03878 Fluorite (CaF2) the values for Madelung’s constant have Cesium Chloride (CsCl) 8:8 1.76267 1.76267 been obtained by taking the contribution 6:3 2.408 4.816 Rutile (TiO2) of all the ions in the crystal lattice. Some ) 6:3 2.4000 4.8000 Anatase (TiO 2 researchers also replace the term z+ z– 4:2 2.2197 4.4394 by z2 taking z as the highest common Quartz (SiO2) ) 2.355 4.71 Cadmium iodide (CdI factor in the ionic charges and describe 2 Madelung’s constant as M = A z+z–/z2. Cuperite (Cu2O) 2:4 2.05776 4.11552 Table 4.19 lists both the values of A as Corundum (Al2O3) 6:4 4.17186 25.03116 well as that of M. This equation holds good and gives the values of lattice energies in good agreement with that of the experimentally obtained values of lattice energies for halides and oxides of alkali and alkaline earth metals. The main conclusions of Born-Lande equation are as follows: 1. Smaller the interionic distances, greater is the lattice energy released (more negative values) and hence more stable is the ionic lattice. 2. Higher is the product of the ionic charges, greater is the lattice energy. For example, the calculated values of lattice energies of some common crystals are given in the table 4.20. It is quite evident that lattice energy also depends upon the value of the Born exponent (n) and the value of Madelung constant. Different researchers have further modified the Born-Lande equation by considering the van der Waal’s forces, vibrational energies and heat-capacity terms. However, a better method used for the indirect

Molecular Symmetry

determination of lattice energies was provided by Born and Haber in terms of Born–Haber cycle.

Table 4.20

2. Born-Haber Cycle

Values of lattice energies calculated from the Born-Lande equation and Born Haber equation

Compound

Born-Haber cycle is based on Hess’s law and is used to calculate the lattice energy of an ionic crystal by considering the various steps involved to make an ionic solid (crystal) from the constituent elements. These steps are as follows: (a) Conversion of the solid/liquid reactants into their gaseous state (if required) (b) Formation of gaseous cation and gaseous anion. (c) Combination of the gaseous ions to give the ionic solid. Consider the formation of an ionic solid, MX, where M is an alkali metal and X is a halogen in gaseous state.

4.61

LiF LiCl LiBr LiI NaF NaCl NaBr NaI MgF2 MgO2 CaF2 MgS

RO (pm) 201 257 275 302 231 281 298 323 199 206 232 250

z+ z– 1 1 1 1 1 1 1 1 2 4 2 4

Uc (kJ mol–1) –1007.1 –811.3 –766.1 –708.4 –902.0 –755.2 –718.8 –668.2 –2883.4 –4042.9 –2641.6 –3400.8

U (kJ mol–1) –1034 –840.1 –781.2 –718.2 –914.2 –770.3 –728.4 –680.7 –2882 –3895 –2581 –3254

1 DH f M(s)+ X 2 (g) æææ Æ MX(s) 2 where Hf is the enthalpy of formation of MX. The above-mentioned steps can be elaborated as follows: (i) Since alkali metals are in solid state, the first step includes the conversion of one mole of metallic alkali metal (M) into gaseous state using sublimation energy ( Hsub) DH

Sub M(s) + æææ Æ M( g ) This is an endothermic process and its value is considered a positive quantity. (ii) Halogens exist in the diatomic state and require dissociation energy ( Hdiss) for the dissociation of one mole of gaseous halogen molecules into gaseous atoms

DH

diss. X 2 (g) æææ Æ 2 ¥ (g )

This is an endothermic process and its value is considered as a positive quantity. (iii) One mole of gaseous alkali metal atoms are converted into cations in the gaseous state by using Ionisation Energy (IE). IE

M(g) → M+ (g)

This is also an endothermic process and its value is considered positive quantity. (iv) One mole of gaseous halogen atoms are converted into gaseous anions with the release of energy known as Electron Affinity (EA). EA X(g) − → X − (g)

This is an exothermic process and its value is considered a negative quantity. (v) One mole of gaseous metal cations and one mole of gaseous anions combine together to form one mole of metal halide

Fig. 4.71

Born-Haber cycle for the formation of alkali halide

4.62

Inorganic Chemistry

crystal with the liberation of a large amount of energy known as lattice energy (U). −U

M+ (g) + X − (g) → MX(s) According to Hess’s law, the enthalpy of formation of alkali halide can be taken as the sum of all these steps. 1 Thus, Hf = Hsub + Hdiss + IE – EA – U 2 Thus, the lattice energy of the ionic crystal can be calculated by using the values of other energy terms. Table 4.20 lists the lattice energy values of some ionic crystals calculated by using the Born–Haber cycle.

3. Applications of Lattice Energy Values Lattice energy values are used to explain the properties of ionic crystals as discussed ahead.

Solubility of Ionic Solids in Various Solvents The solubility of an ionic solid depends upon the two factors, i.e. lattice energy, which holds the constituent ions of the ionic solid tightly in the ionic solid and solvation energy, which corresponds to the amount of energy released when ionic solid dissociates to give ions that interact with the solvent to give solvated ion. In case of water as the solvent, hydrated ions are formed and the solvation energy is known as the hydration energy. Higher magnitude of lattice energy accounts for the lesser tendency of the ionic solid to split into ions, while greater magnitude of solvation energy accounts for the greater tendency of ions to get solvated. This means that these two factors are opposing each other. Hence, if magnitude of the solvation energy is more than the lattice energy, the solid dissolves in the solvent and if it is lesser than that of lattice energy, the solid is insoluble in the solvent. Ionic solids are insoluble in nonpolar solvents, as there is no interaction of the ions with the solvent molecules due to different nature of the polar and nonpolar species. On the other hand, ionic solids are generally highly soluble in polar solvents due to high amount of solvation energy released. M+ (g) + sol $ M+ (sol) + Solvation energy X– (g) + sol $ X– (sol) + Solvation energy Solvation energy depends upon the following factors:

(i) Charge/Size Ratio Greater the charge/size ratio of an ion, more is its hydration energy. Thus, the ions with greater charge and smaller ionic size are more easily hydrated. (ii) Dipole Moment of the Solvent Greater the dipole moment of the solvent, more is the hydration energy.

Table 4.21 Values of lattice energy and hydration energy for some ionic compounds Compound LiF LiCl LiBr LiI NaCl NaBr KCl KBr KI

Lattice energy (kJ mol–1) 1034 840.1 781.2 718.2 770.3 728.4 701.0 621.1 632.2

Hydration energy (kJ mol–1) 1019 883 854 793 775 741 686 657 619

(iii) Dielectric Constant of the Solvent Greater the dielectric constant of the solvent, more is the hydration energy. Table 4.21 lists values of lattice energy and hydration energy of some ionic solids. It is clear from the data that only those ionic solids are soluble in water which have greater hydration energy than the lattice energy.

2. Stability of Ionic Solids Greater the magnitude of the lattice energy, greater is the stability of an ionic solid. For example, CaCl2 (2200 kJ mol–1) is more stable than CaCl (720 kJ mol–1).

Molecular Symmetry

4.63

3. Melting Point of Ionic Solids Greater the magnitude of the lattice energy, greater is the energy required to separate the ions and hence higher is the melting point of the ionic solid. For example, the melting points of alkali metal halides varies as LiF > LiCl > LiBr > LiI LiCl > NaCl > KCl > RbCl > CsCl

4. Conductivity in Solutions Lithium forms hydrated salts while the salts of Rb and Cs are rarely hydrated. This is due to the reason that the extent of hydration of an ion decreases with increase of its size. As the size of the alkali metal increases from Li to Cs, its extent of hydration decreases. Thus, Li+ is most hydrated and Cs+ is least hydrated. As a result, Li+ (aq) is largest in size and Cs+ (aq) is the smallest. Since the mobility and hence the conductance of an ion depends upon its size, the electrical conductivity in dilute solution goes on decreasing from Li to Cs. Same is the case for alkaline earth metals.

4.7.14 Defects in the Ionic Crystals

A+

B–

A+

B–

An ideal ionic crystal has the same unit cell and no lattice B– A+ A+ B– point vacant. The two-dimensional pattern of an ideal crystal can be represented as shown in Fig. 4.72. A+ A+ B– B– However, an ideal crystal exists only at absolute zero with well-ordered arrangement of ions, i.e. has no defect at A+ B– B– A+ all. As the temperature increases, the crystal shows some departure from its ordered arrangement and is known to have a defect. Depending upon the ratio of the constituent Fig. 4.72 Two-dimensional representation particles, the crystals can be classified as stoichiometric and of an ionic crystal nonstoichimetric solids. In a stoichiometric crystal, the ratio of the cations and anions is exactly same as required by the ideal formula, but in non-stoichiometric crystal, the ratio of the cations and anions is different to that required by the ideal formula. However, electrical neutrality is maintained in both types of the crystals. The defects induced in the crystal by the missing of the constituent particles are known as point defects or atomic defects. On the other hand, the defects induced in the crystals due to presence of foreign particles are known as impurity defects. Accordingly, we can discuss the defects in ionic crystals under three headings: (a) Point defects (b) Impurity defects (c) Thermal defects

1. Schottky Defect This defect is created by the missing of an equal number of cations and anions from their positions in the crystal. It results in the formation of holes or lattice vacancies as shown in Fig. 4.73. However, the crystal retains its electrical neutrality, as the number of the remaining cations and anions are the same. This defect is generally produced in the ionic compounds with high coordination number and almost same size of the cations and anions. For example, NaCl and CsCl type crystals with coordination number 6 and 8 respectively show Schottky defect. Consequences of Schottky Defects Due to presence of holes (missing of constituent particles), the density of the crystal decreases and conductance increases. It also results in decrease of lattice energy and hence, lower stability of the crystal.

4.64

Inorganic Chemistry

2. Frenkel Defects This defect is created by missing of a constituent particle from its lattice position and occupation of an interstitial site. Since cations are smaller in size, hence generally cations occupy the interstitial site and create cation vacancy or hole in the crystals as shown in Fig. 4.74. Here too, the electrical neutrality is maintained due to presence of equal number of cations and anions. This defect is generally produced in the ionic compounds with low coordination number and with size of anion much larger that of cation, For example, AgBr and ZnS generally show Frenkel defects. In case of AgBr, Ag+ occupies the interstitial sites and is responsible for the production of photographic images when silver bromide crystals are exposed to light. Similarly, in ZnS, Zn2+ are present in the interstitial sites of the crystal.

Consequences of Frenkel Defects Due to the creation of holes (shifting of the constituent particles in the interstitial sites), the electrical conductance of the crystal increases. However, there is no change in the density of the crystal. At the same time, due to the closeness of the similar charges, the dielectric constant of the crystal increases and stability of the crystal decreases.

3. Metal-Excess Defect This defect is produced due to the presence of extra cations. This may arise in two ways and hence is of the following two types:

A+ Cation vacancy

B–

B–

anion vacancy

A+

B–

B–

A+

A+

B–

A+

B–

B–

A+

A+

B– A+ B–

A+

Fig. 4.73 Representation of Schottky defects

A+

B–

A+

B–

B–

A+

A+ B–

A+

B–

A+

B–

B–

A+

B–

A+

Fig. 4.74 Representation of Frenkel defect

(a) Metal Excess Defect Due to Missing of Anions This defect is produced by the missing of anions from their lattice positions. As a result, the anion vacancy or hole is produced. The electrical neutrality is maintained by the occupation of the anion vacancies by electrons as shown in Fig. 4.75. Thus, there is an excess of cations (metal ions) and the crystal is neutral. This defect is generally found in the crystals which are likely to have Schottky defect. For example, when alkali metal halides are heated in presence of vapours of alkali metals, the halide ions move towards the surface and combine with the alkali Fig. 4.75 Representation of metal excess defect due to presence of anion vacancy metal atom and release the electrons. These electrons diffuse into the crystal and occupy the anion vacancies and are known as F-centres or colour centres, as they impart colour to the crystals. Thus, heating of sodium chloride crystal in presence of sodium vapours produces a pale yellow nonstoichiometric form while the potassium chloride crystal in presence of potassium vapours gives a lilac coloured form.

4.65

Molecular Symmetry

Consequences 1. The crystals with metal-excess defects due to anion vacancy are generally coloured due to the presence of F-centres. The electrons present in the F-centres easily get excited and emit radiations of visible light and the crystal appears coloured. 2. Due to presence of free electrons, the electrical conductance of such crystals increases to considerable extent and these are termed n-type semiconductors.

O2–

Zn2+

Zn2+ O2–

O2–

Zn2+ e–

Zn2+

O2–

Zn2+

Zn2+

O2–

e– O2–

Zn2+

(b) Metal-excess Defect due to Presence of Extra Cations This defect is produced due to the presence of extra cations in the interstitial sites and the electrical Fig. 4.76 Metal-excess defect in zinc neutrality is maintained by the presence of an equal oxide crystal number of electrons in the other interstitial sites as shown in Fig. 4.76. Thus, there is an excess of cations (metal ions) in the crystal. This defect is found in the crystals with Frenkel defects. For example, zinc oxide(ZnO) loses oxygen reversibly, when hot. The excess of Zn2+ get occupied into the interstitial sites and the extra electrons occupy the other interstitial sites (Fig. 4.76).

1 O2 + 2e– 2 Hence, the electrical neutrality is maintained. Thus, zinc oxide is colourless at ordinary temperatures but turns yellow when hot, due to the presence of extra electrons in the interstitial sites which imparts colour to the crystal. The presence of free electrons increases the electrical conductivity of these substances and these are also termed as n-type semiconductors. ZnO $ Zn2+ +

4. Metal-Deficiency Defect This defect is produced by missing of cations from their lattice positions and formation of cation vacancies. The electrical neutrality of the crystal is maintained by a neighbouring cation by acquiring higher oxidation state (Fig. 4.77). Thus, there is a deficiency of cations (metal ions) in the crystal and the crystal is neutral. This defect is generally found in the crystal of the metals which can show variable oxidation states. For example, crystals of ferrous oxide (FeO), iron pyrite(FeS) and nickel oxide show this type of defect. In case of ferrous oxide, one Fe2+ ion is missing and two Fe2+ ions are converted to Fe3+ ions. The conversion takes place by the transfer of electrons and produces a shiny colour in the crystals.

A+

B–

A+

B–

B–

A2+

B–

A+

A+

B–

Fig. 4.77

B–

Representation of metal deficiency defect

Consequences The crystals with metal deficiency defects are lustrous and nearly conducting in nature. The conductance is due to movement of the electrons from one cation to the other by converting these into higher oxidation states. In other words, a positive hole is moving and the substance is called p-semiconductor.

5. Impurity Defects This defect is produced in the crystals due to presence of impurities. Addition of a foreign substance (impurity) to a crystal is known as doping. This is quite useful in case of silicon and germanium crystals which are poor

4.66

Inorganic Chemistry

conductors, but act as semiconductor on addition of elements belonging to Group 13 or Group 15. It results in the formation of n type semiconductors and p-type semiconductors as discussed below. (a) n-type Semiconductors In this case, doping of the silicon or germanium crystal is done with the Group 15 element resulting in substitution of some of the parent crystal atom by that of the Group 15 element, say arsenic. This takes place by the covalent-bond formation using four electrons each from the arsenic atom and the parent atom. As a result one electron is left free with each arsenic atom and is involved in conduction. Since the current is carried in the normal way by these excess electrons, this is known as n-type semiconduction. (b) p-Type Semiconductor In this case, the doping of the silicon or germanium crystal is done with the Group 13 element resulting in the substitution of the some of the parent crystal atoms by that of the Group 13 element, say aluminium. This takes place by covalent-bond formation using three electrons each from the aluminium atom and the parent atom. Thus, the bonding is short of one electron creating some of the empty sites which are otherwise occupied by the electron. These empty sites, or the electron vacancies, are known as positive holes and create new positive holes. The process continues and increases the electrical conductivity, though to a lesser extent. This conduction is known as p-type semiconduction due to the participation of positive holes. The conduction increased due to impurity defect is known as extrinsic conduction and maintains the electrical neutrality of both types of semiconductors, because the added impurity itself is neutral as shown in Fig. 4.78. As the temperature is increased, extra electrons or the positive holes (depending upon the case), bound with the crystals, get free by absorption of heat and increase the electrical conduction. As a result, the conductivity of a semiconductor increases with increase in temperature. Application Semiconductors are mainly used to manufactue transistors by the combination of n- and p-type semiconductors to form np junctions. As the result, the electric current flows in one particular direction and then in the reverse direction. Hence, it is used as a rectifier to change alternating current into direct current.

Fig. 4.78 Impurity defects

On applying an external voltage, current is readily conducted by the flow of electrons from left to right and flow of positive holes from right to left. However, if the direction of the voltage is reversed, the conduction stops due to cancellation of n and p-currents. Hence, an n-p junction can be used to conduct the small current from an outside source in one direction only.

6. Thermal Defects The defects in the crystals produced due to the effect of large temperature are known as thermal defects, e.g. the lattices of silicon and germanium atoms are constituted by covalent bonding involving four electrons. As a result, the conductance is very poor due to absence of any free electron. But if temperature is increased,

Molecular Symmetry

4.67

some of the covalent bonds break, ejecting the electrons and creating a positive hole at the site of the missing bond. When an electric field is applied, the negatively charged electrons migrate to one side and the positive holes migrate to the other side. This is known as intrinsic conduction due to increase of conductance without the increase of any foreign substance.

Density of Crystals The density of a crystal can be determined by the following equation: Mass of the cubic unit cell Density of the crystal, r = ________________________ Volume of the cubic unit cell The volume of the cubic unit cell with the edge length a = a3 (if a is in pm) Mass of the cubic unit cell = Number of atoms in the unit cell × mass of one atom

M No where No is the Avagadro’s number and M is the atomic mass of the element (a molecular mass of the compound) = Z×

Thus,

r=

Z¥M No ¥ a

3

Z¥M =

N o ¥ a 3 ¥ 10 -30

g cm–3

For a single cubic structure, Z = 1 For a body-centered cubic structure, Z = 2 and for a face centred cubic structure, Z = 4

Solids are classified into two broad categories, i.e. crystalline solids with definite geometrical configuration of the constituent particles and amorphous solids with a nondefinite geometrical configuration of the constituent particles. The crystalline solids can be represented by means of 14 distinct types of lattices known as Bravais’s lattices and 7 crystal systems, namely cubic, rhombohedral, tetragonal, hexagonal, orthorhombic, monoclinic and triclinic. The constituent particles in a crystal are packed either in the hexagonal close- packed arrangement, face centred cubic close packing, cubic close-packed arrangement or body-centred cubic arrangement with percentage of occupied space as 68% for bcc and 74% for the rest. During such packing, some empty spaces are left, known as the interstitial voids, which can be of four types depending upon the radius ratio and the coordination number. These are trigonal void (CN = 3, r+/r–= 0.155–0.225), tetrahedral void (CN = 4, r+ /r– = 0.225–0.414), octahedral void (CN = 6, r+ /r– = 0.732–1.0). The structure of some common ionic crystals are as follows: 1. Rock-salt type structure: Cl– — in ccp arrangement and Na+ in octahedral voids 2. Zinc blende type structure: S2– — in ccp arrangement and Zn2+ in half of the tetrahedral voids 3. Wurtzite type structure: S2– — in hcp arrangement & Zn2+ in half of the tetrahedral voids 4. CsCl type structure: Cl–— in simple cubic arrangement and Cs+ at the cubic voids 5. Calcium fluorite structure: Ca2+ in ccp arrangement and F– in the tetrahedral voids 6. Rutile structure: Tl4+ in distorted bcc arrangement and O2– in octahedral voids

4.68

Inorganic Chemistry

The symmetry point group of a molecule can be determined using the following flow diagram:

EXAMPLE 1 Write down the term symbol for LiH molecule. The molecular electronic configuration for LiH can be represented as K(s2s)2 = 0 + 0 = 0, state is S and S = 0 , 2S + 1 = 1 1 + The term symbol is S . The subscript g or u will not be added as heteronuclear diatomic molecule has no center of symmetry.

EXAMPLE 2 On the basis of Walsh diagram, predict the geometry of following molecules: (a) LiH2

(b) BH2

(c) NH2

The molecular electronic configuration of these molecules for both cases of symmetry can be written as follows: (a) LiH2 – 1s2g 1s1u (D h) Destabilisation of 1su, hence linear 1a21 2a11 (C n)

Molecular Symmetry

(b) BH2 – (D h) 1s2g 1s2u 1p1u

Stabilisation of pu, hence bent

(C n) 1a21 2a21 1b12 1s2g 1s2u 1p3u 1a21 2a21 1b22 1b11

Stabilisation of pu, hence bent

(c) NH2 –

4.69

EXAMPLE 3 Discuss the splitting of f-orbitals containing a single electron under the effect of octahedral ligand environment. We will reduce the characters of reducible representation as follows: For the set of seven f orbitals, the variant part of f-orbital wave functions are eimf, where m = +3 to –3. (i) For identity operation, E c(E) = 7 (ii) For rotation operation, C(a) with rotation by angle a

Èe3if ˘ Èe3if Í 2if ˙ Í0 Íe ˙ Í Íeif ˙ Í0 Í ˙ Í C (a ) Íe0 ˙ = Í0 Íe - if ˙ Í0 Í ˙ Í Íe -2f ˙ Í0 Í -3f ˙ Í0 Îe ˚ Î

e3if 0 0 0

0 0

0 0

0 0

0 0

e3if 0 0

0

0 0

0 0

0 0

0

e-1if 0

0 0 0

0

0

0

0

e 0

e-2if 0

Ê 1ˆ sin Á l + ˜ a Ë 2¯ and cC(a) = a sin 2 a sin 7 2 For f orbitals, l = 3 and c C(a ) = a sin 2 For two-fold rotation, a = p p sin 7 2 = -1 = -1 c(C2) = p 1 sin 2 For three-fold rotation, a = 2p/3 7p 14p p sin sin sin 3 = 3 =1 6 = c(C3) = 2p p p sin sin sin 3 3 6 For four-fold rotation, a = p/2 7p 4 = -1 2 = -1 c(C4) = p 1 2 sin 4 sin

˘ ˙ ˙ 0 ˙ ˙ 0 ˙ 0 ˙ ˙ 0 ˙ ˙ e -3if ˚ 0 0

Èe3if ˘ Í 2if ˙ Íe ˙ Íeif ˙ Í 0 ˙ Íe ˙ Íe - if ˙ Í ˙ Íe -2f ˙ Í -3f ˙ Îe ˚

4.70 Thus,

Inorganic Chemistry

O G red

E 6C4 1

3C2 (= C42 ) 8C3

-1

-1

1

6C2 -1

The reducible representation can be reduced using character table for point group O as follows:

1 [1.1.7 + 6.1.(–1) + 3.1.(–1) + 8.1.1 + 6.1.(–1)] = 0A1 24 1 n(A2) = [1.1.7 + 6.(–1).(–1) + 3.1.(–1) + 8.1.1) + 6.(–1).(–1)] = 1A2 24 1 n(E) = [1.2.7 + 6.0.(–1) + 3.2.(–1) + 8.(–1).1 + 6.0.(–1)] = 0E 24 n(A1) =

1 [1.3.7 + 6.1.(–1) + 3.(–1).(–1) + 8.0.1) + 6.(–1).(–1)] = 1T1 24 1 n(T2) = [1.3.7 + 6.(–1).(–1) + 3.(–1).(–1) + 8.0.1) + 6.1.(–1)] = 1T2 24 Thus Gred = A2 + T1 + T2 Similarly for the point group Oh and for u character of all f orbitals, we get n(T1) =

Gred = A2u + T1u + T2u

EXAMPLE 4 Determine the strong field term for t62g and eg4 configuration. t26g and e4g configurations has all paired electrons in completely filled orbitals. Hence, only one term 1A1g is generated.

EXAMPLE 5 Calculate the density of an element containing 12 ×1023 atoms in 100 g and crystallising in a structure with bcc unit cell with an edge of 200 pm.

Atomic mass of the element =

6.22 1023 100 = 50.18 g mol–1 12 1023

For bcc, Z = 2

Z×M g cm–3 N 0 × a 3 × 10 −30 2 × 50.18 r= = 20.83 g cm–3 6.022 × 1023 × (200)3 × 10 −30

Density of the crystal, r = or

EXAMPLE 6 A mixed oxide is crystallised in cubic close packing of oxide ions and contains

one half of the tetrahedral voids occupied by A2+ ions and one half of the octahedral voids occupied by B3+ ions. Determine the formula of the oxide.

Let the number of oxide ions = 100 Number of tetrahedral voids = 2 × Number of oxide ions = 200 Number of octahedral voids = Number of oxide ions = 100 Thus, number of A2+ ions occupying one half of the tetrahedral voids = 100

Molecular Symmetry

4.71

And number of B3+ ions occupying one half of the octahedral voids = 50 Thus stoichiometric ratio of the ions is A2+:B3+: O2– ::100: 50:100 or 2:1:2 and the formula of the oxide is A2BO2.

EXAMPLE 7 Determine the Miller indices of a crystal plane which cuts through the crystal axes at a, 3b, 2c. Fractional intercepts

a 1

b 3

c 2

Reciprocals

1/1

1/3

1/2

Clear fractions

6

2

3

Hence, the Miller indices are 6, 2, 3.

QUESTIONS Q.1 Assign the symmetry point group to the following molecules: (a) Benzene (b) H2O (c) C2H4 (d) Acetylene Q.2 Determine the number of symmetry elements for the following molecules: (a) SO2 (b) CCl4 (c) Fe(CN)63– (d) H2O2 Q.3 Determine the number of symmetry operations for the following molecules: (a) S8 (b) PF5 (c) Ferrocene (d) PtCl42– Q.4 Determine the number of symmetry operations and assign the symmetry point group to the following molecules: (a) Ruthacene (b) N2O (c) PCl5 (d) trans – N2F2 (e) BF3 (f) CO2 (g) N2O4 (h) POCl3 (i) Acetaldehyde (j) Diborane Q.5 Write the multiplication table for C2n point group. Q.6 Using the character table for D4h point group, reduce the following representation to Girred E C4 C2 2C2 2C2 i 2S4 sh 2sn 2sd Gred 4 0 0 2 0 0 0 4 2 0 Use this irreducible representation to discuss the type of hybridisation in [Pt(Cl)4]2– Q.7 Discuss the rules for writing Mulliken’s symbols. Q.8 Determine the irreducible representation for the following: (a) C2n point group (b) Set of p-orbitals Q.9 Discuss the significance of group theory in the determination of probability of electronic transitions. Q.10Discuss the splitting of d-orbitals in octahedral ligand field environment on the basis of group theory. Q.11 Discuss the three laws of crystallography. Q.12 What are the elements of symmetry? What do you understand by diad and tetrad rotation axis? Q.13 What do you mean by Bravais lattices and crystal systems? Q.14 Discuss the methods for designation of planes of a crystal. Q.15 Write short notes on the following: (a) fcc arrangement (b) Schottky defect (c) Metal-excess defect (d) Limiting radius ratio

4.72

Inorganic Chemistry

Q.16 Calculate the percentage of occupied space in bcc and hcp arrangements. Q.17 How does coordination number vary with radius ratio? How does it affect the geometry of the ionic crystal? Q.18 Discuss the structure of the following ionic crystal types: (a) Rock salt type structure (b) Fluorite-type structure (c) Zinc-blende-type structure Q.19 Derive Born-Lande equation and hence prove that lattice energy of an ionic crystal is inversely proportional to the interionic distances. Q.20 Discuss Born-Haber cycle with the help of a suitable example.

MULTIPLE-CHOICE QUESTIONS 1. The molecule belonging to D6h point group is (a) H2O (b) PF5 (c) Benzene (d) Ruthacene 2. The symmetry point group of BF3 is (a) D2h (b) D3h (c) C2n (d) C3n 3. The symmetry of px atomic orbital of oxygen in H2O is (a) a1 (b) b1 (c) b2 (d) a2 4. The Miller indices for a plane with intercepts a, b, 2c on the crystallographic axes are (a) 1 1 1 (b) 1 1 2 (c) 1 2 1 (d) 2 1 1 5. The example of an n-type semiconductor is (a) silicon (b) silicon doped with arsenic (c) silicon doped with phosphorus (d) silicon doped with carbon

chapter

Redox Reactions

5

After studying this chapter, the student will be able to

5.1

INTRODUCTION

Redox reactions, an important class of chemical reactions, are often encountered in inorganic systems. The earlier terminology defined oxidation as the process involving addition of oxygen (or an electronegative element) or the removal of hydrogen (or any electropositive element). Similarly, reduction was defined as the process involving removal of oxygen (or an electronegative element) or the addition of hydrogen (or any electropositive element). However, nowadays, a more general concept describes oxidation as the loss of electrons or an increase in oxidation state of the atom; and reduction is defined as the gain of electrons or a decrease in oxidation state of the atom. Since oxidation and reduction reactions proceed simultaneously, such reactions are termed redox reactions. The alone oxidation and the alone reduction reaction are called half reactions and these are combined to get the whole reaction. The substances which undergo oxidation are said to be oxidised and act as reducing agents, while the substances which undergo reduction are said to be reduced and act as oxidising agents. The reaction between H2 and O2 is an example of redox reaction. 2H2 + O2 $ 2H2O The overall reaction can be described in terms of two half reactions as (oxidation) 2H2 $ 4H+ + 4e– O2 + 4e– $ 2O2– (reduction) 2H2 + O2 $ 4H+ + O2–

5.2

Inorganic Chemistry

The electrons are cancelled, when the two half reactions are added. Thus, in this reaction hydrogen is being oxidised and acts as a reducing agent, while oxygen is being reduced and acts as an oxidising agent. The redox reaction can be a direct redox reaction (oxidation and reduction take place in the same beaker) or an indirect redox reaction (oxidation and reduction takes place in two separate beakers).

5.2

ELECTROCHEMICAL CELL

The chemical energy produced during an indirect redox reaction can be converted into electrical energy with the help of a device known as electrochemical cell. It is also known as galvanic cell or voltaic cell, after the name of the two scientists Luigi Galvani and Alessandro Volta, the pioneers to perform the experiment on conversion of chemical energy into electrical energy. It differs from an electrolytic cell in which the electrical energy is converted into chemical energy. In a typical electrochemical cell, the electrodes are immersed in suitable electrolytes taken in two different containers. The electrodes are connected internally through a salt bridge and externally through a copper wire. Thus, flow of ions through the salt bridge maintains the flow of current in the inner circuit and the flow of electrons completes the external circuit. When the circuit is closed, electrons are produced on the anode (-ve electrode) due to oxidation. These electrons move to the cathode via an external circuit and are used in reduction taking place at the cathode (+ve electrode). Thus, the flow of electrons is from anode to cathode. The Daniell cell is a typical example of an electrochemical cell (Fig. 5.1). In this cell, a zinc rod dipped in a solution of zinc sulphate acts as the anode and the copper rod dipped in a solution of copper sulphate acts as the cathode. The solutions are connected through a salt bridge filled with an aqueous solution of potassium chloride. The current starts flowing on closing the circuit and is indicated by deflection in the ammeter. The reactions take place as At anode

Zn(s) $ Zn2+(aq) + 2e– (oxidation)

At cathode Cu2+(aq) + 2e– $ Cu

(reduction)

The overall chemical reaction or the cell reaction can be written as Zn(s) + Cu2+(aq) $ Zn2+(aq) + Cu(s) Thus, Cu2+ ions discharge as Cu, on the copper electrode by extracting two electrons from the copper rod 2+ leaving behind unpaired SO2– 4 ions which migrate through the salt bridge on the other side and pair with Zn ions. As a result, there is depletion of Cu2+ ions and excess of Zn2+ ions and a state of equilibrium is reached. The cell can be rejuvenated by addition of CuSO4 and/or replacement of ZnSO4 by H2SO4.

Fig. 5.1 The diagrammatic representation of Daniell cell

Redox Reactions

5.3

5.2.1 Electrode Potential Electrode potential of an electrode can be defined as the tendency of an electrode to lose or gain electrons when it is in contact with its own ions in solution. Thus, tendency of an electrode to lose electrons or get reduced is called its reduction potential, while the tendency to gain electrons or get oxidised is called its oxidation potential. Oxidation and reduction potentials are simple reverses of each other, i.e. if oxidation potential of an electrode is E volts, then its reduction potential will be –E volts. However, it is not possible to determine the potential of a single electrode, but the difference of potentials between the two electrodes constituting the electrochemical cell can be measured. The electrode potential of one electrode is arbitrarily fixed at zero and the potential of the other electrode is determined. The electrode potential of an electrode depends upon the nature of the metal of the electrode, temperature, and the concentration of the surrounding ions in the solution. If the electrode is dipped in the solution of its ions with 1M concentration at 25°C, the potential of the electrode is termed standard electrode potential (E°). Standard reduction potential is represented as E°red and standard oxidation potential is represented as E°ox. Conventionally, standard reduction potential is taken as the standard electrode potential.

5.2.2 Determination of Standard Electrode Potential The standard electrode potentials are determined by combining the electrode with the Standard Hydrogen H2 gas Electrode (SHE). It is a reversible hydrogen electrode at 1 atm containing a solution of H+ ions with 1M concentration and H2 gas at 1 atm bubbled through it (Fig. 5.2). Platinum foil Conventionally, its potential has been fixed as zero. coated with SHE can be represented as Pt black Hg Pt; H2(g) (1atm), H+(aq) (c = IM) 1MHCl The standard reduction potentials of a number of electrodes have been determined and arranged in the decreasing order in a series known as electrochemical Fig. 5.2 SHE series (Table 5.1). As can be seen, the half-cell reaction is written as reduction reaction, and the species involved are termed as redox couples (oxidised + reduced species). The negative sign of the E° value of a given electrode indicates that the electrode will act as an anode when combined with SHE to form a galvanic cell. Likewise, the positive sign of the E° value of a given electrode indicates that the electrode will act as a cathode when combined with SHE to form a galvanic cell. For example, E°Cu2+/Cu is positive; thus, a copper electrode acts as a cathode w.r.t. SHE. E°Zn2+/Zn is negative; thus, a zinc electrode acts as an anode w.r.t. SHE. The complete cells can be represented as H+(aq) H2(g) Cu2+(aq) Cu(s) Cathode E°Cu2+/Cu = + 0.34 V Anode E°H+(aq)/H2(g) = 0.0 V Zn(s) | Zn2+(aq) H+(aq) | H2 (g) E°Zn2+/Zn = – 0.76 V

5.2.3 EMF of the Cell The Electromotive Force (EMF) of the cell or cell potential (Ecell) is defined as the difference in electrode potentials of the two electrodes constituting the cell and is calculated as Ecell = Ecathode – Eanode and on the similar terms, the standard cell potential is calculated as E°cell = E°cathode – E°anode

5.4

Inorganic Chemistry

Table 5.1 Electrochemical series Electrode Li+/Li K+/K Ba2+/Ba Ca2+/Ca Na+/Na Mg2+/Mg Al3+/Al Mn2+/Mn Zn2+/Zn Cr3+/Cr Fe2+/Fe Cd2+/Cd Co2+/Co Ni2+/Ni AgI(s)/I–, Ag Sn2+/Sn Pb2+/Pb 2H+/H2 AgBr/Br–, Ag Hg2Br2/Br–, Hg Cu2+/Cu+ Sn4+/Sn2+(Pt) AgCl/Cl, Ag Hg2Cl2/Cl–, Hg Cu2+/Cu Cu+/Cu I2/2I–(Pt) Hg2SO4/SO2– 4, Hg Fe2+/Fe2+(Pt) Ag+/Ag Hg2+/Hg2+(Pt) Br2(l)/2Br–(Pt) Cr3+/Cr2O2– 7 Cl2(g)/2Cl–(Pt) Au3+/Au Mn2+/MnO4– (Pt) Ce4+/Ce3+(Pt) Co3+/Co2+(Pt) F2(g)/2F–

Electrode reaction Li+ + e– Li + – K +e K Ba2+ + 2e– Ba Ca2+ +2e– Ca Na+ + e– Na Mg2+ + 2e– Mg Al3+ + 3e– Al Mn2+ + 2e– Mn Zn2+ + 2e– Zn Cr3+ + 3e– Cr Fe2+ + 2e– Fe Cd2+ + 2e– Cd Co2+ + 2e– Co 2+ – Ni + 2e Ni AgI(s) + e– Ag + I– 2+ – Sn + 2e Sn Pb2+ + 2e– Pb 2H+ + 2e– H2(g)(1 atm) AgBr(s) + e– Ag(s) + Br– Hg2Br2(s) + 2e– 2Hg + 2Br– 2+ – + Cu + e Cu Sn4+ + 2e– Sn2+ – Ag(s) + Cl– AgCl(s) + e – Hg2Cl2(s) + 2e 2Hg + 2Cl– 2+ – Cu + 2e Cu Cu+ + e– Cu I2(s) + 2e– 2I– – Hg2SO4(s) + 2e 2Hg + SO42– 3+ – 2+ Fe + e Fe Ag+ + e– Ag 2Hg2+ + 2e– Hg2+ 2 – Br2(l) + 2e 2Br– Cr2O72– + 14H+ + 6e– 2Cr3+ + 7H2O – Cl2(g)(1 atm) + 2e 2Cl– 3+ – Au + 3e Au MnO4– + 8H+ + 5e– Mn2+ + 4H2O 4+ – 3+ Ce + e Ce Co3+ +e– Co2+ F2 + 2e– 2F–

E°(V) – 3.04 –2.92 –2.90 –2.87 –2.71 –2.52 –1.66 –1.18 –0.76 –0.74 –0.44 –0.40 –0.27 –0.24 –0.15 –0.15 –0.13 0.00 +0.07 +0.13 +0.15 +0.15 +0.22 +0.28 +0.34 +0.52 +0.53 +0.61 +0.77 +0.80 +0.92 +1.06 +1.33 +1.36 +1.50 +1.51 +1.71 +1.82 +2.87

Redox Reactions

5.5

5.2.4 Nernst Equation Consider a metal M, immersed in the solution of its own ions Mn+(aq). Mn+(aq) + ne– M(s) The electrode potential, E can be related with its standard electrode potential, E° by the Nernst equation:

RT [ M(s)] ln nF [ M n ] RT E = E° + ln[Mn+] nF E = E° –

Or

(For pure solids and liquids, [M(s)]is taken as unity)

where R is the gas constant, T is the absolute temperature, F is the Faraday constant, n is the number of electrons and [Mn+] is the concentration of the metal ions in the solution. By substituting the values for R, T and F at 25°C, we get

0.0591 log [Mn+] n Thus, for a zinc electrode, the potential is given by 0.0591 EZn2+/Zn = E°Zn2+/Zn + log [Zn2+] (at 25°C) 2 and for a hydrogen electrode, the potential is given by EH+/H2 = E°H+/H2 + 0.0591 log [H+] (at 25°C) ° EMn+/M = EM + n+ /M

Or

EH+/H2 = – 0.0591 pH

Similarly, for non-standard conditions, emf of the cell is given by

RT ln Q nF 0.0591 Or Ecell = E°cell – log Q n where, Q is the reaction quotient for the cell reaction and becomes equal to K, equilibrium constant of the redox reaction at equilibrium. Ecell = E°cell –

cC + dD

Thus, for a general cell reaction, aA + bB Q=

[C ]c [ D]d [ A]a [ B]b [C ]c [ D]d 0.0591 log n [ A]a [ B]b c d [C ] [ D]

Ecell = E°cell – At equilibrium, Ecell = 0 Thus,

and

K = log E°cell =

[ A]a [ B]b

0.0591 log K n

5.2.5 Applications of Electrochemical Series 1. Relative Tendencies of Metals Greater the reduction potential of an atom, greater is its tendency to get reduced and hence greater is its oxidising power. Thus, fluorine serves as a very strong oxidant. Conversely, greater the oxidation potential of an atom, greater is its tendency to get oxidised and

5.6

Inorganic Chemistry

greater is its reducing power. Thus, lithium serves as a very strong reductant. 2. Displacement of Metals from Solution A metal lying high in the electrochemical series is more reducing and can displace another metals lying lower from their solution. Thus, iron can displace nickel and copper from their solutions. 3. Liberation of H2 from Acid Solution It is now clear that all metals lying above hydrogen in the electrochemical series can liberate H2 from its acid solutions while those lying below cannot. 4. Determination of Feasibility of Chemical Reaction The decrease in the free energy of the cell reaction (DG°) is related to the standard electrode potential (E°) of a cell reaction involving n number of electrons as, – DG° = n FE° or G° = –n FE° As we know that for a spontaneous reaction, DG° should be negative, likewise if E° is positive, DG° will be negative and the reaction will be feasible. A thermodynamically possible reaction might not take place. For example, if a sheet of galvanised iron is scratched, the following half reactions are possible in contact with water. Fe $ Fe2+ + 2e– 2+

E° = +0.44 V

–

Zn $ Zn + 2e E° = +0.76 V Thus, either metal might undergo oxidation. However, due to large positive E° value for Zn/Zn2+, the associated DG0 will be highly negative and hence energetically more favourable for the corrosion of zinc, and the iron is protected. Small Fe2+, even if produced, is immediately reduced and reaction does not occur. Zn + Fe2+ $ Fe + Zn2+ Thus, the coating of zinc an over iron surface not only provides a surface coating but also provides anodic protection. Similar is the action of magnesium blocks attached with underground ship hulls and steel pipelines. There may be chances to reverse the reaction by modifying the potential values, if DG0 is very small. This is because potential values depend on the temperature and concentration or pH of the medium. For example, arsenious acid, HAsO2, is oxidised with triiodide ion, I3– at pH 7 and arsenic acid can be reduced with iodide ions in 5 M acid. The half-reactions can be represented as HAsO2 + 2H2O $ H3AsO4 + 2e– + 2H+ I3– Overall reaction The corresponding

HAsO2 +

–

E° = 0.56 V

–

+ 2e $ 3I I3–

E° = +0.54 V –

+ 2H2O $ H3AsO4 + I + 2H

+

E°cell = 0.54 – 0.56 = –0.02 V

DG° = –2F(–0.02)

= + 0.04 F Thus, the reaction is energetically not favourable, but by controlling the pH cell potential can be modified.

5.3

KINETICS OF REDOX REACTIONS

It has been a matter of common observation that the reactions which involve release of gases are kinetically slower than expected and require some extra potential for discharge of ions. This extra potential is called overpotential and is defined as the difference of potential between an electrode at which the gas is being evolved (Emeasured) and the reversible gas electrode at equilibrium in the same solution. Thus, overpotential ( ) is given by = Emeasured – Etheoretical

Redox Reactions

5.7

During the passage of small current, the equilibrium is easily maintained. But if strong currents are passed through the solution, the equilibrium is not established and reaction proceeds at a slower rate. To proceed a given reaction at an appreciable rate, the reaction is provided with an overpotential. The overpotential is considered to be variable with metal, current density and environment. In case of reversible hydrogen electrode, the reaction taking place at cathode is 2H+ + 2e– " H2 and the hydrogen overpotential = Emeasured – (–0.059 pH). For most of the metals, this overpotential in between 0.4–0.8 V. Thus, Zn, Fe, Ni and Pb do not evolve hydrogen gas when treated with water. However, when hydrochloric acid is added to the solution, the reaction takes place. The hydrogen overvoltage on zinc is 0.70 V. Thus, at pH = 7, the discharge potential for hydrogen = (–0.059 × 7) – 0.70 = – 1.1137 V, which is greater than the discharge potential for zinc (–0.76 V). But in 1 M acid solution the discharge potential for hydrogen = (–0.059 × 0) – 0.70 = – 0.70 V, and hence hydrogen is liberated. This means that higher the hydrogen overpotential, more difficult is the liberation of hydrogen on the metal surface. Thus, more electropositive metals such as Li (E° = –3.04 V), Na (E° = – 2.71 V) and Ca (E° = –2.87 V) are easily oxidised by water or H+ ions (at pH = 1). By applying Nernst equation to the overall reactions, the spontaneity of a reaction can be determined as well. In case of oxidation of 1 M aqueous solution of ZnSO4, the half reactions are at pH = 7

2H+(aq) + 2e–

H2

E = E° – 0.0591 pH = 0 – 0.0591 × 7 = – 0.4137 V

Zn2+(aq) + 2e–

Zn(s)

E = E° = – 0.76 V

For the overall reaction,

Zn(s) + 2H+(aq)

Zn(aq) + H2(g)

Ecell = (– 0.4137) – (–0.76) = + 0.3463 V This otherwise thermodynamically feasible reaction does not take place as the cell potential is lesser than the hydrogen overpotential. However, it 1 M acid solution, 2H+(aq) + 2e– 2+

–

Zn (aq) + 2e For the overall reaction,

H2

E = E° = 0

Zn(s)

E = E° = –0.76 V

Ecell = 0 – (– 0.76) = + 0.76 V

This reaction easily takes place as the hydrogen overpotential is lesser than the cell potential. Similar is the case of oxygen overpotential. For the reduction reaction, O2(g) + 4H+(aq) + 4e–

2H2O(l)

E°O2,H+/H2O = 1.23 V

E = 1.23 V – (0.0591 pH)V A solution of Co3+ slowly gets reduced by water and liberates oxygen. 4Co3+(aq) + 4e–

4Co2+(aq)

E° = +1.92 V

The Nernst equation for the overall reaction is 4Co3+(aq) + 2H2O(l) At pH = 0,

4Co2+(aq) + 4H+(aq) + O2(g)

Ecell = 1.92 – 1.23 = + 0.69 V

The reaction crosses the oxygen overpotential boundary and is thus favoured.

5.8

Inorganic Chemistry

REDOX REACTIONS IN AQUEOUS SYSTEMS

5.4

The stability of an ion, molecule or atom in aqueous solutions depends upon a number of factors such as presence of other solute, solvent, dissolved oxygen and its own nature. The participation of either the solvent or any other species present in the solution may cause oxidation of the reduced form or reduction of the oxidised form of the species. In case of aqueous solution, the redox stability of a species can be discussed as follows:

1. Reaction with Water Water can act as an oxidising agent and itself is reduced to H2 H2O(l) + e– $ ½ H2(g) + OH–(aq) or

H+(aq) + e– $ ½ H2(g)

E°H+/H2 = 0

The Nernst equation gives E = – (0.0591 × pH) V On the other hand, if water acts as a reducing agent, oxygen is liberated. 2H2O(l) $ O2(g) + 4H+(aq) + 4e–

E°O2, H+/H2O = +1.23 V

The Nernst equation gives E = 1.23 – (0.0591 × pH) V Thus, both the cases are pH dependent and the electrode potential for many reactions varies with pH. Case I: Oxidation by Water (The oxidation of metals by H2O or H+ ions) the reactions with s-block metals, metals of 3d series from groups 3 to 9 or a lanthanoid are thermodynamically favoured and the result is liberation of H2 due to high standard oxidation potentials of these metals. M(s) + H2O(l) $ M+(aq) + ½ H2(g) + OH–(aq) Or

M(s) + H+(aq) $ M+(aq) + ½ H2(g)

Thus, according to the nernst equation, the standard reduction potentials for a metal to be oxidised by water should be more negative than 0V at pH = 0 (1 molar H+ solution), – 0.0591 × 7 = – 0.4137 V at pH = 7 (pure water) and – 0.0591 × 14 = – 0.8274 V at pH = 14 (1 molar OH– solution). Hence, metals such as Na, K, Mg, Al and Zn are thermodynamically active to oxidation by water. However, Mg, Al and Zn do not generate hydrogen in water and can remain in water for long. If we consider the cell potential for oxidation of Mg, Mg + 2H2O $ Mg2+ + 2OH– + H2 2H+ + Mg2+ $ Mg2+ + H2 At pH = 7,

E°Mg2+/Mg = – 2.37 V

E°H+/H2 = 0 – 0.0591 × 7 = – 0.4137 V Ecell = (– 0.4137) – (–2.37) = + 1.9563 V

Since the cell potential overcomes the hydrogen overpotential, fresh surface of Mg can be oxidised with water, but the surface gets protected due to formation of protective hydroxide coating. As a result, the metal get passivated. If this layer is dissolved by acid or any other solvent, hydrogen liberation continues. On the other hand, alkali and other alkaline earth metals (except Be) do not form such protective layers and hence are vigorously oxidised by water. Case II: Reduction by H2O It is clear from the high value of the E°O2, H+/H2O that reduction by acidified water can take place only in presence of strong oxidising agents. For example, Co3+(aq) (E°Co3+/Co2+ = 1.92 V) is reduced by water and oxygen is liberated.

Redox Reactions

5.9

H2O $ 4H+ + O2 + 4e– According to the Nernst equation, E = 1.23 V – (0.0591 pH) V Thus, thermodynamically, water can be oxidised by metals with standard reduction potentials more positive than + 1.23 V at pH = 0 (1 molar H+ solution), 1.23 – (0.0591 × 7) = + 0.8163 V at pH = 7 (pure water) and 1.23 – (0.0591 × 14) = 0.4026 V at pH = 14 (1 molar OH– solution). Thus, species such as Ag2+ (E°Ag2+/Ag+ = 1.980 V), MnO–4 (E°MnO–4/Mn2+ = +1.51 V), Au+ (E°Au+/Au = +1.83 V), Au3+ (E°Au3+/Au = +1.52 V) should oxidise water in acid media. However, due to kinetic reasons (transfer of 4 electrons and a high activation barrier for the formation of O2), very few species can be reduced effectively by H2O. It is clear from the above discussion that water is a poor reducing agent at pH = 0 (better oxidising agent) and better reducing agent at higher pH (poor oxidising agent). Thus, a good reducing agent or a good oxidising agent (w.r.t. H2O) cannot prevail in water. However, both reduction and oxidation of water are pH dependent with same slope of – 0.059 V (in the plot of E° vs pH) and the pH dependence of reduction and oxidation of water can be combined to give the range of potential and pH values indicating the redox stability of water. This plot is known as the stability field of water as shown in Fig. 5.3.

Fig. 5.3 Redox stability diagram of water

It is evident that the stability field of water is confined within the boxed region of the two parallel lines of the slope of the redox reactions. The upper line corresponds to the high potential limit for the redox couples that would oxidise water to O2, while the lower line corresponds to the lowest potential limit for the redox couples that would reduce H2O to H2. Thus, the redox couples that are thermodynamically stable in H2O lie within this boxed region and can neither reduce nor oxidise water. Further, two vertical lines mark the pH limit values (at pH = 4 and pH = 9) commonly found in natural water. The stability field of water is extended on the upper side and lower side if oxygen overpotential and hydrogen over potentials are considered respectively (as shown by dotted lines in the figure 5.3). The redox stability diagram has special relevance to the corrosion of metals as the metals which lie below the low-potential diagram tend to undergo corrosion (oxidation) and reduce H2O or H+ ions.

2. Reaction with Atmospheric Oxygen If the solution is present in an open beaker or somehow in contact with the atmospheric oxygen, there is a considerable possibility of reaction of the solute with the atmospheric oxygen. For example, an aqueous Fe2+

5.10

Inorganic Chemistry

solution in the absence of oxygen is stable as E°Fe3+/Fe2+ (+ 0.77 V) lies within the redox stability field of water under standard conditions. But as soon as this solution comes in contact with O2, Fe2+ gets oxidised to Fe3+. Thus, 4Fe2+(aq) + O2(g) + 4H+(aq) $ 4Fe3+(aq) + 2H2O(l) At pH = 0, At pH = 7,

E°Fe3+/Fe2+ = + 0.77 V E°O2, H+/H2O = + 1.23 V – (0.0591 pH) V Ecell = [ + 1.23 – (0.0591 × 0)] – 0.77 = + 0.46 V Ecell = [ + 1.23 – (0.0591 × 7)] – 0.77 = + 0.0463 V

Thus, the oxidation of Fe2+(aq) is spontaneous, though slow due to oxygen overpotential. This is the reason of occurrence of iron as Fe3+ in the earth’s crust and sediments deposited from aqueous environments. Similar is the case of atmospheric oxidation of copper to the green basic copper carbonate in a damp environment. In this case, the cell potential of the reaction can be determined as

At pH = 0, At pH = 7,

2 Cu(s) + O2(g) + 4H+(aq) $ 2Cu2+(aq) + 2H2O(l) E°Cu2+/Cu = + 0.34 V EO2, H+/H2O = +1.23 V – (0.059 pH) V Ecell = [+ 1.23 – (0.0591 × 0)] – 0.34 = + 0.89 V Ecell = [+ 1.23 – (0.0591 × 7)] – 0.34 = + 0.4763 V

Thus, the oxidation of Cu to Cu2+ is spontaneous in neutral (pH = 7) as well as acidic (pH = 0) solutions. Further, in the presence of atmospheric carbon dioxide and sulphur dioxide, basic copper carbonate is formed with the involvement of anions in the redox chemistry. When SO2 is emitted into fogs or clouds, the atmospheric oxidation of SO2 leads to the formation of aqueous solutions of SO42– and H+ ions which are precipitated as acid rain. SO42–(aq) + 4H+(aq) + 2e– $ SO2(g) + 2H2O(l)

E° = + 0.16 V

Thus, the reaction is thermodynamically favoured.

3. Disproportionation and Comproportionation Some species which are expected to be stable in aqueous solution because of their standard potentials, turn out to be unstable due to disproportionation. It is a type of redox reaction in which the oxidation number of an element gets simultaneously increased and decreased. Thus, the element itself acts as its own oxidising and reducing agent. For example, the standard potentials of the redox couples Cu+/Cu (+ 0.52 V) and Cu2+/Cu+ (+ 0.16 V) lie within the redox stability field of water. In other words, Cu+ ions can neither oxidise nor reduce water. Even then, the Cu+ ion is unstable in aqueous solution. This fact can be illustrated with the help of the reaction 2Cu+(aq) $ Cu2+(aq) + Cu(s) as the difference of the two half-reaction Cu+(aq) + e– $ Cu(s)

E° = +0.52 V

Cu2+(aq) + e– $ Cu+(aq)

E° = + 0.16 V

The positive value of E°cell (0.52 V – 0.16 V = + 0.36 V) makes the disproportionation of Cu+ highly spontaneous leading to unstability of Cu+ ion in aqueous solution. The disproportion of Fe2+ in aqueous solutions can be represented as 3Fe2+(aq) $ 2 Fe3+(aq) + Fe(s) and is the difference of the two half-reactions Fe2+ $ Fe + 2e– 3+

–

E° = – 0.41 V 2+

2Fe + e $ 2Fe

E° = + 0.77 V

Redox Reactions

5.11

The value of E°cell (– 0.41V – 0.77 V = – 1.18 V ) makes the disproportionation of Fe2+ in aqueous solution unfavorable. Rather, the reverse reaction is spontaneous (E°cell = + 1.18 V) and an aqueous Fe3+ salt would react with Fe(s) to produce Fe2+ ions. The reverse of disproportionation, i.e. the combination of the two species of an element in different oxidation states to form a species of the same element in an intermediate oxidation state, is known as comproportionation. Both the redox couples Ag2+/Ag+ (+ 1.980 V) and Ag+/Ag (+ 0.80 V) lie outside the redox stability field of water so that Ag+2 and Ag+ completely convert to Ag+ in aqueous solution. This is due to spontaneous comproportionation reaction Ag2+(aq) + Ag(s) $ 2Ag+ (aq) E°cell = + 1.18 V

4. Effect of Complex Formation The oxidising action of a species can be altered by complex formation. If the complex formation leads to the formation of a thermodynamically more stable complex with the metal in its higher oxidation state, the metal undergoes oxidation and hence is more reducing. On the other hand, if the thermodynamically more stable complex is formed with the metal in its lower oxidation state, the reduction is favoured and hence the species is more oxidising. For example, [Fe(CN)6]3–(aq) + e– $ [Fe(CN)6]4–(aq) E° = 0.36 V [Fe(H2O)6]3+(aq) + e– $ [Fe(H2O)6]2+(aq) and

3–

–

4–

[Ru(CN)6] (aq) + e $ [Ru(CN)6] (aq) 3+

–

E° = 0.80 V E° = +0.85 V

2+

[Ru(H2O)6] (aq) + e $ [Ru(H2O)6] (aq) E° = +0.25 V Thus, the standard reduction potential of iron increases when H2O acts as a ligand leading to stability of Fe2+ ions in the solution. On the other hand, CN– ligands are comparatively increasing the stability of Fe3+ ions. On the contrary, in case of ruthenium, the stability of Ru+2 ions is increased by CN– ligands in comparison to the increase of the stability of Ru+3 ions by H2O molecules. Similarly, the oxidising action of a species is altered by precipitation which leads to decrease in its concentration and hence decrease in its potential. For example, 1 mol/L Ag+(aq) is a mild oxidising agent (E°Ag+/Ag = 0.80 V, DG° = – 77.2 kJ/mol). But complexation with Cl– ions leads to precipitation of AgCl, thereby decreasing the ion concentration of Ag+ ions. The decrease in potential can be calculated as Ag+(aq) + e– $ Ag(s) DG° = – 77.2 kJ/mol –(Ag+(aq) + Cl–(aq)) $ AgCl(s) DG° = – 55.6 kJ/mol) AgCl(s) + e– $ Ag(s) + Cl–, DG° = (– 77.2) – (–55.6) = –21.6 kJ/mol Thus

E°AgCl/Ag, Cl– =

−∆G° −21600 = = 0.22 V nF 1 × 96500

The decrease in potential can also be illustrated if we consider the decrease in concentration of Ag+ ions by precipitation, say [Ag+] = 0.1 mol/L

 1  +  Ag 

E = 0.80 – 0.0591 × log 

= 0.80 – 0.0591 = 0.741 V Thus, more the decrease in concentration of ions, lesser is the potential, i.e lesser is the tendency of the species for reduction and lesser is its oxidising action.

5.12

5.5

Inorganic Chemistry

DIAGRAMMATIC REPRESENTATION OF POTENTIAL DATA

The relative stabilities of various oxidation states exhibited by an element can be depicted with the help of diagrammatic representation of the redox potential data. These diagrams help in the summarisation of quantitative data for various elements as described in this section.

5.5.1 Latimer Diagrams (Reduction Potential Diagram) A Latimer diagram represents the redox potential data of an element in its different oxidation states written above a horizontal line connecting its various species. The species present at the extreme left represents the most highly oxidised form, while the species on the right side represents the successively lower oxidation states. Thus, the oxidation number of the element in its various oxidation states decreases from left to right and are mentioned under the species, while the E° values (in volts) are mentioned above the connecting line between the species. For example, the Latimer diagram for chlorine in acidic solutions (pH = 0) is represented as +1.20

+1.18

+1.65

+1.67

+1.36

ClO 4− → ClO3− → HClO2  → HClO → Cl2 → Cl − +7

+1

+3

+5

−1

0

The Latimer diagram can be converted to half-reactions by considering the predominant species and balancing the equation using the procedure for balancing redox reactions. – 1.20 Thus, ClO4– + → ClO3 represents the half-reaction ClO4–(aq) + 2H+(aq) + 2e– $ ClO3–(aq) + H2O(l)

E° = + 1.20 V

+1.67

and HClO → Cl2 represents the half-reaction 2HClO (aq) + 2H+(aq) + 2e– $ Cl2(g) + 2H2O(l)

E° = +1.67 V

Since both these half-reactions contain H+ ions, their potentials are pH dependent. Similarly, in basic solution (pH = 14 or pOH = 0), the Latimer diagram for chlorine can be represented as +0.37

+0.30

+0.68

+0.42

+1.36

ClO 4− → ClO3− → ClO2− → ClO − → Cl2 → Cl − +7

+5

+3

+1

0

−1

Applications of Latimer Diagrams (a) Determination of Potential for a Redox Reaction Involving Non-adjacent Species. +0.42 +1.36 Suppose, we want to determine ClO– $ Cl2 $ Cl– . The two half reaction can be written ? as ClO– $ Cl2 ; (+1) (0)

E 10 = + 0.42 V; DG10 = – n1FE10

Cl2 $ Cl– ; (0) (–1)

E 20 = + 1.36 V; DG20 = – n2FE20

ClO– $ Cl–; E30 =?; DG03 = – n3FE03 where n3 = n1 + n2 (+1) (–1) The standard potentials cannot be simply added, rather the equation DG° = – nFE° is used and the overall DG03 is determined as the sum of the individual values of the two linked half-reactions, i.e. DG03 = DG01 + DG02

Redox Reactions

5.13

– n3FE03 = (– n1FE01) + (– n2FE02)

or

n1 E10 + n2 E20 n1 E10 + n2 E20 = n3 n1 + n2 1 × 0.42 + 1 × 1.18 Thus, E03 (ClO–/Cl–) = = 0.89 V 2 (b) Prediction of Disproportionation of a Species The disproportionation reaction of a species can be expressed as the difference of the two half-reactions involved. E03 =

If i.e. where

E0

E0

2 1 M 2 +  → M +  →M

2M+(aq) $ M(s) + M2+(aq) ; E0 = E01 – E02 (i) M+(aq) + e– $ M(s) ; E01 (ii) M2+(aq) + e– $ M+(aq) ; E02

The disproportionation of M+ to M and M2+ will be spontaneous if E0 is positive, i.e. E01 is greater than E02. Thus, if the potential on the right of a species in the Latimer diagram is greater than on the left, the species undergoes disproportionation. Suppose we want to determine the tendency of H2O2 to disproportionate in acidic solutions, i.e.

2H2O2(aq) $ 2H2O(l) + O2(g)

The Latimer diagram corresponding to this reaction is +0.70

+1.76

O2 → H 2 O2 → H 2 O 0

−1

−2

The half-reactions involved are 2H+(aq) + 2e– + H2O2(aq) $ 2H2O

E01 = +1.76 V

O2 + 2H+(aq) + 2e– $H2O2

E02 = +0.70 V

The standard potential for overall reaction, E° = (+ 1.76) – (+ 0.70) = +1.06 V Thus disproportionation of H2O2 is spontaneous in acidic solutions.

5.5.2 Frost Diagram (Oxidation State Diagram) The Frost diagram of an element represents the plot of nE° for a redox couple against the oxidation number of the species involved. Frost diagrams are also termed free energy plots, because – DG° = nFE°, hence nE° = – DG°/F. In other words, these plots indicate the stability of a particular oxidation state of an element. Thus, the species occupying the lowest position in the Frost diagram with lowest nE° value i.e. the most negative DG° value is the most stable as shown in Fig. 5.4.

nE° Most stable oxidation state Oxidation number(n)

Fig. 5.4

Representation of Frost diagram

1. Important Features of Frost Diagram (a) The standard potential of the redox couple represented any two points is given by the slope of the connecting line. Thus, steeper the slope of the line, higher is the standard potential of the concerned couple (Fig. 5.5).

5.14

Inorganic Chemistry

(b) More positive the slope, more positive the E° and hence greater is the tendency of the oxidising agent in the redox couple to reduce and hence it acts as a better oxidising agent under standard conditions. Thus, NO3– (Fig. 5.7) is a good oxidising agent. On the other hand, a less steeper line with less positive slope indicates the lesser standard potential of the redox couple and hence the reducing agent of the couple has more tendency to oxidise and acts as a better reducing agent. (c) A species with its point lying above the connecting line of the two adjacent species it is unstable and undergoes disproportionation (Fig. 5.6).

Higher reduction potential nE° n2 E20 Lower reduction potential

n1 E10 n2

n1 Oxidation number (n)

Fig. 5.5 Frost diagram showing reduction potential tendencies

nE°

nE°

Tendency for disproportionation

Tendency for comproportionation Oxidation number, n

Oxidation number, n

Fig. 5.6 Frost diagram showing tendency for disproportionation and comproportionation

Thus, NH2OH is thermodynamically less stable (Fig. 5.7) than NH3 and N2 and disproportionates as 3NH2OH $ NH3 + N2 + 3H2O On the other hand, N2O is thermodynamically more stable than NH4+ and NO3– ions which can undergo comproportionation as NH4+ (aq) + NO3– (aq) $ N2O(g) + 2H2O(l) However, this reaction is thermodynamically favoured but kinetically unfavoured in solution. But in the solid state, the reaction is both kinetically and thermodynamically favoured. Thus, solid ammonium nitrate undergoes fast explosion once initiated. However, if three adjacent points lie nearly on the same line, these are almost equally stable. Therefore, NO, HNO2 and N2O4 are almost equally stable. In Fig. 5.7, the frost diagram for nitrogen in basic solution (pH = 14) is represented by solid line and in acidic solution (pH =0) represented by dotted line. The NO2– ion is stable to disproportionation in basic solutions but in acidic solutions, NO, HNO2 and N2O4 are present in equilibrium (being equally stable). Thus, nitrites are stable in basic solutions but evolve NO, on acidification, which gets oxidised to the brown NO2. As a result, the disproportionation of NO to N2O and HNO2 is prevented. NO3– is a stronger oxidising agent in acidic solution than in basic solution due to more steeper slope in acidic solutions. Thus, Cu is oxidised by NO3– in acidic medium as E = +0.463 V Cu + 2NO3– + 4H+ $ Cu2+ + N2O4 + 2H2O – If we compare the slopes of the two redox couples, we find that NO3 /N2O4 has more steeper slope with more positive value than the slope of Cu/Cu2+. Thus, NO3– would oxidise Cu and since Cu2+ is thermodynamically more stable than Cu+, Cu2+ is formed instead of Cu+ (Fig. 5.8).

Redox Reactions

5.15

7 NO3– 6 N2O4 5 HNO2 4 NH2OH

NO N2O4

3 NH3 N 2 H4

2

NH3OH+

N2O

NO

1 NO2–

N2O

NO3–

+

N 2 H5

0

N2 NH4+ –3

–2

–1

0

+1

+2

+3

+4

+5

Fig. 5.7 Frost diagram for N2

2. Construction of Frost Diagram The Frost diagrams can be constructed using Latimer diagrams, as a plot of nE versus oxidation number ranging from n " 0. The predominant redox couples are considered and the nE° values are calculated and plotted. Suppose we want to construct a Frost diagram for oxygen from the Latimer diagram +0.70 +1.76 H2O 2 $ H2O 2 O2 $ 0

–1

–2

+1.23 V Fig. 5.8 Frost diagram for copper and HNO3

Thus, corresponding to O2 $ H2O2;

n1 = –1

and n1E01 = –0.70 V

and

n2 = –2

and n2E02 = – 2 × 1.23 = – 2.46 V

O2 $ H2O;

Now we can plot the values, as shown in fig. 5.9.

5.5.3 Pourbaix Diagram (Potential-pH Diagram) A Pourbaix diagram indicates the range of potential and pH in which a given species is thermodynamically stable. This diagram is used for depiction of redox stability of various metals in natural water in context of corrosion. Since iron is the most important metal of concern, we will illustrate the construction of Pourbaix diagram for iron as shown in Fig. 5.10 1. The reduction half-reaction Fe3+(aq) + e– $ Fe2+ (aq) ; E° = +0.77 V is independent of pH due to no participation of H+ ions and is represented by a horizontal line AB which separates the dominating

5.16

Inorganic Chemistry

Fig. 5.9 Frost diagram for oxygen in acidic medium

FeO42–

B´ Fe3+

O2/H2O

+1.0 +0.8 A

B

+0.6 Fe(OH)3(s)

+0.4

Fe2+

C

E (V) +0.2 0

D

–0.2 –0.4

H2O/H2

–0.6

C´

Fe 0

2

4

6 pH

8

Fe(OH)2(s) 10

12

14

Fig. 5.10 Pourbaix diagram for some iron species present in natural water

region of Fe3+ and Fe2+ ions. Hence, in presence of a redox couple with standard potential above this line, will oxidise iron and Fe3+ will be the major species. 2. The reaction Fe3+(aq) + 3H2O (l) $ Fe(OH)3 (s) + 3H+(aq) is not a redox reaction due to no change in oxidation number of any element and hence does not depend on the potential represented by a vertical line BB . This reaction involves H+ ions and hence depends upon pH, Fe3+(aq) exists in low pH while Fe(OH)3(s) prevail in high pH. Thus, at pH = 3, Fe3+ becomes dominant and as the pH is increased, Fe(OH)3 become dominant. 3. The reaction Fe(OH)3(s) + 3H+(aq) + e– $ Fe2+(aq) + 3H2O(l) is also pH dependent and its potential is given by E = E° – (0.0591) log [Fe2+] – (3 × 0.0591) pH As can be seen from the figure 5.10, as pH is increased, potential falls linearly, as shown by the line BC. Thus, above the line BC, Fe(OH)3(s) is stable while below the line BC, (FeOH)2(s) is stable.

Redox Reactions

5.17

4. The reaction Fe2+(aq) + 2H2O(l) $ Fe(OH)2(s) + 2H+(aq) is again not a redox reaction but it depends upon pH change as there is liberation of H+ ions. The change is shown by the vertical line CC (at pH = 9) which divides the regions, as low pH region favours the formation of Fe2+ and the high pH region favours the formation of Fe(OH)2(s). 5. The redox reaction Fe(OH)3(s) + H+(aq) + e– $ Fe(OH)2(s) + H2O(l) also depends upon pH due to consumption of H+ ions. Thus, the variation of potential with pH can be given by Nernst equation as E = E° – (0.0591) pH The change is depicted by the line CD separating the region of Fe(OH)3(s) and Fe(OH)2(s) dominance. The position of the graph below the H2O/H2 line contains Fe in stable form while the portion of the graph above the O2/H2O lines contains FeO42– species. Further, it is evident from the graph that Fe3+ ions can prevail in water below pH = 4 and in presence of plentiful O2. However, the pH of natural water varies between 4 and 9, hence iron exists mainly as Fe(OH)3 in natural water. However, the rich organic-matter and water-logged soils are deficient of oxygen as a lot of oxygen is consumed in biodegradation of organic matter. Thus, organic matter has a strong reducing character and it converts any Fe(OH)3 into Fe2+ at pH around 4.5. Thus, organic-matter rich and water-logged soils mainly contain Fe2+ ions.

5.5.4 Ellingham Diagram The Ellingham diagram represents the plot of DG° of formation of oxide per mole of oxygen consumed versus temperature. It helps in the identification of the temperature required for reduction of a metal oxide by carbon or carbon monoxide. This diagram has been illustrated in Chapter 7.

Redox reactions consist of simultaneous loss of electrons (oxidation) and gain of electrons (reduction). The oxidised substance acts as a reducing agent while the reduced substance is known as oxidising agent. The chemical energy produced during an indirect redox reaction can be converted into electrical energy with the help of an electrochemical cell or galvanic cell. The electrode potential of an electrode can be defined as the tendency of an electrode to lose or gain electrons when it is in contact with its own ions in solution and is represented in terms of reduction potential by convention. The standard electrode potential is determined with the help of standard hydrogen electrode. The standard reduction potential of all the elements can be arranged in the form of a series known as electrochemical series. The electrode potential of an electrode can be related with its standard electrode potential by the Nernst equation.

0.0591 log Q n [C ]c [ D]d where E°cell = E°cathode – E°anode and Q = [ A]a [ B]b The feasibility of a chemical reaction can be determined by using the relation DG° = – nFE° and the equilibrium constant for a reaction at equilibrium can be determined by using a relation 0.0591 log K E°cell = n The reactions which involve release of gases are kinetically slower than expected due to requirement of some extra potential known as hydrogen overpotential and oxygen overpotential in case of H2 and O2 gases. Ecell = E°cell –

5.18

Inorganic Chemistry

The standard reduction potentials for a metal to be oxidised by water should be more negative than 0V at pH = 0, –0.4137 at pH = 7 and –0.8274 V at pH = 14. On the other hand, the standard reduction potential for a metal to be reduced by water should be more positive than +1.23 V at pH = 0, +0.8163 V at pH = 7 and +0.4026 V at pH = 14. A species may be unstable in aqueous solution due to disproportionation while two unstable species of an element may combine by comproportionation to give a stable species. The plot of redox potential data of an element in its different oxidation states links a horizontal line known as Latimer diagram and contains the most oxidised state on the extreme left while the successively lower oxidation states on the right side. The Frost diagram represents the plot of nE° for a redox couple against the oxidation number of the species involved. The Frost diagram contains the most stable oxidation state at the lowest position while the least stable oxidation state at the highest position. The Pourbaix diagram indicates the range of potential and pH in which a given species is thermodynamically stable. This diagram is used for depiction of redox stability of various metals in natural water.

EXAMPLE 1 An electrochemical cell is prepared by placing a copper rod in 1 M aqueous solution of CuSO4 and a nickel rod is 1 M aqueous solution of NiSO4. Determine the standard electrode potential and give the representation of the cell. Given, E°Cu2+/Cu = 0.34 V and ENi2+/Ni = –0.25 V. For a spontaneous reaction, the standard electrode potential should be positive. Thus, a copper electrode will act as the cathode and a nickel electrode will act as the anode. E°cell = E°cathode – E°anode = 0.34 – (–0.25) = + 0.59 V. The cell can be represented as Ni | NiSO4 (1 M) || CuSO4(1 M) | Cu (It should be noted that since reduction potential of nickel is more negative, it can displace copper from its solution).

EXAMPLE 2 Can an aqueous solution of ferrous ions be oxidised with H2O2? Given E°Fe3+/ Fe2+

= +0.77 V and E°H2O2/H2O = +1.77 V.

If the reaction can take place, the half-reaction can be represented as At anode:

Fe2+(aq)

Fe3+(aq) + e–

At cathode:

H2O2 + 2H+(aq) + 2e–

Overall reaction

2Fe2+(aq) + H2O2 + 2H+(aq)

E°Fe2+/Fe3+ = + 0.77 V 2H2O

E°H2O2/H2O = +1.77 V 2Fe3+(aq) + 2H2O

E° = E°Cathode – E°anode = 1.77 – 0.77 = +1.00 V Since the E°cell is positive, ferrous ions can be oxidised with H2O2.

Redox Reactions

5.19

EXAMPLE 3 Calculate the cell potential for the following cell reaction at 298 K 2 Ag+ (0.1 M) + Cu(s) $ Cu2+ (4.0 M) + 2Ag(s) Given E°Cu2+/Cu = + 0.34 V and E°Ag+/Ag = + 0.80 V

The Nernst equation can be written as

[ 2+ ][ Ag] 0.0591 log Cu 2 n  Ag +  [ Cu ] 2

Ecell = E°cell –

0.0591 (4)(1)2 [Ag(s)] = [Cu(s)] = 1] log n (1) (0.1)2 = 0.46 – 0.0295 log 400 = 0.46 – 0.0295 × 2.6020 = 0.46 – 0.0767 = 0.3833 = (+ 0.80 – 0.34) –

EXAMPLE 4

Calculate the equilibrium constant and the decrease in Gibb's free energy for the following cell reaction at 298 K.

2Fe3+ + 3I–

2Fe2+ + I3–

Given

E°Fe3+/Fe2+ = 0.77 V and E°I3–/ I– = 0.54 V 2Fe2+ + I3–

2Fe3+ + 3I– The half-reactions are

2Fe3+ + 2e– $ 2Fe2+ (+3)

(+2)

3I+ $ I3– + 2e–

(–1)

(- 13 )

Thus, n = 2 The value of equilibrium constant can be determined from the relation E°cell = or or \

0.0591 log K n

0.0591 log K 2 0.23 × 2 log K = = 7.783 0.0591

(0.77 – 0.54) =

K = Antilog (7.783) = 6.62 × 107

EXAMPLE 5 Calcuale the value of E° for the following cell reaction Au3+ + 3e– $ Au. Also construct the Frost diagram and latimer diagram for the reactions given below: Au+ + e– $ Au E° = 1.7 V

Given:

Au3+ + 2e– $ Au+

E° = 1.4 V

Au+ + e– $ Au

E°1 = 1.7 V,

n1 = 1

Au3+ + 2e– $ Au+

E°2 = 1.4 V,

n2 = 2

5.20

Inorganic Chemistry

Required: Using the relation

Au3+ + 3e– $ Au DG03 –n3E03 –3 × E03

=

DG01

=

–n1E01

+

E°3 = ?,

n3 = 3

DG02 + (–n2E02)

= 1 × 1.7 – 2 × 1.4 = – 4.5

−4.5 = 1.5V −3 The Latimer diagram can be represented as E03 =

or

+1.4 V +1.7 V Au3+ $ Au+ $ Au

+1.5 V

Fig. 5.11

The Frost diagram can be represented as shown in Fig. 5.11.

QUESTIONS Q.1 Discuss electrochemical cell in detail with the help of a suitable example. Q.2 Define the following terms: (a) Oxidation (c) Overpotential

(b) Standard electrod potential (d) Redox couple

Q.3 What do you mean by electrochemical series? Discuss its important applications. Q.4 Write short notes on (a) Redox stability field of water (b) Hydrogen overpotential (c) Nernst equation(d) Disproportionation Q.5 Give reasons for (a) Sodium reacts with water to liberate hydrogen while aluminium remains safe with water. (b) Co3+ oxidises water (c) Organic-matter rich water-logged soil contains Fe2+ ions while oxygen-rich water contains Fe3+ ions Q.6 Draw a Frost diagram for mercury in acid solution and comment on the tendency of the involved species to undergo disproportionation. Given E°Hg2+/Hg2+ = 0.911 V and E°Hg2+ /Hg = 0.796 V 2

2

Q.7 Determine the reduction potential E°Sn4+/Sn if E°Sn2+/Sn = – 0.136 V and E°Sn4+/Sn2+ = 0.15 V. Also draw the corresponding Latimer diagram. Q.8 Discuss the Pourbaix diagram for iron in detail. Q.9 Calculate the equilibrium constant for the possible reaction between Fe2+(aq) ions and Ce4+(aq) ions if E°Ce4+/Ce3+ = 1.44 V and E°Fe3+/Fe2+ = 0.68 V. Also comment on the spontaneity of the reaction. Q.10 Calculate the concentration of silver ions if the concentration of Cu2+ ions is 0.01 M in a copper–silver system. Given E°Cu2+/Cu = 0.337 V and E°Ag+/Ag = 0.799 V.

Redox Reactions

5.21

MULTIPLE-CHOICE QUESTIONS 1. If E°Cu2+/Cu = 0.34 V and E°Cu2+/Cu+ = 0.15 V then E° for the reaction 2Cu+(aq) equal to (a) 0.38 V (b) – 0.38 V (c) + 0.49 V 2. The potential of hydrogen electrode in a solution of pH 7 at 25°C is (a) 0 (b) – 0.4137 V (c) – 0.8274 V 3. The element which can oxidise H2O is (a) sodium (b) beryllium (c) silver 4. The species which can act as an oxidising agent for water is (a) Ag2+ (b) Au3+ (c) both (a) and (b) 5. Iron is found in natural water as (a) Fe2+ (b) Fe3+ (c) Fe(OH)3

Cu2+(aq) + Cu(s) is (d) – 0.49 V (d) + 1.23 V (d) gold (d) none (d) Fe(OH)2

chapter

Non-aqueous Solvents

6

After studying this chapter, the student will be able to

6.1

INTRODUCTION

Most of the inorganic reactions are studied in aqueous solution, water being the best-known solvent due to its high value of dielectric constant. However, many non-aqueous solvents have also been studied extensively in the last few years. A large number of non-aqueous polar solvents such as liquid ammonia, glacial acetic acid, sulphuric acid, sulphur dioxide and hydrogen halides have been introduced. The behaviour of a solvent is determined on the basis of several physical properties as discussed below:

1. Dielectric Constant For an ionic crystal, coulombic force of attraction F between the cation and anion is given by the expression : F=

q1q2

Dr 2 where q1 and q2 are the charges on the cations and anions respectively, r is the distance between the cation and the anion and D is a constant, known as dielectric constant. D depends on the nature of the solvent used to dissolve the ionic crystal. The expression shows that if a solvent has higher value of D, it will reduce the F, i.e. it will weaken the attractive forces in the crystal and will dissolve it quickly. Thus, solvents with a high value of dielectric constant such as anhydrous hydrogen fluoride and water can easily dissolve an ionic compound as compared to a solvent with low value of dielectric constant (suitable for dissolving nonpolar compounds).

6.2

Inorganic Chemistry

2. Dipole Moment Greater is the polarity of a solvent or higher the dipole moment of the solvent, greater is the solvation energy released and hence greater is the solubility of a solute. In general, an ionising solvent has high values of both dipole moment and dielectric constant, as evident from Table 6.1.

3. Melting Point and Boiling Point

Table 6.1 Dipole moment and dielectric constants of some ionising solvents Solvent

Dipole moment (Debye units) 2.93 1.91 1.84 1.69 1.63 1.46

HCN HF H2O C2H5OH SO2 NH3

Dielectric constant 106.8 (25°C) 83.6 (0°C) 78.5 (25°C) 24.2 (25°C) 17.4 (–19°C) 22.0 (–34°C)

The range of temperature in which a solvent can exist as a liquid is indicated by its melting point and boiling point. A good solvent must have a wide range of existence as a liquid. It is evident from Table 6.2 that water and sulphuric acid hold a wide range of temperature and exist as liquid while the other listed solvents can be used only at low temperatures. Table 6.2 Melting points, boiling points and critical constants of some solvents Solvent H2O

m. pt. (°C)

b. pt. (°C)

Critical Temperature (°C)

Critical Pressure (atm)

0

100.0

374.0

217.7

10.4

300.0

–

–

HF

–89.4

19.5

230.2

91.5

SO2

–75.5

–10.1

157.5

77.7

NH3

–77.7

–33.5

132.4

112.0

N2O4

–11.2

21.1

158.2

100

H2SO4

4. Heat of Fusion and Vaporisation The nature and strength of the intermolecular forces in the Table 6.3 Molar heats of fusion and vaporisation of various solvents solid and the liquid state are indicated by their molar heats of fusion and vaporisation respectively. The high values of Molar heat of Solvent Molar heat these parameters indicate the high intermolecular forces. vaporisation of fusion The values of these parameters have been given in Table 6.3 (kJ mol–1) (kJ mol–1) for some solvents. It can be concluded from the table that the 7.40 24.93 SO2 intermolecular forces are weakest in HF and greatest in SO2. 6.02 40.65 H2O The Trouton constant is normally used to account for the 5.65 23.34 NH3 intermolecular forces in normal liquids. It is the ratio of the HF 4.58 30.28 heat of vaporisation (J) to the boiling point (K) of a solvent. Its value comes out to be 90 JK–1 mol–1 for normal liquids, i.e. the liquids without any association. The polar liquids undergo association and have a high value of the Trouton constant. Such solvents include H2O, HF, NH3 and alcohols.

6.2

CLASSIFICATION OF SOLVENTS

The solvents can be classified in a number of ways as follows:

Non-aqueous Solvents

6.3

6.2.1 Depending upon the Proton–Donor and Proton–Acceptor Ability 1. Protonic or Protic Solvents The solvents which can either donate and or accept protons depending upon the nature of the other species are known as protonic or protic solvents. All these solvents have hydrogen atoms in their formula. These are further of three types: (a) Protogenic or Acidic Solvents These solvents can donate protons readily. For example, HF, H2SO4, CH3COOH, etc. (b) Protophilic or Basic Solvents These solvents can accept protons readily. For example, NH3, C5H5, NH2NH2, etc. (c) Amphi-protic or Amphoteric Solvents These solvents can show dual character, i.e. can donate or accept protons depending upon the nature of the other species. For example, H2O, CH3COOH, NH3, etc.

2. Aprotic or Non-protonic Solvent These solvents can neither donate nor accept protons irrespective of presence/absence of hydrogen in their formula. For example, C6H6, SO2, CCl4, CHCl3, etc.

6.2.2 Depending upon the Polarity of the Solvent 1. Polar, Ionising or Ionic Solvent The solvents which have high polarity and can undergo auto-ionisation, are called polar, ionising or ionic solvents. These solvents also have high dielectric constants and dissolve ionic compounds. These solvents have high tendency to undergo association which leads to increase in boiling point and hence the liquid-state range of the solvent. For example, H2O, NH3, HF, SO2, etc. The auto ionisation of these solvents can be shown as H2O + H2O

H3O+ + OH–

NH3 + NH3

NH4+ + NH2–

HF + HF SO2 + SO2

HF2+ + F– SO2+ + SO32–

2. Nonpolar or Non-ionising Solvents The solvents which are nonpolar and hence do not ionise at all, are called nonpolar or non-ionising solvents. These solvents have dielectric constants and very little associating tendency. These solvents cannot undergo auto-ionisation and cannot dissolve ionic solutes but dissolve nonpolar solutes. For example, C6H6, CCl4, etc.

6.2.3 Third Classification 1. Aqueous Solvent Water is known as aqueous solvent. 2. Non-aqueous Solvent All other solvents are known as non-aqueous solvents. For example, NH3, SO2, HF, CHCl3, C6H6, etc. Some important non-aqueous solvents will be discussed in this chapter.

6.3

LIQUID AMMONIA

Liquid ammonia is one of the most extensively used non-aqueous solvents in various inorganic and organic reactions. It exhibits many characteristics similar to those of water except that its dielectric constant is lower

6.4

Inorganic Chemistry

than that of water. Thus, it is a poor solvent for ionic substances but at the same time, a better solvent for nonpolar molecules. Some substances show a higher solubility in liquid ammonia due to formation of stable ammine complexes.

6.3.1 Comparative Account of H2O and Liquid NH3 as Solvents 1. Dielectric constant of liquid NH3 (22) is lower than that of water (78.5). 2. Its freezing point (–77.7°C) and boiling point (–33.5°C) are much lower than the freezing point (0°C) and boiling point of H2O (100°C). Thus, the liquid range of NH3 is very less. 3. Dipole moment of NH3 (1.47) is lower as compared to that of water (1.85). 4. Viscosity of liquid NH3 (2.65 millipoise) is much less than that of water (10.08 millipoise). However, this effect promotes the ionic mobilities and compensates the effect of low dielectric constant to some extent. The auto-ionisation of H2O and liquid ammonia can be compared as H2O + H2O

H3O+

+

Hydronium ion

NH3 + NH3

NH4+ 2NH4+

Kw = 1.0 × 10–14 (25°C)

Hydroxyl ion

+

Ammonium ion

NH3 + 2NH3

OH–; NH2–

Kb = 1.9 × 10–33 (–50°C)

Amide ion

+ NH2– Imide ion

2NH3 + 2NH3

3NH4+

+ N3– Nitride ion

Thus, auto-ionisation of NH3 is analogous to that of H2O; however the extent of ionisation is very less. Any substance which increases the concentration of NH4+ ion in liquid NH3 is known as an ammono acid, while the substance which increases the concentration of NH2–, NH2– or N3– ions in liquid NH3 is known as an ammono base. The ammono acid and ammono base combine together to form un-ionised NH3. This is known as the process of neutralisation.

1. Ammono Acids The examples include ammonium salts, organic amide, acetic acid, sulphamic acid etc. NH4X NH+4 + X– RCONH2 + NH3 $ RCONH– + NH4+ CH3COOH + NH3 $ CH3COO– + NH4+ H2N . SO2 . OH + 2NH3 $ –HN . SO2 . O– + 2NH4+

2. Ammono Bases The examples include amides, imides and nitrides – + 3 KNH2 liq.NH  → K + NH2 3 PbNH + 2NH4Cl liq.NH  → PbCl2 + 3NH3 3 BiN + 3NH4Cl liq.NH  → BiCl3 + 4NH3 In aqueous medium, a compound is said to be amphoteric if it dissolves both in acids and bases. Similar is the case in liquid ammonia too. For example, zinc hydroxide is amphoteric in aqueous medium. 2O Zn(OH)2 + 2HCl H → ZnCl2 + 2H2O

Non-aqueous Solvents

6.5

2O Zn(OH)2 + 2NaOH H → Na2ZnO2 + 2H2O Similarly, zinc amide is amphoteric in liquid NH3 3 Zn(NH2)2 + 2NH4Cl liq.NH  → ZnCl2 + 4NH3

liq.NH

3 Zn(NH2)2 + 2NaNH2  → Na2Zn(NH)2 + 2NH3

Thus, amphoterism in liquid NH3 is analogus to that in H2O.

6.3.2 Chemical Reactions in Liquid Ammonia The chemical reactions that take place in liquid NH3 are as follows:

1. Metathetical or Precipitation Reaction The precipitate formation depends upon the solubility of a substance in the solvent. As already discussed, due to difference in physical properties of liquid NH3 and H2O, various substances differ in their solubilities in these solvents. Thus, many precipitation reactions that are normally not possible in water, may take place in liquid NH3. Some particular examples are the following: (a) In aqueous medium, KCl + AgNO3 $ AgCl . + KNO3 In liquid NH3 liq.NH3 KNO3 + AgCl  → KCl . + AgNO3 liq.NH

Similarly,

3 KI + NH4Cl  → KCl . + NH4I

and

3 AgCl + Ba(NO3)2  → BaCl2 . + 2AgNO3

liq.NH

This is due to the reason that most of the Chlorides (except NaCl, BeCl2 and NH4Cl) are insoluble in liquid. NH3. (b) Metal iodides and bromides can be precipitated by treatment of solutions of various nitrates in liquid NH3 with ammonium halides. liq.NH

3 → ZnI2 . + 2NH4NO3 Zn(NO3)2 + 2NH4I 

liq.NH

3 Sr(NO3)2 + 2NH4Br  → SrBr2 . + 2NH4NO3

(c) (NH4)2S in liquid NH3 is used to precipitate the sulphides of many metals such as Mg, Ba, Pb, Bi, Mn, Cu, Zn, Cd, Hg from their nitrate solutions. liq.NH

3 → Ag2S . + 2NH4NO3 2AgNO3 + (NH4)2S 

liq.NH

3 Cd(NO3)2 + (NH4)2S  → CdS . + 2NH4NO3

2. Reactions of Ammono Acids (a) Neutralisation KNH2 + NH4Cl $ KCl. + 2NH3 (b) Reaction with Active Metals The solutions of ammonium salts in liquid NH3 can react with active metals to give hydrogen analogous to that in aqueous medium. In aqueous medium, Co + H2SO4 $ CoSO4 + H2 In liquid NH3,

liq.NH

3 Co + 2NH4NO3  → Co(NO3)2 + 2NH3 + H2

6.6

Inorganic Chemistry

Blue solutions of alkali metals in liquid NH3 get decolourised on treatment with ammonium salts. + 3 Na + NH4+ liq.NH  → Na + NH3 + ½ H2 (c) Protololysis Some compounds are incapable of donating protons to water but can readily donate protons in liquid NH3 (protolysis). Examples include urea, acetamide, acetamidine and sulphamide. These are weakly basic in aqueous solution but behave as acid in liquid NH3

NH2CONH2 + NH3 $ –HNCONH2 + NH+4 CH3CONH2 + NH3 $ CH3CONH– + NH+4 CH3C=NHNH2 + NH3 $ CH3C=NHNH– + NH+4 H2NSO2NH2 + NH3 $ H2NSO2NH– + NH+4 The basicity of sulphamic acid is increased in liquid NH3. In aqueous medium, H2NSO2OH + H2O $ H2NSO2O– + H3O+ In liquid NH3, H2NSO2OH + 2NH3 $ –HNSO2O– + 2NH+4 CH3COOH, a weak acid in aqueous solution, behaves as a strong acid in liquid NH3. CH3COO– + NH+4 CH3COOH + NH3 AgNH2, a weak base in aqueous solution, exhibits acidic property in liquid NH3. [Ag(NH2)2]– + NH+4

AgNH2 + 2NH3

3. Reactions of Ammono Bases Ammono bases are used to precipitate amides, imides and nitrides of many metals. The general reaction can be represented as – NH2 + NH3 $ – NH+3 + NH2– Amide

= NH + NH3 $ = NH2+ + NH2– Imide

NH+ + NH2–

N + NH3 $ Nitride

It is evident that due to release of NH2– in all these reactions, the amide, imide and nitride are all treated as ammono bases. liq.NH

3 → KNO3 + AgNH2. KNH2 + AgNO3 

liq.NH

3 KNH2 + PbI2  → KI + HI + PbNH.

liq.NH

3 KNH2 + BiI3  → KI + 2HI + BiN .

4. Solvolysis Reactions in Liquid NH3 In a solvolysis reaction, the solvent splits into its characteristic ions and concentration of either of these ions increases due to interaction with cations or anion of a salt. Solvolysis in H2O, alcohol and liq.NH3 are known as hydrolysis, alcoholysis and ammonolysis respectively. For example, in hydrolysis, concentration of either H3O+ or OH– is increased by auto-ionisation of H2O. Consider the hydrolysis of CuSO4 2H3O+ + 2OH–

Auto-ionisation of H2O:

2H2O + 2H2O

Ionisation of CuSO4:

CuSO4 $ Cu2+ + SO42–

Non-aqueous Solvents

Interaction of cation with characteristic ion of solvent:

6.7

Cu2+ + 2OH– $ Cu(OH)2 _________________________________________ Cu(OH)2 + SO42– + 2H3O+ weak base _________________________________________

CuSO4 + 4H2O

Similarly, in ammonolysis, concentration of either NH4+ in or NH2– ion is increased. Auto-ionisation of H2O: Ionization of SiCl4: Interaction of cation with characteristic ion of solvent:

4NH3 + 4NH3 4NH+4 + 4NH2– 4+ SiCl4 $ Si + 4Cl– Si4+ + 4NH2– $ Si(NH2)4 + 4NH4+ + 4Cl– _____________________________________ SiCl4 + 8NH3 $ Si(NH2)4 + 4NH4+ + 4Cl–

Weak base _____________________________________ Some another examples of ammonolysis are given here:

(a) Ammonolysis of Inorganic Halides Ionic halides do not undergo ammonolysis in liq.NH3, but many covalent halides tend to ammonolyse. AlCl3 + 2NH3 $ AlCl2(NH2) + NH4+ + Cl– BX3 + 6 NH3 $ B(NH2)3 + 3NH4+ + 3X– Hg2Cl2 + 2NH3 $ Hg(NH2)Cl + Hg + NH4+ + Cl– SO2Cl2 + 4NH3 $ SO2(NH2)2 + 2NH4+ + 2Cl– (b) Ammonolysis of Organic Halides Alkyl halides undergo slow ammonolysis with liquid NH3 at its boiling point to give a mixture of primary, secondary and tertiary amines. (X = Cl, Br, I) RX + 2NH3 $ RNH2 + NH2X 2RX + 3NH3 $ R2NH + 2NH4+ + 2X– 3RX + 4NH3 $ R3N + 3NH4+ + 3X– Similarly,

NH

NH +4

CH3 – CONH2 + NH3  → CH3 — C

+ H2 O NH2

(c) Ammonolysis of Cl2 and POCl3 Cl2 + 2NH3 $ Cl(NH2) + NH4+ + Cl–

POCl3 + 6NH3 $ PO(NH2)3 + 3NH4 + 3Cl– (d) Ammonolysis of Alkali Metal Hydrides and Oxides NaH + NH3 $ NaNH2 + H2 Na2O + 2NH3 $ 2NaNH2 + H2O

Analogous to

NaH + H2O $ NaOH + H2 Na2O + H2O $ 2NaOH

5. Solvation Reactions in Liquid NH3 In a solvation reaction, the solute and solvent species get attached to each other by a coordinate or H-band to form an addition compound or adduct known as solvate (Lewis acid-base reaction). Solvation in H2O and NH3 are known as hydration and ammoniation respectively. Likewise, their adducts are known as hydrate (for H2O) and ammoniate (for NH3). For example,

6.8

Inorganic Chemistry

BF3 + NH3 $ BF3.NH3 SO + 2NH $ SO .2NH

(1 : 1 adduct)

3 3 3 3 (1 : 2 adduct) SiF4 + 2NH3 $ SiF4.2NH3 Some other examples of ammoniation reaction are:

Cu2+ + 4NH3 $ [Cu(NH3)4]2+ Co2+ + 4NH3 $ [Co(NH3)6]2+ Ag+ + 2NH3 $ [Ag(NH3)2]+ + 2H2O

6. Complex Formation in Liquid NH3 In aqueous system, many compounds react with the excess of alkalies to form soluble complexes. For example, 2O Zn(NO3)2 + 4NaOH H → Na2[Zn(OH)4] + 2Na NO3 2O Zn(OH)2 + 2NAOH H → Na2[Zn(OH)4] 2O AlCl3 + 4NaOH H → Na[Al(OH)4] + 3NaCl

Similarly, in liquid NH3, many compounds react with the excess of an ammino base to from soluble complexes—amido and/or imido. For example : 3 (a) Zn(NO3)2 + 4KNH2 liq.NH  → K2[Zn(NH2)4] + 2KNO3 3 Zn(NO3)2 + 4KNH2 liq.NH  → K2[Zn(NH2)] + 2KNO3 + 2NH3 3 (b) Zn(OH)2 + 2NaNH2 liq.NH  → Na2[Zn(NH2)4] 3 Zn(OH)2 + 2NaNH2 liq.NH  → Na2[Zn(NH2)] + 2NH3 3 (c) AgNH2 + KNH2 liq.NH  → K[Ag(NH2)2] 3 AgNH2 + KNH2 liq.NH  → K[Ag(NH)] + NH3 3 (d) Zn(NH2)2 + 2NaNH2 liq.NH  → Na2[Zn(NH2)4] 3 Zn(NH2)2 + 2NaNH2 liq.NH  → Na2[Zn(NH2)2] + 2NH3 3 (e) AlCl3 + 4 NaNH2 liq.NH  → Na[Al(NH)2]4 + 3NaCl

7. Redox Reaction in Liquid NH3 Many oxidising agents act as weaker oxidising agents in liquid ammonia than in aqueous solutions e.g. KMnO4 is reduced to K2MnO4 in presence of KNH2 as 3 6KMnO4 + 6KNH2 liq.NH  → 6K2MnO4 + 4NH3 + N2

However, if instead of KNH2, potassium in liquid ammonia is used, KMnO4 is reduced to MnO. 6KMnO4 + 30K + 20NH3 $ 6MnO + 18KNH2 + 18KOH + 3H2 + N2 Excess of K in liquid NH3 can reduce K2[Ni(CN)4] and [Pt(NH3)4]Br2 as 3 K2[Ni(CN)4] + 2K liq.NH  → K4[Ni(CN)4] 3 [Pt(NH4)]Br2 + 2K liq.NH  → [Pt(NH3)4] + 2KBr

Non-aqueous Solvents

6.9

K in liquid NH3 also reduces nitrous oxide to N2 as 3 2K + NH3 + N2O liq.NH  → KNH2 + KOH + N2

liq.NH

3 3 Reaction with oxygen takes place as K + O2  → KO2 ; 2K + O2 liq.NH  → K2O2 Oxidising action of iodine is also weaker in liquid NH3 3 I2 + K4[Sn(NH2)6] liq.NH  → K2[Sn(NH2)6] + 2KI

Sodium in liquid NH3 is able to reduce various substances as liq.NH

3 2Na + S  → Na2S 3 Na + CuI liq.NH  → Cu + NaI 3 2Na + O2 liq.NH  → Na2O2 Liquid NH3 itself acts as a reducing agent for many subsances

2NH3 + 3Mg $ Mg3N2 + 3H2 2NH3 + 3CuO $ N2 + 3Cu + 3H2O

6.3.3 Solubility of Various Substances in Liquid NH3 Liquid NH3 is a poor solvent for ionic substances due to its low dielectric constant. However, most ammonium salts are soluble in water. Such examples include CH3COONH4, NH4NO3 etc. Similarly, most of the perchlorates, cyanides, thiocynates, nitrites and nitrates are soluble in liquid ammonia. Amongst the halides, the fluorides are least soluble and solubility increases from fluorides to iodides. Thus, chlorides (except BeCl2 and NaCl) are insoluble, whereas the iodides are freely soluble. Most of the bromides are slightly soluble in liquid NH3. Sulphides, phosphates, oxides, hydroxides and carbonates are insoluble in liquid NH3. Most of the metal amides are insoluble, except those of alkali metals. Many metal salts dissolve in liquid NH3 due to the formation of ammoniates. Some particular examples are [Cu(NH3)4]2+, [Ni(NH3)6]2+, [Fe(NH3)6]2+, [Cr(NH3)6]3+, [Hg(NH3)2]2+ and [Pt(NH3)4]2+, etc Many organic compounds such as halogen compounds, alcohols, ketones, esters, phenols, simple ether, amines are soluble in liquid NH3. However, alkanes are insoluble, whereas alkenes, alkynes and aromatic hydrocarbons are sparingly soluble. The nonmetals dissolve in liquid NH3 by reacing with it. 16NH3 + 45S8 $ 6(NH4)2S + N4S4 -35∞C to - 75∞C 7NH3 + 3I2ææææææ Æ NI3.3NH3 + 3NH4I

The metals with low ionisation energies, high energies of solvation and low energies of sublimation, alkali and alkaline earth metals (except Be) readily dissolve in liquid ammonia. In case of alkali metals, the solubility increases lower the group as Li < Na < K < Cs (Table 6.4). Table 6.4 Solubility of alkali metals in liquid ammonia Metal

Temperature (°C)

Concentration of saturated solution (m)

Temperature (°C)

Concentration of saturated solution (m)

Li

0

16.31

–33.2

15.60

Na

0

10.00

–33.5

10.93

K

0

12.4

–33.2

11.86

Cs

0

–

–50.0

25.1

6.10

Inorganic Chemistry

The solutions of alkali metals in liquid NH3 have very extraordinary properties. This dissolution of alkali metals in liquid NH3 probably does not involve a chemical change in the metal, as the metal is regenerated when the solution is evaporated. (The alkaline earth metals are recovered as hexa-amoniates). All the solutions are blue in dilute form, but bronze coloured when the concentration exceeds 1 M. On further increase of concentration, the colour disappears. The dissolution can be represented in terms of the production of solvated metal ions and electrons produced by ionisation of the metal atoms. M + (x + y) NH3 $ M(NH3)x+ + e–(NH3)y It is believed that the solvated electrons reside in the cavities surrounded by the ammonia molecules (with their protons oriented towards the electrons (Fig. 6.1). This representation attributes to the low density of these solutions as compared to that of the solvent. The Fig. 6.1 Representation of solvated electrons blue colour of the solution is due to the presence of these solvated cations and electrons. The dilute solution has very high electrical conductivity in the range of fully ionised salts. As the concentration is increased, the conductivity decreases and is characteristic of the metal. The dilute solutions are paramagnetic due to presence of solvated electrons, but as the concentration of the solution is increased, the magnetic succeptibility decreases and finally the solution becomes diamagnetic. It is believed that with an increase in concentration, the solvated cations tend to form aggregates such as M2, M3, etc. The very dilute solutions are metastable and undergo decomposition in the presence of a catalyst.

n H + M(NH2)n 2 2 The decomposition reaction also takes place slowly upon long standing of the solutions. Due to presence of ammoniated electrons, the blue solution of alkali metals in liquid NH3 are strongly reducing. M + nNH3 $

6.3.4 Advantages of Liquid Ammonia as a Solvent 1. It is quite clear with the above discussion that use of liquid NH3 as a solvent for dissolution of alkali metals is highly useful as the dissolved alkali metals can be recovered easily from the solution by simple evaporation. 2. The dilute blue solutions of alkali metals are very strong reducing agents and can be used for reducing the substances soluble in liquid NH3. 3. Many metallic salts that cannot be precipitated in aqueous medium are easily obtained. 4. Liquid ammonia has very less tendency for solvolysis of dissolved solutes.

6.3.5 Disadvantages of Using Liquid Ammonia as a Solvent 1. Liquid range for ammonia lies in between –33.5°C to –77.7°C. Hence, low temperature or high pressure has to be maintained while working with it. 2. Liquid ammonia is hygroscopic in nature, hence the reactions are carried out in sealed tubes so as to avoid contact with moisture. 3. Due to offensive odour of liquid NH3, special techniques are required to use it as a solvent and reaction medium.

6.4

LIQUID SULPHUR DIOXIDE

Liquid SO2 is a nonprotonic solvent as it contains no hydrogen atoms and hence does not give a proton (H+)

Non-aqueous Solvents

on ionisation. The concept of acids and bases based upon the solvent system was first given for sulphur dioxide. The auto-ionisation of SO2 was suggested as SO2 + SO2

SO2+

+

(Thionyl ion)

SO2– 3 (Sulphite ion)

6.11

Table 6.5 Physical properties of liquid sulphur dioxide Property

Value

Trouton’s constant

22.7

Dipole moment

1.61 Debye

Dielectric constant

17.4 (–20°C)

Freezing point

–75.5°C

The thionyl ion is analogous to the hydronium Boiling point –10°C ion, while the sulphite ion is analogous to the Specific condutance 3.4 × 10–8 ohm–1 cm–1 (–10°C) hydroxyl ion and the amide ion. Although SO2 is a gas under normal Density 1.46 g ml–1 (–10°C) temperature and pressure, yet it is extensively Viscocity 0.428 centipoise (–10°C) used in industries as it can be readily liquified. Enthalpy of fusion 1.97 kcal mol–1 Its liquid range is sufficiently high (–10°C to Enthalpy of vaporisation 5.96 kcal mol–1 –75.5°C) and hence it can be used as a solvent for many covalent substances. Due to its low dielectric constant (17.4 at –20°C), it is a poor solvent for ionic compounds, however, the highly charged ionic species are soluble in this solvent. This is due to the presence of the electrons in the SO2 molecules which leads to formation of ven der Waals interactions between the solute and solvent molecules.The physical properties of liquid SO2 are given in Table 6.5.

6.4.1 Solubility of Substances in Liquid SO2 Iodides and thiocyanates of alkali metals are the most soluble. The solubilities of other halides decreases in the order MI > MBr > MCl > MF. Most of the ammonium, thallium and mercuric salts are insoluble. Most of the sulphites, cyanides and acetates are moderately soluble. Many covalent halides such as PBr3, BCl3, AlCl3, AsCl3, CCl4, SiCl4, GeCl4, SnCl4, IBr are soluble in liquid SO2.PCl3, POCl3, and SOCl2 are highly soluble in accordance with the hard soft interaction principles. Aliphatic hydrocarbons are insoluble but the aromatic hydrocarbons and alkenes are highly soluble. In fact, liquid SO2 is used as a reaction medium in a number of organic reactions and for solvent extraction method to separate the aromatic and aliphatic hydrocarbons.

6.4.2 Conductivity of Salt Solutions in Liquid SO2 The dissociation and conductivity of electrolytes in liquid SO2 is roughly related to the size of cation as Na+ < NH4+ < K+ < Rb+ < (CH3)3S+ < (CH3)4N+ and – for various anions as SCN– < ClO4– < Cl– < Br– < I– < SbCl6 . Solutions of many highly soluble covalent compounds are highly conducting in nature.

6.4.3 Chemical Reactions in Liquid SO2 1. Acid–Base Reactions In accordance with the auto-ionisation of liquid SO2, the compounds which contain or increase the concentration of SO32– ions act as bases, while the compounds which contain or increase the concentration of SO2+ act as acids in liquid SO2. SO2 + SO2

SO2+ + SO32–

Thus, SOCl2, SOBr2, SO(SCN)2 etc. behaves as acid in liquid SO2 while the substances such as K2SO3, Cs2SO3, [N(CH3)4]2 SO3 act as a base.

6.12

Inorganic Chemistry

The acid base reaction in liquid SO2 can be represented as SO2 SOCl2 + Cs2SO3 liq.  → 2CsCl + 2SO2 SO2 SOBr2 + [N(CH3)4]2SO3 liq.  → 2[N(CH3)4]Br + 2SO2

2. Precipitation Reactions In accordance with the specific solubility relationships, several insoluble materials can be precipitated in liquid SO2 as mentioned below: (a) Metal chlorides are precipated from the solutions of soluble metallic salts and SOCl2 in liquid SO2. SO2 2KI + SOCl2 liq.  → 2KCl . + SOI2 SO2 2Ag(CH3COO) + SOCl2 liq.  → 2AgCl . + SO(CH3COO)2

From these reactions, many new thionyl derivatives are obtained. (b) Precipitates of other compounds have also been obtained. SO2 AlCl3 + 3NaI liq.  → 3NaCl -+AlI3 SO2 SbCl3 + 3LiI liq.  → SbI3 . + 3LiCl SO2 PbF2 + Li2SO4 liq.  → PbSO4 . + 2LiF SO2 BaI2 + Zn(CNS)2 liq.  → Ba(CNS)2 . + ZnI2

3. Solvolysis Reactions in Liquid SO2 The solvolysis reactions are quite typical and differ from the true solvolysis reactions. (a) Many covalent halides form oxyhalides during solvolysis and increase the concentration of SO2+ ions. PCl5 + SO2 (liq.) $ POCl3 + SOCl2 AsCl5 + SO2 (liq.) $ AsOCl3 + SOCl2 UCl6 + 2SO2 (liq.) $ UO2Cl2 + 2SOCl2 WCl6 + SO2 (liq.) $ WOCl4 + SOCl2 (b) CH3COONH4 increases the concentration of SO32– ion during solvolysis in liquid SO2. 2CH3COONH4 + 2SO2 (liq.) $ (NH4)2SO3 + (CH3CO)2O + SO2 (c) Zinc diethyl reacts in a typical way. Zn(C2H5)2 + SO2 (liq.) $ ZnO + (C2H5)2SO

4. Solvation Reactions in Liquid SO2 Liquid SO2 can act as a lewis base and forms a variety of solvates containing one or more molecules of sulphur dioxide. Some typical examples are LiI . 2SO2, KBr . 4SO2, NaI . 4SO2, KI . 4SO2, RbI . 4SO2, SrI . 4SO2, AlCl3.2SO2, K(SCN) . 2SO2, Cs(SCN) . SO2, [(CH3)4N] . 3SO2, Dioxane . 2SO2.

5. Complex Formation in Liquid SO2 Due to presence of lone pair of electrons on both the S and O atoms, it can form many complexes, (a) Many compounds react with a compound containing SO32– ions to form precipitates, which dissolve in excess of SO32– due to formation of soluble complexes.

Non-aqueous Solvents

Reaction with AlCl3

6.13

SO2 2 AlCl3 + 3K2SO3 liq.  → Al2(SO3)3 . + 6KCl SO2 Al2(SO3)3 + 3K2SO3 liq.  → 2K3[Al(SO3)3]

Similarly, soluble sulphito complex, [N(CH3)4]3[Al(SO3)3], is obtained by treating AlCl3 with excess of [N(CH3)4]2SO3. Reaction with ZnCl2 SO2 ZnCl2 + K2SO3 liq.  → ZnSO3 + 2KCl SO2 ZnSO3 + K2SO3 liq.  → 2KCl + K2[Zn(SO3)2]

Similarly, SO2 SbCl3 + KCl liq.  → K3[SbCl6] SO2 SbCl5 + KCl liq.  → K[SbCl6] SO2 SbCl5 + NOCl liq.  → [N(CH3)4][SbCl6]

(b) Solubility of many compounds in SO2 is increased greatly on addition of KI or PbI due to formation of soluble comlexes. SO2 KI + I2 liq.  → KI3 SO2 RbI + I2 liq.  → RbI3 SO2 2KI + HgI2 liq.  → K2[HgI4]

6. Amphoteric Behaviour in Liquid SO2 Many compounds whose hydroxides are amphoteric in aqueous medium behave in an analogous way in liquid SO2. AlCl3 reacts with NaOH to give gelatinous precipitate of Al(OH)3. 2O AlCl3 + 3NaOH H → Al(OH)3 . + 3 NaCl

gelatinous ppt.

The precipitates dissolve in excess of NaOH to form the soluble complex. 2O Al(OH)3 + NaOH H → Na[Al(OH)4]

(excess)

Soluble complex

Al(OH)3 can be reprecipitated from the soluble complex by the addition of HCl. 2O Na(Al(OH)4) + HCl H → Al(OH)3 . + NaCl + H2O Similarly, in liquid SO2, the reaction takes place as

SO2 2AlCl3 + 3K2(SO3) liq.  → Al2(SO3)3 + 6KCl SO2 Al2(SO3)3 + 3K2 SO3 liq.  → 2K[Al(SO3)3]

Al2(SO3)3 can be reprecipitated as SO2 2K3[Al(SO3)3] + 3SOCl2 liq.  → 6KCl + Al2(SO3)3 + 6SO2

7. Redox Reactions in Liquid SO2 Liquid SO2 does not have redox properties like gaseous SO2 and is used only as a medium for redox reactions. Interestingly, I2 is reduced by a sulphite in liquid SO3.

6.14

Inorganic Chemistry

SO2 I2 + K2SO3 liq.  → 2KI + K2SO4 + SO2 SO2 I2 + 2R2SO3 liq.  → 2RI + R2SO4 + SO2

While, a soluble iodide is oxidised by SbCl5 or FeCl3 in liquid SO2. SO2 6KI + 3SbCl5 liq.  → 3I2 + SbCl3 + 2K3[SbCl6] SO2 2KI + 2FeCl3 liq.  → I2 + 2FeCl2 + 2KCl

Table 6.6 Comparison of H2O, liquid NH3 and liquid SO2 as solvents. Property Nature

H2O

Liquid NH3

Protonic

Protonic

Liquid SO2 Non protonic

Liquid range

0°C to 100°C

Conditions for use

Can be used at ordinary temperature and pressure

Can be used only at low temperature and high pressure

Can be used at low temperature and high pressure

Solubility of Alkali metals

No solubility

High solubility

No solubility

Amphoteric behaviour

Zn(OH)2 and Al(OH)3 are amphoteric

Zn(NH2)2 and Al(NH2)3 are amphoteric

Al2(SO4)3, AlCl3 are amphoteric

6.5

–33.5°C to –77.3°C

–10°C to –75°C

ANHYDROUS HYDROGEN FLUORIDE

Anhydrous hydrogen fluoride is an excellent ionising solvent. Because to its special characteristics, HF persists as (HF)6 in the vapour phase and forms chains and rings of various sizes due to H-bonding. Its boiling point is 19.4°C and freezing point is –89.4°C. Thus, it has a wide liquid range compared to other non-aqueous solvents. It has high dielectric constant (83.6 at 0°C) and a dipole moment (1.90 D) quite close to that of water. However, due to its low specific conductance, poisonous character and ability to dissolve only a few substances without any chemical reaction limits its use as a non-aqueous solvent. Table 6.7 lists the characteristic physical properties of HF. Table 6.7 Physical properties of Anhyd. HF.

Auto-ionisation of HF Liquid HF is highly ionised as HF + HF $ H2F+

Fluoronium ion

+

F–

K = 8 × 10–12

Fluoride ion

Due to H-bonding in HF, the reaction can be represented as 2HF + HF $ H2F+ + HF2–

Property

Value

Freezing point Boiling point Dipole moment Dielectric constant Density Viscocity Specific conductance

–89.4ºC +19.4°C 1.90D 83.6 (0°C) 0.99 g ml–1 0.256 centipoise (15°C) 1.4 × 10–5 ohm–1 cm–1 (15°C)

Thus, any substance which increases the concentration of H2F+ will act as an acid and the substance which increase the concentration of F– will act as a base in HF. However, even acids which appear strong in aqueous solution, behave as bases in HF. For example,

HNO3 + HF $ H2NO3+ + F– H2SO4 + HF $ H2SO4+ + F –

Non-aqueous Solvents

6.15

HClO4, the strongest acid in aqueous solution, acts as amphoteric in HF. Thus, As a base,

HClO4 + HF $ HClO4+ + F–

As an acid,

HClO4 + HF $ ClO4+ + H2F+

Electron-aceptor fluorides such as BF3, AsF3, SnF4, PF5 and SbF5 act as an acid in HF. BF3 + 2HF $ H2F+ + BF4– SnF4 + 4HF $ SnF62– + 2 H2F+ SbF5 + 2HF $ H2F+ + SbF6– These solutions can dissolve electropositive metals such as Mg. Ionic fluorides such as the fluorides of alkali metals, alkaline earth metals, silver and thallium (I) act as bases in HF as the concentration of F– ions is increased and that of H2F+ ions is decreased. NaF + HF $ Na+ + HF2– (HF and F–) KH2F3 has also been formed with HF. The acids which are weak in aqueous solutions behave as bases in HF. For example : CH3COOH + HF $ CH3COOH2+ + F–

6.5.1 Chemical Reactions in Liquid HF 1. Precipitation Reactions Sulphates, periodates and perchlorates of non-alkali metals are easily precipitated in liquid HF. HF 2AgF + Na2SO4 liq.   → Ag2SO4 . + 2NaF HF TlF + KClO4 liq.   → TlClO4 . + KF HF AgF + KIO4 liq.   → AgIO4 . + NaF

AgNO3 + BF3 + 2HF $ AgBF4 + H2NO3F

2. Protonation of Organic Compounds Organic compounds such as alkanes, benzene, ethers, ethanol, aldehydes, ketones, carboxylic acids, etc. give conducting solutions in HF due to protonation on dissolution. The solubility is increased further by addition of fluoride acceptors (BF3, SbF5). Alkane

(CH3)3CCH3 + HF $ (CH3)3CCH4+ + HF2–

Benzene

C6H6 + 2HF

C6H7+ + HF2–

Ether

R2O + 2HF

R2OH+ + HF2–

Ethanol

C2H5OH + 2HF $ C2H5OH2+ + HF2–

Ketones

R2CO + 2HF $ R2COH+ + HF2–

Carboxylic Acids

HCOOH + 2HF $ HCOOH2+ + HF2–

3. Formation of Addition Compounds Metallic fluorides form a number of addition compounds in the HF. For example, KF.2HF, KF.3HF and NH4F.HF.

6.16

Inorganic Chemistry

4. Redox Reaction Metals and hydrofluoro acids undergo redox reactions in HF. HF 3HAsF6 + 2Ag liq.   → 2AgAsF6 + AsF3 + 3HF

These reactions are analogous to those taking place between metal and HNO3 in aqueous solutions. 4HNO3 + 3Ag æaq æÆ 3AgNO3 + NO + 2H2O

5. Solvolysis Reactions Simple salts undergo solvolysis as KCN + HF $ HCN - + K+ + F– KCl + HF $ HCl - + K+ + F– Some salts undergo solvolysis which is followed by further reaction as follows: (a) KNO3 + HF $ HNO3 + K+ + F– HNO3 + 2HF $ H2NO3+ + F– H2NO3+ + HF $ NO2+ + H3O+ + F– (b) H2SO4 + 2HF $ H2SO4 + 2K+ + 2F– H2SO4 + HF $ HSO3F + H2O H2O + HF $ H3O+ + F–

6.5.2 Solubility of Some Compounds in Liquid HF AlF3 is slightly soluble in liquid HF. However, the solubility increases on addition of NaF. liq. HF

AlF3 + NaF  →Na[AlF4] AlF3 is precipitated back on the addition of BF3. Na[AlF4] + BF3 $ AlF3 . + NaBF4 Similarly, solubility of CrF3 in HF is enhanced in presence of NaF. CrF3 + 3NaF $ Na[CrF6] CrF3 is precipitated back on the addition of RF3. Na3[CrF6] + 3BF3 $ CrF3 . + 3NaBF4

6.6

ANHYDROUS SULPHURIC ACID

Anhydrous sulphuric acid is a highly associated solvent due to presence of H-bonding. As a result, it has a high boiling point and high viscosity, even 25 times more than that of water. Thus, the solutes dissolve very slowly in this solvent and are not easily crystallised. Hence, it is not preferred as a solvent for usual reactions. It is rather used as a dehydrating agent to extract water from chemical compounds. 2SO 4 C12H22O11 H → 12C + 11H2O

1. Auto-protolysis of H2SO4 H2SO4 undergoes auto-protolysis as H2SO4 + H2SO4

H3SO4+ + HSO4–

K = 2.7 × 10–14

Thus, any substance which increases the concentration of H3SO4+ acts as an acid and the substance which increases the concentration of HSO4– acts as a base in H2SO4. However, there are other species formed in the following dissolution equilibria:

Non-aqueous Solvents

H2SO4

6.17

H2O + SO3 H3O+ +HSO4–

H2O + H2SO4

SO3 + H2SO4

K=1

H2S2O7 H3SO4+ + HS2O7–

H2S2O7 + H2SO4 +

H2SO4 + H2SO4

H3O + HS2O7–

K = 5.1 × 10–7

H2SO4 + H2SO4

H3SO4+ + HSO4–

K = 1.4 × 10–2

2. Substances as Base in H2SO4 The substances which behave as a base in water are also basic in H2SO4. NH3 + H2SO4 $ NH4+ + HSO4– OH– + 2H2SO4 $ H3O+ + 2HSO4– The substances which dissociate to produce HSO4– ions act as a base in H2SO4. KHSO4 $ K+ + HSO4– The substances with lone pairs of electrons such as amides behave as bases in H2SO4, as they accept protons from it. H2NCONH2 + H2SO4 $ H2NCONH3+ + HSO4– The substances which act as an acid in aqueous solutions (including weak acids and strong acids) act as bases in H2SO4. H3PO4 + H2SO4 $ P(OH)4+ + HSO4– HNO3 reacts with H2SO4 as HNO3 + H2SO4 $ HSO4– + H2NO3+ H2NO3+ + H2SO4 $ NO2+ + H3O+ + HSO4– _____________________________________________ HNO3 + 2H2SO4 $ 2HSO4– + NO2+ + H3O+ _____________________________________________ Acetic acid readily accepts a proton from H2SO4. CH3COOH + H2SO4 $ CH3COOH2+ + HSO4– However, a number of carboxylic acids decompose in H2SO4 to give acyl ion. RCOOH + 2H2SO4 $ RCO+ + H3O+ + 2HSO4– For example,

(CH3)3C6H2.COOH + H2SO4 $ (CH3)3C6H2.COOH2+ + HSO4– Mesitioic acid

(CH3)3C6H2.COOH2+ + H2SO4 $ (CH3)3C6H2CO+ + H3O+ + HSO4– Mesitoic acid is recovered back if water is added, while addition of methanol gives methylester (impossible to obtain via direct reaction). Some carboxylic acids decompose to give carbon monoxide in H2SO4. HCOOH + H2SO4 $ CO + H3O+ + HSO4– COOH + H2SO4 $ CO + CO2 + H3O+ + HSO4– COOH Except alkanes, most organic compounds dissolve in H2SO4 and are easily protonated. Water easily accepts protons from H2SO4. H2O + H2SO4 $ HSO4– + H3O+

6.18

Inorganic Chemistry

3. Substances as Acids in H2SO4 Perchloric acid, chlorosulphuric acid and fluorosulphuric acid behave as weak acids in sulphuric acid. HClO4 + H2SO4 $ ClO4– + H3SO4+ HClSO3 + H2SO4 $ ClSO3– + H3SO4+ Sulphur dioxide and borontris(hydrogensulphate) are moderately strong acids in H2SO4. SO3 + 2H2SO4 $ H3SO4+ + HS2O7– B(HSO4)3 + 2H2SO4 $ H3SO4+ + B(HSO4)4– Similar is the case of fuming sulphuric acid (oleum) and hydrogen tetrakis(hydrogensulphato)borate. H2S2O7 + H2SO4 $ H3SO4+ + HS2O7– HB(HSO4)4 + H2SO4 $ H3SO4+ + B(HSO4)4– HB(HSO4)4 can be obtained in solution form by dissolving boric acid in sulphuric acid followed by addition of excess of fuming sulphuric acid. H3BO3 + 6H2SO4 $ B(HSO4)4– + 2HSO4– + 3H3O+ B(HSO4)4– + 3H2S2O7 $ HB(SO4)4 + 8H2SO4 The extremely strong acids which are even more acidic than H2SO4, are known as super acids. Mixtures of very strong Lewis acids such as SbF5 and very acidic protonic solvents such as HSO3F, HF, CF3SO3H and H2S2O7 are the particular examples of such super acids. A mixutre of SbF5 in HSO3F is termed magic acid and it can protonate even an extremely weak base.

H3C

H3 C

H 3C

+

–

C — CH3 + SO3F – SbF5

C = CH2 + HSO3F + SbF5 $ H3C

6.6.1 Chemical Reactions in Anhydrous H2SO4 1. Redox Reactions in H2SO4 Anhydrous sulphuric acid acts as a mild oxidising agent. Chalcogens give colored solutions with H2SO4 2+ (red) and Se82+ (green). Similarly, I3+ is obtained as containing cations such as S82+ (blue), S16 HIO3 + 7I2 + 8H2SO4 $ 5I3+ + 3H3O+ + 8HSO4–

2. Solvolysis in H2SO4 SeO2 undergoes slow solvolysis in H2SO4 as SeO2 + H2SO4 $ SeO(OH).HSO4 $ SeO.OH+ + HSO4– Some organic compounds such as nitriles, nitro compounds, sulphones and sulphoxides also undergo slow solvolysis in H2SO4 as RCN + H2SO4 $ RCNH+ + HSO4– RCNH+ + HSO4– $ RCONH2 + SO3

6.7

ACETIC ACID

Acetic acid is a commonly used laboratory solvent. It is non-ionic, nontoxic and stable in ordinary conditions. It has a wide liquid range between 16.6 and 118°C. Acetic acid exists as a dimer due to formation of H-bond and is an associated solvent. This association results in its high boiling point and unique properties. It has

Non-aqueous Solvents

6.19

zero dipole moment and very low dielectric constant (7.1), yet a large number of ionic compounds are soluble in this solvent. Many compounds solubilise in acetic acid due to H-bond formation.Thus, it is a good solvent for many organic and inorganic compounds. Auto-ionisation of acetic acid can be represented as CH3COOH2+ + CH3COO–

2 CH3COOH

Thus, substances giving or increasing the concentration of CH3COOH2+ are regarded as acids and those giving or increasing the concentration of CH3COO– are considered bases. All strong acids in aqueous medium act as acids in acetic acid; HClO4 being the strongest. HClO4 + CH3COOH $ CH3COOH2+ + ClO4– Weak organic bases such as acetamide, amines, acetanilide and heterocyclic bases behave as strong bases in acetic acid. B + CH3COOH $ CH3COO– + BH+ Weak organic base

RNH2 + CH3COOH $ RNH3+ + CH3COO– C5H5N + CH3COOH $ C5H5NH+ + CH3COO– Acetic acid is commonly used as a solvent medium for titration of weak organic bases, which is not possible in aqueous medium. A typical acid base reaction in acetic acid can be represented as CH3COONa + HCl $ NaCl + CH3COOH Salts of alkali and alkaline earth metals with organic acid act as bases in acetic acid. CH3COONa $ CH3COO– + Na+ Zinc acetate shows amphoteric behaviour in acetic acid. 2– 3COO Zn2+ + 2 CH3COO– $ Zn(CH3COO)2 CH  → [Zn(CH3COO)4] −

6.7.1 Chemical Reactions in Acetic Acid 1. The precipitates of Many compounds can be obtained in acetic acid as Acetic acid

BaI2 + 2NaNO3  → Ba(NO3)2. + 2NaI Acetic acid

CaCl2 + 2AgNO3  → Ca(NO3)2 + 2AgCl. Acetic acid

2CH3COONa + Cu(NO3)2  → Cu(CH3COO)2 + 2NaNO3 Acetic acid

Pb(CH3COO)2 + 2KCl  → PbCl2. + 2 CH3COOH Acetic acid

Cd(CH3COO)2 + H2S  → CdS. + 2 CH3COOH 2. Complex formation reactions in acetic acid take place as CH COOH

3−

3 Fe3+ + 6CNS−  → [ Fe (CNS)6 ]

Insoluble red-coloured complex

3. Solvolysis reactions in acetic acid can be exemplified as : CN– + CH3COOH $ CH3COO– + HCN SO2Cl2 + 4CH3COOH $ SO2(CH3COO)2 + 2CH3COOH2+ + 2Cl– Acetic acid forms solvates such as CH3COOK . 2CH3COOH.

6.20

6.8

Inorganic Chemistry

LIQUID DINITROGEN TETROXIDE, N2O4

Liquid dinitrogen tetroxide, N2O4, has been used extensively as a protic non-aqueous solvent, because of its convenient liquid range (–11.2°C to 21.1°C) and the ease of its preparation. It has a very low dielectric constant (2.42) and hence is a poor solvent for polar substances but good solvent for nonpolar substances. The low specific conductance of the solvent (1.3 × 10–13 ohm–1 cm–1 at 10°C) indicates that the self-ionisation of N2O4 is extremely small. However, the specific conductance increases on addition of a polar solvent such as nitromethane or by the addition of a donor, thereby increasing its self-ionisation. The chemistry of N2O4 can be rationalised in terms of the equation corresponding to its self-ionisation as 2 N2O4

2NO+ + 2NO3–

Nitrosyl ion Nitrate ion.

Thus, any substance furnishing the NO+ ion would behave as an acid and those furnishing NO3– would behave as a base in liquid N2O4. Hence, NOCl, NOBr, etc., behave as an acid in N2O4, while AgNO3, [Et2NH2]NO3, etc., behave as a base in N2O4. The acid-base neutralisation reaction in liquid N2O4 can be represented as NOCl + AgNO3 $ AgCl + N2O4 (Analogous to the neutralisation reactions in NH3 and H2O) liq.NH

3 KNH2 + NH4Cl  → KCl + 2NH3

HO

2 HCl + NaOH  → NaCl + 2H2O

Chemical Reactions in Liquid N2O4 1. Reaction with Metals The behaviour of liquid N2O4 with metals is analogous to that of water. Very active metals (liquid Na, K) react with liquid N2O4 readily. M + N2O4 (l) $ MNO3 + NO M + H2O $ MOH + ½ H2 Less active metals (Zn, Sn, Fe) are less reactive with liquid N2O4 and the reactivity increases with the addition of NOCl. Likewise the reactivity of these metals with H2O increases with the addition of HCl. N2 O4 M + 2NOCl liq.  → MCl2 + 2NO 2O M + 2HCl H → MCl2 + H2 (M = Zn, Sn, Fe)

The behaviour of the amphoteric metals such as Zn and Al towards liquid N2O4 is analogous to that in aqueous solutions and ammonia Zn + x[Et2NH2]NO3 + 2N2O4(l) $ [Et2NH2]x [Zn(NO3)x+2] + 2NO (undetermined composition)

Similarly, Zn(NO3)2 reacts with diethylammonium nitrate to form nitratozincates. N2 O4 Zn(NO3)2 + x[Et2NH2]NO3 liq.  → [Et2NH2]x [Zn(NO3)x+2] N2 O4 Zn + 2NOCl liq.  → ZnCl2 + 2NO

Non-aqueous Solvents

6.21

In aqueous solutions, Zn + 2NaOH + 2H2O $ Na2[Zn(OH)4] + H2 2O Zn(OH)2 + 2NaOH H → Na2[Zn(OH)4]

In liquid ammonia, Zn + 2KNH2 + 2NH3 $ K2[Zn(NH2)4] + H2 3 Zn(NH2)2 + 2KNH2 liq.NH  → K2[Zn(NH2)4]

2. Solvolytic Reactions Some solvolytic reactions in liquid N2O4 are [Et2NH2]Cl + N2O4(l) $ [Et2NH2]NO3 + NOCl [Mg(H2O)6]Cl2 + N2O4(l) $ [Mg(H2O)6](NO2)2 + 2NOCl Li2CO3 and KCl are solvolysed in presence of traces of water. of H 2O KCl + N2O4(l) traces  → NOCl + KNO3. of H 2O Li2CO3 + 2N2O4(l) traces  → 2LiNO3 + N2O3 + CO2Thus, liquid N2O4 is used to prepare anhydrous nitrates of metals which are otherwise difficult to obtain in aqueous systems. Metal carbonyls are also solvolysed with liquid N2O4

Mn2(CO)10 + N2O4(l) $ Mn(CO)5NO3 + Mn(CO)4NO + CO Some solvolysis reaction results in the formation of complexes. Ca + 3N2O4 $ [Ca(NO3)2.N2O4] + 2NO AlCl3 + 4N2O4 $ NO[Al(NO3)4] + 3NOCl Decomposition of these complexes yields the otherwise inaccessible anhydrous nitrates.

3. Solvate Formation Many metals and nitrates of Zn, Fe and U are known to form solvates in N2O4. These solvates are formulated as complex salts. N O (l )

2 4 → Cu(NO3)2 . N2O4 Cu

N2 O4 (l )

Zn(NO3)2 → Zn(NO3)2 . 2N2O4 N2 O4 (l )

Fe(NO3)3 → Fe(NO3)3 . N2O4

or

NO[Cu(NO3)3]

or

(NO)2[Zn(NO3)4]

or

NO[Fe(NO3)4]

N2O4 is a powerful oxiding agent and reacts explosively with organic substances. Hence, these reactions are carried out with extreme care.

6.9

MOLTEN SALTS AND IONIC LIQUIDS

Molten salts have been used extensively as non-aqueous solvents since 1960s. The term molten salt was initially used for a salt which is solid at standard temperature and pressure but exists in liquid phase at elevated temperature. But some molten salts exist as liquid at standard temperature and pressure and were called ambient molten salts while those earlier known were called high temperature molten salts. Later on, the definition was changed to include all the salts which melt without any decomposition or vapourisation.

6.22

Inorganic Chemistry

Depending upon the melting point and working range of temperature, molten salts have been classified into two broad categories : high-temperature molten salts and ionic-liquids.

1. High-temperature Molten Salts This category includes those salts and eutectics which melt above 100°C to give liquids, without any salt decomposition or vaporisation. These salts and eutectics may be ionic or covalent compounds and can be divided into two broad categories: (a) Ionic High-temperature Molten Solids The category includes those salts and eutectics whose melts are ionic. Examples of such salts are fused alkali metal salts and the ionic metal halides of Group IIA, Al and the nitrates of transition metals with comparatively much lower m.pt. than the corresponding individual alkali metal salt. For example, eutectic of LiF (830°C), NaF (880°C) and KF (912°C) melts at 454°C. Solution of alkali metal salts in liquid AlCl3 also gives an ionic hightemperature molten salt, such as NaAlCl4 (152°C). The ionic melt behaves as a good electrolyte and behaves normally to cryoscopy. Thus, the number of ions of BaF2 and CaBr2 is 3 in molten NaCl. NaCl 2+ – ���� � BaF2 � ��� � Ba + 2F NaCl 2+ – ���� � CaBr2 � ��� � Ca + 2Br

However, the salt, with common ion as that of the ionic salt behaves anomalously. For example, the number of ions given by NaF in NaCl is 1, while CaBr2 in CaCl2 is 2. NaCl + – ���� � NaF � ��� � Na + F CaCl2 2+ – ���� CaBr2 � ���� � Ca + 2Br

Ionic melts dissolve the solutes appreciably and this feature can be accounted either on the basis of inclusion of solutes in the ‘hole’ of the ionic structures as exemplified by the noble gases in molten nitrates or due to chemical interaction as exemplified by HF in NaF/ZnF4 (60: 40 mole %). It has also been found that metals dissolve appreciably in the ionic melts and tend to form otherwise unstable species. (b) Covalent High-temperature Molten Salts These molten salts are covalently bonded compounds and ionize as the aprotic solvents. The particular examples are covalent halides of Group IIIA, IVA, transition and inner transition metals. The solubility of metals in these melts is accounted either on the basis of coordination bond formation or on the basis of formation of subhalides. For example, Cd + CdCl2 $ Cd2Cl2 (c) Reactions in High-temperature Molten Salts

(i)

Auto-ionization and Acid-base Reactions Some molten salts undergo auto-ionisation up to some extent. The auto-ionisation in covalent high-temperature molten salts can be represented as 2HgX2

HgX+ + HgX3–

As per the solvent concept, any species which increases the concentration of HgX+ ion will give an acidic solution while the species which increases the concentration of HgX3– ion will give a basic solution. For example, solution of Hg(ClO4)2 in HgX2 is acidic while that of KX is basic and these two neutralise in presence of HgX2 melts.

Non-aqueous Solvents

6.23

2HgX+ + 2ClO4–

Hg(ClO4)2 + HgX2

K+ + [HgX2]–

KX + HgX2

The ionic melt of sodium tetrachloroaluminate is considered to auto-ionise as 2AlCl4–

Al2Cl7– + Cl–

The concentration of basic chloride ion is considered to increase by the addition of the weakly basic fluoride ion, strongly basic oxide ion and water, while the concentration of the acidic polymeric ion Al2Cl–7 is increased with the addition of protons or AlCl3.

(ii)

2+ Reactions Involving Participation of the Metals a Reactant Bi forms Bi+, Bi3+ 5 and Bi8 in 2+ 2+ 3+ 2+ 2+ 2+ 2+ 2+ + 2+ molten NaCl/AlCl3. Similarly, Hg2 , Hg3 , Pb , Pb2 , Cd2 , Se4 , Se8 , Te , Te2 , Te3 , Te2+ 4 , 2+ Te2+ 6 , Te8 have been formed. Many unstable compounds have also been obtained.

For example,

NaNO

KNO (l )

3− 3 NpO22+  → NpO65− NaOH + KBrO 3

Fe + KNO3 (l) $ K2FeO4 K2FeO4 + KOH (l) 600°  → K3FeO4 The most important use of these reactions is in the slag formation which takes place only in molten form. CaO + B2O3 $ Ca(BO2)2

(slag formation)

CdCl2(l) + Zn(l) $ Cd + ZnCl2

(exchange reaction)

HgCl2 (l) + Zn(l) $ Hg + ZnCl2 Due to excellent solvent property of molten salts, many reactions have been carried out which are otherwise not possible in aqueous medium. 2– l) ZrCl4 + Cl– LiCl( → [ZrCl6]

(halide complex)

CoCl2 + KSCN LiCl/KCl → Co(NCS)2 –

(pseudohalogen)

CCl4 + F KF/KCl  → CCl3F + C2Cl3F + C6Cl4F2

(halo compounds)

2. Ionic Liquids Ionic liquids are defined as those molten salts whose melting points lie below 100°C. Ionic liquids are based on cationic moities such as substituted imidazolium, pyridinium, pyrrolidinium, piperidinium, ammonium, phosphonium, pyrazolium, thiazolium and sulfonium, etc. The anions vary from halides (AlCl–4) to coordinates such as NO3–, SO42–, BF4–, PF6–, SbF6–, CuCl2– and organics such as RSO3–, RCO2–, Ots, CF3SO3–, (CF3SO2)2N, etc. Ethanolammonium nitrate (m.pt. 50–56°C) was the first ionic liquid reported by Gabriel and Weines in 1888, and Walden was the pioneer in the synthesis of first ionic liquid, ethylammonium nitrate (m.pt. 12°C). Two important categories of ionic liquids are the Room Temperature Ionic Liquids (RTILs) which exist as liquid even at room temperature due to their very low melting point, and the ionic liquids which have melting points above the room temperature (but below 100°C). Ionic liquids act as a green solvent as well as catalysts for organic synthesis of many compounds. A recent analogues of ionic liquids have been released in 2003 in the form of Deep Eutectic Solvent (DES), an ionic solvent which is simply a mixture of metal or organic

6.24

Inorganic Chemistry

salt (or its hydrate) and hydrogen donors. The first DES was prepared by mixing choline chloride and urea in a 1:2 molar ratio. DES has much lower melting point than either of its components. For example, eutectic mixture of choline chloride (302°C) and urea (133°C) melts at 12°C. Some other examples of DES are ZnCl2/choline chloride, COCl2.6H2O/chlorine chloride and ZnCl2/urea. DES resemble ionic liquids in its proporties and are also able to dissolve many metals and their oxides. Thus, DES provide a better solvent medium due to cheaper synthesis, less toxicity and biodegradability.

6.10

CONCEPT OF ACID–BASE

Concept of acid–base chemistry has developed since the time of alchemists. During ancient times, acids were characterised as sour substances which turned blue litmus paper red, while bases were characterised as bitter substances which turned red litmus paper blue and reacted with acids to form salts. First attempt for acid–base theory was made by Lavoisier in 1776. He defined acid as an oxide of N, P, S and base as a substance that reacted with acids to form salts. However, the first satisfactorily explanation for acid–base chemistry was produced by Ostwald and Arrhenius in 1884 which won Nobel Prize for Arrhenius. Some important theories are the following:

6.10.1 Arrhenius Theory (Water Ion System) Arrhenius concept was based on the existence of ions in aqueous solution. He proposed the self – ionization of water as H+ + OH– H 2O Acid was defined as the substance that produced H+ and base was defined as the substance that produced OH– in aqueous solution. Thus, HCl was characterised as an acid and NaOH as a base according to the following reactions. HCl NaOH

H2O H2O

H+ + Cl– Na+ + OH–

The neutralization process was represented as a reaction that involves the combination of H+ and OH- to produce H2O.

1. Applications of Arrhenius Concept (a) This concept helped to explain the acidic nature of aqueous solution of non-metallic oxides such as CO2, SO2, SO3, N2O3, N2O5, P4O6, P4O10, etc. SO3 + H2O

H2SO4

2H+ + SO42–

CO2 + H2O

H2CO3

2H+ + CO32–

N2O5 + H2O

2H+ + NO3–

2HNO3

Thus, due to release of H+ ions these solutions are acidic. (b) This concept helped to explain the basic nature of aqueous solution of metallic oxides such as CaO, Na2O and the compounds derived from NH3. CaO3 + H2O NH3 + H2O N2H4 + H2O

Ca2+ + 2OH–

Ca(OH)2 NH4OH

NH4+ + OH–

[N2H5]+ [OH]–

N2H5+ + OH–

Non-aqueous Solvents

6.25

(c) This concept explains the neutralisation of acid–base in aqueous solutions. (d) This concept helps to understand the acid–base equilibria of weak acids and bases. Strong acids and strong bases ionise completely in the aqueous solutions, while weak acids and weak bases do not dissociate completely in the aqueous solutions. Thus, the dissociation of weak acid and bases can be represented as an equilibrium process, for aqueous solutions: H+(aq) + A– (aq), Ka =

HA + aq

and

[ H + ][ A- ] [ HA]

B+(aq) + OH– (aq), Kb =

BOH + aq

[ B+ ][OH - ] [ BOH ]

Where HA is a weak acid and BOH is a weak base, Ka and Kb stand for ionization constant of weak acid and base respectively. According to Arrhenius, greater the values of Ka, stronger is the acid and greater the value of Kb, stronger is the base. Generally, strong acids have Ka greater than 10–2 and strong bases have Kb greater than 10–2. This concept was widely accepted due to its simplicity and extensive use of water as a solvent for reactions. However, due to some limitations the need of new concept was soon realized.

2. Limitations of Arrhenius Concept (a) The main limitation of this concept was its application for aqueous solutions only. For example, NH3 + HCl $ NH4Cl is an acid–base reaction to produce NH4Cl, a salt but there is no involvement of H2O as a solvent. (b) This concept is not applicable for non-aqueous solvents. For example, NH4NO3 acts as an acid in liquid NH3 but does not produces H+ ion. (c) This concept cannot explain the acidic characters of many salts such as AlCl3 in aqueous solutions.

6.10.2 Bronsted–Lowry Concept (Proton–Donor–Acceptor System) Bronsted and Lowry in 1923 independently defined acid as proton donors and bases as proton acceptors. For example,

HCl

+ H2O

Acid

Base

H3O+ + Cl– Acid

Base

In this reaction, HCl donates a proton to H2O and thus acts as an acid, water accepts a proton from HCl and thus acts as a base. In the reverse reaction, H3O+ donates a proton to Cl– ion and therefore acts as an acid, Cl– ion accepts a proton from H3O+ and, therefore, acts as a base, i.e. –H+ HCl

+H+

+

Cl– and H3O

–H+ +H+

H2O

Such pairs of substances which can be formed from each other by the gain or loss of a proton are called Conjugate acid–base pairs. Thus, HCl is the congujate acid of Cl– ion and Cl– ion is the conjugate base of HCl. Similarly, water is the conjugate base of H3O+ ion and H3O+ ion is the conjugate acid of water. This reaction can be written more accurately as conjugate acid – base pair 1 HCl Acid1

+

H2O Base2

H3O+ Acid2

Conjugate acid – base pair 2

+

Cl– Base1

6.26

Inorganic Chemistry

1. Monoprotonic, Polyprotonic and Amphiprotonic Substances The substance which can lose one proton is called monoprotonic acid, while the substance which can lose two or more protons is called polyprotonic acid. For example, HF and H2O are monoprotonic acids, and H2S and H2SO4 are polyprotonic acids. Similarly, the substance which can accept one proton is called monoprotonic base, while the substance which can accept two or more protons is called polyprotonic base. For example, Cl– and H2O are monoprotonic bases, while PO43– and S2– are polyprotonic bases. The substance which can lose as well as accept proton(s) is called amphiprotonic substance. For example, H2O can accept H+ to behave as a base and can lose H+ to behave as an acid as evident from the following reaction: Acid1 Base2 Acid2 Base1 H2O + NH3 NH4+ + OH– + HCl + H2O H 3O + Cl–

2. Uses of Bronsted Concept 1. This concept confirmed the existence of H+ as H3O+ and the amphiprotic nature of H2O as I

HCl + H2O

H3O+ + Cl–

NH3 + H2O

NH4+ + OH–

Acid Base

II

Base Acid

In the first reaction, H2O is proton acceptor, but in the second reaction H2O is proton donor. Hence, H2O is amphiprotic, i.e. it can donate as well as accept protons. 2. According to this concept, even ion can act as acids or bases. 3. According to this concept strength of an acid depends upon its tendency to lose protons and the strength of a base depends upon its tendency to gain protons. It means as the reaction I & II are favoured in forward direction, HCl is a stronger acid and H3O+ is a weaker acid. Similarly H2O is a stronger base and Cl- is a weaker base. It means that every strong acid has a weak conjugate base and vice versa. 4. This concept is extended to the use of non-aqueous solvents such as liquid ammonia, glacial acetic acid, anhydrous sulphuric acid and all hydrogen containing solvents. NH3 + HCl $ NH4+ + Cl– NH3 is acting as a base, as it accepts H+ from HCl. The presence of OH– group is not essential for a substance to act as a base. 5. According to this concept, tendency of an acid to lose protons depends upon the tendency of a base to gain protons. For example, acetic acid acts as an acid in water since it loses proton to water. However in benzene, acetic acid does not act as an acid, since benzene does not accept the proton. When two acids are mixed, the weaker acid acts as a base for the stronger acid. For example HClO4 + HF H2F+ + ClO4– HClO4 + H2SO4

H3SO4+ + ClO4–

HClO4 + HNO3

NO3– + ClO4–

In all these reactions, HClO4 is acting as an acid, while the so – called stronger acids HF, H2SO4 and HNO3 being less acidic than HClO4 are acting as a base.

3. Limitations of Bronsted–Lowry Concept 1. This concept cannot explain the acidic character of non-protonic acids (which cannot give a proton). For example BF3, AlCl3, etc.

Non-aqueous Solvents

6.27

2. This concept cannot explain a number of acid – base reaction which take place in the absence of the solvent or without proton transfer. For example CaO + SO3 $ CaSO4

4. Levelling Effect of Solvents It has been seen that all strong acids appear almost equally strong in aqueous solutions. The ionisation of a strong acid is almost complete in water. Thus, the relative strengths of strong acids cannot be compared in aqueous solutions. Therefore, HClO4, HBr, H2SO4, HCl and HNO3 all appear equally strong in water. HA + H2O H3O+ + A– Acid1 Base2 Acid2 Base1 HClO4 + H2O $ H3O+ + ClO4– HCl + H2O $ H3O+ + Cl– HNO3 + H2O $ H3O+ + NO3– The solvent in which strong acids ionise to same extent and behave equally strong is known as levelling solvent and the phenomenon is called levelling effect of solvent. Thus water acts as a levelling solvent for strong acid. However, if CH3COOH is used as a solvent, the ionization can be represented as HA + CH3COOH Acid1 Base2

CH3COOH2+ Acid2

+ A– Base1

Acetic acid has a little tendency to accept protons. Thus, the equilibrium is not much towards right, even the strong acids are only weakly ionised. The solvent in which strong acids ionise to different extents and donot behave equally strong is known as a differentiating solvent. Thus acetic acid acts as a differentiating solvent for strong acids such as HClO4, HBr, H2SO4, HCl and HNO3. These acids have been found in the order of their decreasing strength as HClO4 > HBr > H2SO4 > HCl > HNO3. On the other hand, weak acids do not ionise completely in presence of water and hence can be compared for their relative order of strengths in their aqueous solutions. This concept is applicable for bases as well. However, same solvent may not act as levelling solvent for both acids and bases. For example, acetic acid acts as a levelling solvent for all bases, so that sodium hydroxide, a very strong base and aniline, a very weak base behave equally strong in comparatively stronger acid, acetic acid. On the other hand, these bases ionise to different extents in presence of a weak acid, water. Thus water acts as a differentiating solvent for bases. Liquid ammonia is a better proton acceptor than proton donor. Hence, even a weak acid behaves like a strong acid due to considerable ionisation in liquid ammonia. NH4+ + CH3COO– NH3 + CH3COOH Base1 Acid2 Acid2 Base2 It has been observed that all strong acids and weak acids appear to be almost equally strong when liquid ammonia is used as the solvent. Quite interestingly, all acids ionise as bases in hydrogen fluoride and acetic acid appears to be stronger base than nitric acid in presence of hydrogen fluoride. This is due to the reason that hydrogen fluoride is too strong to show any proton accepting properties. Hence, the normal acids ionise as bases when present in solutions of hydrogen fluoride as

Similarly,

HF + HNO3 Acid1 Base2

H2NO3+ Acid2

+

F– Base1

HF + H2SO4 Acid1 Base2

H3SO4+ Acid2

+

F– Base1

6.28

Inorganic Chemistry

However, a weaker acid acetic acid which is comparatively better proton accepter, readily ionises as a base in solution of hydrogen fluoride as CH3COOH Base1

+

CH3COOH2+ Acid1

HF Acid2

+ F– Base2

Levelling Effect in Terms of Solvent–System Concept The levelling effect can be better justified in terms of solvent – system concept. According to this concept, the acids and bases which ionise completely in the solvent to give its characteristic cation and anion behave as a strong acid and base. On the other hand, the acid and base which do not ionise completely in the solvent to give its characteristic cation and anion behave as a weak acid and base. If H2O is used as a solvent, its autoionization can be represented as H3O+ + OH–

H2O + H2O

Any substance which ionizes completely in water to give H3O+ , would be strong acid and any substance which ionizes completely in water to give OH– would be a strong base. Since all strong acids ionise completely in water to give H3O+ ions, hence they appear equally strong in aqueous solutions. However all bases do not ionise completely in water to give OH- ions and thus can be differentiated according to their relative strength. If CH3COOH is used as a solvent, its auto-ionisation can be represented as CH3COOH + CH3COOH

CH3COOH2+ + CH3COO–

The normally strong acids HCl, H2SO4 and HNO3 do not dissociate completely in acetic acid and the concentration of CH3COOH2+ varies accordingly. HCl + CH3COOH

CH3COOH2+ + Cl–

Similarly, CH3COOH ionises almost completely in ammonia and leads to the formation of NH4+ ion, characteristic cation of the solvent, ammonia. Thus, acetic acid behaves as a strong acid in ammonia. CH3COOH + NH3 $ NH4+ + CH3COO–

6.10.3 Lux–Flood Concept This concept was originally proposed by Lux and was extended by Flood. According to this concept, acids were defined as oxide ion acceptor and base as oxide-ion donor. For example: CaO + CO2 $ CaCO3 Base Acid SiO2 + CaO $ CaSiO3 Acid Base This concept is applicable for anhydrous reaction and high temperature reaction involving molten oxides.

6.10.4 Lewis Concept (Electron Donor Acceptor System) In 1923, Lewis developed a broader concept of acid–base reactions in terms of the electronic structure of the compounds. According to this concept, acid was defined as electron acceptor and base as electron donor. The acid–base reaction was defined as the process of neutralisation involving the formation of a coordinate bond. For example, H+ + : NH3 $ [H3N:"H]+ Acid Base H+ accepts electron pair from NH3 molecule and thus acts as an acid, while NH3 donates an electron pair to H+ and thus acts as a base. This concept can be elaborated as given ahead:

Non-aqueous Solvents

6.29

1. Lewis Acids (a) Simple Cations All cations are regarded as lewis acids. The strength of these cations in general, increase with decrease in ionic radius and increase in the positive charge carried by the cation. For eg. Na+, K+ are weak lewis acids and H+, Fe2+ are strong Lewis bases. 3+

:

OH2 OH2

H2O

Co3+ + 6H2O : $

Co H2O

OH2 OH2

(b) Electron Deficient Compounds All molecules which have a central atom with an incomplete octet act as lewis acids For example – BF3, AlCl3 etc. (c) Molecules with Central Atom Containing Vacant d-orbitals All molecules which have central atom with vacant d-orbitals, can extend their valence. For e.g. SiF4, SnCl4, SF4, TeCl4, PX3, PF5 etc.

SO32–

For e.g.

Lewis base

:

(d) Elements with Sextent of Electrons Two elements in the periodic table, S and O with sextent of electrons are regarded as lewis acids. :O: $ [O ! SO3]2–

+

Lewis acid

(e) Molecules Containing Multiple Bond between Atoms of Dissimilar Electro-negativity In such compounds, the electron – density of p-electrons is displaced towards more electronegative atom and the less electronegative atom becomes electron deficient. Hence, it can accept an electron pair from a lewis base. OH d+

d–

O C O + OH– $ O Lewis acid Lewis base

C

O–

2. Lewis Bases : :

: :

(a) Neutral molecules with lone pair of electrons on atleast one of the atoms. For example, :NH3, R – O – H, H2O, etc. (b) All anions like F–, OH–, CN–, etc.

3. Limitations of Lewis Concept This concept includes the reactions in which no protons are involved and the reactions which take place in absence of solvent. But there are some limitations of this concept – (a) This concept is based on the formation of coordinate bond. But many acid – base reactions donot involve any coordinate bond formation and are too rapid. (b) This concept does not help in determination of relative strength of acids and bases.

6.10.5 Solvent System Concept In 1928, Cady and Elsey proposed a general concept applicable to protonic and non-protonic solvents for acid – base reactions. According to this concept, acid was defined as a substance which, either by direct dissociation or by reaction with the solvent gives the anion characteristic of that solvent and base was defined as a substance, which either by direct dissociation or by reaction with the solvent gives the cation characteristic of that solvent.

6.30

Inorganic Chemistry

For example, the characteristic cation and anion of H2O are H3O+ and OH– respectively: H3O+ +

H2O + H2O

solvent cation

OH– solvent anion

Thus, all the compounds which produce H3O+ ions in H2O will act as acids and all the compounds which produce OH- ions in H2O will act as bases. For example,

HCl + H2O Acid

H3O+ +

OH–

solvent cation

Neutralisation reaction has been defined as the combination of solvent characteristic cations and anions. This concept can explain many acid–base reactions in protonic and non-protonic solvents. However, it is not applicable to acid–base reactions which take place in absence of solvents. It can also not explain the acid–base reactions which take place in those solvents which donot autoionize. For example, liquid SO2 cannot autoionize because of very low dielectric constant. Reaction between Cs2SO3 and SOCl2 in liq. SO2 takes place as liq. SO

2 Cs2SO3 + SOCl2 ææææ Æ 2CsCl + 2SO2

There is no explanation of such cases in this concept.

6.10.6 Usanovich Concept (Positive Negative System) This concept defines acid as any species which can neutralise a base to form salt, either by giving cations or combining with anions or electrons. Conversely, a base is a species which reacts with acids & is capable of giving anions or electrons or combining with cations. In short, this concept includes all the earlier concepts of acids and bases as well as redox reactions

Examples 1. 2Na + Cl2 $ 2NaCl Explanation: 2Na $ 2Na+ + 2e–

(Loss of e by Na)

Base

2Cl + 2e– $ 2Cl–

(gain of e by Cl)

Acid

2Na+ + 2Cl– $ 2NaCl 2. Fe(CN)2 + 4KCN $ K4[Fe(CN)6] Explanation: 4KCN $ 4K+ + 4CN–

(KCN gives CN– ion)

Base

Fe(CN)2 + 4CN– $ [Fe(CN)6]4–

(FeCN2 reacts with anion)

Acid

4K+ + [Fe(CN)6]4– $ K4[Fe(CN)6] This concept however considers all the reactions as acid–base reactions, which can be better considered from some other point of view.

6.10.7 Pearson Concept (HSAB Concept of Hard and Soft Acids and Bases) In coordination chemistry, metal ions and ligands have been seen to have preferential affinity for particular ligands and metal ions respectively. Depending upon this tendency, Pearson classified metal ions and ligands into hard and soft acids and bases:

Non-aqueous Solvents

6.31

(a) Hard Acids These are the metal ions which are small in size, have a high positive charge and a noble gas electronic configuration and are not very polarisable. These metal ions include cations from groups 1, 2 and light transition and inner transition metals. (b) Hard Bases These are the anions or neutral molecules which are having high electronegativity and are not easily polarisable. These preferentially react with hard acids. (c) Soft Acids These are the metal ions which are large in size, have low positive or zero charge and do not possess noble gas electronic configuration. These are easily polarised and include heavy metal ions. (d) Soft Bases These are the anions or neutral molecules with low electronegativity and easily polarisable anions. These react with soft acids. However, there are some cations of transition and inner transition metals which behave as soft and hard acids in different cases, depending upon their oxidation state. Similarly, some bases also lie in intermediate category. The table below shows complete classifications: Table 6.8 Classifications of hard/soft acids and bases Hard acids

Border line or intermediate acids

Soft acids

H+, Li+, Na+, K+, Be2+, Ca2+, Sr2+, Mn2+, Al3+, Ga3+, In2+, La3+, Lu3+, Cr3+, Co3+, Fe3+, As3+, Si4+, Ti4+, Zr4+, Th4+, U4+, Ce3+, Sn4+, VO2+, UO22+, MoO3+, BF3, SO3, Cr6+, I7+, I5+, CO2, (HF)n

Fe2+, Co2+, Ni2+, Cu2+, Zn2+, Pb2+, Sn2+, Sb3+, Bi3+, Rh3+, SO2, NO+, GaH3

Cu+, Ag+, Au+, Tl+, Hg+, Pd2+, Cd2+, Pt2+, Hg2+, Tl3+, BH3, GaCl3, InCl2, I+, Br+, I2, Br2 and metals in zero oxidation state

H2O, OH–, F–, CH3COO–, PO43–, SO42–, Cl–, CO32–, ClO4–, NO3–, RO–, O2–, NH3, RNH2, N2H4, ROH, R2O

C6H5NH2, C5H5N, Br–, NO2– SO32–, N2

CO, C2H4, C6H6, (RO)3P, R2S, RSH, RS–, S–, SCN–, S2O32–, H–

1. HSAB Principle—Principle of Hard and Soft Acids and Bases According to this principle, hard acids prefer to combine with hard bases and soft acids prefer to combine with soft bases to form stable products. Hard – soft acid base or soft – hard acid base combinations yield comparatively less stable product.

2. Nature of Bonding Bonding between hard acids and hard bases is predominantly ionic due to transfer of electrons from hard bases to the vacant d-orbitals of hard acids. In case of soft acids and soft bases, covalent bond is formed by p-interaction between the easily polarisable cations and anions.

3. Applications of HSAB Principle (a) This principle is mainly used to explain the relative stability of complexes. For example , AgI2– is stable, but AgF2– does not exist. It can be explained as Ag+

+

Soft acid +

Ag

Soft acid

2I– $ AgI2– Soft base

+

Stable

2F $ AgF2– –

Hardbase

Unstable

As already discussed, soft acid–soft base interaction yields stable complex.

6.32

Inorganic Chemistry

(b) This principle is used to predict the feasibility of reaction between two compounds. For example, Hg(OH)2 dissolves readily in acidic aqueous solution, but HgS does not. Hg(OH)2 is a product of soft acid–hard base combination, so it is unstable and dissociates readily. Whereas HgS is a product of soft acid–soft base combination. Hence it is stable & does not dissociate easily. Similarly, LiI and CsF react easily as LiI + CsF $ LiF + CsI Hard Soft

Hard soft Less stable

Soft hard Hard soft More stable

(c) This principle is used to predict the existence of certain metal ores. For example, hard acids like Mg2+, Ca2+ and Al3+ exist as oxides or carbonates, because oxides and carbonates are hard bases. Where as soft acids like Ag+, Cu+ and Hg2+ exist as sulphides, because sulphide is a soft base. (d) It has been observed that a strong and or base can displace a weaker one , irrespective of the HSAB principle. For example, SO32– + HF HSO3– + F– Soft base

Hard-hard

Hard base

In this case, very strong but soft base, sulphite ion is displacing the weak but hard base from the hard acid, H+. Similar is the case for displacement of weaker soft base, SO32– by the very strong hard base, OH– ion from the soft acid, methyl mercury cation: OH–

+ CH3HgSO3–

Hard base

CH3HgOH + SO32–

Soft-Soft

Soft base

(e) The hardness or softness of a site can be changed with the addition of substituent. Addition of hard (electronegative) atoms can harden an otherwise soft site and vice -versa. R2SBF3

+ R2O

Soft-Hard

$ R2OBF3

Hard

R2OBH3

+

Hard-Soft

R 2S

+

R2S

Hard-Hard

Soft

$ R2SBH3 +

Soft

R2O

Soft-Soft

Hard

Similarly, BF3H–

+ BH3F–

Hard-Soft

CF3H Hard-Soft

$ BF4– +

Soft-Hard

+

CH3F Soft-Hard

BH4–

Hard

$ CF4 + Hard

Soft

CH4 Soft

This tendency of a centre which is attached with soft ligands to favour further coordination by a soft ligand and vice – versa is termed as ‘symbiosis’.

4. Theoretical Basis to HSAB Principle Several explanations have been put forth to provide the theoretical basis for hard–hard and soft–soft interactions. The most simple explanation is based on the type of bonding involved between the groups. It is believed that the hard–hard interactions involve the electrostatic interactions and thus bonding between hard acid–hard base must be predominantly ionic. It is due to the reason that most of the typical acids include the small sized and highly charged positive metal ions. Such metal ions would preferably favour ionic bonding with the non–polarisable hard bases and the resulting compound would be highly stable. On the other hand, hard–soft interactions would be unfavourable and the resulting species are comparatively less stable.

Non-aqueous Solvents

6.33

Electrostatic explanation cannot be accounted for soft–soft interactions, due to inclusion of the large sized and easily polarisable species which prefer covalent bonding instead of ionic bonding. On this basis, Misons and co-workers (1967) have proposed an equation to correlate the hard–hard or soft–soft interaction as: pK = – log K = aX + bY + g

Table 6.9 Values of g for some hard

where K is the dissociation constant of the acid – base complex, X and and soft acids Y are the parameters of the acids, and a and b are the parameters for Hard acids Soft acids the bases. g is the constant for adjustment of all the values on the same + 2+ 0.36 Sn 3.17 Li scale. It has been found that for hard acids, g < 2.8 and for soft acids, 3+ 3+ 0.70 Ti 3.2 Al g > 3.2. The value of g lies in between 2.8 and 3.2 for borderline acids. 0.87 Cu+ 3.45 Mg2+ The value of b increases with increase in the softness of the base. For hard bases, b < 3 and for soft bases, b > 5. Table 6.8 lists the values of 0.93 Pb2+ 3.58 Na+ 2+ parameter for some of the hard-soft acids. 1.62 Tl+ 3.78 Ca It can be simplified in terms of electronegativity. Most of the species 2.37 Hg2+ 4.25 Fe3+ with high electronegativities are hard, while the species with low 2.56 Au+ 5.95 Co2+ electronegativities are soft. Thus, trifluoromethyl group is harder as 2.73 Cs+ compared to methyl group. Similarly boron trifluoride is harder than borane. The concept can be elaborated on the basis of HOMO – LUMO approach. The polarizability of soft species can be accounted due to small HOMO – LUMO gap as compared to non – polarizability of hard species due to large HOMO – LUMO gap. Further, due to presence of large number of d-electrons in metal and empty low lying un-occupied acceptor orbitals on the ligands, p interactions take place. For example, the small size and presence of low lying p*acceptor orbitals in CO makes it as the best soft ligand.

6.11

ACID STRENGTH BEHAVIOUR IN THE PERIODIC TABLE

1. Hydrides of Group 17 When we consider the acid strength of hydrides of Group 17 elements, HX, one would except the increase in acid strength with increase in the electronegativity of X. Because higher the electronegativity of X, greater is the electron pulling capacity and easier is the release of H+. Thus, the order of acid strength should be HF > HCl > HBr > HI However, the actual order is HF < HCl < HBr < HI This is due to the reason that there are a number of other factors involved in the process as discussed below with the help of a Born–Haber type cycle: Acid strength can be represented as the tendency of hydrated HX molecule to form hydrogen ions, H+. The energy cycle involves different stages such as dissociation, ionisation and hydration. ∆Hdehydration

HX (g)

HX (aq.)

DE Acid strength

H(g) + X(g) I. E.

H+(g)

+ X(g)

E.A. –

H+(g) + X (g)

DHhydration

H+(aq) + X–(aq)

6.34

Inorganic Chemistry

Applying Hess’s Law, Acid strength = DHdehydration + DE + IE + EA + DHhydration Table 6.10 lists the values of all the energies associated with the cycle. Table 6.10 Energy values (KJ mol–1) DHdehydration

DE

IE

DHhydration

EA

H+

X–

Acid strength

HF

48

574

1311

–338

–1091

–513

–18

HCl

18

428

1311

–355

–1091

–370

–68

HBr

21

363

1311

–331

–1091

–339

–75

HI

23

295

1311

–302

–1091

–394

–167

It is evident from the table 6.9 that the acid strength value for HF is small and increases with maximum for HI. This is due to the high dissociation energy, high heat of dehydration of HF and low electron affinity of F-.

2.Oxoacids of Same Elements The acid strength of oxoacids of halogens (for a particular halogen) increases in the order +1

+3

+5

+7

HXO < HXO2 < HXO3 < HXO4 This is due to the reason that higher the oxidation state of the central atom, higher is its electronegativity, greater is the tendency to pull the electrons and easier is the removal of H+. Now, the oxidation state of halogen in these oxoacids is +1, +3, +5 and +7 respectively. Thus, the halogen with highest oxidation state is most electronegative and hence most acidic. Similarly, acid strength order of oxyacids of other elements varies as H2SO3 < H2SO4 and HNO2 < HNO3 However, in case of oxyacids of phosphorus, the order of acidic strength is +1

+3

+4

H3PO2 > H3PO3 > H3PO4 This is due to the reason that in case of oxyacids of phosphorus, acidic strength depends on the number of dissociable protons (protons attached to oxygen atom). The structure of these oxyacids reveal that the number of dissociable protons (number of OH groups) increases from H3PO2 (one) to H3PO4 (three). Thus acidity of these oxyacids increases in the order H3PO4 < H3PO3 < H3PO2. O H

P

O OH

H (H3PO2) Monobasic

H

P

OH (H3PO3) Dibasic

O OH

HO

P

OH (H3PO4) Tribasic

3. Hydrides of Non–metals of Second Period The acid strength of hydrides of non–metals of 2nd period varies in the order as CH4 < NH3 < H2O < HF

OH

Non-aqueous Solvents

6.35

This is due to increasing electronegativity of the non-metal in the order C < N < O < F. As a result, the stability of their conjugate bases increases in the order. CH3– < NH2– < OH– < F– Similarly, on moving down a group (except Group 17) the acid strength increases (basic strength decreases) with decrease in electronegativity of the elements in the same group. Thus, the order for increasing acid strength is NH3 < PH3 < AsH3 < SbH3 < BiH3 H2O < H2S < H2Se < H2Te

4. Oxoacids of Similar Structures The acid strength of oxoacids of similar structures increases with increase in the electronegativity of the central atom as HIO3 < HBrO3 < HClO3 HNO3 > HPO3 > HAsO3 H3PO4 > H3AsO4 > H3SbO4 H2SO4 > H2SeO4 > H2TeO4 The solvent behaviour is determined by its dielectric constant, dipole moment, melting point, boiling point, and heat of fusion and vaporisation. A solvent which can donate or accept protons is known as protic solvent while which cannot donate or accept protons is known as aprotic solvent. A solvent which can donate as well accept protons is known as an amphiprotic solvent. The solvent which can undergo auto ionisation is known as an ionising solvent, and which cannot ionise at all is known as a non-ionising solvent. In general terms, the solvents other than water are known as non aqueous solvents while water is known as the aqueous solvent. Due to low dielectric constant and low dipole moment, liquid ammonia acts as a better solvent for nonpolar molecules. It auto-ionises to give ammonium ion and amide ion (imide and nitride ions can also be formed). Any substance which increases the concentration of ammonium ion in liquid ammonia is known as ammono acid, while the substance which increases the concentration of amide, imide or nitride ions in liquid ammonia is known as ammono base. Most of the chlorides can be precipitated in liquid ammonia while many active metals react with solutions of ammonium salts in liquid ammonia to liberate hydrogen. Alkali metals dissolve in liquid ammonia to give blue solutions which are very strong reducing agents. Liquid SO2 is an aprotic solvent which dissociates to give thionyl ion and sulphite ion. Thus, accordingly, SO2+ producing substances act as acids while SO32– producing substances acts as bases in liquid SO2. Anhydrous HF is an excellent ionising solvent which ionises to give fluoronium ion and fluoride ion. The strong acids in water behave as bases in HF while the electron acceptor fluorides act as acids in HF. However, the weak acids in water behave as bases in HF. Anhydrous sulphuric acid is a highly associated non-aqueous solvent and is used as a dehydrating agent. Only some substances act as acids in H2SO4; these are chlorosulphuric acid, fluorosulphuric acid, perchloric acid, sulphuric dioxide, trisboron (hydrogen sulphate) and mixture of strong lewis acids and highly acidic protonic solvents. Acetic acid is a commonly used laboratory solvent. All strong acids in aqueous media act as acids in acetic acid while weak organic bases and salts of ‘s’ -block elements with organic acids act as a base. Liquid N2O4 is also an excellent aprotic non-aqueous solvent. Nitrosyl ion producing substances act as acids while

6.36

Inorganic Chemistry

nitrate ion producing substances act as a base in N2O4. It is also a powerful oxidising agent and reacts vigorously with organic substances. Molten salts have proved to be excellent non-aqueous solvent and refer to the salts which melt without any decomposition or vaporisation. These are classified into high temperature molten salts with m.pt. above 100°C and ionic liquids with m.pt. below 100°C. Molten salts are good conductors and are highly stable thermally as well as w.r.t. oxidation.

EXAMPLE 1 Predict the behaviour of (NH4)2SO4 and urea in water, liquid NH3 and anhydrous H2SO4.

(a) Behaviour of (NH4)2SO4 and urea in water (NH4)2SO4 dissociates in water to give NH+4 and SO2+ 4 . + 2– 2O (NH4)2SO4 H → 2NH4 + SO4

NH4+ ions further react with H2O to give H3O+ ions and NH4OH is formed. NH4+ + 2H2O Overall reactions:

(NH4)2SO4 + 4H2O

NH4OH + H3O+ 2 NH4OH + 2H3O+ + SO42–

Urea behaves as a non-electrolyte in water. (b) Behaviour of (NH4)2 SO4 and urea in liquid NH3 (NH4)2SO4 dissociates in liquid NH3 and concentration of NH+4 ion is increased, the characteristic cation of liquid NH3. Hence, the solution is acidic. + 2– 3 (NH4)2SO4 liq.NH  → 2NH4 + SO4

Urea gives NH+4 ion in liquid NH3 due to protolysis. NH2CONH2 + NH3(l) $ NH2CONH– + NH4+ (c) Behaviour of (NH4)2SO4 and urea in anhydrous H2SO4 (NH4)2SO4 dissociates in anhydrous H2SO4 and concentration of SO42– is increased, the characteristic anion of H2SO4 and the solution is basic. + 2– 2 SO 4 anhyd. (NH4)2SO4 N → 2NH4 + SO4 – Urea accepts a proton from H2SO4 and concentration of HSO 4 is increased.Thus, the solution is basic. NH2CONH2 + H2SO4 $ NH2CONH3+ + HSO4–

EXAMPLE 2 Complete the following equations: 1. CH3C = NHNH2 + NH3(l) $ 3. UCl6 + SO2(l) $

3 (l ) 2. PbI2 + KNH2 NH   → HF anhyd. 4. CrF3 + 3NaF →

of H 2O 5. KCl + N2O4(l) traces  →

1. CH3C = NHNH2 + NH3 (l) $ CH3C = NHNH– + NH4+ Acetamidine undergoes protolysis in liquid NH3 and behaves as an acid in liquid NH3.

Non-aqueous Solvents

6.37

2. KNH2 is an ammono base and forms PbNH, lead imide. 3 (l ) PbI2 + KNH2 NH   → KI + HI + PbNH.

3. UCl6 undergoes solvolysis in liquid SO2 and forms oxylchloride. It also results in an increase in concentration of SO2+ ion. 4. CrF3 dissolves in NaF in presence of anhydrous HF due to the formation of sodium hexafluorochromate (III) CrF3 + 3NaF $ Na3[CrF6] 5. KCl gets solvolysed by N2O4(l) in presence of traces of water. of H 2O KCl + N2O4 traces  → NOCl + KNO3 .

EXAMPLE 3 What happens when (a) Potassium is treated with nitrous oxide in presence of liquid NH3? (b) Mercury (II) iodide is treated with potassium iodide in presence of liquid SO2? (a) Potassium reduces nitrous oxide to N2 in presence of liquid NH3. 2K + N2O + NH3(l) $ KNH2 + KOH + N2 (b) Mercury (II) iodide dissolves in potassium iodide due to the formation of potassium tetraiodomercurate (II). 3 HgI2 + 2KI liq.NH  → K2[HgI4]

EXAMPLE 4 Write conjugate acids for the following (a) NH2CONH2 (a) NH2CONH3+

(b) HNO3

(b) H2NO3+

(c) H2SO4

(c) H3SO4+

EXAMPLE 5 Identify conjugate acid base pair in the following reaction CH3COOH2+ + Cl–

HCl + CH3COOH Conjugate acid base pair – I;

Acid I – HCl; Base I – Cl-

Conjugate acid base pair II ;

Acid II – CH3COOH2+; Base II – CH3COOH

EXAMPLE 6 Arrange the following in their increasing basic strength & justify your answer. NH3, H2O, HF and Ne. HF :

: :

:

H2O

: :

NH3

:

:

All these substances contain lone pair on at least one atom as shown below: : Ne :

Thus, all the substances can act as lewis base. Now as the number of lone pair increases, the tendency of the substance to donate the electron pair decreases & hence basic strength decreases. Thus, the order of basic strength is Ne < HF < H2O < NH3

EXAMPLE 7 Give reasons for the following:

(a) [Co(NH3)5F]2+ is stable while [Co(NH3)5I]2+ is unstable. (b) Copper and silver occur in nature as their sulphides while Magnesium and calcium occur as carbonates.

6.38

Inorganic Chemistry

(a) NH3 and F– are hard ligands, while I- is a soft ligand. Thus, [Co(NH3)5F]2+ with all hard ligands is more stable than [Co(NH3)5I]2+. (b) Mg2+ and Ca2+ are hard acids, while Cu+ and Ag+ are soft acids. On the other hand CO32– is a hard base and S2– is a soft base. Thus, according to HSAB principle Cu2S and Ag2S are stable. Similarly MgCO3 and CaCO3 are stable to occur in nature.

EXAMPLE 8 LiI reacts with CsF, but CsI cannot react with LiF. Justify the statement. LiI

+

Hard-Soft

CsF $ Soft-hard

LiF

+ CsI

Hard-hard

Soft-Soft

According to HSAB principle, LiF is more stable than LiI, while CsI is more stable than CsF. Thus LiI and CsF can react together, while LiF and CsI do not react.

QUESTIONS Q.1 Discuss the important characteristics of solvents. Q.2 Discuss the classification of solvents in detail. Q.3 Give an account of each of the following in liquid NH3. (a) Acid-base reaction (b) Metathesis reaction (c) Ammonolysis (d) Complex formation reaction Q.4 What do you mean by auto-ionisation? Illustrate your answer with the help of suitable examples. Q.5 Complete the following equations. 3 (l ) 3 (l ) (a) SO2Cl2 NH (b) AgCl NH   →   → 2 (c) C2H5OH HF (d) Cr2(SO3)3 SOCl →  → Q.6 Give reasons for the following: (a) Acetamide is basic in water but acidic in liquid NH3. (b) SbF6 behaves as an acid in HF. (c) Alkali metals dissolve in liquid NH3 to give reducing solutions. (d) Solution of KNH2 in liquid NH3 gives pink colour to phenolphthalein. (e) Solution of Hg(ClO4)2 in HgCl2 is acidic. Q.7 Predict the behaviour of each of the following in water, liquid NH3 and liquid SO2. (a) NH4Br (b) CH3COONH4 Q.8 Discuss the following reactions in liquid HF and anhydrous H2SO4. (a) Acid-base reactions (b) Auto-ionisation Q.9 Discuss the use of molten salts as a non-aqueous solvents in brief. Q.10 Discuss the use of Ionic liquids as a reaction medium. Q.11. Discuss Arrhenius theory of acids and bases and its application. Q.12 Discuss Bronsted–Lowry concept of acids and bases. What are its uses? Q.13 Discuss Lewis concept of acids and bases. Q.14 Discuss solvent concept of acids and bases. What are its advantages? Q.15 What do you mean by levelling and differentiating solvents? Q.16 Explain your answer with the help of suitable examples. Q.17 What is HSAB principle? What are its applications?

Non-aqueous Solvents

6.39

Q.18 Arrange the following in increasing order of acid strength with proper explanation: (a) HIO2, HBrO2, HClO2 (b) HF, H2O, NH3, CH4 Q.19 Give reason for the following : (a) AgI2– is stable but AgF2– is unstable. (b) CF4 is more stable than CH3F.

MULTIPLE-CHOICE QUESTIONS 1. NH4Br is reduced by Na in liquid NH3 to give (a) NH3 (b) H2 (c) NaBr (d) all of these 2. Cr2SO3 acts as a base in (a) H2SO4 anhyd. (b) H2O (c) SO2 (l) (d) all of these 3. HClO4 in liquid HF acts as (a) acid (b) base (c) amphoteric (d) none of these 4. Iron dissolves in molten potassium nitrate to give (a) Fe(NO3)2 (b) K2FeO4 (c) K3FeO4 (d) Fe2O3 5. The number of ions obtained by cryoscopic method in case of BaF2, CaBr2 and NaF in molten NaCl is (a) 1, 2, 3 (b) 3, 2, 1 (c) 3, 3, 2 (d) 3, 3, 1 6. The conjugate base of NH2CONH2 is (a) NH2CONH3+ (b) NH2CONH– (c) H2O (d) none of these 7. Which of the following is a Lewis base? (a) Ne (b) NH3 (c) H2O (d) All of these 8. Which of the following is a soft base according to HSAB principle? (a) I2 (b) NH3 (c) Br– (d) C6H6 9. Which of the following is least acidic? (a) HF (b) HI (c) NH3 (d) CH4 10. Which of the following is most acidic? (a) HIO2 (b) HBrO2 (c) HClO2 (d) HClO3

chapter

Extraction of Elements

7

After studying this chapter, the student will be able to

7.1

INTRODUCTION

The elements constituting matter have been broadly classified into metals, nonmetals and metalloids. Metals are elements with low ionisation energies, low electronegativities and low electron affinities. These are generally solids, malleable, ductile, lustrous and good conductors of heat and electricity. Nonmetals are elements with high ionisation energies, high electronegativities and high electron affinities. These are generally brittle, nonlustrous and poor conductors of electricity. The elements with common characteristics of metals and nonmetals are termed metalloids. In this chapter, we will discuss the general methods used for metal extraction and purification.

7.2

OCCURRENCE OF ELEMENTS

Elements occur in nature in either native state or combined state, depending upon their standard electrode potential.

1. Nature or Uncombined (Free) State Elements, which have positive standard electrode potentials, are found in their elementary form, known as native state. For example, metals such as Cu, Au, Ag, Pt. Such metals are known as noble metals. Non-metals such as O, N and noble gases occur in the free state.

7.2

Inorganic Chemistry

2. Combined State Elements which have negative standard electrode potentials are found to occur in the form of their compounds. Such metals are known as active metals and generally occur as oxides, sulphates, sulphides, carbonates, chlorides, nitrates, etc. These compounds of the metals are known as minerals. The minerals from which the metals are extracted economically and conveniently are known as ores. Thus, all ores are minerals but all minerals are not ores. The ores can be classified into the following groups: (a) Free Ore Noble elements are generally found in native state in association with alluvial impurities such as sand and clay. Sometimes pure metals are also found in lumps known as nuggets. Copper, silver, gold, mercury, platinum and osmium are examples of such cases. (b) Oxide Ores Al, Mn, Fe, Cu, Zn, Sn, Cr, etc., occur as their oxides For example, bauxite (Al2O3.2H2O), pyrolusite (MnO2), haematite (Fe2O3), cuprite (Cu2O), zincite (ZnO), tinstone or cassiterite (SnO2), chrome (Cr2O3), etc. (c) Carbonate Ores Na, Ca, Mg, Ba, Fe, Cu, Zn, Pb, etc., occur as their carbonates. For example, limestone (CaCO3), magnesite (MgCO3), siderite (FeCO3), cerussite (PbCO3), etc. (d) Sulphate Ores Ca, Mg, Ba, Pb, etc., are found as sulphates. For example, Gypsum (CaSO4.2H2O), kieserite (MgSO4.H2O), barytes (BaSO4), anglesite (PbSO4), etc. (e) Sulphide Ores Fe, Hg, Cu, Co, Ni, etc., occur as their sulphides For example, iron pyrite (FeS2), cinnabar (HgS), copper glance (Cu2S), etc. (f) Halide Ores Na, K, Ca, Mg, Al, Ag, etc., occur as their halide ores. For example, common salt (NaCl), sylvine (KCl), fluorspar (CaF2), carnallite (KCl.MgCl2.6H2O), cryolite (Na3AlF6), etc. (g) Phoshate Ores Fe, Li, Ca, etc., occur as phosphates. For example, phosphorite [Ca3(PO4)2] and triphylite [(Li Na)3PO4] [(FeMn)3(PO4)2], etc.

7.3

METALLURGY

It is the process of extracting metals in their free states from their ores. This process depends upon the nature of the ore and the impurities associated with the minerals. Hence, different procedures are used for different metals. The steps involved in the metallurgical processes are discussed below:

1. Pulverisation This step involves the crushing and grinding of ores by a jaw crusher, stamp mill or ball mill (Fig. 7.1).

3. Concentration of Ore The ore is often found to be mixed with impurities such as quartz, mica and other silicates, collectively known as gangue. These impurities are removed from the ore by the operation known as concentration or benefication of the ore. The most common methods are the following:

(a) Hand-Picking Large pieces of impurities can be removed from the ore by hand-picking. (b) Magnetic Separation The magnetic ore can be separated from nonmagnetic impurities or vice versa by using a magnetic separator. For example, nonmagnetic ore of tin, tin stone (SnO2), contains magnetic impurity of wolframite (Fe, Mn) WO4. Similarly, the magnetic

Jaw crusher

Fig. 7.1

Ball mill

Stamp mill

Pulverisation of ore by jaw crusher, ball mill and stamp mill

7.3

Extraction of Elements

ore of chromium, chromite ore (FeO.Cr2O3), contains nonmagnetic silicious impurities. A magnetic separator consists of a leather belt which moves over two rollers, one magnetic and the other, nonmagnetic, (Fig. 7.2). The crushed ore is dropped over the belt at one end. The magnetic particles in the ore are attracted by the magnetic roller and fall nearer to the roller. The nonmagnetic particles fall away and hence are removed.

Crushed ore

Roller Nonmagnetic particles

Leather belt Magnetic particles

Fig. 7.2 Magnetic separator Powered ore

(c) Gravity Separation Method This method involves the treatment of crushed and powdered ore particles with a strong and running current of water to wash away the lighter gangue particles, leaving the heavier ore particles behind. Gravity separation can be done by two ways: (i)

(ii)

Hydraulic Classifier Method In this method, finely powdered ore is dropped through the top of a conical reservoir called a hydraulic classifier. A powerful current of water is introduced through the bottom, which carries away the lighter gangue particles. The heavier ore particles are collected at the base due to the conical shape of the reservoir Fig. 7.3.

Gangue particles

Water

Ore particles

Fig. 7.3 Hydraulic classifier

Wilfley Table Method In this method, crushed ore is dropped on the top of a vibrating and slopping table attached with wooden strips called cleats or riffles. The powerful current of water is passed over the table carrying away the lighter gangue particles.

(d) Leaching or Hydrometallurgical Process In this process, the powdered ore is treated with a suitable reagent which dissolves only the metal and the impurities are left behind. The solution is filtered to remove the impurities and the metal is extracted by using chemical methods. (e) Froth Flotation Process This method is mainly liked for metal sulphides which gets preferentially wetted by pine oil while the associated impurities (gangue) are preferentially wetted by water (Fig. 7.4). In this method, the metal sulphide ore is mixed with water containing a small quantity of pine oil, sodium ethyl xanthate and a depressing agent. The mixture is agitated violently with air to form the froth at the air-water interface. The ore particles (preferentially wetted with oil) rise up with the froth while the gangue particles (preferentially wetted with water) remain in water.

Froth

Ore

Air

Gangue

Fig. 7.4 Froth-flotation process

3. Calcination In this process, the concentrated ore is heated strongly, in absence or limited supply of air, just below its melting point. As a result, the carbonate ore gets converted to its oxide and water is removed from the hydrated oxide ores. The volatile impurities and organic matter are expelled from the ore, making it porous and easily workable for further process.

7.4

Inorganic Chemistry

For example, calcination of limestone: CaCO3 ∆ → CaO + CO2 Calcination of malachite: CuCO3 . Cu(OH)2 ∆ → 2CuO + CO2 + H2O

4. Roasting This process involves the heating of concentrated ore strongly in excess of air just below its melting point. This process is carried out in a reverberatory furnace usually with addition of other material and is mainly used for sulphide ores. The sulphides are converted into their oxides, or sulphates as ∆

→ 2ZnO + 2SO2 2ZnS + 3O2  ZnS + 2O2 $ ZnSO4 The nonvolatile impurities are removed is the form of volatile oxides which easily escape through the chimney. 4As + 3O2 $ 2As2O3 S + O2 $ SO2 P4 + 5O2 $ 2P2O5

5. Pyrometallurgical Process The calcined or roasted ore is reduced further by various reducing agents as discussed below:

(a) Reduction by Metals Highly active metals are used as reducing agents for many metal oxides, e.g. B2O3 + 6Na $ 2B + 3Na2O ∆

→ 2Rb + 3MgO Rb2O3 + 3Mg  ∆

→ 2V + 5CaO V2O5 + 5Ca  (b) Reduction of Aluminium—Goldschmidt’s Aluminothermite Process In this process, metal oxides are reduced by aluminium powder by an exothermic reaction. The large amount of heat evolved melts the metal produced, e.g. 3MnO2 + 4Al $ 3Mn + 2Al2O3 Cr2O3 + 2Al $ 2Cr + Al2O3 This process is also used for welding by using a mixture of Fe2O3 and aluminium powder (3:1) known as thermite mixture, covered with a mixture of BaO2 and Mg powder, known as ignition mixture. In this mixture a piece of Mg ribbon is inserted which on ignition catches fire resulting in reduction of Fe2O3 to Fe (molten state). This molten iron fills the gap between the broken pieces (Fig. 7.5). Fig. 7.5 Aluminothermite process

(c) Reduction by Hydrogen Hydrogen is used as a reducing agent for the oxides of metals which are less electropositive than hydrogen, such as Cu, Ag, Cr, Fe, Co, etc. CuO + H2 $ Cu + H2O

Extraction of Elements

7.5

∆

→ 3Fe + 4H O Fe3O4 + 4H2  2 (d) Reduction by Carbon Smelting or Carbon Reduction Process This process is used for the oxides of the metal which do not form carbides with carbon, such as Zn, Sn, Fe, Pb, etc. Smelting is carried out by heating a mixture of roasted ore and coal or coke in a reverberatory furnace or a blast furnace, above the melting point of the metal, e.g. ∆

→ Zn + CO ZnO + C  ∆

→ Pb + CO PbO + C  ∆

→ Pb + CO2 PbO + CO  Some oxide ores may contain infusible impurities, which are converted into fusible product (slag) by addition of another substance known as flux, i.e. Flux + Infusible impurities ∆ → Fusible product (slag) Depending upon the nature of impurities, flux can be of two types: Acidic Flux If the impurities are basic like CaO, FeO, MgCO3, then acidic oxides like SiO2, B2O3, P2O5 are used as flux. SiO2 + Acidic flux

FeO

$

Basic impurity

FeSiO3 Slag

Basic flux If the impurities are acidic like P2O5, SiO2, then basic oxides like MgCO3, CaO, Fe2O3, etc. are used as flux. CaO

+

Basic flux

SiO2

$

Acidic impurities

CaSiO3 Slag

(e) Reduction by Air Oxides of less active heavy metals like Cu, Hg, Pb, Sb, etc., are unstable. Roasting of the sulphide ores of such metals directly gives the metal. Cu2S + O2 $ 2Cu + SO2 PbS + O2 $ Pb + SO2 Sb2S3 + 3O2 $ 2Sb + 3SO3 HgS + O2 $ Hg + SO2 This process is known as auto-reduction, self-reduction or air reduction.

(f) Electrometallurgy Oxides of highly active metals such as Na, K, Mg, Ca, Al, etc., are very stable and thus cannot be reduced by simple reduction processes. Reducing these oxides with carbon requires very high temperature, but the metal produced reacts with carbon to form metallic carbide. These metals are extracted by electrolytic reduction of their suitable molten salts such as oxides, hydroxides, chlorides, etc. This process is known as electrometallurgy and librates metal at the cathode.

7.4

PURIFICATION OF IMPURE METALS OR REFINING

The metal obtained by the reduction process is usually impure and needs to be purified. The impurities may consist of unreduced sulphides and oxides of the metals, other metals produced by the simultaneous reduction of their corresponding oxides already present in the ore, nonmetals, dissolved gases, slags, etc. The method of purification depends on the nature of the metal obtained and the impurities to be removed. The methods used are the following:

7.6

Inorganic Chemistry

1. Fractional Distillation This method is used to purify the volatile metals containing nonvolatile impurities and vice versa. In this method, the impure metal is converted into its vapours and condensed in a receiver, while the nonvolatile impurities are left behind and removed. This method is used for purification of Zn, and Hg. 2. Liquation Method This method is used for those metals which are more readily fusible than the impurities, e.g. Sn, Bi and Pb. The impure metal is heated on the sloping hearth of a furnace. The metal melts leaving behind the infusible impurities. 3. Zone-Refining Method This method is used to obtain metals in very high purity. In this method, the impure metal rod is heated by using a moving circular heater. The impure metal melts to form a molten zone. The pure metal crystallises earlier while the impurities shift to the adjacent molten zone (Fig. 7.6). This method is used for purification of Si, Ge and Ga.

Circular heater Impure metal

Recrystallised pure metal

Molten zone

Fig 7.6 Zone-refining method

4. Thermal Decomposition Method In this method, the impure metal is treated with a suitable reagent in a sealed vessel. It forms a volatile compound which decomposes on heating to give the pure metal, while the impurities remain unreacted. For example, crude nickel is treated with CO to form Ni(CO)4, which decomposes on heating to give pure nickel and CO is recycled again. This process is termed as Mond process. Similarly, impure Zr and Ti are heated with iodine in a sealed vessel to form their volatile iodides. These iodides are passed over a thin and hot tungsten or tantalum filament to deposit the metal on its surface. This process is termed Van Arkel process. 5. Oxidative Method In this method, the impurities are oxidised into their oxides by passing air through the impure molten metal. These oxides may by removed as gases (if volatile) or as slag or scum. 6. Electrolytic Refining This process is used for many metals such as Zn, Cu, Pb, Sn, Ag, Ni, Cr, etc. The impure metal bar is made the anode and a thin plate of pure metal is made the cathode. An aqueous solution of the suitable salt of the metal is used as the electrolyte. When electric current is passed, the pure metal from the anode gets deposited on the cathode. The insoluble impurities settle down at the bottom as anode mud. For some particular cases, specialised techniques such as ion-exchange chromatography, solvent extraction and amalgamation are required. These techniques have been discussed in the later chapters.

7.5

THERMODYNAMICS OF THE METALLURGY: ELLINGHAM DIAGRAM

The reduction of metal during extraction depends upon the standard electrode potential of the metal. However, one more criterian is important, i.e. the free energy change taking place during the reduction process. In general, a spontaneous reaction takes place with decrease in free energy of the system, i.e. DG should be –ve. Thus, the reduction of an oxide, sulphide or halide ore, at a given temperature and pressure, takes place spontaneously only if it occurs with decrease in the free energy of the system. If we consider a general oxidation reaction of a metal as 2M(s) + O2(g) $ 2MO(s)

Extraction of Elements

7.7

the free-energy change, DG is given by the relation DG = DH – TDS where DH is the enthalpy change and DS is the entropy change at the temperature T. Oxide formation takes place with decrease in entropy due to the fact that the gaseous reactant, oxygen with higher entropy is consumed, while the metal oxide with lesser entropy is formed. Thus, DG of the system becomes negative, so that the term TDS gets more negative, with increase in temperature and consequently, DG becomes increasingly less negative. As the temperature is increased further, DG would become zero and may become positive so that the oxidation reaction would get nonspontaneous. The plots of DG° for oxidation of metals to the oxides using one mole of oxygen against temperature are known as Ellingham diagrams. Figure 7.7 shows the Ellingham diagram following a straight line in most of the cases. However, some significant features are seen as follows. 1. An Ellingham diagram is a straight line up to the melting point of the metal. At the melting point or boiling point, there is a large change in entropy resulting in change of slopes of the lines, e.g. the slopes of Ca–CaO, Mg–MgO and Hg–HgO lines change at their boiling points, i.e. 1440°C, 1107°C and 356°C respectively. The marked increase in slope, at the boiling point of the metal, is due to the disappearance of metal vapour and gaseous oxygen leading to large entropy loss. 2. As the temperature is increased, DG becomes less and less negative and a stage is reached when DG becomes zero. Below this particular temperature, the reduction process is spontaneous as DG is negative and the enthalpies of formation of the oxides are negative, i.e. the oxides are stable. However, above this temperature, DG is positive and oxides are unstable, i.e. decompose to give back the metals and oxygen. Such is seen in case of mercury and silver and hence these metals can be obtained by the thermal decomposition of their corresponding oxides at an easily attainable temperature. This process is known as hydrometallurgical process.

Fig. 7.7 Ellingham diagram for some metals

7.8

Inorganic Chemistry

3. The Ellingham diagram can be used to predict the tendency of a metal to reduce other metal oxides. In general, any metal can be used to reduce the oxides of other metals lying above it in the Ellingham diagram, e.g. aluminum, lying below Cr–Cr2O3, can be used to reduce Cr2O3 to Cr. This can be illustrated by considering the free-energy change of formation of Al2O3 and Cr2O3

4 2 Al + O2  → Al2 O3 DG = – x kJ 3 3 4 2 Cr + O2  → Cr2 O3 DG = – y kJ 3 3 4 2 2 4 Al + CrO2  → Al2 O3 + Cr DG = (y – x) kJ 3 3 3 3 It is clear from the Ellingham diagram that free energy of formation (DG) of Al2O3 is more negative that of Cr2O3 as the Al–Al2O3 line lies below the Cr–Cr2O3 line. That means (x – y) is negative, i.e. the free energy change for the reduction of Cr2O3 by Al is negative and Al can reduce Cr2O3 to Cr. 4. The Ellingham diagram is used to predict the reduction behaviour of carbon and carbon monoxide (Fig. 7.8). Carbon can react with oxygen in two ways: C + O2 $ CO2 and

2C + O2 $ 2CO

∆G° (kJ/mol)

2Z

n+

O

2

→

2Z nO

It can be seen from the Ellingham diagram that free energy change for the C–CO2 line is practically independent of temperature due to the formation of same volume of CO2 as that of the O2 used. On the otherhand the C–CO line is sloping downwards with increase in temperature. This is because of the reason that in this reaction, one volume of oxygen is used and two volumes of CO are produced. Thus, there is an increase in entropy, i.e. DS is positive and DG becomes more and more negative as the temperature is increased. Now, the two lines intersect at about 710°C. It is clear that the first reaction is thermodynamically more feasible below 710°C, so that carbon can reduce oxides of metals lying below the Ellingham C–CO2 line and itself is oxidised to CO2. However, above 710°C, the second reaction is more favourable, i.e. carbon can reduce all the oxides Zn – ZnO 200 and is itself is oxidised to CO. CO 2 CO – CO2 But it is usually difficult to attain 0 →2 2 O + very high temperatures and hence, 2CO –200 other methods are preferred for the C + O2 → CO2 reduction of metal oxides. Since C – CO2 –400 below 710°C, CO–CO2 line lies below 2C + O 2→2 the C–CO2 line, it can be concluded –600 CO that in this temperature range, CO is C – CO –800 a better reducing agent than C, that is why Fe2O3 is believed to be reduced –1000 with CO in this temperature range. –1200 In case of zinc (boiling point 1180°C), reduction by C becomes 1140 thermodynamically favoured only 500 710 1000 1500 2000 2500 after 1000°C. Hence, reduction of Zn Temp. (°C) with C produces metal in the vapour Fig. 7.8 Ellingham diagram for carbon

Extraction of Elements

7.9

form. Thus, the metal oxide can be reduced by carbon at the temperature at which C–CO line lies below the metal-metal oxide line. 5. Ellingham diagram also accounts for the conversion of the sulphide ores into their oxides before reduction with carbon. There is no compound CS analogous to CO, with negative value of free energy of formation. Thus, it is necessary to convert sulphide ores into their oxides before the reduction can be carried out with carbon.

Elements occur in nature in either native state (metals with positive standard electrode potential) or combined state (metal with negative standard electrode potential). The process of extracting metals in their free states from their ores is known as metallurgy. There are various steps involved in a metallurgy process. The crushing and grinding of ones is known as pulverisation. The removal of impurities from the ore its known as concentration of ore. It can be done by handpicking, magnetic separation, gravity separation method, leaching or froath floation process. The concentration ore is heated strongly either in absence of air (calcination) or in excess of air (roasting) just below the melting point of the metal. The calcined or roasted ore is reduced by various reducing agents such as sodium, potassium, calcium, aluminium, H2, carbon, carbon monoxide or oxygen. Highly active metals are extracted by the electrolytic reduction of their suitable molten salts. The fluid metal obtained after reduction is purified by using various methods depending upon the nature of the metal and impurities. These methods are fractional distillation (volatile metal with nonvolatile impurities), liquation method (metals more readily fusable than the impurities), zone- refining (metal melts as well as crystallises earlier than the impurities), thermal decomposition (formation of a volatile compound by a metal while impurities remain as such) and electrolytic refining.

EXAMPLE 1 Give the chemical reactions for roasting of sulphide ores of zinc, lead, mercury and antimony. ∆

ZnS + 3O2  → 2ZnO + 2SO2 ∆

2Cu2S + 3O2  → 2Cu2O + SO2 Cu2S + O2 $ 2Cu + SO2 PbS + O2 $ Pb + SO2 HgS + O2 $ Hg + SO2 Sb2S3 + 3O2 $ 2Sb + 3SO2

EXAMPLE 2 reduced by H2.

Give the reason for the statement: The oxides of s-block elements cannot be

7.10

Inorganic Chemistry

The reaction for the reduction of a metal oxide can be represented as ∆

MO3 + 3H2  → M + 3H2O i.e.

Mn+ $ M + ne–

For a spontaneous reaction, the standard potential of the metal should be positive. Since the standard potential of the alkali metals and alkaline earth metals is highly negative, they cannot be reduced by H2.

EXAMPLE 3 Give an example of each for acidic flux and basic flux. Acidic flux—SiO2 Basic flux—CaO

EXAMPLE 4 Give the principle of the liquation process. The liquation process is based on the principle of difference in melting points of the metal and the impurities. Thus, the metal is readily fusible while the impurities are infusible.

QUESTIONS Q.1 Discuss the occurrance of elements in nature. Q.2 Write short notes on calcination and aluminothermite process. Q.3 Discuss the following in brief. (a) Froth-flotation process (b) Roasting (c) van Arkel method Q.4 Discuss the method for the extraction of a metal from its oxide and sulphide ores with the help of suitable examples. Q.5 Discuss smelting process is brief. Give the role of acidic and basic flux in metallurgy. Q.6 What is an Ellingham diagram? How is it constructed? Q.7 How will you use Ellingham diagram (a) To predict the reduction behaviour of carbon and carbon monoxide? (b) To predict the tendency of a metal to act as a reducing agent? Q.8 Discuss various methods used for concentration of ores. Q.9 What is pyrometallurgical process? Discuss in detail. Q.10 Discuss various methods used for refining of crude metals.

MULTIPLE-CHOICE QUESTIONS 1. The metal which is extracted by leaching of its ore with dilute cyanide solution: (a) Na (b) Be (c) Zn (d) 2. The compound used as acidic flux is (a) CaO (b) MgO (c) CO2 (d) 3. The metal which cannot be extracted by reduction with hydrogen is (a) Na (b) Cu (c) Ag (d) 4. Zone refining is used in the purification of (a) Ag (b) Zn (d) Si (d)

Ag SiO2 Fe Cu

chapter

Periodic Table and Periodic Properties

8

After studying this chapter, the student will be able to

8.1

INTRODUCTION

Since the discovery of elements, attempts have been made to classify and group the elements into groups or families. Dobereiner classified some elements into triads, groups of three elements, but couldn’t apply this classification to all elements. Newland proposed the law of octaves stating that if elements are arranged in the order of their increasing atomic weights, every eighth element will have similar properties as that of the first element. But this method of classification failed with the discovery of noble gases. Lothar Meyer plotted a graph of atomic volumes of elements versus their atomic weights. However, the most noteworthy work was made by a Russian chemist, Mendeleef, which led to the formation of first significant periodic table.

8.2

MENDELEEF’S PERIODIC TABLE

Mendeleef in 1869 enunciated a law, known as Mendeleef’s Periodic Law. According to this law, the physical and chemical properties of elements are periodic functions of their atomic weights, i.e. if the elements are arranged in the increasing order of their atomic weights, elements with similar properties are repeated at regular intervals. He arranged the then known elements in the form of a table called Mendeleef’s

8.2

Inorganic Chemistry

periodic table. His table not only made the study of elements easier, but also predicted the existence as well as properties of unknown elements. However, it suffered from some serious drawbacks as listed below: 1. The position of hydrogen was not justified as it resembles both alkali metals and halogens. 2. He placed many elements with dissimilar properties in the same group (e.g. coinage metals with alkali metals) and many elements with similar properties in different groups. (i.e. Cu and Hg). 3. The positions of lanthanides and actinides was not justified. 4. He placed four pairs of elements against his own law, i.e. elements of lower atomic weights preceeded the elements of higher atomic weights (e.g. Ar and K). 5. Isotopes of elements were not considered.

8.3

MODERN PERIODIC LAW AND PERIODICITY

Henry Moseley, a British scientist, observed that the physical and chemical properties of elements depend upon the number of electrons and their arrangement in the atom. This idea led Moseley to put forth the modern periodic law which states that the physical and chemical properties of the elements are periodic functions of their atomic number, i.e. if the elements are arranged in the increasing order of their atomic numbers, elements with similar properties are repeated at regular intervals. This repetition of the elements with similar properties at regular intervals of atomic numbers is called periodicity.

Cause of Periodicity The elements in a particular group have similar outer-shell electronic configuration as evident from Table 8.2. As already stated, physical and chemical properties of elements depend upon the outer-shell electronic configuration. Hence, elements with similar outer-shell electronic configuration have similar physical and chemical properties. Now elements with similar outer-shell electronic configurations are repeated after certain regular intervals of atomic numbers as 2, 8, 8, 18, 18, 32 and 32. These numbers are called magic numbers. More discussion on magic numbers will be done later on.

8.4

LONG FORM OF PERIODIC TABLE

After Mendeleef’s work, many scientists suggested different forms of tables based on the modern periodic law. However, the most widely used table is Bohr’s table, commonly known as long form of periodic table. The main features of this table have been discussed below. 1. The long form of the periodic table is divided into four blocks—s, p, d and f block, depending upon the type of orbital in which the last electron is filled up. The vertical columns of the periodic table are called groups and the horizontal rows are called periods. 2. There are seven periods in the table: (i) First period (shortest period): This period has only two elements—H and He. (ii) Second period (short period): This period has eight elements—Li to Ne (known as true representative elements.) (iii) Third period (short period): This period also has eight elements—Na to Ar (known as typical elements) (iv) Fourth period (long period): This period has eighteen elements—K to Kr. (v) Fifth period (long period): This period has eighteen elements—Rb to Xe.

I

Ag 107.87

Cu 63.54

B

Sr 87.62

Be 9.012 Mg 24.31 Ca 40.08

A

II

B 10.81 Al 26.98

A

Zn Ga 65.37 69.72

B

III

Y 88.91

Sc 44.96

B

IVa

Ge 72.59

B

Zr 91.22

Ti 47.90

C 12.011 Si 28.09

A

Ce 140.12

Th 232.04

**Actinide elements

Pa (231)

U Np 238.23 (237)

Pr Nd Pm 140.91 144.24 (145)

V

Nb 92.91

Bi 208.98

Ta 180.95

Sb 121.75

As 74.92

B

V 50.94

N 14.007 P 30.974

A

Po (210)

Te 127.60

Se 78.92 Mo 95.94

Cr 52.20

B

W 183.85

VI

O 15.999 S 32.06

A

At (210)

I 126.90

Br 78.96

B

VIII Transition triads

Re Os Ir Pt 186.85 190.2 192.2 195.9

Rh Pd Tc Ru (99) 101.07 102.91 106.4

Co Ni Mn Fe 54.94 55.85 58.33 58.71

VII

F 18.998 Cl 35.453

A

Rn (222)

Xe 131.30

Kr 83.80

He 4.0036 Ne 20.183 Ar 39.948

Zero Noble gases

Pu (244)

Am (247)

Cm (245)

Bk (247)

Cf (249)

Es (254)

Fm (252)

Md (256)

No (254)

Lw (257)

Sm Eu Gd Tb Dy Ho Er Tm Yb Lu 150.35 151.96 157.25 158.92 162.90 164.93 167.26 168.93 173.03 174.97

Sn Cd In 118.69 112.40 114.82 * Rare earths Ba Cr Hf (Lanthanide 137.34 132.90 178.49 element) La Pb 138.91 207.19 Hg Tl Au 200.59 204.37 196.97 Fr Ra * Actinide ele223 226 ments Ac (227) Eka H 112

Rb 85.47

H 1.008 Li 6.939 Na 22.99 K 39.102

A

Rare earths *(lanthanide elements)

Groups

Table 8.1 Modified form of Mendeleef’s periodic Table

Periodic Table and Periodic Properties

8.3

8.4

Inorganic Chemistry

(vi) Sixth period (longest period): This period has 32 elements—Cs to Rn including lanthanides. (vii) Seventh period (Incomplete period): This period is an incomplete period and has at present 27 known elements—Fr to Cn, Fl and Lv including actinides (all are radioactive). Out of these, only six elements from Fr to U are naturally occurring while the remaining elements have been prepared artificially through nuclear reactions and are called transuranic elements.

Table 8.2

Electronic configuration of elements

Group

Electronic configuration

1

H–1s1 Li–[He] 2s1 Na–[Ne] 2s1 K–[Ar] 4s1 Rb–[Kr] 5s1 Cs–[Xe] 6s1 Fr–[Rn] 7s1

ns1

Be–[He]2s2 Mg–[Ne] 3s2 Ca–[Ar] 4s2 Sr–[Kr] 5s2 Ba–[Xe] 6s2 Ra–[Rn] 7s2

ns2

3. There are 18 groups in the table. (i) Group 1

(Group IA)

– Alkali metals (H to Fr)

(ii) Group 2

(Group IIA)

– Alkaline earth metals (Be to Ra) (active metals)

(iii) Group 13

(Group IIIA)

– Boron family (B to Tl)

(iv) Group 14

(Group IVA)

– Carbon family (C to Pb)

(v) Group 15

(Group VA)

– Pnicogens or nitrogen family (N to Bi) (metals and nonmetal)

(vi) Group 16 (Group VIA) family (O to Po)

– Chalcogens

(vii) Group 17

– Halogens (F to At)

(Group VIIA)

or

(viii) Group 11 Au)

(Group IB)

– Coinage metals (Cu, Ag,

(ix) Group 12

(Group IIB)

– Zn, Cd, Hg

(x) Group 3 Lw

(Group IIIB)

– Sc, Y, La to Lu and Ac to

(xi) Group 4

(Group IVB)

– Ti, Zr, Hf and Ku

(xii) Group 5

(Group VB)

– V, Nb, Ta and Ha

(xiii) Group 6

(Group VIB)

– Cr, Mo, W and Unh

(xiv) Group 7

(Group VIIB)

– Mn, Tc and Re – Fe, Ru and Os

(xvi) Group 9

(Group VIIIB) – Co, Rh and Ir

(xvii) Group 10

13

B–[He] 2s22p1 Al–[Ne] 3s23p1 Ga–[Ar] 4s24p1 ns2np1 In–[Kr] 5s25p1 Tl–[Xe] 6s26p1

14

C–[He] 2s22p2 Si–[Ne] 3s23p2 Ge–[Ar] 4s24p2 Sn–[Kr] 5s25p2 Pb–[Xe] 6s26p2

ns2np2

N–[He] 2s22p3 P–[Ne] 3s23p3 As–[Ar] 4s24p3 Sb–[Kr] 5s25p3 Bi–[Xe] 6s26p3

ns2np3

O–[He] 2s22p4 S–[Ne] 3s23p4 Se–[Ar] 4s24p4 Te–[Kr] 5s25p4 Po–[Xe] 6s26p4

ns2np4

F–[He] 2s22p5 Cl–[Ne] 3s23p5 Br–[Ar] 4s24p5 I–[Kr] 5s25p5

ns2np5

oxygen

These elements are collectively called normal or maingroup elements.

(xv) Group 8

2

– Ni, Pd and Pt

These elements of group IB to VIII B and Sc and Y (from group IIIB) are collectively known as transition elements while the remaining elements in Group IIIB are called inner transition elements. (iii) Group 18 (zero group): This group has noble gases— He, Ne, Ar, Kr, Xe and Rn.

15

16

17

d-block elements p-block elements

107 Bh

60 Nd 92 U

106 Sg

59 Pr 91 Pa

105 Db

58 Ce 90 Th

89–103 104 Ac–Lr Rf

Lanthanoids

Actinoids

88 Ra

87 Fr

75 Re

74 W

73 Ta

72 Hf

57–71 La–Lu

62 Sm 94 Pu

61 Pm 93 Np

95 Am

63 Eu

f-block elements

110 Ds

96 Cm

64 Gd

97 Bk

65 Tb

98 Cf

66 Dy

113 Uut

112 Cn

109 Mt

111 Rg

108 Hs

81 Tl

80 Hg

79 Au

78 Pt

77 Ir

76 Os

99 Es

67 Ho

114 Fl

82 Pb

100 Fm

68 Er

115 Uup

83 Bi

51 Sb

56 Ba

49 In

48 Cd

47 Ag

46 Pd

45 Rh

44 Ru

50 Sn

55 Cs

43 Tc

42 Mo

41 Nb

40 Zr

39 Y

38 Sr

30 Zn

29 Cu

37 Rb

22 Ti

21 Sc

33 As

7 N

VA

32 Ge

28 Ni

6 C

IVA

31 Ga

27 Co

IIIA

20 Ca

IIB

19 K

IB

15 P

26 Fe

VIIIB

14 Si

25 Mn

VIIB

13 Al 24 Cr

VIB

12 Mg 23 V

VB

11 Na

IVB

5 B

IIIB

4 Be

IIA

3 Li

1 H

IA

101 Md

69 Tm

116 Lv

84 Po

52 Te

34 Se

16 S

8 O

VIA

102 Nb

70 Yb

117 Uus

85 At

53 I

35 Br

17 Cl

9 F

VIIA

103 Lr

71 Lu

118 Uuo

86 Rn

54 Xe

36 Kr

18 Ar

10 Ne

2 He

Group Group Group Group Group Group Group Group Group Group Group Group Group Group Group Group Group Group 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

s-block elements

Table 8.3 Long form of Periodic Table

Periodic Table and Periodic Properties

8.5

8.6

Inorganic Chemistry

Concept of Magic Numbers

Table 8.4 Filling of elements in periods

Period Corresponding No. of elements The entire periodic table is based on the electronic valence orbitals configuration of elements. Period number which 1 2 1s indicates the number of shell last occupied by 2 8 2s, 2p the electron in the atom of the elements. Each 3 8 3s, 3p period contains the number of elements equal 4 18 4s, 3d, 4p to the maximum number of electrons that can be accommodated in that last shell. This can be 5 18 5s, 4d, 5p illustrated as follows: 6 32 6s, 5d, 4f, 6p The first period corresponds to elements with 7 32 (incomplete yet) 7s, 6d, 5f, 7p valence (last) shell n = 1. For n = 1 (only 1s-orbital), only two electrons can be added, hence there are only 2 elements in the first period. The second period corresponds to elements with valence shell n = 2. For n = 2 (2s and 2p orbitals), a total of eight electrons can be added; hence there are 8 elements in the second period. The third period corresponds to elements with valence shell n = 3. For n = 3 (3s, 3p and 3d orbitals), total eighteen electrons can be added. But the 3d orbital has higher energy as compared to the 4s orbital. Hence, only eight elements corresponding to filling of 3s and 3p orbitals are present in the third period. Similar is the case of remaining periods as shown in Table 8.4.

8.5

PERIODIC PROPERTIES

The properties of the atoms of the elements are called atomic properties. On proceeding from left to right, in a period, and from top to bottom, in a group of the periodic table, atomic properties show a regular variance. Hence, these properties, are also called periodic properties. Certain important periodic properties have been discussed in the chapter.

8.5.1 Atomic Radius The distance between the nucleus and the outermost shell of electrons of an atom and an ion is generally named as the atomic radius and ionic radius respectively. However, it is not possible to isolate an individual atom or ion. Hence, the exact position occupied by an electron cannot be defined with certainty. Further, the probability distribution of an electron in an atom is influenced by the presence of other atoms in its environment resulting in the change of atomic size. Thus, the absolute size of an atom is difficult to be defined and arbitrary concepts are discussed here.

1. Covalent Radius This concept is used for nonmetals. One half of the distance between the nuclei of two atoms bonded together by a single covalent bond is called the covalent radius. X-ray diffraction or spectroscopic measurements are used to determine the distance between the nuclei of any two neighbouring atoms. For example, the internuclear distance in Cl2 molecule is 198 pm. This means covalent radius of Cl atom is 99 pm. In general, for a homonuclear diatomic molecule A2, linked by a single covalent bond, the covalent radius of the atom A is given by the relation d rA = A A 2 where dA–A is the internuclear distance of the A2 molecule, also known as bond length of the molecule. Spectroscopic studies reveal that in case of a heteronuclear diatomic molecule, AB, the length of the bond is shorter than the sum of the covalent radii of the individual atoms. Due to difference in the

Periodic Table and Periodic Properties

8.7

electronegativities of the two bonded atoms, covalent bond gets some ionic character, i.e. polarisation of the atoms takes place resulting in additional electrostatic attraction between the two atoms. As a result, the bond length gets shortened. Schomaker and Stevenson suggested the following relation for heteronuclear diatomic molecule, AB: rA–B = rA + rB – 0.09 (cA – cB) where rA and rB are the normal covalent radii of the atoms A and B respectively, and cA and cB are the electronegativities of the atoms A and B respectively. This equation was modified by Porterfied as rA–B = rA + rB – 0.07 (cA – c )2 The atomic radii of s and p-block elements are listed in Table 8.5. It has been seen that covalent radius of an atom linked by a double or triple bond is shorter than its normal covalent radius (determined from a single bond). For example, Single bond covalent radii (pm)

B–B 82 B=B 76 B B 68

Double bond covalent radii (pm) Triple bond covalent radii (pm)

C–C 77 C=C 67 C C 60

N–N 75 N=N 60 N N 55

This is because of the reason that single covalent bond is formed by overlap of sigma orbitals, whereas double and triple bonds involve overlapping of sigma ( ) as well as pi ( ) orbitals. As a result, the atoms come more closer reducing the bond length i.e. why the covalent radius of the C atom for C–C, C=C and C C (bond distances are equal to 154 pm, 134 pm and 120 pm respectively) goes on decreasing. This indicates that magnitude of the bond length decreases with increase in the bond multiplicity.

2. Van der Waal’s Radius This concept is used for noble gases and for molecules of nonmetallic elements. It is defined as one half of the distance between the nuclei of two nonbonded and identical neighbouring atoms of two adjacent Table 8.5 The atomic radii of s and p-block elements

IA H 37 Li 125 Na 154 K 203 Rb 216 Cs 235

IIA

Be 90 Mg 136 Ca 174 Sr 192 Ba 198

Covalent bond radii (pm) IIIA IVA VA

B 82 Al 118 Ga 126 In 144 Tl 148

C 77 Si 111 Ge 122 Sn 141 Pb 147

N 75 P 106 As 119 Sb 138 Bi 146

VIA

VIIA

O 74 S 109 Se 116 Te 135 Po 146

F 72 Cl 99 Br 114 I 133 At 145

Vander Waals radius (pm) Zero He 120 Ne 160 Ar 191 Kr 200 Xe 220

8.8

Inorganic Chemistry

molecules in the solid state. Since this radius corresponds to van der Waal’s forces of attraction between the molecules, its magnitude is always greater than that of covalent radius, e.g. van der Waal’s radius of hydrogen atom is 120 pm while its covalent radius is 37 pm as illustrated in Fig 8.1. The Van der Waal radii of some common elements are shown in Table 8.6.

3. Metallic Radius This term is used for radius of metallic atoms assumed to be closely packed spheres in the metallic crystals. In case of metals, covalent radius is not determined since most of the metals don’t form covalent compounds (exception—metallic hydrides and organometallic compounds) The radius of metallic atoms is determined from the atomic volumes of metallic phases which are determined by dividing their atomic masses by their respective densities. The metallic radii are also smaller than van der Waal’s radii because metallic bonds are stronger than the van der Waal’s forces. Metallic radii of some s and p block elements are shown in Table 8.7.

van der waal’s radii of some elements (pm)

Table 8.6 VA N 150 P 190 As 200 Sb 220

VIA O 140 S 185 Se 200 Te 220

VIIA F 136 Cl 180 Br 195 I 215

Zero He 120 Ne 160 Ar 191 Kr 200

van der Waal’s radius 120 pm

Covalent radius (37 pm)

Fig. 8.1 Two adjacent molecules

Table 8.7 Metallic radii of s and p-block elements (pm) IA Li Na K Rb Cs

IIA 155 190 235 248 267

Be Mg Ca Sr Br

IIIA 112 160 197 215 222

B Al Ga In Tl

IVA 98 143 141 166 171

C Si Ge Sn Pb

VA 91 132 137 162 175

N P As Sb Bi

VIA 92 128 139 159 170

O S Se Te Po

– 121 140 160 176

4. Factors Affecting the Atomic Size (a) Number of Shells Atomic size of an atom directly depends upon the number of shells present in that atom. As the number of shells increases, the outermost-shell electrons in the atom get farther away from the nucleus increasing the atomic size. (b) Effective Nuclear Charge Atomic size of an atom is inversely proportional to the effective nuclear charge of the atom. As the effective nuclear charge increases, the force of attraction between the nucleus and the outermost-shell electron increases resulting in moving of electron cloud closer to nucleus. Hence, it results in decrease of atomic size of the atom. (c) Bond Multiplicity As already discussed, atomic size of an atom is inversely proportional to the bond multiplicity.

5. Periodic Trends In general, atomic radii shows an increase in magnitude on moving from top to bottom in any group and decrease in magnitude on moving from left to right in any given period. (a) Variation in a Group As we move from top to bottom in any group, the number of shells and the effective nuclear charge goes on increasing, but the effect of addition of new shells dominates over

Periodic Table and Periodic Properties

8.9

the effect of effective nuclear charge and hence the atomic size goes on increasing from top to bottom in any group. For example, Elements of Group IA Li Na K Rb Cs Outer-shell electronic configuration [He]2s1 [Ne]3s1 [Ar]4s1 [Kr]5s1 [Xe]6s1 Covalent radius (pm) 125 154 203 216 235 (b) Variation in a Period As we move from left to right in any period, the number of shells remains the same as the electrons are added in the same shell. However, the effective nuclear charge goes on increasing and pulls the electron cloud closer to the nucleus. As a result, the atomic size of an atom goes on decreasing on moving from Li to F. It can be seen from the table 8.5 that atomic size of noble gases are greater than the preceeding elements in their respective periods. This is because of the reason that in case of noble gases, van der Waal’s radius is determined, whose magnitude is larger than the covalent radius as discussed earlier. Table 8.8 Varification of atomic size for elements of second period Li Be Outer-shell [He]2s1 [He]2s2 elements 1.30 Zeff by the 1.95 (2s1) outermost (2s1) shell electron Atomic size 125 90 (pm)

B C [He]2s22p1 [He]2s22p2

N [He]2s22p3

O [He]2s22p4

F [He]2s22p5

Ne 1s22s22p6

2.60 (2p1)

3.25 (2p1)

3.90 (2p1)

4.55 (2p1)

5.20 (2p1)

5.85 (2p1)

82

77

75

74

72

160

8.5.2 Ionic Radius The radius of an ion corresponds to the ionic radius. Like atomic radius, ionic radius is also defined in the same way as the distance from the nucleus of the anion up to the point where the nucleus has influence on its electron cloud. Ionic radii are determined from X-ray measurements. The distance between the nuclei of the two ions is determined and is taken as the sum of the radii of the two ions, i.e. d(A+B–) = rA+ + rB–

Na Mg

150 Li

Al Si 50

Be H

0

Fig. 8.2

BC NO

F

P

S Cl

Ar

Ne

10 Atomic number $

20

Covalent radii of the elements of first and second periods

Atomic and ionic radius (pm)

Covalent radius (pm) $

where d(A+ B–) is the internuclear distance between the nucleus of cation (A+) and nucleus of anion (B–) and rA+ and rB– are the radii of cation (A+) and anion (B–) respectively. Cs Rb

200

K

Cs+

Rb+ Na

Li

K+

Na+

100 Li+

3

11

19 37 Atomic number

55

Fig. 8.3 Atomic and Ionic radii of Group 1 elements

8.10

Inorganic Chemistry

1. Radius of Cation A cation is formed by the loss of one or more electrons from an atom. Sometimes it results in the removal of the whole of the outer shell of the electrons. But the nuclear charge remains the same resulting in an increase in effective nuclear charge. As a result, the electron cloud is pulled more towards the nucleus resulting in decrease of radius of cation than the parent atom. For example, atomic radius of sodium is 154 pm and ionic radius of Na+ is 95 pm. Figure 8.3 shows the comparison of atomic and ionic radii of group 1 elements. (One shell is removed) Electronic configuration Nuclear charge: Radius (pm):

Na Na+ 2 6 1 2 2 6 1s 2s 2p 3s 1s 2s 2p 11 11 154 95

Table 8.9 Comparison of Ionic radii

2

Mn 126 pm Mn2+ 80 pm Mn 46 pm

2. Radius of Anion

Pb 147 pm Pb2+ 120 pm Pb4+ 84 pm

Fe 126 pm Fe2+ 76 pm Fe3+ 64 pm

Table 8.10 Ionic radii for halogens

(No change in last shell) Electronic configuration: Nuclear charge: Radius (pm):

F 1s 2s22p5 9 72 2

F– 1s 2s22p6 9 133 2

3. Radii of Iso-electronic Species The atoms or ions having the same number of electrons but different nuclear charges are known as iso-electronic species. It is quite clear from the above discussion that as the nuclear charge increases, the electrons are held more tightly by the nucleus resulting in a decrease in the ionic radius. Periodic Trends in Ionic Radii In general, the ionic radii shows an increase in moving from top to bottom in a group and a decrease in moving from left to right in a period. It is because of the reason that the atomic size of the elements (from which the ion is derived) increases from top to bottom in a group and decreases in a period.

Atomic and ionic radius (pm)

An anion is formed by the addition of one or more electrons to the F 72 pm F– 133 pm outermost shell of an atom but the nuclear charge and the number of Cl 99 pm Cl– 181 pm shell remains the same resulting in a decrease in the effective nuclear Br 114 pm Br– 195 pm charge. Hence, the electron cloud expands resulting in increase of radius I 133 pm I– 216 pm of anion, more than its parent atom. For example, atomic radius of F is 72 pm and ionic radius of F– is 133 pm. Fig. 8.4 shows the comparison of atomic and ionic radii of group 17 elements. 200

F

I–

Br–

Cl–

I

–

Br 100 F

9

Cl

17

35 Atomic number

53

Atomic and ionic radii of Group 17 elements

Fig. 8.4

Isoelectronic species Nuclear charge Number of electrons Ionic radius (pm)

N3– +7 10 171

O2– +8 10 145

F– +9 10 133

Na+ +11 10 95

Mg2+ +12 10 60

Table 8.11 Comparison of Ionic radii of some elements +

Li

+

Na +

60 pm 95 pm

Be

2+

Mg

2+

2+

31 pm 65 pm 99 pm

B3+ 3+

Al

3+

Ga

20 pm

C4+

50 pm

4+

Si

4+

15 pm

N3–

171 pm O2–

41 pm

3–

2–

P

3–

212 pm S

2–

145 pm F– 184 pm Cl

133 pm –

181 pm

–

186 pm

K

133 pm Ca

61 pm

Ge

53 pm

As

222 pm Se

202 pm Br

Rb+

148 pm Sr2+

110 pm In3+

81 pm

Sn4+

71 pm

Sb3–

245 pm Te2–

222 pm I–

Cs+

167 pm Ba2+ 129 pm Tl3+

91 pm

Pb4+

84 pm

Bi3–

120 pm

219 pm

Periodic Table and Periodic Properties

8.11

8.5.3 Ionisation Energy (IE) Ionisation energy of an element is defined as the minimum amount of energy required to remove the outermost electron of an isolated gaseous atom resulting in the formation of the gaseous cation, i.e. I.E.

M(g) ææÆ M + (g) + e The process is called ionisation and the energy required is represented as IE measured in kJ/mole or electronvolts per atom. Successive Ionisation Energies The ionisation energy corresponding to the removal of first electron from the outermost shell of an atom is called IE1. Similarly, IE2 corresponds to the removal of an electron from M+(g) and IE3 corresponds to removal of an electron from M2+(g) and so on. + – M(g) IE 1 → M (g) + e 2+ – M+(g) IE 2 → M (g) + e 3+ – M2+(g) IE 3 → M (g) + e and so on

1. Factors Affecting the Magnitude of Ionisation Energy (a) Number of Shells and Atomic Size Ionisation energy is inversely proportional to the number of shells present in an atom of an element, i.e. more the number of shells, greater is the atomic size, farther is the electron from the nucleus and easier is its removal, i.e. lesser is the IE. (b) Nuclear Charge Ionisation energy is directly proportional to the nuclear charge of the atom. Greater the nuclear charge, greater is the energy required to remove an electron from the atom, i.e. more is the IE. (c) Screening Effect or Shielding Effect The inner-shell electrons act like a screen or shield for the outermost-shell electron. As a result, the outermost shell electron experiences less attraction from the nucleus. This is known as screening effect. Now, more the number of electrons in the inner shells, greater is the screening effect and lesser is the energy required to remove an electron from the outermost shell of an atom resulting in a decrease in the ionisation energy. (d) Penetration Effect Ionisation energy also depends upon the penetration effect, i.e. the type of electron to be removed. The penetration orders of the orbitals in a given shell is s > p > d > f. This means that ‘s’ electron is most penetrating towards the nucleus and hence rquires more energy for its removal followed by p, d and f electrons. Thus, ionisaion energy for an electron in any shell varies in the order of s > p > d > f electrons. (d) Electronic Configuration It is well known that half filled and fully filled electronic configurations are more stable and hence require more amount of energy for removal of an electron. Hence, atom with fully filled or half filled electronic configuration has higher ionisation energies than expected.

2. Periodic Trends In general, the ionisation energy decreases in moving from top to bottom in a group and increases from left to right in a period. (a) Variation in a Group As we move from top to bottom in any group, there is an increase in the atomic size, nuclear charge and shielding effect. However, the combined effect of increase in atomic size and shielding effect dominates over the effect of increase in the nuclear charge. Hence, ionisation energy decreases from top to bottom in any group with some exceptions to be discussed in the later chapters.

8.12

Inorganic Chemistry

IE1 (kJ mol–1)

(b) Variation in a Period As we move from left He 2000 to right in a period, there is no change in the Ne number of shells but atomic size decreases 1500 F N and nuclear charge increases, thereby pulling the electron cloud towards the nucleus. O H C 1000 Hence, more energy is required to remove the Be B electron from the outermost shell of the atom 500 Li resulting in an increase of ionisation energy. However, this charge is not smooth as clear 1 2 3 4 5 6 7 8 9 10 from Fig. 8.5. We can see that first ionisation Atomic number energy of ‘Be’ is higher as compared to ‘B’ Fig. 8.5 Variation of IE of some elements because of the reason that ‘Be’ has stable outer-shell electronic configuration and also the penetration effect of the ‘s’ electron is more than the ‘p’ electron. Similarly, ionisation energy of ‘N’ is higher than ‘O’ because of exactly half-filled and stable electronic configuration.

3. Order of Successive Ionisation Energy The order of successive ionisation energies is IE1 > IE2 > IE3 ... due to the fact that it is relatively more difficult to remove an electron from a cation having smaller size and greater electrostatic force of attraction.

8.5.4 Electron Affinity (EA) Electron affinity of an element is defined as the amount of energy released when an electron is added to an isolated gaseous atom to form an anion, i.e. X(g) + e– $ X–(g) + EA It is measured in kJ/mol or electronvolts per atom. The electron affinity corres-ponding to the addition of one electron to the outermost shell of an atom is called EA1, while the energy corresponding to the addition of the electron to an anion is called EA2. EA1 is positive while EA2 is negative, because energy has to be supplied to overcome the electrostatic forces of repulsion felt by the second electron and hence energy has to be supplied for its addition. For example, in case of oxygen, EA1 is positive (+ 140.9 kJ mol–1), but EA2 is negative (–842.9 kJ mol–1) as 702 kJ of energy is required to add an extra electron to the O– ion so as to convert it into the O2– ion.

First ionisation energies of s and p-block elements (kJ mol–1)

Table 8.12 IA H 1312 Li 520 Na 495 K 418 Rb 403 Cs 375

IIA

IIIA

IVA

VA

VIA

VIIA

Be 899 Mg 737 Ca 509 Sr 545 Ba 502

B 800 Al 577 Ga 579 In 558 Tl 589

C 1086 Si 786 Ge 760 Sn 708 Pb 715

N 1402 P 1011 As 944 Sb 831 Bi 703

O 1314 S 999 Se 945 Te 869 Po 813

F 1681 Cl 1255 Br 1140 I 1008 At 917

Table 8.13

Zero He 2372 Ne 2080 Ar 1520 Kr 1350 Xe 1170 Rn 1037

Successive ionisation energies (kJ mol–1)

Element Li

IE1 520

IE2 7298

IE3 11815

Be B C N O F Ne

899 800 1086 1402 1314 1681 2080

1757 2427 2353 2856 3388 3375 3952

14848 3660 4620 4578 5300 6050 6122

Periodic Table and Periodic Properties

8.13

1. Factors Affecting the Electron Affinity (a) Number of Shells and Atomic Size Electron affinity is inversely proportional to the number of shells present in an atom of an element, i.e. more the number of shells, more the atomic size and lesser the attraction of the nucleus for the electron to be added, i.e. lesser the electron affinity. (b) Nuclear Charge Ionisation energy is directly proportional to the nuclear charge of an atom. Greater the nuclear charge, more the force of attraction of the nucleus for the electron to be added and hence more the electron affinity. (c) Screening Effect or Shielding Effect As already discussed, more the number of electrons in the inner shell, more the screening effect and hence lesser the force of attraction of the nucleus for the electron to be added, i.e. lesser the electron affinity. (d) Electronic Configuration Exactly half-filled and fully filled electronic configurations are more stable and hence such atoms have no need to accept electrons and have almost zero and highly negative electron-affinity values.

2. Periodic Trends In general, electron affinity decreases on moving from top to bottom in a group and increases from left to right in a period. Table 8.14

Electron affinities of s- and p- block elements

(a) Variation in a Group As we move (kJ mol–1) from top to bottom in any group, there IA IIA IIIA IVA VA VIA VIIA VIIIA is an increase in the atomic size, nuclear Li Be B C N O F Ne charge and shielding effect. However, the 0 26.7 122.3 0 140.9 332.6 0 59.8 combined effect of increase in atomic size Na Mg Al Si P S Cl Ar and shielding effect dominates over the 0 42.6 133.6 72.3 200.7 348.5 0 53.1 effect of increase in the nuclear charge. K Ca Ga Ge As Se Br Kr Hence, force of attraction of the nucleus for 0 28.7 120 77.2 195 324.7 0 48.4 the electron to be added decreases resulting Rb Sr In Sn Sb Te II Xe in a decrease in electron-affinity values. 0 29.0 107.9 103 190.1 295.5 0 46.9 However, it has been seen that the first Cs Ba Tl Pb Bi Po At Rn elements of groups 15, 16 and 17 have lower electron affinities than the corresponding 0 11.2 35.1 31 180 270 0 45.5 next element in their groups (Table 8.14). This is because of the reason that the first elements of groups 15, 16 and 17 have very small size and high charge density which opposes the next electron to be added, resulting in lower electron affinity. (b) Variation in a Period As we move from left to right in a period, there is no change in the number of shells but atomic size decreases and nuclear charge increases, thereby increasing the force of attraction of the nucleus for the electron to be added and hence electron affinity increases. However, elements of Group 2, particularly ‘Be’ and ‘Mg’ have almost zero electron affinity because they have stable electronic configuration and addition of an extra electron to the 2p orbital having considerably higher energies would make the electronic configuration highly unstable. Similar is the case of Group 15 and 18 elements which have exactly half filled and completely filled electronic configurations respectively.

8.5.5 Electronegativity In case of a nonpolar covalent bond, i.e. bond between two identical atoms, the bonded pair of electrons is shared equally by the two atoms. However, in case of a polar covalent bond, i.e. the bond between two different atoms, the bonded pair of electrons is not shared equally by the two atoms. The atom which has a

8.14

Inorganic Chemistry

greater tendency to attract the shared pair of electrons towards itself, aquires partial negative charge while the other atom acquires partial positive charge. This relative tendency of an atom to attract the pair of electrons towards itself when combined in a compound was termed as the electronegativity of the atom by Pauling (1931). It is generally represented by c.

1. Scales of Electronegativity Different researchers have used different methods to measure electronegativity values. Some important methods are described here: (a) Pauling’s Method Pauling devised an electronegativity scale based on relationship between bond energies of the molecules and the difference of electronegativities of the bonded atoms. According to Pauling, the difference in electronegativity values of A and B atoms, A – B (for a diatomic molecule AB) is related to the ionic resonance energy of A–B bond, A–B as A

–

B

= kκ ∆ A − B

(8.1)

where, ∆ A − B is the measure of the ionic character in A-B covalent bond and is related to the bond dissociation energies of A–A and B–B as (

½ A–B)

½ = E  A − B − ( EA − A × EB− B ) 

½

(8.2)

½

½ = k E (8.3)  A − B − ( EA − A × EB− B )  The value of the constant is taken as 0.182, when the bond dissociation energies are expressed in kcal/ mole, ½ ½  i.e. (8.4) A – B = 0.182  EA − B − ( EA − A × EB− B ) 

i.e.

A

–

B

This equation is known as Pauling equation. Pauling determined the electronegativity values of elements by assuming an arbitrary value of electronegativity for H-atom equal to 2.1. These values are listed in Table 8.15. (b) Mulliken’s Method Mulliken (1934) proposed that the average of the ionisation energy (IE) and electron affinity (EA) of an atom is related to the electronegativity of the atom, M as

 IE + EA  = 0.187  (8.5)  + 0.17  2  here, IE and EA are measured in electronvolts. The main advantage of Mulliken’s method is that it helps in the determination of the electronegativities of an element in different oxidation states or hybridisation states. For example, electronegativity of C for all its molecules has been reported by Pauling as equal to 2.5. But according to Mulliken’s method, electronegativity of C in sp3, sp2 and sp hybridisation states are 2.48, 2.66 and 2.99 respectively. However, since only a few electron affinities are known, the method is not in much use.

Table 8.15

M

IA H 2.1 Li 1.0 Na 0.9 K 0.8 Rb 0.8 Cs 0.7

Pauling values of electronegativity of s- and p- block elements

IIA

IIIA

IVA

VA

VIA

VIIA

Be 1.5 Mg 1.0 Ca 1.2 Sr 1.0 Ba 0.9

B 2.0 Al 1.5 Ga 1.6 In 1.7 Tl 1.8

C 2.5 Si 1.8 Ge 2.0 Sn 1.9 Pb 2.3

N 3.0 P 2.1 As 2.0 Sb 1.8 Bi 1.7

O 3.5 S 2.5 Se 2.4 Te 2.1 Po 2.0

F 4.0 Cl 3.0 Br 2.8 I 2.4 At 2.2

(c) Allred–Rochow Method Allred and Rochow proposed that the electrostatic force of attraction, between an atom and its bonding electrons, is a measure of its electronegativity and is given by the relation

Periodic Table and Periodic Properties

=

0.359 Z eff

0.744 r2 where r is the covalent radius of the atom A in Å and Zeff is effective nuclear charge at the periphery of the atom. The calculated values of electronegativities of s- and p- block elements are listed in Tables 8.16. The values obtained are in close agreement with those obtained by Pauling and Mulliken. A

8.15

Table 8.16 Allred-Rochow values of electronegativity IA H 2.20 Li 0.97 Na 1.01 K 0.91 Rb 0.81 Cs 0.86

IIA

IIIA

IVA

VA

VIA

VIIA

Be 1.47 Mg 1.23 Ca 1.04 Sr 0.99 Ba 0.97

B 2.01 Al 1.47 Ga(III) 1.82 In 1.78 Tl(III) 1.44

C 2.50 Si 1.74 Ge(IV) 2.02 Sn(IV) 1.72 Pb(IV) 1.55

N 3.07 P 2.06 As(III) 2.2 Sb 1.82 Bi 1.67

O 3.50 S 2.44 Se 2.48 Te 2.01 Po 1.76

F 4.0 Cl 2.83 Br 2.74 I 2.21 At 1.90

Zero He 3.2 Ne 5.1 Ar 3.3 Kr 3.1 Xe 2.4 Rn –

2. Factors Affecting Electronegativity (a) Atomic Size Electronegativity is inversely proportional to atomic size as the smaller atom has more tendency to attract the shared pair of electrons towards itself and has higher electronegativity. (b) Effective Nuclear Charge As evident from the Allred and Rochow method, electronegativity is directly proportional to the effective nuclear charge of the atoms. Hence, electronegativity goes on increasing with the increase in effective nuclear charge. (c) Oxidation State As already discussed, size of the atom decreases with the increase in its oxidation state and the tendency of a cation to attract the electrons is more than a neutral atom. Thus, electronegativity increases with increase in oxidation state. Hence, electronegativity of iron in its various oxidation state changes as Fe(1.80) < Fe2+(1.83) < Fe3+ (1.96). It is similar with anion, i.e. electronegativity of an anion is lesser than its parent atom because of decrease in size. That is why electronegativity of F–(0.8) is lesser than that of F atom (4.0). (d) Hybridisation Electronegativity of an atom increases with increase in s-character of its hybrid orbitals because s-orbital is most penetrating towards the nucleus. Thus, electronegativity of carbon atom increases with change in hybridisation state as C2H2> C2H4> CH4 (sp) (sp2) (sp3) s-character 50% 33% 25% (e) Screening Effect As already discussed, more the number of electrons in the inner shell, more the screening effect and hence lesser the force of attraction of the nucleus for shared pair of electrons, i.e. lesser the electronegativity.

3. Periodic Trends In general, electronegativity decreases on moving from top to bottom in a group and increases from left to right in a period. (a) Variation in a Group As we move from top to bottom in a group, there is an increase in the atomic size, nuclear charge and shielding effect. However, the combined effect of increase in atomic size and shielding effect results in decrease of effective nuclear charge due to which the force of attraction of the nucleus for the shared pair of electron decreases i.e. the small atoms are more electronegative than the large atoms.

8.16

Inorganic Chemistry

(b) Variation in a Period As we move from left to right in a period, there is no change in the number of shells but effective nuclear charge increases due to decrease in atomic size and hence electronegativity goes on increasing. Hence, F is the most electronegative element and Cs is most electropositive element in the periodic table.

4. Application of Electronegativity (a) Prediction of Nature of Bond The Nature of bond between two bonded atoms depends upon difference of electronegativities of these two atoms. A–B bond would be nonpolar if ( A– B) is equal to zero and would be polar if ( A– B) is not equal to zero. (b) Calculation of Percentage Ionic Character According to Pauling, percentage ionic character of a covalent bond can be calculated by using the following correlations. Percentage ionic character = [1–e–0.25( – B) ] × 100 He gave another empirical equation as Percentage ionic character = 18 ( A– B)1.4 This equation gives more accurate results. It can be said that more the difference in the electronegativities of the two atoms, more is percentage ionic character of a bond. (c) Calculation of Bond Length Bond length of A–B bond can be calculated by using the correlation proposed by Schomaker and Stevenson as dA–B = rA + rB – 0.09 (

A–

B)

where is dA–B is the bond lengths of A–B bond; rA and rB are the covalent radii of A and B and ( A– B) is the electronegativity difference between the two atoms. The factor 0.09( A– B) accounts for the shortening of the bond length due to difference in electronegativity. (d) Explanation for Bond Angles Bond angle of a molecule depends upon the electronegativity of the central atom. More the electronegativity of the central atom, more is its tendency to hold the bonding pair of electrons towards itself and more is the bond angle due to increase in the bond pair-bond pair repulsion, e.g. the bond angle of trichlorides of Group 15 elements varies as NF3 > PF3 > AsF3 > SbF3. (e) Explanation of Diagonal Relationship When we move diagonally from first group to second group, there is no marked change in electronegativity due to cancellation of increase in electronegativity in a period by the decrease in electronegativity in a group. As a result, the first element of Group 1 has similar chemical properties as that of second element of Group II. This is called diagonal relationship.

Compound Electronegativity of central atom 3.0 NF3 2.0 PF3 2.2 AsF3 1.8 SbF3

Bond angle 102° 97° 96° 88°

(f) Calculation of Enthalpy of Formation of a Molecle Enthalpy of formation of a molecule can be calculated by using the relation proposed by Pauling as Hf = 23 ( A– B)2 – 55.4 nN – 26.0 nO where nN and nO are the numbers of nitrogen and oxygen atoms in the moledule and indicaties the number of bonds in the molecule. For example, for HCl, = 1 and for BeCl2, = 2, and so on. (g) Prediction of Acidic or Basic Nature of Compounds Consider the case of Be(OH)2 and HOCl. The electron pair of Be–O bond would shift towards the more electronegative oxygen atom.

Periodic Table and Periodic Properties

8.17

This would facilitate the release of OH– group due to cleavage of the Be–O bond. Thus, Be(OH)2 acts as a base. On the other hand, the electron pair of O–Cl bond would shift towards the more electronegative Cl (due to +1 oxidation state). This would make oxygen electron deficient and it would pull the electron pair of O–H bond. As a result, the O–H bond will be ruptured with the release of H+ ion. Thus, HOCl behaves as an acid. (h) Prediction of Reactions It has been observed that treatment of the nucleophile [Mn(CO)5]– with CH3I yields CH3Mn(CO)5, but with CF3I, Mn(CO)5I is obtained. It is due to the reason that in case of CH3I, the more electronegative iodine pulls the electron charge cloud towards itself and creates a partial positive charge on the C atom. The nucleophile attacks the positively charged C atom to form CH3Mn(CO)5. In case of CF3I, the most electronegative F atoms pull the electron charge cloud and create the electron deficiency at the C atom which now pulls the electron charge cloud of the easily polarisable I atom. The nucleophile now attacks the positively charged I atom to form Mn(CO)5I. CH3I + [Mn(CO)5]– $ CH3Mn(CO)5 + I– 2 CF3I + [Mn(CO)5]– $ Mn(CO)5I + I– + C2F6

8.5.6 Metallic Character of Elements Metallic character, or electropositive character, of an element is characterised by its ionisation energy. The elements with low ionisation energies are highly electropositive and vice versa as evident from Fig. 8.6. It can be seen that the elements located in the extreme left and the lower half of the left portion of the periodic table are strongly metallic while those located in the right portion of the periodic table are strongly nonmetallic. Some elements which lie in between these elements show characteristics of both metals and nonmetals and are called semi-metals or metalloids.

Alkaline earth metals

Alkali metals

s-block elements

p-block elements

Transition metals

Metals

Metalloids

B

C

N

O

F

Ne

Al

Si

P

S

Cl

Ar

Ga

Ge

As

Sc

Br

Kr

In

Sn

Sb

Te

I

Xe

Tl

Pb

Bi

Po

At

Rn

Nonmetals

Fig. 8.6 Metallic character of elements

8.6

SHIELDING OR SCREENING EFFECT

In multielectron atoms, the inner-shell electrons, i.e. the electrons present in the shells between the nucleus and the outermost shell, are known as intervening electrons. As a result, the electrons present in the outermostshell do not feel the actual nuclear charge. This reduction in the effect of nuclear charge due to the presence of intervening electrons is known as screening effect or shielding effect. The reduced nuclear charge, which is

8.18

Inorganic Chemistry

actually felt by the outermost-shell electrons, is known as the effective nuclear charge and is denoted by Zeff. The effective nuclear charge can be determined by the correlation Zeff = Zactual – S where S is the measure of the extent of shielding by the inner-shell electrons and is known as screening constant or shielding constant. It may be recalled that ‘s’ orbitals are most penetrating towards the nucleus and hence tend to shield the most and as the penetrating effect decreases (s > p > d > f), the shielding effect also decreases. As a result, the electrons present in different orbitals are attracted towards nucleus, by different extents and have different energies. That is why the energy-level diagram of multielectron atoms is different from the energy-level diagram of hydrogen or hydrogen-like species. Slater has proposed a set of empirical rules to estimate the extent of shielding based upon the average behaviour of the electrons present in various orbitals. According to these rules, the shielding constant ‘S’ can be calculated as follows: (a) The electronic configuration of an element is written in the following order and groups: (1s) (2s 2p) (3s 3p) (3d) (4s 4p) (4d) (4f) (5s 5p) etc. (b) Electrons present in the group to the right side of the group containing the electron for which ‘S’ is to be determined, contribute nothing to ‘S’. (c) Each of the remaining electrons in the group shield the valence electron to an extent of 0.35. (d) Each of the electrons in the (n – 1) shell shield to an extent of 0.85. (e) Each of the electrons in the next inner shell shield to the extent of 1.0. For an electron present in a group of ‘d’ or ‘f’ electrons, the rules (b) and (c) are the same but instead of rule (d) and (e) the rule (f) is used. (f) Each of the electrons present to the left side of the ‘d’ or ‘f’ group shield to an extent of 1.00. In case the electron belongs to the ‘1s’ orbital for which ‘S’ is to be determined, the contribution of the other electron present in the 1s orbital is taken as 0.30. Thus, the screening constant for an electron present in the 1s orbital, S = 0.30 × (Number of remaining electrons in the 1s orbital) The screening constant for an electron present in ns or np orbital, S = 0.35 × (Number of remaining electrons in n shell) + 0.85 × (Number of electrons in (n – 1) shell) + 1.0 × (Number of electrons in the inner shells) The screening constant for an electron in (n – 1)d orbital, S = 0.35 × [Number of remaining electrons in (n – 1)d orbitals] + 1.0 × (Number of electrons in inner shells) The effective nuclear charge experienced at the periphery of an atom or an ion is determined by imagining an ‘extra’ electron in the last occupied orbital of the atom or the ion. Thus, the effective nuclear charge experienced by 1s electron at the periphery of Cu atom is calculated as Zeff = 2 – [(0.30 × 5) + (2 × 0.85)] = 7 – 3.45 = 3.55

Applications of Slater’s Rules (a) A Cation is Always Smaller in Size than its Parent Atom This fact can be explained by considering the example of Sodium atom (1s2 2s2 2p6 3s1) Zeff experienced by last electron (3s) of sodium atom = 11 – [(0.35 × 0) + (0.85 × 8) + (2 × 1.0)] = 2.20 Zeff experienced by last electron (2p) of sodium ion = 11 – [(0.35 × 7) + (0.85 × 2)] = 6.85 It is evident that as an electron is removed from the sodium atom, its Zeff increases and thus the force of attraction between the nucleus and the remaining electrons increases. As a result, the size of the sodium ion is smaller than the parent sodium atom.

Periodic Table and Periodic Properties

8.19

(b) An Anion is Always Larger in Size than its Parent Atom Now, consider the case of fluorine atom (1s22s22p5) Zeff experienced by last electron (2p) of fluorine atom = 9 – [(0.35 × 6) + (0.85 × 2)] = 5.2 Zeff experienced by last electron (2p) of fluorine ion = 9 – [(0.35 × 7) + (0.85 × 2)] = 4.85 It is evident that when an extra electron is added in the F atom, Zeff decreases and thereby decreases the force of attraction between the nucleus and the electrons. Consequently, the size of an anion is always larger than its parent atom. (c) Atomic Size Goes On Decreasing on Moving from Left to Right in a Period Consider the element of 2nd periods. The Zeff goes on decreasing in the order (Li < Be < B < C < N < O < F) and thereby the nuclear force of attraction on the electrons goes on increasing. Thus, the atomic size goes on decreasing. (d) Formation of Cations from the Transition Metal Atoms by the loss of ns Electrons A transition metal atom loses ‘ns’ electrons instead of (n – 1) d electrons. This fact can be explained by considering the case of formation of Cu+ from Cu. The electronic configuration of Cu is 1s22s22p63s23p63d104s2. Zeff experienced by the 3d electron = 29 – [(0.35 × 4) + (1.0 × 18)] = 7.85. Zeff experienced by the 4s electron = 29 – [(0.35 × 0) + (0.85 × 18) + (1.0 × 10)] = 3.70. Thus, 4s electron experiences lesser force of attraction towards the nucleus as compared to the 3d electrons and hence is removed easily. (e) 4s is Filled Earlier than 3d Orbital This fact can be explained by considering the case of the potassium atom. Here, after filling of 18 electrons up to 3p, the nineteenth (19th) electron can enter either in 4s or in 3d orbital. Zeff experienced by 4s1 electron in the potassium atom = 19 – [(0.35 × 0) + (0.85 × 8) + (1.0 × 10)] = 2.20 Zeff experienced by 3d1 electron in the potassium atom = 19 – [(0.35 × 0) + (1.0 × 18)] = 1.00 Since Zeff for 3d1 electron is lesser than that for 4s1 electron, a 4s1 electron would be more attracted towards nucleus making the electronic configuration more stable. Thus, 4s is preferably filled earlier than the 3d orbital. Mendeleef’s Periodic Table was based on the periodic law that the physical and chemical properties of the elements are periodic function of their atomic masses. With the help of this table, he even predicted the properties of some undiscovered elements and corrected the doubtful masses of many elements. The long form of the periodic table is based on the modern periodic law which states that the physical and chemical properties of the elements are periodic functions of their atomic numbers. The main features of the long form of the Periodic Table are as follows: (i) It is divided into four blocks, viz. s, p, d and f. (ii) It has seven horizontal rows (periods) and 18 vertical columns (groups). (iii) s-block and p-block elements are collectively known as representative elements while d-block and f-block elements are known as transition and inner transition elements respectively. (iv) The outer-shell electronic configuration of these elements can be represented as, s-block elements: ns1–2, p block elements: ns2 np1–6, d-block elements: (n – 1)d1–10 ns0–2 and f-block elements: (n – 2) f1–14 (n – 1) d1–2 ns2

8.20

Inorganic Chemistry

The properties which show a gradual change on moving along a period and on moving down the group are known as periodic properties. The reoccurrence of similar properties is known as periodicity of properties. It is due to the repetition of similar electronic configurations after some regular interval. These properties are atomic radius, ionisation energy, electron affinity and electronegativity, etc. The atomic radius of an element is the distance from the centre of the nucleus to the outermost shell of the atom and is classified as covalent radius, ionic radius, metallic radius and van der Waal’s radius (largest out of all). The radius of a cation is always smaller than its parent atom, while the radius of an anion is always larger than its parent atom. In case of isoelectronic series, the radius decreases with increase of effective nuclear charge. Ionisation energy of an isolated gaseous atom is the energy required to remove the outermost electron from an isolated gaseous atom in its ground state. For all cases, IE1 < IE2 < IE3. Ionisation energy increases on moving from left to right in a period and decreases on moving from top to bottom in a group. Group 2 and 15 elements have higher IE than their neighbouring elements due to stable electronic configurations. Electron affinity of an isolated gaseous atom is the energy released when an electron is added to a gaseous atom in its ground state. In general, electron affinity increases on moving from left to right in a period and decreases on moving from top to bottom in a group. Group 2, 15 and 18 elements have negligible electron affinity due to stable electronic configuration. Electronegativity is the tendency of an atom to attract the shared pair of electrons towards itself in a molecule. Electronegativity of an element increases from left to right in a period (up to F) and decreases from top to bottom in a group.

EXAMPLE 1 Compare the acidic character of ethylene and acetylene. C2H4 2

C2H2

hybridisation

sp

sp

% s-character

33%

50%

The electronegativity of carbon atom increases with increase in percentage s-character. Accordingly, the carbon atom in ethylene is less electronegative as compared to that of acetylene. The more electronegative carbon atom in acetylene pulls the electron pair of the C–H bond more towards itself and facilitates the release of H+ ion. Thus, acetylene is more acidic than ethylene.

EXAMPLE 2 Compare the basic character of aniline and pyridine. hybridisation % s-character

C6H5NH2 sp3 35%

C5H5N sp2 33%

The more electronegative N-atom in pyridine has lesser electron-donating power as compared to the comparatively less electronegative N-atom in aniline. Thus, aniline is a stronger base than pyridine.

EXAMPLE 3 If the bond dissociation energies of H–H, F–F and H–F bonds are 104.2, 36.6 and 134.6 k cal mol–1 respectively, calculate the electronegativity of fluorine.

Periodic Table and Periodic Properties

Using Pauling’s method, H

–

F

F

8.21

= 0.182 [EH–F – (EH–H × EF–F)½]½ = 0.182 [134.6 – (104.2 × 36.6)½]½ = 0.182(72.85)½ = 1.55 = H + 1.55 = 1.55 + 2.1 = 3.65

EXAMPLE 4 Calculate the percent ionic character of the H-Cl bond. cCl – cH = 0.91 Percentage ionic character of H-Cl bond = [1–e–0.25(0.9)] × 100 = 20.14 Alternatively, Percentage ionic character of H-Cl bond = 18 (

Cl –

H)

= 18(0.9)1.4

= 15.53

EXAMPLE 5 Calculate the bond length of the C-O bond. The covalent radius of C and O atoms are 0.77 and 0.74 Å respectively.

Using the corelation

dC–O = rC + rO – 0.09 (

O–

C)

dC–O = 0.77 + 0.74 – 0.09 (3.5 – 2.5) = 1.42 Å

EXAMPLE 6 Calculate the enthalpy of formation of HCl. Using the Pauling’s equation, Hf = 23 e (

H–

Cl)

2

– 55.4 nN – 26.0 nO

Hf = 23 × 1 [2.1 – 3.0)2] – 0 – 0 = 18.63 kcal mol–1

EXAMPLE 7 Calculate the electronegativity of the Nitrogen atom using Allred- Rochow’s approach. The covalent radius of the Nitrogen atom is 0.75 Å. Electronic configuration of N is 1s22s22p3 Using Slater’s rules Zeff = Zactual – S = 7 – [(5 × 0.35) + (2 × 0.85)] = 7 – 3.45 = 3.55 Using Allred-Rochow’s approach, N

= 0.359 ×

Z eff

+ 0.744 rN2 = [0.359 × 3.55/(0.75)2] + 0.744 = 3.0 N

Calculate the effective nuclear charge for the 1s electron in the (i) helium atom, and (ii) boron atom.

EXAMPLE 8

(i) The electronic configuration of helium atom (1s2) The screening constant, S for 1s1 electron of the He-atom = 0.30 – 1 = 0.30 Zeff = Z – S = 2 – 0.30 = 1.70

8.22

Inorganic Chemistry

(ii) The electronic configuration of boron atom. (1s2) (2s2 2p1) There will be no contribution from the electrons present in (2s 2p) electrons as per Rule (b). The screening constant, S for 1s1 electron of B-atom = 0.30 × 1 = 0.30 Zeff = Z – S = 5 – 0.30 = 4.70

EXAMPLE 9 Calculate the effective nuclear charge for 4s and 3d electrons in the copper atom.

The electronic configuration of copper atom (1s2) (2s2 2p6) (3s2 3p6) (3d10) (4s1) Zeff for 4s electrons: The screening constant , S for 4s electron of Cu atom S = [(0.35 × 0) + (0.85 × 18) + (1.0 × 10)] = 25.30 Zeff = 29 – 25.30 = 3.70 Zeff for 3d electron: The screening constant , S for 3d electron of Cu atom S = [(0.35 × 9) + (1.0 × 18)] = 21.15 (There will be no contribution from 4s electron). Zeff = 29 – 21.15 = 7.85

EXAMPLE 10 Calculate the effective nuclear charge at the periphery of a Cu atom. The electronic configuration of Cu atom (1s2) (2s2 2p6) (3s2 3p6) (3d0) (4s1) The screening constant, S at periphery of Cu atom S = (0.35 × 1) + (0.85 × 18) + (1.0 × 10) = 25.75 Zeff = 29 – 25.75 = 3.35

QUESTIONS Q.1 Q.2 Q.3 Q.4

Enumerate the defects of Mendeleef’s Periodic Table. Discuss the main characteristics of the long form of the Periodic Table. What is periodicity? Discuss its cause with the help of suitable examples. What are isoelectronic species? Arrange the following in increasing order of atomic size: C4–, N3–, O2–, F–, Ne Q.5 Give reasons for the following: (a) size of a Cation is smaller than its parent atom. (b) Size of an anion is larger than its parent atom. (c) Radius of argon is greater than that of chlorine. (d) Be and Al show similar properties. (e) The first electron affinity of oxygen is positive but the second electron affinity is negative.

Periodic Table and Periodic Properties

Q.6 Q.7 Q.8

Q.9 Q.10 Q.11 Q.12

8.23

(f) The difference of atomic size between lithium and beryllum is much greater than between sodium and magnesium. Discuss the trends across the third period for the following properties: (a) Ionisation energy (b) Electron affinity (c) Electronegativity Which one has higher ionisation energy and why? (a) Al or Mg (b) S or P (c) Al or Ga Give reasons for the following: (a) First ionisation energy of nitrogen is larger than that of oxygen but the reverse is true for its second ionisation energy. (b) First ionisation energy of arsenic is higher than that of selenium. (c) Electron affinity of flourine is lesser than that of chlorine. How does electron affinity depend upon (a) Atomic size (b) Nuclear charge (c) Electronic configuration How have different investigators used different approaches to determine electronegativity? Discuss the factors affecting electronegativity. Discuss the application of electronegativity w.r.t. prediction of chemical reactions.

MULTIPLE-CHOICE QUESTIONS 1. The correct order of increasing atomic size is (a) Ne > F– > Na+ > Mg2+ (b) F– > Ne > Na+ > Mg2+ (c) Mg2+ > Na+ > Ne > F– (d) F– > Ne > Mg 2+ > Na+ 2. The element with highest second ionisation energy is (a) V (b) Cr (c) Mn (d) Fe 3. The correct order of electron affinity is (a) F > Cl > Br>I (b) F Br > I (d) Cl < F < Br 111°).

16.6

Inorganic Chemistry

This is because of the reason that in OF2, the bond pair of electrons lie nearer to the more electronegative fluorine atom resulting in lesser BP–BP repulsion, reducing the bond angle from 109° 28 to 103°. In Cl2O and Br2O, the bond pair of electrons lie closer to the more electronegative oxygen atom, thereby increasing the bond pair–bond pair repulsion resulting in increase the bond angle from 109° 28 . In Br2O, the bond angle is further increased due to steric crowding of the larger bromine atoms. O2F2 has a similar structure to that of H2O2 but with shorter O–O bond length. ClO2 is paramagnetic with an odd number of electrons and is highly reactive. The molecule is angular and contains a three electron bond. The odd electron is delocalised and hence, two resonating forms can be shown as in Fig. 16.2. The oxides have been discussed Fig. 16.2 Resonating structure of ClO2– in detail later in this chapter.

16.3.3 Reaction with Water All halogens are soluble in water, but vary in their actions. Fluorine oxidises water vigorously as F2 + 3H2O $ 2H3O+ + 2F– + ½O2 But Cl2 shows a disproportionation reaction as Cl2 + H2O $ HCl + HOCl Aqueous solutions of Br2 and I2 contain a negligible amount of OBr– and OI– respectively. Oxoacids The halogens form four series of oxoacids as. Oxidation state of the halogen

+1 HOF HOCl HOBr HOI

+3 — HClO2 — —

Acidic strength Stability All these oxiacids have a tetrahedral based geometry formed by sp3 hybridisation as shown below in Fig. 16.3. It should be noted that in all these oxides, hydrogen is not directly attached to the halogen atom and forms a part of OH group. Some structural characteristics of the anions of these oxyacids are listed in Table 16.6.

16.4

+5 — HClO3 HBrO3 HIO3

+7 — HClO4 HBrO4 HIO4

Table 16.6 Some structural characteristics of oxyacids Anion ClO– ClO2– ClO3– ClO4–

Bond length Cl–O (pm) 170 164 157 147

Bond energy kJ/mol 209 245 244 364

Bond angle O–Cl–O – 111° 106° 109.5°

FLUORINE (F)

16.4.1 Occurrence and Extraction of Fluorine Fluorine (from Latin word fluere, meaning to flow) is the thirteenth most abundant element and occurs to an extent of 544 ppm by weight in the earth’s crust. It does not occur free in nature due to its extremely reactive nature and is found only in the combined state. The main source of fluorine is fluorspar, CaF2. It is also found as cryolite (Na3AlF6) and fluorpatite [CaF2.3Ca3(PO4)2]. Fluorine is also found in small quantities as fluoride in sea water, springs, soil, bones, teeth, plants, etc.

Chemistry of Group 17 Elements

Fig. 16.3

16.7

Structure of oxoacids

Fluorine is very difficult to obtain due to the following problems: 1. Fluorine is extremely reactive and catches fire. It reacts with most metals and even glass; hence suitable materials are required for its production. 2. HF is highly corrosive and etches glass. Anhydrous HF is very toxic and slightly ionised; hence, the electrolysis is very difficult. 3. The evolved fluorine reacts with water and rather dioxygen is produced. 4. Hydrogen and fluorine liberated at the respective electrodes must be separated by the diaphragm to prevent explosive reaction. The first successful preparation of fluorine was done by Moissan in 1886. He prepared fluorine by the electrolysis of a solution of KHF2 in anhydrous HF in a U-shaped tube. The tube was made of platinumiridium alloy with electrodes of the same materials and its end was closed with fluorspar stoppers. The method was further explored by Dennis, Veeder and Rochow in 1931. They used a V-shaped tube made of copper using graphite electrodes fixed with bakelite cement. The ends of the tube were sealed with copper caps and a current of 5 A at 12 volts was passed through fused sodium or potassium hydrogen fluoride. The reactions taking place can be represented as KHF2 $ KF + HF; HF $ H+ + F– At cathode: 2H+ + 2e– $ H2 At anode:

2F– $ F2 + 2e–

The liberated fluorine was passed through U-shaped copper tubes containing NaF to remove the HF vapour (Fig. 16.4). Fig. 16.4 Manufacture of fluorine NaF + HF $ NaH + F2 Nowadays fluorine is manufactured by the electrolysis of a fused mixture of HF and KF in a rectangular steel tank. The tank itself serves as a cathode while graphite anode is fitted in an airtight removable copper cap. A metal cylinder diaphragm is

16.8

Inorganic Chemistry

used to separate the hydrogen evolved at the cathode and the fluoride evolved at the anode. The electrolyte is kept in the fused state by passing steam through the stream jacket enveloping the steel tank. The evolved fluorine is freed from HF and is stored in steel tank under pressure (Fig. 16.5).

16.4.2 Physical Properties Fluorine is a pungent-smelling, yellow gas which is Fig. 16.5 Fluorine production by modern method poisonous in nature. On condensing, it changes to a pale yellow liquid with boiling point of 86 K and solidifies as pale yellow crystals with melting point of 53 K.

16.4.3 Chemical Properties of Fluorine Fluorine is very reactive as indicated by the following discussion. (a) Reaction with Hydrogen Fluorine has a strong affinity for hydrogen as it combines explosively with hydrogen even in the dark. H2 + F2 $ 2HF

H = –536 kJ

(b) Reaction with Water Fluorine can decompose water even in the dark, at very low temperature. It fumes in moist air and liberates O2 or O3. 2F2 + 2H2O $ 4HF + O2 3F2 + 3H2O $ 3HF + O3 (c) Reaction with Metals Fluorine combines with most of the metals. The s-block elements burn spontaneously in its presence and form the corresponding fluorides. Mg, Al, Ni and Ag react with fluorine on little warming while gold and platinum require heating up to 300ºC. Cu, Hg, Ni and steel get covered by a protective coating of the fluoride which resists the further attack of fluorine. (d) Reaction with Nonmetals and Metalloids Nonmetals and metalloids react with fluorine to form the corresponding fluorides. C + 2F2 $ CF4 S + 3F2 $ SF6 2Sb + 3F2 $ 2SbF3 (e) Action as Oxidising Agent Fluorine is a strong oxidising agent and oxidises all the common reducing agents. It can displace other halogens from their corresponding halides. KClO3 + F2 + H2O $ KClO4 + 2HF 2KHSO4 + F2 $ K2S2O8 + 2HF 2KCl + F2 $ 2KF + Cl2 (f) Reaction with Ammonia Fluorine reacts with ammonia to form nitrogen and NF3. 2NH3 + 3F2 $ N2 + 6HF NH3 + 3F2 $ NF3 + 3HF (g) Reaction with Hydrogen Sulphide H2S burns in the presence of F2 to give SF6 and HF. H2S + 4F2 $ SF6 + 2HF

Chemistry of Group 17 Elements

(h) Reaction with Alkalis O2.

16.9

Fluorine reacts with dilute alkalis to form OF2 and with conc. alkalis it gives 2F2 + 2NaOH (dil.) $ 2NaF + CF2 + H2O

2F2 + 4NaOH (conc.) $ 4NaF + O2 + 2H2O (i) Action with Organic Compounds Fluorine reacts explosively with organic compounds to form fluorocarbons. Thus, fluorination is carried in a nitrogen atmosphere and in the presence of a catalyst. Cu-guaze

→ C6F2 + 6HF C6H6 + 9F2 N ,265°C 2

(j) Reaction with Glass Fluorine reacts with glass above 100°C to form SiF4.

16.4.4 Compounds of Fluorine (a) Hydrofluoric Acid or Hydrogen Fluoride (HF) Pure anhydrous hydrogen fluoride can be prepared in the laboratory either by heating pure, dry KHF2 in a platinum retort or by passing hydrogen gas over dry AgF. ∆ → KF + HF KHF2  An aqueous solution of hydrogen fluoride is manufactured by heating fluorspar with conc. H2SO4 in a lead retort. Ca + F2 + H2SO4 $ CaSO4 + 2HF The dilute aqueous solutions are stored in Gutta–Percha bottles and the concentrated solutions are stored in wax bottles. Anhydrous hydrogen fluoride is a colourless liquid with boiling point 292.5 K and freezing point 170.5 K. It fumes strongly in air and is highly soluble in water. It exists as a dimeric molecule. It is a weak acid which give two types of ions. 2HF HF2–

HF2– + H+ + F2– 2 +H

Thus, it forms normal salts containing F-ions and acidic salts containing HF2– ions. Anhydrous HF is extremely stable and does not react with metals under ordinary conditions. But aqueous conc. HF shows many reactions like the following: (i) Action with metals Zn + 2HF $ ZnF2 + H2 Mg + 2HF $ MgF2 + H2 (ii) Action with halogens 2HF + X2 $ 2HX + F2

(X = Cl, Br, I)

(iii) Action with silica and glass With silica:

SiO2 + 4HF $ SiF4 + 2H2O 3SiF4 + 3H2O $ H2SiO3 + 2H2SiF6

With glass:

Na2SiO3 + 6HF $ Na2SiF6 + 3H2O CaSiO3 + 6HF $ CaSiF6 + 3H2O

Thus, HF cannot be kept in glass vessels; rather it is used for etching glass. (iv) Action with AgNO3 solutions AgNO3 + HF $ AgF

+

(Soluble in water)

HNO3

16.10 Inorganic Chemistry (v) Action with BaCl2, SrCl2 and CaCl2 MCl2 + 2HF $ MF2 + 2HCl

(M = Ca, Sr, Ba)

(ppt. sol. in conc. HCl)

(vi) It acts as a nonaqueous solvent and dissolves alkali metal fluorides, nitrates and sulphate KNO3 + 2HF $ KHF2 + HNO3 (b) Oxygen Difluoride (OF2) It is prepared by passing F2 through dilute NaOH solution. 2F2 + 2NaOH $ OF2 + 2NaF + H2O It is a pale yellow highly poisonous gas and is slightly soluble in water. It rather decomposes in presence of water as H2O + OF2 $ 2HF + O2 It also decomposes into its elements, on heating. OF2 ∆ → ½O2 + F2 It is a strong oxidising agent and oxidises many metals and non metals. 4KI + OF2 + H2O $ 2KOH + 2KF + 2I2 It gets hydrolysed in presence of sodium hydroxide. OF2 + 2NaOH $ O2 + 2NaF + H2O (c) Dioxygen Difluoride (O2F2) It is prepared by passing an electric discharge through a mixture of O2 and F2 at a very low temperature and pressure. It is an orange yellow solid which is highly unstable and decomposes into F2 and O2 above 173 K. It is also a strong fluorinating and oxidising agent. (d) Hypofluorous Acid (HOF) It is prepared by passing F2 over ice at 0°C and the product is removed into a cold trap. 0 C ���� � F2 + H2O � ��� � HOF + HF It is a colourless, unstable gas which decomposes to HF and O2 by its own. It is a strong oxidising agent and acidic in nature. (e) Fluorocarbons (RF) Fluorocarbons are derived from hydrocarbons by the substitution of H by F atoms. These compounds are very useful organic compounds and can be represented as CnF2n+2 (perfluoro compounds) if they are completely substituted by F atoms. Preparation (i) Fluorocarbons can be prepared by the direct treatment of an organic compound with F2 in presence of an inert atomosphere and a suitable catalyst. Cu / N

2 → C6F12 + 6HF C6H6 + 9F2  265°C

Instead of F2, a fluorinating agent can also be used. RCOR + SF4 $ R2CF2 + SOF2 (ii) The organic compounds can be electrolysed in liquid HF using Ni anode and steel cathode in a steel cell. HF  → C8F18 C8H18 electrolysis (C4H9)3N $ (C4F9)3N (iii) Organic halides can be treated with fluorinating agents to obtain fluorocarbons.

Chemistry of Group 17 Elements

16.11

RCl + NaF $ RF + NaCl 5 2CCl4 + 3HF SbF  → CCl2F2 + CCl3F + 3HCl

Properties (i) Fluorocarbons are low-boiling compounds due to very weak intermolecular forces of attraction. (ii) These compounds are highly inert and can be heated in air without burning. This is due to the reason that presence of the most electronegative fluorine atom develops a partial positive charge on the C atom and hence it cannot be oxidised easily. On the otherhand in the hydrocarbons, C develops a partial negative charge and hence can be readily oxidised. (iii) Fluorocarbons are highly stable compounds. The stability of these compounds can be accounted as follows: (a) High bond dissociation energy of C–F bond [486 kJ mol–1] as compared to that of C–H bond [415 kJ mol–1] (b) Absence of vacant d-orbital in carbon protects it from hydrolysis. (c) Replacement of F by any other atom produces strain in the compound, while F and H with similar size pose no strain in fluorocarbons. (d) The F atoms shield the C atoms from the attacks by other reagents. Thus, fluorocarbons are inert even to conc. acids and are strong oxidising as well as strong reducing agents. The pyrolysis of fluorocarbons at very high temperature results in cleavage of C–C bond rather than C–F bond. Uses Fluorocarbons are industrially important compounds. These are used as lubricants due to their very low coefficients of friction. Freons, mixed chlorofluorocarbons (CClF3, CCl2F2 and CCl3F, etc.) are used as nontoxic aerosol propellants and refrigerators. CHClF2 is used to prepare tetrafluorothene. −1000° 2CHClF2 500  → CF2 = CF2 + 2HCl It is used to prepare Teflon by polymerisation. Polymerisation

→ (CF2–CF2)n nCF2 = CF2  CF3CHBrCl is used as an anaesthetic under the name fluothane. (f) Halogen Oxide Fluorides These are the compounds of halogens (Cl, Br or I) in which the halogen is bonded to both O and F to give FnXOm.

(a)

Chlorine Oxide Fluorides These are strong oxidising and fluorinating agents. Five wellcharacterised oxide fluorides of chlorine are known viz. FClO, F3ClO, FClO2, F3ClO2 and FClO3. Out of these, FClO is the most unstable while FClO3 is the most stable.

(i)

Chlorine Oxide Monofluoride (FClO) It can be prepared the following reactions ClF3 + H2O $ FClO + 2HF Ar ClF + O3   → FClO + O2 4 −15 K

Cl2O + F2 $ FClO + ClF It is a colourless and thermally unstable gas which decomposes in about 25 s at room temperature. 2FClO $ FClO2 + ClF

Fig. 16.6 Structure of FClO

Structure of FClO In FClO, the Cl atom is sp3 hybridized and forms one bond with F atom and one bond with the oxygen atom. Due to presence of two pairs of electrons, it has a V-shaped structure (Fig. 16.6).

16.12 Inorganic Chemistry

(ii) Chlorine Oxotrifluoride (F3ClO) It can be prepared as follows. NaF

→ F3ClO + ClF Cl2O + 2F2 − 78°C C Cl2O + N2O5 0°  → 2ClONO2 −35°C

ClONO2 + 2F2 → F3ClO + FNO2 F3ClO is a colourless gas freezing at –43°C and boiling at 28°C. It is kinetically more stable than FClO at room temperature but decomposes above 300°C to liberate oxygen. >300°C

→ ClF3 + ½O2 F3ClO  It is hydrolysed even with a small amount of water. F3ClO + H2O $ FClO2 + HF F3ClO + HF $ [F2ClO]+ [HF2]– It readily reacts with glass or quartz and can be handled in Teflon or well-passiviated metal. It acts as a strong fluorinating agent as shown here: C F3ClO + Cl2 200°  → 3ClF + H2O2 25°C

F3ClO + Cl2O  → 2ClF + FClO2 F3ClO + ClOSO2F $ ClF + FClO2 + SO2F2 F3ClO + 2ClOSO2F $ 2ClF + FClO2 + S2O5F2 It also exhibits combined oxygenating and fluorinating capacity. F3ClO + MoF5 $ MoF6 + MoF4O + FClO2 F3ClO + SF4 $ SF6 + SF5Cl + SF4O + FClO2 C F3ClO + N2F4 100°  → NF3 + FNO + ClF

It also acts as a reducing agent in presence of extremely strong oxidising agents. F3ClO + PtF6 $ [F2ClO]+ [PtF6]– + ½ F2 It can act as a Lewis base in presence of strong fluoride ion acceptors. F3ClO + MF5 $ [F2ClO]+ [MF6]– (M = P, As, Sb, Bi, V, Nb, Ta, Pt, O)

It can also act as a Lewis acid in presence of strong fluoride-ion donors. F3ClO + MF $ M+ [F4ClO]– (M = K, Rb, Cs)

Fig. 16.7 Structure of F3ClO

Structure of F3ClO In F3ClO, the Cl atom is sp3d hybridised and forms one sigma bond with oxygen atom and one sigma bond each with the fluorine atoms. It is -bonded with the oxygen atom and has one lone pair of electrons. Thus, the structure is distorted trigonal pyramidal (Fig. 16.7).

(iii) Chlorine Dioxofluoride (FClO2) It is prepared either by the low-temperature fluorination of ClO2 or by the interaction of NaClO3 with ClF3 at room temperature for 24 hours. °C −30°C 6NaClO3 + 4ClF3 25  → 6FClO2 + 6NaF + 2Cl2 + 3O2

Chemistry of Group 17 Elements

16.13

It is a colourless gas which freezes at –115°C to –123°C and boils at –6°C. It is thermally stable at room temperature but decomposes above 200°C in monel metal and at 300°C in quartz. FClO2 $ ClF + O2 It is chemically more reactive than FClO3. It hydrolyses slowly with water but reacts rapidlywith anhydrous HNO3. 2FClO2 + H2O $ 2HF + 2ClO2 + ½ O2 2FClO2 + 2HNO3 $ 2HF + 2ClO2 + N2O5 + ½ O2 It also reacts with other protonic reagents such as FClO2 + 2OH– $ ClO3– + F– + H2O −78°C

FClO2 + NH3 → NH4Cl + NH4F −110°C

→ ClO + HF + ½ Cl FClO2 + HCl  2 2 It reacts explosively with strong reducing agents such as HBr (–110°) and SO2 (–40°C). FClO2 is a strong oxidising and fluorinating agent. C FClO2 + SF4 50°  → SF6 + SF4O + SF2O2 30°C FClO2 + N2F4  → NF3 + FNO2 + FNO

Some oxides form fluorocomplexes with FClO2. FClO2 + B2O3 $ [ClO2]+ [BF4]–

F

It acts as a Lewis base with fluoride-ion acceptors. FClO2 + AsF5 $ [ClO2]+ [AsF6]– FClO2 + SbF5 $ [ClO2]+ [SbF6]– FClO2 + 2SbF5 $ [ClO2]+ [Sb2F11]– It can also act as a Lewis acid with some fluoride-ion donors. FClO2 + CsF $ Cs+ [F2ClO2]–

Cl O O

Fig. 16.8 Structure of FClO2

Structure of FClO2 It is pyramidal due to presence of one lone pair of electrons on the sp3 hybridised Cl atom (Fig. 16.8).

(iv) Chlorine Dioxotrifluoride (F3ClO2) It is prepared by fluorine transfer reactions: 2FClO2 + 2PtF6 $ [F2ClO2]+ [PtF6]– + [ClO2] + [PtF6]– FNO2 F3ClO2 + FClO2 + 2[NO2]+ [PtF6]– It is a colourless gas which boils at –21.6ºC and freezes at –81.2°C. It is highly corrosive and is thus stored in Teflon or sapphire apparatus. It also acts as a strong oxidising agent and acts as a Lewis base with fluoride-ion acceptors. F3ClO2 + BF3 $ [F2ClO2]+[BF4]– F3ClO2 + AsF5 $ [F2ClO2]+[AsF6]– Structure of F3ClO2 16.9).

Its structure is trigonal bipyramidal with sp3d hybridised Cl atom (Fig.

16.14 Inorganic Chemistry

(v) Chlorine Trioxofluoride (FClO3) It is prepared by fluorination of KClO3 at low temperature or by the action of F2 on an aqueous solution of NaClO3. 40°C KClO3 + F2 − → FClO3 + KF

NaClO3 + F2 $ FClO3 + KF It is prepared on an industrial scale by the fluorination of a perchlorate with HOSO2F/SbF5.

Fig. 16.9 Structure of F3ClO2

HOSO F −SbF

2 5 KClO4  → FClO3 25°C −30°C

It is a colourless gas which freezes at –148°C and boils at –47°C. It is highly stable but decomposes above 465ºC. It is hydrolysed very slowly, only in the presence of alkaline medium. FClO3 + 2NaOH $ NaClO4 + NaF + H2O It reacts with liquid NH3 in presence of NaNH2. FClO3 + 3NH3 $ NH4F + [NH4]+[HNClO3]– It doesn’t form adducts like other oxofluorides of chlorine. However, it is highly reactive for nucleophitic attack at Cl and is thus used in organic synthesis. FClO3 + Li+Ph– $ PhClO3 + LiF

FClO3 + Li+[C4H3S]– $

S

F + LiClO3 F

+ FClO3 $

+ HClO2

O It can also fluorinate a reactive ethylene group.

Fig. 16.10 Structure of FClO3

3 CH2(CO2R)2 FClO  → CF2(CO2R)2

Structure of FClO3

Its structure is tetrahedral with sp3 hybridised Cl atom (Fig. 16.10).

(b) Bromine Oxide Fluorides Oxide fluorides of bromine are less numerous and not much characterised. Only three bromine oxide fluorides, viz. FBrO2, F3BrO and FBrO3 are well known. These are comparatively thermally less stable than the corresponding chlorine compounds (but have similar structure) and are more reactive.

(i)

Bromine Dioxofluoride (FBrO2) It is prepared by the fluoride transfer reactions between BrF5 and FIO2 or F3IO or I2O5 BrF5 + FIO2 $ FBrO2 + IF5 BrF5 + 2F3IO $ FBrO2+ 2IF5 5BrF5 + 2I2O5 $ 5FBrO2 + 4IF5 It is a colourless liquid which freezes at –9°C. It readily attacks glass and undergoes rapid decomposition. ∆ → BrF3 + Br2 + 3O2 3FBrO2 

Chemistry of Group 17 Elements

16.15

It undergoes hydrolysis in alkaline medium even at 0°C FBrO2 + 2NaOH $ NaBrO3 + NaF + H2O It reacts with Lewis acids and acts as a fluoride-ion donor. FBrO2 + AsF5 $ [BrO2]+[AsF6]– It also acts as a fluorine-ion acceptor. FBrO2 + KF $ [K]+[F2BrO2]–

(ii)

Bromine Oxotrifluoride (F3BrO) acids.

It is prepared by the treatment of K[F4BrO] with weak Lewis

72°C K[F4BrO] + HF (anhyd.) − → F3BrO + KHF2

K[F4BrO] + [O2]+[AsF6]– $ F3BrO + KAsF6 + O2 + ½F2 It is a colourless solid and melts at –5°C to give a clear liquid. It decomposes slowly even at room temperature to liberate oxygen. F3BrO $ BrF3 + ½O2 It can also participate in fluoride transfer reactions. F3BrO + BF3 $ [F2BrO2]+[BF4]– F3BrO + KF $ K+[F4BrO] –

(iii)

Bromine Trioxofluoride (FBrO3) It is prepared by the fluorination of KBrO4 with AsF5, BrF5 or SbF5 in the presence of HF. KBrO4 + 2AsF5 + 3HF $ FBrO3 + [H3O] +[AsF6]– + KAsF6 KBrO4 + BrF5 + 2HF $ 2FBrO3 +FBrO2 + 2KHF2 KBrO4 + [BrF6]+[AsF6]– $ FBrO3 + BrF5 + KAsF6 + ½ O2 It is a highly reactive gas at room temperature and converts to a white solid at –110°C. It condenses to form a colourless liquid which boils at 2.4°C. It reacts rapidly with water and alkaline solutions. FBrO3 + H2O $ BrO4– + H+ + HF FBrO3 + 2OH– $ BrO4– + F– + H2O

(c) Iodine Oxide Fluorides Five iodine oxide fluorides of iodine are characterised. These are FIO2, F3IO, FIO3, F3IO2 and F5IO

(i)

Iodine Dioxofluoride (FIO2) It is prepared by the direct fluorination of iodine pentoxide in anhydrous hydrogen fluoride at room temperature. HF I2O5 + F2  → 2FIO2 + ½O2 20°C

It is also prepared by the thermal dismutation of F3IO. 110°C 2F3IO − → FIO2 + IF5 It is a colourless polymeric solid which decomposes above 200°C. It undergoes alkaline hydrolysis and shows fluoride transfer reactions.

FIO2 + 2OH– $ IO3– + F– + H2O + – FIO2 + KF HF → K [F2IO2]

16.16 Inorganic Chemistry

(ii)

Iodine Oxotrifluoxide (F3IO) It is prepared by the treatment of IF5 with I2O5. C I2O5 + 3IF5 105°  → 5F3IO It is a colorless solid which dismutates above 110°C into FIO2 and IF5.

(iii)

Iodine Dioxotrifluoride (F3IO2) It is prepared by the partial fluorination of barium periodate with fluorosulphuric acid followed by treatment with SO3. HSO F

SO

3 3 Ba 3 H 4 (IO6 )2  → [ HIO2 F4 ]  → F3 IO2

It is a yellow solid which melts at 41°C and exists as polymeric structure with coordination number of I as 6 (Fig. 16.11).

(iv)

Fig. 16.11 Structure of polymeric F3lO2

Iodine Trioxofluoride (FIO3) It is prepared by the action of liquid HF or F2 on HIO4. HIO4 + F2 $ FIO3 + HF + ½O2 It is a white crystalline solid which decomposes on heating to liberate oxygen. C FIO3 100°  → FIO2 + ½O2 It forms adducts with Lewis acids.

FIO3 + BF3 $ [IO3]+[BF4]– FIO3 + AsF5 $ [IO3]+[AsF5]– It is reduced by SO3 and oxygen is liberated. FIO3 + SO3 $ IO2.SO2F + O2

(v)

Iodine Oxopentafluoride (F5IO) It is prepared by the treatment of IF5 with H2O, silica, I2O5 or glass. IF5 + H2O $ F5IO + H2 5IF5 + I2O5 $ F5IO + I2 It is a colourless liquid which melts at 45°C and is quite stable. It is octahedral with 6-coordinate geometry (Fig. 16.12).

Fig. 16.12 Structure of F5lO

2. Uses of Fluorine and Its Compounds Fluorine is used as an insecticide and to prepare derivatives which are very useful as discussed here. 1. Freon-12 is used in refrigeration, air conditioning and cold storage plants. It is also used as a solvent for insecticides. 2. UF6 is used for the separation of uranium isotopes. 3. Cryolite and fluorspar are used in metallurgy. 4. HF is used for etching of glass and for removal of silica from iron castings. 5. Oxyfluorides are used as powerful oxidising and fluorinating agents.

16.4.5 Anomalous Behaviour of Fluorine Being most reactive, fluorine directly combines with metals and nonmetals. However, its properties are different from the rest of the halogens due to its small size, very high electronegativity and absence of vacant d-orbitals in the valence shell. Some important differences are the following:

Chemistry of Group 17 Elements

16.17

(a) Fluorine can’t show the oxidation states like other halogens. (b) Fluorine is the most reactive halogen due to very low F–F bond dissociation energy whereas other halogens are comparatively less reactive. (c) Fluorine can form hydrogen bond as in HF and exists as a liquid while the hydrides of other halogens do not form hydrogen bonds and exists as gas under ordinary conditions. (d) Calcium fluoride is insoluble in water while the corresponding calcium halides are soluble in water. (e) Fluorine cannot form polyhalides like other halogens.

16.5

CHLORINE (Cl)

Chlorine was discovered by Scheele in1774. and named it oxymuriatic gas (oxide of HCl—muriatic acid). In 1800, Davy named it chlorine because of its colour (Greek, chlorios, greenish yellow)

16.5.1 Occurrence and Extraction of Chlorine Chlorine is the twentieth most abundant element and occurs to an extent of 126 ppm by weight in the earth’s crust. It is found only in the combined state, mostly as chlorides. The most important chloride is common salt, NaCl, found as rock salt. Chlorine was prepared by Scheele in 1774 by oxidation of HCl with MnO2. MnO2 + 4HCl $ MnCl2 + Cl2 + 2H2O Chlorine can be prepared in the lab by the oxidation of HCl with an oxidising agent (KMnO4, K2Cr2O7, CaOCl2, NaOCl, PbO2 and Pb3O4). 2KMnO4 + 16HCl $ 2KCl + 2MnCl2 + 5Cl2 + 8H2O The cheapest laboratory method is the heating of manganese dioxide with HCl. However instead of HCl, a mixture of common salt and conc. H2SO4 can also be used. MnO2 + 2NaCl + 3H2SO4 $ MnSO4 + 2NaHSO4 + Cl2 + 2H2O Chlorine is manufactured commercially by the electrolysis of brine solutions (in the manufacture of NaOH) as an important by-product.

16.5.2 Properties It is a greenish yellow gas with pungent smell and causes headache. It gets liquefied to a yellow liquid when cooled under pressure. This liquid has boiling point 235.4 K and freezes to a pale yellow solid at temperature 172 K. It is slightly less reactive than fluorine and combines with a number of elements to form the corresponding chlorides. It acts as an oxidising agent and oxidises Fe2+ to Fe3+, S2– to S, SO32– and S2O32– to SO42–, etc. It is used as a bleaching and disinfecting agent as it releases nascent oxygen when treated with water. Cl2 + H2O $ HCl + HClO HClO $ HCl + [O] Vegetable colouring matter + [O] $ Vegetable colourless matter. It can displace Br and I but cannot displace F in their corresponding salts. 2KBr + Cl2 $ 2KCl + Br2 2NaI + Cl2 $ 2NaCl + I2 It gives nitrogen with excess of ammonia, but an explosive substance, nitrogen trichloride, is obtained when chlorine is used in excess.

16.18 Inorganic Chemistry

8NH3(Excess) + 3Cl2 $ 6NH4Cl + N2 NH3 + 3Cl2(Excess) $ 3HCl + NCl3

16.5.3 Compounds of Chlorine (a) Hydrochloric Acid or Hydrogen Chloride (HCl) It can be prepared by any of the following methods: H2 + Cl2 Burn  → 2HCl

(Excess)

Sunlight

H2 + Cl2  → 2HCl 2NaCl + H2SO4 ∆ → Na2SO4 + 2HCl PCl3 + 3H2O $ H3PO3 + 3HCl Properties Pure HCl is a colourless, pungent-smelling gas and fumes in moist air. It gets liquefied to a colourless liquid with boiling point 189 K, which freezes to a white crystalline solid with melting at 162 K. It is extremely soluble in water and forms an azeotropic mixture boiling at 382 K (20.4% HCl). Aqueous solutions of HCl are acidic in nature and turn blue litmus red. It gives white fumes of NH4Cl with NH3. It reacts with metals and their compounds to form chlorides. M + 2HCl $ MCl2 + H2

(M = Zn, Fe)

MOH + HCl $ MCl + H2O

(M = Na, K)

4Ag + 4HCl + O2 $ 4AgCl + 2H2O (White ppt.)

AgNO3 + HCl $ AgCl + HNO3 (b) Oxides of Chloride Chlorine forms different oxides with the more electronegative oxygen atom as discussed below:

(i)

Chlorine Monoxide (Cl2O) It is prepared by passing dry Cl2 over freshly precipitated HgO heated at 400ºC. 2HgO + 2Cl2 $ HgCl2.HgO + Cl2O It is a brownish yellow gas which condenses to an orange liquid which boils at 2ºC. It is highly unstable and explodes on heating or sparking or in the presence of ammonia. 2Cl2O $ 2Cl2 + O2 3Cl2O + 10NH3 $ 6NH4Cl + 3H2O + 2N2 It is considered an anhydride of hypochlorous acid. Cl2O + H2O

2HOCl

It is a powerful oxidising agent and oxidises HCl to Cl2 2HCl + Cl2O $ H2O + 2Cl2

(ii)

Chlorine Dioxide (ClO2) It is prepared by the action of conc. H2SO4 or oxalic acid on powered KClO3. 2KClO3 + 3H2SO4 $ 3KHSO4 + HClO4 + 2ClO2 + H2O 2KClO3 + 2(COOH)2 $ K2C2O4 + 2CO2 + 2ClO2 + 2H2O

Chemistry of Group 17 Elements

16.19

Pure ClO2 is prepared by passing dry Cl2 over heated AgClO3. 2AgClO3 + Cl2 $ 2AgCl + 2Cl2O + O2 It is a yellow gas which condenses on cooling to give a deep red liquid, boiling at 11°C and freezes to give orange-red crystals at –59°C. It is highly unstable and decomposes in the presence of light. ClO2 $ ClO + O It explodes violently on heating with a glass rod or sparking or in contact with alcohol. 2ClO2 $ Cl2 + 2O2 It dissolves in water and alkalis as 2ClO2 + H2O $ HClO2 + HClO3 2ClO2 + 2KOH $ KClO2 + KClO3 + H2O Hence, it is considered a mixed anhydride of HClO2 and HClO3

(iii)

Chlorine Hexoxide (Cl2O6)

It is prepared by the action of ozone on ClO2 or Cl2.

C 2ClO2 + 2O3 0°  → Cl2O6 + 2O2 Cl2 + 4O3 $ Cl2O6 + 3O2

It is a dark red liquid which freezes at 3.5°C. It decomposes at its melting point as Cl2O6 $ 2ClO2 + O2 It dissolves in water and alkalis as Cl2O6 + 2H2O $ HClO4.H2O + HClO3 Cl2O6 + 2NaOH $ NaClO4 + NaClO3 + H2O Hence, it is a mixed anhydride of HClO3 and HClO4. It reacts reversibly with anhydrous HF. Cl2O6 + HF

ClO2F + HClO4

Liquid Cl2O6 is diamagnetic but its vapours are paramagnetic due to the formation of ClO3 with one unpaired electron. ��� � Cl2O6 � �� � 2ClO3 Hence, its structure is considered as shown in Fig. 16.13. O O

O Cl

O

O O or

Cl O

Cl O

O

O Cl O

+

–

or ClO2 ClO4 O

Fig. 16.13 Structure of Cl2O6

(iv)

Chlorine Heptoxide (Cl2O7) It is prepared by dehydration of perchloric acid by P2O5 at –10°C. PO

2 5 Æ Cl2O7 + H2O 2HClO4 æææ

It is a colourless oily liquid boiling at 82°C and is comparatively more stable than other oxides of chlorine. It dissolves in water and alkalis to form HClO4.

16.20 Inorganic Chemistry

Cl2O7 + H2O $ 2HClO4 Cl2O7 + 2NaOH $ 2NaClO4 + H2O Hence, it is an anhydride of HClO4. In ClO7, two tetrahedra are linked by oxygen atom as revealed by electron diffraction analysis as shown in Fig. 16.14.

(v)

Chlorine Tetraoxide (ClO4)

AgClO4.

Fig. 16.14 Structure of Cl2O7

It is prepared by the treatment of I2 with an ethereal solution of

2AgClO4 + I2 $ 2AgI + 2ClO4 The solution is filtered and shaken with water to produce HClO4. 4ClO4 + 2H2O $ 4HClO4 + O2

3. Oxy-acids of Chlorine and its Salts Chlorine forms four types of oxy-acids, viz. hypochlorous acid (HClO), chlorous acid (HClO2), chloric acid (HClO3), and perchloric acid (HClO4). (a) Hypochlorous Acid, HOCl It is prepared by the treatment of chlorine water with freshly precipitated HgO or by the treatment of chlorine with suspension of CaCO3 or KClO or CaOCl2. 2Cl2 + H2O + 2HgO $ HgCl2.HgO + 2HOCl Cl2 + H2O + CaCO3 $ CaCl2 + CO2 + 2HOCl Properties The dilute solution of HOCl is colourless and fairly stable but the concentrated solution is yellow in colour and decomposes in light to give O2. 2HOCl $ 2HCl + O2 It is a monobasic acid and reacts with alkalis to form hypochlorides. HOCl(aq)

H+(aq) + OCl–(aq)

NaOH + HOCl $ NaOCl + H2O We will discuss two important hypochlorites, viz. sodium hypochlorite and bleaching powder. (b) Sodium Hypochlorite, (NaOCl) It is a prepared by electrolysis of cold brine with stirring. During electrolysis, H+ ions are discharged at the cathode and the concentration of OH– ion increases. The Cl2 formed at the anode reacts with OH– ions to give OCl–. At anode: At cathode

2Cl– $ Cl2 + 2e– 2H+ + 2e– $ H2

2OH– + Cl2 $ OCl– + Cl– + H2O Properties NaOCl decomposes on heating to give NaClO3 and NaCl and on standing it liberates O2. 3NaOCl ∆ → NaClO3 + 2NaCl 2NaOCl $ 2NaCl + O2 It acts as a strong oxidising and bleaching agent in acidic medium. NaOCl $ NaCl + [O] or OCl– + 2H+ + 2e– $ Cl– + H2O

Chemistry of Group 17 Elements

Eg:–

16.21

NaOCl + 2HCl $ NaCl + Cl2 + H2O

(c) Calcium Chloro Hypochlorite or Calcium Oxychlorite (CaOCl2) It is manufactured by the action of chlorine gas on dry slaked lime, Ca(OH)2 in Bachmann’s plant as shown in Fig. 16.15. It is soluble in water and undergoes slow autooxidation on long standing and decomposes in the presence of COCl2. 6CaOCl2 $ Ca(ClO3)2 + 5CaCl2 2 2CaOCl2 CoCl  → 2CaCl2 + O2

When Cl2 or dry CO2 gas is passed through the aqueous solution of CaOCl2, HOCl is obtained while Cl2 is obtained on passing moist CO2. CaOCl2 + H2O + Cl2 $ CaCl2 + 2HOCl 2CaOCl2 + H2O + CO2 (dry) $ CaCl2 + 2HOCl +CaCO3 CaOCl2 + H2O + CO2 (moist) $ CaCO3 + 2H2O + Cl2 Fig. 16.15 Manufacture of bleaching powder

It acts as a strong oxidising and bleaching agent in presence of insufficient dilute acids. When bleaching powder is treated with excess of dilute acid or CO2, chlorine is evolved, known as available chlorine. CaOCl2 + 2HCl $ CaSO4 + H2O + Cl2 CaOCl2 + CO2 $ CaCO3 + Cl2 (d) Chlorous Acid (HClO2) It is obtained by the action of dilute sulphuric acid on barium chloride or by the action of H2O2 on ClO2 Ba(ClO2)2 + H2SO4 (dil.) $ BaSO4 + HClO2 2ClO2 + H2O2 $ 2HClO2 + O2 Properties The aqueous solution of the acid is colourless when freshly prepared but starts decomposition with time and turns yellow. 5HClO2 $ HClO2 + HCl + 2H2O 4HClO2 $ 2ClO2 + 4HClO3 + HCl + H2O 3HClO2 $ 2HClO3 + HCl 2HClO2 $ HOCl + HClO3 HClO2 $ HCl + O2 It is more acidic than hypochlorous acid and is a strong oxidising agent. The important salt of chlorous acid is sodium chlorite (NaClO2). It is prepared by the action of ClO2 with an alkali peroxide or hydroxide. 2ClO2 + Na2O2 $ 2NaClO2 + O2 2ClO2 + 2NaOH $ NaClO2 + NaClO3 + H2O

16.22 Inorganic Chemistry Pure NaClO2 can be obtained by the action of H2O2 on a concentrated solution of ClO2 in alkaline medium. 2ClO2 + H2O2 + 2NaOH $ 2NaClO2 + 2H2O + O2 Sodium chlorite is comparatively more stable than HClO2 but decomposes on boiling in presence of sunlight. 3NaClO2 ∆ → 2NaClO3 + NaCl hn

æææ Æ 2NaClO3 + 2NaClO4 + 6NaCl+ 3O2 10NaClO2 æpH = 4.0 hn 6NaClO2 æpH æææ Æ 2NaClO3 + 4NaCl + 3O2 = 8.4

It is used as a bleaching agent and oxidising agent. ClO2– + 4H+ + 4I– $ 2I2+ Cl– + 2H2O (e) Chloric Acid (HClO3) It is prepared by the action of dilute sulphuric acid on Ba(ClO3)2. Ba(ClO3)2 + H2SO4 $ BaSO4 + 2HClO3 Properties It exists only in solution and gets decomposed on heating. ∆

→ HClO4 + Cl2 + 2O2 + H2O 3HClO3  The solution is colourless in the dark and becomes yellow in the presence of light due to decomposition. 3HClO3 $ HClO4 + 2ClO2 + H2O It is a strong oxidising and bleaching agent. 2HClO2 + I2 $ 2HClO3 + Cl2 It is a monobasic acid and forms only one type of salt, known as chlorates. The most important salt is potassium chlorate (KClO3). It is prepared by electrolysis of hot brine solution, which is vigorously stirred. The obtained NaOH and Cl2 react together to form NaClO3, which is heated with KCl to obtain KClO3. 3Cl2 + 6NaOH $ NaClO3 + 5NaCl + 3H2O NaClO3 + KCl $ KClO3 + NaCl Potassium chlorate is a white crystalline solid with melting point of 370ºC. It decomposes on heating to give oxygen at low temperature and perchlorate at high temperature. C 2KClO3 400°  → 2KCl + 3O2 400°C 4KClO3 > → 3KClO4 + KCl

It is a strong oxidising agent as indicated by the following reactions. KClO3 + 3H2SO3 $ KCl + 3H2SO4 2KClO3 + 4HCl $ 4KCl + 2ClO2 + Cl2 + 2H2O (f) Perchloric Acid (HClO4) It is prepared either by the heating of HClO3 or by the action of conc. H2SO4 on potassium perchlorate. 3HClO3 $ HClO4 + Cl2 + 2O2 + H2O KClO4 + H2SO4 $ KHSO4 + HClO4

Chemistry of Group 17 Elements

16.23

Properties It is a colourless, hygroscopic and oily liquid which strongly fumes in moist air. It is the strongest and most dangerous acid and decomposes with explosion on heating or on standing for a few days. It dissolves metals like Zn, Fe, etc. to liberate hydrogen. Zn + 2HClO4 $ Zn(ClO4)2 + H2 It is a strong oxidising agent and explodes on contact with organic materials. It has a strong affinity for water and forms a number of hydrates like HClO4.H2O, HClO4.2H2O and HClO4.3H2O. It’s most important salt is potassium perchlorate which is obtained by heating potassium chlorate. ∆

→ 3KClO4 + KCl 4KClO3  KClO4 is a colourless crystalline solid which liberates O2 on heating and forms anhydrous HClO4 on distillation with H2SO4 under reduced pressure. ∆

→ KCl + 2O2 KClO4  KClO4 + H2SO4 conc. $ KHSO4 + HClO4

16.5.4 Uses of Chlorine and its Compounds 1. 2. 3. 4.

16.6

It is used to prepare cholro derivatives of many organic and inorganic compounds. It is used for disinfecting water and as a bleaching agent Hydrogen chloride is an important laboratory reagent and is used as ‘aqua regia’ to dissolve metals. Oxides and oxy acids of chlorine are used as oxidising agents. Salts of oxy acids of chlorine like, bleaching powder, are used as bleaching agents.

BROMINE (Br)

16.6.1 Occurrence and Extraction of Bromine Bromine is the forty-seventh most abundant element and occurs to an extent of 2.5 ppm by weight in the earth’s crust. It mainly occurs in sea water, mineral springs and deposits in the form of bromides. It was discovered by A J Balardin in 1826, by the treatment of chlorine on the mother liquor left after the crystallisation of NaCl from sea water. He named the liquid bromine (Greek bromos, stench) due to its strong odour. Cl2 + 2Br– $ 2Cl– + Br2 The liberated bromine is removed by a stream of air and is passed through a solution of Na2CO3 to obtain a mixture of NaBr and NaBrO3. 3Br2 + 3Na2CO3 $ 5NaBr + NaBrO3 + 3CO2 The solution is acidified and distilled to obtain pure bromine. 5NaBr + NaBrO3 + 3H2SO4 $ 3Na2SO4 + 5HBr + HBrO3 5HBr + HBrO3 $ 3Br2 + 3H2O

16.6.2 Properties It is a redish brown liquid which boils at 58.8°C and freezes at –7.3°C. It is highly volatile and gives strong fumes which are irritating to the throat and lungs. Bromine is less reactive than chlorine but more active than iodine and combines with many metals, nonmetals and their compounds.

16.24 Inorganic Chemistry Aqueous solutions of bromine are used for oxidising, bleaching and disinfection. Some examples showing its oxidising action are Na2S2O3 + Br2 + H2O $ Na2SO4 + S + 2HBr Na3AsO3 + Br2 + H2O $ Na3AsO4 + 2HBr Bromine reacts with cold solution of NaOH to give hypobrominates. With hot solutions, bromates are produced. Br2 + 2NaOH (cold) $ NaBr + NaOBr + H2O 3Br2 + 6NaOH (hot) $ 5NaBr + NaBrO3 + 3H2O

16.6.3 Compounds of Bromine (a) Hydrobromic Acid, Hydrogen Bromide (HBr) It is prepared by passing a mixture of hydrogen and bromine over an electrically heated platinum spiral . HBr cannot be prepared by the treatment of conc. H2SO4 on metal bromides because the produced HBr gets oxidised to Br2 by conc. H2SO4. However, H2SO4 can be replaced by conc. H3PO4. ∆

→ Na3PO4 + 3HBr 3NaBr + H3PO4  HBr can also be produced by the treatment of bromine with H2S on benzene in the presence of iron or red phosphorous with water. Properties HBr is a colourless and pungent-smelling gas which is highly soluble in water and fumes in moist air. It forms an azeotropic mixture with water with boiling point 399 K (48% HBr). It gets liquefied on cooling under high pressure to form a colourless liquid which boils at 206 K and freezes at 187 K. Anhydrous HBr is not acidic but turns blue lithium red when in solution. Dilute aqueous solutions of HBr are very reactive and turn yellow due to oxidation of HBr to Br2. (b) Oxides of Bromine Bromine forms many oxides which decompose even below room temperature. Some important oxides of bromine are discussed here:

(i)

Bromine Monoxide (Br2O) It is prepared by the action of bromine vapours on dry HgO at

323–343 K.

2HgO + 2Br2 $ HgO.HgBr2 + Br2O It is a dark brown liquid (freezing point –17.5°C). It undergoes disproportionation in presence of alkalis. 6NaOH + 6Br2O $ 5NaBrO3 + NaBr + 3H2O It dissolves in water to give hypobromous acid and hence is considered an anhydride of HBrO. It acts an a oxidising agent and oxidises iodine to iodine pentoxide. 5Br2O + I2 $ I2O5 + 5Br2

(ii)

Bromine Dioxide (BrO2)

It is prepared by ozonolysis of bromine at low temperature.

78°C Br2 + 4O3 − → 2BrO2 + 4O2 It is a yellow solid stable only below –40°C and decomposes at 0°C. It dissolves in alkalis and gives a mixture of bromides and bromates.

6BrO2 + 6NaOH $ 5NaBrO3 + NaBr + 3H2O It structures is same as that of ClO2.

Chemistry of Group 17 Elements

(iii)

16.25

Bromine Trioxide (BrO3) It is prepared by ozonolysis of bromine at temperature between –5°C to 10°C. 5°C to 10°C Br2 + 2O3− → 2BrO3 It is a white crystalline compound stable only below –70°C and is a strong oxidising agent.

(c) Oxyacids of Bromine Bromine forms two oxyacids, viz. hypobromous acid (HOBr) and bromic acid (HBrO3). However salts of bromous acid (HBrO2) are known as bromites.

Hypobromous Acid (HOBr) It is prepared by the treatment of bromine water with freshly precipitated mercuric oxide or silver oxide. Properties Its aqueous solution is straw yellow in colour and decomposes rapidly at 50°C to give a mixture of bromic acid and bromine.

16.6.4 Uses of Bromine and its Compounds 1. It is used as an oxidising agent and in the manufacturing of disinfectants and other important organic compounds such as tetraethyl lead. 2. Silver bromide is used in photography. 3. HBr in used as a laboratory reagent. 4. Oxides and oxyacids of bromine are used as oxidising agents.

16.7

IODINE (I)

16.7.1 Occurrence and Extraction of Iodine Iodine is the sixty-second most abundant element and occurs to an extent of 0.46 ppm in the earth’s crust (by weight). It occurs in traces as iodides in sea water and seaweeds. Its another main source is caliche which contains traces of sodium iodate and sodium periodate. Iodine was discovered by a French chemist, Bernard Courtois in 1911, when he noticed purple vapour coming out from the ashes of sea weeds. It was named by Davy in 1930’s from the Greek word ioeides, meaning violet. In common, the seaweeds are dried and burnt in shallow pits to obtain ash, known as kelp, containing 0.4 to 1.3% of iodine as iodides. The ash is dissolved in hot water and its solution is concentrated to separate out the sulphates and chlorides. The mother liquor left behind is treated with MnO2 and hot conc. H2SO4 to obtain iodine. Iodine is prepared from the crude Chile saltpetre by treating the mother liquor, left after crystallisation of nitre, with sodium bisulphate. 2NaIO3 + 5NaHSO3 $ 2Na2SO4 + 3NaHSO4 + I2 + H2O Iodine is prepared in the lab by heating an iodide with MnO2 in presence of conc. H2SO4 or with Cl2 or Br2. MnO2 + 2NaI + 3H2SO4 $ I2 + 2NaHSO4 + MnSO4 + 2H2O 2NaI + X2 $ 2NaX + I2

(X = Cl, Br)

16.7.2 Iodine as Metalloid—Electropositive Character of I2 We have discussed earlier that all the halogens have high electronegativities and high ionisation energies which decrease down the group. As a result, their metallic or basic character increases from F to I so that iodine is the least electronegative and exhibits metalloid properties. Thus, iodine can form I+ and I3+ as discussed ahead.

16.26 Inorganic Chemistry

1. Evidences for the Formation of I+ (a) The electrolysis of the molten or aqueous solution of ICl, I2SO4, ICl3, etc. gives iodine at the cathode indicating the dissociation of these compounds to give I+. I+ + ICl2–

2ICl

2I+ + SO42–

I2SO4

(b) I2 dissolves in H2O to give species containing unipositive iodine such as [I(H2O)]+ and [I(OH)]. [I(H2O)]+ + I–

I2 + H2O

H2O [I(OH)] + H3O+ (c) Hypoiodous acid undergoes ionisation to give I+ ion. OH– + I+

HOI

(d) Many stable complexes of unipositive iodine have been prepared. For example, AgNO3 + I2 + 2py Chloroform  → AgI + [I(py)2]NO3 The electrolysis of these complexes in chloroform also liberates iodine at the cathode, again supporting the existence of I+. (e) Iodonation of acetanilide and salicylic acid with the strong electrophilic iodinating agent, ICl also supports the existence of I+. NHCOCH3

NHCOCH3 ICl

Acetanilide

I 4-iodoacetanilide

OH

OH COOH

COOH + 2HCl I I 3,5-diiodosalicyclic acid

ICl

Salicyclic acid

2. Evidences for the Formation of I3+ (a) Electrolysis of molten ICl3 liberates a mixture of I2 and Cl2 at both the electrodes. This indicates the ionisation of ICl3 into ICl2+ and ICl4– which contain iodine as the tripositive iodine I3+. ICl2+ + ICl4–

2ICl3

(b) Iodine is oxidised by fuming nitric acid in the presence of acetic anhydride to give iodine triacetate while iodine phosphate is formed in the presence of H3PO4. Similarly, oxidation with ozone in presence of HClO4 yields iodine triperchlorate. HNO3 I2  → I(CH3COO)3 ( CH CO ) O 3

2

HNO3 I2   → H3 PO 4

IPO4

O3 I2  → I(ClO4)3 HClO 4

Chemistry of Group 17 Elements

16.27

All these compounds contains the I3+ ion. (c) Electrolysis of molten iodine triacetate in presence of silver electrodes consumes 3 faradays of electricity for the liberation of one equivalent of AgI at the cathode. This supports the ionisation of iodine triacetate as I3+ + 3(CH3COO)– I(CH3COO)3 I3+ + Ag + 3e– $ AgI

16.7.3 Properties of Iodine Iodine is a steel grey coloured solid with metallic lusture and sublimes on heating to give violet coloured vapours with irritating smell. It dissolves in CS2 and CHCl3 to form violet solution (as free I2). However it is only slightly soluble in water. Solubility in water increases in presence of KI due to the formation of KI3. KI + I2 $ KI3 It is chemically less reactive as compared to other halogens. Thus it reacts with hydrogen in reversible manner and cannot decompose water. It reacts violently with many metals and non-metals to form their corresponding iodides. It disproportionates in presence of cold and hot dilute solutions of alkalis. I2 + NaOH (cold) $ NaI + HOI 3I2 + 6NaOH (hot) $ 5NaI + NaIO3 + 3H2O It cannot decompose any halide; however, it reacts with KClO3, in presence of little nitric acid, on heating to liberate Cl2. 2KClO3 + I2 $ 2KIO3 + Cl2 It can oxidise S2– to S, SO2 and SO32– to SO42–, NO2– to NO3– and AsO33– to AsO43–. It forms an explosive black powder, nitrogen tri-iodide, on reacting with liquor ammonia, which explodes due to decomposition in air. 2NH3 + 3I2 $ NI3.NH3 + 3HI 8NI3.NH3 $ 5N2 + 6NH4I + 9I2

16.7.4 Compounds of Iodine (a) Hydroiodic Acid or Hydrogen Iodide (HI) It is prepared by the treatment of I2 on hydrogen in presence of finely divided platinum. The mixture is heated in a tube. Pt H2 + I2 � ��� � �� � 2HI Just like HBr, it cannot be prepared by heating metal iodide with conc. H2SO4, because the formed HI gets oxidised to I2. Hence, H2SO4 is replaced by H3PO4.

NaI + H3PO4 $ Na3PO4 + 3HI HI is prepared in the laboratory by slowly pouring water from a dropping funnel, on the mixture of red phosphorous and iodine in a flask. Properties It is a colourless and pungent-smelling gas which gets easily liquefied and solidified (boiling point 237.5 K and melting point 222.2 K). It is highly soluble in water and fumes in moist air. Aqueous solution of HI is acidic and turns blue litmus red. The solution turns brown in air due to the liberation of free I2. It is the strongest reducing agent among the halogen acids. Table 16.7 list the comparison of properties of halogen acids.

16.28 Inorganic Chemistry Table 16.7 Comparison of properties of halgen acids Property HF HCl HBr HI State (15°C) Liquid Gas Gas Gas Boiling point (°C) 19.4 – 85 –67 –35.5 Melting point (°C) –83 – 111 –86 50.8 574 428 362 295 Bond energy (kJ mol–1) Solubility in water 85.3 42.0 49.0 57.1 at 0°C (g/litre) Boiling point of 120 110 126 127 azeotropic solution HCl gas evolved HBr and Br2 evolved, HI and I2 evolved, Action with conc. HF vapours evolved H2SO4 on heating which turn starch which turn starch paper and oily globules are which gives white fumes with NH3. paper yellow blue seen in the solution Reddish brown vapour Violet vapour of I2 Action with MnO2 Same as above Greenish yellow vapours of Cl2 evolved of Br2 evolved which evolved which turns in presence of conc. H2SO4 on hearting starch paper blue which bleach indigo turns starch paper solution yellow. White ppt. of AgCl Action of AgNO3 Pale yellow ppt. of No ppt. as AgF is Yellow ppt. of AgI formed which is AgBr are formed solution formed which is formed which is soluble insoluble in HNO3 but which is soluble in soluble in water. in both HNO3 and HNO3 but sparingly soluble in NH4OH NH4OH soluble in NH4OH No ppt. No ppt. No ppt. Action of CaCl2 White ppt. of CaF2, solution which are soluble in conc. HCl Action of lead No ppt White ppt. of PbCl2, White ppt. of PbBr2, Yellow ppt. of PbI2, solacetate solution soluble in hot water soluble in hot water uble in hot water to give colourless solution Action of Cl2 water No action No action Br2 gas liberated I2 liberated which diswhich dissolves in solves in CS2 to give CS2 to give orange co- violet solution. loured solution

(b) Oxides of Iodine Iodine forms oxides, viz. I2O4, I4O3, I2O5 and IO4, which are described below: (i) Iodine Peroxide (I2O4) It is prepared by the treatment of hot H2SO4 (conc.) with iodic acid. .H 2 SO 4 4HIO3 Conc  → 2I2O4 + 2H2O + O2

Properties It is a lemon-yellow solid which is slightly soluble in cold water; it however, dissolves in hot water to liberate I2. 5I2O4 + 4H2O $ 8HIO3 + I2 It dissolves in KOH solution slowly to liberate I2. 3I2O4 + 6KOH $ 5KIO3 + KI + 3H2O It reacts with hydrochloric acid to form iodine monochloride. I2O4 + 8HCl $ 2ICl + 3Cl2 + 4H2O

Chemistry of Group 17 Elements

16.29

It decomposes on heating above 130°C to liberate I2. ∆

→ 4I2O5 + I2 5I2O4  It is an ionic compound and regarded as [IO]+[IO3]–

(ii)

Iodine Iodate (I4O9)

It is prepared by ozonolysis of dry I2. 2I2 + 9O3 $ I4O9 + 9O2

Properties It is a pale yellow, deliquescent solid, which reacts with H2O to liberate I2. 4I4O9 + 9H2O $ 18HIO3 + I2 It decomposes on heating above 75°C. ∆

→ 6I2O5 + 2I2 + 3O2 4I4O9  It is an ionic compound and is regarded as I+[IO3]3–.

(iii)

Iodine Pentoxide (I2O5) presence of conc. HNO3.

It is prepared by heating I4O9 or HIO3 or paraperiodic acid or I2 in

C 4I4O9 75°  → 6I2O5 + 2I2 + 3O2 C 2HIO3 240°  → I2O5 + H2O C 2H5IO6 138°  → I2O5 + O2 + 5H2O

I2 + 10HNO3(conc) $ I2O5 + 10NO2 + 5H2O Properties It is a colourless, odourless, deliquescent, crystalline solid which decomposes above 300°C. 300°C 2I2O5 Above  → 2I2 + 5O2 It dissolves in water to form HIO3. I2O5 + H2O $ 2HIO3 It is a strong oxidising agent as evident from the following reactions: I2O5 + 5H2S $ I2 + 5S + 5H2O I2O5 + 10HCl $ 2ICl3 + 2Cl2 + 5H2O I2O5 + 5CO $ I2 + 5CO2

Fig. 16.16 Structure of I2O5

Its structure has been established by IR spectroscopic studies and is found as O2I–O–IO2, with two pyramidal IO3 units linked through a common oxygen atom (Fig. 16.16). (c) Oxyacids of Iodine Iodine forms hypoiodous acids (HOI), iodic acid (HIO3) and periodic acid like metaperiodic acid (HIO4) and paraperiodic acid (H5IO6) as discussed here.

(i)

Hypoiodous Acid (HOI) It is prepared by shaking iodine with a freshly prepared suspension of mercuric oxide. Properties It is unstable and undergoes disproportionation in standing or in presence of acids. 3HOI $ HIO3 + 2HI HIO3 + 5HI $ 3I2 + 3H2O

16.30 Inorganic Chemistry

Sodium hypoiodite (NaOI) is an important salt of HIO, though less stable than hypochlorites and hypobromites. It is formed by the action of cold and dilute solution of NaOH on I2. 2NaOH + I2 $ NaOI + NaI + H2O The freshly prepared aqueous solution of NaOI acts as an oxidising and bleaching agent like NaOCl.

(ii) Iodic Acid (HIO3)

It can be prepared by the following methods: I2 + 10HNO3 $ 3HIO3 + 10NO2 + 4H2O I2 + 6H2O + 5Cl2 $ 2HIO3 + 10HCl Ba(IO3)2 + H2SO4 $ BaSO4 + 2HIO3 5KClO3 + 3H2O + 3I2 $ 6HIO3 + 5KCl 2HClO3 + I2 $ HIO3 + Cl2

Properties It is a colourless solid which dissolves in water to form reddish aqueous solution and bleaches litmus paper. It decomposes on heating above 240°C and inflames on heating with podwered S, P, charcoal and organic matter. It acts as on oxidising agent and itself is reduced to I2. Its forms potassium iodate, a crystalline solid which is soluble in hot water and decomposes on heating above 560°C to give KI. 4KIO3 $ 3KIO4 + KI and 2KIO3 $ KI + 3O2 Hence, it is also used as an oxidising agent.

(iii) Paraperiodic Acid (H5IO6)

It is prepared by heating HClO4 in suspension of I2.

2HClO4 + I2 + 4H2O $ 2H5IO6 + Cl2 It can also be prepared by the electrolytic oxidation of 50% HIO3 solution Properties It is a colourless and deliquescent crystalline solid which is soluble in water. It decomposes on heating to 413 K. 353°C °C °C 2H 5 IO6 − → H 4 I 2 O9 373  → 2HIO 4 413  → 2HIO3 + O2 3H O −H O Paraperiodic acid

2

Dimesoperiodic acid

2

Metaperiodic acid

Iodic acid

It can oxidise KI to I2 in acidic medium. H5IO6 + 7KI + 7HCl $ 7KCl + 4I2 + 6H2O Its structure has been determined by X-ray studies to be octahedral with sp3d2 hybridisation of I atom and is shown in Fig. 16.17.

(iv) Metaperiodic Acid (HIO4) of H5IO6 at 100°C.

It is prepared by the careful decomposition Fig. 16.17 Structure of H5IO6

C H5IO6 100°  → HIO4 +2H2O

Properties

It converts back to H5IO6 when dissolved in water. HIO4 + 2H2O $ H5IO6

It is used as a strong oxidising agent in acidic medium. HIO4 + 7KI + 7HCl $ 7KCl + 4I2 + 4H2O

Chemistry of Group 17 Elements

16.31

Table 16.8 Comparative behaviour of the halogens Properties

Fluorine

Chlorine

Bromine

Iodine

Physical state

Gas

Gas

Liquid

Solid

Vapour

Pale yellow coloured

Greenish yellow coloured

Dark reddish brown coloured

Violet coloured

Smell

Very pungent

Irritating

Very irritating

Intensely irritating

Action of H2

Very fast even in dark with explosion

Slow in dark but explosively in sunlight

Only on heating

Very slow and requires catalyst

Action of H2O

Decomposes even in the dark and cold to give HF, O2/O3

Decomposes slowly to give HCl and O2

Decomposes only in sunlight to give HBr and O2

No action

Action of metals

Metals burn to form fluorides

Many metals burn to form chlorides

Metals slowly react to Very slowly iodides form bromides are formed

Action of nonmetals

Except N2, O2 and noble gases, direct action

Except N2, O2, noble gases and C, direct action

Except N2, O2, noble Direct action only gases, C and Si, direct with P, As, H and other halogens action

Action of alkalis

NaF, O2 and H2O are formed

In cold, NaCl and NaOCl; and in hot, NaCl and NaClO3 are formed

In cold, NaI and NaI and NaIO3 are NaOBr; and in hot, formed NaBr and NaBrO3 are formed

Oxidising action

Very strong

Strong

Slow

Very slow

Bleaching action

Destroys organic matter

Bleaches vegetable colouring matter readily

Bleaches slowly

No action

Action with hydro carbons

Decomposition of hydrocarbons

Substitution/addition

Slow reaction

Very slow in presence of catalyst

16.7.5 Uses of Iodine and its Compounds 1. It is used as a reagent in the laboratory for analysis of compounds. It is also used in estimating reducing substances by iodimetry and oxidising substances by iodometry. 2. It is used as a disinfectant under the name tincture of iodine and as an analgesic in the form of ‘Iodex’. 3. HI is used as a reducing agent. 4. Oxides and oxyacids of iodine are used as oxidising agents.

16.8

ASTATINE (At)

16.8.1 Occurrence and Synthesis Astatine is a radioactive and the rarest naturally occurring terrestrial element with all its natural isotopes of half lives less than 1 minute. However, many synthetic isotopes of astatine have been prepared. Corson, Mackenzie and Segre synthesised the first synthetic isotope, 211At(t½ = 7.21h) in 1940 by bombarding 209Bi with -particles.

16.32 Inorganic Chemistry 209 83Bi

1 + 42 $ 211 85At + 2 0n

Its isotopes occur in the uranium decay series.

16.8.2 Chemistry of Astatine 1. At–1 is obtained by the action of moderately powerful reducing agents. (Zn/H+, SO4, AsIII) on At or AtI. 2. At(0) is obtained by the action of oxidising agents (dil. HNO3 or AsIV) on As–1. – IV 3. AtO3– is obtained by the action of powerful oxidising agents (S2O2– 8 , IO4 or Ce ) on At(0). – – 4. AtO4 is obtained by the treatment of AtO3 with periodate or XeF2/NaOH. 5. At reacts with halogens to form interhalogen species and halide ions to form polyhalide ions. At + ½ I2 $ AtI I→ AtI2– −

Br −

I / IBr

2 At + ½Br2  → AtBr  → AtBr2–

– At + ½Cl2 $ AtCl Cl  → AtCl2 AtI is used to prepare many organic derivates. −

Cl

PhI

Hg Ph

2 AtI  → PhAt 2 → PhAtCl2  → Ph 2 AtCl

16.9

INTERHALOGEN COMPOUNDS

Two different halogens may react with each other to form covalent compounds known as interhalogen compounds. The more electropositive halogen is written and mentioned first. These are formed by direct combination of the two halogens under specific conditions. These compounds can be grouped into specific categories as mentioned in Table 16.9. As fluorine is the most electronegative halogen, the interhalogen compounds are usually formed as fluorides. Due to small difference in electronegativity of the two halogens, the bond is essential covalent and is comparatively weaker than B—B bond. Hence, interhalogen compounds are very reactive. We will discuss these compounds one by one.

16.9.1 AB Types These interhalogens are diatomic and usually the difference in the electronegativities of the two halogens is not much. The fluorides have been found to be more reactive as compared to chlorides and bromides. These are linear molecules. Table 16.9 Categories of interhalogens Group

AB

AB3

AB5

Oxidation state of A atom

(+1)

(+3)

(+5)

Compounds of fluorine

ClF, colourless

ClF3, colourless

ClF5, colourless

Compounds of bromine

BrF, Pale brown

BrF3, Pale yellow

BrF5, colourless

Compounds of iodine

(IF), unstable; ICl, ruby-red; IBr, black

(IF3), yellow; ICl3, Bright yellow

IF5, colourless

AB7 (+7)

BrCl, Red-brown IF7, colourless

Chemistry of Group 17 Elements

16.33

1. Chlorine Monofluoride (ClF) It is prepared by the action of chlorine on fluorine or chlorine trifluoride. 200°C

Cl2 + F2 → 2ClF C Cl2 + ClF3 300°  → 3ClF Properties It is a colourless gas with melting point 117 K and boiling point 373 K. It decomposes on heating and is more reactive than F2. 2ClF $ Cl2 + F2 As fluorinating reagent: Se + 4ClF $ SeF4 + 2Cl2 As chlorine fluorinating reagent:

W + 6ClF $ WF6 + 3Cl2 SF4 + ClF $ SF5Cl SO2 + ClF $ SO2FCl CO + ClF $ COFCl

Structure It is a linear molecule with a bond length of 162.8 pm.

2. Bromine Monofluoride (BrF) It is prepared by the reaction of BrF3 or BrF5 with bromine but it cannot be isolated. It is a pale brown gas which condenses to form a dark-red liquid and freezes as yellow crystalline solid with a boiling point of 20°C and and a melting point of –34°C. BrF3 + Br2 $ 3BrF BrF5 + Br $ 5BrF It is highly unstable and decomposes through dismutation. 3BrF $ BrF3 + Br2 Structure It is a linear molecule with a bond length of 175.6 pm.

3. Bromine Monochloride (BrCl) It is formed by the combination of an equimolar mixture of Br2 and Cl2 at room temperature. Br2 + Cl2 ? 2BrCl It is a dark red irritating liquid with a boiling point of 5°C and melting point of –66°C. It acts as a brominating agent and oxidising agent. It is a linear compound with a bond length of 213.8 pm.

4. Iodine Monofluoride (IF) It can be prepared by the action of iodine with fluorine or iodine trifluoride or silver fluoride. 3F I2 + F2 CCl  → 2IF −45°C 3F I2 + IF3 CCl  → 3IF −78°C

I2 + AgF $ IF + AgI It is a dark brown solid melting at –45°C and decomposing at 0°C. 5IF $ 2I2 + IF5 It is a linear molecule with bond length of 190.9 pm.

16.34 Inorganic Chemistry

5. Iodine Monochloride (ICl) It is prepared by the following methods: C I2 + Cl2 35°  → 2ICl

KClO3 + I2 $ KIO3 + ICl C ICl3 68°  → ICl + Cl2

Properties It is a dark red liquid with a boiling point of 974°C. It exists in two solid forms which are formed depending upon the method of cooling, viz. stable form and metastable form. The stable -form (m.pt. 272°C) is obtained as red needle-like crystals when the liquid is cooled rapidly. The metastable -form (m.pt. 14°C) is obtained as a black solid when the liquid is cooled slowly at –10°C. It changes back to stable form on standing. It is hydrolysed in water as HOI + HCl ICl + H2O 3ICl + 3H2O

HIO3 + 3HCl + 3HI

5ICl + 3H2O

HIO3 + 5HCl + 2I2

However, addition of N/5 HCl prevents the hydrolysis. It also dissolves in excess of alkalis as 5ICl + 6NaOH $ NaIO3 + 5NaCl + 2I2 + 3H2O On electrolysis, molten ICl liberates I2 at the cathode and a mixture of I2 and Cl2 at anode. It undergoes auto-ionisation in the liquid state. 2ICl

I+

ICl2–

+

(Solvent cation)

(Solvent anion)

The metal chlorides which give I ions in liquid ICl behave as acids and those giving ICl2– ions behave as bases in the presence of ICl. +

MCl + ICl MCl4 + ICl

M+ + ICl2–

(M = K, Rb, NH4)

I + MCl5– MCl6– (M –

(M = Sb, Nb)

I+ + = Si, Sn, Ti and V) MCl5 + ICl Thus, it acts as a nonaqueous solvent for the reaction of such salts. I+[SbCl6]– + K+[ICl2]– Acid

Base

K+[SbCl6]– + 2ICl Salt

Solvent

It is used as Wiz’s solution (ICl in glacial acetic acid) for the determination of iodine value of oils. Structure It is a linear molecule in the bond length of 237.07 pm.

6. Iodine Monobromide (IBr) It is prepared by the direct combination of the elements I2 + Br2 $ 2IBr It is a steel grey hard, crystalline solid melting at 42°C. It conducts electricity in the molten state due to ionisation and boils at 116°C. IBr $ I+ + Br– It is a linear molecule with a bond length of 249 Å.

Chemistry of Group 17 Elements

16.35

16.9.2 AB3 Type These are tetra-atomic interhalogens. Spectroscopic and structural studies reveal that their geometry is distorted triangular bipyramidal with sp3d hybridisation. The distortion is due to lone pairs and the molecule acquires T-shape with bond angles 90° (Fig. 16.18).

Fig. 16.18 Structure of AB3 type interhalogen

1. Chlorine Trifluoride (ClF3) It is prepared by the action of F2 on Cl2 in a Cu vessel at 200–300°C. 200 −300°C

3F2 + Cl2  → 2ClF3 Cu vessel Properties It is a colourless gas which on condensation gives a pale green liquid boiling at –12°C. It is the most reactive interhalogen of AB3 type and forms other interhalogens. ν ClF3 + F2 h → ClF5 C ClF3 + Br2 10°  → BrF3 + BrCl

It gets hydrolysed by water as ClF3 + H2O $ 2HF + ClOF It produces HF on treatment with NH3 and N2H4. 2ClF3 + 2NH3 $ 6HF + Cl2 + N2 It undergoes self-ionisation as

4ClF3 + 2N2H4 $ 12HF + 2Cl2 + 3N2 2ClF3

ClF2+ (solvent cation)

Hence, the salts which produce

ClF2+

+

ClF4– (solvent anion)

ion act as acids in this solvent.

MF5 + ClF3

ClF2+ + MF6–

(M = As, Sb, V)

It is used as a fluorinating reagent. 2AgCl + ClF3 $ 2AgF + ClF + Cl2 6NiO + 4ClF3 $ 6NiF2 + 2Cl2 + 3O2 2Co3O4 + 6ClF3 $ 6CoF3 + 3Cl2 + 4O2 − 90°C U + 3ClF3 50  → UF6 + 3ClF

2. Bromine Trifluoxide (BrF3) It is prepared by the following methods:

16.36 Inorganic Chemistry (a) By the action of Br2 vapour on F2 in the presence of N2 atmosphere. In N 2 Br2 + 3F2  → 2BrF3 atmosphere

(b) By the action of Br2 on ClF3

10°C

Br2 + ClF3  → BrCl + BrF3 (c) By the treatment of F2 with HBr 3F2 + 2HBr $ 2BrF3 + H2 It is a straw coloured, corrosive liquid which boils at 125.8°C and solidies at 8.8°C. It gets hydrolysed in water as BrF3 + H2O $ BrF3 + H2 It is a nonprotonic ionising solvent and is used for preparation of many inorganic compounds. 2BrF3

BrF2+ + BrF4– MF4– + BrF2+

MF3 + BrF3

(M = Au)

MF62– + 2 BrF2+

MF4 + 2BrF3 MF5 + BrF3

MF6–

MF + BrF3

M+ + BrF4–

+

2+

CaF2 + 2BrF3

(M = Ge, Sn, Ti)

BrF2+

Ca +

(M = P, Sb, Bi, Nb, Ta) (M = Li, K, Ag)

2BrF4–

NO+ + BrF4–

NOF + BrF3

3 GeF4 + 2NOF BrF  → [NO]2[GeF6] BrF3 is a very good fluorinating agent.

2WO3 + 4BrF3 $ 2WF6 + 2Br2 +3O2

3. Iodine Trichloride (ICl3) It is prepared by the treatment of I2 or ICl with excess of chlorine at 100°C. 3Cl2 + I2 $ ICl3 Cl2 + ICl $ ICl3 It can also be prepared by the action of dry HCl gas on heated I2O5. I2O5 + 10HCl $ 2ICl3 + 2Cl2 + 5H2O Properties It is a pale yellow crystalline solid which is hydrolysed by H2O and KOH as 3ICl3 + 12KOH $ 2KIO3 + 9KCl + KI + 6H2O 2ICl3 + 2H2O $ HIO3 + 5HCl + ICl It decomposes on heating at 68°C.

68°C

→ ICl + Cl2 ICl3  It acts as an ionising solvent like BrF3. It acts as a chlorinating reagent.

2ICl3

MF + ICl3

[ICl2]+ + [ICl4]– MICl3F

(M = K, Rb, Cs)

Chemistry of Group 17 Elements

16.37

16.9.3 AB5 Type These are hexa-atomic interhalogens with distorted octahedral geometry and have a square pyramidal shape due to repulsions from the lone pairs (Fig. 16.19).

Fig. 16.19 Structure of AB5 type interhalogen

1. Chlorine Pentafluoride (ClF5) It is prepared by the action of F2 on ClF3 or Cl2. hν

ClF3 + F2 → ClF5 350°C

 → 2ClF5 Cl2 + F2 250 atm It is a gas (m.pt. –103ºC, b.pt. –13.1ºC) and gets hydrolysed by water. ClF5 + 2H2O $ 4HF + FClO2 It undergoes self-ionisation and acts as a fluorine donor 2ClF5

ClF4+ + ClF6–

MF5 + ClF5 $ [ClF4]+[MF6]–

[M = As, Sb]

2. Bromine Triflouride (BrF3) It is prepared by the action of F2 on KBr or Br2. 25°C

→ KF + BrF5 KBr + 3F2  150 °C Br2 + 5F2 > → 2BrF5

It is very reactive and gets hydrolysed as BrF5 + 3H2O $ 5HF + HBrO2

3. Iodine Pentafluoride (IF5) It is prepared by the action of I2 on F2 or AgF. I2 + 5F2 $ 2IF5 I2 + 10AgF $ 2IF5 + 10Ag It can also be prepared by the treatment of F2 with I2O5. I2O5 + 10F2 $ 4IF5 + 5O2 It is a colourless liquid with a melting point of 9.6°C. It undergoes self-ionisation as 2IF5 IF4+ + IF6–

(Solvent cation) (Solvent anion)

16.38 Inorganic Chemistry Thus, it is a good ionising solvent It gets hydrolysed as

+ – SbF5 + KF IF 5 → K [SbF6]

2IF5 + 3H2O $ 5HF + HIO3 It forms iodine oxy-fluoride on reacting with I2O5. 3IF5 + I2O5 $ 5IOF3

16.9.4 AB7 Type IF7 is the only one compound in this group. Iodine can accommodate seven small atoms of F due to its large size. IF7 is prepared by the treatment of F2 with KI, IF5 or PbI2 C KI + F2 250°  → KF + IF7 250 −300°C

IF5 + F2  → IF7 PbI2 + 8F2 $ PbF2 + 2IF7 It is a gas and gets hydrolysed by water as IF7 + 6H2O $ H5IO6 + 7HF It can form adducts like IF7 + MF5 $ [IF6]+[MF6]– [M = Sb, As] It reacts with silica as 250°C 2IF7 + SiO2 → 2IOF5 + SiF4 IF7 has a pentagonal pyramidal structure formed by sp3d3 hybridisation (Fig. 16.20).

Fig. 16.20 Structure of IF7

16.10

POLYHALIDES

The diatomic halogen molecules and interhalogens can combine with halide ions to form the ions known as polyhalide ions. The compounds c-ontaining these polyhalide ions are called polyhalides, e.g. K+[Cl3]– contains Cl3– and [I5]+[AlCl4]– contains [I5]+ as the polyhalide ions.

16.10.1 Synthesis of Polyhalides These compounds can be prepared by the following methods: 1. By the action of halogens on metal halides Cl2 + KCl $ K[Cl3] Cl2 + KI $ K[ICl2] Br2 + CsI Alc.  → Cs[IBr2] 2I2 + NH4I $ NH4[I5]

Chemistry of Group 17 Elements

16.39

2. By the action of interhalogens with metallic hydrides ICl + MCl $ M[ICl2]

(M = K, Rb, NH4) +

KCl + ICl3 $ K (ICl4)– KBr + ICl $ K+(BrICl)– KF + BrF3 $ K+(BrF4)– 3. Preparation of polyhalide acids by addition of halogen or interhalogen to an acid HCl + ICl3 $ H+(ICl4)– BrF3 + MF

M[BrF4]

(M = Li, K, Ag)

ICl + KBr $ K[ClBrI] ICl + K[ClBrI] $ K[ICl2] + IBr

16.10.2. Properties The polyhalides dissociate on heating to form the metal monohalides (corresponding to the smaller of the halogen molecule) and the halogen molecule or interhalogen molecule. For example, CsICl2 undergoes thermal dissociation to give CsCl and ICl, rather than CsI and Cl2. The thermal stability of a particular polyhalides increases with the increase in size of cation Na+ < K+ < NH4+ < Rb+ < Cs+ The thermal stability of a polyhalides ion increases as: [BrI]– < [FIBr]– < [ClIBr]– < [I3]– < [BrIBr]– < [ClICl]– Polyhalides react with halogens to give another polyhalides. ICl2– + Cl2 $ ICl4– Some polyhalides also form complexes with organic molecules, e.g., LiI3.4C6H5CN and NaI3. 2C6H5CN.

16.10.3 Classification of Polyhalides Polyhalides can be classified as follows. Thihalide ions

Br–3, I–3, Cl–2, ICl+2, etc.

Tetrahalide ions

I42–

Pentahalide ions

ICI4–, CIF4–, I5–, IF4–, CIF4+, etc.

Heptahalide ions

I7+, I7–, IF6+, etc.

Octahalide ions

I8–, etc.

Others

I9–, Br9–, I–11, etc.

16.10.4 Structure of Polyhalides The structure of many polyhalides have been established by X-ray studies. Examples of some linear and nearly linear triatomic polyhalide ions are ClF2–, Cl3–, BrF2–, BrCl2–, Br2Cl–, Br3–, IF2–, IBrF–, IBrCl–, ICl2–, IBr2–, I2Cl–, I2Br– and I3–. The linear shape results from sp3d hybridisation of the central atom. The two halogen atoms occupy the axial positions and the lone pairs occupy the basal positions (Fig. 16.21).

16.40 Inorganic Chemistry –

xl

X

–

Electronic structure in excited state

180° X ns

np

s

nd

s

Xl Xm sp3d hybridisation

xm

Fig. 16.21 Structure of trihalide ions

The trihalide cations like ICl2+, BrF2+ and ClF2+ are angular in shape due to sp3 hybridisation of the central atom (Fig. 16.22). + +

X

Electronic structure in excited state

X ns

nd

nd

s Xl

s Xm

sp3 hybridisation

Xm

Xl

Fig. 16.22 Structure of trihalide cations

The penta-atomic polyhalide anions like ClF4–, BrF4–, ICl4– etc. favour the square planar geometry due to sp3d2 hybridisation of the central atom (Fig. 16.23). –

Xl

–

Xl

X

Electronic structure in excited state

.

ns

nd

s

s

s

s

Xl

Xl

Xl

Xl

X

nd Xl

Xl

sp3d2 hybridisation

Fig. 16.23 Structure of Pentahalides anions

However, the I–5 ion in Me4NI5 is formed in planar V-shaped geometry as shown (Figure 16.24).

Fig. 16.24 Structure of I5+ ion

Chemistry of Group 17 Elements

16.11

16.41

PSEUDOHALOGENS AND PSEUDOHALIDES

Brickenbach and Kellerman found that a large number of inorganic radicals possess properties similar to those of halogens, when in free state and to those of halide ions, when in ionic state. These are termed as pseudohalogens and pseudohalides accordingly. Table 16.10 lists some important examples. Some important similarities in properties of halide ion and pseudo-halides ions along with halogens and pseudohalogens have been discussed here:

Table 16.10 Some important pseudohalide and Psedohalogens Pseudohalide ions CN–, Cyanide OCN–, Cyanate SCN–, Thiocyanate SeCN–, Selenocyanate N3–, Azide SCSN3–, Azidothiocarbonate ONC–, Isocyanate

1. Formation of Ionic Compounds Ionic compounds of pseudohalides Ionic compounds of halides

AgCN, AgCl,

Pseudohalogens (CN)2, Cyanogen (OCN)2, Oxocyanogen (SCN)2, Thiocyanagen (SeCN)2, Selenocyanogen (SCSN3)2, Azidocarbon disulphide

Pb(CNS)2, PbCl2,

Hg(CNS)2 HgCl2

2. Formation of Covalent Compounds Covalent Compounds of pseudohalides Covalent Compounds of halides

ICN, ICl,

3. Formation of Monobasic Hydracids Hydracid of pseudohalides Hydracid of halide 4. Oxidation of Hydracids Oxidation of HY to give Y2 : (Y = SCN, CN etc.) Oxidation of HY to give X2

Si(NCS)4, SiCl4,

Co(N3)2, CoCl2,

SO2(N3)2 SO2Cl2

H2 + 2CN– $ 2HCN + 2e– H2 + 2Cl– $ 2HCl + 2e– 4HSCN + MnO2 $ Mn(SCN)2 + (SCN)2 + 2H2O 4HCl + MnO2 $ MnCl2 + Cl2 + 2H2O

5. Thermal Decomposition of Salts Decomposition of Pb(IV) salts of pseudohalides to give Pb (II) salt and free pseudohalogen ∆

Pb(SCN)4  → Pb(SCN)2 + (SCN)2 Decomposition of Pb(IV) halides to give Pb (II) halides and free halogen PbCl4 ∆ → PbCl2 + Cl2 6. Formation of Insoluble Salts with Ag+, Pb2+ and Hg+ Insoluble salt of pseudohalide Ag+ + CN– $ AgCN Insoluble salt of halide Ag+ + CN– $ AgCl 7. Formation of Compounds Interpseudohalogen Compounds Interhalogen compounds

CN.N3, CN.SCN ClF, ICl, IBr

Some important dissimilarities in properties of halide ions and pseudohalide ions are the following: (a) Hydracids of pseudohalide ions are comparatively weaker acids than the corrsponding hydracids of the halides ions.

16.42 Inorganic Chemistry (b) Pseudohalide ions act as stronger coordinating ligands than the halide ions due to ability to form back bond. (c) In contrast to the monodentate halide ions, pseudohalide ions act as ambidentate ligands. Important similarities between halogens and pseudohalogens: (a) Both are dimeric and volatile in the free state. (b) Both can add to ethylenic double bond (c) Both can react with alkali in a similar manner Action of halogens: Cl2 + 2KOH(cold and dil) $ KOCl + KCl + H2O Action of pseudohalogens: (SCN)2 + 2KOH (cold and dilute) $ KOSCN + KSCN + H2O (d) Both combine with H2 to give monobasic hydracids. However, unlike halogens, pseudohalogens form polymerised series. n(Y)2 $ 2(Y)n where Y= CN, SCN, etc.

Some Important Pseudohalogens 1. Cyanogen, C2N2 or (CN)2 It is prepared by heating a mixture of Hg(CN)2 and HgCl2. Hg(CN)2 + HgCl2 $ Hg2Cl2 + (CN)2 It can also be prepared by the action of KCN on CuSO4 solution. CuSO4 + 2KCN $ K2SO4 + 2CuCN + (CN)2 Dehydration of ammonium oxalate also yields cyanogen. It is a colourless, flammable gas with a smell of bitter almonds. It is extremely poisonous and freezes to a solid which melts at –27.9°C. It burns explosively in air to give a violet flame. (CN)2 + 2O2 $ 2CO2 + N2 It slowly hydrolyses in water to form a number of products such as urea and ammonium oxalate.

It readily reacts with alkalis and alkali metals. (CN)2 + 2KOH $ KCNO + KCN + H2O It is readily reduced with H2 to give ethlenediamine.

2. Thiocyanogen (SCN)2 It is prepared by the oxidation of HSCN with lead tetracetate or manganese dioxide. It can also be prepared by passing bromine gas through the solution of lead thiocyanate at 0°C.

Chemistry of Group 17 Elements

16.43

Pb(SCN)2 + Br2 $ PbBr2 + (SCN)2 It is a yellow solid which polymerises even at room temperature to give brick-red coloured parathiocyanogen, (SCN)n. The polymeric form is insoluble in water but (SCN)2 is hydrolysed rapidly to give thiocyanic acid and hydrocyanic acid. (SCN)2 + H2O $ HSCN + HCN It acts as an oxidising agent as supported by the following reactions in which it is converted to the thiocyanate ion. 2I– $ I2; CuSCN $ Cu2+ + 2SCN– 2(S2O3)2– $ (S4O6)2–; H2S $ 2H+ + S (AsO3)3– + H2O $ 2H+ + AsO43– Its structure is linear and can be represented as N C–S–S–C N.

3. Selenocyanogen (SeCN)2 It is prepared by the action of bromine with silver selenocyanate in presence of ether. 2AgSeCN + I2 $ (SeCN)2 + 2AgI It is a yellow crystalline solid and turns red on standing. It hydrolyses with water to give a mixture of hydroselenic acid and hydrogen cyanide 2(SeCN)2 + 3H2O $ 3HSeCN + H2SeO3 + HCN Its structure is similar to that of thiocyanogen

4. Azidocarbon Disulphide (SCSN3)2 It is prepared in a number of ways as shown below

It is a white crystalline substance which is highly unstable and decomposes violently. (SCSN3)2 $ (SCN)2 + 2N2 + 2S It reacts slowly with water and dilute acids but reacts rapidly with concentrated acids and alkalis.

5. Cyanogen Halide or Pseudohalogen–Halogen Compounds These are the compounds of cyanogens and a halogen and are colourless, volatile and highly poisonous. ClCN and BrCN have been prepared as a liquid by reacting the respective halogen with sodium cyanide or hydrogen cyanide. (X = Cl, Br) NaCN + X2 $ XCN + NaCl These are slowly hydrolysed by water XCN + H2O $ HOX + HCN

16.44 Inorganic Chemistry

6. Inter Halogenoids or Inter–pseudohalogen Compounds These compounds are formed by the combination of two pseudohalogens and are analogous to the inter halogens. Some of these have been prepared as follows: Hg(CN)2 + 2(SCN)2 $ Hg(SCN)2 + 2CNSCN KSCN + CNBr $ KBr + CNSCN SeCN2 $ Se + CNSeCN CNBr + NaN3 $ NaBr + CNN3 Group 17 is constituted by five elements, viz. F, Cl, Br, I and At, collectively known as the halogens. Their general electronic configuration can be written as ns2np5. These elements exist as diatomic covalent molecules and are coloured. F2 and Cl2 exist as gases, Br2, as a liquid and I2 as a black flaky solid. These elements have the smallest atomic and ionic sizes in their periods, while their IE, and EA and electronegativity are the highest. F is the most electronegative element and can exhibit only (–I) oxidation state while other elements can exhibit (+I), (+III), (+V) and (+VII) in addition to (–I) oxidation state. Except I2, other halogens are nonmetallic while At is radioactive. The halogens form ionic compounds with metals while covalent compounds are formed with the nonmetals. F2 is the strongest oxidising agent. HF shows hydrogen bonding and has the highest boiling point among the halogen acids. The acidic strength of halogen acids varies as HF < HCl < HBr < HI. Halogens form a number of oxides and oxyacids. The acidic strength of oxyacids increases in the order HOX < HXO2 < HXO3 < HXO4. Two different halogens may react with each other to form covalent compounds known as interhalogens. The AB type of interhalogens are linear, AB3 type are T shaped (sp3d hybridisation), AB5 type are square pyramidal (sp3d2 hybridisation) and AB7 type are pentagonal pyramidal (sp3d3 hybridisation). The interhalogens and the diatomic halogens may combine with halide ions to form polyhalide ions. The AB2– type are linear shaped (sp3d hybridisation) and AB2+ type are angular shaped (sp3 hybridisation), AB4– types are square planar in shape (sp3d2 hybridisation), AB4+ are distorted tetrahedral (sp3d hybridisation), AB6– are distorted octahedral (sp3d3 hybridisation). The pseudohalogens and pseudo halides have similar properties as that of halogens and halide ions, while the interhalogenoids are analogous to that of interhalogens. Halogens also formed halogen oxide fluorides in which the halogen is bonded to both O and F. These are strong oxidising and fluorinating agents.

EXAMPLE 1

Complete the following reactions.

(a) AgClO4 + I2 $ (a) 2AgClO4 + I2 $ 2AgI + 2ClO4

(b)

(b) Cl2O7 + 2NaOH $ Cl2O7 + 2NaOH $ 2NaClO4 + H2O

Chemistry of Group 17 Elements

EXAMPLE 2

16.45

What is the action of cold and hot NaOH on

(a) Br2?

(b) I2?

(a) Br2 reacts with cold NaOH to give hypobromide while with hot NaOH, bromate is formed. Br2 + 2NaOH (cold) $ NaBr + NaBrO3 + H2O 3Br2 + 6NaOH (hot) $ 5NaBr + NaBrO3 + 3H2O (b) I2 reacts with cold NaOH to give hypoiodous acid while with hot NaOH sodium iodate is formed I2 + NaOH (cold) $ NaI + HOI 3I2 + 6NaOH (hot) $ 5NaI + NaIO3 + 3H2O

EXAMPLE 3

What is the action of HF with silica and glass?

HF reacts with silica to give hydrofluorosilicic acid while sodium fluorosilicate is formed with glass. SiO2 + HF $ H2SiO3 + H2SiF6 Na2SiO3 + 6HF $ Na2SiF6 + 3H2O

EXAMPLE 4

Write the balanced chemical reactions for the attack of

(a) Conc. HCl on K2Cr2O7 (b) Acidified KMnO4 with HBr solution (a) K2Cr2O7 + 14HCl $ 2KCl + 2CrCl3 + 3Cl2 + 7H2O (b) 2KMnO4 + 3H2SO4 + 10HBr $ K2SO4 + 2MnSO4 + 8H2O + 5Br2

QUESTIONS Q.1 Discuss the special properties of HF in brief. Q.2 Give reasons for the following: (a) Fluorine exhibits only (–I) oxidation state while the other halogens can show higher oxidation states. (b) HF has the highest boiling point while lowest acidic strength among the halogen acids. (c) F2 is the strongest oxidising agent. Q.3 Account for the following: (a) HF is not stored in glass bottles. (b) SiF4 reacts with HF to give H2SiF6. (c) Electrolysis of aqueous HF does not give F2 at anode (d) Fluorine does not form any oxyacid except HOF. (e) I2 is insoluble in H2O but dissolves in aqueous solution of KI. Q.4 How will you prepare the following? (a) Bleaching powder (b) Iodine pentoxide (c) Potassium chlorate

16.46 Inorganic Chemistry

Q.5 Q.6 Q.7 Q.8

Discuss the bleaching action of bleaching powder. What are pseudohalogens and what are their important characteristics? Discuss the characteristics of interhalogens and interhalogenoids. Discuss the structures of the following compounds. (a) IF5 (b) IO4– (c) ClO3– (d) Cl2O (e) I2O5 (f) IF7 Q.9 Discuss the electropositive character of iodine with the help of suitable examples. Q.10 Discuss the action of bleaching powder with the following compounds: (a) CO2 (b) Cl2 (c) COCl2 (d) PbO (e) H2SO4

MULTIPLE-CHOICE QUESTIONS 1. The most reactive halogen is (a) F2 (b) Cl2 (c) Br2 (d) 2. The strongest acid is (a) HF (b) HCl (c) HBr (d) 3. The gases evolved by the action of hot conc. H2SO4 on KBr are (a) HBr (b) HBr + SO2 (c) HBr + Br2 (d) 4. The structure of BrF5 is (a) T-shaped (b) triangunlar bipyramidal (c) square pyramidal (d) Pentagonal bipyramidal 5. The structure of ICl4– is (a) tetrahedral (b) square planar (c) T-shaped (d)

I2 HI SO2 + Br2

see-saw shaped

chapter

Chemistry of Group 18 Elements

17

After studying this chapter, the student will learn about

17.1

INTRODUCTION

Group 18 elements (helium, neon, argon, krypton, xenon and radon) have been characterised as inert gases for many years and believed incapable of forming any chemical compounds with other elements. Zero group was another term that was used for them because of their closed-shell electronic configuration with zero valence. However, these labels were deprecated with the discovery of some xenon compounds. These elements have also been called rare gases, but this term proved a misnomer because argon makes up a considerable part of the earth’s atmosphere (1.3% by mass). The most appropriate term is noble gas, used first by Erdmann in 1898, translated from the German word edelgas to indicate that these gases have extremely low levels of reactivity.

17.2

HISTORY AND DISCOVERY

The first attempt in the discovery of noble gases was initiated by Cavendish in 1784. He noticed a discrepancy between the density of nitrogen extracted from air and that obtained from chemical reactions, but was unable to analyse the reason. In 1868, Janssen and Lockyer independently identified a new element in the spectrum of the sun and named it helium (from the Greek word helios meaning sun). After many years, Rayleigh repeated the experiment performed by Cavendish. Finally in 1895, in assistance with Ramsay, he was successful in isolating a new element, argon (from the Greek word argos meaning inactive). In later studies, they isolated helium by treating the mineral cleveite with acids. In 1898, Ramsay and Travers obtained three new elemens by fractional distillation of liquefied air. These elements were named krypton, neon and xenon from the Greek words kryptos (hidden), neos (new) and

17.2

Inorganic Chemistry

xenos (stranger) respectively. Radon, the radioactive noble gas, was obtained by Dorn in 1900 from the radioactive disintegration of radium. Recently, in 2006, scientists have synthesised the seventh element, ununoctium (Uuo) by bombarding californium with calcium (Ca). Interestingly, a new element of Group 14, ununquadium (Uuq), now named as flerovium (F1), has been found with abnormal noble-gas- like properties.

17.3

OCCURRENCE AND ISOLATION OF NOBLE GASES

All noble gases, except radon, occur in the free state as monoatomic gases in the atmosphere. The adundance of these gases generally increases with increase in atomic number. However, argon is the most abundant noble gas constituting about 0.93% by volume of air and krypton is the least abundant constituting about 0.001% by volume of air. Thus, neon, argon, krypton and xenon can be isolated from air by removing other gases. This isolation is done first by removal of N2, O2, CO2 and other gases by chemical methods to get a mixture of noble gases. The individual gas is obtained from this mixture by the methods as given ahead:

1. Chemical Methods Three methods have been used as described below: (a) Ramsay–Rayleigh First Method In this method, air is passed over sodalime or caustic potash to remove CO2 followed by heated copper and magnesium to remove O2 and N2. The reactions involved are CO2 + KOH $ K2CO3 + H2O 2Cu + O2 $ 2CuO 3Mg + N2 $ Mg3N2 (b) Ramsay–Rayleigh Second Method Here, an electric discharge is passed through a mixture of air and oxygen to convert gases into oxides of nitrogen. CO2 and oxides of nitrogen are absorbed in sodium hydroxide solution and excess oxygen is removed by passing the mixture through alkaline pyragallol. N2 + O2 $ 2NO 2NO + O2 $ 2NO2 2NO2 + 2NaOH $ NaNO2 + NaNO3 + H2O CO2 + NaOH $ Na2CO3 + H2O (c) Fischer–Ringe Method In this method, free air is circulated through an iron retort, over a hot mixture of 90% calcium carbide and 10% anhydrogous calcium chloride at 800°C to remove N2 and O2. CaC2 + Na $ CaCN2 + C C + O2 $ CO2 2C + O2 $ 2CO 2CO + O2 $ 2CO2 2CaC2 + 3CO2 $ 2 CaCO3 + 5C The evolved gases are passed over hot cupric oxide followed by passing through caustic potash. CO + CuO $ Cu + CO2 2KOH + CO2 $ K2CO3 + H2O

Chemistry of Group 18 Elements

17.3

2. Physical Methods The mixture of noble gases obtained from chemical methods is separated by using two physical methods: Claude’s method and Dewar’s method. (a) Claude’s Method This method is mainly used for isolation of helium from natural gas. Helium is found in natural-gas deposits located in some parts of USA and Canada. In this method, hydrocarbons are liquified followed by fractional distillation to separate the individual noble gas. The schematic representation is given: Liquid air (77 K) Fractional distillation

Liquid nitrogen, helium and neon passed over heated CaC2

CaCN2

Liquid oxygen, argon, krypton and xenon cooled in liquid nitrogen

Mixture of helium and neon cooled in liquid H2 (20 K)

Oxygen, krypton and xenon Evaporation

Oxygen and argon Passing over heated copper Argon

Liquefied neon

Gaseous helium

Oxygen Krypton

Fractional distillation Xenon

(b) Dewar’s Method This method is based on the principle that adsorption of noble gases on charcoal increases with increase of the atomic weight and decreases with increase of temperature. The schematic representation is as follows. Mixture of noble gases Passing over Charcoal at 173 K

Unadsorbed helium and neon

Adsorbed argon, krypton and neon

Passing over Charcoal at 93 K

Unadsorbed helium

Adsorbed neon Heating

Passing over Charcoal at 77K

First charcoal (adsorbed argon) heating

Second charcoal (adsorbed krypton) and xenon

Evolved neon

183 K Evolved argon Evolved krypton

Adsorbed xenon Heating Evolved xenon

17.4

Inorganic Chemistry

Helium is also formed in the earth’s crust as a radioactive decay product. 238 U $ 234Th + 4He Helium is the second most abundant element in the universe after hydrogen. Some naturally occurring argon is produced during decay of 40K by electron capture. 40

K $ 40Ar

Radon is produced in the radioactive decay series of 235U, 238U and 232Th. In 238U series

223

In 235U series

226

In 232U series

224

219

 → Ra 11 .2 d  → Ra 1620 y → Ra 3 .6 d

Rn + 4He

222

Rn + 4He

220

Rn + 4He

However, all these isotopes of radon are unstable and undergo further decay.

17.4

USES OF NOBLE GASES

Table 17.1 Isotopes of noble gases The extremely low reactivity of noble gases makes them very useful in many industrial, medical and domestic He Ne Ar Kr Xe Ra applications. Insolubility of helium in blood even at high 20 4 36 78 124 (32 10 Ne 2He 18Ar 36Kr 54Xe pressure makes it a component of breathing gases used by isotopes 3 21 38 80 126 sea divers and asthma patients. Due to its nonflammable 2He 10Ne 18Ar 36Kr 54Xe with mass nature, helium is filled in meterological balloons and 22 40 82 128 10Ne 18Ar 36Kr 54Xe numbers tyres of airships. The very low boiling temperature of 83 129 199–226) Kr Xe 36 54 liquid helium (4.2 K) makes it a cryogenic refrigerant. 219 Ra, 84 130 It is used in the strudy of low-temperature phenomena 220 36Kr 54Xe Ra, 86 131 such as superconductivty and high-resolution NMR 222 Ra 36Kr 54Xe spectroscopy. Helium is used as a heat-transfer agent in occurs in 132 54Xe gas-cooled nuclear reactors because of its high thermal nature 134 conductivity and resistance to radiation. An important use 54Xe 136 of noble gases is to provide an inert atmosphere. Helium 54Xe and argon are used in the high- temperature metallurgical processes and welding of easily oxidisable metals like Mg, Ti, Zr and Al, etc. Another important use of noble gases is in electric lights. Argon is used as a filling gas for incandescent lamps because it does not react with the tungston filament and prolongs the bulb’s life. Krypon mixed with small amounts of compounds of iodine or bromine are used in halogen lamps. Noble gases show distinctive light emission when excited by an electric discharge in presence of certain other gases, mercury and phosphors inside gas-discharge tubes, called neon lights. These lights are used for advertising purposes. Xenon is used in xenon arc lamps for projection lights. Noble gases have direct application in medicine. These gases are a component of excimer lasers used in surgery. Because of high solubility in lipids and rapid removal from the body, xenon is used as an anaesthetic. Radon is used in radiotheraphy for cancer treatment. Radon is also used to detect defects in steel castings and earthquake prediction.

17.5

PHYSICAL PROPERTIES

The general outer-shell electronic configuration of Group 18 elements can be represented as ns2np6, Due to completely filled electronic configuration, these elements are chemically inactive. Table 17.2 shows the electronic structures of these elements.

17.5

Chemistry of Group 18 Elements

Noble gases are colourless, odourless, tasteless and monoatomic Table 17.2 Electronic structures of Group 18 elements gases. All noble gases, except helium, have a closed octet of electrons Element Electronic structure in their outer shell. Helium has only two electrons in its shell. The closed-shell electronic structures are highly stable, as shown by the He 1s2 high ionisation potentials and negligible electron affinities. Atomic Ne [He] 2s2 2p6 radii of noble gases are larger than Group 17 elements in their Ar [Ne] 3s2 3p6 corresponding periods. This is because, in case of noble gases, the Kr [Ar] 3d10 4s2 4p6 atomic radii correspond to the van der Waal’s radii which are higher than the covalent radii. Ionisation potentials decrease down the group Xe [Kr] 4d10 5s2 5p6 with increasing atomic radii due to addition of one new shell at each Rn [Xe] 4f14 5d10 6s2 6p6 successive step. The atoms of these elements are held together by very weak van der Wall’s forces as indicated by enthalpies of vapourisation and boiling points of liquid noble gases. The enthalpies and boiling points increase monotonically with increasing polarisability of the atom as the atomic size increases. Noble gases are only slightly soluble in water due to dipole-induced dipole interactions. The solubility increases down the group with increase in magnitude of such interactions with increase in atomic size. Due to low value of enthalpies of vapourisation and weak van der Wall’s force, these gases are not easily liquefied. However, down the group, the case of liquifaction goes on increasing. All noble gases, except helium, are adsorbed by active wood charcoal at low temperature. The adsorption capacity further increases with increase of atomic size. The general physical properties of the noble gases have been listed in Table 17.3 Table 17.3 Physical properties of noble gases He

Ne

Ar

Kr

Xe

Rn

–1

4.0026

20.183

39.948

83.80

131.3

222

% by volume in air

–4

–3

0.934

–3

–6

6 × 10–18

Mol. mass (gmol )

5.24 × 10

1.82 × 10

1.14 × 10

8.7 × 10

Atomic size (pm)

120

160

191

200

220

—

First ionisation energy (kJ mol–1)

2372

2081

1520

1351

1170

1037

1.8 × 10–4

9.0 × 10–4

1.8 × 10–3

3.7 × 10–3

5.9 × 10–3

3.7 × 10–3

Melting point (K)

0.95

25

84

116

161

202

Electron affinity (kJ mol–1)

–54

–99

Boiling point (K)

4.2

27

87

120

165

211

0.082

1.74

6.53

9.70

12.7

18.1

5.25

44.39

150.86

209.38

289.74

Density (g cm–3)

Enthalpy of vaporisation (kJ mol–1) Critical temp. (K)

17.6

CHEMICAL PROPERTIES

Noble gases, especially, He and Ne, are very less reactive due to their stable octets, high ionisation energies, almost zero electron affinities and absence of any vacant d-orbitals. Argon, krypton and xenon have so far been induced to enter into chemical combinations with other atoms and only bonds with highly electronegative atoms like F and O are stable. Chemical activity of noble gases increases from Ar to Rn with decrease of ionisation potentials.

17.6

Inorganic Chemistry

Radon is very reactive but its chemistry is dificult to assess because, the most stable isotope of radon, Rn, has a half-life of only 3.825 days.

222

Prior to 1940, some unstable species were formed by sparking helium at low pressure in the presence of mercury and tungsten. These molecular species in excited conditions were called helides like HgHe2, He2+, HeLi+, etc. Later on, some species with argon were also formed and detected spectroscopically. However, these species were not considered as true chemical compounds due to lack of bonding. Later on, Booth and Wilson reported the formation of a number of unstable coordination compounds of argon with varying number of BF3 molecules. However, the existence of such compounds could not be verified. In 1940, the first so-called chemical compounds containing noble gases were formed. These were known as clathrates or cage compounds in which the noble-gas atoms were trapped in cavities in the crystal lattice of other compounds. These compounds were obtained by crystallisation of water or a solution containing quinol, 1,4–C6H4(OH)2 in the presence of the noble gas at a pressure of 10–40 atm. The gas molecules were trapped in the large cavities of hydrogen-bonded lattices with only weak van der Wall’s interactions between host and guest molecules. Because of absence of any chemical bond, these compounds are nonstoichiometric with approximately 3 quinol : 1 gas molecule coordination ratio. Helium and neon do not form clathrates because of their small size and escape from the cavities. These compounds are considerably stable but the gas is released on dissolution or melting of clathrates. Hence, these are used to separate helium and neon gas from a mixture of noble gases and separate argon, krypton and xenon from mixtures of other gases. These compounds are also used for storing radioactive isotopes of Xe and Kr. The true noble-gas compounds were discovered in 1962, by three groups of researchers independently. The first credit goes to Bartlett and Lohmann. They obtained a solid ionic compound of the composition O2+ [Pt F6]– by reacting oxygen with platinum hexafluoride. O2(g) + PtF6(g) $ O2+PtF6–(g). After realising the similarities in the first ionisation potential of the oxygen molecule (1165 kJ mol–1) and xenon (1169 kJ mol–1), the experiment was repeated with xenon. They were successful in obtaining a yellow solid and reported its composition as Xe+[PtF6]–. + – Xe + PtF6 278K → Xe [PtF6]

Later studies revealed that the reaction was actually more complicated and the product was a mixture of several xenon compounds. After this initial break through, other noble gas compounds were prepared and characterised. However, no compound of helium has been discovered till now.

17.7

CHEMISTRY OF XENON (Xe)

Xenon reacts direcly with fluorine to form xenon fluorides. Other compounds in oxidation state from +2 to +8 can be prepared by using xenon fluorides as starting materials and opting the following methods. 1. Hydrolysis to yield oxides, oxofluorides and xenates 2. Combination with fluoride-ion acceptors to yield fluorocations of xenon 3. Combination with fluoride-ion donors to yield fluoroanions of xenon 4. Metathesis between xenon fluorides and an anhydrous acid

1. Xenon Difluoride (XeF2) It is prepared by combining xenon and fluorine in 2.1 molar ratio at room temperature under the influence of ultraviolet light or by heating the mixture in a nickel tube at 400°C. 400°C Xe(g)+ F2(g)  → XeF2(g) Ni tube

Chemistry of Group 18 Elements

17.7

Properties (a) XeF2 is a colourless solid and sublimes at room temperature. (b) It reacts immediately with hydrogen and NH3. XeF2 + H2 $ 2HF + Xe 3XeF2 + 2NH3 $ 3Xe + N2 + 6HF (c) It acts as a strong fluorinating agent for organic compounds and is used for selective fluorination. XeF2 + CH3I $ CH3IF2 + Xe (d) It dissolves in liquid HF without any chemical action. In presence of anhydrous HF, its reactivity increases probably due to the formation of XeF+. (e) It is a strong oxidising agent. XeF2(aq) + 2H+ + 2e– $ Xe + 2HF(aq.); E0 = +2.64 V Aqueous acidified solution of XeF2 oxidises HCl to Cl2, I–, and IO3– to IO4–, Ag(I) to Ag(II), Co(II) to Co(III) and Ce(III) to Ce(IV). (f) XeF2 is used as an etchant for silicon. 2XeF2 + Si $ 2Xe + SiF4 (g) XeF2 is used to prepare fluoroalkanes from carboxylic acids by oxidative decarboxylation. RCOOH + XeF2 $ RF + CO2 + Xe + HF (h) It is slowly hydrolysed in dilute aqueous acids. 2XeF2 + 2H2O $ 2Xe + 4HF + O2 In aqueous solutions base hydrolysis is rapid and complete. XeF2 + 2OH– $ Xe + ½ O2 + H2O + 2F– or Xe + H2O2 + 2F– (i) XeF2 acts a fluoride-ion donor and forms complexes with fluoride-ion acceptors to form cationic species. XeF2 + MFn $ [XeF]+ [MFn + 1]– n = 3 M = As, Sb, Bi, Pt n = 4 M = Zr, Cr, Mn n = 5 M = Al, Fe, Co (j) XeF2 acts as a ligand coordination complex in presence of HF. xM(AF6)2 + nXeF2 $ [Mx(XeF2)n](AF6)x M = Ca, Ba, Sr, Pb, Ag

A = As, Sb, P

(k) Unlike XeF4 and XeF6, it is the weakest fluoride-ion acceptor as XeF3– is not formed. (l) It yields higher xenon fluorides on reacting with fluorine. XeF2 + F2 $ XeF4 Structure The XeF2 molecule is a linear symmetrical molecule with Xe–F bond length of 197.73 pm. Its shape can be predicted with the help of VSEPR model, hybridisation theory and molecular orbital theory. According to VSEPR model, out of the five electron pairs, three lone pairs occupy the equatorial positions of the trigonal bipyramidal geometry resulting into linear structure. The structure can be explained using the

17.8

Inorganic Chemistry

concept of hybridisation with promotion of an electron from the 5p level to the 5d level and mixing of orbitals to undergo sp3d hybridisation.

ground state of Xe

F

Xe

excited state of Xe 5s

5p sp3d hybridisation

F

5d

Fig. 17.1 XeF2

2 Xenon Tetrafluoride (XeF4) It is obtained by combing xenon and fluorine in 1.5 molar ratio at 400°C in a nickel tube at 6 atm. 400°C

 → XeF4(g) Xe(g) + 2F2 (g) Ni tube Properties (a) XeF4 is a colourless crystalline solid, stable in the absence of moisture. It sublimes on heating in a nitrogen atmosphere to give colourless vapours. (b) It acts as a stronger fluorinating agent than XeF2 for both organic and inorganic compounds. XeF4 + 2H2 $ Xe + 4HF XeF4 + 2SF4 $ Xe + 2SF6 XeF4 + Pt $ Xe + PtF4 (c) XeF4 is a stronger oxidising agent than XeF2. It oxidises Pt(0) to Pt(IV), Hg(0) to Hg(I) and I– to I2. XeF4 + Pt $ PtF4 + Xe XeF4 + 4Hg $ 2Hg2F2 + Xe (d) XeF4 reacts with Xe at 400°C to form XeF2 and with F2 at 300°C to form XeF6. (e) XeF4 reacts violently with water to produce Xe and O2 or disproportionates to give XeO3 and O2. 3XeF4 + 6H2O $ 2XeO + XeO4 + 12 HF Xe + O2

XeO3 + ½ O2

(f) XeF4 is the weakest fluoride-ion donor and reacts only with strong Lewis acid metal fluorides like SbF5 and BiF5 to form cationic species. XeF4 + MF5 $ [XeF3]+ [M2F11]–

(M = Sb and Bi)

(g) XeF4 is a stronger fluoride-ion acceptor than XeF2 but weaker than XeF6 due to the formation of XeF5– ions. (M = Na to Cs) XeF4 + MF $ M+[XeF5]– XeF4 + NR4F $ NR4+ [XeF5]– (h) XeF4 reacts with O2F2 at 143 K to yield XeF6. XeF4 + O2F2 $ XeF6 + O2 Structure The XeF4 molecule is square planar with Xe–F bond length of 193 pm. Its geometry can be predicted just like XeF2.

Chemistry of Group 18 Elements

17.9

According to the VSEPR model, two lone pairs occupy two positions of the octahedron with minimum repulsions and the four bond pairs occupy the remaining positions resulting into a square planar structure. The hybridisation concept explains the structure by promotion of two electrons and mixing of orbitals to undergo sp3d2 hybridisation. F

F

Ground state of Xe Xe

Second excited state of Xe 5s

5p

5d

F

F

Fig. 17.2 XeF4

sp3d2 hybridisation

3. Xenon Hexafluoride (XeF6) It is obtained by reacting xenon with fluorine in a 1:5 molar ratio at 400°C and 200 atm. Xe(g) + 3 F2(g) $ XeF6(g) 1:5

Properties (a) XeF6 is a colourless crystalline solid but turns yellow on heating to give a yellow liquid and vapour. (b) XeF6 is stable at room temperature. It is stored in nickel containers as it reacts with silica and cannot be stored in glass vessels. 2XeF6 + SiO2 $ 2XeOF4 + SiF4 (c) XeF6 is the stronget fluorinating agent of all the xenon fluorides. It converts metallic mercury to HgF2 and reduces to xenon on heating with hydrogen. XeF6 + 3H2 $ Xe + 6HF (d) It reacts violently with water to give an explosive solid, XeO3, in the following steps. XeF6 + H2O $ XeOF4 + 2HF XeOF4 + H2O $ XeO2F2 + 2HF XeO2F2 + H2O $ XeO3 + 2HF (e) XeF6 acts as a fluoride donor and forms complexes with fluoride-ion acceptors to form the cationic species XeF5+ or Xe2F+11 depending upon the concentration of XeF6. XeF6 + MFn $ [XeF5]+ [MFn+1]– n=5 M = As, Sb, Pt, Pu, Ru n=4 M = Pt, Mn n=3 M = B, Al If MF5 is in excess, [XeF5]+[M2DF11]– is formed and if XeF6 is in excess, [Xe2F11]+[MF6]– is formed. (f) XeF6 acts as a fluoride acceptor with alkali fluorides to form [XeF8]2– species. Cesium and rubidium salts first form heptafluoroxenate salts. CsF + XeF6 $ CsXeF7 (Yellow) RbF + XeF6 $ RbXeF7 (Colourless) 2MF + XeF6 $ M2XeF8 (M = Na, K) Sodium and potassium octafluoroxenates decompose even at 120°C. These methods are used to purify XeF6.

17.10 Inorganic Chemistry

(g) XeF6 acts as a ligand in coordiation complexes like XeF2 and XeF4. HF

M(AsF6)2 + nXeF6 22°  → MF2 + 2Xe2F11AsF6 + (n – 4)XeF6 C

(M = Mg, Ca, Sr)

Structure The XeF6 has been an issue of contradiction for its strurcture. The VSEPR theory predicts its geometry as a capped octahedron or a pentagonal bipyramid with one F lone pair of electrons (Fig. 17.3). Same geometry is predicted on the basis of hybridization by promoting three electrons form 5p to 5d. F

Xe in ground state

Xe

F

F F

5s

5p

5d

Xe in third excited state

F

Fig. 17.3 XeF6

sp3d3 hybridisation

4. Xenon Trioxide (XeO3) It is obtained by hydrolysis of XeF4 and XeF6 to yield a colourless, orthorhombtic and hygroscopic explosive solid. It is soluble in water but remains in unionised form.

Properties (a) The colourless aqueous solution of XeO3 has strong oxidising properties. It oxidises Pu(III) to Pu(IV) and I– to I3–. E° = 2.10 V XeO3 + 6H+ + 6e– $ Xe + 3H2O; XeO3 + 6H+ + 9I– $ Xe + 3H2O + 3I3– 6PuCl3 + XeO3 + 6HCl $ 6 PuCl4 + Xe + 3H2O (b) In basic solutions, XeO3 exists as xenates. XeO3 + OH– $ HXeO4–

K = 1.5 × 10–3

HXeO4– slowly disproportionates to produce perxenate and elemental xenon. 2 OH −

→ 2XeO42– $ XeO64– + Xe + O2 2 HXeO4– + 2OH– − 2H O 2

(c) It explodes to yield xenon and oxygen gas. 2XeO3 $ 2Xe + 3O2 Structure The structure of xenon trioxide can be predicted by VSEPR theory and hybridisation concept. According to the VSEPR theory, due to lone pair–bond pair repulsions, the tetrahedral geometry is distorted as pyramidal strcture (Fig. 17.4). The hybridisation concept also predicts the same geometry.

O

O O

Fig. 17.4 XeO3

5. Xenon Tetraoxide (XeO4) It is formed by the action of concentrated H2SO4 on barium perxenate at 233 K. Ba2XeO6 + 2H2SO4 $ 2 BaSO4 + 2H2O + XeO4 It is a highly unstable and explosive gas. It sometimes explodes at –40°C. XeO4(g) $ Xe + 2O2

Xe

H = 643 kJ mol–1

Chemistry of Group 18 Elements

Structure It is tetrahedral as predicted by the VSEPR theory and hybridisation concept, and confirmed by electron-diffraction studies (Fig. 17.5).

17.11

O

6. Xenon Dioxide (XeO2)

Xe

This compound has been synthesised recently in 2011 by hydrolysis of xenon tetrafluoride in presence of conc. H2SO4 at 0°C.

O

O O

XeF4 + H2O $ XeO2 + HF

Fig. 17.5 XeO4

Structure XeO2 has a square planar geometry consistent with the VSEPR theory. But it exists as an extended chain or network structure in which the coordination numbers of xenon and oxygen are four and two respectively.

7. Xenon Tetrafluoride Oxide (XeOF4) It is prepared by the partial hydrolysis of XeF6 or by the reaction of XeF6 with sodium perxenate, Na4XeO6, silica and NaNO3. −80°C

XeF6 + H2O → XeOF4 + 2HF 2 XeF6 + SiO2 $ 2 XeOF4 + SiF4 XeF6 + NaNO3 $ XeOF4 + NaF + FNO2

Properties (a) It is a colourless volatile liquid which solidifies at 245 K. (b) It is more volatile than XeF6 and is metastable at room temperature. (c) Just like XeF6 it forms complexes with both fluoride-ion acceptors and fluoride-ion donors. O

XeOF4 + SbF5 $ [XeOF3]+[SbF6]– or [XeOF3]+[Sb2F11]– F

XeOF4 + CsF $ Cs[(XeOF4)3F]–

F Xe

(d) It combines with XeO3 to give xenon difluoride dioxide. XeOF4 + XeO3 $ 2XeO2F2

F

F

(e) It reacts with cesium nitrate immediately. Fig. 17.6 XeOF4

XeOF4 + CsNO3 $ XeO2F2 + CsF + FNO2 (f) It reacts with water and quartz to give XeO3.

Structure It has a square pyramidal arrangement in which oxygen atom is at the apex of the pyramid and the xenon atom is almost coplanar with the fluorine atoms with one lone pair at the sixth position of the octahedron (Fig. 17.6).

8. Xenon Dioxide Difluoride (XeO2F2) It is prepared by the reaction of XeO3 with XeOF4.

Properties

F O

(a) It is a colourless crystalline solid and can be stored for several days at room temperature. (b) It is much less stable than XeOF4. (c) It yields a 1:2 adduct with SbF5 which decomposes to give O2. XeO2F2 + SbF5 $ [XeO2F]+[Sb2F11]–

Xe O F

Fig. 17.7 XeO2F2

17.12 Inorganic Chemistry

Structure Single-crystal neutron diffraction data reveal that XeO2F2 exists as layers based on trigonal bipyramids with the fluorides at the axial positions and weak Xe = O----Xe bridges in the structure (Fig. 17.7).

9. Xenon Oxide Difluoride (XeOF2) It is prepared by the reaction of OF2 with xenon gas at low temperatures. It can also be obtained by the partial hydrolysis of XeF4 at low temperature. F

Xe + OF2 $ XeOF2

Properties (a) It is a bright yellow, nonvolatile solid stable up to –25°C. (b) It disproportionates at –25°C into XeF2 and XeO2F2.

Xe

2XeOF2 $ XeO2F2 + XeF2

F

Fig. 17.8 XeOF2

(c) It reacts with CsF at low temperature. CsF + XeOF2 $ Cs[XeOF3]

F

Xe

O

10. Xenon Trioxide Difluoride (XeO3F2)

O

Structure Its geometry is trigonal bipyramidal as predicted by the VSEPR theory and exists in distorted T-Shape due to presence of two lone pairs and is validated by Raman spectra (Fig. 17.8).

O

It is prepared by room-temperature reaction of XeF2 with sodium perxenate or XeO4. It is more volatile than XeO2F2. The structure of XeO3F2 has been verified by NMR spectroscopy and is consistent with as predicted by the VSEPR theory, i.e. trigonal bipyramidal (Fig. 17.9).

17.8

O

F

Fig. 17.9 XeO3F2

COMPOUNDS OF KRYPTON (Kr)

The chemistry of krypton is limited to that of krypton difluoride and its derviatives. The formation of KrF4 has not been verified yet and krypton monofluoride, KrF an important species in excimer laser systems, is short lived.

1. Krypton Difluoride (KrF2) It can be prepared by passing an electric discharge or by irradiating a mixture of krypton and fluorine at low pressure.

Properties (a) It is a white crystalline solid and sublimes at temperature well below 273 K. (b) It decomposes spontaneously at room temperature. It can be stored for indefinite periods of time at –78°C. (c) It is a powerful oxidative fluorinating agent. It oxidizes and fluorinates Xe to XeF6 and Au to AuF5. It is an aggressive fluorinating agent even at low temperature. (d) It reacts with strong and weak fluoride ion acceptors to give complex salts analogous to those of XeF2. KrF2 + MF5 $ [KrF]+ [MF6]–

(M = Bi, Sb, As, Ta, Pt, Au, As)

Chemistry of Group 18 Elements

17.13

These salts are stable at room temperature for appreciable amounts of time. KrF+ is the most powerful chemical oxidant and oxidises gasesous xenon to XeF5+, O2 to O2+, NF3 to NF4+ and ClF5 to ClF6+. KrF2 + MOF4 $ FKr – F – MOF4 KrF2 + MF5 $ [KrF] [M2F11] KrF2 + MF5 $ [Kr2F3] [MF6]

(M = Cr, Mo, W) (M = Sb, Ta, Nb) (M = As, Sb, Ta)

(e) Some salts of KrF2 with organic compounds have also been prepared. KrF2 + RCN $ [RCN – KrF]+

(R = H, CF3, C2H5, HC3H7)

Compounds of krypton and their derivatives have geometries similar to that of corresponding compounds of xenon and their derivatives.

17.9

COMPOUNDS OF RADON (Rn)

Since the ionisation energy of radon is lower than that of xenon, it should also form many similar compounds. But due to the short lifetime of its most stable isotope, its chemistry is limited. The most stable compound of radon obtained is radon fluoride. Radon difluoride can be formed by any of the following methods: 1. By heating a mixture of fluorine gas and trace amounts of radon-222 up to a temperature of 400°C. 2. By irradiating radon-222 with gaseous fluorine at room temperature and with liquid fluorine at –196°C. 3. By oxidising radon-222 in HF with IF7, NiF62– or bromine fluorides. Radon also reacts with solid oxidants at room temperature to form involatile complex salts containing RnF+. – Rn + O+2 SbF6– $ RbF+Sb2F11 The chemistry of radon and its compounds have been found similar to that of xenon and krypton. Low temperature, matrix-isolation techniques are being used to study the chemistry of noble gas compounds to explore the field further.

Group 18 elements, or noble gases, include helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe) and radon (Rn). The outer-shell electronic configuration is ns2 np6 (1s2 for He) with completely filled orbitals. Hence, these elements have very high ionisation energies, zero electron affinities and low values of enthaltpies of vapourisation. These gases are not easily liquified due to weak van der Wall’s forces. All noble gases, except helium, are absorbed by active wood charcoal at low temperature. Chemical activity of noble gases increases down the group due to decrease in ionisation potentials. Clathrates were the first so-called noble- gas compounds obtained by crystallisation of water in the presence of the noble gas (except He and Ne) at a pressure of 10–40 atm. The true noble gas compound, XePtF6, was discovered in 1962. Later on, many compounds of xenon were obtained.

17.14 Inorganic Chemistry

EXAMPLE 1

Complete the following reactions:

(a) XeF6 + SiO2 $ (c) XeF4 + C6H6 $

(b) XeF2 + CH3I $ (d) RbF + XeF6 $

(a) XeF6 + SiO2 $ XeOF4 + SiF4

(b)

XeF2 + CH3I $ CH3IF2 + Xe

(c) XeF4 + C6H6 $ Xe + C6H5F + HF

(d)

RbF + XeF6 $ RbXeF7

EXAMPLE 2

What will happen if

(a) Xenon trioxide is treated with hydrogen iodide? (b) Xenon tetrafluoride reacts with water? (a) Xenon trioxide oxidises I– to I3–. XeO3 + 6H+ + 9I– $ Xe + 3I3– + 3H2O (b) Xenon tetrafluoride reacts violently with water to produce either Xe and O2 or undergoes disproportionation to give XeO3 and O2. 3XeF4 + 6H2O $ 2XeO + XeO4 + 12HF Xe + O2

XeO3 + ½ O2

QUESTIONS Q.1 Give reasons for the following: (a) Noble gases have very high ionisation energies and almost zero electron affinities. (b) He does not form clathrates. (c) Noble gases are monoatomic. (d) Clathrates are not true chemical compounds. Q.2 Discuss the formation of clathrates by noble gases. Q.3 Describe the isolation process of noble gases. Q.4 Complete the following reactions: (a) XeF4 + SF4 (b) XeF6 + H2O (c) XeF6 + NaF (d) KrF2 + SbF5 (e) HXeO4– + OH– Q.5 Discuss the hybridisation and sructure of (a) XeF4 (b) XeO2F2 (c) XeO3 (d) XeO3F2 Q.6 Discuss the preparation and properties of xenates and perxenates. Q.7 Describe the discovery of the first true noble-gas compound.

Chemistry of Group 18 Elements

Q.8 Write complete balanced equation for the preparation of the following compounds: (a) XeO3 (b) XeO4 (c) XeOF2 Q.9 Compare the fluoride-ion donor acceptor behaviour of xenon fluorides. Q.10 Explain why the noble-gas compounds of Xe, Kr and Rn exist as fluorides and oxides.

MULTIPLE-CHOICE QUESTIONS 1. The ease of liquefication of the noble gases increases in the order as (a) Ar < He < Xe < Kr < Ne (b) Ne < He < Xe < Kr < Ar (c) He < Ne < Ar < Kr < Xe (d) Xe < Kr < Ar < Ne < He 2. The product obtained by complete hydrolysis of XeF6 is (a) Xe (b) XeO (c) XeO2 (d) XeO3 3. XeF6 reacts with SiO2 to give (a) XeOF2 + SiF4 (b) XeF4 + SiF4 (c) XeOF4 + SiF4 (d) none of these 4. The least polarisable noble gas is (a) He (b) Ne (c) Xe (d) Kr 5. The hybridisation and geometry of the XeO64– ion is (a) sp3, tetrahedral (b) sp3d2, octahedral (c) sp3, pyramidal (d) sp3d3, pentagonal bipyramidal 6. The liquid possessing the property of a super fluid is (a) liq. O2 (b) liq. N2 (c) liq. He (d) liq. NH3 7. The product obtained by treatment of XeO3 with metal hydroxides is (a) XeO2F2 (b) XeO4 (c) HXeOF4– (d) none of these 3 8. The compound with sp d hybridisation is (a) XeF2 (b) XeF4 (c) XeF6 (d) XeO4 9. The maximum number of lone pairs on the central atom are present in (a) XeF2 (b) XeF4 (c) XeF6 (d) XeO3 10. The most reactive noble gas is (a) Xe (b) Ne (c) Kr (d) Rn

17.15

chapter

Chemistry of d-block Elements

18

After studying this chapter, the student will learn about d-block elements d-block elements d-block elements d-block elements

18.1

INTRODUCTION

The element of groups IB–VIII B (groups 3–12) are known as d-block elements. The last electron in the atoms of these elements enters in d-subshell of the penultimate, i.e. (n–1)th, shell. Since the properties of these elements are intermediate to that of s- and p-block elements, these elements are also called transition elements. Generally these elements have partly filled (n – 1)d subshell in their elementary form or in their common oxidation states. The electronic configuration of the d-block elements is given in Table 18.1. Their valence shell electronic configuration can be represented as (n – d)d1–10 ns0–2. Table 18.1 Electronic configuration of d-blcok elements Group III B

IVB

VB

3d series

5d series

6d series

39Y

57La

89Ac

[Ar]18 3d1 4s2

[Kr]36 4d1 5s2

[Xe]54 4f 0 5d1 6s2

[Rn]86 5f 0 6d1 7s2

22Ti

40Zr

72Hf

104Rf

[Ar]18 3d2 4s2

[Kr]36 4d2 5s2

[Xe]54 4f14 5d2 6s2

[Rn]86 5f14 6d2 7s2

23V

41Nb

73Ta

105Db

3

2

5

1

[Ar]18 3d 4s VIB

4d series

21Sc

24Cr

4

1

5

1

[Kr]36 4d 5s 42Mo

[Ar]18 3d 4s

[Kr]36 4d 5s

14

3

2

[Xe]54 4f 5d 6s

[Rn]86 5f14 6d3 75s2

74W 14

4

2

106Sg

[Xe]54 4f 5d 6s

[Rn]86 5f14 6d4 75s2

18.2

Inorganic Chemistry

Group VIIB

3d series 25Mn

43Tc 5

2

6

2

[Ar]18 3d 4s VIII

VIII

1

76Os

[Xe]54 4f14 5d6 6s2

27Co

45Rh

77Ir

[Kr]36 4d8 5s1

[Xe]54 4f14 5d7 6s2

28Ni

46Pd

78Pt 10

0

10

1

[Xe]54 4f14 5d10 6s1

30Zn

48Cd

80Hg

[Ar]18 3d10 4s2

[Kr]36 4d10 5s2

[Xe]54 4f14 5d10 6s2

[Kr]36 4d 5s

29Cu

47Ag 10

[Ar]18 3d 4s

18.2

8

[Kr]36 4d 5s

2

1

6d series

[Xe]54 4f14 5d5 6s2

[Ar]18 3d7 4s2 [Ar]18 3d 4s

IIB

2

44Ru

8

IB

5d series 75Re

5

[Kr]36 4d 4s

26Fe

[Ar]18 3d 4s VIII

4d series

[Xe]54 4f14 5d8 6s1 79Au

[Kr]36 4d 5s

CLASSIFICATION OF d-BLOCK ELEMENTS

Depending on the filling of 3d, 4d, 5d and 6d orbitals, these elements are classified into four series as follows: 1. 3d-series (First Series) This series contains ten elements from 21Sc to 30Zn, present in the fourth period. In these elements, 3d orbitals are progressively filled up, as shown in Table 18.1. It can be seen that Cr and Cu have anomalous configuration, the reason already been discussed in chapter 1. 2. 4d-series (Second Series) The series also contains ten elements from 39Y to 48Cd. This series involves the progressive filling of 4d orbitals (Table 18.1). The anomalous configuration of Mo (4d5 5s1) and Ag(4d10 5s1) can be justified on the bases of extra stability of exactly half-filled and completely filled d-subshell. However, the anomalous configurations of 41Nb, 44Ru, 45Rh and 46Pd are explained on the basis of electron-electron interaction and nuclear electron interactions existing in these atoms. 3. 5d-series (Third Series) This series also contains 10 elements starting from 57La and 72Hf to 80Hg, present in the sixth period. The elements of this series involve the progressive filling of 5d-orbitals. The 14 elements from 58Ce to 71Lu, involve the gradual filling of 4f-orbitals and hence all are not included in 5d-series. Two elements, viz. 78Pt and 79Au, show anomalous configuration which is explained by the combined effect of extra stability of exactly half-filled and completely filled subshells along with electron-electron and nuclear electron interactions. 4. 6d-series (Fourth Series) This series is still uncomplete and contains elements present in seventh period with progressive filling of the 6d-subshell. The 14 elements from 90Th to 103Lw, viz. progressive filling of 5f-orbitals are not included in this series. Six elements, viz. Cu, Ag, Au, Zn, Cd and Hg, should be excluded from d-block elements as these elements, both in their atomic and most common oxidation state, do not have partially filled (n – 1) d-orbitals, e.g. Cu 3d10 4s1 Cu+ 3d10 10 1 + Ag 4d 5s Ag 4d10 10 1 + Au 5d 6s Au 5d10 Zn 3d10 4s2 Zn2+ 3d10

Chemistry of d-block Elements

18.3

Cd 4d10 5s2 Cd2+ 4d10 10 2 2+ Hg 5d 6s Hg 5d10 However, in order to maintain rational classification and their ability of complex formation, these elements are studied as d-block elements.

18.3

GENERAL CHARACTERISTICS OF d-BLOCK ELEMENTS

1. Atomic Radii In each transition series, the atomic radii of the elements decrease gradually till middle elements and then become constant and increase gradually for the last elements. This is due to the reason that as the atomic number increases, the gradual increase in nuclear charge shrinks the atomic size till the middle element. After that, due to screening effect of d-electrons, the nuclear charge does not change much, hence the size remains constant. At the end of each series, due to the electron–electron repulsion in d-orbitals, the size increases. The atomic radii of elements of the first three transition series are listed in Table 18.2. Table 18.2 Atomic radii of d-block elements (pm) Group/ Series 1st 2nd 3rd

IIIB Sc 162 Y 180 La 187

IVB Ti 147 Zr 160 Hf 158

VB V 134 Nb 146 Ta 146

VIB Cr 127 Mo 139 W 139

VIIB Mn 126 Tc 136 Re 137

VIII Co 125 Rh 134 Ir 136

Fe 126 Ru 134 Os 135

IB Cu 128 Ag 144 Au 144

Ni 124 Pd 137 Pt 138

IIB Zn 138 Cd 154 Hg 157

In moving down any group, the radii of elements of the second transition series are higher than the corresponding elements of the first transition series. However, the atomic radii of elements of second and third transition series are almost similar due to lanthanide contraction, to be discussed later on.

2. Ionic Radii Ionic radii follow the same trend for the cations of different elements with same oxidation state, as the atomic radii. However, different cations of the same elements (with different oxidation states) show a decrease with increase in oxidation state (Table 18.3). The ionic radii of divalent cations are comparable with the ionic radii of Ca2+ ions. However, they are less basic and less soluble in aqueous media as compared to Ca2+. Table 18.3 Ionic radii (pm) of d-block elements Group/Series

III B

IV B

VB

VI B

VII B

Sc2+

Ti2+

V2+

Cr2+

Mn2+

81 Sc3+ 88.5

1st

91

Ti3+ 76

Ti4+ 74.5

88 V3+ 74 V4+ 72 V5+ 68

84

Cr3+ 62

80 Mn3+ 66

Cr4+ 68

Cr5+ 63

Cr6+ 58

Mn7+ 60

VIII Fe2+ 76 Fe3+ 64

Co2+ 76 3+

Ce 63

Ni2+ 72

IA

II B

Cu+

Zn2+

91 Cu2+ 69

74

18.4

Inorganic Chemistry

Group/Series

III B

IV B

VB

VI B

VII B

Y 3+ 104

Zr4+

Nb3+

Mo3+

Tc4+

36

2nd

La3+

86 Nb4+ 82 Nb5+ 78

83 Mo4+ 79 Mo5+ 75 Mo6+ 73

78.5 Tc5+ 74 Tc+7 70

VIII Ru3+

Rh3+

82

Ru4+ 76

Ru5+

70.5

Ta3+

117.2

Hf 4+ 85

3rd

86 Ta4+ 82 Ta5+ 78

Pd2+

Re4+

Os4+

W5+

Re5+

Os5+

W6+

Re6+

80

76

74

77

72

77

Ag+

Cd+2

100 Pd3+ 90 Pd4+ 75.5

129 Ag+2

Ir3+

Pt2+

Hg+

Ir4+

Pt4+

Hg2+

94

76.5 Ir5+ 71

71.5

II B

80.5 Rh4+ 74 Rh5+ 69

82

W4+

IA

76.5 Pt5+ 71

109

108

Ag3+ 89

133 116

69

Re7+ 67

3. Ionisation Energy The ionisation energies of d-block elements are intermediate to that of s and p-block elements. Thus, transition elements are less electropositive than s-block elements but more electropositive than p-block elements. In a given series, the ionisation energy does not show a regular behaviour. It can be seen from Table 18.4 that with increase in atomic number, the increase in nuclear charge results in increase in ionisation energy. However, the increase is rather slow and gradual as the effect of increase of nuclear charge is opposed by the increase in the shielding effect due to expansion of d-subshell. The exceptionally high ionisation energies of the last elements can be accounted to the extra stability of fully filled d-orbitals. On moving down the group in IIIB and IVB, on moving from first to second transition series, there is more or less decrease in ionisation energies, due to lanthanide contraction. However, in other groups, there is a irregular increase in ionisation energies. Table 18.4 Ionisation energies (kJ mol–1) of d-block elements Groups/ Series Ist 2nd 3rd

IIIB Sc 632 Y 616 La 538

IVB Ti 659 Zr 660 Hf 654

VB V 650 Nb 664 Ta 761

VIB Cr 652 Mo 685 W 770

VIIB Mn 716 Tc 702 Re 760

VIII Fe 762 Ru 711 Os 840

Co 758 Rh 720 Ir 880

IB Ni 736 Pd 805 Pt 870

Cu 745 Ag 731 Au 890

IIB Zn 906 Cd 868 Hg 1007

4. Oxidation State The transition elements show variable oxidation states due to almost similar energies of (n – 1) d and ns orbitals of the atoms of d-block elements. Thus, in addition of ns electrons, (n – 1) d-electrons can also be removed leading to variable oxidation states. Table 18.5 shows the variable oxidation states of these elements. The most common oxidation state of the transition elements is generally +2 except Cr, Cu, Ag, Au and Hg with +1 oxidation state. On moving from left to right in a period, the number of oxidation states increases till midway and then starts decreasing. The maximum oxidation states shown by most of the elements is +6 except Ru and Os which shows the maximum oxidation state as +8.

Chemistry of d-block Elements

18.5

The relative stability of variable oxidation states depend on the extra stability of d0, d5 and d10 electronic configuration. For example, Ti4+ is more stable thanTl3+, while Mn2+ is more stable than Mn4+. The lower oxidation states are found in complexes with -acid ligands by the formation of d –p back bonding. On the other hand, the higher oxidation states are found in complexes with highly electronegative ligands. It should be noticed that the compounds with lower oxidation state of a transition metal are ionic, while the compounds with higher oxidation states of the same metal are covalent. Table 18.5 Oxidation states of transition metals Group/ Series 1st

IIIB

IVB

VB

Sc +2, (+3)

Ti (0), +2, +3, (+4)

V (0), +2, +3, +4, (+5)

2nd

Y (+3)

Zr (0), +3, (+4)

Nb (0, + 1, + 2, + 3, +4), + 5

3rd

La +3

Hf (+3), +4

Ta (+1), (+2), (+3), (4), (5)

VIB

VIIB

VIII

IB

Cr (0), +1, +2, (+3), +4, +5, (+6) Mo (0), (+1), +2, +3, +4, +5, +6

Mn (0), (+2), + 3, (+ 4), +6, +7

Fe (0), (+2), (+3), +4, (+6)

Co (+2), (+3)

Tc (0), +2, (+3), (+4), (5), (+6), (+7)

Rh Pd Ag (0), (+1), (0), +2, +1, (+2), (+3), +4, (+3) +2 +3, +4, (+6)

Cd +2

W (0), (+1), +2, (+3), +4, +5, +6

Re (0), (+1), (+2), +3, +4, +5, (+6), +7

Ru (0), +2, +3, +4, (+5), (+6), (+7), (+8) Os (0), +2, +3, +4, (+5), +6, (+7), +8

Ir (0), (+1), +2, +3, +4, (+6)

Ag +1, +2

Ni Cu (0) (+2), +1, (+2) +3, +4

Pt Pu (0), +2, +1, +3 (+3), +4, (+5), (+6)

IIB Zn (+2)

5. Atomic Volumes and Densities The atomic volumes of transition elements are lower as compared to s and p-block elements. This is due to the reason that in these elements, the electrons are present in (n – 1) d-orbitals with poor shielding effect. As a result, the ns electrons are strongly pulled towards the nucleus, resulting in decrease of atomic volume and hence, increase of densities. In a particular transition series, the atomic volumes of elements follow the same trend as their atomic radii.

6. Metallic Character The transition elements are typically metallic in nature due to their low ionisation energies and presence of only one or two ns electrons, which can be easily lost. These elements are hard, malleable and ductile, except mercury (liquid at room temperature). Further, greater the number of unpaired electrons, stronger the metallic bond and more the hardness. Thus, Cr, Mo and W are very hard metals due to presence of maximum number of unpaired electrons. On the other hand, Zn, Cd and Hg are soft metals due to absence of any unpaired electrons. The transition metals are good conductors of heat and electricity, with exceptionally high thermal and electrical conductivity of Cu, Ag and Au.

7. Melting and Boiling Points The melting and boiling points of transition metals are very high as compared to s- and p- block elements, due to presence of strong metallic bond, except Cd and Hg which have comparatively very low melting

18.6

Inorganic Chemistry

points, 419.5°C, 320.9°C and –38.4°C respectively. These metals have no unpaired electrons available for metallic bond and hence are moderately volatile.

8. Standard Reduction Potential and Reducing Character The standard reduction potentials of the transition metals, except Cu, are negative and hence, these metals can evolve H2 gas on treatment with acid solutions. M(s) + 2H+(aq) $ M2+(aq) + H2(g) (from acid)

Table 18.6

Standard reduction potentials of some transition metals

Elements Sc

Sc + 3e $ Sc

Eored (volts) – 2.10

Ti

Ti2+ + 2e– $ Ti

– 1.60

V

V2+ + 2e– $ V

– 1.18

Cr

Cr3+ + 3e– $ Cr

Electrode Reactions 3+

2+

–

–

– 0.74 – 1.18

Mn

Mn + 2e $ Mn

Fe

Fe2+ + 2e– $ Fe

– 0.44

Co

Co2+ + 2e– $ Co

– 0.28

Ni

Ni2+ + 2e– $ Ni

– 0.25

The standard reduction potential of elements of the + 0.34 Cu Cu2+ + 2e– $ Cu first transition series are given in the table 18.6. – 0.76 Zn Zn2+ + 2e– $ Zn Thus, these metals, except copper, should act as a good reducing agent. But, in actual practice, the reducing capacity of transition metals is very poor due to high ionisation potentials, high heats of vaporization and low heats of hydration. As a result, these metals are not easily converted into their aqueous ions. Especially, Cu has extremely low reducing character. Other metals also react with acids at a very slow rate as they get protected by a thin non-reactive oxide layer. For instance, chromium gets coated with Cr2O3 and hence can be used as a protective metal.

9. Tendency to form Complex Compounds The transition metals have a unique tendency to form complexes with neutral molecules (CO, NO, NH3 H2O, etc.) and ions (CN–, F–, Cl–, etc.), known as ligands. The tendency to form complex compounds is due to three important reasons: (a) The transition metal ions are very small in size and have high effective nuclear charge. Hence, they have a high positive charge density to attract the lone pair of electrons from the ligands. (b) The transition metals and their ions have vacant (n – 1) d-orbitals to accept the lone pair of electrons. (c) The transiton metals show variable oxidation states.

10. Catalytic Properties Many transition metals and their compounds act as good catalysts. The catalytic property can be due to one of the following reasons: (a) Due to presence of vacant d-orbitals and tendency to show variable oxidation states, the transition metals form unstable intermediate compounds which can readily decompose to give products and the original catalyst. (b) Transition metals also provide a large surface area for reactants to get absorbed and react together. Some examples of catalysts are Vanadium is used in contact process, (for the oxidation of SO2 to SO3) as V2O5. The mechanism can be shown as 2V2O5 $ 2V2O4 + O2 SO2 + ½O2 $ SO3 V2O4 + ½O2 $ V2O5 Nickel is used in hydrogenation reactions, as finely divided nickel.

Chemistry of d-block Elements

Table 18.7

Ni

CH2 = CH2 + H2 $ CH3–CH3 410

Ion

11. Colour of Transition Metal Ions

Sc3+

18.7

Colour and outer electronic configuration of transition metal ions Outer electronic configuration 3d0 3d1 3d0 3d2 3d3 3d5 3d4 3d6 3d5 3d7 3d8 3d10 3d9 3d10

Colour Colourless

The transition metal ions and oxy-anions are Purple Ti3+ generally coloured in the solid or in solution from. 4+ Colourless Ti When white light falls on a compound, some portion Green V3+ is absorbed and the remaining portion is transmitted 3+ Violet Cr or reflected back. If the transmitted light is associated Light pink Mn2+ with wavelength in the visible region, the compound appears coloured. For example, hydrated Cu2+ ions Violet Mn3+ transmit blue radiation and hence appear blue. On Light yellow Fe2+ the other hand, hydrated Ti3+ ions, transmit purple Yellow Fe3+ radiation and appear purple. The colour and outer Pink Co3+ electronic configurations of transition metal ions of Green Ni2+ the first transition series are given in Table 18.7. It Colourless Cu+ is quite evident that the metal ions with completely 2+ Blue Cu filled or empty (n – 1) d-orbitals are colourless. The Colourless Zn2+ colour of transition metal ions can be justified on the basis of crystal field theory. In the transition metal ions containing partially filled (n – 1) d-orbitals, the electrons in lower energy d-orbitals get excited to the higher energy d-orbitals and transmit the corresponding radiation when it falls back. However, such transition is not possible in the transition metal ions with completely filled or empty d-orbitals and hence they appear colourless.

12. Formation of Interstitial Compounds Transition metals can accommodate small atoms (H, B, C, N, etc.) in their interstitial sites and form nonstoichiometric compounds known as interstitial compounds. These compounds are similar to their parent metal in chemical behaviour, but possess different physical properties and are generally more rigid and hard. Some important examples are, TiH1.73, TiC, Hg3N2 and NbO1.9-2.09, etc.

13. Formation of Alloys Due to similar size of atoms of many transition metals, the atoms of one transition metal can be substituted by the atoms of the other metal.

14. Magnetic Behaviour An electrons behaves like a tiny magnet associated with a particular value of magnetic moment. The magnetic moment of an electron is a contribution of spin magnetic moment, ( s) due to its spin motion and orbital magnetic moment ( l), due to its orbital motion, i.e. = s+ l In case of transition metal ions, the orbital motion of unpaired electrons are restricted by the ligand fields, thereby the orbital angular momentum is quenched. As a result, the contribution of s is more significant in comparison to the contribution of l. Thus, for a transition metal ion, with n unpaired electrons, the effective magnetic moment can be expressed as eff

=

spin

= n(n + 2) BM

18.8

Inorganic Chemistry

Thus, the effective magnetic moment of a transition metal ion depends on the number of unpaired electrons. Greater the number of unpaired electrons, greater is the value of effective magnetic moment. The transition metal ions with no unpaired electrons ( eff = 0) are diamagnetic, while those with some unpaired electrons ( eff 0) are paramagnetic. It is evident from the table that for most of the transition metal ion, there is a correlation between calculated and experimental values of the magnetic moment. However, the discrepancy observed in some cases is due to the lesser quenching of orbital angular momentum to be discussed in chapter 26.

The electronic configuration of transition metals can be represented as (n – 1) d1-10 ns0-2. The transition elements are classified into four series as 3d series (Sc to Zn), 4d series (Y to Cd), 5d series (La, Hf to Hg) and 6d series. The general characteristic of transition metals are 1. High melting point and high boiling point 2. Very hard and high electrical conductivity 3. Variable oxidation state and complex-forming ability 4. Malleability and ductility 5. Formation of coloured salts and complexes 6. Ability to act as catalyst and formation of interstitial compounds The magnetic behaviour of transition metal ions can be calculated as eff = spin = n(n + 2) BM The transition elements have their properties in between that of s and p elements. However, presence of d-orbitals has a prominent influence on the bonding and properties of transition elements.

EXAMPLE 1

Determine the number of unpaired electron in the ground states of the

following ions:

(a) Sc3+ (b) Cu2+ (c) which of these will be colorless

Fe2+

(d)

Mn2+

(e)

Cr3+

(a)

21Sc – 3+

[Ar]18 3d10 4s2 Sc – [Ar]18 – no unpaired electrons

(d)

[Ar]18 3d5 4s2 Mn – [Ar]18 3d5 – five unpaired electrons

(b)

[Ar]18 3d10 4s1 Cu – [Ar]18 3d9 – one unpaired electrons

(e)

[Ar]18 3d5 4s1 Cr – [Ar]18 3d3 – three unpaired electrons

(c)

[Ar]18 3d6 4s2 Fe – [Ar]18 3d6 – four unpaired electrons

29Cu – 2+

25Mn – 2+ 24Cr – 3+

26Fe – 2+

Only Sc3+ will be colourless due to absence of any unpaired electrons

EXAMPLE 2

Calculate the spin-only magnetic moment of the following ions:

(a)

Cu+

(b) Ni2+

(c)

V3+

(d)

Co3+

Chemistry of d-block Elements

18.9

(a) Cu+ – [Ar]18 3d0 – no unpaired electrons; n = 0 =0 (b) Ni2+ – [Ar]18 3d8 – two unpaired electron; n = 2 = n(n +2) = 2(2 +2) = 2.84 BM 3+

(c) V – [Ar]18 3d2 – two unpaired electrons; n = 2 = 2.84 BM (d) Co3+ – [Ar]18 3d3 – three unpaired electrons; n = 3 = 3(3 +2) = 3.87 BM

QUESTIONS Q.1. Write the electronic configurations for the following. (a) Fe3+ (b) Co+ (c) Mn4+ (d) Ni2+ Q.2. Give reasons for the following: (a) Scandium forms colourless and diamagnetic compounds (b) Transition metals show variable oxidation state (c) Transition metals form complexes (d) Transition metals act as a catalyst Q.3. Account for the statement: Zinc, cadmium and mercury are not strictly transition elements. Q.4 Explain the following: (a) Transition elements form alloys. (b) Transition metals are usually hard solids but mercury exists as liquid at room temperature. (c) VCl5 doesn’t exist but VF5 exists. (d) What are the characteristic properties of transition elements? Q.5. Describe the trend of atomic and ionic radii of first, second and third transition series. Q.6. What are the characteristic properties of transition elements? Q.7. Discuss the variation of ionisation potential across a transition series. Q.8. Explain the following properties of transition elements: (a) Catalytic properties (b) Magnetic properties Q.9 Give reasons for the following (a) Mn2+ shows the maximum magnetic character among the divalent ions of 3d-transition series. (b) Cu+ is diamagnetic but Cu2+ is paramagnetic. (c) Transition metals have high electrode potential but they are not good reducing agents. (d) Alkali metals are more reactive than transition metals Q.10. Why do copper and chromium have exceptional electronic configuration. Q.11. Give a brief account of magnetic behaviour of transition elements. Q.12. How do transition elements differ from s and p-block elements.

18.10 Inorganic Chemistry

MULTIPLE-CHOICE QUESTIONS 1. The greatest number of oxidation states is exhibited by (a) Ti (b) V (c) Ni 2. The diamagnetic ion is (a) Ni2+ (b) Cr3+ (c) Mn2+ 3. Transition metal ions act as (a) Lewis acids (b) Lewis bases (c) amphoteric 4. The element with largest atomic radius is (a) Cu (b) Cr (c) Mn 5. Which of the following ion is highly paramagnetic (a) [Cr(H2O)6]3+ (b) [Cu(H2O)6]3+ (c) [Fe(H2O)6]2+ 6. The transmition metal compounds show (a) diamagnetism (b) paramagnetism (c) ferromagnetism 7. The pair of elements with almost similar radii is (a) Ti, Zr (b) Mo, W (c) Ni, Pd 8. The atom with largest first ionisation energy (a) Mn (b) Ni (c) Fe 9. Transition elements show highest oxidation state in (a) oxides (b) sulphides (c) fluorides 10. The element with maximum number of unpaired electrons is (a) V3+ (b) Cr2+ (c) Fe3+

(d) Mn (d) Cu+ (d) none of these (d) Ti (d) [Zn(H2O)6]2+ (d) all of these (d) Cr, Mo (d) Zn (d) Iodides (d) Co3+

chapter

Chemistry of Elements of 3d Series

19

After studying this chapter, the student will learn about the occurrence, extraction and chemistry of the following 3d elements:

19.1

INTRODUCTION

The general characteristics of transition elements have been discussed in the previous chapter. In this chapter, we will discuss the chemistry of elements of the first transitions series, viz. Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu and Zn. The first series elements differ significantly from the elements of the second and third transition series. These elements have quite similar energies of 3d and 4s obitals and have their outer-shell electronic configuration as 3d1–10 4s2, except Cr (3d5 4s1) and Cu (3d10 4s1). Now, we will discuss individual elements and their chemistries in references to their oxidation states.

19.2

SCANDIUM (Sc)

19.2.1 Occurrence and Extraction Scandium is the first element of the 3d transition series and Group IIIB (Group 3)–Scandium group. It is the thirty-first most abundant element (by weight) in the earth’s crust and is found only in the combined state. It is found in the rarely occurring minerals monazite and thortveitite [Sc2(Si2O7)]. It is also obtained as a byproduct from the extraction of uranium. Scandium cannot be easily extracted from its compounds. Scandium oxide is highly stable, even more than alumina, hence the thermite process cannot be employed for the extraction of scandium from its

19.2

Inorganic Chemistry

oxide. Scandium is highly electropositive and readily reacts with water to liberate hydrogen. Hence, electrolysis of aqueous solutions of its compounds cannot be carried out for the extraction of the metal. Therefore, electrolysis of fused scandium chloride is carried out in presence of a small amount of sodium chloride (to lower the melting point). Scandium gets deposited on the cathode (zinc) and forms Zn–Sc alloy. Zinc is volatilised from the alloy at low pressure to leave behind scandium in a reasonably pure state.

19.2.2 Properties 1. Scandium is a silvery white metal which tarnishes in air due to the formation of scandium oxide, Sc2O3. 2. Scandium crystallises with hcp structure. Its melting point is 1812 K and boiling point is 2457 K. 3. The stable isotope of scandium is 45Sc, while twelve radioactive isotopes are also known. 4. Scandium resembles aluminium in several respects due to similar sum of the three ionisation energies.

19.2.3 Chemistry of Scandium Outer-shell electronic configuration of scandium is 3d14s2. It can lose three electrons to form the Sc3+ ion. Thus, in various compounds of scandium, its oxidation state is +III.

1. Scandium Oxide (Sc2O3) It is obtained by heating of the metal or its salts like scandium nitrate, scandium carbonate, scandium hydroxide, etc. D Æ 2Sc2O3 + 12NO2 + 3O2 4Sc2(NO3)3 ææ D

Sc2(CO3)3 ææ Æ Sc2O3 + 3CO2 It is also obtained by the hydrolysis of scandium or scandium halides. D

Æ Sc2O3 + 3H2 2Sc + 3H2O ææ D

2ScF2 + 3H2O ææ Æ Sc2O3 + 6HF Sc2O3 is a white powder which melts at 3370 K.

2. Scandium Hydroxide Scandium hydroxide appears not to have a definite formula. However, the basic oxide (ScO.OH) is well established. It is less basic than Ca(OH)2 but is amphoteric like Al(OH)3 and readily dissolves in NaOH. ScO.OH + 3NaOH + 3H2O $ Na3[Sc(OH)6]2.2H2O

3. Scandium Halides The anhydrous halides are usually prepared by heating the metal in a current of halogen in non-aqueous media. Scandium chloride has also been obtained by dehydration of the oxide with ammonium chloride. However, scandium iodide, if prepared by this method, decomposes readily. The hydrated salts produce basic salts on thermal dehydration and hence cannot be used to prepare anhydrous salts. These halides are hydroscopic solids and are soluble in water, except ScF3. However, ScF3 dissolves in excess of fluoride ions to form the soluble complex [ScF6]3–. ScF3 + 3NH4F $ 3NH+4 + [ScF6]3–

Chemistry of Elements of 3d Series

19.3

4. Scandium Hydride Scandium reacts with hydrogen to form a somewhat nonstoichiometric hydride. The number of hydrogen atoms per atoms of Sc are nearer to 2; hence, the formula is generally represented as SrH2. However, there is a lot of controversy regarding the nature and type of bonding involved in SrH2. It is considered that this compound contains one Sc3+ ion and two H– ions. The extra electron resides in the conduction band and makes SrH2 a highly conducting compound. Due to the presence of H– ion, it liberates hydrogen on reacting with water.

5. Scandium Carbide (SrC2) It is obtained by heating the oxide with carbon in an electric furnace. C Sr2O3 + 7C æ1000∞ æææ Æ 2SrC2 + 3CO

It reacts with water to yield acetylene. Hence, it is considered an acetylide. Initially, SrC2 was supposed to contain Sc2+ and (C C)2– ions. However, magnetic measurements indicate the presence of Sc3+ and (C C)2– ions and the extra electron in the conduction band. Hence it shows some metallic conduction.

6. Complexes of Scandium Due to small size and high charge, Sc3+ readily forms a number of complexes. The complexes are generally octahedral with a coordination number of six and are formed with strong complexing agents like EDTA, acac, DMSO, oxalic acid and citric acid. Some examples are [Sc(acac)3], [ScF6]3–, [Sc(DMSO)3]3+, [Sc(C2O4)2], etc. Complexes with bidentate ligands are also known with high coordination numbers. For example, in [Sc(NO)5]2–, one NO3– group is unidentate and four NO3– groups are bidentate, leading to the coordination number nine.

19.3

TITANIUM (Ti)

19.3.1 Occurrence and Extraction Titanium is the ninth most abundant element and constituties about 0.6% by mass of the earth’s crust. The main ores are rutile, TiO2 and ilmenite, FeTiO3. Titanium has very high melting point (1667°C) and it readily reacts with air, oxygen, nitrogen and hydrogen at elevated temperatures. Hence, extraction of titanium is not an easy task. Reduction of scandium oxide with carbon, the most common method of extraction of a metal, cannot be employed for Ti, due to the formation of very stable carbide. Further, the oxide itself is highly stable; hence it is converted into titanium tetrachloride by passing a current of chlorine over heated rutile or ilmenite ore. C TiO2 + 2C + 2Cl2 900°  → TiCl4 + 2CO C 2FeTiO3 + 6C + 7Cl2 900°  → 2TiCl4 + 6CO + 2 FeCl3

TiCl4 (b.pt.137°C) is freed from impurities (such as FeCl3) by fractional distillation and is reduced with molten magnesium at about 800°C, in presence of an atmosphere of argon. This process is known as Kroll process. C TiCl4 + 2 Mg 800°  → Ti + 2MgCl2 MgCl2 is removed by leaching with water or dilute HCl. Alternatively, the product is subjected to vacuum distillation which gives metallic titanium as a spongy mass. This spongy mass is fused in an electric arc (in an atmosphere of helium or argon) under high vacuum and cast into ingots.

19.4

Inorganic Chemistry

Extremely pure titanium can be produced in small amount by the van Arkel–de Boer method. In this method, impure Ti is heated in an evacuated vessel with I2 to form TiI4, which decomposes at 1400 K, on a white hot tungsten filament, to give pure metal. − 520 K 1700 K Ti + 2I2 230  → TiI4 tungsten  → Ti + 2I2 filament

Electrolysis of fused TiCl3 in presence of NaCl and KCl also provides pure titanium metal.

19.3.2 Properties Ti is popularly known as ‘the wonder metal’ due to its unique and useful properties. 1. It is a shining white metal with high melting point (1668°C) and is much lighter (density = 4.51 g cm–3) and stronger than steel. 2. It is a good conductor of heat and electricity, even better than group 3 metals and is better corrosion resistant than stainless steel. 3. It is quite unreactive and remains stable in air even at 673 K. This results due to the formation of a thin protective oxide film at the surface, which is impermeable and prevents further attack. 4. At temperature above 873 K, titanium is quite reactive and combines with many nonmetals. Thus, interstitial nitrides (TiN), interstitial carbides (TiC), halides and oxides (TiO2) are formed. These compounds are very hard and have refractory properties. 5. Titanium remains unaffected by acids and alkalis, at room temperature, due to formation of a thin oxide layer at the surface. However, the metal dissolves slowly in hot concentrated HCl, to form Ti3+ and H2. It is oxidised by hot HNO3 to form the hydrated oxide, TiO2.(H2O)n. It dissolves in HF to form hexafluoro complexes. Ti + 6HF $ H2[TiF6] + 2H2.

19.3.3 Uses Because of their unique properties, titanium and its alloys find great use in technical applications. The most important use is in the aircraft industry and marine equipments. TiO2 is used as a filler and pigment. TiCl3 is used as Ziegler–Natta catalyst. TiC is uses as a refractory material.

19.3.4 Chemistry of Titanium Titanium is known to form compounds in –I, 0, +II, +III, +IV oxidation states, as discussed below: Compounds of Ti (–I) and (0) are rare and are known only with bipy as ligands. Examples are [Ti(biby)3]– and [Ti(biby)3] with octahedral geometry and coordination numbers equal to six.

19.3.5 Chemistry of Titanium(II) Compounds of divalent titanium are less stable and do not have aqueous chemistry due to its oxidation by water. The well-known compounds are TiO, TiCl2, TiBr2 and TiI2.

1. Titanium(II) Oxide (TiO) It is prepared by heating Ti and TiO2. It has NaCl type structure and is normally non-stoichiometric. Its composition has been expressed as TiO0.75. It gets readily oxidised to TiO2 and hence acts as a strong reducing agent. For example, it readily reduces water as TiO + H2O $ TiO2 + H2

2. Titanium(II) Chloride (TiCl2) It is best obtained by the reduction of titanium tetrahalide with titanium.

Chemistry of Elements of 3d Series

19.5

TiCl4 + Ti $ 2TiCl2 It can also be prepared by the disproportionation of its trichloride. 2TiCl3 $ TiCl2 + TiCl4 Titanium tetrachloride being volatile is readily removed by distillation. TiCl2 is unstable and undergoes disproportionation as 2TiCl2 $ TiCl4 + Ti

19.3.6 Chemistry of Titanium(III) Chemistry of Ti+3 species is more extensive than Ti+2 species and their compounds are more ionic than Ti+4 compounds. Ti+3 compounds are coloured and paramagnetic with d1 configuration. The most important compounds formed by Ti+3 are TiCl3 and Ti2O3.

1. Titanium(III) Chloride (TiCl3) Anhydrous TiCl3 exists in several crystalline forms. It can be obtained in the violet a-form by the reduction of TiCl4 vapour with H2 in a red-hot tube at about 920 K. TiCl4 + H2 $ TiCl3 + HCl

la di

TiCl3

[Ti(H2O)6]Cl3 Violet D

co cid

.a

nc

Hydrated TiCl3 exists in two forms with different colours which disproportionate on heating to give TiCl4 and TiCl2.

cid

While the brown b-form is obtained by the reduction of TiCl4 with aluminium alkyls in presence of inert solvents.

TiCl4 + TiCl2

[Ti(H2O)5Cl]Cl2 Green

2. Titanium(III) Oxide (Ti2O3) The anhydrous form is obtained by reducing titanium(IV) oxide wih hydrogen at about 1300 K. 2TiO2 + H2 $ Ti2O3 + H2O It can also be prepared by heating titanium(IV) oxide with titanium metal at 1873 K. 3TiO2 + Ti $ 2 Ti2O3 It has corundum (Al2O3) structure and is reactive only with oxidising acids. It is strongly reducing due to its tendency to change to TiO2. Ti2O3 + [O] $ 2TiO2 Ti(III) compounds are extensively used for the estimation of organic nitro compounds, in presence of methylene blue as an indicator. Methylene blue gets reduced and decolorised even in slight excess of Ti3+. RNO2 + 6Ti3+ + 4H2O $ RNH2 + 6TiO2+ + 6H+ Ti(III) compounds are also used for the volumetric estimation of Fe3+, in presence of NH4SCN as an indicator. The endpoint is detected by the presence of red colouration during the presence of Fe3+. TiCl3 + FeCl3 + H2O $ TiOCl2 + FeCl2 + 2 HCl

3. Complexes of Ti(III) Ti(III) forms a wide variety of cationic as well as anionic complexes. Some examples are [TiCl5H2O]2–, [TiF6]3–, [TiBr4(dipyryidyl)2], etc.

19.6

Inorganic Chemistry

19.3.7 Chemistry of Ti(IV) Titanium forms a number of stable Ti(IV) compounds with d0 configuration. These compounds are colourless and diamagnetic due to absence of any unpaired electrons.

1. Titanium(IV) Oxide (TiO2) It is the most important oxide due to its extensive use as a white pigment. It is used for whitening paper and as an opacifier in paints. Pure TiO2 is obtained either by hydrolysis of TiOSO4 or by vapour phase oxidation of TiCl4 with oxygen. TiO2 is insoluble in water, but dissolves in concentrated alkalis, to form titanates and yields titanyl compounds in acidic solutions. TiO2 + 2NaOH $ Na2TiO3 + H2O TiO2 + H2SO4 $ TiOSO4 + H2O Titanates can be represented as TiO4M2 (orthotitanates) and TiO3M (metatitanates) where M is a divalent element and are termed as mixed oxides. Sodium titanate, Na2TiO3 is usually obtained by fusing TiO2 with NaOH, Na2CO3 or Na2O. When Na2TiO3 is reduced with H2 at high temperature, nonstoichiometric materials of formula Na0.2–0.25TiO2 are produced. These have a metallic bluish-black lustre, are highly conducting, and are known as titanium bronzes. Ca Calcium titanate, CaTiO3, is a naturally occurring mineral O commonly known as perovskite. It lends its name to ABX3 Ti compounds where A and B are the cations from two different groups (differing much in size) and X is the anion, bonded to both cations. In perovskite structure, A atoms occupy cube corners, B atoms occupy body centre and X atoms occupy face centres. Thus, in CaTiO3, Ca and O form ccp arrangement, while Ti occupies ¼th of the octahedral voids bounded completely by O atoms Fig. (19.1). Fig 19.1 Structure of CaTiO3 Iron titanate, FeTiO3, occurs naturally as ilmenite and has similar structure as that of corundum, with one Ti4+ and one Fe2+ in place of 2Al3+ ions. Thus, O2– ions are present in hcp arrangement, Ti4+ ions occupy 1/3rd of the octahedral sites and another one third of the octahedral sites are occupied by Fe2+ ions. Magnesium titanate, Mg2TiO4, has a spinel structure with ccp arrangement of O2– ions while half the octahedral sites are occupied by Mg2+ ions and the 1/8th of the tetrahedral sites are occupied by Ti4+ ions. Barium titanate, Ba2TiO4, is an important compound because of its ferroelectric behaviour. Due to the large size of Ba2+ ions, Ti4+ can ‘rattle’ in its octahedral site of the perovskite structure and hence is drawn to one side, in presence of an electric field. As a result, the crystal gets highly polarised.

2. Titanium(IV) Chloride (TiCl4) TiCl4 is the most important titanium compound which can be prepared by heating titanium dioxide and carbon in presence of chlorine. TiO2 + Cl2 + C $ TiCl4 + CO2 TiCl4 is used to prepare other titanium compounds, e.g. TiCl4 + 4HX $ TiX4 + 4HCl It is a colourless, diamagnetic, fuming liquid which gets readily hydrolysed by water and fumes in moist air. TiCl4 + 2H2O $ TiO2 + 4HCl

Chemistry of Elements of 3d Series

19.7

However, hydrolysis with aqueous HCl gives titanium oxochloride, TiOCl2. TiCl4 + H2O HCl  → TiOCl2 + 2HCl

3. Aqueous Chemistry of Titanium(IV) Compounds In aqueous solutions, the simple aquated Ti4+ ion does not exist, rather the titanyl ion, TiO2+, is considered to exist as (TiO)n2n+ chain (Fig. 19.2). It forms a number of crystalline salts such as TiOSO4.H2O. However, in dilute perchloric acid solutions, [Ti(OH)2(H2O)4]2+ is known to exist. Species like Ti(OH)3HSO4 and Ti(OH)2HSO4+ have been characterised in presence of sulphuric acid. Aqueous Ti(IV) solutions develop an intense orange colour on treatment with hydrogen peroxide due to the formation of peroxo complexes with most probable composition as [Ti(O2)(OH)]+(aq).

4. Titanium(IV) Complexes Titanium tetrahalides form hexa coordinate complex ions in presence of aqueous acidic solutions. TiF4 + 2HF(conc.) $ H2TiF6 Stable

Ti

Ti O Ti

O

O Ti

Fig 19.2 Structure of aqueous titanyl ion

TiCl4 + 2HCl (conc.) $ H2TiCl6

Very unstable

The tetrahalides form a large number of octahedral complexes with the wide variety of electron donors such as R3P, R3As, R2O, NH3, etc., of type TiX4L or TiX4L2. Some are unusual five-coordinated complexes such as [TiCl4.AsH3] and Et4N[TiCl5]. One eight-coordinated complex, Ti(NO3)4, is known with regular triangulated dodecahedron shape with two oxygen atoms from each nitrate group bonded to Ti (Fig. 19.3).

19.4

Fig. 19.3

Structure of anhydrous Ti(NO3)4

VANADIUM (V2)

Vanadium was discovered by Del Rio in 1801, who named it erythronium. The name vanadium was given by N G Sefstrom in 1830 and its properties were discovered by Berzelius in 1831.

19.4.1 Occurrence and Extraction Vanadium is widely spread on the earth, but with few concentrated deposits and forms about 0.02% by mass of the earth’s crust. Out of the 90 minerals containing vanadium, the most important are patronite [V2S5.3CuS2], vanadinite [Pb5(VO4)3Cl or 3Pb(VO4)2.PbCl2] and carnolite [K(VO)2VO4.3/2 H2O]. The last mineral is more important as a uranium ore. Vanadium is also isolated from crude oil found in Venezuela and Canada. Vanadium is not extracted in pure state from its ore, due to its high reactivity towards carbon, oxygen and nitrogen at the high temperature acquired in the conventional thermometallurgical processes. Vanadium is however obtained from reduction of its compounds—vanadates, tetrachlorides and ferrovanadium (an alloy of vanadium and iron) as discussed below:

1. Preparation of Vanadate The vanadate ores are used to prepare vanadate which on heating yields vanadium pentaoxide as shown in the flow chart.

19.8

Inorganic Chemistry

(a) From Vanadinite (powerded) Pb5(VO4)3Cl (Vanadinite) (i) Heating with conc. HCl (ii) Cooling PbCl2

[V(OH)4]Cl (in solution)

(ppt.)

NH4Cl

V2O5

∆ NH4VO3 NH3 (orange ppt.)

HCl

NH3

2NH4Cl

(b) From Patronite V2S5.3CuS2 (powdered) (i) Roasting with air (ii) Fused with Na2CO3 and NaNO3 V2O5

∆ NH3

NH4VO3

(i) Extraction with water (ii) Treatment with NH4Cl

NaVO3

2. Preparation of Ferro–vanadium Alloy This alloy is obtained by heating the mixture of crude vanadate and iron ore in presence of coke in an electric furnace. V2O5 + 5C $ 2V + 5COFe2O3 + 3C $ 2Fe + 3COThe alloy can also be obtained by using Goldschmidt’s alumino-thermite process. 3V2O5 + 10Al $ 6V + 5Al2O3 Fe2O3 + 2Al $ 2Fe + Al2O3

3. Preparation of Pure Vanadium Metal Pure vanadium metal can be obtained satisfactorily by the following processes: (a) Jantesh Process In this process ferrovanadium alloy is heated in a current of chlorine to obtain a dark red liquid (VCl4), which is distilled and reduced with hydrogen to get vanadium. V + 2Cl2 $ VCl4 2Fe + 3Cl2 $ 2FeCl3

Distillation Vapours of VCl 4

H2 600°C

2V + 8HCl

(b) Beard and Crook’s Process In this process vanadium pentoxide is reduced with calcium in a steel bomb. V2O5 + 5Ca $ 2V + 5CaO (c) Reduction of Tetra or Trichloride H

V + 6HCl ←2 VCl3 Na → V + NaCl Mg

VCl4  → V + 2 MgCl2

19.4.2 Properties Vanadium is a greyish white infusible metal which appears silvery white in its compact state. It has the highest melting point (1900°C) in the first transition series due to maximum participation of d electrons in the metallic bond formation. The melting point further increases on the addition of carbon. It is a light metal, corrosion resistant and a good conductor of electricity.

Chemistry of Elements of 3d Series

19.9

Vanadium is unreactive at ordinary temperature, but reacts appreciably at high temperature. It forms vanadium pentoxide on heating in air. 4V + 5O2 $ 2V2O5 It combines with carbon, on heating, to form carbides, VC, VC2 and VnC3. It combines with halogens to form covalent volatile compounds like VF5, VCl4, VBr3 and VI3. It reacts with nitrogen at high temperature to form the nitride, VN. It fuses in hot alkalis to form the corresponding vanadate and liberates hydrogen. 2V + 6NaOH + 2H2O $ 2Na3VO4 + 5H2 It does not react with cold mineral acids. It, however, dissolves in ammonium persulphate and perchloric acid to form metavanadic acid, HVO3. It dissolves in hot mineral acids and aqua regia to give HVO3. V + 5HNO3 $ 2H2O + 5NO2 + HVO3.

19.4.3 Uses of Vanadium The most important application of vanadium is in the form of ferrovanadium alloy which is used in the steel industry. Addition of small amounts of this alloy (0.1–0.2%) improves the quality of steel. It makes the steel tougher, stronger and shock resistant by acting as a scavenger for atmospheric gases. It also improves the tensile strength and ductility of the steel. This type of steel is used for making high-speed tools and parts for motor vehicles. Vanadium pentoxide is used as a catalyst in the manufacture of sulphuric acid (contact process) and in the oxidation of naphthalene. Table 19.1 Oxidation states of vanadium

19.4.4 Compounds of Vanadium

Oxidation state

Vanadium exhibits a number of oxidation states in its compounds as listed in Table 19.1. The compounds with lower oxidation states are ionic, coloured and good reducing agents. The maximum oxidation state exhibited by vanadium is (V) with completely empty ‘d’ subshell and hence these compounds are generally colourless.

1. Compounds of Vanadium(II) This is the least stable oxidation state of vanadium and not much compounds are known.

Example

–I

[V(CO6)]–

0

[V(CO)6]

+I

[V(bipy)3]+

+II

[V(H2O)6]2+

+III

[VCl4]–

+IV

VO(acac)2

+V

V2O5

(a) Vanadium(II) Oxide (VO) It is obtained by the reduction of higher oxides with suitable reducing agent. V2O5 + 6K $ 2VO + 3K2O V2O5 + 3H2 $ 2VO + 3H2O It is a black lustrous oxide with rock-salt type structure but exhibits marked tendency for nostoichiometry leading to lattice defects. Hence, it is conducting in nature. The oxide dissolves in alkalis to yield a precipitate of hypovanadous hydroxide and forms soluble hypovanadous salts containing the ion [V(H2O)6]2+ in acidic medium. (b) Vanadium(II) Chloride (VCl2) It is prepared by heating vanadium trichloride in an inert atmosphere at 1070 K. 1070 K  → VCl2 + VCl4 2VCl3  N 2

19.10 Inorganic Chemistry

VCl4 + H2 $ VCl2 + 2HCl V3O(OH)(OCOCH3)8

2VCl2 + 2H2O $ 2VOCl + 2HCl + H2 Thus, it acts as a strong reducing agent.

CH

3C

OO H

e3

NM

VCl3 MeCN

It dissolves in water to give a violet-coloured solution, containing the ion [V(H2O)6]2+, which soon turns green due to oxidation by water.

acac

V(acac)3 VCl2(NH2).4NH3 NH

3

Et (Li OH OE t)

VCl3(NMe3)2 V(OEt)3 VCl3 Me(CN)3

(c) Vanadium(II) Sulphate (VSO4.7H2O) It is Fig. 19.4 Complex formation by VCl3 crystallised from the solutions obtained by the electrolytic reduction of V2O5 in H2SO4 in the presence of SO3. It dissolves in water to give a violetcoloured aqueous solution containing the ion [V(H2O)6]2 +. The red-violet monoclinic crystals are isomorphous with FeSO4.7H2O and CrSO4.7H2O. It forms double salts of type M2SO4.VSO4.6H2O (M = alkali metal or NH4+) which are isomorphous with Mohr’s salt, FeSO4.(NH4)2 SO4.6H2O. (d) Aqueous Chemistry and Complexes of Vanadium(II) As discussed above, all vanadium(II) compounds dissolve in water to form violet air-sensitive solutions containing the ion [V(H2O)6]2 +, which is strongly reducing and turn green in air due to formation of [V(H2O)6]3+ ion indicating the V2+/V3+ conversion with the evolution of hydrogen. [V(H2O)6]2+ + H+ $ [V(H2O)6]3+ + ½ H2. Some other complexes of vanadium(II) are K4[V(CN)6], [V(en)3]Cl2, [V(acac)3], etc. These are however not much stable.

2. Compounds of Vanadium(III) (a) Vanadium(III) Oxide (V2O3) It is obtained by the reduction of vanadium pentoxide, V2O5 with carbon monoxide or hydrogen. V2O5 + 2H2 $ V2O3 + 2H2O. It has a strong affinity for oxygen and is found as a nonstochiometric composition varying from V2O2.75 to V2O3. Thus, it acts as a strong reducing agent. It is less basic than VO and dissolves in acids to form green-coloured solutions containing the ion [V(H2O)6]3+. Addition of alkalis to these solutions yield green precipitates of V(OH)3. (b) Vanadium(III) Chloride (VCl3) It is obtained by either by heating or by reduction of vanadium tetrachloride. 2VCl4 + H2 $ 2VCl3 + 2HCl VCl4 $ 2VCl3 + Cl2 It is a violet-coloured hygroscopic substance which crystallises as VCl3.6H2O in acidic solutions. It is nonvolatile and yields acidic solutions on hydrolysis with water. It forms a number of complexes with coordinating solvents as shown in Fig. 19.4. It is considered to exist in polymeric form, with each V atom octahedrally surrounded by six Cl atoms. (c) Vanadium(III) Fluoride (VF3) It is crystallised from the green solution obtained by addition of V2O3 to hydrofluoric acid. It can also be obtained by heating of VCl3 in dry HF. It exists as a greenish yellow crystalline powder of the composition VF3.3H2O. On heating, it forms a white solid, vanadyl trifluoride. D Æ 2VOF3 2VF3 + O2 ææ

Chemistry of Elements of 3d Series

19.11

(d) Vanadium(III) Iodide (VI3) It is obtained by reduction of vanadium tetraiodide with hydrogen. It is thermally less stable and disproportionates as follows 2VI3 $ VI4 + VI2 (e) Vanadium(III) Bromide (VBr3) It is obtained by heating vanadium in presence of Br2. It is a hygroscopic black solid and forms hexahydrate in acidic solution. (f) Vanadium(III) Sulphate [V2(SO4)3] It is obtained by the reduction of acidic solution of V2O5 with zinc, having the compostion V2(SO4)3.H2SO4.12H2O. On heating at 180°C, it loses water to form a yellow powder of the composition V2(SO4)3 with the liberation of SO3. (g) Aqueous Chemistry of V(III) and Complexes All V(III) compounds dissolve in aqueous solution to form [V(H2O)6]3+, a blue aquo ion, which readily undergoes aerial oxidation and acts as powerful reducing agent. Vanadium (III) forms a large number of adducts like [VF6]3–, [V(CN)6]3–, [V(C2O4)3]3– and [VX3(py)3], etc.

3. Chemistry of Vanadium(IV) This is the most stable and important oxidation state of vanadium under ordinary conditions. Thus aqueous solutions of vanadium(V) are readily reduced to vanadium(IV) by mild reducing agents and vanadium(III) are oxidised by air to vanadium(IV). The important compounds are the following. (a) Oxovanadium (IV) Compounds These are the most important vanadium(IV) compounds containing the VO unit and occur as a number of mostly bluecoloured salts. Some particular examples are [VO(H2O)5]2+, [VO(acac)]2 (Fig. 19.5), [VOCl2(NMe3)]2 (Fig. 19.6), etc. In all these compounds there is strong VO p bonding generally represented as V = O and these compounds are considered as complexes of VO2+ ion (with magnetic moment 1.73 BM).

CH3

C

C HC

O C

O

V

O

O

CH C CH3

CH3

Fig. 19.5 Structure of [VO(acac)2]

Me

Me

N Me Cl V

O

Cl

(b) Vanadium (IV) Oxide (VO2) It is obtained by reduction of vanadium pentoxide with mild reducing agents such as H2S, SO2, Fe2+ or oxalic acid. V2O5 + SO2 $ 2VO2 + SO3

CH3 O

N Me

Me Me

Fig. 19.6 Structure of [VOCl2(NMe3)2]

V2O5 + H2C2O4 $ 2VO2 + 2CO2 + H2O It is a dark blue oxide with rutile structure. It is amphoteric, but more basic than acidic. Thus, it forms hypovanadites with alkalis and with acids, blue solution containing VO2+ ion in the form of vanadyl salts are formed. VO2 + 2HCl $ VOCl2 + H2O VO2 + 2NaOH $ Na2VO3 + H2O

19.12 Inorganic Chemistry

(c) Vanadium(IV) Chloride (VCl4) It is obtained by passing Cl2 over heated vanadium metal or by passing a mixture of Cl2 and VOCl3 over heated carbon. D

V + 2Cl2 ææ Æ VCl4

VCl6 VCl3(py)3 and VCl4(py)2

It is a highly unstable reddish brown oily liquid which decomposes even at room temperature. 2VCl4 $ 2VCl3 + Cl2 It is readily hydrolyzed by water and fumes in moist air. VCl4 + H2O $ VOCl2 + 2HCl

Cl–

py

Me

OH

NH

VCl4

VCl2(OMe)2MeOH

NM

(diars)2

2VOCl3 + Cl2 + 2C $ 2VCl4 + 2CO

2–

3

VCl (NH3) 2

e3

VCl3(NMe3)2

VCl4(diars)2

Fig. 19.7 Reactions of VCl4

It behaves like a lewis acid and forms a large number of compounds as shown in Fig. 19.7. (d) Vanadium (IV) Oxosulphate (VOSO4) It is also known as vanadyl sulphate. The solution of V2O5 in conc. H2SO4 is reduced with SO2 to obtain a blue solution, which on evaporation yields bright blue crystals of VOSO4.5H2O. On further heating, grey-green coloured crystals of anhydrous vanadyl sulphate, VOSO4, are obtained. Vanadyl sulphate forms dark blue double salts with alkali sulphates, having the composition M2SO4.VOSO4.xH2O (where M is an alkali metal).

4. Chemistry of Vanadium(V) Vanadium forms a number of V(V) compounds which are moderately strong oxidising agents and convert themselves to V(IV). (a) Vanadium(V) Oxide (V2O5) It is the most important oxide of vanadium and is obtained on heating of ammonium metavanadate as an orange powder or by treatment of ammonium vanadate solutions with dilute sulphuric acid as brick-red precipitates. 2 NH4VO3 $ V2O5 + 2NH3 + H2O It can also be obtained by the oxidation of the metal or the lower vanadium oxides. It also results from the hydrolysis of the oxohalide VOCl3. 3H 2 O 2VOCl3  → V2O5 −6 HCl

It is springly soluble in water and forms pale yellow acidic solution. It forms a series of vanadates with alkalis. On addition of V2O5 to NaOH, colourless solutions are obtained. V2O5 + 6NaOH $ 2Na3VO4 + 3H2O The addition of acid results into polymerisation of VO43– as follows. pH = 10 VO43– pH = 12 HVO42– Colourless Colourless

VO2 + Pale yellow

pH < 1

V10O28

6– pH = 2.2

pH = 9 HV2O3– V3O9 3– 7 Orange Colourless pH = 7

V2O5(H2O)n Brown precipitate

pH = 6.5

V5O14 3– Red

These species are known as isopolyvanadates. Vanadate ions also form complexes with the anions of other acids. Due to condensation of more than one type of acid unit, these are known as heteropolyacids.

Chemistry of Elements of 3d Series

19.13

V2O5 dissolves in acid solutions to form the pale yellow vanadyl ion, VO2+. V2O5 + 6HCl $ 2VOCl2 + 3H2O + Cl2 Thus, production of V(IV) and evolution of chlorine indicate that V2O5 can act as a strong oxidising agent. It can also be reduced by warm sulphuric acid as follows:

VO + 2H2O V2O3 + VO2

4C

H 2

V2O5 + 2H2SO4 $ 2VOSO4 + 2H2O Some reactions shown by V2O5 are given in Fig. 19.8. V2O5 exists as distorted trigonal bipyramids of VO5 units which share edges to form zig-zig double chains. V2O5 is used as a catalyst probably due to its reversible dissociation.

NH

V2O5

SO

NH4O – OVO2

O H2

2

H

4O

Cl2

2

Cl2 C

VOCl3

SO3 + VO2 VOCl3 + CO

Fig. 19.8 Reactions of V2O5

���� � 2V2O5 � ��� � 2V2O4 + O2 Thus, it is used in the contact process and oxidation of alcohols. (b) Vanadium(V) Fluoride (VF5) It is obtained by heating, either the elements at high temperature or vanadium tetrafluoride in a current of nitrogen. K 2V + 5F2 573  → 2VF5 K 2VF4 870  → VF3 + VF5

The mixture of products is sublimed to collect VF5 as a white sublimate, leaving behind non volatile residue of VF3. It readily hydrolyses and liberates fluorine on heating. Hence, it is used as a powerful fluorinating agent. VF2 + H2O $ VOF3 + 2HF

F

F

F

V F

F F F

F V F

F F

Fig. 19.9 Structure of VF5

2VOF3 + 6H2O $ V2O5.3H2O + 6HF O

2 2VF5  → 2VOF3 + 2F2

VF5 exists as a monomer in gaseous phase with trigonal bipyramidal structure. However, in liquid phase, it exists as linked octahedral units forming a linear chain which polymerise in the solid phase (Fig. 19.9). VF5 forms complexes with other metal fluorides. VF5 + 2KF $ K2[VF7] (c) Vanadium(V) Complexes—the Dioxovanadium(V) Ion As discussed earlier, VO2+ is formed in strong acid solutions, which forms complexes with a number of ions. Some particular examples of octahedral complexes formed as [VO2Cl4]3–, [VO2EDTA]3– and [VO2ox2]3–, etc. Solutions of vanadium(V) give a red colour on addition of H2O2 due to the formation of peroxo complexes known as peroxovanadates such as [KV(O2)3bipy].4H2O. Some donor adducts are also formed such as VOCl3(NEt3)2, VOCl3(MeCN)2, VOF4–, etc.

19.5

CHROMIUM (Cr)

Chromium was discovered by L N Vauquelin, a French chemist, in 1797 and named it so because of the colour of its compounds (Greek word, chroma, colour).

19.14 Inorganic Chemistry

19.5.1 Occurrence and Extraction Chromium is as common as chlorine and forms about 0.037 percent of the earth’s crust. Its commercially important ore is chromite (chrome iron ore), FeO.Cr2O3. It also occurs as chrome ochre, Cr2O3 and crocoisite, PbCrO4. Chromium can be produced in two forms depending upon its use, i.e. ferrochrome and pure Cr. Ferrochrome, an alloy of Fe, Cr and C is produced by reduction of chromite with C. electric furnace

→ Fe + 2C + 4CO FeCr2O4 + 4C  Pure Cr is obtained by reducing chromite oxide which is prepared from the chromite ore in several steps as discussed here:

1. Formation of Sodium Chromate The finely powdered ore is concentrated by the gravity separation process and is roasted in a reverberatory furnace in presence of excess of sodium carbonate so as to obtain sodium chromate. 4FeO.Cr2O3 + 8Na2CO3 + 7O2 $ 8Na2CrO4 + 2Fe2O3 + 8CO2 Chromite can be also be fused with NaOH in air. 1100°C

→ 8Na2CrO4 + 2Fe2O3 + 8H2O 4FeO.Cr2O3 + 16NaOH + 7O2 

2. Formation of Sodium Dichromate Na2CrO4, being soluble, is removed by dissolving it in water and is treated with sulphuric acid to give sodium dichromate. Na2CrO4 + H2SO4 $ Na2Cr2O7 + Na2SO4 + H2O

3. Formation of Chromium Oxide The solution is concentrated to crystallise Na2SO4 followed by Na2Cr2O7, which is reduced to yield Cr2O3 by heating with C. Na2Cr2O7 + 2C $ Cr2O3 + Na2CO3 + CO

4. Reduction of the Oxide Finally, Cr2O3 is reduced with Al powder (alumino-thermite process) to obtain the metal, which is refined electrolytically. Cr2O3 + 2Al $ 2Cr + Al2O3 The metal is brittle and hence is used to make nonferrous alloys.

19.5.2 Properties Chromium is a brilliant bluish-white metal which melts at 2173 K. It is very hard, malleable and brittle. It is passive at low temperature due to formation of a surface oxide coating. Hence, it does not tarnish in air and is extensively used in corrosion prevention of iron and other metals. However, on heating in the oxy-hydrogen flame, it burns with brilliance and forms green chromic oxide, Cr2O3. It forms chromium nitride on reacting with nitrogen, at elevated temperature. Similarly, it does not react with water under ordinary conditions, but decomposes steam at red heat. 2Cr + 3H2O $ Cr2O3 + 3H2

Chemistry of Elements of 3d Series

19.15

The redox potential relationship for chromium with dilute acids can be correlated as

Cr

–0.91 V

Cr2+

–0.41 V

Cr3+

–0.74 V It means that Cr2+/Cr is more feasible than Cr3+/Cr and hence the metal dissolves in dilute acids to form the chromium (II) salts, and H2 gas is evolved. However, these salts are readily oxidised to chromium (III) salts as indicated by the redox potential. Cr + 2HCl(dil.) $ CrCl2 + H2 Cr + H2SO4(dil.) $ CrSO4 + H2 Treatment with hot concentration nitric acid results into formation of chromium (III) sulphate and sulphur dioxide is evolved. However, the metal is not attacked by cold aqua regia or nitric acid and is rendered passive by hot nitric acid as a result of formation of an insoluble surface coating. Chromium combines directly with halogens like F2 or dry Cl2 to form chromium (III) fluoride or chloride. It can readily displace nickel, copper and tin from aqueous solutions of their salts.

19.5.3 Uses Due to the corrosion-resistant nature of chromium, it is extensively used for chrome plating. Chromium is used in the manufacture of alloy steels such as stainless steel, chrome steel, etc. It is used in the manufacture of many industrially important alloys. For example nichrome, an alloy of chromium and nickel, is used in making resistance coils of electric furnaces due to its high melting point and oxidation resistance. Stellite, an alloy of chromium and tungsten, is used in making high-speed tools and surgical instruments.

19.5.4 Oxidation States of Chromium The ground-state outer electronic configuration of chromium is represented as 3d54s1 and hence, it can exhibit oxidation states varying from (I) to (VI). In addition, some compounds such as carbonyl complexes, dipyridyl complexes, with lower oxidation state of chromium, are also formed. The higher oxidation states are strongly oxidising, while the lower oxidation states are strongly reducing. The most important and stable oxidation state is Cr(III), known literally in thousands and hexacoordinate compounds, with a few exceptions. The oxidation states of chromium have been summarised in Table 19.2.

19.5.5 Chemistry of Chromium(II)

Table 19.2 Oxidation states of chromium Oxidation state – II –1 0 I II III IV V VI

Example Na2[Cr(CO)5] Na2[Cr2(CO)10] Cr(CO)6 [Cr(bipy)3]+ CrF2 [Cr(NH3)6]3+ K2CrF6 CrF5 CrO2Cl2

Chromium(II) compounds are not much stable and are readily oxidised. Cr3+ + e– $ Cr2+

E° = – 0.41 V

CrCl2 is the most important dihalide of chromium and dissolves in water to give blue solutions. It is rapidly oxidised to CrCl3.

1. Aqueous Chemistry of Chromium(II) Sky-blue coloured aqueous solutions of chromium(II) ions are obtained either by electrolytic dissolution of the metal in dilute mineral acids or by reduction of Cr(III) solutions with zinc amalgam. From these

19.16 Inorganic Chemistry

solutions, various hydrated salts can be obtained. The examples are Cr(ClO4)2.6H2O, CrCl2.4H2O and CrSO4.7H2O. These solutions are strongly reducing in nature.

2. Complexes of Chromium(II) A few chromium(II) complexes are also known, such as CrCl2.nNH3, CrCl2.2CH3CN, KCrF3 and K2CrCl4. Some dinuclear species are also reported. For example, chromium(II) acetate, Cr2(CH3COO)4.2H2O, is obtained as a stable compound on addition of chromium (II) solution to a solution of sodium acetate (Fig. 19.10).

CH3 C

CH3

O

O

C

O

H2O

O

Cr 236 p

m Cr

O O C

O

OH2 O

CH3 C

19.5.6 Chemistry of Chromium(III)

CH3

Chromium forms a large number of ionic compounds in this oxidation Fig. 19.10 Structure of chromium (II) acetate state. It is the most stable oxidation state of chromium, hence the compounds in other oxidation states readily convert to (III) oxidation state. The hydrated compounds are generally green in colour, while the anhydrous forms are generally violet.

1. Chromium(III) Halides Chromium(III) halides are generally obtained by halogenation of the metal. CrCl3 is the most important chromium(III) chloride, which can also be obtained by passing a current of dry chlorine through the mixture of chromic oxide and carbon. Cr2O3 + 3C + 3Cl2 $ 2CrCl3 + 3CO It sublimes at 600°C in presence of a stream of chlorine and decomposes to yield CrCl2 and Cl2 at this temperature, in absence of chlorine.

���� � 2CrCl3 � ��� � 2CrCl2 + Cl2 It is known to form four crystalline hexahydrates: Violet hydrated chloride

[Cr(H2O)6]Cl3

Pale green form

[Cr(H2O)5Cl]Cl2.H2O

Dark green form

[Cr(H2O)4Cl2]Cl.2H2O

Covalent hydride

[Cr(H2O)3Cl3]

2. Chromium(III) Oxide (Cr2O3) It is the most important oxide of chromium with corundum structure and is usually obtained by igniting the metal in oxygen, roasting of the hydrous oxide or by the thermal decomposition of ammonium dichromate. 2Cr(OH)3 $ Cr2O3 + 3H2O It is also obtained by heating potassium dichromate with starch, ammonium chloride or sulphur. It is amphoteric, dissolves in acid to give aquo ions and forms chromites in concentrated alkalis. However, it becomes inert towards acids as well as bases, on heating too strongly. In presence of an oxidising agent, it gets fused with an alkali to form chromate. 2Cr2O3 + 8NaOH + 3O2 $ 4Na2CrO4 + 4H2O

Chemistry of Elements of 3d Series

3. Chromium(III) Hydroxide [Cr(OH)3]

OH2

OH2 H O

H2O

It is obtained as hydrous oxide on treating a solution of chromium(II) salt with alkali. The hydrous oxide exists as a polynuclear polymer with a layered structure (Fig. 19.11). It readily dissolves in acids to form chromium (III) salts. Cr(OH)3 + 3HCl $ CrCl3 + 3H2O While in presence of an alkali and an oxidising agent, soluble chromates are formed. 2Cr(OH)3 + 3Na2O2 $ 3NaCrO4 + 2NaOH + 2H2O

19.17

Cr

Cr HO

O H H O

OH

OH2 Cr

Cr HO

Fig. 19.11

2Cr(OH)3 + 10 NaOH + 3Br2 $ 2Na2CrO4 + 6NaBr + 8H2O

OH2

O H

OH

OH2

Structure of hydrous Cr(OH)3

4. Aqueous Chemistry and Complexes of Chromium(III) Chromium(III) forms a large number of complexes with cationic, anionic as well as neutral ligands. Some particular examples are [Cr(H2O)6]3+, [Cr(NH3)6]3+, [CrX6]3– (X = F–, Cl–, CN–, NCS–). Complexes with polydentate anions are also known such as, [Cr(ox)3]3–. Neutral complexes like Cr(acac)3 and Cr(OCOCF3)3 are also formed. Chromium (III) also forms basic acetate complex with an unusual structure. Its composition is [Cr3O(CH3COO)6L3]+, where L is a molecule like H2O, py, etc. Three chromium atoms are arranged in a triangular plane and are linked to an O atom which is present in the centre of the plane. The acetate groups act as bridging ligands. Each Cr atom is surrounded octahedrally by the central O, O atoms form four acetate groups and the ligand L (Fig. 19.12).

H3C

O C H 3C

C O

H 2O O Cr

C

O

O

Cr H 2O

Cr O

OH2

O C CH3 C CH3

O

Fig. 19.12

O

O

O

CH3

CH3 C

O

O

Structure of basic chromium acetate

19.5.7 Chemistry of Chromium(IV) This state of chromium is not much stable and the compounds are very rare. Chromium(IV) fluoride is obtained by fluorination of the metal at 350°C. The chloride and bromide are obtained by heating the corresponding trihalides in excess of the halogen and appear to exist only in the vapour form.

Chromium(IV) Oxide (CrO2)

It is obtained by the reduction of CrO3 with H2. It is a black solid with undistorted rutile structure and is ferromagnetic.

19.5.8 Chemistry of Chromium(V) This state of chromium is highly unstable and a few compounds are known. For example, CrF5 is obtained by fluorination of the metal at 350–500°C. Chromium does not form normal oxide, but forms peroxo compound, the deep blue chromium peroxide, CrO5. It is obtained by shaking the solution of acidified dichromate and hydrogen peroxide with ether. This is the only evidence of Cr(V) in solution. Another peroxo species, M3CrO8, a red-brown compound is obtained by treating an alkaline solution of chromate with hydrogen peroxide. It is formulated as [Cr(O2)4]3– (Fig. 19.13).

3–

O

O

O

O Cr O

O

O

O

Fig. 19.13 Structure of [Cr(O2)4]3–

19.18 Inorganic Chemistry

19.5.9 Chemistry of Chromium(VI) Only a few compounds of chromium(VI) are known. These compounds are very strong oxidising agents.

1. Chromium(VI) Oxide (CrO3) It is obtained by treatment of a saturated solution of potassium dichromate with concentrated H2SO4. K2Cr2O7 + 2H2SO4 $ 2KHSO4 + 2CrO3 + H2O Potassium bisulphate gets crystallised and addition of more concentrated H2SO4 to the solution yields bright orange needle-like crystals of CrO3. When heated above 250°C, it decomposes in stages and eventually forms green-coloured Cr2O3. 2CrO3 $ 2CrO2 + O2 2CrO2 $ Cr2O3 + ½ O2 Thus, it acts as a powerful oxidising agent. It can oxidise alcohols, oxalic acid, glycerine, sugar, etc. It is used as a powerful oxidising agent in organic chemistry. 2CrO3 + 3H2S + 3H2SO4 $ Cr2(SO4)3 + 3S + 6H2O 2CrO3 + 6 FeSO4 + 6H2SO4 $ Cr2(SO4)3 + 2 Fe2(SO4)3 + 6H2O CrO3 + 2HCl $ CrO2Cl2 + H2O It is a strong acid and dissolves in alkalis to form chromium. CrO3 + 2NaOH $ Na2CrO4 + H2O It reacts with F2 in different conditions to give different compounds. K 2CrO3 + 2F2 423  → 2CrO2F2 + O2 K CrO3 + 2F2 493  → CrOF2 + O2 673 K CrO3 + 3F2  → CrF6 200 atm

It dissolves in water to give acidic solution. The solution contains a number of chromic acids. CrO3 + H2O $ H2CrO4

Monochromic acid

2H2CrO4 $ H2Cr2O7 + H2O Dichromic acid

3H2CrO4 $ H2Cr3O10 + 2H2O Trichromic acid

Hence, CrO3 is considered a chromic anhydride.

2. Chromates and Dichromates These are the salts of H2CrO4 and H2CrO7 respectively. Sodium chromate, Na2CrO4, is obtained as a yellow solid by fusion of chromite with Na2CO3 or NaOH and oxidising with air. 4FeCr2O4 + 8Na2CO3 + 7O2 $ 8Na2CrO4 + 2Fe2O3 + 8CO2 4FeCr2O4 + 16NaOH + 7O2 $ 8Na2CrO4 + 2Fe2O3 + 8H2O It is a strong oxidising agent and is quite soluble in water. Sodium dichromate, NaCr2O7, is obtained as an orange-coloured solid by acidifying the chromate solution. 2Na2CrO4 + H2SO4 $ Na2Cr2O7 + Na2SO4 + H2O It is less soluble in water and is hygroscopic. It is widely used as an oxidising agent but is not preferred for volumetric analysis. Instead, K2Cr2O7 is used as a primary standard as it is not hygroscopic. It is obtained

Chemistry of Elements of 3d Series

19.19

as an orange-red solid by treating the hot saturated solution of sodium dichromate with potassium chloride. Na2Cr2O7 + 2KCl $ 2NaCl + K2Cr2O7 It is soluble in water and melts at 671 K. It decomposes on strong heating to liberate oxygen. 4K2Cr2O7 $ 4K2CrO4 + 2Cr2O3 + 3O2 The chromates and dichromates show pH-dependent equilibria. – 2– ���� � Cr2O42– + OH– � ��� � HCrO4 + CrO4 2– ���� � HCrO4– + OH– � ��� � CrO4 + H2O

���� � CrO42– + H+ � ��� � HCrO4– ���� � HCrO–4 + H+ � ��� � H2CrO4 ���� � 2H2CrO–4 � ��� � Cr2O72– + H2O Thus, addition of an alkali changes the orange colour of K2Cr2O7 solution to yellow. K2Cr2O7 + 2KOH $ 2K2CrO4 + H2O Orange

Yellow

While, addition of acid to the yellow solution causes the change from yellow to orange. 2K2CrO4 + 2H2SO4 $ K2Cr2O7 + K2SO4 + H2O. However, on treatment with conc. acids, the products are K2Cr2O7 + 4NaCl + 6H2SO4 $ 2KHSO4 + 4NaHSO4 + 2CrO2Cl2 + 3H2O K2Cr2O7 + 14HCl $ 2KCl + 2CrCl3 + 7H2O + 3Cl2 Acid solutions of dichromates are powerful oxidants: Some particular examples are: (a) Fe2+ to Fe3+

Cr2O42– + 14H+ + 6e– $ 2Cr3+ + 7H2O

Cr2O72– + 6Fe2+ + 14H+ $ 2Cr3+ + 6Fe3+ + 7H2O

(b) I to I2

Cr2O72– + 6I– + 14H+ $ 2Cr3+ + 3I2 + 7H2O

(c) SO32– to SO42–

Cr2O72– + 3SO32– + 8H+ $ 2Cr3+ + 3SO42– + 4H2O

(d) H2S to S

Cr2O72– + 3H2S + 8H+ $ 2Cr3+ + 3S + 7H2O

(e) SO2 to H2SO4

Cr2O72– + 3SO2 + 2H+ $ 2Cr3+ + 3SO42– + H2O.

–

The structure of the chromate and dichromte ions are given in Fig. 19.14.

3. Chromyl Chloride (CrO2Cl2) It is obtained by heating potassium dichromate and potassium chloride in the presence of conc. sulphuric acid. K2Cr2O7 + 4NaCl + 6H2SO4 $ 2KHSO4 + 4NaHSO4 + 2CrO2Cl2 + 3H2O.

CrO2Cl2 + 2H2O $ H2CrO4 + 2HCl This reaction is shown by only electrovalent chlorides.

O 1.9A

O

The dark red vapours of CrO2Cl2 on condensation give a dark red liquid boiling at 389.7 K. These vapours dissolve in water as

O

Cr 115°

° Cr

Cr O

O O

O

1.6 A° O

O O

O

Fig. 19.14 Structure of chromate and dichromate ions

19.20 Inorganic Chemistry

19.6

MANGANESE (Mn)

Manganese was obtained by Gahn in 1774, in its metallic form, and the pure form was obtained in 1870 by John. It was named from the Latin word magnese, meaning magnet.

19.6.1 Occurrence and Extraction Manganese is the 12th most abundant element in the earth’s crust forming about 0.085% by weight. It does not occur free in nature. The principal ore of manganese is pyrolusite, MnO2. Some other ores of manganese are manganite (Mn2O3.H2O), braunite (Mn2O3) and hausmanite (Mn3O4). It can be extracted from its oxide ores by reduction with charcoal (carbon-reduction process) or aluminium (aluminothermite process). Mn3O4 + 4C $ 3Mn + 4CO MnO2 + 2C $ Mn + 2CO 3Mn3O4 + 8Al $ 9Mn + 4Al2O3 3MnO2 + 4Al $ 3Mn + 2Al2O3 However, these reactions are violent and hence nowadays it is obtained by electrolysis of aqueous MnCl2 or MnSO4 solution. Manganese is obtained as amalgam which is distilled to get pure manganese.

19.6.2 Properties It is a greyish metal which resembles iron roughly in its physical and chemical properties. It differs mainly being more brittle and harder, but less refractory than iron. Its melting point is 1245 K and boiling point is 2170 K. The standard reduction potential of manganese supports its electropositive nature. E° = – 1.19 V Mn2+ + 2e– $ Mn Hence, it readily dissolves in dilute, non-oxidising acids to evolve hydrogen. Mn + H2SO4 $ MnSO4 + H2 Mn + 2HNO3 $ Mn(NO3)2 + H2 It is less reactive at room temperature, but reacts vigorously at elevated temperatures. Thus, it decomposes boiling water or steam to liberate hydrogen. Mn + 2H2O $ Mn(OH)2 + H2 The massive metal, on heating strongly, reacts with many nonmetals. It reacts with F2 to form a mixture of MnF2 and MnF3. It combines with O2, N2 and Cl2 to form Mn3O4, Mn3N2 and MnCl2 respectively. It also combines with sulphur and carbon to form MnS and Mn2C respectively.

19.6.3 Uses Manganese is mainly used to produce alloys which are used in the steel industry. The most important alloy, ferromanganese, contains 80% Mn. Silicomanganese contains 65% Mn and spiegeleisen contains 5–25% Mn. These alloys are used in the Bessemer process during manufacture of steel. Mn acts as a scavenger by removing both oxygen and sulphur. This prevents bubble formation and increases the hardness of steel. The steel containing about 13% Mn, called Hadfield steel, is very hard, wearing and shock resistant. It is used for making rock crushing machinery and excavators. Mn is also used in smaller amount for making nonferrous alloys. Manganese bronze (Cu = 55–65%, Zn = 30–45%, Fe = 0.5–1%, Mn = 0.5–3%, Al = 0.2–4%) is water resistant and is used in the shipping industry. Manganin (Cu = 82–84%, Mn = 12–15%, Ni = 2–4%, Fe = 0.1%) is used for making resistance wires due to its very low temperature coefficient of resistance.

Chemistry of Elements of 3d Series

Table 19.3 Oxidation states of mangenese

19.6.4 Oxidation States of Manganese The highest oxidation state of Mn corresponds to the outer electronic configuration, 3d54s2, i.e. (VII). Mn exhibits a large number of oxidation states as summarised in Table 19.3. The lower oxidation states are found in compounds with carbonyl.

19.21

Oxidation state

Example

– III

[Mn(NO)3CO]

– II

[Mn (phthalocyanine)]2

–I

[Mn(CO)5]–

0

Mn2(CO)10, [K6Mn(CN)6]

19.6.5 Chemistry of Mn(II)

+I

[Mn(CO)5]Cl, K5[Mn(CN)6]

This is the most stable and most important oxidation state of the element. Manganese forms an extensive series of compounds in this oxidation state.

+ II

MnCl2, MnS

+ III

[MnCl5], K3[Mn(CN)6]

+ IV

MnO2, Mn(SO4)2

1. Manganous Salts

+V

K3MnO4

Manganese forms a number of Mn(II) salts known as manganous salts. These salts are pinkish in colour and are generally obtained by dry reactions.

+ VI

K2MnO4, MnO3

+ VII

KMnO7, MnO3F, Mn2O7

(a) Manganese (II) Chloride (MnCl2) It is obtained by boiling concentrated HCl with the metal, the oxide or the carbonate. It is soluble in water and forms hydrates like MnCl2.H2O with a polymeric chainlike structure and MnCl2.4H2O containing cis MnCl2(H2O)4 units (Fig. 19.15). (b) Manganese (II) Sulphate (MnSO4) It is obtained by boiling MnO2 with concentrated sulphuric acid. It is quite soluble in water and also forms a number of hydrates such as MnSO4.4H2O, MnSO4.5H2O and MnSO4.7H2O.

Cl H2O

OH2 Mn

H2O

OH2 Cl

Fig. 19.15

Structure of MnCl2.4H2O

2. Aqueous Chemistry of Mn(II) The manganous salts dissolve in water to form a pinkish solution containing the complex ion, [Mn(H2O)6]2+ and are quite resistant to oxidation in the neutral or acidic solutions as evident by the potentials: 0.56 V .5 V 1.18 V 2.26 0.95 MnO −4 + → MnO24 − + → MnO2 + → Mn 3+ 1 → Mn 2 + − → Mn

On treatment with dilute nitric acid, they form colourless deliquescent crystals of the nitrates. The nitrates are highly soluble in water and decompose on heating. Mn2+ + 2NO3– $ Mn(NO3)2 Mn(NO3)2 $ MnO2 + 2NO2 Likewise, buff-coloured precipitates are obtained on addition of sodium carbonate to the Mn2+ solutions. It is sparingly soluble in water and decomposes on heating giving off carbon dioxide. With (NH4)2S manganous salts form buff-coloured precipitates of MnS. When the solutions of manganous salts are treated with alkali, pale pink gelatinous precipitates of the hydroxides are formed. Mn2+ + 2OH– $ Mn(OH)2 The hydroxides readily turn brown-black due to oxidation to the dioxide.

19.22 Inorganic Chemistry

3. Complexes of Mn(II) Many complexes of Mn(II) are known, mostly in octahedral grometry with high-spin arrangement giving zero crystal field stabilisation energy. Hence these complexes are not stable except in solution. Such examples are [Mn(NH3)6]2+, [MnCl6]4– and [Mn(H2O)6]2+. Many hydrated salts such as Mn(ClO4)2.6H2O, MnSO4, etc., contain the complex ion [Mn(H2O)6]2+. Anhydrous salts of Mn(II) form ammoniates, by the direct action of ammonia, containing the complex ion [Mn(NH3)6]2+. Some low-spin octahedral complexes like [Mn(CN)6]4–, [Mn(CNR)6]2+ and [Mn(CN)5 NO]3– have also been formed. These are relatively less stable and can be reduced readily. Zn

3– O2  [Mn(CN)6]4– ← [Mn(CN)6]5– ←  [Mn(CN)6] reduction oxidation

Some square planar complexes containing [Mn(H2O)4] has also been reported such as [Mn(phthalocyanine)] and MnSO4.5H2O. Some tetrahedral halogeno complexes are known, which are less stable in aqueous solutions but are comparatively more stable in solvents like acetic acid or ethanol. These complexes are greenish yellow in colour and change to pink-coloured octahedral complexes in solutions.

19.6.6 Chemistry of Mn(III) Manganese forms a number of mixed oxide systems in this oxidation state. Black crystals of haussmannite, Mn3O4 are obtained on heating manganese oxide or hydroxide at high temperatures. It exists in spinel structure, with the composition, MnIIMn2IIIO4. Braunite, Mn2O3, is also believed to contain Mn(III). A small number of Mn3+ salts have also been obtained. Red-purple solid MnF3 has been obtained by fluorination of MnCl2. The black trichloride, has been obtained by the treatment of HCl on Mn(III) acetate at 173 K. it decomposes above – 40°C.

1. Aqueous chemistry of Mn(III) These salts hydrolyse in water and disproportionate in acid. Mn3+ + 2H2O $ MnO.OH + 3H+ 2+ + 2Mn3+ + 2H2O acid  → Mn + MnO2 + 4H

Thus, Mn(III) state is unstable in aqueous solutions, but can be stabilised to some extent by complexing ions. Generally, the hydrated manganic ion [Mn(H2O)6]3+ is obtained by electrolysis or persulphate oxidation of Mn2+ or by reducing MnO4–. However, due to reduction of Mn3+ by water, strong concentrations are not obtained. Mn3+ + e– $ Mn2+

2. Complexes of Mn(III) Most of the Mn(III) complexes are octahedral and high spin, but some low-spin complexes are also formed with strong ligands. The complex [MnIII(C2O4)2]3– is formed during the titration of KMnO4 with oxalate ion. Mn3+ + 3(C2O4)2– $ [Mn(C2O4)3]3–

High-spin complex

This complex is thermally unstable above 60°C. Thus, the titrations are performed above 60°C so as to decompose the complex as it would upset analytical calculations based on the reduction of MnVII to MnII. Low spin complex, [MnIII(CN)6]3–, is obtained on bubbling air through the KCN containing solutions of Mn2+. Mn3+ + 6CN– $ [Mn(CN)6]3– An unusual basic acetate has been obtained on oxidation of Mn2+ with KMnO4 in glacial acetic acid. It is a deep red-brown substance with the stoichiometry [Mn3O(CH3COO)6]+ [CH3COO]–.

Chemistry of Elements of 3d Series

19.23

Its structure consists of a triangle of Mn atoms with an oxygen atom at the centre and the acetate group atoms as bridging between the Mn atoms. The sixth position of the octahedral is occupied by another ligand such as water.

19.6.7 Chemistry of Mn(IV) Not many compounds of Mn(IV) are known. The most important compound is greyish-black oxide, MnO2 which exists as pyrolysite. It has the rutile structure, when made by heating Mn in O2. However, the nonstoichiometric form is obtained on heating Mn(NO3)2.6H2O in air. MnO2 is not much stable and is used as an oxidising agent. It is insoluble in water and inert to most acids but acts as an oxidant on heating. Thus, it liberates Cl2 and O2 on treatment with hot conc. HCl and H2SO4 respectively. MnO2 + 4HCl $ MnCl2 + 2H2O + Cl2 2MnO2 + 2H2SO4 $ 2MnSO4 + 2H2O + O2 It is used in organic chemistry for oxidising compounds. It is extensively used in dry batteries and for the production of potassium permanganate. fusion

 → K2MnO4 + NO MnO2 + KNO2 in NaOH electrolytic

2K2MnO4 + H2O  → 2KMnO4 + 2KOH + H2 oxidation It is used in the glass industry to prepare red or purple glass. It is also used in the brick industry to produce red or brown bricks. It is also used as a catalyst in the preparation of O2 from KClO3. 2 2KClO3 MnO  → 2KCl + 3O2 150°C

Aqueous Chemistry and Complexes of Mn(IV) Aqueous chemistry of Mn(IV) is very limited and only a few complexes are known including K2[MnF6], K2[MnCl6], K2[Mn(IO3)6], K2[Mn(CN)6] and Na2[Mn(C3H5O3)2]. No complexes of Mn are known in further higher oxidation states.

19.6.8 Chemistry of Mn(V) Mn(V) is the little known and highly unstable oxidation state. Only one compound, hypomanganate, K3MnO4, has been obtained by the reduction of an aqueous solution of KMnO4 with an excess of K2SO3. KMnO4 + K2SO3 + H2O $ K3MnO4 + H2SO4 It is also formed by dissolving MnO2 in concentrated KOH. It is a bright blue compound and tends to disproportionate.

19.6.9 Chemistry of Mn(VI) This state is represented only by the deep green manganate ion, MnO42–, and the compounds formed are called manganates. These are obtained by oxidation of MnO2 in fused KOH by air, KNO2, NaBiO3 or any other oxidising agent. Treatment of KMnO4 with alkali also yields manganates. 4MnO–4 + 4OH– $ 4MnO2– 4 + O2 + H2O Only two dark greenish black salts, K2MnO4 and Na2MnO4, have been prepared. These are stable only in highly alkaline solutions. In acidic, slightly alkaline or neutral solutions, these compound disproportionate and hence act as strong oxidants. 3MnO42– + 4H+ $ 2MnO4– + MnO2 + 2H2O

19.24 Inorganic Chemistry

19.6.10 Chemistry of Mn(VIII) This state is also not common and is best known in the form of purple-coloured permanganate ion, MnO4–. Its potassium salt, KMnO4, is widely used as a strong oxidising agent. KMnO4 is prepared on a large scale, from MnO2 by fusion with caustic potash or potassium carbonate in contact with air or any other oxidising agent. 2MnO2 + 4KOH + O2 $ 2K2MnO4 + 2H2O 2MnO2 + 2K2CO3 + O2 $ 2K2MnO4 + 2CO2 The greenish fused mass of potassium manganate is extracted with water and is oxidised with chlorine, ozone or carbon dioxide to yield a purple solution. This solution is concentrated to deposit dark purplish, lustrous, needle-like crystals of potassium permanganate. 3K2MnO4 + 2CO2 $ 2KMnO4 + MnO2 + 2K2CO3 2K2MnO4 + Cl2 $ 2KMnO4 + 2KCl 2K2MnO4 + O3 + H2O $ 2KMnO4 + 2KOH + O2 However, the electrolytic oxidation of the manganate solution is preferred nowadays. The solution is passed through an iron pipe acting as cathode and is poured on rotating wire-melting anode made up of nickel. Dilute alkali solution is calculated around the anode and the manganate ion is oxidised during electrolysis. At anode

MnO42– $ MnO4– + e–

At cathode 2H+ + 2e– $ 2H $ H2 Potassium permanganate is isomorphous with potassium perchlorate, KClO4, and is moderately soluble in water. Aqueous solutions of permanganate are purple in colour intrinsically unstable in acidic solution due to slow decomposition. 4MnO–4 + 4H+ $ 4MnO2 + 3O2 + 2H2O Addition of a small quantity of KMnO4 solution to conc. H2SO4 gives a green solution due to the formation of MnO3+ ions. KMnO4 + 3H2SO4 $ K+ + MnO3+ + 3HSO4– + H3O+ It excess of KMnO4 is added, an explosive oil, Mn2O7, is formed which decomposes into manganese dioxide and oxygen. 2KMnO4 + H2SO4 $ Mn2O7 + K2SO4 + H2O Mn2O7 $ 4MnO2 + 3O2 KMnO4 decomposes on heating alone or in presence of H2 or in alkaline solution. KMnO4 æDæ Æ K2MnO4 + MnO2 + O2 2KMnO4 + 5H2 $ 2KOH + 2MnO + 4H2O 4KMnO4 + 4KOH $ 4K2MnO4 + 2H2O + O2 KMnO4 is a strong oxidising agent and the product depends on the pH.

1. In Acidic Medium The reaction is best carried out in the presence of dilute sulphuric acid. KMnO4 + 3H2SO4 $ K2SO4 + 2MnSO4 + 3H2O + 5O MnO4–

is reduced to Mn2+, represented in the ionic form as MnO–4 + 8H+ + 5e– $ Mn2+ + 4H2O; E° = 1.51 V.

Chemistry of Elements of 3d Series

19.25

This involves the five-electron change and thus the equivalent mass of KMnO4 in acidic medium is onefifth of its molecules mass. Some particular examples of oxidation by KMnO4 in acidic medium are the following. (a) Ferrous salts to ferric salts 2MnO–4 + 16H+ + 10Fe2+ $ 2Mn2+ + 8H2O + 10Fe3+ (b) Sulphites to sulphates 2MnO–4 + 5SO32– + 6H+ $ 5SO42– + 2Mn2+ + 3H2O (c) Hydrogen sulphide to sulphur 2MnO–4 + 5S2– + 16H+ $ 2Mn2+ + 5S + 8H2O (d) Sulphur dioxide to sulphuric acid 2MnO4– + 5SO2 + 2H2O $ 2Mn2+ + 5SO42– + 4H+ (e) Oxalic acid to carbondioxide 2MnO–4 + 5C2O42– + 16H+ $ 2Mn2+ + 10CO2 + 8H2O (f) Haloacids to the corresponding halogens 2MnO–4 + 10X– + 16H+ $ 2Mn2+ + 5X2 + 8H2O (X = Cl, Br, I) (g) Potassium chloride to iodine 2MnO–4 + 10I– + 16H+ $ 2Mn2+ + 5I2 + 8H2O (h) Ethyl alcohol to acetaldehyde acidified

→ CH3CHO + H2O C2H5OH + O  KMnO 4

2. In Alkaline Medium In alkaline medium, the permagnate ion gets reduced to manganate ion which further reduces to manganese (IV) oxide, i.e. 2MnO4– + 2OH– $ 2MnO2– 4 + H2O + O – 2MnO2– 4 + 2H2O $ 2MnO2 + 4OH + 2O

Thus, overall three nascent oxygen atoms are released and the equivalent mass of KMnO4 in alkaline solution is 1/3rd of its molar mass, so the complete reaction can be written as alkali

→ 2MnO2 + 2OH– + 3O 2MnO4– + H2O  The ionic equation can be written as MnO4– + 2H2O + 3e– $ MnO2 + 4OH– Some typical example of oxidation by KMnO4 in alkaline medium are (a) Iodides to iodates.

2MnO–4 + H2O + I– $ 2MnO2 + IO3– + 2OH–

(b) Ethylene to Glycol

C2H4 + H2O + O $ CH2 — CH2

(c) Toluene to Benzoic acid

OH OH C6H5CH3 + 3O $ C6H5COOH + H2O

(d) Nitrotoluene to Nitrobenzoic acid

C6H4

CH3 + 3O $ C6H4 NO2

NO2

+ H2O COOH

19.26 Inorganic Chemistry

3. In Neutral Medium In neutral medium the reaction is same as that of alkaline medium. 2MnO4– + H2O $ 2OH– + 2MnO2 + 3[O] Thus, the equivalent mass of KMnO4 in neutral medium is also 1/3rd of its molar mass: Some typical examples of oxidation by KMnO4 in neutral medium are the following: (a) Hydrogen sulphide to sulphur 2MnO4– + 3H2S $ 2MnO2 + 3S + 2OH– + 2H2O (b) Sodium thiosulphate to sulphate and sulphur 2MnO–4 + 3S2O32– + H2O $ 2MnO2 + 3SO42– + 3S + 2OH–

19.7

IRON (Fe)

Iron was discovered about 4000 years ago in south east Asia and was considered even more valuable than gold. The Iron Age began by around 1200 BC and Indians were specialists in its metallurgy and manufacture of iron articles as proven by the glory of Indian history.

19.7.1 Occurrence It is the fourth most abundant element in the earth’s crust and forms much of the outer and inner core of the earth. Most of the iron is found in the combined state with oxygen, in the form of iron oxide minerals. For example, the reddish brown ore haematite (Fe2O3); the hydrated oxide limonite (2Fe2O3.3H2O) and the magnetic oxide magnetite (Fe3O4). Some other minerals are siderite (FeCO3) and iron pyrite (FeS2)—the yellow metallic, lustrous mineral, commonly known as fool’s gold. Iron is also found in meteorites as iron nickel mineral, kamacite (an alloy of iron with 5–10% of nickel) and taenite (an alloy of iron with 20–65% of nickel.)

Commercial Varieties of Iron Iron is worked mainly for the manufacture of three commercial varieties, viz. cast iron, wrought iron and steel. These varieties of iron differ from each other in their carbon content. Cast iron is the least pure form containing 2.5–4.5% of carbon. Wrought iron is the purest form containing about 0.5% of carbon while steel contains carbon content varying from 0.4–2%.

19.7.2 Extraction Iron is extracted mainly from its oxide ore, haematite. Other ores are also used to some extent. The process is carried out as follows: The ore is crushed and then washed with water to remove impurities. The concentrated ore is roasted in shallow kilns in presence of excess of air. It removes moisture, carbon dioxide and volatile impurities of arsenic and sulphur. It also oxidises ferrous oxide to ferric oxide. The roasted ore is reduced by heating with coke and limestone in a blast furnace. Limestone decomposes to form lime and CO2. The lime combines with silica to form calcium silicate (slag) and floats over the molten iron, thus protecting it from oxidation. The molten iron dissolves some carbon from coke and is known as pig iron or cast iron. Pig iron and slag are collected from the separate openings. The various reactions taking place in the blast furnace are summarised in Fig. 19.16. Pig iron contains about 2.6–4.3% carbon, 1–2% silicon, and other impurities such as phosphorus, sulphur and manganese. The carbon in pig iron is partly in the form of graphite and partly as iron carbide, Fe3C

Chemistry of Elements of 3d Series

19.27

Ore, limestone and coke Waste gases 400°C Zone of reductions 500-600°C

2Fe3O4 + CO2 2FeO + CO2

3Fe2O3 + CO Fe3O4 + CO 2CO

C + CO2

800°C

FeO + CO

Fe + CO2

900°C Zone of heat absorption 1000°C

CaCO3

CaO + CO2

1800°C Zone of fusion

FeO + CO CO2 + C

Fe + CO2 2CO

CaO + SiO2

CaSiO3 Fe + CaS + CO

FeS + CaO + C MnO + C

Mn + CO (in Fe)

Hot air

2000°C

Iron

SiO2 + 2C Slag 2C + O2

Si + 2CO (in Fe)

2CO

Fig. 19.16 Blast furnace

(cementite). The molten pig iron gets softer when slowly cooled and is known as grey pig iron. It contains most of the carbon in the form of graphite. However, when molten pig iron is quickly cooled, it gets hard and brittle and is known as white pig iron. It contains most of the carbon in the form of cementite. Pig iron is hard, but brittle and expands on solidification. It cannot be welded and is used for casting a number of articles like pipes, toys, toolstones, cooking ranges and agricultural implements etc. Pig iron melts in the range of 1150–1250°C depending upon the amount of impurities. It is used in the making of wrought iron and steel.

19.7.3 Manufacturing of Wrought Iron It is the purest form of iron and is manufactured by the puddling process. In this process, pig iron is heated in a reverberatory furnace lined with iron oxide, Fe2O3 (haematite). The hot gases and flames are deflected on the pig iron and the carbon and the other impurities present get oxidised to form their corresponding oxides. These oxides combine to form slag and is removed by puddling (stirring). Wrought iron contains about 0.2% of carbon and traces of other impurities. It is extremely tough and corrosion resistant with a high melting point. It is used in the making of chains, wires, bolts, nails and cores of electromagnets.

19.7.4 Manufacturing of Steel Steel contains about 0.2 to 2% of carbon and manganese. Other elements such as nickel, chromium, silicon, tungsten, vanadium and molybdenum are used to prepare alloy steel. Steel is manufactured from pig iron by using several processes. The processes involve removal of impurities from pig iron and addition of a requisite amount of carbon and other required elements. The most common processes used for the manufacturing of steel are discussed ahead.

19.28 Inorganic Chemistry

1. Bessemer Process In this process, molten pig iron is poured in a Besemer converter. It is a pear-shaped furnace made of steel plates and lined with silica or lime depending upon the nature of the impurities. Molten pig iron is poured in the horizontally positioned converter and a hot-air blast is blown through the holes (tuyerers) present at the base. The converter is tilted into vertical position whilst containing the air blast. As a result, the Si and Mn are oxidised to form manganous silicate slag.

Flame

Converter Air

Molten iron

2Mn + O2 $ 2MnO Si + O2 $ SiO2 MnO + SiO2 $ MnSiO3

Tuyers

(slag)

Fig. 19.17 Bessemer converter These reactions produce a large amount of heat leading to oxidisation of carbon to CO which starts burning with a blue flame. After the oxidation of carbon, the blue flame dies out. Now the requisite amount of C is added in the form of spiegeleisen, an alloy of iron, manganese and carbon. The converter is tipped to pour the molten steel into cast-iron moulds. Usually, a little amount of aluminium or ferrosilicon alloy is added to avoid ‘blow holes’ in the castings (Fig. 19.17).

2. Open-Hearth Process In this process, pig iron mixed with scrap iron, lowgrade wrought iron, haematite and lime is taken in an open-hearth furnance lined with silica or magnesia depending upon the impurities. The furnace is fired with producer gas and works on the regenerating principle of heat economy. The impurities are oxidised by haematite and carbon is removed as CO. The carbon content is checked from time to time and the requisite metals are added to obtain alloy steels. This process is very slow and takes about 10 hours.

Charge

Fire-brick checker

Fig. 19.18 Open-hearth furnace

3. Electric-Arc Furnace Process This process is used to manufacture alloy steel and high-quality steel. Most commonly Heroult’s furnace is used. It is a steel shell lined with dolomite or magnesite and is attached with vertically held water jacketed electrodes. The charge consists of pig iron, scrap iron and haematite mixed with a small amount of lime. The electrodes are struck with an electric arc to produce high temperature which initates the reactions instantaneously. The impurities are slagged off almost completely to yield high-quality steel. The requisite metals are added to obtain the alloy steels (Fig. 19.19). Composition, properties and uses of some alloy steels are given in Table 19.4.

Charge

Basic Lining

Fig. 19.19 Heroult’s furnace

Chemistry of Elements of 3d Series

19.29

Table 19.4 Composition, properties and uses of alloy steels S. No 1.

Alloy steel Stainless steel

Composition 18% Cr and Ni

Properties Corrosion resistant

2.

Nickel steel

3.5% Ni

3.

Chrome steel

2–4% Cr

Corrosion resistant, hard and resistant High tensile strength

4.

Manganese steel

15% Mn

5.

Tungsten steel

6.

Silicon steel

15–20% W, 3–8% Cr 15% Si

7.

Invar

36% Ni

Hard and wear resistant Retains hardness at high temperture Resistant to acids and extremely hard No coefficient of expansion

Uses Utensils, automobile parts, ornamental pieces Cables, gears, armour plates, automobile and aeroplane parts Cutting tools, crushing machinery, ball bearing and cutlery Grinding machinery, railroad tracks, safes High speed tools Pumps and pipes for carrying acids Precision instruments, clocks, antimagnetic watches

Properties of steel can be controlled by heat-treatment processes. When steel is heated to bright red heat and is cooled suddenly by plunging in oil or water, it becomes very hard and brittle. This process is known as hardening. However, if it is cooled slowly, it becomes very soft and pliable. This heat treatment is called annealing. If the hardened steel is heated to a temperature much below redness and is cooled slowly, the steel obtained is neither so brittle nor so hard. This heat treatment is known as tempering. The surface of the steel can be further hardened by nitriding process. In this process, steel is heated in an atmosphere of ammonia resulting into formation of iron nitride coating on the surface. Such steel is commonly used for cylinder bores. The surface of wrought iron and wild steel can be hardened by heating in contact with charcoal or potassium ferrocyanide. This process is called case-hardening. It leaves a thin coating of hardened steel over the surface of the treated wrought iron and mild steel. Case hardened wrought iron is used for making armour plates and the parts of the machinery that can withstand shock. Case-hardened steel is used for making locomotive axes. Depending upon the carbon content and the heat treatment process, steel can be classified into three types according to the difference in its properties. (a) Mild Steel This type of steel contains a small percentage of carbon. It is soft and ductile. (b) Hard Steel As the carbon percentage increases, the hardness of steel increases and it becomes brittle like glass. Hardness can also be increased by heat treatment as discussed earlier. (c) Alloy Steel Properties of steel can be changed by the addition of impurities or alloying with other metals. The steel obtained by this process is called alloy steel. For example, addition of manganese imparts elasticity and tensile strength, while chromium makes the steel corrosion resistant. Various industrially important alloy steels have been listed in Table 19.4.

19.7.5 Properties of Iron Pure iron is a silver – white lustrous metal with a very great tensile strength. It melts at 1536°C and boils at 3000°C. It is strongly ferromagnetic but becomes paramagnetic when heated above its Curie point (771°C).

19.30 Inorganic Chemistry

It is malleable and ductile. Iron exists in three allotropic forms at atmospheric pressure. 910°C a-iron Ferrite (bcc)

g-iron Austenite (fcc)

1394°C

d-iron Delta iron (bcc)

These forms differ from each other in crystalline shape. One more form named b-iron is also produced at 766°C. It is nonmagnetic and has the same crystalline form as that of a-iron. Hence, it is not considered a separate allotrope. Iron is quite reactive and easily gets oxidised with air, water and acids. Massive Fe is almost unreactive with dry air. However, moist air reacts quickly and oxidises iron to form hydrated iron(III) oxide, 2Fe2O3.3H2O and a small quantity of iron(III) hydroxide, Fe(OH)3. This product is commonly known as rust of iron. Electrochemical Theory Rusting of iron is an electrochemical phenomenon and involves the formation of any electrochemical cell at the metal surface. When iron comes in contact with water containing dissolved O2 and CO2, a mixture of ferric hydroxide and ferric oxide, known as rust, is formed at its surface. The water containing dissolved gases acts as the electrolyte. At the anodic areas, ferrous ions are formed while H+ ions are discharged at the cathode in the acidic medium. In neutral aqueous solutions, the electrons are used by the dissolved oxygen to form OH– ions, which diffuse towards the anode and ferrous hydroxide is formed. In excess of air, ferrous ions are oxidised to ferric ions and rust is formed. At anode: At cathode:

Fe $ Fe2+ + 2e– In acidic medium,

In neutral medium

2H+ + 2e– $ H2-

O2 + 4e– + 2H2O $ 4OH– H O

2 Fe2+ + 2OH– $ Fe(OH)2  → Fe2O3 . xH2O

Rust

4Fe(OH)2 + O2 + 2H2O $ 4Fe(OH)3 Rusting can be prevented if the iron surface is not allowed to come in contact with air and moisture. This can be done by providing a metallic or nonmetallic coating on the iron surface. A film of oil, grease, paint, lacquer and enamel can prevent the direct contact with air. This method of protection is known as barrier protection method. If the coating is made of a metal which is more active than iron, the method is known as sacrificial protection as the more active metal (anodic) gets sacrificed and the iron surface is protected. Generally, zinc is used for protecting iron surfaces and the process is known as galvanisation. Sometimes a less active metal is also coated on the iron surface, but it is not so durable and if broken, exposes the iron surface for a rapid corrosion. Generally, tin is used for this purpose and the process in known as tinning. Red hot iron liberates hydrogen on coming in contact with steam. 3Fe + 4H2O $ Fe3O4 + 4H2 It reacts with dilute HCl and H2SO4 to liberate hydrogen but forms nitrates with dilute HNO3. Fe + 2HCl $ FeCl2 + H2 Fe + H2SO4 $ FeSO4 + H2 4Fe + 10HNO3 $ 4Fe(NO3)2 + NH4NO3 + 3H2O

Chemistry of Elements of 3d Series

19.31

It gives sulphur dioxide with hot concentrated H2SO4 and a mixture of ferrous and ferric sulphates are formed. 2Fe + 6H2SO4 $ 2FeSO4 + Fe2(SO4)3 + 3SO2 + 6H2O It liberates nitrogen oxide with moderately conc. nitric acid but becomes passive in highly conc. HNO3 due to formation of a protective layer at its surface. Iron does not react with alkalis but forms salts with many nonmetals. 2Fe + 3Cl2 $ 2FeCl3 Fe + S $ FeS Iron is fairly electropositive and can displace less electropositive metal from their salt solution. Fe $ Fe2+ + 2e–

E = +0.44 V

Fe + CuSO4 $ FeSO4 + Cu

19.7.6 Chemistry of Iron The outer-shell electronic configuration of iron is 3d64s2, which suggests the maximum oxidation state as (+VIII). However, the highest oxidation state of iron is (+VI), a rare oxidation state of little importance. The most common oxidation state are (+II) and (+III). These oxidation states are almost nearly stable. Other oxidation states exhibited by iron are listed in Table 19.5. Iron also shows (0) and (–II) oxidation states with carbonyls, the strong pi-acis acid ligands.

Table 19.5 Oxidation states of Iron Oxidation state (– II)

Example [Fe(CO)4]2–

(0)

[Fe(CO)5]

(+ I)

[Fe(H2O)5NO]2+

(+ II)

[Fe(H2O)6]2+, FeO

(+ III)

[FeCl4]–, Fe2O3

(+ IV)

[FeH4(PR3)3]

(+VI)

[FeO4]2–

1. Chemistry of Iron(II) It is one of the most common oxidation states of iron and is easily oxidised to the (+III) state. The compounds exist mostly as pale green crystals and are commonly known as ferrous salts. Except the double salt, FeSO4.(NH4)2SO4.6H2O, other compounds are mostly easily oxidized and are not obtained in pure form. (a) Iron(II) Oxide (FeO) It is commonly known as ferrous oxide and is obtained as a pyrophoric black powder by the thermal decomposition of iron(II) oxalate in vacuum. FeO is nonstoichiometric and metal deficient with a typical composition as Fe0.95O. It has rock salt type structure. It is insoluble in water, but forms ferrous salts with non-oxidising acids. (b) Iron(II) Halides The anhydrous FeF2 and FeCl2 are obtained by heating the metal with gaseous HF and HCl respectively. The bromide and iodide can be obtained by direct halogenation of the metal. (c) Iron(II) Sulphate (FeSO4.7H2O) It is prepared from Kipp’s apparatus containing ferrous sulphate and dilute sulphuric acid. Ferrous sulphate forms double salts with sulphates of ammonium and alkali metal ions. For example, FeSO4.(NH4)2SO4.6H2O and K2SO4.FeSO4.6H2O. Ferrous sulphate is mainly used in the industry as a precursor to other iron compounds. When a pure ferrous salt solution is treated with a soluble base in anaerobic conditions, white precipitate of ferrous hydroxide are obtained. It is readily oxidized by air to form reddish brown hydrous iron(III) oxide. (d) Aqueous Chemistry of Iron(II) In the absence of any complexing agents, the aqueous solutions of iron(II), invariably, contain the bluish green complex ion, [Fe(H2O)6]2+. It is readily oxidised in presence of air.

19.32 Inorganic Chemistry 3+ ���� � In acidic medium, 2Fe2+ + ½O2 + H+ � ��� � 2Fe + H2O – ���� – � In basic medium, ½Fe2O3 . 3H2O + e ���� � Fe(OH)2. + OH

E° = –0.56 V

Thus, basic solution of ferrous ion oxidises more rapidly and freshly precipitated Fe(OH)2 immediately becomes dark in presence of air.

(e) Complexes of Iron(II) Iron(II) forms a number of complexes, mostly octahedral and high spin. Some tetrahedral halide complexes are also known (FeX4)2–. The bluish green complex ion [Fe(H2O)6]2+ is paramagnetic and is easily oxidised to ferric complex. 3+ – ���� � [Fe(H2O)6]2+ � E° = 0.77 V. ��� � [Fe(H2O)6] + e 2+ The H2O molecule in [Fe(H2O)6] can be easily replaced by strong ligands such as CN– and phenanthroline to form low-spin complexes, [Fe(CN)6]4– and [Fe(phen)]3+. The brown complex [Fe(H2O)5NO]2+ is the basis of the ‘ring test’ for NO–2 and NO3–. It contains iron in the (+I) oxidation state. The presence of NO+ has been proved by IR studies. Similarly, a series of complexes are formed with chelating amine ligands. 2+ ���� � [Fe(H2O)6]2+ + en � K = 104.3 ��� � [Fe(en)(H2O)4] + 2H2O 2+ 2+ ���� � [Fe(en)(H2O)4] + en � K = 103.3 ��� � [Fe(en)2(H2O)2] + 2H2O 2+ ���� � [Fe(en)2(H2O)2]2+ + en � K = 102 ��� � [Fe(en)3] + 2H2O Anhydrous ferrous compounds absorb gaseous ammonia to form ammoniates, stable more in saturated aqueous ammonia than in aqueous media. The most common example is [Fe(NH3)6]2+.

Two best known complexes of iron (II) are potassium ferrocyanide and sodium nitroprusside.

(f) Potassium Ferrocyanide, K4[Fe(CN)6] It is a lemon-yellow coloured crystalline compound and is prepared by the action of cyanide ions on ferrous salts in solution. FeSO4 + 2KCN $ Fe(CN)2 + K2SO4 Fe(CN)2 + 4KCN $ K4[Fe(CN)6] For large scale production, coal gas (containing about 0.2% of HCN) is passed through a washer containing KOH and FeSO4. HCN + KOH $ KCN + H2O 6KCN + FeSO4 $ K4[Fe(CN)6] + K2SO4 The coal gas is sometimes passed through hydrated iron oxide [a mixture of iron(II) and (III) hydroxide] to form Prussian blue, Fe4[(FeCN)6]3, which is boiled with lime and the product is treated with potassium carbonate. The resulting solution is evaporated to give monoclinic crystals of potassium ferrocyanide. 2HCN + Fe(OH)2 $ Fe(CN)2 + 2H2O 3HCN + Fe(OH)3 $ Fe(CN)3 + 3H2O 3Fe(CN)2 + 4Fe(CN)3 $ Fe4[Fe(CN)6]3 Ferric ferrocyanide (Prussian blue)

Fe4[Fe(CN)6]3 + 6Ca(OH)2 $ 3Ca2[Fe(CN)6] + 4Fe(OH)3 Ca2[Fe(CN)6] + 2K2CO3 $ 2CaCO3 + K4[Fe(CN)6] It is also obtained by fusing nitrogenous organic matter (blood clots, leather scraps, horns, hoofs, etc.) with scrap iron and K2CO3. The mass is digested with water to separate out the crystals of potassium ferrocyanide. 8C + 3N2 + Fe + 4KOH $ K4[Fe(CN)6] + 2CO + 2H2O

Chemistry of Elements of 3d Series

19.33

Properties It is a trihydrate and highly soluble in water. On heating up to 140°C, it loses its water of crystallisation and decomposes on further heating. K4[Fe(CN)6] $ 4KCN + FeC2 + N2 On heating with dilute H2SO4, it liberates HCN. 2K4[Fe(CN)6] + 2H2SO4 $ 3K2SO4 + K2FeII[FeII(CN)6] + 6HCN (white ppt.) Potassium ferrous ferrocyanide

While CO is evolved with hot and conc. H2SO4. K4[Fe(CN)6] + 6H2SO4 + 6H2O $ 2K2SO4 + 3(NH4)2SO4 + FeSO4 + 6CO It gives precipitates with metal salts and hence is used for detection of metal ions in solutions. K4[Fe(CN)6] + 2CuSO4 $ Cu2[Fe(CN)6] + 2K2SO4 (reddish brown)

K4[Fe(CN)6] + FeCl2 $ K2FeII[FeII(CN)6] + 2KCl (white)

III

K4[Fe(CN)6] + FeCl3 $ KFe [FeII(CN)6] + 3KCl Prussian blue

Potassium ferrocyanide is used in the laboratory for qualitative analysis. It is used for the preparation of HCN, CO and Prussian blue. It is good reducing agent, for example 2K4[Fe(CN)6] + Cl2 $ 2K3[Fe(CN)6] + 2KCl 2K4[Fe(CN)6] + O3 + H2O $ 2K3[Fe(CN)6] + 2KOH + O2 2K4[Fe(CN)6] + H2O2 $ 2K3[Fe(CN)6] + 2KOH II

2+

Ferroin, [Fe (phen)3] is an intense red-coloured complex and is used as a redox indicator. It usually contains anions such as sulphate or chloride. Its intense colour persists till the entire oxidation process takes place and there is excess of oxidising agent. 3+ – ���� � [Fe(phen)3]2+ � ��� � [Fe(phen)3] + e

E° = 1.12 V

Deep red

Light blue

2+

It is due to this reason that [Fe(phen)3] is highly stable and is not easily oxidised. Sodium nitroprusside (sodium nitrosopentacyanoferrate (II) Na2[Fe(CN)5(NO)]. It is obtained by boiling potassium ferrocyanide with 30% HNO3 and then neutralising the solution with sodium carbonate. The resulting solution is filtered and concentrated to separate out brown-red crystals of sodium nitroprusside. 3K4[Fe(CN)6] + 4HNO3 $ 3K3[Fe(CN)6] + NO + 2H2O + 3KNO3 2K3[Fe(CN)6] + 3Na2CO3 $ 2Na3[Fe(CN)6] + 3K2CO3 Na3[Fe(CN)6] + NO $ Na2[Fe(CN)5(NO)] + NaCN It is highly soluble in water and is used to detect sulphide ions in solutions. H2S reacts with sodium nitroprusside in alkaline solutions to give an intense violet colouration. Na2[Fe(CN)5NO] + 2NaSH $ Na4[Fe(CN)5NOS] + H2S Violet colour

Addition of acids or excess of alkali destroys the colour. H2S gives no colouration in absence of alkali but soluble sulphides produce the colour directly by the reaction [Fe(CN)5NO]2– + S2– $ [Fe(CN)5NOS]4–

19.34 Inorganic Chemistry

Similarly, silver nitrate gives a flesh-coloured precipitate with sodium nitroprusside. Na2[Fe(CN)5NO] + 2AgNO3 $ Ag2[Fe(CN)5NO] + 2NaNO3 H4[Fe(CN)6], the free acid can also be precipitated as white powder in the strongly acidic solution. It is a strong tetrabasic acid in aqueous medium. In the solid form, the hydrogen atoms are hydrogen bonded to the nitrogen atoms of the cyanides.

2. Chemistry of Iron(III) It is the most important oxidation state of iron. Iron(III) forms crystalline salts with almost every common anion except (I–), as FeI3 dissociates in aqueous solution to form FeI2 and I2. The salts are commonly known as ferric salts and are obtained by the oxidation of the corresponding ferrous salts. Most of the salts exist in both hydrated and anhydrous states. (a) Iron(III) Oxide, Ferric Oxide or Fe2O3 It exists naturally in three forms. The most common form, known as haematite, is reddish brown, antiferromagnetic and rhombohedral a-Fe2O3. It is obtained by oxidation of the metal with pressurised oxygen, by dehydration of iron(III) hydroxide or by heating iron(III) salts of volatile acids. Iron(III) oxide exists in several hydrated forms, commonly described as hydrous ferric oxide Fe2O3 . xH2O, and is obtained as reddish brown gelatinous precipitates on addition of alkali to the solution of soluble Fe(III) salts. Hydrous iron(III) oxide is slightly amphoteric and dissolves more readily in acids. It forms white [Fe(OH)6]3– in strongly basic solutions and [Fe(H2O)6]3+ in acidic solution. Stoichiometric combinations with alkali-metal hydroxides yield ferrites of the composition MFeO2. On electrolytic oxidation in basic medium, a red-purple solution containing FeO42– is obtained. At anode:

Fe + 8OH– $ FeO42– + 4H2O + 6e–

At cathode:

2H2O + 2e– $ H2 + 2OH–

(b) Aqueous Chemistry of Iron(III) Due to small size and high charge density of Fe3+ ion, it gets readily hydrolysed in aqueous solutions. Fe3+ ion exists as pale purple [Fe(H2O)6]3+ in aqueous solutions and at pH around zero. As the pH is increased up to 2–3, the solution turns yellow. 2+ + ���� � [Fe(H2O)6]3+ � ��� � [Fe(H2O)5OH] + H ;

K = 10–3.05

A binuclear brownish species is formed at pH 4–5. 2[Fe(H2O)5(OH)]2+ $ [Fe(H2O)4(OH)2Fe(H2O)4]4+ + 2H+

K = 10–2.91

With further increase of pH, a reddish brown gelatinous mass of hydrous ferric oxide is precipitated. Fe3+ is readily reduced by many reducing agents. For example, Fe3+ + H2S $ 2Fe2+ + S + 2H+ 2Fe3+ + Sn2+ $ 2Fe2+ + Sn4+ Fe3+ + H $ Fe2+ + H+ (c) Complexes of Iron(III) Iron(III) forms a large number of complexes which are generally more stable than complexes of iron(III). The complexes are mostly octahedral but some tetrahedral complexes are also known (FeCl4–). Fe3+ has more affinity for ligands which coordinate through O as compared to N. The complexes with NH3 and chelating N ligands are less stable than the corresponding Fe(+II) forms. Thus, it is easier to oxidise [Fe(H2O)6]2+ than to [Fe(phen)3]2+. On the contrary, it is easier to oxidise

Chemistry of Elements of 3d Series

19.35

[Fe(CN)6]4– than [Fe(H2O)6]2+, owing that Fe3+ forms more stable complexes with CN– than Fe2+. This is due to large negative entropy of hydration associated with highly charged [Fe(CN)6]4–. Thus, [Fe(H2O)6]2+ $ [Fe(H2O)6]2+ + e– 4–

3–

–

[Fe(CN)6] $ [Fe(CN)6] + e

E° = – 0.77 V E° = – 0.36 V

–

Complexes with halide ions and SCN are also known. The Fe3+ ion gives blood-red colour with ammonium sulphocyanide solutions due to formation of a mixture of [Fe(SCN)(H2O)5]2+, Fe(SCN)3 and [Fe(SCN)4]–. The colour is discharged with fluoride ion due to the formation of colourless FeF63–. This serves as a sensitive test for estimation of iron(III). Similarly, intense coloured [Fe(dike)3] complexes are formed on addition of b-diketone to the solutions of ferric ion. This is also a useful diagnostic test for Fe3+. (d) Potassium Ferricyanide, K3[Fe(CN)6] One of the most important complex is the hexacyanoferrate ion, [Fe(CN)6]3–. The free acid H3[Fe(CN)6] is also well known. The dark red salt is crystallized from the solution obtained by the oxidation of potassium ferrocyanide with H2O2 or Cl2. 2K4[Fe(CN)6] + H2O2 $ 2K3[Fe(CN)6] + 2KOH 2K4[Fe(CN)6] + Cl2 $ 2K3[Fe(CN)6] + 2KCl It is a good oxidising agent in alkaline medium. For example, 2K3[Fe(CN)6] + 2KOH + SO2 $ 2K4[Fe(CN)6] + H2SO4 2K3[Fe(CN)6] + 2KOH + PbO $ 2K4[Fe(CN)6] + PbO2 + H2O 3K3[Fe(CN)6] + 8KOH + CrCl3 $ 3K4[Fe(CN)6] + K2CrO4 + 3KCl + 4H2O 2K3[Fe(CN)6] + 2KOH + H2S $ 2K4[Fe(CN)6] + S + 2H2O On addition of Fe2+, a blue precipitate called Turnbull’s blue is obtained. [Fe(CN)6]3– + Fe2+ $ [Fe(CN)6]4– + Fe3+ $ {Fe[Fe(CN)6]}– In this case, the ferrous ion is oxidised to ferric ion which reacts with the ferrocyanide ion to form the complex ion containing both Fe(II) and Fe(III). Turnbull’s blue is identical with Prussian blue in composition. However, its colour is less intense due to the presence of a white compound of the composition K2{Fe[Fe(CN)6]}, formed initially in the solution.

3. Chemistry of Iron(IV) Only a few compounds are known, the best known being Sr2FeO4 and Ba2FeO4, obtained by the reaction − 900°C M3[Fe(OH)6]2 + M(OH)2 + ½ O2 800  → 2M2FeO4 + 7H2O

4. Chemistry of Iron(VI) The discrete, tetrahedral reddish purple ferrates, FeO42– have been obtained. These are strongly oxidising and stable only in basic solution and decompose in acid or water to liberate oxygen. Ferrous and ferric salts can be distinguished by the following tests. (a) Addition of Ammonium Hydroxide Ferrous salts give greenish white precepitates of ferrous hydroxide while ferric salts give reddish brown precepitates of ferric hydroxide. (b) Addition of Ammonium Sulphocyanide There is no action with ferrous salts while blood-red colouration is obtained with ferric salts.

19.36 Inorganic Chemistry

(c) Addition of Potassium Ferrocyanide Solution Ferrous salts give bluish white precepitates of ferrousferricyanide which turn blue after some time. Ferric salts produce dark blue precepitates of Prussian blue. (d) Addition of Potassium Ferricyanide Ferrous salts give blue precepitates of turnbull’s blue while reddish brown colouration is obtained with ferric salts.

19.8

COBALT (Co)

Cobalt was discovered by Brandt in 1735 and was isolated in 1742. It derives its name from the German word kobaltd, evil spirit.

19.8.1 Occurrence and Extraction Cobalt is very less abundant in the earth’s crust, about 30 ppm by weight. Cobalt ores are always found in association with nickel ores and to some extent with ores of copper and lead. The most important ores are cobaltite (cobalt glance), CoAsS; smaltite (speiess cobalt), CoAs2 and linnaeite. Extraction of cobalt is very complex due to its association with several other metals. The powdered ore is roasted to remove sulphur as SO2 and arsenic as As4O10. Iron sulphide (often present as impurity) is oxidised to ferrous oxide. The roasted ore is smelted with limestone and coke in a small blast furnace. Here iron is slagged off and the residue, containing a mixture of arsenites of cobalt, nickel, iron and copper, called speisses, and impure silver is separated. The speisses is ground and roasted with sodium chloride in a reverberatory furnace to convert the metals into their chlorides while most of the arsenic and sulphur is driven off. The product is heated with conc. sulphuric acid and is extracted with water. Iron, arsenic and antimony are precipitated by treatment with limestone and removed. The filtrate is treated with sodium carbonate to separate precipitates of copper as basic carbonates. Cobalt and nickel are precipitated as their hydroxides on the addition of lime and bleaching powder. These hydroxides are ignited to form a mixture of cobalt and nickel oxides, which are reduced either by Goldschmidt process or heating with carbon and limestone in an electric furnace. Finally, the metal is refined electrolytically using ammonical cobalt sulphate as an electrolyte.

19.8.2 Properties Cobalt is the bright silvery metal which is hard and has a high tensile strength, even more than steel. It is highly malleable and ductile and lustrous in appearance. It is ferromagnetic and retains this property up to 1000°C. Cobalt is relatively unreactive and is not affected by air, water, H2 or N2. However, it is oxidised to Co3O4 on heating in air and decomposes steam to form CoO. It dissolves slowly in dilute sulphuric acid, more rapidly in dilute nitric acid and is rendered passive by conc. nitric acid. Co + H2SO4 $ CoSO4 + H2 3Co + 8HNO3(dil.) $ 3Co(NO3)2 + 2NO + 4H2O It shows a little action even with conc. caustic soda solution but combines readily with halogens.

19.8.3 Uses Cobalt is used in the manufacture of high-temperature alloys with steel. For example, stellite (50% cobalt, 27% chromium, 12% tungsten, 5% iron and 2.5% carbon) is used for making rock drills and surgical instruments. Cochrome (60% cobalt, 14.16% chromium and 24% iron) is used for making electrical goods. Alnico (aluminium, nickel and cobalt) is used for making powerful permanent magnets. Cobalt is used to

Chemistry of Elements of 3d Series

19.37

make pigments for the glass and paint industry. For example, smalt, a blue- coloured substance, obtained by fusion of CoO with K2CO3 and silica, is used to impart blue colour to glass. Cobalt is also used in electroplating of metals.

19.8.4 Oxidation States

Table 19.6 Oxidation states of cobalt

The outer-shell electronic configuration of cobalt is 3d74s2. It follows the same trend in oxidation states like that of iron. Thus, (+II) and (+III) are the most important oxidation sates of cobalt. (+I) oxidation state has been observed in many complexes and especially in the reduced form of Vitamin B12. (+IV) is the highest oxidation state attained by cobalt in certain binuclear complexes and in [CoF6]2–. A compound K3[CoO4] is believed to contain cobalt in (V) oxidation state but it is not confirmed yet.Table 19.6 lists the oxidation states of cobalt with examples.

Oxidation State (–I) (0) (+I) (+II) (+III) (+IV) (+V)

Example [Co(CO)4]–, [Co(CO)3NO] [Co2(CO)8], K4[Co(CN)4] [Co(bipy)3]+, [Co(CNR)5]+ CoCl2, Co3O4, [CoCl4]2– CoF3, (CoF6)3–, [Co(CN)6]3– [CoF6]2–, [(NH3)5Co – NH2 – Co(NH3)5]5+ (K3CoO4)

19.8.5 Chemistry of Lower Oxidation States Just like iron, the lower oxidation states are stabilised with p-bonding ligands such as CN–, CO, NO and PF3. The (–I) oxidation state has been observed in [Co(CO)4]– and [Co(CO)3NO] with tetrahedral geometry. While complexes such as Co2(CO)8, Co4(CO)12 and K4[Co(CN)4] are the zero-valent compound of cobalt.

19.8.6 Chemistry of Cobalt(+I) The Co(+I) species has been observed with p-acid ligands. The reduced form of Vitamin B12 and cobalt oxines also contain Co(+I).

19.8.7 Chemistry of Cobalt(+II) It is the most important state for simple compounds of cobalt and a large number of compounds are known. Most of these compounds are water soluble, except the carbonates, and are known as cobaltous compounds. (a) Cobalt(II) Oxide (CoO) is obtained as an olive-green powder by heating the metal with oxygen or the cobaltous compounds such as hydroxide, nitrate or carbonate in absence of air. (b) Cobalt(II) Nitrate, Co(NO3)2.6H2O is obtained by evaporation of the solution of CoCO3 in dilute nitric acid. It is used in the charcoal cavity test. Co(NO3)2 decomposes on heating to give cobalt oxide which combines with other metal oxides on strong heating to give coloured compounds. 2Co(NO3)2 $ 2CoO + 4NO2 + O2 CoO + Al2O3 $ Co(AlO2)2 blue

CoO + ZnO $ CoZnO2 green

CoO + MgO $ CoO.MgO pink

19.38 Inorganic Chemistry

CoCl2 . 6H2O is used for moisture detection as cobalt chloride paper and as an indicator along with silica gel. CoCl2 . 6H2O is pink coloured and turns blue on heating. 2+ ���� � [Co(H2O)6]2+ � ��� � [Co(H2O)4] + 2H2O

pink

blue

Thus, addition of water reverses the reaction and the moist cobalt chloride paper is pink in colour. Similarly, CoCl2 . 6H2O added as an indicator in silica gel turns pink when the drying agent becomes ineffective.

Aqueous Chemistry and Complexes of Cobalt(II) Co(+II) is very stable and in the absence of any complexing agents, the oxidation of aqueous cobalt(II) is very unfavourable.

19.8.8 Chemistry of Cobalt(III) Co(+III) is evidenced in only a few simple compounds which are unstable as compared to that of Co(+II). Thus, the compounds of Co(+III) easily get reduced to Co(+II) and are used as strong oxidising agents. There is no evidence of pure Co2O3. However, the mixed black oxide Co3O4 or CoIICo2IIIO4 is obtained either by heating the metal or cobalt (II) oxide in air at 400–500°C. Co2O3 is a normal spiral like Fe3O4.

Aqueous Chemistry of Cobalt(III) and Complexes The hydrated Co(+III) salts are blue coloured and invariably contain the aqua complex [Co(H2O)6]3+ which is strongly oxidising. Co (+III) oxidises water rapidly and evolves oxygen. 4Co3+ + 2H2O $ 4Co2+ + O2 + 4H+ However, Co (+III) is stabilised on complexation and forms more complexes than any other element. The treatment of K3[CoII(CN)5] with KCN in presence of air yields K6[(CNIII)5Co—O—O—Co(CNIII)5], a brown complex which on further oxidation yields K5[(CN)5 Co—O—O—Co(CN)5]. The product on boiling yields the yellow-coloured K3[Co(CN)6]. The complex [Co(CN)6]3– is very stable as the CN– ligands are strongly linked by p-back bonding. The complexes of particular interest are the cobalinitrites of alkali metals. Sodium cobaltnitrite, Na3[Co(NO2)6], is obtained on addition of a cobaltous salt to the solution of sodium nitrite in acetic acid. The reaction involves the oxidation of cobalt(II) by the nitrite ions. The cobalt(III) ion formed reacts with excess of nitrite ions to give the complex. Co2+ $ Co3+ + e– + 2H + NO2– + e– $ NO + H2O Co3+ + 6NO2– $ [Co(NO2)6]3– Sodium cobaltinitrite or sodium hexanitritocobaltate(III) reacts with potassium salt in aqueous solution to give a yellow precipitate of potassium cobaltinitrite. Thus, it is used as a laboratory regent for detection and estimation of potassium salts.

19.9

NICKEL (Ni)

Nickel was discovered in 1751 by Cronstedt. It was isolated from the kupfernickel ore (false copper) and was named nickel.

19.9.1 Occurrence and Extraction Nickel occurs to an extent of 90 ppm in the earth’s crust and is the 28th most abundant element. Elemental nickel is found in many iron meteors. It is found in nature mainly in combination with sulphur, arsenic and

Chemistry of Elements of 3d Series

19.39

antimony namely, millerite (NiS), nickel glance (NiAsS), niccolite (NiAs), garnierite [(NiMg)6Si4O10(OH)8], nickeliferrous limonite [(FeNi)O(OH)(H2O)n] and pentlandite [(Fe,Ni)9S8]. Nickel is extracted mainly from the sulphide ore pentlandite containing about 3% of nickel. The process of extraction involves the following slages:

1. Concentration of the Ore The powdered sulphide ore is usually concentrated by froth floatation process to yield a concentrate containing NiS, CuS and FeS.

2. Production of Matte Containing Nis and CuS The concentrated ore is roasted in air so as to convert FeS into FeO. 2FeS2 + 5O2 $ 2FeO + 4SO2 The roasted ore is mixed with coke and silica and is smelted in a blast furnace to slag off the iron and the matte is left behind. The matte may contain the impurities of iron and is now heated in a Bessemer converter lined with a siliceous lining. The hot blast of air removes the remaining iron as ferrous silicate slag and is skimmed off. The bessemerised matte is cooled slowly and crystallised with an upper silvery layer (Cu2S) and lower black layer (Ni2S3). A copper–nickel alloy is also formed which is used to dissolve the platinum group metals.

3. Production of Pure Nickel Pure nickel can be extracted by either the electrolytic process or Mond’s process. (a) Electrolytic Process In this process, the matte is cast into rods and made as anode in the solution of nickel (II) ammonium sulphate. Iron is made as a cathode on which nickel is deposited. Copper is recovered from the electrolytes. (b) Mond’s process This process provides an alternative method for obtaining high-purity Ni. Here, the matte is roasted to turn sulphides into oxides which are digested with dilute H2SO4 at 80°C. As a result, cuprous oxide converts into the sulphate while NO remains as such. CuO + H2SO4 $ CuSO4 + H2O. The mixture is treated with water gas at 50°C so as to reduce the oxides to the metallic form. NiO + H2 $ Ni + H2O NiO + CO $ Ni + CO2 Excess of carbon monoxide is passed which converts nickel into volatile nickel carbonyl Ni(CO)4, while the impurities remain as such. The vapours of nickel carbonyl are passed over nickel balls heated to 230°C, where it decomposes to deposit metallic nickel over the nickel balls and CO is recycled. 50°C

C → Ni(CO)4 230° Ni + 4CO   → Ni + 4CO

19.9.2 Properties Nickel is a silvery white metal with high thermal and electrical conductivities. It melts at 1452°C and can be rolled, forged, drawn and polished. It can be magnetised though much less than iron. Nickel is highly unreactive in massive state and does not tarnish in air or water at ordinary temperature. However, Raney Ni (Finely divided metal) is pyrophoric and is readily oxidised in air. It is used as a catalyst. It decomposes steam at red heat. Ni + H2O $ NiO + H2 It is moderately electropositive and dissolves in dilute mineral acids to liberate hydrogen.

19.40 Inorganic Chemistry

���� � Ni2+(aq) + 2e– � ��� � Ni(s)

E° = – 0.24 V

Ni + 2H+ $ Ni2+ + H2 It reacts more rapidly with dilute nitric acid and yields crystals of Ni(NO3)2 . 6H2O on evaporation. However like iron, it is rendered passive by conc. nitric acid. It readily dissolves in aqua regia and yields deliquescent green crystals of NiCl2 . 6H2O on crystallisation. Nickel is unaffected by alkalis, even in their fused state. Nickel crucibles are, therefore, used for alkali fusion experiments. Nickel reacts with halogens but very slowly with fluorine. Therefore, the metal and its alloy, Monel, are used to handle F2 and the corrosive fluorides. Finely divided nickel can occlude 17 times its own volume of H2 which is evolved on heating and is used to purity H2.

19.9.3 Uses Most of the nickel is used to manufacture ferrous and nonferrrous alloys. It imparts strength and resistance to the steel. Invar steel (35% Ni, 0.3% C) has very low coefficient of expansion. It is used in making surveying instruments, pendulums and other precision instruments. Nickel steel (2.5–5% Ni) is extremely tough, hard but elastic and rustproof. It is used in armour plates, underground cables, aeroplane and automobile parts. Alnico steel (15–26%Ni) is used to manufacture very strong magnets. Monel metal (70% Ni, 30% Cu) is corrosion resistant. It is used in chemical plants, boilers and decoration particles. Constantan (Ni 40%, Cu 60%) is uses in electrical resistances. Nichrome (60% Ni, 40% Cr) is used in making heating element in electric devices. Cupro-nickel (20% Ni, 80% Cu) is used to make imitation silver articles. German silver (20% Ni, 60% Cu and 20% Zn) is used for making ornaments, domestic utensils and tableware. Coinage alloy (25% Ni, 75% Cu) is used for making silver coins. Small amounts of Raney Ni are used as important hydrogenation catalyst.

19.9.4 Electronic Configuration The outer-shell electrionic configuration of nickel is 3d84s2. It shows a wide range of oxidation states from (–I) to (+IV). However, (+II) is the only stable and important state. Table 19.7 lists the oxidation states of nickel.

19.9.5 Chemistry of Lower Oxidation States of Nickel

Table 19.7 Oxidation states of nickel Oxidation state (–I) (0) (+I) (+II) (+III) (+IV)

Examples [Ni2(CO)6]2– [Ni(CO)4], [Ni(CN)4]4– [Ni2(CN)6]4– NiO, [NiCl4]2– [NiF6]3– K2[NiF6]

The lower oxidation states of Nickel, i.e. (–I), (0) and (+I), are found with strong p-acid ligands, Ni(0) being the most numerous. The preeminent known carbonyl is Ni(CO)4, used in the Mond process. The molecule is tetrahedral, easily oxidised and is much less stable than the carbonyls in earlier transition metal groups. Other NiL4 species (L = P(OR)3, PF3, PCl3, etc.) are relatively more stable. [Ni(CN)4]4– is extremely reactive and is obtained by reduction of [Ni(CN)4]2– with potassium in liquid ammonia. [(–I) oxidation state is represented by the carbonyl anion [Ni2(CO)6]2–. Nickel in the (+I) state is rare and is best represented by [Ni2(CN)6]4– formed by reduction of K2[Ni(CN)4] with hydrazine sulphate in aqueous medium. It contains short Ni–Ni bonds.

19.9.6 Chemistry of Nickel(II) It is the most important state of nickel in both aqueous and non-aqueous state. Nickel forms a wide series of compounds in this state, including the oxides, halides, sulphide and salts of all the common acids.

Chemistry of Elements of 3d Series

19.41

1. Nickel(II) Oxide, NiO It is obtained as a green solid on heating the hydroxide, nitrate, carbonate or oxalate of nickel(II). It has rocksalt type structure. It is a basic oxide and readily dissolves in acids, but is insoluble in water. It is reduced by hydrogen at about 470 K. The anhydrous halides are prepared by direct halogenation of the metal, except for the fluoride.

2. Aqueous Chemistry of Nickel(II) and Complexes The aqueous solutions of Ni(+II) are invariable green in colour due to presence of the octahedral hexaqua ion [Ni(H2O)6]2+. Some or all of the H2O molecules present in [Ni(H2O)6]2+ can be replaced by different ligands to form complexes such as [Ni(NH3)2(H2O)4]2+, [Ni(NH3)6]2+, [Ni(en)3]2+ etc. These complexes are octahedral, paramagnetic and have a blue or purple colour. Ni(II) forms diamagnetic, square planar complexes with strong field ligands. These complexes are generally reddish brown or yellow in colour. One particular example is yellow [Ni(CN)4]2– formed by the action of excess of cyanide ions on Ni(II) salts. It is the most stable anionic complex of Ni(II) (Ks = 3 × 1031). CN -

Æ [Ni(CN)4]2– Ni2+ + CN– $ Ni(CN)2 æææ Another square planar, well known complex is bis(dimethylglyoximato)nickel(II). The red coloured complex is precipitated from ammoniacal solution of Ni(II) salt by the addition of dimethylglyoxime. Most interestingly, in the solid form, the square planar molecules lie stacked on top of each other to form Ni-Ni bonding links and thus make nickel as octahedrally coordinated. OH

OH H3C — C

H3C—C=N

N

N=C—CH3

+ Ni2+ H3C — C

O

N

+ 2H+

Ni H3C—C=N

N=C—CH3

OH O

HO

Ni(II) also forms several tetrahedral complexes. These complexes are paramagnetic and typically intense blue coloured. The best known tetrahedral complexes are [Ni(X)4]2– (X = Cl, Br, I) obtained from ethanolic solutions of [NR4]+. Other examples include [NiCl2(Ph3P)2], [NiBr2(Ph3AsO)2] and [Ph4As]2+ [Ni(Cl)4]2–.

19.9.7 Chemistry of Higher Oxidation States of Nickel The higher oxidation states of Ni(III) and (IV) are not important and only a few examples are known.

1. Oxides and Hydroxides Anhydrous oxides of Ni(III) and (IV) are not known. On oxidation of alkaline Ni(OH)2 with hypochlorite solution a black powder b–NiO(OH) is obtained, which on further oxidation yields Ni(II)–Ni(III) hydroxide, Ni3O2(OH)4. However, if Br2 is used as an oxidant, the black NiO2.nH2O is obtained. Oxides and hydroxides of nickel(IV) are not yet reported.

2. Complexes The (+III) state is more stabilised in complexes. Na[NiO2] is obtained by bubbling oxygen through fused nickel in sodium hydroxide solution at about 800°C. Fluorination of NiCl2 in presence of KCl at a high temperature yields a strongly oxidising violet soild, K3[NiF6]. It liberates O2 with water or dilute H2SO4. Some other complexes include [NiCl2(en)2]Cl and Ni[Br2(PEt3)2]Br.

19.42 Inorganic Chemistry Ni(+IV) is stabilised with ions of highly electronegative elements. The examples include [NiMo9O32]6–, [NiNb12O38]12–, Na(K)NiIO6 . nH2O and K2NiF6.

19.10

COPPER (Cu)

Romans named this metal cuprum, as they used to obtain it from the island of Cyprus. Like iron, copper has also been known since ancient time.

19.10.1 Occurrence and Extraction Copper is found to the extent of 68 ppm by weight in the earth’s crust. It is found native in Russia, China, USA, Mexico and Chile. It occurs in the combined state as sulphides (copper pyrites, CuFeS2; Copper glance, Cu2S and bornite, Cu5FeS4), oxides (cuprite, Cu2O) and basic carbonate (malachite, CuCO3 . Cu(OH)2 and azurite, Cu(OH)2 . 2CuCO3) Copper pyrite is the principal ore of copper and may contain only 0.4–1% Cu. The process of extraction involves the following steps:

1. Removal of Iron The sulphide ore is crushed and concentrated by the froth-floatation process. The ore now contentrated up to 15% of Cu is roasted with air in a reverberatory furnace. Volatile impurities of arsenic and antimony are removed while a mixture of copper (I) and iron (II) sulphides is left behind. 2CuFeS2 + O2 $ Cu2S + 2FeS + SO2 This mixture is smelted in a water-jacketed blast furnace in the presence of coke and sand. Air is blown through the base and most of the iron sulphide is slagged off. 2FeS + 3O2 $ 2FeO + 2SO2 2Cu2S + 3O2 $ 2Cu2O + 2SO2 Cu2O + FeS $ Cu2S + FeO FeO + SiO2 $ FeSiO3

2. Production of Blister Copper The molten mass called matte is transferred to a Bessemer converter lined with a basic lining. A blast of air mixed with sand is blown through the matte and the remaining iron is slagged off. Cuprous oxide converts cuprous sulphide to copper. Cu2S + 2Cu2O $ 6Cu + SO2 The molten mass is allowed to solidify in sand moulds. Sulphur dioxide escapes out leaving a blister-type appearance on the surface of copper, which is now called blister copper. It is about 98% pure and can be refined further by electrolytic method.

19.10.2 Properties Copper is a heavy and tough metal with reddish brown colour and melting point of 1356 K. It is highly malleable, ductile and tenacious. It is the best conductor of heat and electricity, next to silver. However, presence of impurities decreases its conductance. Copper tends to be unreactive due to its higher enthalpy of sublimation, higher ionisation energy and negative reduction potential.

Chemistry of Elements of 3d Series

Cu2+ + 2e– $ Cu

19.43

E° = – 0.337 V

Copper is replaced from its solution by the metals positioned above it in the electrochemical series. Whereas, less electropositive metals such as Ag, Hg and Au are displaced from their salt solution by copper. 2Ag+ + Cu $ Cu2+ + 2Ag Cu2+ + Fe $ Fe2+ + Cu Copper does not react with water and is inert for non-oxidising acids in absence of air. However, slow reaction takes place in presence of air. Cu + 2HCl + ½O2 (air) $ CuCl2 + H2O Cu + H2SO4 + ½O2 (air) $ CuSO4 + H2O. It readily reacts with oxidising acids. Cu + 2H2SO4 (conc.) $ CuSO4 + 2H2O + SO2 3Cu + 8HNO3 (dilute) $ 3Cu(NO3)2 + 2NO + 4H2O Cu + 4HNO3 (conc.) $ Cu(NO3)2 + 2NO2 + 2H2O Copper is inert towards dry air but is slowly oxidised in moist air and is covered with a protecting green layer of basic copper carbonate, CuCO3 . CuOH)2 and/or basic copper sulphate, CuSO4 . 3Cu(OH)2. 2Cu + CO2 + H2O + O2 $ CuCO3 . Cu(OH)2 8Cu + 2SO2 + 6H2O + 5O2 $ [2CuSO4 . 3Cu(OH)2] Copper reacts at red heat with dioxygen to form CuO and Cu2O (at higher temperature). heat Above 2Cu + O2 red  → 2CuO 1370  → Cu2O + ½O2 K

Copper is attacked by halogens. Cu + Cl2 $ CuCl2 On heating with sulphur, copper gives a nonstoichiometric compound with composition Cu1.87S. Copper is soluble in aqueous ammonia solutions in the presence of oxygen. 2Cu + 8NH3 + O2 + 2H2O $ 2[Cu(NH3)4]2+ + 4OH– It also dissolves in potassium cyanide solutions in presence of oxygen and gives the complex, [Cu(CN)4]2–.

19.10.3 Uses Copper is a very useful metal and is used in enormous ways. Due to its high electrical conductivity, it is used in the electrical industry. Due to its inertness towards air and water, it is used for water pipes, utensils, coins, etc. Copper is used for making alloys with many metals. Some important alloys are listed in Table 19.8. Table 19.8 Alloys of copper Name of the Alloy Brass Bronze German silver Gunmetal Nickel silver Bell metal

Composition 80% Cu, 20% Zn 80% Cu, 10% Zn, 10% Sn 60% Cu, 20% Zn, 20% Ni 87% Cu, 10% Sn, 3% Zn 60% Cu, 20% Zn, 20% Ni 80% Cu, 20% Sn

Uses Machinery parts, sheets and utensils Statues, utensils and coins. Silverware, resistance coils and plating Gun barrels, gears and castings Coins Gongs, bells

19.44 Inorganic Chemistry

Basic copper hydroxide is used as a fungicide under the name Bordeaux mixture. Basic copper acetate is a constituent of the insecticide Paris green. Mixed oxides of copper such as La(2-x)Ba(x)CuO(4-y) and YBa2Cu3O7-x behave as superconductors.

19.10.4 Electronic Configuration The outer-shell electronic configuration of copper is 3d104s1. It shows the oxidation states ranging from (+I) to (+III).(+I) state exists only in the insoluble state and disproportionates in aqueous solution. Cu(+III) is strongly oxidising and is also stablised in insoluble compunds and complexes. Cu(+II) is the only state stable in hydrated form. Table 19.9 lists the various oxidation states of copper.

Table 19.9 Oxidation states of copper Oxidation state

Examples

(+I)

Cu2O, CuI, [Cu(CN)4]3–

(+II)

CuO, CuCl2, [CuCl4]2–

(+III)

KCuO2, K3CuF6

19.10.5 Chemistry of Copper(+I) Copper(+I) compounds are mostly diamagnetic and colourless due to absence of any unpaired electron. However, some coloured compounds such as Cu2O (yellowish red), Cu2CO3 (yellow) and CuI (brown) are also known. These compounds are coloured due to the presence of charge-transfer bonds.

1. Copper(I) Oxide (Cu2O) It is obtained by the reduction of Cu(+II) by mild reducing agents and forms the basis of Fehling’s test. Fehling solution A (CuSO4 + Rochelle salt, sodium potassium tertarate) and Fehling solution B (NaOH) are mixed in equal quantities and sugar is added. The solution is warmed to give yellowish red precipitate of Cu2O. Cu2O is insoluble in water, but dissolves in aqueous ammonia to form colourless [Cu(NH3)2]+ in the absence of oxygen and the blue coloured [Cu(NH3)4]+ in presence of oxygen. Cu2O + NH4OH $ [Cu(NH3)2]OH $ [Cu(NH3)4]OH It is readily oxidised in air to form blue copper (II) hydroxide. Cu2O + ½O2 + 2H2O $ 2Cu(OH)2 It forms dark brown [CuCl2]– with hydrochloric acid. Cu2O + 4HCl $ H[CuCl2] + H2O

2. Copper Halides The bromide is obtained by boiling Cu2O or an acidic solution of Cu(II) with copper (in excess) and the corresponding halogen acid. The reaction mixture on dilution gives the corresponding halide. CuO + Cu + 2HX $ 2CuX + H2O (X = Cl, Br) CuSO4 + Cu + 2HX $ 2CuX + H2SO4 The iodide is obtained by addition of potassium iodide solution to copper(II) solution. The initially formed cupric iodide decomposes to give cuprous iodide and iodine is released. 2Cu2+ + 4I– $ 2CuI2 2CuI2 $ 2CuI + I2

Chemistry of Elements of 3d Series

19.45

Copper(I) halides exist in zinc blende structure in the solid state and become polymeric in the vapour state. These are insoluble in water but dissolve in the presence of excess of halide ions due to the formation of soluble halide complexes [CuCl2]–, [CuCl3]2– and [CuCl4]3–. They also dissolve in aqueous solution of ammonia and strong mineral acids. CuCl + 2NH3 $ [Cu(NH3)2]Cl CuCl + HCl $ H[CuCl2] On addition of potassium cyanide, cuprous cyanide is precipitated and cyanogen is evolved. Cuprous cyanide dissolves in excess of potassium cyanide due to the formation of soluble potassium cuprocyanide. 2Cu2+ + 4CN– $ 2CuCN + C2N2 CuCN + 2CN– $ [CuCN4]3–

2. Aqueous Chemistry of Copper (+I) and Complexes Copper(I) has no unpaired electron (d10 configuration) and should be very stable. However, it is highly unstable in aqueous solution and undergoes disproportionation. 2Cu+ * Cu2+ + Cu K = 1.6 × 106 Some copper(I) compounds are stable to water because they are either insoluble in water or form complexes, e.g. CuCl, CuCN and CuSCN. Solutions of CuCl in ammonia and conc. HCl are worth mentioning. Copper(I) forms tetrahedral complexes with simple ligands. However in solid [CuCN2]–, copper is bonded to three CN– ions in a planar triangular fashion and a spiral polymeric structure is obtained.

19.10.6 Chemistry of Copper(II) It is the most important state of copper and most of the cuprous compounds are readily oxidised to cupric compounds. Copper(II) contains one unpaired electron and therefore, the compounds are coloured and paramagnetic.

1. Copper(II) Oxide (CuO) It is obtained by pyrolysis of the nitrate, hydroxide or by the calcination of malachite. It is a black basic oxide which is insoluble in water and forms salts with acids. CuO + H2SO4 $ CuSO4 + H2O It decomposes on heating above 800°C to give Cu2O. It is reduced to copper on heating in a current of hydrogen.

2. Copper(II) Hydroxide [Cu(OH)2] It is obtained as a blue precipitate when aqueous copper(II) solution is treated with alkali hydroxide. Cu2+ + OH– $ Cu(OH)2 It is readily soluble in strong acids. It also dissolves in conc. alkali hydroxides to give deep blue [Cun(OH)2n–2]2+. It forms the deep blue tetra-ammine complex [Cu(NH3)4]2+ in ammoniacal solutions.

3. Copper(II) Halides The halides are obtained either by direct halogenation of the metal or by dissolving cupric oxide in the corresponding halogen acids. CuI2 is unstable and decomposes to give copper(I) iodide and iodine.

19.46 Inorganic Chemistry

CuF2 is colourless and loses elemental fluorine on melting. It forms the fluoro complexes with excess of fluoride ions, such as [CuF3]–, [CuF6]4– and [CuF6]4–. CuCl2 is brownish in anhydrous form and greenish in hydrated form. It is soluble in water. The dilute aqueous solution is blue due to the presence of the complex [Cu(H2O)4]2+ while the concentrated solution is green due to the presence of a yellow complex [CuCl4]2– along with the blue complex [Cu(H2O)4]2+. The reactions can be represented as 2+ – ���� � CuCl2 � ��� � Cu + 2Cl

���� � Cu2+ + 4H2O � ��� � [Cu(H2O)4]2+ ���� � Cu2+ + 4Cl– � ��� � [CuCl4]2– It dissolves in conc. hydrochloric acid to give a brownish yellow solution. 2+ 2– ���� � CuCl2 HCl  → Cu[CuCl4] � ��� � Cu + [CuCl4]

It dissolves in ammoniacal solutions to give precipitates of cupric hydroxide which dissolve in excess of ammonia to give deep blue solutions. – ���� � CuCl2 + OH– � ��� � Cu(OH)2 + 2Cl 2+ – ���� � Cu(OH)2 � ��� � Cu + 2OH

Cu2+ + 4NH4OH $ [Cu(NH3)4]2+ + 4H2O All copper(II) halides have a distorted rutile structure. The halides are volatile at higher temperatures and impart an intense green colour to the Bunsen flames. It forms the basis of Beilstein test. In this test, if a small amount of a halogen-containing organic compound is heated on a copper wire, a green flame indicates the presence of halogen.

4. Copper(II) Sulphate (CuSO4.5H2O) It is prepared by the action of the dilute sulphuric acid on copper scrap placed in a perforated lead basket.

5. Aqueous Chemistry of Copper(II) and Complexes Most of the copper(II) salts are soluble in water and give the blue aquo ion, [Cu(H2O)6]2+. This ion is a distorted octahedral due to Jahn Teller distortions. Copper also forms polynuclear compounds in which the Cu–Cu distance is short but there is no actual Cu–Cu bond. An important compound is green copper (II) acetate monohydrate, Cu2(CH3COO)4 . (H2O)2. In this compound, the four acetate groups act as bridging ligands between the two roughly octahedrally coordinated copper atoms positioned at a distance of 2.64 Å. This distance is significantly more than the Cu–Cu distance found in metallic copper (2.55 Å) and rules out the possibility of Cu–Cu bond.

R C O

O O

H 2O

Cu

O R

R C O

Cu

OH2

O C O

O C R

Fig. 19.20 Structure of copper (II) acetate monohydrate

19.10.7 Chemistry of Copper(III) This state is quite uncommon. However, if a fused mixture of KCl and CuCl2 is fluorinated, K3[CuF6] is obtained. Oxidation of Cu2+ in alkaline solution gives KCuO2, while with periodic acid, K7Cu(IO6)2. 7H2O is obtained.

Chemistry of Elements of 3d Series

19.11

19.47

ZINC (Zn)

Zinc is commonly called ‘Jast’ and has been mentioned in Ayurvedic treatises as yashade.

19.11.1 Occurrence and Extraction Zinc is the 24th most abundant element and occurs in the earth’s crust up to the extent of 132 ppm by weight. Zinc is found mainly in the combined state. The chief ores of zinc include sphalerite (ZnFeS), zinc blende (ZnS), calamine (ZnCO3), hemimorphite [Zn4(OH)2(Si2O7) . H2O] and zincite (ZnO). Zinc can be extracted from its ore by two processes:

1. Reduction Process Zinc is extracted mainly from zinc blende and calamine. In case of zinc blende, the crushed ore is concentrated by gravity process and froth floatation process. The ore is roasted at about 1200 K in a current of air. Any zinc sulphate present in the ZnS gets oxidised to oxide. 2ZnS + 3O2 $ 2ZnO + 2SO2 Zn + 2O2 $ ZnSO4 2ZnSO4 $ 2ZnO + 2SO2 + O2 In case of calamine, the concentrated ore is calcined to form the oxide. ZnCO3 $ ZnO + CO2 The oxide obtained is heated strongly in fire-clay retorts in presence of crushed coke and is reduced to the metal. ZnO + C $ Zn + CO ZnO + CO $ Zn + CO2 The retort is sprayed with droplets of molten lead to yield 99% pure zinc. Zinc can be further refined by using fractional distillation or electrolysis.

2. Electrolytic Process In this process, the concentrated ore is roasted at moderate temperatures to yield ZnO and ZnSO4. The roasted ore is leached with dilute sulphuric acid so as to convert the oxide into the sulphate. ZnO + H2SO4 $ ZnSO4 + H2O The solution obtained contains impurities like iron, copper, aluminium, manganese, cadmium, arsenic and antimony. These impurities are precipitated by treatment with milk of lime and Zn dust. The purified ZnSO4 solution is electrolysed using aluminium sheets as cathode. High current density is used to deposit zinc on the aluminium sheets and is stripped off by melting, as the melting point of zinc (421°C) is lower than that of aluminium (660°C). The zinc obtained is 99.9% pure and the solution left in the electrolytic bath is used for leaching process.

19.11.2 Properties Metallic zinc is bluish- white and brittle at room temperature. When heated between 370 K and 420 K, it becomes malleable and ductile. On heating further, it again becomes brittle and finally melts at 692 K. The molten zinc is poured in cold water to yield granulated zinc.

19.48 Inorganic Chemistry

Dry air has no action on zinc at room temperature, but it tarnishes rapidly in moist air to form basic zinc carbonate which acts as a protective coating on its surface. 4Zn + 3H2O + 2O2 + CO2 $ ZnCO3 . 3Zn(OH)2 On heating in air at about 500°C, it burns with a bluish white flame and forms a white woolly layer of zinc oxide, generally known as philosopher’s wool. Pure zinc has no action with water but impure zinc decomposes boiling water and steam. Zn + H2O $ ZnO + H2 Zinc is strongly electropositive and precipitates less electropositive metals from their salt solutions. Zn2+ + 2e– $ Zn

E° = – 0.76 V

2+

Zn + Cu $ Zn2+ + Cu Zn + Pb2+ $ Zn2+ + Pb Pure zinc is almost unreactive with dilute mineral acids due to the very large hydrogen overvoltage at its surface. However, if it is coated with metals with low hydrogen overvoltage such as copper, silver and dilute platinum, it reacts rapidly and liberates H2 from non-oxidising acids. Zn + 2H3O+ $ Zn2+ + H2 + 2H2O However, with hot conc. sulphuric acid, SO2 is evolved. Zn + 2H2SO4 $ ZnSO4 + SO2 + 2H2O It reacts with very dilute nitric acid to give ammonium nitrate. Zn + 10HNO3 (very dil.) $ 4Zn(NO3)2 + NH4NO3 + 3H2O Nitric oxide is formed with moderately strong nitric acid. 3Zn + 8HNO3 $ 3Zn(NO3)2 + 4H2O + 2NO With conc. nitric acid, nitrogen dioxide is the main product obtained. Zn + 4HNO3 $ Zn(NO3)2 + 2H2O + 2NO2 Zinc is amphoteric and dissolves in alkalis to form zincate formulated as Na2[Zn(OH)4] or simply as Na2ZnO2. Zn + 2NaOH $ Na2ZnO2 + H2 Zinc combines directly with halogens and sulphur.

19.11.3 Uses Zinc is used in large quantities for protecting iron from rusting. A thin layer of zinc is deposited on the iron surface electrolytically (galvanisation). A thick layer can be deposited by dipping the metal in molten zinc (hot dipping). The metal can be coated with zinc dust (sherardizing). It is also used to make alloys such as brass, bronze and German silver. Zinc is also used as the cathode container in dry cells. It is used in the cyanide process used in the metallurgy of gold and silver. It is used in the Parke’s process for desilverisation of lead. Zinc dust is used as a reductant in the manufacture of chemicals, perfumes and drugs. Zinc oxide is used as a white pigment in paints.

Chemistry of Elements of 3d Series

19.49

19.11.4 Chemistry of Zinc Zinc has two s-electrons after a complete d shell. Hence, zinc shows preferably (+II) oxidation state in all its compounds.

1. Zinc Hydroxide, Zn(OH)2 It is obtained as a white precipitate when the solution of a zinc salt is treated with alkalis. ZnCl2 + 2NaOH $ Zn(OH)2 + 2NaCl It is also amphoteric in nature like the oxide. Zn(OH)2 + 2HCl $ ZnCl2 + 2H2O Zn(OH)2 + 2NaOH $ Na2ZnO2 + 2H2O It also dissolves in ammonia to form the complex [Zn(NH3)4]2+. It converts into yellowish peroxide on treatment with hydrogen peroxide. ZnO + H2O2 $ ZnO2 + H2O ZnO2 is decomposed by dilute acids. ZnO2 + 2HCl $ ZnCl2 + H2O2

2. Halides All the dihalides are known. ZnF2 is white and has higher melting point than the other halides. This is due to its higher lattice energy. ZnF2 has rutile structure and is not much soluble in water. ZnCl2, ZnBr2 and ZnI2 exist as close-packed lattices of halide ions in which Zn2+ ions occupy one quarter of the tetrahedral voids. These are hygroscopic and highly soluble in water due to their weak crystal lattice. The most important halide is ZnCl2, obtained by dissolving the metal, the oxide, hydroxide or carbonate in hydrochloric acid. The solution on evaporation in excess of hydrochloric acid yields the crystals of colourless ZnCl2 . H2O. However, in absence of hydrochloric acid, hydrolysis takes place.

3. Aqueous Chemistry of Zinc(II) Zn

Zn(II) is colourless due to absence of any unpaired electron. In water, slight hydrolysis takes place which is very less than that of Cu(II) ion. Thus, solution of same concentration of Zn(II) is less acidic than that of Cu(II) + + ���� � Zn2+ + H2O � ��� � Zn(OH) + H Zinc forms a large number of complexes, mostly tetrahedral. Hydrated Zn(II) contains invariably the complex ion, [Zn(H2O)4]2+. The water molecules can be substituted with other ligands, for example, [Zn(NH3)4]2+ and [Zn(CN)4]2+. However, the complexes with cyanide ligands are more stable than with ammonia as a ligand.

ac

ac ac O ac

Zn

Zn

ac

ac Zn

Fig. 19.21 Structure of basic zinc acetate

���� � [Zn(CN)4]2– � ��� � Zn4+ + 4CN–

K = 1.2 × 10–18

���� � [Zn(NH3)4]2+ � ��� � Zn2+ + 4NH3

K = 3.4 × 10–10

Some octahedral complexes such as [Zn(H2O)6]2+, [Zn(NH3)6]2+, [Zn(en)3]2+ are also known, but these are not much stable. Zinc is also known to form [Zn(NO3)4]2–, with the coordination number 8. Basic zinc acetate, (CH3COO)6 . Zn4O, resembles basic beryllium acetate but hydrolyses more readily.

19.50 Inorganic Chemistry

The first transition series contains ten elements, Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu and Zn. Scandium is prepared by electrolysis of fused scandium chloride in presence of sodium chloride. Its oxidation state is (+III), mainly represented by Sc2O3 as all its salts yield Sc2O3 on heating. Titanium is extracted by the Kroll process from its rutile ore. Extremely pure titanium is obtained by van Arkel method. TiO2 is the most important oxide of titanium, existing mainly in its rutile form. It dissolves in conc. alkalis to form titanates. Sodium titanate, when reduced with hydrogen at high temperature, gives a lustrous metallic and nonstoichiometric compound known as titanium bronze. Barium titanate is ferroelectric with perovskite structure. TiCl4, a diamagnetic compound of titanium, is used to prepare other titanium compounds. Aqueous solution of titanium (IV) contains (TiO)n2n+ chains. Vanadium cannot be extracted from its ore due to its high reactivity towards carbon, oxygen and nitrogen at high temperature. Hence, it is obtained from reduction of its ferrovanadium alloy. All vanadium (II) compounds dissolve in water to give strongly reducing, violet, air-sensitive solutions containing the ion [V(H2O)6]2+. This solution turns green in air due to the formation of blue [V(H2O)6]3+. (+IV) is the most important oxidation state of vanadium, represented by oxides, halides and oxo salts. Vanadium forms complexes of VO2+ with many ligands. V2O5 is the most important oxide of vanadium which forms the series of vanadates with NaOH. Chromium is extracted mainly from its chromite ore, FeO.Cr2O3. The most important and stable oxidation state of chromium is (+III) found in a large number of compounds. Cr2O3 is amphoteric and forms chromates when fused with an alkali. (+V) is the most unstable oxidation state of chromium represented mainly by the deep blue chromium peroxide CrO5. CrO3 is a very strong oxidising agent which dissolves in water to give acidic solution containing a number of chromic acids. Potassium dichromate is an important oxidising agent, obtained from iron chromate. Manganese cannot be extracted from its ores by the usual reduction processes due to their violent nature; hence, it is prepared by electrolysis of aqueous MnCl2 or MnSO4 solutions. (+II) is the most stable oxidation state of Mn exhibited in a series of compounds. This state is highly stable due to the low oxidation potential towards oxidation. The (+VII) oxidation state is best represented by the purple coloured potassium permagnate, a strong oxidising agent. Iron is extracted mainly from its oxide ore, haematite. Iron is highly reactive towards moist air resulting in the formation of rust The most common oxidation state of iron is (+II), easily oxidisable to (+III) oxidation state. The important oxides of iron are FeO, Fe2O3 and Fe3O4. Iron (II) sulphate, green vitriol, is a good reducing agent. It forms a brown complex with NO and is used for the test of nitrites and nitrates. Cobalt is extracted mainly from its sulphide ore. Its most important oxidation states are (+II) and (+III). It also forms binuclear and polynuclear complexes with strong p-acid ligands. Cobalt(II) nitrate is used in the charcoal cavity test. Cobalt(II) chloride is used as a moisture indicator. Cobalt(III) nitrate is used as an important nitrating agent. Sodium cobaltinitrite is used for estimation of potassium salts. Nickel is extracted usually from its sulphide ore. Pure nickel is extracted mainly by the Mond process. (+II) is the only stable oxidation state of nickel. It also forms an important compound with zero (0) oxidation state, Ni(CO)4, used in Mond’s process. Dimethylglyoxime is used for estimation of nickel(+II) due to the formation of red complex with ammoniacal solution of nickel (II) salts. Copper is extracted mainly from copper pyrite. It is inert towards dry air but is covered with a protecting green coating of basic copper carbonate in moist air. (+II) is the only stable state in the hydrated form while (+I) and (+III) states are stabilised in insoluble compounds. Copper (II) sulphate is an important compound of copper which dissolves in excess of ammonia to give a blue solution. It reacts with more reactive metals to

Chemistry of Elements of 3d Series

19.51

deposit copper on their surface. Copper also forms polynuclear compounds such as green copper(II) acetate monohydrate, with absence of any actual Cu–Cu bond. Zinc is extracted either form zinc blende or from calamine. Metallic zinc tarnishes rapidly in moist air due to the formation of a protective coating of basic zinc carbonate on its surface. Zinc is amphoteric and gives zincates when dissolved in alkalis. It shows preferably (+II) oxidation state in its compounds. ZnO is commonly known as philosopher’s wool, an amphoteric oxide. ZnSO4.7H2O, white vitriol, is a colourless crystalline solid which hydrolyses slightly in aqueous medium to give the complex ion [Zn(H2O)4]2+. Basic zinc acetate is an important polynuclear complex of zinc which resembles basic beryllium acetate.

EXAMPLE 1 What happens when (a) Scandium nitrate is heated in air?

(b) Titanium is treated with HF?

(a) Scandium nitrate converts to scandium oxide on heating in air. Sc2(NO3)3 æDæ Æ 2 Sc2O3 + 12NO2 + 3O2 (b) Titanium dissolves in HF to form hexafluoro complexes. Ti + 6HF $ H2[TiF6] + 2H2

EXAMPLE 2 Write balanced chemical equations for the reactions of copper with (a) dilute nitric acid

(b) silver nitrate solution

(a) 3Cu + 8HNO3(dilute) $ 3Cu(NO3)2 + 4H2O + 2NO (b) Cu + 2AgNO3 $ Cu(NO3)2 + 2Ag

EXAMPLE 3 What happens when (a) NaOH is added to K2Cr2O7 solution? (b) Potassium ferrocyanide is heated with conc. H2SO4? (a) Addition of NaOH to K2Cr2O7 solution results in the formation of yellow coloured K2CrO4. Hence, the orange coloured solution becomes yellow. K2Cr2O7 + 2NaOH $ 2K2CrO4 + H2O Yellow

(b) When potassium ferrocyanide is heated with conc. H2SO4, CO gas is liberated. K4[FeCN]6 + 6H2SO4(conc.) + 6H2O $ 2K2SO4 + FeSO4 + 3(NH4)2SO4 + 6CO.

QUESTIONS Q.1. Discuss the extraction of the following elements: (a) Zinc (b) Nickel (c) Cadmium

19.52 Inorganic Chemistry

Q.2. Discuss the uses and properties of the following elements: (a) Titanium (b) Manganese (c) Cobalt Q.3. How will you prepare the following compounds? (a) Potassium ferrocyanide (b) Sodium nitroprusside Q.4. Give reasons for (a) Compounds of Ti(+III) are strong reductants (b) CuSO4 dissolves in ammonia. Q.5. Discuss the action of the following elements with conc. H2SO4. (a) Zinc (b) Copper (c) Iron Q.6. Discuss the preparation, properties and uses of potassium permanganate. Q.7. Write short notes on (a) Alloy steel (b) Titanium alloys (c) van Arkel method (d) Ring test Q.8. What are the different oxidation states exhibited by the following elements: (a) Titanium (b) Nickel (c) Vanadium Also comment on the colour and magnetic properties of the compounds in these oxidation states. Q.9. Discuss the structures of the following compounds. (a) Vo(acac)2 (b) Ni(CO)4 Q.10. Give balanced chemical reactions for: (a) Treatment of acidified K2Cr2O7 with KI (b) Treatment of acidified K2Cr2O7 with H2O2 (c) Treatment of acidified KMnO4 with oxalic acid (d) Treatment of K4Fe(CN)6 with conc. H2SO4 (e) Treatment of zinc chloride with sodium hydroxide Q.11. How will you prepare the following? (a) Chromyl chloride (b) Potassium chromate (c) Pure MnO2 (d) Sodium cobaltinitrite (e) Prussian blue Q.12. Discuss the manufacture of steel. Q.13. Write a short note on the oxidising action of KMnO4. Q.14. Write short note on electrochemical theory of corrosion. Q.15. Name the important alloys of the following elements: (a) Vanadium (b) Iron (c) Nickel Q.16. Discuss the metallurgy of copper. Also give an account of alloys of copper. Q.17. Write short note on aqueous chemistry of following elements: (a) Copper (b) Zinc (c) Manganese Q.18. Discuss the general chemical behaviour of following elements: (a) Chromyl chloride (b) Potassium chromate (c) Pure MnO2 (a) Chromium (b) Cobalt (c) Zinc Q.19. Describe the preventive methods of corrosion. Q.20. Comment on the statement: Scandium does not exhibit the oxidation state of +II as exhibited by other transition elements.

MULTIPLE-CHOICE QUESTIONS 1. The element with highest tendency to form complexes is (a) Sc (b) Ti (c) Na 2. The purest form of iron is (a) Wrought iron (b) pig iron (c) stainless steel

(d) Mg (d) steel

Chemistry of Elements of 3d Series

3. The compound formed by treatment of cupric sulphate solution with hypo solution is (a) CuS2O3 (b) NaCuS2O3 (c) [Cu(NH3)2]SO4 (d) Na2SO4 4. Zinc is (a) amphoteric (b) basic (c) acidic (d) none of these 5. The compound used in the charcoal cavity test is (a) CuSO4 (b) Co(NO3)2 (c) FeSO4 (d) HgO 7. The purest form of iron is (a) wrought iron (b) pig iron (c) stainless steel (d) steel 8. The compound formed by treatment of cupric sulphate solution with hypo solution is (a) CuS2O3 (b) NaCuS2O3 (c) [Cu(NH3)2]SO4 (d) Na2SO4 9. Zinc is (a) amphoteric (b) basic (c) acidic (d) none of these 10. The compound used in the charcoal cavity test is (a) CuSO4 (b) Co(NO3)2 (c) FeSO4 (d) HgO

19.53

chapter

Chemistry of Elements of 4d Series

20

After studying this chapter, the student will learn about the occurrence, extraction and chemistry of the following 4d elements:

20.1

INTRODUCTION

Ten elements, namely, yttrium (Y), zirconium (Zr), niobium (Nb), molybdenum (Mo), technetium (Te), ruthenium (Ru), rhodium (Rh), palladium (Pd), silver (Ag) and cadmium (Cd) constitute the 4d or second transition series. We have already discussed in chapter 18 about the general characteristics of these elements. These elements show pronounced differences from their light congeners. This chapter illustrates the chemistry of these elements in detail.

20.2

YTTRIUM (Y)

20.2.1 Occurrence and Extraction Yttrium is the twenty-ninth most abundant element and occurs in the earth’s crust by up to 31 ppm by weight. It is found along with lanthanides in several minerals. The important minerals are bastnaesite (YCO3F) and monazite (YPO4). Extraction of yttrium is very difficult and has been discussed in chapter.

20.2.2 Properties Yttrium is a silvery white metal which melts at 1509°C. The element always exists as Y3+ ion with d0 configuration and its compounds are colourless. It is quite reactive but is unreactive to air even at high temperature due to formation of a protective oxide coating on its surface. The hydroxide Y(OH)3 is a weak

20.2

Inorganic Chemistry

base but more basic than Sc(OH)3 and is slightly amphoteric. It reacts with carbon dioxide and forms salts with acids. 2Y(OH)3 + 3CO2 $ Y2(CO3)3 + 3H2O All the oxo salts of yttrium are known but they undergo decomposition to give oxides even on slight heating. It reacts with the halogens to form YX3. The fluoride is insoluble but the chloride is soluble and crystallises as hydrated salt. On heating, the hydrated salt decomposes to give oxochloride. YCl3.(H2O)7 ∆ → YOCl + 2HCl + 6H2O Due to its fairly large size, it shows little tendency to form complexes. [Y(acetylacetone)3 H2O] exists in a capped trigonal prism structure with the coordination number 7. In [Y(NO3)5]2–, the coordination number is 10. Y is used in red phosphor and synthetic garnets.

20.3

ZIRCONIUM (Zr)

20.3.1 Occurrence and Extraction Zirconium is the eighteenth most abundant element and occurs to an extent of 162 ppm by weight in the earth’s crust. It is found mainly as baddeleyite (ZrO2) and zircon (ZrSiO4). It is extracted from its ore mainly by the Kroll process. The ore is treated with carbon and chlorine, at red heat to give ZrCl4 which is purified by fractionation. The purified ZrCl4 is reduced with molten magnesium at about 800°C to give zirconium. Extremely pure zirconium is obtained by decomposition of purified ZrI4 (van Arkel method).

20.3.2 Properties Zirconium resembles stainless steel in its appearance. It is hard but softer than titanium. However, it is more corrosion resistant than titanium. Its melts at 1855°C. The massive form is passive towards air, acids and alkalis at room temperature. Due to formation of a protective thin impermeable oxide film at its surface. However, the finely divided form is pyrophoric. It becomes reactive at high temperatures and burns in air to give a mixture of nitride, oxide and oxide nitride (ZrO2, ZrN and Zr2ON2). It dissolves in hot concentrated H2SO4 and aquq regia. Like titanium, it is best dissolved in HF due to the formation of hexafluoro complex. It forms interstitial compounds with hydrogen, carbon and nitrogen. These compounds are chemically inert, hard and have refractory properties.

20.3.3 Uses Due to its high melting point, corrosion resistance and low absorption of neutrons, it is used to make the cladding for UO2 fuel in water-cooled nuclear reactors. Its alloy with Nb is also used as an important superconductor. It is also used to remove traces of oxygen and nitrogen from thermoionic valves. Zirconium nitrate, ZrO(NO3)2, is used to remove interferring PO43– ions in qualitative analysis.

20.3.4 Chemistry of Zirconium (+IV) The outer-shell electronic configuration of zirconium is 4d25s2. The most common and most important oxidation state is (+IV); however, some unstable compounds in lower oxidation states have been authenticated. Zirconium resembles titanium in its chemistry. The important compounds of zirconium are described here.

Chemistry of Elements of 4d Series

1. Oxides and Hydroxide White gelatinous hydrous oxide, ZrO2.nH2O, is precipitated on addition of a hydroxide to Zr(+IV) solutions. However, no true hydroxide exists. On strong heating, the hydrous oxide gives a very stable white, hard and nonvolatile solid, ZrO2. It is insoluble and has very high melting point (2700°C). It is exceptionally resistant to acids and alkalis and has good mechanical properties. These properties make it useful for making furnace linings and high-temperature crucibles. ZrO2 is more basic than TiO2. The metal atom is seven coordinate in the baddeleyite form of ZrO2. Zirconium forms mixed metal oxides called zirconates just like titanium form titanates.

20.3

O O O

Zr O O

O O

Fig. 20.1 Structure of ZrO2

2. Halides The halides are obtained by direct combination of the elements just like titanium halides. ZrCl4 is best obtained by chlorination of a heated mixture of ZrO2 and charcoal. It is a white solid and sublimes at 331°C. It undergoes vigorous hydrolysis and fumes in moist air. ZrCl4 + 9H2O $ ZrOCl2.8H2O + 2HCl It is highly soluble in donor solvents and forms complexes. ZrBr4 (yellow) and ZnI4 (reddish brown) are similar to ZrCl4 . ZrF4(white) sublimes at 903°C and is insoluble in donor solvents.

3. Aqueous Chemistry of Zirconium and Complexes The existance of Zr4+ aquo ions is very doubtful even in strong acid solutions due to its extensive hydrolysis. The hydrolysed zirconyl ion ZrO2+, is also not directly identified due to its lesser tendency towards complete hydrolysis. In concentrated HF, unusual complexes such as [ZrF7]3– (trigonal prismatic or pentagonal bipyramidal) and [ZrF8]4– (bisdisphenoid or square antiprismatic) with coordination numbers 7 and 8 respectively are formed. In other acidic solutions, partially hydrolysed polymeric species are present. ZrOCl2.8H2O, the most important zirconyl salt is crystallised from solutions in dilute hydrochloric acid.

4. Chemistry of Lower Oxidation States Some compounds of Zr are known in oxidation states lower than (+IV). Reduction of the tetrahalides ZrCl4, ZrBr4 and ZrI4, yields the highly reactive trihalides. The trihalides form complexes with unknown structures such as ZrX3.2py and 2ZrX3.5CH3CN.

20.4

NIOBIUM (Nb)

20.4.1 Occurrence and Extraction Niobium is a very less abundant element and comes at the thirty-second place in relative abundance on the earth’s crust. Its importance lies in the cluster compounds, formed in lower oxidation states. Niobium is generally found together with tantalum in the columbite–tantalite series of minerals generally written as (Fe/Mn)(Nb/Ta)2O6. The mineral is called tantalite if it contains more of tantalum than of niobium. If it contains more of niobium than of tantalum, the mineral is called columbite. The most important mineral pyrochlorite, CaNaNb2O6F, is a mixed calcium sodium niobate. The process of extraction is very complex and involves the decomposition of the mineral by fusion with KHSO4 as shown in Fig. 20.2.

20.4

Inorganic Chemistry

(Fe/Mn) (Nb/Ta)2O6

Fe(OH)2+Mn(OH)2 (Insoluble)

Fusion with KHSO4

H2O CO2

K3NbO4 + K3TaO4 (Soluble) Nb Ta

Reduction with Al

Nb2O5 + Ta2O5 KF Conc. HF

K2NbOF5 (More soluble)

Electrolysis

K2TaF7 (Less soluble)

K2NbOF5 Fractional + Crystallisation K T F 2 a 7

Fig. 20.2 Representation of extraction process

20.4.2 Properties Niobium is a bright silvery metal with a very high melting point (2468°C). The pure metal is ductile and moderately soft, but presence of even traces of impurities makes it brittle and harder. It is highly resistant to attack of air, water and acid (except HF) due to formation of a protective surface film of oxide and thus is corrosion resistance. Niobium dissolves in conc. alkalis and HF. It reacts with many nonmetals (such as O2, N2, S, Cl2) on heating to form interstitial compounds.

20.4.3 Uses Niobium is added in steel to encapsulate the fuel elements in some nuclear reactors. It is also used in surgical instruments and manufacturing plants. Its alloy (Nb/Zr) is a superconductor at low temperature. Table 20.1 Oxidation states of niobium

20.4.4 Chemistry of Niobium (+V) The outer-shell electronic configuration of niobium is 4d35s2. Its most common and important oxidation state is (+V). Some lower oxidation states are also known in aqueous solutions. Table 20.1 lists the oxidation states of niobium. Niobium resembles typical nonmetals in its chemistry of (+V) oxidation state and forms numerous anionic species.

1. Oxides

Oxidation state

Example

(–I)

[Nb(CO)6]–

(+I)

(h5 – C5H5)Nb(CO)4

(+II)

NbO

(+IV)

NbX4, Nb(NEt2)4

(+V)

NbCl5, NbOCl3, K2NbOF6

Nb2O5 is obtained as a dense white powder by ignition of other Nb compounds in air. It is chemically inert but gets dissolved in concentrated HF to form fluoro complexes and forms niobates when fused with NaOH, Na2CO3 or NaHSO4. Nb2O5 + 3Na2CO3 $ 2Na3NbO4 + 3CO2 If the melt is dissolved in water, isopolyion [Nb6O19]8– is found in the solution.

2. Halides NbF5 is obtained as a volatile white solid by directly heating the metal or the pentachloride in presence of F2 or HF. It exists in cyclic tetrameric structure in which fluorine acts as a bridging group (in Fig. 20.3). Other pentahalides can be obtained by direct reaction of the metal with excess of the halogens or by heating the pentaoxide with carbon in the presence of the corresponding halogen. 670 K

Nb2O5 + 5C + 5Cl2 → 2NbCl5 + 5CO

Chemistry of Elements of 4d Series

F

Cl

Cl

Nb

Cl

Cl

Nb

56 Å

2.2

F

Nb

101.3°

Nb 5Å

F F

F

Cl

F

F F

Nb F

F F

F

F

Cl

F Nb

F F

F

20.5

2.

Cl

Cl

F

F

Cl

Cl

F

Fig. 20.4 Structure of NbCl5

Fig. 20.3 Structure of NbF5

These halides are yellow to brown or purple-red solids and are soluble in organic solvents such as CCl4, ether, etc. They are readily hydrolysed in water to give hydrous pentoxide and hydrochloric acid. The pentahalides can be sublimed in an atmosphere of the corresponding halogens, without any decomposition. NbCl5 exists as a dimeric structure in the solid state but exists as a monomer in the vapour state (in Fig. 20.4).

3. Oxohalides Four oxohalides, NbOCl3, NbOBr2, NbO2I and NbOI3 are known and are prepared by controlled oxidation of the pentahalides. O 2NbCl5 + O2 $ 2NbOCl3 + 2Cl2 O These are less volatile than their corresponding pentahalides due to the presence of oxygen-bridged infinite chains of the planar Nb2Cl6 groups in the solid state. However, the vapour state is monomeric. They undergo hydrolysis in water to form hydrous pentoxides. They form complex oxide halides such as MNbOCl4 and M2NbOCl5 in presence of alkali metal cations in concentrated hydrohalic acids (Fig. 20.5).

4. Complexes of Niobium (+V)

Cl

Cl

Nb Cl

Cl

Nb Cl

Cl O

O

Fig. 20.5 Structure of NbOCl3

Addition of metal fluoride to niobium yields [NbF6]– in 50% HF solutions. [NbOF5]2–.H2O can be isolated in weakly acidic solutions. [NbF7]2– can be crystallised from solutions of very high acidity and high F– concentration.

20.4.5 Chemistry of Niobium (+IV) 1. Oxide NbO2 is obtained as a dark grey powder by heating the pentoxides in a current of hydrogen. Nb2O5 + H2 $ 2NbO2 + H2O It is insoluble in water and acids, but dissolves in hot aqueous alkalis.

2. Halides All the tetrahalides are obtained by reduction of the pentahalides with H2, Al or the metal itself at elevated temperatures. All of these are brown-black or black solids. NbF4 is paramagnetic and the metal atoms are present at the centre of the octahedra linked by their edges. Other tetrahalides are diamagnetic and the metal atoms are displaced from the centre of the octahedra to form metal pairs due to formation of weak Nb–Nb bonds.

20.6

Inorganic Chemistry

20.4.6 Chemistry of Niobium in Other Lower Oxidation States Niobium also forms nonstoichiometric halides in its lower oxidation states (+II) and (+III). The halides of niobium in (+III) oxidation state are NbF3, NbCl2.67, NbCl2.33 and NbI3, while those in (+II) oxidation states are NbF2.5, NbCl2.67, NbCl2.33, NbBr2.67, NbBr2, NbI2.67 and NbI2.

20.5

MOLYBDENUM (Mo)

20.5.1 Occurrence and Extraction Molybdenum was discovered by Scheele in 1778 and was isolated in 1790 by Hjelm. Molybdenum is quite rare in its abundance (42nd) in the earth’s crust (10–4%). It occurs chiefly as molybdenite (MoS2), molybdite (MoO2) and as molybdates. Molybdenum is extracted from its sulphide ore which is concentrated by froth floatation process. The concentrated ore is roasted to obtain the oxide, MoO3. 2MoS2 + 7O2 $ 2MoO3 + 4SO2 The crude oxide is dissolved in ammonia, the product is purified by heated crystallisation and is ignited strongly to get relatively pure MoO3. MoO3 + 2NH3 + H2O $ (NH4)2MoO4 Ammonium molybdate

(NH4)2MoO4 $ MoO3 + 2NH3 + H2O MoO3 is reduced, either by hydrogen or by alumino-thermite process, to the metal. MoO3 + 3H2 $ Mo + 3H2O MoO3 + 2Al $ Mo + Al2O3 Carbon cannot be used for reduction of MoO3 as it yields the carbide rather than the metal.

20.5.2 Properties Molybdenum is dull grey in powder form but turns into a lustrous, silvery white metal when converted into the massive form. It is fairly hard and melts at 2620°C. It is inert to attack of air at normal temperatures but forms trioxide when heated to redness. It is almost inert to aqueous acids and alkalis but dissolves in fused KNO3–NaOH. It is initially attacked by concentrated HNO3 but is soon passivated. However, it readily dissolves in HNO3/HF mixtures. It reacts with halogens to form the corresponding halides.

20.5.3 Uses Molybdenum is mainly used in the manufacture of alloy steels as it imparts hardness and strength to the steel. It is used in making acid-proof steels (2–3% Mo, 60% Cr). It is also used in making magnetic cores (4% Mo, 79% Ni, 17% Fe). It is used for support of tungsten filament in electric lamps. MoS2 is an excellent lubricant due to its layered lattice.

20.5.4 Electronic Configuration and Oxidation States The outer-shell electronic configuration of Mo is 4d55s1 and shows the oxidation states ranging from (+I) to (+VI). In addition some lower oxiation states are also found in complexes with strong -acid ligands. The most important oxidation state is (+ VI) but (+ V) is equally stable in aqueous solutions. The (+III) state is strongly reducing. The oxidation states with examples are listed in Table 20.2.

Chemistry of Elements of 4d Series

20.7

Table 20.2 Oxidation states of molybdenum

20.5.5 Chemistry of Molybdenum (+VI) It is the most important and stable oxidation state of molybdenum. A number of compounds are known.

1. Oxide

Oxidation State

Examples

(–II)

[Mo(CO)5]2–

(–I)

[Mo2(CO)10]2–

MoO3 is obtained as a white solid by heating the metal (0) [Mo(CO)5I]– or its sulphide in air. It turns yellow on heating due to (+I) [(C6H6)2Mo]+, [(C5H5)Mo(C6H6)] the formation of defects in the lattice. It is insoluble in water and is not attached by acids (expect HF). It (+II) [Mo2Cl8]4–, Mo6Cl12 2– dissolves in NaOH to form [MoO4] ions. It differs (+III) [MoCl6]3–, [Mo(NCS)6]3– from CrO3 in being non-oxidizing. When MoO3 is fused (+IV) MoS2, [Mo(CN)8]4– with alkali or alkaline earth metal oxides; mixed oxides such as K2Mo4O13 are formed. They exist as chain or (+V) MoCl5, [Mo(CN)8]3– ring structures with linked MoO6 polyhedra. Acidified (+VI) MoO6, [MoO4]2–, [MoF8]2–, MoF6 solutions or suspensions of MoO3 in water on mild reduction with Sn, SO2 or H2S acquires a blue colour due to formation of blue oxides. These are considered with composition as MoO2.0(OH) and MoO2.5(OH)0.5 and commonly known as molybdenum blue.

2. Halides MoF6 is obtained by direct combination of the elements. It is colourless, volatile and diamagnetic. It is very reactive and is readily hydrolysed. It is easily reduced under suitable conditions. ~ 660 K

→ 3MoF2 MoF6 + 2Mo  MoCl6 is obtained as a black powder by the action of SoCl2 on MoO3. It is highly sensitive to water.

3. Oxohalides Two oxohalides of type MOX4 and MO2X2 are known. MoOF4 is obtained by the fluorination of MoO3 as a colourless and volatile solid. MoO2F2 is also colourless and volatile. It is obtained by the treatment of MoO2Cl2 with HF.

4. Oxo Acids The simplest oxo acid is H2MoO4. It is obtained by heating the yellow hydrate MoO3.H2O. The hydrate is obtained from the strongly acidic solutions of molybdates obtained by treatment of MoO3 in aqueous alkali metal hydroxide.

5. Simple Molybdates These are crystallised from the solutions of MoO3 dissolved in aqueous alkali metal hydroxides. The general formula of the simple molybdates can be represented as M2MoO4 containing the discrete tetrahedral ions, MoO42–. Ammonium molybdate is obtained from the ammoniacal solutions of MoO3. The nature of the molybdate obtained depends on the pH of the solutions. In excess of hot concentrated ammonia, crystals of ammonium molybdate, (NH4)2 MoO4 are obtained after cooling, but in near neutral solutions, (NH4)6Mo7O24.4H2O is obtained. Ammonium molybdate in dilute nitric acid is used as a laboratory reagent for the detection and determination of phosphates under the name molybdate reagent.

20.8

Inorganic Chemistry

6. Polyacids A predominant features of the chemistry of molybdenum (+VI) is the formation of polyacids. The polyacids are of two types, i.e. isopoly and heteropoly acid. The polyacid containing only molybdenum along with oxygen and hydrogen are called the isopolyacids, while the polyacids which contain one or two atoms of another element along with molybdenum, oxygen and hydrogen are called hetropolyacids. The salts of isopolyacids and hetteropoly acids of molybdenum are known as isopolymolybdates and hetetropolymolybdates respectively. All the polymolybdates contain octahedral MoO6 groups which share corners and edges in a variety of ways. A series of isopolymolybdates are obtained, when the basic solutions of MoO42– and alkali metals or ammonium ions are acidified. The product formed depends upon the pH of the solutions. pH=1

6– 4– 6 pH= 2 → MoO3.2H2O MoO42– pH= → [ Mo7O24] → [Mo8O26] 

Heteropolymolybdates are obtained when the basic solutions of molybdates containing other oxo anions such as PO43–, SiO44– and metal ions are acidified. Some particular examples are Na3[PMo12O40] and H3[PMo12O40].

7. Complexes of Molybdates (+VI) A large number of complexes of molybdates (+VI) are known. MoO3 dissolves in aqueous 12 M HCl to form [ MoO2Cl4]2– and [ MoO2Cl2(H2O)2] is obtained in aqueous 6 M HCl solutions

20.5.6 Chemistry of Molybdates (+IV) 1. Pentoxides Mo2O5 is obtained by heating the finely divided metal with MoO3 at 750°C. Mo + 5MoO3 $ 3Mo2O5 It is also obtained by heating the ammoniacal solutions of Mo (+V) salts. It is a violet solid which dissolves in warm acids.

2. Pentahalides MoF5 is obtained by the action of molybdenum with fluorine or the hexafluoride. Mo + 5MoF6 $ 6MoF5 It is also obtained by the reduction of the hexafluoride with the metal carbonyl. C 5MoF6 + Mo(CO)6 25°  → 6MoF5 + 6CO

Crystalline MoF5 has a tetrameric structure. MoCl5 is monomeric in the vapour state with a trigonal bipyramidal structure and dimeric in the solid phase as Mo2Cl10. It is obtained by direct chlorination of the metal. Mo2Cl10 is paramagnetic and contains chlorine-bridged Mo atoms without any Mo–Mo bond. It is moderately volatile and is readily hydrolysed by water. It is soluble in benzene and polar organic solvents. It is used to prepare lower chlorides and oxochlorides.

3. Oxochloride Black MoOCl3 and MoOBr3 are obtained by reduction of MoOCl4 and MoOBr4 respectively.

4. Complexes Reduction of molybdates or MoO3 in acid solution yields the emerald – green ion [MoOCl5]2–. It is also

Chemistry of Elements of 4d Series

20.9

obtained on addition of KCl to a solution of MoCl5 in concentrated hydrochloric acid. However, if SO2 is added as a solvent, a mixture of [MoOCl4]– and [MoCl6]2– is obtained. Treatment of MoCl5 with Et4NCl in CH2Cl2 yields black crystals of [Et4N]+[MoCl6]–.

20.5.7 Chemistry of Molybdenum (+IV) 1. Oxides MoO2 is obtained by heating the trioxide with H2 or NH3 below 470°C. 2MoO3 + 2H2 $ 2MoO2 + 2H2O It is also obtained by the reaction of the metal with steam at 800°C. It is a violet solid with coppery luster. It is insoluble in non-oxidising mineral acids but is oxidised by concentrated nitric acid to MoO3.

2. Tetrahalides MoF4 is obtained by reduction of the hexafluoride with benzene at 110ºC. It is a nonvolatile solid. Dark red MoCl4 is obtained by refluxing the solution of the pentachloride in benzene. Re flux 2MoCl5  → 2MoCl4 + Cl2 Benzene

It is also obtained by treatment of MoO2 with CCl4 at 250°C. CCl 4 MoO2 + → MoCl4 + CO2 250°C

3. Complexes Dark green [MoCl6]2– is obtained by treatment of the pentachloride with alkali metal chloride in ICl as solvent. It is also obtained by treatment of the tetrachloride and tetraethyl ammonium chloride in MeCN. KCl

R 4 NCl → MoCl62– ← Mo2Cl10   MoCl4 ICl Me CN

20.5.8 Chemistry of Molybdenum (+III) The oxide of molybdenum (+III) is not known. However, Mo(OH)3 has been obtained by addition of alkali to the solution of reduced molybdate. It is soluble in acids. MoF3, a yellowish brown non-volatile solid, is obtained by heating Mo with MoF6 at about 400°C. MoCl3 is obtained either by treatment of the trihydroxide Mo(OH)3 with HCl or by reduction of the pentachloride with hydrogen at 250°C. It is a dark red solid which is insoluble in cold water. Molybdenum forms a large number of complexes. The solution of MoO3 in concentrated HCl, on prolonged electrolytic reduction, gives [MoCl6]3–. The complex ion is readily aquated in dilute solution and gives the yellow air-sensitive [Mo(H2O)6]3+. Some mixed complexes of Mo(+III) are also known such as [MoCl3py3] and [MoCl4bipy]–.

20.5.9 Chemistry of Molybdenum (+II) The most common compound is brown dichloride obtained by heating the trichloride in an atmosphere of CO2. 2MoCl3 $ MoCl2 + MoCl4 It can also be prepared by passing chlorine over molybdenum carbonyl or the metal itself. Mo(CO)6 Cl 2 → Mo(CO)4Cl2 ∆ → MoCl2 Molybdenum carbonyl reacts with carboxylic acids to give yellow Mo2(OOCR)4 crystals which on dissolving in very concentrated HCl yield [Mo2Cl8]4– ion containing an Mo–Mo quadruple bond.

20.10 Inorganic Chemistry

20.6

TECHNETIUM (Tc)

20.6.1 Occurrence and Preparation Technetium is the first human made element and does not occur in nature. Tc97 was made in 1937 by irradiation of molybdenum with deuterons and was named technetium by their discoverers—Perrier and Serge. All isotopes of technetium are radioactive. Tc99( , 2.12 × 105 years) has been obtained on macroscale as one of the fission products of uranium. The metal can also be obtained by reduction of NH4TcO4 with H2. The metal crystallises in hcp arrangement. Technetium finds no commercial use although some Tc compounds are used in radiographic scanning. The TcO4– ion can be used as an excellent corrosion inhibitor for steels.

20.6.2 Properties The metal resembles platinum in its appearance. It is quite unreactive and has no action with H2O or nonoxidising acids but dissolves in oxidising acids such as concentrated nitric acid and concentrated sulphuric acid to form the oxo acid HTcO4. The massive metal gets tarnished in moist air and forms the oxo acid. It burns in oxygen to give Tc2O7 and forms TcF5 with F2.

20.6.3 Chemistry of Technetium The metal differs considerably from manganese in its chemical properties. It has no cationic chemistry like Mn and does not form any compounds in the characteristic (+II) oxidation state of Mn. Rather (+VII) is the most common and most stable oxidation state for Tc and is only slight oxidising. The (+VI) state is unstable and undergoes disproportionation.

1. Oxides Three oxides of technetium are known, viz. TcO2, TcO3 and Tc2O7. TcO3 is unstable as supported by the standard reduction potential data and is obtained by heating TcO3Br. -0.737 V

TcO4–

+0.698 V

(TcO3)

+0.272 V +0.757 V

+0.144 V

TcO2

(Tc2+)

+0.400 V

Tc

+0.50 V

The heptoxide, Tc2O7, is obtained by heating the metal or the acidic solution of the pertechnate ion, TcO4–. It is a pale yellow volatile solid and consists of linear Tc–O–Tc chains formed by sharing of one oxygen atom of the TcO4 tetrahedra. It dissolves in water to form pertechnic acid. The dark red crystals of the anhydrous acid can be obtained by evaporation of these solutions. Tc2O7 + H2O $ 2HTcO4 It gives crystals of ammonium and potassium pertechnate with ammonium hydroxide and potassium hydroxide respectively. The pertechnates are more stable than the permanganates and are not oxidising. The solution of TcO4– in hydrochloric acid can be reduced with zinc to give hydrated dioxide TcO2.2H2O. The anhydrous dioxide can be obtained by thermal decomposition of NH4TcO4 or by heating the heptoxide in presence of the metal at 200–300°C. It is the most stable oxide and sublimes at 1000°C. It has a distorted rutile structure. Saturation of concentrated hydrochloric acid solution of TcO4– with H2S yields the black heptasulphide, Tc2S7. The heptasulphide can be heated with sulphur in vacuum to yield the disulphide, TcS2.

Chemistry of Elements of 4d Series

20.11

2. Halides The golden yellow hexafluoride, TcF6, has been obtained by the fluorination of the metal at 400°C. It hydrolyses to give a black precipitate of hydrous dioxide. The paramagnetic red crystals of TcCl4 are obtained by the treatment of Tc2O7, with CCl4 or by the direct chlorination of the metal. Some oxohalides such as TcOF4 (blue), TcO3F(yellow), TcOCl3(brown), TcOCl4(purple), TcO3Cl(colourless) and TcOBr3(brown) are also known.

3. Complexes The most important complex is K2[TcCl6], obtained by reduction of TcO4– with KI in presence of excess of concentrated HCl. It exists as yellow octahedral crystalline solid. TcCl4 reacts directly with suitable reagents to form complexes such as TcCl4(PPh3)2, [TcCl4dipy] and [TcCl2dipy2] Cl2.

20.7

RUTHENIUM (Ru)

20.7.1 Occurrence and Extraction Ruthenium is a 44th most abundant element with an abundance of the order of 10–7% in the earth’s crust. It is found mainly with the coinage metals and platinum metals. They are also found as alloys in the form of osmiridium. It is extracted mainly from the nickel-copper sulphide ores containing trace amounts of ruthenium. The ore is concentrated by gravitation and froth floatation process. The concentrated ore is smelted in the presence of lime, coke and sand and is bessemerised to yield Ni–Cu sulphide matte. The matte is cast into anodes and is refined by electrolysis in sulphuric acid solution. Ni is electrodeposited and the anode slime containing the mixture of platinum metals is further processed to give the massive metal using special techniques. The metal can also be extracted from its alloy. The alloy is heated with a mixture of KOH and KNO3 to obtaine potassium ruthenite, K2RuO4. The mixture is acidified and boiled to remove insoluble OsO4. The solution is made alkaline and is distilled to give RuO4. This is reduced in a current of hydrogen to give Ru.

20.7.2 Properties Ruthenium is a greyish white, fairly hard metal which melts at 2310°C. It is unaffected by acids or alkalis but dissolves on fusion with alkalis in presence of an oxidising agent such as KClO3, Na2O2, KNO3, etc. It combines with nonmetals on heating to give RuO2, RuS2, Table 20.3 Oxidation states of ruthenium RuF3 and RuCl3. Oxidation State

Examples

20.7.3 Uses

(–II)

[Ru(CO)4]2–

Ruthenium is used mainly to alloy with Pd and Pt so as to increase hardness and inertness. These alloys are used in fountain-pen tips and gramophone needles.

(0)

[Ru(CO)5]

(+I)

[h5(C5H5)Ru(CO)2]2

(+II)

RuCl2(PPh3)3

20.7.4 Chemistry of Ruthenium

(+III)

[Ru(NH3)5Cl]2+, K3RuF6

The outermost electronic configuration of ruthenium is 4d75s1. It shows oxidation states ranging from (–II) to (+VIII) in its compounds. Ruthenium also forms a large number of complexes and organometallic compounds with -acid ligands. The various oxidation states are listed in Table 20.3.

(+IV)

RuO2, K2RuCl6

(+V)

RuF5, KRuF6

(+VI)

[RuO4]2–, RuF6

(+VII)

[RuO4]–

(+VIII)

RuO4

20.12 Inorganic Chemistry

1. Oxides Blue-black RuO2 is obtained by the action of oxygen on Ru at 1250ºC or RuCl3 at 500–700°C. Orange-yellow RuO4 is obtained when acidified ruthenium solutions are treated with oxidising agents (Cl2 or MnO4–). The hydrous oxide, Ru2O3.nH2O, is precipitated when RuO4 is reduced or an aqueous metal solution is treated with alkalis.

2. Halides The most important and most stable is the hexafluoride obtained by direct reaction of the elements. It is extraordinarilly reactive and corrosive and reacts with glass even at room temperature. Lower fluorides are obtained either by the thermal dissociation or by irradiation of the hexafluoride. Other halides are normally obtained by direct interaction of the elements under selected conditions. The lower halides are usually obtained by thermal dissociation of the higher halides.

3. Complexes Ruthenium forms a vast array of complex compounds with almost all types of ligands; however, the complexes with -acid ligands are worth mentioning. There is an extensive and important chemistry of higher oxidation states yet the lower oxidation states are equally important. Ru forms mononuclear and polynuclear carbonyls in the lower oxidation states (0), (–I), (+I) which undergo further substitution and protonation reactions. The (+2) state is represented by the aqua-ion [Ru(H2O)6]2+ which is readily oxidised to [Ru(H2O)6]3+. A number of complexes with -acid ligands are formed such as [Ru(NH3)6]2+ and [Ru(CN)6]4–. Ru shows the particular affinity towards NO and forms a number of mixed complexes such as [Ru(NO)Cl5]2– and [Ru(NO)CN5]2–. Ru halides dissolved in strong ammoniacal solution on reduction with zinc dust yield the orange hexammine, [Ru(NH3)6]3+. It is strongly reducing and undergoes various substitution reactions, e.g.

RuO4 is converted to per-ruthenate and ruthenate on reduction with hydroxides. 4RuO4 + 4OH– $ 4RuO4– + H2O + O2 (Green)

4RuO4 + 4OH $ 4RuO42– + 2H2O + O2 –

(Deep orange)

20.8

RHODIUM (Rh)

20.8.1 Occurrence and Extraction Rhodium is a rare element and occurs with an abundance of 10–7ppm in the earth’s crust. Trace amounts of rhodium are found alloyed with platinum metals and the coinage metals. It is also extracted from the anode slime obtained during electrolytic refining of nickel. It is also found in nature as an alloy of Ru, Os and Ir and is extracted by fusion of the alloy with KHSO4 to dissolve Rh as KRh(SO4)2. This compound is extracted with hot water and is separated by crystallisation. The crystals are decomposed in air at about 470 K to obtain the metal.

Chemistry of Elements of 4d Series

20.13

20.8.2 Properties Rhodium is the silvery white metal which is soft and ductile. It is much more noble and unreactive in composition to other platinum metals. It is resistant towards acids but dissolves in hot concentrated HCl and hot concentrated H2SO4 at about 425 K. It can combine with oxygen or chlorine only at red heat.

20.8.3 Uses Due to its high inertness, it is used to alloy with platinum for the manufacturing of various scientific instruments and apparatus. Table 20.4 Oxidation states of rhodium

20.8.4 Chemistry of Rhodium It does not from any oxo anion or volatile oxides and shows the oxidation states ranging from –I to +VI but +V state is not known. Various oxidation states with examples for rhodium are listed in Table 20.4.

Oxidation State

Examples

(–I)

[Rh(CO)4]–

(0)

[Rh6(CO)16]

(+I)

[Rh(CO)2Cl]2

1. Halides

(+II)

[Rh(OCOR)2]2

RhF6 is the highly unstable and most reactive halide of rhodium with +VI oxidation state. However, the trihalides are comparatively more stable.

(+III)

[Rh(H2O)6]3+, RhF3, RhCl63–

(+IV)

K2RhF6

(+VI)

RhF6

2. Complexes The (0) oxidation state is exhibited in [Rh4(CO)12] with Rh–Rh bond in a cluster of four Rh atoms. The (+I) state is exhibited in a number of complexes with -acid ligands such as CO, PPh3 and alkenes. The most important compound is Wilkinson’s catalyst, [Rh(Cl)(PPh3)], obtained by refluxing hydrated trichloride with triphenyl phosphene. It is red violet in colour and has a square planar structure. It is used as an important catalyst in the hydrogenation of alkenes and Oxo process. The other (+I) complexes are also obtained by reduction of hydrous trichloride in presence of complexing ligands or by reduction of rhodium (III) complexes with suitable complaxing legends. The complexes of rhodium (+III) are very stable and are mostly diamagnetic with octahedral geometry. Rhodium gives a stable yellow aquo-ion [Rh(H2O)6]3+ obtained by dissolution of the trioxide in cold mineral acids. Rhodium is quite stable in aqueous solution even more than [Rh(H2O)6]3+ indicating that aqueous solution of rhodium are more stable than the Co(+III) solutions. Reduction of hydrated trichloride in concentrated sulphuric acid solutions by zinc in presence of NH4OH give the white crystalline salts, [RhH(NH3)5] SO4 containing hydride as a ligand. [RhF6]2– and [RhCl6]2– are obtained when RhCl3 and alkali chlorides are treated with F2 or Cl2.

20.9

PALLADIUM (Pd)

20.9.1 Occurrence and Extraction Palladium is a rare and expensive element but more abundant than the other platinum group metals (Pt, Ru, Os, Rh and Ir). It occurs naturally in the alloyed form with metals such as gold, copper, nickel and iron. It is also found as traces in the sulphide ores of copper and nickel. The chief source is the Sudbury nickel ore. It contains platinum and palladium in equal quantities. The metal can be extracted from its alloys. The method is based on the principle that palladium forms an insoluble cyanide which can be separated out. The alloy is fused with potassium cyanide and is extracted with water to obtain the precipitation of Pd(CN)2. This is ignited to get Pd of almost 99.5 percent purity.

20.14 Inorganic Chemistry

The metal is generally extracted from the Sudbury ore. The ore is concentrated to remove the sulphides of iron, copper and nickel. The concentrated ore is digested with dilute aqua regia under pressure to obtain the chlorides of platinum and palladium. The solution obtained is treated with ammonium chloride to obtain the yellow precipitates of ammonium chloroplatinate, (NH4)2PtCl2. The precipitates are separated and ignited to obtain spongy platinum while the solution left behind containing palladium dichloride is concentrated in presence of NH3 and cooled to yield crystals of tetrammine palladium (II) chloride, [Pd(NH3)4]Cl2.H2O. The crystals are dissolved in water and the solution is acidified with a small amount of HCl to obtain a yellow precipitate of dichloroamminepalladium (II), [PdCl2(NH3)2]. The precipitates are again washed with water and dissolved in NH3 to form soluble tetramine palladium (II) chloride. It is crystalline and the process is repeated a number of times to obtain a pure product which is decomposed by heating strongly to give palladium as a spongy metal or fine powder. HCl ���� � [Pd(NH3)4]Cl2.H2O � ��� � [PdCl2 (NH3)2] NH 3

20.9.2 Properties Palladium is a grey-white metal which melts at 1825 K. The metal is malleable and ductile. It is unreactive in the massive state and doesn’t tarnish with air or water at ordinary temperatures. However, it reacts with air on heating and forms a blue film of PbO at its surface. It combines with chlorine at red heat and with sulphur at high temperature. It dissolves slowly in concentrated HCl in presence of O2 or Cl2 but readily in concentrated HNO3 to give [Pd(NO3)2(OH)2]. It forms Pd(NO3)2 in dilute HNO3 and PdSO4.2H2O in dilute H2SO4 in presence of a little HNO3. It forms H2PdCl4 with aqua regia which on heating gives PdCl2. It reacts rapidly with fused alkali metal oxides and peroxides. It absorbs large volumes of gaseous H2. When red hot Pd is cooled in H2, the volume of Pd absorbs 935 volumes of H2, which is more than any other metal. It cannot absorb other gases and hence is used to purify H2.

20.9.3 Uses It is used alloyed with gold as a substitute for platinum under the name white gold. It is extensively used in chemical processes as catalyst. Nowadays it is used in three way catalylic convertors to convert unburnt fuel and exhaust gases into CO2 and N2. PdCl2 is used in the wacker process for conversion of C2H4 into CH3CHO. It is used in many hydrogenation and dehydrogenation reactions.

20.9.4 Chemistry of Palladium The outer-shell electronic configuration of Pd is 4d10. It shows the oxidation state of (0), (+II) and (+IV). The most common oxidation state is (+II). Pd(0) does not form any simple carbonyl complexes but many triphenylphosphine complexes are known. This is due to the poorer tendency of Pd to - bonding, but presence of a better -donor triphenylphosphine ligand makes the compounds relatively stable, for example [Pd(CO)(PPh3)3]. Some cluster compounds are also known such as [Pd3(CO)3(PPh3)]. The most important complexes are with triphenylphosphine formed by action of hydrazine on ethanolic solutions of K2PdCl4 to obtain white crystalline Pd(PPh3)3 or Pd(PPh3)4. These are extensively used in the oxidation–addition reaction. The (+II) oxidation state is characterised by many compounds such as oxides, halides, nitrates and sulphates. These compounds are anhydrous and covalent, except PdF2 which is ionic. Palladium (II) oxide, PdO, is obtained as a black powder by heating palladium nitrate or the spongy metal at 1070 K in oxygen. It is insoluble in all acids and aqua regia. It decomposes on heating to liberate oxygen and thus is used as a strong oxidising agent. It forms the aqua ion in dilute noncomplexing acids and brown deliquescent salts such as [Pd(H2O)4](ClO4)2 can be obtained. The aqua ion is diamagnetic and is believed to have a square planar structure. All palladium dihalides are known, PdCl2 being the most important. The

Chemistry of Elements of 4d Series

20.15

halides are obtained by direct interaction of the elements and are polymeric. PdCl2 exists in two forms:

Pd + Cl2

> 550°C

β-(PdCl2)n

Cl

Dark red

α-(PdCl2)6 The -form has a flat polymeric structure, but the -form is based on the Pd6Cl12 unit. PdCl2 dissolves in aqueous HCl to form yellowish PdCl42– which is used in many catalytic reactions. PdCl2 can be directly dissolved in many ligands to obtain complexes such as [PdCl2(C6H5CN)2], [Pd(NH3)4]Cl2 and [PdCl2(NH3)2]. PdCl2 is used for oxidation of alkenes. Two important examples are the formation of acetaldehyde from ethane and acetone from propane. It also catalyses the formation of ethanoic acid from ethane (Fig. 20.6 and 20.7). CH2 = CH2 + CO + H2 $ CH3CH2COOH

Cl Pd

Pd

< 550°C

Cl

Pd Cl

Fig. 20.6 Structure of ( -PdCl2)n Cl Cl

Cl Cl Pd

Cl

Cl Pd

Pd Cl

Cl

Pd Cl

If KCN is added to the aqueous solution of PdCl2, Pd(CN)2 Cl Cl is obtained. The precipitate dissolves in excess of KCN to form Cl K2[Pd(CN)4].3H2O. Fig. 20.7 Structure of ( -PdCl2)6 The (+IV) state is characterised by PdF4. It is obtained in combination with PdII[PdIVF6] [also reported as PdF3) by direct reaction of Pd and F2 at 500°C. PdO2 is known to exist only in the hydrated form. Pd(+IV) complexes are found to be more stable than simple compounds. Except [Pd(NO3)2(OH)2], the other complexes are mainly octahedral. Pd dissolves in aqua regia to give the red [PdCl6]2– ion which is also obtained by passing Cl2 through the [PdCl4]2– solutions. [PdCl6]2– decomposes Cl Cl H3N H3N in water to give [PdX4]2– and halogens. K2[Pd(CN)6] is obtained by oxidation of K2PdCl4 by persulphate in Pd Pd presence of KCN.[Pd(NH3)4]Cl2 is a yellow solid and NH3 Cl Cl exhibits two isomeric forms. The cis-form is commonly H3N Fig. 20.8 cis-form and trans-form of platin known as cis-platin and is used in cancer treatment.

20.10

SILVER (Ag)

Silver has been used since ancient times as a coinage metal, since it is found in native state in the nature.

20.10.1 Occurrence and Extraction Silver is a rare element and is the 66th most abundant element in the earth’s crust occurring to the extent of 0.08 ppm by weight. Native silver is usually found in alluvial sands in association with copper and gold. It is also found as sulphide ores, argentite or silver glance (Ag2S), pyragyrite or ruby silver ore (Ag2S.Sb2S3) and as the chloride ore, chlorargyrite or horn silver (AgCl). Silver is extracted from its ores by the cyanide process. In this process, the crushed ore is concentrated by froth floatation process and is leached with a dilute solution of sodium cyanide for several hours. The mixture is agitated by a current of air so that all the silver present in the solution forms a complex cyanide, sodium argentocyanite, Na[Ag(CN)2].

���� � Ag2S + 4NaCN � ��� � 2Na[Ag(CN)2] + Na2S The reaction proceeds to completion by oxidising the sodium sulphide to sodium sulphate.

20.16 Inorganic Chemistry

2Na2S + 2O2 + H2O $ Na2S2O3 + 2NaOH Na2S2O3 + 2NaOH + 2O2 $ 2Na2SO4 + H2O The solution is filtered and treated with zinc dust or scrap to precipitate the metal. 2Na[Ag(CN)2] + Zn $ Na2[Zn(CN)4] + 2Ag The precipitated silver is filtered and fused with borax to get a compact mass. The impurities (traces of zinc, copper and gold) can be removed by electrolytic refining.

20.10.2 Properties Silver is a white lustrous, heavy metal with highest electrical and thermal conductivity. It is highly malleable ductile and melts at 961°C. Molten silver absorbs oxygen which is expelled out on cooling and results in violent spurting of silver globules known as spitting of silver. That is why the molten metal is covered by a layer of charcoal so as to avoid this phenomenon. Silver is less reactive than copper. Ag+ + e– $ Ag

Eº = +0.80 V

The electron potential data reflects that silver is weakly electropositive and cannot displace hydrogen from dilute acids. Thus, silver is inert towards dilute HCl and H2SO4 but it dissolves in conc. hydrochloric acid, in the presence of oxidising agents to form silver chloride, while SO2 is liberated with hot conc. H2SO4. It dissolves readily in dilute nitric acid to liberate nitric oxide, while nitrogen dioxide is evolved with conc. nitric acid. 3Ag + 4HNO3(dil.) $ 3AgNO3 + NO + 2H2O Ag + 2HNO3(conc.) $ AgNO3 + NO2 + H2O Silver is inert towards dioxygen, but tarnishes slowly in air due to reaction with traces of H2S present in the air. That is why polished silver articles turn black after some time. Silver also reacts with sulphur, hence sulphur-containing food (mustard and egg yolk) tarnish silver black. Silver is unaffected by water and aqua regia. It dissolves in the solution of sodium cyanide or potassium cyanide in presence of air due to formation of the complex ion. – – ���� � 2Ag + H2O + O2 + 4CN– � ��� � 2[Ag(CN)2] + 2OH

Silver directly combines with halogens to form halides.

20.10.3 Uses Silver is alloyed with copper for making coins, jewellery, silverware and decoration pieces, etc. It is used for silver plating of metallic articles for silvering mirrors. Silver compounds are used in photographic emulsions. Table 20.5 Oxidation states of silver

20.10.4 Chemistry of Silver 10

1.

The outer-shell electronic configuration of Ag is 4d 5s It shows the oxidation states from (+ I) to (+III) in its simple compounds and complexes. However, only the (+I) state is the most stable state and the other two states are reduced readily to (+ I) state. Table 20.5 lists the oxidation states of silver with examples.

Oxide state

Examples

(+I)

[Ag(CN)2]–, [Ag(NH3)2]+, AgF, AgCl, Ag2O

(+II)

[Ag(py)4]2+, AgO, AgF2

(+III)

[AgF4]–, (Ag2O3)

Chemistry of Elements of 4d Series

20.17

The (+I) state is the most dominant oxidation state and the argentous ion, Ag+ dissolves in aqueous solution. However, the aqua ion is not found in salts and silver salts are anhydrous, except AgF.4H2O. Practically, only AgNO3, AgF, AgClO3 and AgClO4 are water soluble. Other Ag (+I) salts are insoluble in water. Ag2O is obtained as a dark brown precipitate, when alkali hydroxides are added to Ag (+I) solutions. It is a basic oxide and dissolves in acids to form corresponding silver salts. Its aqueous suspensions are alkaline. It is more soluble in strongly alkaline solutions and results in the formation of AgOH and [Ag(OH)2]–. Ag2O dissolves in solutions of alkali cyanides, ammonia and thiosulphates and results in the formation of complex ions [Ag(CN2)]–, [ Ag(NH3)2]+ and [ Ag(S2O3)2 ]2– respectively. All argentous halides are well known and precipitated by the addition of a soluble halide to Ag (+I) solutions, except AgF, which is prepared by treatment of the oxide or carbonate of the metal in HF. AgF is soluble in water due to its very low lattice energy. On the other hand, the precipitates of AgCl (white), AgBr (pale-yellow) and AgI (yellow) are insoluble in water. The solubility in water increases as Cl < Br < I. This is due to the reason that as the van der Waal’s attraction increases, number of electrons increase and results in increase in lattice energy and decrease in solubility of the compound. Thus, AgCl dissolves in aq. NH3 to form [Ag(NH3)2]+ while AgBr and AgI cannot dissolve in aq. NH3. However, all three dissolve in liquid ammonia, KCN solution and hypo-solution to form complex salts. + ���� � Ag+ + NH3 (liquid) � ��� � [Ag(NH3)2] – + ���� � Ag+ + 2KCN � ��� � [Ag(CN)2] + 2K 3– + ���� � Ag+ + 3Na2S2O3 � ��� � [Ag(S2O3)2] + 3Na

AgCl reacts with hypo-solution to form a series of complexes such as [Ag(S2O3)]3– and [Ag(S2O3)2]3–. Similarly, with NH3, a series of salts are formed such as AgCl.NH3, 2AgCl.3NH3 and AgCl.3NH3. AgNO3 is a very important Ag (+I) compound. It is obtained by dissolving silver in dilute nitric acid. It dissolves in ammonia to form the complex [Ag(NH3)2]NO3. When AgNO3 is treated with a very dilute solution of Na2S2O3, a white precipitate is obtained which changes to yellow, brown and finally turns black due to the formation of silver sulphide. 2O 2Ag+ + S2O32– $ Ag2S2O3 H → Ag2S + H2SO4

(White)

(Black)

However, with conc. solutions of Na2S2O3, the initially formed silver thiosulphate dissolves in excess of Na2S2O3 to form a complex salt. 2− 3– 2 O3 Ag2S2O3 S → [Ag(S2O3)2] The (+II) oxidation state is characterised by two compounds, AgF2 and AgO. The dark brown AgF2 is obtained by heating AgF or other silver compounds in F2. It is readily hydrolysed by moisture and is used as a fluorinating agent. AgO is obtained by passing ozone over the metal. It readily decomposes to liberate oxygen and acts as a strong oxidising agent. 2AgO $ Ag2O + O The paramagnetic argentic ion, Ag2+, is obtained by dissolution of AgO in acid or by oxidation of Ag+ with ozone in HNO3 or HClO4 solution. Many complexes of Ag (+II) are known such as [Ag(py)4]2+, [Ag(dipy)2]2+ and [Ag(phen)2]2+. These are obtained by persulphate oxidation of Ag+ solutions in presence of the complexing ligands. The (+III) oxidation state is represented by some complexes. The most readily obtained is K6H[Ag(IO6)2] obtained by the persulphate oxidation of Ag2O in presence of periodate ion in strongly alkaline medium. The less readily obtained is the yellow fluoro complex, KAgF4, obtained by treatment of a stoichiometric mixture of the KCl and AgNO3 with F2 at 300°C.

20.18 Inorganic Chemistry

20.11

CADMIUM (Cd)

20.11.1 Occurrence and Extraction Cadmium is a rare element and is the 65th most abundant element, occurring to the extent of 0.16 ppm by weight in the earth’s crust. Cadmium does not occur free in nature and its ore greenockite, CdS, is very rare. It is found in trace amounts (2–3 parts per thousands) in the zinc ores. It is extracted as a by-product in the metallurgy of zinc. The cadmiferrous zinc obtained is distilled to obtain the more volatile cadmium first, partly in the form of oxide. This sample is purified by electrolytic refining using dilute H2SO4.

20.11.2 Properties Cadmium is a soft silvery solid which is heavier than zinc and is slightly more malleable and ductile. It is less reactive than zinc being less electopositive. Cd2+ + 2e– $ Cd

Eº= 0.403V

Cd resembles Zn in many properties. It gets tarnished in moist air and forms oxides, sulphides and halides on heating like zinc. It is however not amphoteric like zinc. Its sulphide can be precipitated from a weakly acidic solution unlike that of zinc.

20.11.3 Uses Cadmium is used to prepare many fusible alloys such as rose metal and wood’s metal. It is added to steel to prevent it from rusting. It is used to make control rods to absorb neutrons in nuclear reactors. It is used in Ni/ Cd storage batteries. CdS is used in paint as an expensive yellow pigment.

20.11.4 Chemistry of Cadmium The outer-shell electronic configuration of Cd is 4d105s2. It shows the only oxidation state as (+II). CdO is obtained by igniting the metal or its hydroxide, carbonate or the nitrate. It may be yellow, green or brown at room temperature, but is white at liquid air temperature. It is mainly basic but dissolves in strongly alkaline medium to form [Cd(OH)4]2–. On addition of a base to the solution of the cadmium salts, white precipitates of Cd(OH)2 are obtained. When Cd(OH)2 is treated with H2O2, peroxides are formed with variable compositions. It is insoluble in bases, but readily dissolves in excess of concentrated NH3 to form the complex [Cd(NH3)4]2+ All the dihalides are known and are colourless except for CdBr2 which is pale yellow. Except CdF2, other halides dissolve in aqueous medium with self-complexing. Thus, solution of CdX2 is regarded as the system containing a mixture of hydrated Cd2+, CdX+, [CdX3]– and [CdX4]2– all in equilibrium. The salts of oxo acids are soluble in water. In absence of any complexing agent and in presence of perchlorate, CdOH+ ion is formed (in solution below 0.1 M). + + ���� � Cd2+ + H2O � ��� � CdOH + H

In more concentrated solutions, Cd2OH3+ is formed. 3+ + ���� � 2Cd2+ + H2O � ��� � Cd2OH + H

In presence of complexing anions, species such as Cd(OH)Cl or CdNO3+ may be formed. Except fluoro, the complexes of other halides have been obtained as [CdX4]2–. [CdCl5]3– has also been formed with trigonal bipyramidal geometry.

Chemistry of Elements of 4d Series

20.19

Unstable species of Cd(+I) can be obtained by irradiating the aqueous solutions of Cd2+ with the formula Cd+.Cd22+ can be obtained by adding AlCl3 to the Cd–CdCl2 melt. The second transition series is constituted by ten elements, yttrium (Y), zirconium (Zr), niobium (Nb), molybdenum (Mo), technetium (Tc), ruthenium (Lu), rhodium (Rh), Palladium (Pd), silver (Ag) and cadmium (Cd). Yttrium forms oxo salts which decompose on slight heating to its oxide, Y2O3. It shows little tendency to form complexes due to its fairly large size. Zirconium is extracted from its ore by Knoll process and extremely pure zirconium is obtained by the van Arkel method. The most common oxidation state of zirconium is (+IV), represented by ZrO2 and ZrX4 (X = F, Cl, Br and I). ZrO(NO3)2 is used to remove interferring PO43– ions in qualitative analysis. Niobium is extracted from columbite–tantalite series of minerals. Its most important oxidation state is (+V), represented by Nb2O5, NbX5 and NbOX3 (X = F, Cl, Br, I). The pentahalides exist in polymeric structure in the solid state. Niobium forms a series of complexes such as [NbF6]–, [XbF7]2–, [NbF6]9– and [NbCl6]–. Molybdenum is extracted from its sulphide ore, molybdonite (MoS2). Its most common oxidation state is (+VI) but (+V) is the most stable oxidation state in aqueous solutions. MoO3 is not attacked by acids, expect HF. It forms blue oxides with the composition MoO2.0(OH) and MoO2.5(OH)0.5. MoF6 is readily hydrolysed and yields MoOF4 on partial hydrolysis, while MoO3 is obtained on complete hydrolysis. Molybdenum forms molybdates and polyacids. Ammonium molybdate is used to test phosphate ions. Reduction of molybdates or MoO3 in acid solution yields the emerald green ion (MoOCl5)2– which gives dinuclear species on addition of solid NaOH. Technetium is the first human-made element. It was made by irradiation of molybdenum with deuterons. Its most common oxidation state is +VII represented by Tc2O7. Tc2O7 dissolves in water to form pertechnic acid. Its most important complex is K2[TcCl6]. Ruthenium exists in the form of osmiridium and in trace amounts along with nickel–copper sulphide ores. It is mainly extracted from the anode slime obtained during electrolysis of nickel copper sulphide matte. Its most important compound is RuF6, highly reactive and corrosive. It forms a number of complexes such as [Ru(NH3)6]2+, [Ru(H2O)6]2+ and [Ru(CN)6]4–. Rhodium exists in nature as an alloy of Ru, Os and Ir. RhF6 is the highly unstable and most reactive halide of rhodium. Rh forms diamagnetic and highly stable complexes in (+III) oxidation state, mainly represented by the yellow aquo ion [Rh(H2O)6]3+. Palladium is extracted mainly from the Sudbury nickel ore. Its most common oxidation state is (+II). PdO is used as a strong oxidising agent as its decomposes on heating to liberate oxygen. PdCl2 exists in two forms: [ -PdCl2]n and [ -PdCl2]6. PdCl2 is used as an important catalyst in many organic reactions. cisplatin is used in cancer treatment. Silver is extracted from its sulphide ore by the cyanide process.. Its most common oxidation state is (+I) and the argentous ion (Ag+) is highly solvated in aqueous solutions, but is not found in salts. Cadmium is extracted as a by product in the metallurgy of zinc.. Its most common oxidation state is (+II) mainly represented by CdO and CdX2 (X = F, Cl, Br and I). It also forms [CdOH]+ and [Cd2OH] 3+.

20.20 Inorganic Chemistry

EXAMPLE 1 Complete the following reactions: (a) Nb2O5 + H2 $ (d) RuF5 + I2 $

(b)

MoF6 + H2O $

(e) CH2 = CH2 + CO + H2

(c)

MoF6 + Mo(CO)6 $

PdCl2

(a) Nb2O5 + H2 $ 2NbO2 + H2O (b) MoF6 + H2O Partialhydrolysis → MoOF4 + 2H (c) MoF6 + Mo(CO)6 $ MoO3 + xH2O.4HF 250°C

(d) 5RuF5 + I2 → 5RuF3 + 2IF5 Pd Cl

2 (e) CH2 = CH2 + CO + H2  → CH3CH2COOH

EXAMPLE 2 What happens when RuO4 is treated with alkali hydroxide? RuO4 is reduced to per ruthenate and ruthenate on treatment with alkali hydroxides. 4RuO4 + 4OH– $ 4RuO4– + H2O + O2 Perruthenate (Green) 4RuO4 + 4OH– $ 4RuO42– + 2H2O + O2 Ruthenate (Deep orange)

EXAMPLE 3 Give reason why PdCl2 dissolves in excess of KCN. PdCl2 gives precipates of Pd(CN)2 when treated with KCN which dissolves in excess of KCN to give K2[Pd(CN)4].3H2O. PdCl2 + 2KCN $ Pd(CN)2 + 2KCl Pd(CN)2 + 2KCN(aq.) $ K2[Pd(CN)4].3H2O (excess)

EXAMPLE 4 Discuss the action of silver with dilute and concentrated nitric acid. Silver dissolves readily in dilute and concentrated nitric acid to liberate nitric oxide. 3Ag + 4HNO3(dil.) $ 3AgNO3 + NO + 2H2O However, with conc. nitric acid, nitrogen dioxide is evolved. Ag + 2HNO3(conc.) $ AgNO3 + NO2 + H2O

EXAMPLE 5 What happens when AgCl is treated with (a) KCN solution?

(b) hypo solution?

(a) AgCl dissolves in aq. KCN solution due to the formation of K[Ag(CN)2].

���� � AgCl + 2KCN � ��� � K[Ag(CN)2] + KCl (b) AgCl dissolves in hypo solution to form a series of complexes such as Na3[Ag(S2O3)] and Na3[Ag(S2O3)2]. ���� � AgCl + 3Na2S2O3 � ��� � [Ag(S2O3)2]3– + 3Na+

Chemistry of Elements of 4d Series

20.21

QUESTIONS Q.1. Discuss the occurrence, extraction and properties of the following elements. (a) Molybdenum (b) Palladium Q.2. Describe the important properties and uses of the following elements. (a) Zirconium (b) Silver Q.3. Write a short note on alloy steel and titanium alloy. Q.4. Give reasons for (a) Titanium cannot be extracted from its oxide by reduction with carbon. (b) Silver chloride is soluble in ammonia but sparingly soluble in water. Q.5. How will you obtain the following elements in ultra pure state from their crude samples? (a) Zirconium (b) Titanium Q.6. Write short note on oxides and halides of technetium. Q.7. How will you prepare the following compounds? (a) Paladium (II) chloride (b) Sodium titanate (c) [Cd(NH3)4](OH)2 Q.8. What happens when (a) Sodium hydroxide is treated with cadmium chloride? (b) Silver nitrate is treated with dilute hypo solution? (c) MoO3 is treated with excess of hot concentrated ammonia? Q.9. Discuss the structure of (a) ZrO2 (b) NbOCl3 (c) β-PdCl2 Q.10. Discuss the aqueous chemistry of the following elements (a) Zirconium (b) Ruthenium

MUTLIPLE-CHOICE QUESTIONS 1. The compound formed by hydrolysis of ZrCl4 is (a) ZrO2 (b) ZrO(OH)2 (c) Zr(OH)3 (d) ZrOCl2 2. The products obtained by ignition of Y2(SO4)3 are (a) Y2O3 (b) SO2 (c) Y2SO4 (d) both a and b 3. The reagent used to test phosphate ions is (a) (NH4)2MoO4 (b) MoF6 (c) Na2S2O3 (d) None of these 4. Nb2O5, when treated with HF in presence of conc. HF, yields ‘A’ which on reduction with aluminium yields ‘B.’ ‘A’ and ‘B’ are (a) K3NbO4 and NbF3 (b) K2NbOF5 and Nb (c) K3NbO4 and Nb (d) K2NbOF5 and NbF3 H2 O 5. AgNO3 + Na2S2O3 $ A  → B. What are A and B? (a) Ag2S2O3 + Ag2S (b) Ag2S2O3 + Ag2O (c) Ag2S2O3 + Na2Ag(S2O3)2 (d) None of these

chapter

Chemistry of 5d Series

21

After studying this chapter, the student will learn about

21.1

INTRODUCTION

Excluding lanthanum (a member of lanthanide series), the other nine elements, namely hafnium (Hf), tantalum (Ta), tungsten (W), rhenium (Re), osmium (Os), iridium (Ir), platinum (Pt), gold (Au) and mercury (Hg) form the 5d series or the third transition series. The elements of the second and third transition series of a particular group have similar chemical properties and show some differences from their light congeners. This will be clear with the discussion of the chemistry of the individual members.

21.2

HAFNIUM (Hf)

21.2.1 Occurrence and Extraction Hafnium is the 45th most abundant element in the earth’s crust and occurs to the extent of 2.8 ppm by weight. Due to Lanthanide contraction, Hf is very similar in size (Zr, 1.45Å and Hf, 1.44Å) and properties of Zr. Hence, hafnium is found to the extent of 1–2% in all zirconium ores. Separation of Hf and Zr is extremely difficult, even more than separating the lanthanides. However, ion exchange chromatography or solvent extraction makes the separation possible quite satisfactorily. In the ion-exchange chromatography, the column of alcoholic solution of the tetrachlorides on silica gel is eluted with an alcohol/HCl mixture to obtain zirconium first. In the solvent extraction method, the nitrates are extracted into tri-n-butyl phosphate.

21.2.2 Chemistry of Hafnium The outer shell electronic configuration of Hf is 4f14 5d2 6s2. There is close resemblance between compounds of zirconium and hafnium.

21.2

21.3

Inorganic Chemistry

TANTALUM (Ta)

21.3.1 Occurrence and Extraction Tantalum is the 53rd most abundant element in the earth’s crust and occurs to the extent of 1.7 ppm by weight. Tantalum and niobium occur together; however, niobium is almost 10–12 times more abundant than tantalum. Tantalum is extracted mainly from the columbite tantalite series of minerals.

21.3.2 Properties Tantalum is identical in its covalent and ionic radius with niobium (covalent radius, 1.34 Å and ionic radius 0.72 Å for M3+). Consequently, the properties of these two elements are very similar. However, tantalum is more dense and melts at higher temperature in comparison to niobium.

21.4

TUNGSTEN (W)

21.4.1 Occurrence and Extraction Tungsten is identical to molybdenum in respect to occurrence (~10% abundance in the earth’s crust), metallurgy and properties. However, tungsten is found exclusively as tungstates. Wolframite (FeWO4 . MnWO4) is the principal ore of tungsten; other ores are scheelite (CaWO4), tungstenite (WS2), solzite (PbWO4) and cuproscheelite (CuWO4). The wolframite ore is concentrated by electromagnetic process to remove the nonmagnetic impurities. The concentrated ore is fused with sodium carbonate in presence of air to form sodium tungstate and the oxides of iron and manganese are precipitated. 4FeWO4 + 4Na2CO3 + O2 $ 4Na2WO4 + 4CO2 + 2Fe2O3 2MnWO4 + 2Na2CO3 + O2 $ 2Na2WO4 + 2MnO2 + 2CO2 The products are extracted with hot water and the insoluble Fe2O3 and MnO2 are filtered off. The remaining solution containing dissolved Na2WO4 is acidified so as to precipitate the hydrated tungstic oxide, WO3.xH2O. It is ignited and the product is reduced with hydrogen to obtain the metal. Na2WO4 + 2HCl $ 2NaCl + H2WO4 ∆

H2WO4  → WO3 + H2O WO3 + 3H2 1470K → W + 3H2O

21.4.2 Properties Tungsten is a silvery white lustrous metal. It is very hard and has the highest melting point (3380°C) of any metal, the next highest to carbon. It is as heavy as gold and has electrical conductance about 30% that of silver. Tungsten is inert towards oxygen, chlorine and dilute acids at ordinary temperature. However, at red heat it readily gives the trioxide and combines with chlorine on heating. It dissolves in HNO3/HF mixtures and also in fused alkalis or fused KNO3/NaOH.

21.4.3 Uses Due to its extremely high melting point and low volatility, tungsten is extensively used for lamp filaments. It is alloyed with steel to increase its hardness and strength. This steel is used to make cutting tools which retain the cutting edge even at red heat. Tungsten carbide, WC, is extremely hard and is used to make the tip for drills as well as for cutting of glass.

Chemistry of 5d Series

21.3

21.4.4 Chemistry of Tungsten The outer-shell electronic configuration of W is 4f145d46s2. It exhibits the oxidation states (0) and from (II) to (+VI), (+VI) being the most stable and (+III) strongly reducing. The lower oxidation states are shown in complexes with strong -acid ligands. Various oxidation states are listed in Table 21.1.

1. Oxides The most important oxide of tungsten is WO3. It is obtained as a yellow powder by heating the metal in oxygen. Its melting point is highest among the oxides of its group (CrO3 197°C, MoO3 795°C and WO3 1473°C). It is almost non-oxidizing and is insoluble in water. It has a slightly distorted ReO3 structure with WO6 octahedra sharing corners in three dimensions. WO3 on fusion with alkali or alkaline earth metal oxides gives the mixed oxide, K2W4O13. It has WO6 octahedra linked by corners resulting in the formation of sixmembered rings and K+ ions are present in the tunnels formed. Like molybdenum blue, tungsten blue is obtained by mild reduction of suspensions of WO3 in water.

2. Tungsten Bronzes Tungsten also forms tungstates like molybdates with Table 21.1 Oxidation states of tungsten the general formula M2WO4. The reduction of sodium Oxidation Examples tungstate, Na2WO4, with hydrogen or the vapour phase (0) [W(CO)6] reaction of alkali metals with WO3 gives a chemically (+II) W6Cl12, [n5 – C5H5W(CO)3Cl] inert and nonstoichiometric substance of general formula MnWO3 (n 1). The substance is deeply coloured with a (+III) [W2Cl9]3– bronzelike appearance and is known as tungsten bronze. (+IV) [W(CN)8]4– The colour of the substance varies with composition and for (+V) [WF6]– sodium compounds: blue-black for x ~ 0.3, red for x ~ 0.5, (+VI) WO42–, WOCl4, WCl6, WO3 orange for x ~ 0.7 and yellow x ~ 0.9. These are regarded + as defective NaWO3 phases with perovskite structure. Na occupies interstitial positions, some of which are vacant leading to defective lattice. As a result, there is slight change from W(+V) to W(+VI). The valence electrons from the alkali metal are distributed throughout the lattice imparting good electrical conductivity to these substances. These substances are insoluble in water and are oxidised by oxygen in presence of alkalis. 4NaWO3 + 4NaOH + O2 $ 4Na2WO4 + 2H2O Isopolytungstates and heteropolytungstates are formed similar to that of the molybdate systems.

3. Halides WF6 is the most stable halide of tungsten and is less active in comparison to MoF6. WCl6 is more stable than MoCl6 and is formed by direct combination of the elements. It reacts with hot water to give tungstic acid. WF5 is obtained by quenching the reaction of the metal with hexaflouride at 1000°C. It disproportionates into WF4 and WF6 above 320°C. Tungsten cannot form the complexes of the type [MoCl6]3 – rather [W2Cl9]3 – is known to be obtained by electrolytic reduction of acidic solutions of tungstates in HCl.

4. Oxohalides Oxohalides of tungsten are more stable than those of molybdenum. WOF4 can be obtained by fluorination of WO3 and has the tetrameric NbF5 structure. WOCl4 and WO2Cl2 are obtained together by heating WO3 in Cl4. The red crystals of WOCl4 are violently hydrolysed by water while the yellow crystals of WO2Cl2 are hydrolysed very slowly.

21.4

21.5

Inorganic Chemistry

RHENIUM (Re)

Rhenium was detected in 1925 by its X-ray spectrum. Later, it was isolated from molybdenite by Nolack, Berg and Tache.

21.5.1 Occurrence and Extraction Rhenium is a very rare element, being the 76th most abundant element by weight in the earth’s crust. It occurs in trace amounts in molybdenum sulphide ores. It is obtained from the flue dust in the roasting of these ores in the form of Re2O7. This is dissolved in NaOH to form the perrhenate ion. The solution is concentrated and precipitates of potassium perrhenate, KReO4, are obtained by the addition of KCl. This compound is reduced with hydrogen to the metal.

21.5.2 Properties Rhenium resembles platinum in appearance and has very high melting point (3180°C), next to W. It is less reactive than Mn. It does not react with water or non-oxidising acids but dissolves in oxidising acids to form perrhenic acid, HReO4. It forms Re2O7 on heating with air. Table 21.2 Oxidation states of rhenium

21.5.3 Uses Due to its very high melting point, it is used as W-Re thermocouples and in electric furnace windings. Pt–Re alloy is used as a catalyst for petroleum reforming.

21.5.4 Chemistry of Rhenium The outer-shell electronic configuration of Re is 4f145d54s2. It exhibits oxidation states from (–I) to (+VII). The lower oxidation states are formed in complexes with strong p-acid ligands. Table 21.2 lists various oxidation states of Re with specific examples.

1. Oxides

Oxidation States

Examples

(–I)

[Re(CO)5]–

(0)

Re2(CO)10

(+I)

[Re(CO)5Cl], [K5Re(CN)6]

(+II)

ReX2TAS

(+III)

Re2X82–

(+IV)

K2ReCl6, ReCl4

(+V)

ReCl5, ReF5, (ReOX4)–, Re2Cl10

(+VI)

ReO3, ReF6, (ReF8)2–

(+VII)

ReO4–, Re2O7, ReF7

V The known oxides of rhenium are Re(+IV)O2 (brown), ReIIIO3 (red), Re2IIIO3.xH2O (black), Re2O5 (blue) and VII Re2 O7 (yellow). Re2O7 is a deliquescent oxide which readily dissolves in water to give acidic solutions. These solutions on evaporation over P2O5 yield yellow crystals of perrhenic acid. Electrolytic reduction of perrhenate solutions in presence of sulphuric acid yields Re2O5. The heptoxide is heated with rhenium to obtain the trioxide at 570 K. However, on strong heating, ReO3 undergoes disproportionation as 3ReO3 $ Re2O7 + ReO2

These lower oxides are also obtained by thermal decomposition of NH4ReO4. The halides yield oxohalides when treated with halogens. ReO2 has distorted rutile structure while ReO3 has a structure closely related to perovskite structure. In this structure, every metal atom is octahedrally surrounded by oxygen atoms.

2. Halides Only one stable heptahalide, ReF7, is known. It is obtained by the direct reaction of the elements at 673 K

Chemistry of 5d Series

21.5

under pressure. It dissolves in liquid HF and reacts with dioxygen or water to give pale yellow or colourless oxohalides. Only two hexahalides, ReF6 and ReCl6, are known. ReF6 is obtained by the direct reaction of the elements at 393 K and ReCl6 is obtained by the treatment of the fluoride with BCl3. Due to strong spin orbital coupling, the magnetic moment is much lower than the spin-only value. The compounds are low melting and are sensitive to hydrolysis. 3ReF6 + 10H2O $ 2HReO4 + ReO2 + 18HF 3ReCl6 + H2O $ 2HReO4 + ReO2 + 16HCl ReCl6 is less stable and decomposes, to give ReCl5 and Cl2 on heating. Reduction of ReF6 with metal carbonyls yields a mixture of ReF5, ReF4 and oxide fluoxides. ReCl5 is obtained by passing chlorine over a mixture of Re and KCl at 500°C. It is dark red-brown in vapour state and condenses to a dark red-brown solid. Two Re atoms occupy the centre of the octahedra with Re-Re distance of 3.74 Å indicating the absence of Re-Re bond (Fig. 21.1). It exists as Re2Cl10 with two octahedra sharing an edge. ReBr5 is obtained by bromination of the metal at 650°C. It decomposes to Re3Br9, on heating. ReCl4 is obtained in several ways as

3.74Å

Fig. 21.1 Structure of ReCl5

2ReCl5 + C2Cl4 $ 2ReCl4 + C2Cl6 2ReCl5 + SbCl3 $ 2 ReCl4 + SbCl5 and

3ReCl5 + Re3Cl9 $ 6ReCl4

93°

3. Perrhenates All rhenium compounds are oxidised by nitric acid or H2O2 to give perrhenates, ReO4– or perrhenic acid, HReO4. It is best obtained by hydrolysis of Re2O7 in H2O. Perrhenic acid is highly acidic, but is Fig. 21.2 Structure of [ReH9]2– a quite weaker oxidising agent as compared to KMnO4. The ReO4– – ion is tetrahedral and quite stable in alkaline solution whilst MnO7 in unstable. ReO4– is colourless in contrast to deep purple MnO4– ion. However, concentrated solutions of HReO4 are yellow or green due to formation of unsymmetrical HO–ReO3. KReO4 yields hydrido complex, K2[ReH9]2– on reduction with excess of potassium in presence of ethylene diamine or ethanol. [ReH9]2– exists in tri-capped trigonal prismatic structure (Fig. 21.2). Alkali fluorides yield MReF7 and M2ReF8 with a square antiprismatic structure. MF

ReF6  → MReF7/M2ReF8

(M = Na, K, Rb, Cs)

The yellow-green [ReCl6]2– has been obtained by reduction of HReO4 with KI in presence of 8–13 M HCl. It is readily hydrolysed to give ReO2.xH2O. [ReCl6]2– + xH2O $ ReO2 . xH2O

21.6

OSMIUM (Os)

Osmium was discovered by Smithson and Wollaston in 1803.

21.6

Inorganic Chemistry

21.6.1 Occurrence and Extraction Osmium is very rare and occurs to an extent of 0.005 ppm in the earth’s crust. It is found naturally together with other platinum metals and the coinage metals. It is extracted commercially from the anode slime obtained during electrolytic refining of nickel. The anode slime is fused with sodium peroxide followed by treatment with aqua regia so as to dissolve Pd, Pt, Ag and Au. The solid residue left behind containing Ru, Os, Rh and Ir is treated with sodium oxide and extracted into water to obtain soluble salts of Rh and Os. These salts are oxidised with chlorine gas and treated with alcoholic sodium hydroxide and ammonium chloride to get precipitates of OsCl2O2(NH3)4. These precipitates are reduced using hydrogen to yield the powdery metal.

21.6.2 Properties Osmium is a blue-grey platinum metal and is the densest stable element. It is hard but brittle and remains lustrous even at very high temperatures. It has very high melting point (3306 K) and high bulk modulus. Thus, the metal is difficult to work on. The solid metal is noble and very resistant to attack by air, water and acids. However, the powdered metal can be oxidised to OsO4 by aqua regia.

21.6.3 Uses of Osmium Osmium alloys are used in the tips of fountain pens, surgical implants and instrument pivots.

21.6.4. Chemistry of Osmium

Table 21.3 Oxidation states of osmium

The outer shell electronic configuration of Os is 4f145d66s2. It exhibits the oxidation states ranging from (–II) to (+VIII). The lower oxidation states are stabilised mainly by strong -acid ligands. Some examples of its compounds in the oxidation states are listed in Table 21.3.

(–II)

Na2[Os(CO)4]

(–I)

Na2[Os4(CO)13]

Oxides

(0)

[Os3(CO)12]

Oxidation State

Example

(+I) OsI OsO4 is obtained either by burning osmium in O2 or by treating the osmium solution with concentrated HNO3. OsO4 (+II) OsI2, [Os(CN)6]4– is sparingly soluble in dilute H2SO4 but readily dissolves in (+III) OsBr3 alkali hydroxide solutions to give [OsO4(OH)2]2–. It gets (+IV) OsO2, OsCl4 reduced to black OsO2 or Os in contact with organic matter and hence is used as a biological stain. It is used in organic (+V) OsF5, NaOsF6 chemistry to give cis-glycols from olefinic compounds. (+VI) OsF6, OsOCl4 OsO2 is obtained by the treatment of osmium with osmium (+VII) OsF7, OsOF5 tetroxide, OsO4 or sodium chlorate. It is readily attacked by (+VIII) OsF4, Os(NCH3)4, OsO3F2 dilute HCl. Perosmates such as K2[OsO4(OH)2] can be obtained from the deep red solutions of [OsO4(OH)2]2 –, while reduction of these solutions yields [OsO2(OH)4]2 –. OSO4 can be reduced by HCl in presence of KCl to give [OSO2Cl4]2– and [OS2OCl10]2 –. However, [OSCl6]2– is obtained when HCl is used in presence of Fe2+ or alcohol.

21.7

IRIDIUM (Ir)

Iridium was discovered by Tennant in 1803 and named after the Greek goddess (Iris). The high- purity iridium was obtained by Robert Hall in 1842.

Chemistry of 5d Series

21.7

21.7.1 Occurrence and Extraction Iridium is also very rare and forms about 0.001 ppm of average mass fraction in the earth’s crust. However, it is relatively more common in meteorites (0.5 ppm). It is found naturally in alloyed form with platinum metals and the coinage metal. It is extracted by the similar process as used for extraction of osmium. The solid residue of platinum metals obtained during extraction of osmium is treated with NaHSO4 and is extracted with water to get an insoluble residue containing iridium. This residue is fused with Na2O2 and extracted with water to obtain the residue of IrO2 which is dissolved in aqua regia to get solutions of pure (NH4)3IrCl6. These solutions are evaporated to dryness and reduced with hydrogen to get pure iridium.

21.7.2 Properties Iridium is a silvery white platinum metal, the most corrosion resistant and second densest element after osmium. It is very hard, brittle, has very high melting point (2739 K) and maintains its mechanical properties even at high temperature. It becomes a superconductor at temperature just below 0.14 K.

21.7.3 Uses Due to its very high melting point, hardness and brittleness, it is not easy to fabricate, yet it is used extensively in its alloyed form. Os-Ir is used for balances and compass bearings. Ir-Ti is used in deep-water pipes and aircraft engines due to its being corrosion resistant. Iridium is also used to harden other platinum alloys. Worth mentioning is Pt-Ir (9:1) used to construct the international prototypes of metre and kilogram mass.

21.7.4 Chemistry of Iridium The outermost electronic configuration of Iridium is 4f145d76s2. It exhibits oxidation states ranging from (–III) to (+VI) however, (+III) and (+IV) are the most common oxidation states. As expected, lower oxidation state occur in compounds with -acid ligands. Such examples are [Ir–III(CO)3]3–, [Ir–I(CO)3PPh3]–, [Ir04(CO)12] and [IrCOCl(PPR3)2]. Table 21.4 lists the various oxidation states of iridium in its compounds.

1. Oxides

Table 21.4 Oxidation states of iridium Oxidation State

Examples

(–III)

[Ir(CO)3]3–

(–I)

[Ir(CO)3(PPh3)]–

(0)

[Ir4(CO)12]

(+I)

[Ir(CO)(Cl)(PPh3)2]

(+II)

IrCl2

(+III)

IrCl3, [IrH3(PR3)2], [IrCl6]3–

(+IV)

IrO , [IrCl ]2–

2 6 Only one oxide IrO2 (brown) is well characterised with (+V) Ir4F20, CsIrF6 suitable structure and is obtained by oxidation of the metal in finelly divided powdery form. IrO4 has been reported to be (+VI) IrF6 prepared under special conditions but is not expected to be stable. Oxidation of IrO2 with HNO3 yields the blue-black sesquioxide, Ir2O3.

2. Halides IrF6 is obtained by direct reaction of the elements. It is comparatively more stable than RhF6. IrF5 exists as (IrF5)4 and is very reactive. It has a tetrameric structure with Ir–F–Ir bridges. It is obtained by reduction of IrF6 with H2 in presence of anhydrous HF. Although (+IV) is the most stable state of Ir, yet IrCl4 is not very stable. However, IrF4 is comparatively more stable and is obtained by reduction of IrF5 with H2 in aqueous HF. All the trihalides of iridium are known. IrF5 is obtained by reduction of IrF6 with Ir. The other trihalides are obtained by direct reaction of the elements. All trihalides are insoluble in water and are unreactive. They have structures seen for AlCl3. Dihalides are uncertain and only IrCl2 is known with polymeric structure.

21.8

Inorganic Chemistry

3. Complexes of Ir The very extensive chemistry of Ir (+I) complexes is confined with p-acid ligands such as CO, PR3 and alkenes, etc. The most important complex is Vaska’s complex, trans [Ir(Cl)(CO)(PPh3)2]. The square planar complex is obtained by refluxing sodium chloroiridate and phosphine in diethylene glycol in presence of CO. However, when the added molecule contains multiple bonds, the added molecule behaves as a bidentate ligands and forms a cyclic structure. The reversible binding of Vaska’s complex with oxygen is analogous to that of haemoglobin. Hence, this complex is used as a model for oxygen-binding ability of biological oxygen carriers such as haemoglobin. A number of typically octahedral, low-spin, diamagnetic and stable complexes of Ir (+III) are known. Yellowish green Na3[IrCl6] is obtained by heating finely divided iridium with NaCl and Cl2. Aqua ion of Na3[IrCl6] yields species such as [Ir(H2O)Cl5]2–, [Ir(H2O)2Cl4]– and [Ir(H2O)3Cl3]. Dark red-brown Na2[IrCl6] can be obtained by adding NaCl to the suspension of hydrous IrO2 in aqueous HCl. The complexes are unstable in basic solutions and undergo spontaneous reduction as pH >11 3– ����� � 2 IrCl62– + 2OH– � ���� � 2RIrCl6 + H2O + ½O2 pH D0 > Dt (= 0.45D0)

24.4.2 Applications of Crystal Field Theory The important applications of crystal field theory have been discussed below:

1. Colour of Transition Metal Complexes When white light falls on a transition metal Table 24.8 Colour of the absorbed and transmitted light complex, the complex may absorb some portion Colour of the Wavelength fo the Colour of the of it and the remaining portion is reflected or absorbed light absorbed light (Å) transmitted light transmitted back.The colour of the transmitted Violet 4000–4500 Yellow green light is different from the colour of the absorbed Blue 4500–4800 Yellow light and is known as the complementary Green-blue 4800–4900 Orange colour of the absorbed light. The colour of the Blue-green 4900–5000 Red transition metal complex is the colour of this Green 5000–5600 Violet transmitted light. Table 24.8 shows the relation Yellow-green 5600–5800 Violet between the colour of the absorbed and the Yellow 5800–5900 Blue transmitted light. Orange 5900–6050 Green-blue Origin of Colour In a transition metal Red 6050–7500 Blue-green complex, the degeneracy of the five d-orbitals is splitted under the influence of the ligands to form two sets of orbitals—lower energy t2g set and higher-energy eg set of orbitals. The difference in energy between these two levels (D0) is very small and can be covered even by the

low-energy radiations of the visible region ÊÁ E = hc ˆ˜ . Thus, when the electrons from the low energy Ë l¯ set of orbitals is excited to the high energy set of orbitals due to absorption of light of suitable wavelength in the visible region, the complex appears coloured. This transition is known as d-d-transition. Since D0

Coordination Compounds-II Theories of Bonding

24.15

Ground state of Absorption Excited state varies with the nature of metal ion, ligands and the of light radiation of [Ti(H2O)6]3+ [Ti(H2O)6]3+ geometries of complexes, the absorption of light of selected wavelength also varies accordingly. Absorption of light radiation eg Consider the case of [Ti(H2O)6]3+ with d1 eg configuration. The single electron is present in t2g orbitals. It absorbs in green and yellow regions t2g t2g t21g e0g t20g e1g (around 5000 Å wavelength), corresponding to about 240 kJ per mole. This amount of energy is close to D0 and the electron gets excited from t2g to eg set of Fig. 24.15 Excitation of electrons in d–d transition orbitals. Thus, the colour of the complex appears purple, the complementary colour of the absorbed light radiation (Fig. 24.15). On the other hand, the complex [Co(H2O)6]3+ absorbs in the blue-green region of the visible light and therefore appears pinkish red due to comparatively smaller D0. Similarly, the colour of the absorbed light varies with the nature of ligands. Thus, as the D0 increases from H2O < NH3 < CN– ligands, the colour of the complexes [Co(NH3)6]3+ and [Co(CN)6]3– varies from blue to yellow respectively. The value of D0 also varies with oxidation state of the central metal atom; thus, [V(H2O)6]2+ appears violet but [V(H2O)6]3+ appears yellow.

The transition metal complexes with empty (d0) or completely filled d-orbitals (d0) are colourless due to absence of any d-d transitions. Thus, complexes of Sc3+ (d 0), Ti4+ (d 0), Cu+ (d 0), Zn2+ (d 10), Ag+ (d10), Cd2+ (d10) and Hg2+ (d10) etc, are colourless.

2. Magnetic Properties of Transition Metal Complexes The number of unpaired electrons in a transition metal complex can be determined with the help of crystal field theory. It can be seen that for a central metal ion with d4 to d8 configuration, the number of unpaired electrons are different in high-spin and low-spin octahedral complexes. As a result, their n(n + 2) BM also comes different magnetic moments calculated by the spin-only formula, ms = (Table 24.4). Thus, it helps in determining the diamagnetic (all paired electrons) and paramagnetic (unpaired electrons) character of the complexes.

3. Stabilisation of Oxidation States This theory helps to explain the preferential stabilisation of some oxidation states by certain ligands. For example, [Co(H2O)6]2+ is more stable than [Co(H2O)6]3+ i.e. Co2+ is more stabilized by the weak-ligand H2O than is Co3+. It is clear from Fig. 24.17 that Co2+ (d7) has a higher value of CFSE than Co3+(d6) in the weak octahedral field leading to greater stability. On the other hand, [Co(NH3)6]3+ is more stable than [Co(NH3)6]2+ i.e. Co3+ is more stabilised by the strong ligand NH3 than is Co2+. This is due to the higher value of CFSE for Co3+ than Co2+ in the strong octahedral field leading to greater stability.

4. Prediction of Stereochemistry of Complexes According to our discussion, it is clear that greater the CFSE values, greater is the stability of the complexes. Since Dsp > D0 > Dt, Cu2+ forms square planar complexes (CFSEsp = 1.22 D0, CFSEo = 0.6 D0 and CFSEt = 0.18 D0). Similarly, most of the four coordinated complexes of Ni2+ (d8) are square planar (CFSE = 1.45 D0) rather than tetrahedral (CFSE = 0.36 D0). Similarly, the structure of spinels can be predicted with the help of this theory. Mn3O4 is a normal spinel while Fe3O4 is an inverse spinel. We can use crystal field theory to predict their stereochemistry. Now, O2– is a weak field ligand. If we calculate the CFSE values for Mn3+(d4), Mn2+(d5), Fe3+(d5) and Fe2+ (d6)

24.16 Inorganic Chemistry

ions for both octahedral and tetrahedral cases (as shown in Table 24.9), we find that Mn3+ and Fe2+ ions have greater CFSE values in octahedral geometry in comparison to that for tetrahedral geometry. It means that Mn3+ and Fe2+ ions preferentially occupy the octahedral sites while the Mn2+ ions occupy the tetrahedral sites. Thus, Mn3O4 is a normal spinel and can be represented as Mn2+[Mn23+]O4. On the other hand, Fe3O4 is an inverse spinel and can be represented as Fe3+[Fe2+Fe3+]O4 due to occupation of octahedral sites by all the Fe2+ ions and half the Fe3+ ions, whereas the tetrahedral sites are occupied by the remaining half of the Fe3+ ions.

Table 24.9 CFSE for octahedral and tetrahedral weak field for Mn3+, Mn2+, Fe3+ and Fe3+ ions Ions

CFSE Octahedral weak field

Tetrahedral weak field

0.60 D0

0.18 D0

Mn

0

0

Fe2+

0.40 D0

0.27 D0

Fe3+

0

0

Mn3+ 2+

24.4.3 Limitations of Crystal Field Theory 1. Crystal field theory considers only the ‘d’ orbitals of the metal ion and gives no consideration to the other orbitals of the metal atom (s, px, py nad pz) and the ligand p-orbitals. Thus, it cannot explain the p-ligand orbital’s dependent properties of the complexes. 2. It does not consider the formation of p-bonding in the complexes. 3. It gives no satisfactory explanation for the relative strength of the ligands in the spectrochemical series. 4. This theory considers the metal and ligand interaction as purely ionic and doesn’t consider the partially covalent nature of the metal ligand bonds as observed by the experimentation, discussed in the next article.

24.4.4 Experimental Evidence of Covalent Bonding in Complexes The metal ligand covalent bonding in complexes is supported by the following experimental evidances.

1. Nuclear Magnetic Resonance (NMR) Spectra The NMR spectral studies of complexes such as KMnF3 and KNiF3 have revealed the interaction of electron spin of the unpaired electron of the paramagnetic metal ion and the nuclear spin of 19F. This is possible only if there is overlapping of the ligand orbitals and the metal ion orbitals (covalent bonding).

2. Electron Spin Resonance (ESR) Spectra The ESR spectrum of (IrCl6)2– ion shows a complex pattern of sub-bands (hyperfine structure). This structure is possible only if the single unpaired d-electron is somewhat localised on each of the ligand i.e. there is overlapping of ligand and metal-ion orbitals.

3. Nuclear Quadrupole Resonance (NQR) The NQR spectrum of square planar complexes such as [PtX4]2– and [PdX4]2– support for the considerable amount of covalency in the M–X bonds.

4. Nephelauxetic Effect It has been found that in case of complexes, the interelectronic repulsions between metal d electrons decreases on complexation. As a result, the effective size of the metal d-orbitals gets increased, known as the nephelauxetic effect (electron cloud expanding). As a result, the extent of overlap of ligand and metal orbitals increases leading to covalent bonding. This concept has been further illustrated in the chapter 26.

Coordination Compounds-II Theories of Bonding

24.5

24.17

THE LIGAND FIELD THEORY–MOLECULAR ORBITAL THEORY

No doubt, crystal field theory is remarkably successful in explaining the spectra and magnetic properties of the complexes; however, there are some deviations observed in the positions and intensities of the spectral bands of some complexes. Further, it considers the metal–ligand attraction as purely electrostatic. However, the metal in zero oxidation state cannot form metal–ligand bond by only electrostatic attration. This theory gives no account of the partly covalent character of the metal–ligand bond. It also fails to provide a satisfactory explanation for the order in spectro-chemical series. Hence, this theory was modified in the form of ligand field theory with the incorporation of molecular orbital theory. This theory considers the overlap of atomic orbitals of the central metal ion with the suitable atomic orbitals of ligands. The appropriate combinations are based on the principles of group theory, discussed in a similar way in this article. +z

24.5.1 MOT for Octahedral Complex 1.

Bonding

B5 p5x

p5y p3y B3

+x

σ5z We have to consider the following points for an octahedral σ3z p3x p2x complex:p4x σ2z 1. For an octahedron, the atomic orbitals in the valence –y B2 B4 +y A σ4z shell of the central metal ion of 3d series are: 3dxy, 3dyz, p4y p2x 3dxz, 3dx2–y2, 3dz2, 4s, 4px, 4py and 4pz. The symmetries σ1z σ6z p1y p6y of these atomic orbitals are: 3dxy, 3dxz, 3dyz"t2g; 3dz2, p1x B1 B 3dx2–y2 "eg 4s"a1g; 4px, 4py, 4pz"t1u p6x 6 The six ligands are designated as L1, L2, L3, L4, L5, –x and L6, with the ligand s orbitals along the positive –z x, y and z axes, represented as s2z, s3z, and s5z and Fig. 24.16 Octahedral complex, ML6 that along the negative x, y and z axes as s1z, s4z and s6z respectively. The numbers 1....6 designate the ligand while the letters x, y and z correspond to the axis. Out of these nine orbitals, only six suitable atomic orbitals are to be identified which may overlap along the axis with the ps orbitals of the ligand to form metal–ligand s-bonds (Fig. 24.16). Now, the three orbitals 3dxy, 3dyz and 3dxz have their lobes orientated in the space between the axes; hence, they cannot participate in s-bonding and remain as nonbonding orbitals. These can participate in p-bonding with the filled or unfilled p-orbitals of the s-bonded ligands. 2. The ps orbitals of the six ligands undergo symmetry adapted linear combination to form Ligand Group Orbitals (LGOs) with suitable symmetries as shown below. (a) Combination (Sa) with a1g symmetry to combine with the 4s orbital of the metal with a1g symmetry and is normalised as,

Sa =

1 6

(s1z + s2z + s3z + s4z + s5z + s6z) with group symmetry a1g.

(b) Combinations (Sx, Sy and Sz) with t1u symmetry to combine with the 4p orbitals of the metal with t1u symmetry and is normalised as

Sx = Sy = Sz =

(s1z - s 3z ) ¸ Ô Ô 1 s s ( ) 2z 4z ˝ 2 Ô 1 s ( 5z - s 6 z ) Ô˛ 2 1 2

with group symmetry t1u

24.18 Inorganic Chemistry

(c) Combination (Sx2–y2) with eg symmetry to combine with the 3dx2 – y2 orbital of the metal with eg symmetry and is normalised as

1 (s1z + s3z – s2z – s4z) with group symmetry eg. 2 (d) Combination (Sz2) with eg symmetry to combine with the 3dz2 orbital of the metal with eg symmetry and is normalised as: 1 Sz2 = (2s5z+ 2s6z – s1z – s2z – s3z – s4z) with group symmetry eg. 2 3 3. Now, the six orbitals of the metal with symmetry a1g, t2g and eg (4s, 4px, 4py, 4pz, 3dx2 – y2 and 3dz2) overlap with the s-orbitals of the six ligands with symmetry a1g + eg + t1u (Sa, Sx, Sy, Sz, Sx2 – y2 and Sz2) to form six sigma BMOs and six sigma ABMOs, as shown in Fig. 24.17. Sx2–y2 =

z

z

z x

+ + +

x +

+ L1

+ + + +

y

+

+

– –

+

y

–

–

–

+ –

+

+ M

y

–

+ +

–

+

–M

y

M

y

Overlap of 4py and Sy to form sy (BMO) and sy* (ABMO)

y –

–

+

Overlap of 4pz and Sz to form sz (BMO) and sz* (ABMO)

z –

+

x

z

x

+

+

– – – – – – – – + +

y

Overlap of 4px and Sx to form sx (BMO) and sx* (ABMO) z z

M

z

–

–

+

–

x +

+

+

z –

x

+

+

Overlap of 4s and Sa to form ss (BMO) and ss* (ABMO)

–

z x

y

+ + – – – – + + + –

Overlap of 3dz 2 and Sz2 to form sz2 (BMO) and sz*2 (ABMO)

z

x

x

–

+ –

–

y +

+ – –

+ +

y

– + – +

+ –

y

Overlap of 3dx2–y2 and Sx2–y2 to form sx2–y2 (BMO) and s*x2–y2 (ABMO)

Fig. 24.17 Overlap of symmetry-matched metal atomic orbitals and LGOs

The MOT energy-level diagram with distribution of electrons for an octahedral complex [Co(NH3)6]3+ has been shown in Fig. 24.18. The MO diagram is unsymmetrical, as ligands are usually more electronegative. Six NH3 ligands contribute 2 × 6 = 12 electrons to the molecular orbitals from their filled ps orbitals of symmetry combination

Coordination Compounds-II Theories of Bonding

σx*

σy*

24.19

σz*

* ) (t1u

σs* (a1g)

4p (t1u)

4s (a1g)

σx*2–y2 σz* 2 (eg* ) ∆0

Energy 3d6 (t2g + eg) [Co3+]

MOT energy level diagram of [Co(NH3)6]3+

3dxy 3dy 2 3dxz (nonbonding) (t2g) Σa σx

2

σz2 (eg)

–y

2

Σx

Σy

Σz Σz2 Σx2–y 2

(a1g + t1u + eg) Six ligands

σx

σy σz (t1u)

σs (a1g)

Fig. 24.18 MO diagram for [Co(NH3)6]3+ showing only σ-bonding

(a1g + t1u + eg). These twelve electrons are considered to be filled in the lower energy MO’s (ss), (sx = sy = sz) and (sx2 – y2 = sz2) of the corresponding symmetry a1g, t1u and eg. This is because these orbitals are nearer in energy to the energy of the ligands orbitals and hence, possess more ligand orbital character i.e. the twelve electrons present in the sigma BMOs are localised mainly on the ligand orbitals. Now, there are six electrons in the 3d orbitals of the central metal ion, Co3+. These are considered to be filled in the non-bonding atomic orbitals 3dxy, 3dyz, 3dxz, of symmetry t2g which do not participate in the MO formation and have energy intermediate to that of BMOs and ABMOs. This corresponds to the electrons occupied in t2g level according to crystal field theory. The orbitals next higher in energy are (sx2 – y2 = sz2) of symmetry eg*, with energy nearer to the metal orbitals and hence, possess more metal orbital character (with major contribution from eg orbitals of the metal). The difference in energy between t2g and eg*, thus, corresponds to the crystal-field splitting energy (D0). It also indicates that, if, there are no ligand orbitals, the electrons in t2g set of orbitals would be of pure metal ion. According to MOT, this energy gap D0 becomes smaller in presence of weak field ligands and the MO diagram becomes more unsymmetrical in presence of highly electronegative ligands. The MO energy- level diagram of [CoF6]3– has been shown in Fig. 24.19. It should be noted that in case of weak ligands, orbitals are occupied according to Hund’s rule of maximum multiplicity

24.20 Inorganic Chemistry

sx* sy* sz* * ) (t 1u 4p (t1u) ss* (a1g)

σx*2–y2 σz* 2 (eg* )

4s (a1g)

3dxy 3dy 2 3dxz (nonbonding)

3d (t2g + eg) Energy

D0

[Co3+]

Σa Σx Σy Σz Σz2 Σx2–y 2 (a1g + t1u + eg)

σx2–y 2 σz2 (eg)

σx

Six ligands

σy σz (t1u)

(a1g)

Fig. 24.19 MO energy level diagram of [CoF6]3–

2.

-bonding

The p-bonding in an octahedral complex is possible when either the ligands or the metal have empty orbitals for the acceptance of electron density. The potential metal–ligand p-interactions can be of four types dp – pp, dp – dp, dp – p* and dp – s*, as shown in Fig. 24.20. The metal orbitals capable of participating in the formation of p-bonds are the three p-orbitals (px, py and pz) of symmetry t1u and three d-orbitals (dxy, dyz and dxz) of symmetry t2g. These orbitals can undergo sideways overlapping with the ligand orbitals of proper symmetry, mutually perpendicular to one another and to the inter nuclear axis. M

+ + +

–

+

L

dp – dp

– + –

+

dp – pp

–

+ L

M

–

–

M

– +

– – + +

+

+ -

L

M

d p – p*

Fig. 24.20 Type of p-interactions between metal and ligand orbitals

–

+

– – +

L +

dp – s*

24.21

Coordination Compounds-II Theories of Bonding

The simplest case of p-bonding is of type-I Table 24.10 Symmetry adapted linear combinations of ligand orbitals for -bonding (dp – pp) interaction in which there are simply two p orbitals on each of the six ligands as shown in Fig. 24.16 Appropriate combination Symmetry (for an octahedral complex). Thus, there are in total 12pp orbitals, viz. p1x, p1y, p2x, p2y, p3x, p3y, p4x, p4y, p5x, p5y, pxy = 1 (p1x + p2y + p3y + p4x) 2 p6x, p6y, present on the ligands. 1 (p + p + p + p ) p = t2g xz 5x 3x 6y These ligand pp-orbitals undergo symmetry-adapted 2 1y linear combinations to form ligand group p-orbitals with pyz = 1 (p2x + p5y + p4z + p6x) 2 symmetry t1g, t1u, t2u and t2g. The overlap of some of these ligand group orbitals with the appropriate symmetry orbitals of the metal has been shown in Fig. 24.22. The t1u set of the metal orbitals has already been used in s bonding, so formation of p-bond with these sets will weaken the s bond, hence is not favored and is considered ineffective (Fig. 24.21). There is no t1g and t2u symmetry set of orbitals on metal atom, thus t1g and t2u set of ligand group orbitals remain nonbonding. The t2g set of ligand-group orbitals are of appropriate symmetry to form p-bonding with the t2g set of metal orbitals. In other words, the p-bonding in an octahedral z

z p5y + –

–

L5 p3y

z + p5x

x + L3

+

p3z +

–

y

L4

– p4x

M

y

L2 –p 2x

–

+ p4z

+

+

+ – M

x

L5

–

+

p6y

Overlap of py LGOs with py orbital of metal

+

–

L2 – p2z

–

y

L1 – p1z

L6

–

L6

+

–

M

L4

+ p6x

p1y – L1 +

L3

+

Less effective Overlap of pz LGOs with pz orbital of metal

Overlap of px LGOs with px orbital of metal

Fig. 24.21 Overlap of LGOs with t1u symmetry with the t1u symmetry metal orbitals z

z L5

–

–

+

L5

x

–

+

–

+ L2 y

+ L4 +

– +

L6

–

Overlap of pp ligand orbital with dyz orbital of metal

+ +

z +

x

– L3

–

– – L1

–

–

–

x

–

+ L5

+

L4

y +

+

–

+ –

+ +

L6

Overlap of pp ligand orbital with dxz orbital of metal

+

y

–

+

– L2

L6 Overlap of pp ligand orbital with dxy orbital of metal

Fig. 24.22 Overlap of ligand group orbitals with the metal orbitals of most appropriate symmetry

24.22 Inorganic Chemistry

t2*g

3d (t2g + eg) Energy

Metal orbitals (eg involved in s-bonding and would not be considered here)

t1g + t2u + t1u

t1g + t2g + t1u + t2u LGO's

t2g

Fig. 24.23 Molecular orbital energy-level diagram for an octahedral complex ML6 with p-bonding

complex, ML6, is limited to the t2g symmetry orbitals, while t1g, t1u and t2u set of ligand group orbitals remain as nonbonding. The schematic molecular orbital energy-level diagram of this case has been shown in Fig. 24.23. It can be seen that by the overlap of t2g metal orbitals and t2g LGO’s, t2g bonding molecular orbitals and t*2g antibonding molecular orbitals are formed. Since the overall complex formation involves both s as well as p-bonding, hence we will discuss the combined molecular orbital energy-level diagram with specific examples. p-bonding affects the energy level of the metal t2g orbitals, while s bonding affects the energy level of the metal eg orbitals. However, it has been seen that depending upon the donor or acceptor tendency of ligand p-orbitals and hence their lower or higher energy than the metal orbitals, the overall effect on splitting of d-orbitals is different as discussed in the next article.

3.

-acceptor and

-donor Ligands

The donation of electrons can take place either from metal to ligand (p-back bonding) or from ligand to metal (simple p-bonding). Accordingly, the ligands are classified as follows: (a)

−acceptor Ligands These ligands have vacant p* molecular orbitals of symmetry t2g, such as in case of CNR, CN, CO or NO+, etc. These orbitals are higher in energy than the metal t2g orbitals and hence, form BMOs lower in energy than the metal t2g orbitals and ABMOs higher in energy than the eg metal orbitals as shown in the figure 24.24. The electrons occupy the BMOs and thereby increase D0 to D 0. As a result, significant energy stabilisation taken place and these ligands are known as strongfield ligands. The particular example is bonding of Fe(CO)5, discussed in chapter 28. This type of interaction is known as metal to ligand or M–L interaction.

(b)

-donor Ligands These ligand have filled pp orbitals capable of electron donation, as in case of halide ions, O–, NR3, h5– C5H5, etc. These ligands have orbitals perpendicular to one another and to the internuclear axis of the metal. It means that for six ligands, there would be twelve such orbitals which combine together to form ligand group p-orbitals of symmetry combinations. The ligand orbitals are lower in energy than the metal t2g orbitals and hence, the BMOs are stabilised as compared to the metal t2g orbitals and the ABMOs are also stabilised as compared to eg* MOs. The electrons are filled in the lower energy t2g BMOs, while the ligand electrons are filled in the higher energy t2g* ABMOs.This means the p-interactions are destabilizing and thereby decreasing the D 0 as compared to D0. These ligands are hence known as weak field ligands and this interaction is known as ligand to metal or L-M p-bonding.

Coordination Compounds-II Theories of Bonding

* ABMOS t2g

24.23

(Dl0 > D )

Dl0

(Dm0 < D0) Dm0

eg* MOs

eg*

e g*

Empty ligand orbitals of higher energy

D0

t2g* ABMOS

t2g t2g MOs Filled ligand orbitals of lower energy

Fig. 24.24 Comparison of MO energy-level diagram for -acceptor and -donor ligands

Consider the case of [CoF6]3–. After s-bonding, the ligands can contribute a total of 12 × 2 = 24 electrons to the MOs of the complex. Since in this case ligand orbitals are lower in energy, hence these electrons are considered to be occupied in t2g BMO and t1g, t1u and t2u ABMOs with more of ligand character. Now the electrons from the metal t2g orbitals are considered to be occupied in t*2g ABMOs and the energy gap between eg* and t*2g is considered as equivalent to that described by crystal field theory (denoted by D0). It is evident from the above discussion that p 0 < P, and that is why these complexes are high spin complexes. The complete MO diagram with representation of both s and p- bonding has been shown in Fig. 24.25. Now the difference between the energy of the Highest Occupied Molecular Orbital (HOMO) and the energy of the Lowest Occupied Molecular Orbital (LOMO) is known as the Ligand Field Stabilisation Energy (LFSE).

24.5.2 MOT for Tetrahedral Complexes Working on the same principles, we can discuss MO theory for tetrahedral complex. Formation of LGOs has already been discussed in chapter 4. The MOT energy–level diagram of a tetrahedral complex is shown in Fig. 24.27 (for only sigma bonding). The t2 orbitals of the metal ion and the t2 orbitals of the ligands combine to form t2, t2* and t2** molecular orbitals with mixed contribution from the set of p orbitals and dxy, dyz, dxz orbitals of the metal ion (Fig 24.27), whereas a orbitals of the metal ion and the a orbitals of the ligands combine to form a1 and a1* molecular orbitals. This means that the energy difference between nonbonding e-orbitals with contribution entirely from the dx2–y2 and dz2 orbitals of the metal ion and t2* orbitals can be compared for the crystal field splitting parameter Dt. Four ligands contribute 2 × 4 = 8 electrons to the a1 and t2 bonding molecular orbitals, while the electrons previously occupied in the d-orbitals of the free metal ion may be considered to be filled in the e–t2* set.

-bonding The -bonding in a tetrahedral complex can be considered by the same approach as used in octahedral complexes. Thus, the four ligands would have eight p-orbitals capable of forming -bond with combination

24.24 Inorganic Chemistry

sx*

sy* sz* (t *1u)

ss* (a*1g) 4p (t1u) sx2*–y2 sz2* (eg*) 4s (a1g) * * pxy pxy pyz * (t *2g)

3d (t2g + eg) Energy t1g + t1u + t2u

t1g + t2g + t1u + t2u

a1g + eg + t1u pxy pyz pxz (t2g)

sx2–y 2 sz2 (eg)

sx s y s z (t1u)

ss (a1g)

Fig. 24.25 Molecular orbital energy-level diagram for [CoF6]3– complex ion

of (t1 + t2 + e) symmetry. However, the metal ion has only e orbitals of appropriate symmetry to combine with e orbitals of the ligand so as to form e and e* molecular orbitals. Since e–e* energy gap is quite less than the pairing energy, tetrahedral complexes are generally high-spin complexes. However, still there is some mixing of p-orbitals as shown in Fig. 24.28.

Coordination Compounds-II Theories of Bonding

z –

z –

L4

L4 +

+ – L1

+

+

y

+ M

+

y

M – +

–

+

– L1

– L2

L2 x

x

+

+ –

L3

–

L3 pz – ps

s – ps

z

z –

– L4

L4 +

+ L1 –

+

– L1

+

+ –

y

M

– + L2

+

–

y +

– L2

x

x

+

+ –

24.25

L3

L3

–

Px – P x

Py – Py

Fig. 24.26 Representation of -overlap between metal orbitals and ligand-group orbitals. these orbitals is difficult to represent and hence is not shown.

a1* (s*) t2**(s*) t2

a1 t2*(s*) t2 + e Metal orbitals

e

Dt

a1 + t 2 s-bonding LGO's

t2(s) a1(s)

Fig. 24.27 MO diagram for ML4, showing only σ bonding

-overlap between

24.26 Inorganic Chemistry

t2* (s*, p*)

s1*(s*)

t2

a1 t2* (s*, p*)

e* (p*)

t2 + e

Metal orbitals t1(p) e(p)

e + t1 + t2 (p)

t2(p)

(s) a2 + t2 LGO's

a1, t2 (s) (s)

Fig. 24.28 Molecular orbital diagram for a tetrahedral complex showing both σ as well as -bonding

24.5.3 MOT for Square-Planar Complexes The molecular orbital theory can be applied to a square planar complex with D4h symmetry point group as discussed ahead:

1.

-bonding 1. The atomic orbitals of the metal for a square planar geometry can be represented as 3s $ a1g 3pz $ a2u 3dx2 – y2 $ b1g 3px, 3py, $ eu

3dxy $ b2g

3dxz, 3dyz $ eg out of which the metal orbitals of symmetry b1g, a1g and eu can form a s-bond. 2. The ligand p-orbitals undergo symmetry-adopted linear combination to give LGOs of appropriate symmetry as listed in Table 24.11.

Coordination Compounds-II Theories of Bonding

Table 24.11

Symmetry adapted linear combination of ligand orbitals LGO

z x p3y

p s3x p3z 2z

p4x L4

L3

M

s3y

L2 s1y

p4z

p2x

p1z

s1x L1

Fig. 24.29

p1y

Coordinate representation for a square planar complex

Y

24.27

Symmetry

1 2

(s1x + s2y + s3x + s4y)

a1g

1 2

(P1y + P2x + P3y + P4y)

a2g

1 2

(P1z + P2z + P3z + P4z)

a2u

1 2

(s1x – s2y + s3x – s4y)

b1g

1 2

(P1y – P2x + P3y – P4x)

b2g

1 2

(P1z – P2z + P3z – P4z)

b2u

1 2

(P2z – P4z)

eg

1 2

(P1z – P3z)

eg

1 2

(s1x – s3x) and

1 2

(s2x – s4x)

eu

1 2

(s2y – s4y) and

1 2

(s1y – s3y)

eu

eu* a2u,eu

a1g*

Table 24.12 Symmetry adapted linear combinations of ligand orbitals for -bonding

a2u a1g b1g*

Symmetry Energy

D

eg, a1g, b1g, b2g

a1g

Combinations

a2g

( 21 ) (p1y + p2x + p3y + p4x)

b2g

( 21 ) (p1y – p2x + p3y – p4x)

eu

(

1 2)

(p1y – p3y)

eu

(

1 2)

(p2x – p4x)

a2u

( 21 ) (p1z + p2z + p3z + p4z)

b1g

b2u

( 21 ) (p1z – p2z + p3z – p4z)

a1g

eg

(

1 2)

(p1z – p3z)

(

1 2)

(p2z – p4z)

eg

b2g

a1g, b1g, eu

eu

Fig. 24.30 MO energy-level diagram for a square planar complex with only σ-bonding

eg

24.28 Inorganic Chemistry

Out of these combinations, s-bond will be formed by combinations with symmetry a1g, b1g and eu, while p-bonds can be formed by combinations with symmetry a2g, a2u, b2g, eu, b2u and eg. The MOT energy level diagram with only sigma bonding has been shown in Fig. 24.30. 3. As already discussed, the MO diagram is unsymmetrical due to more electronegative ligands with considerably lower energy than the metal ion orbitals. Thus, the most stable bonding molecular orbitals (a1g, b1g and eu) are mainly of ligand character. The nonbonding (b2g, eg) and the bonding (a1g) molecular orbitals have also major contribution from the ligands, while the higher energy antibonding molecular orbitals (b*1g, a*1g, e*u) and the nonbonding (a2u) molecular orbitals are mainly of metal ion character. Consider the case of [PtCl4]2– with d4h symmetry point group. Thus, the four ligands will contribute 2 × 4 = 8 electrons to the a1g and b1g molecular orbitals, while the electrons from the metal ion will be occupied in eu, b2g, eg and a1g molecular orbitals. Y

Y

Y

Y

x

+

–

+ M

+

+ +

+ +

+ –

+

–

–

M +

+

x

+

–

M

–

x

M

+

a1g

x

– –

b1g

eu

Fig. 24.31 Overlap of metal orbitals with LGOs of a1g, b1g and eu symmetry for σ-bonding

z +

+

– –

M +

+ – –

+ +

Y

–

+

–

+

–

– M+ +

–

+ dz2

–

x

dxy

z

z +

+ Y –

–

+

–

– M+

+

– +

x

Y

+

x dyz (dxz)

–

+ – M – – + + –

+ Y

dx2 – y 2

Fig. 24.32 Overlap of metal orbitals with LGOs for -bonding

2.

-bonding

The p-bonding in square planar complexes will involve metal orbitals of symmetry b2g, eg, a2u and eu. Thus, the ligand orbital combinations with symmetry a2g and b2u will remain nonbonding because of absence of the metal orbitals with these symmetries. The MOT of square planar (D4h) complex with only p-bonding has been represented in Fig. 24.33. The complete molecular orbital energy-level diagram with both s and p-bonding molecular orbitals has been shown in Fig 24.34. It can be seen that the lowest energy molecular orbitals are s-bonding, while the highest energy molecular orbitals are both s as well as p-antibonding. Now the electrons from the four ligands = 4 × 6 = 24 electrons will be occupied in the bonding and nonbonding molecular orbitals localised mainly on ligands. The metal ion electrons will be occupied in the anti-bonding orbitals up to b*2g. Thus, the energy gap between b*1g and b*2g can be compared equivalent to

Coordination Compounds-II Theories of Bonding

24.29

eu* eu + a2u a2u*

b2g*(dxy)

eg + b2g

eg* (dxz, byz)

Energy a2g + a2u + b2g + eu + b2u + eg b2u

a2g

eu a2u eg (dxz, dyz)

b2g

Fig. 24.33 MO diagram for a square planar complex with only -bonding

the crystal-field splitting parameter. Thus, the results are consistent with the results of crystal field theory.

3. Application of MOT for Charge Transfer Spectra According to MOT, the electrons occupied in the highest occupied molecular orbital can be excited into the lowest unoccupied molecular orbital to produce charge transfer bonds. Thus in a particular complex apart from the d-d transitions, charge transfers named as L–M (if the highest occupied molecular orbital is localised on the ligand) and M-L (if the highest occupied molecular orbital is localized on the metal) may appear. Charge transfer spectra has been noticed in many compounds such as [Fe(CN)6]3–, [Fe(CN)6]4–, Cr(CO)6, [RuCl6]2– etc.

24.5.4 Comparative Account of Bonding Theories 1. Valence Bond Theory and Crystal Field Theory Valence bond theory considers the participation of 3dx2 – y2, 3dz2, 4s, 4px, 4py and 4pz orbitals of the metal atom in bond formation while 3dxy, 3dyz and 3dxz do not participate in bonding and remain nonbonding.

24.30 Inorganic Chemistry

e*u

ea + a2u

a*2u

a*1g

b*1g a1g b*2g

a*1g

e*g

eg + a1g + b2g + b1g a*2g a2g + b2u + eu + a2u b2u

+ eg + b2g

eu

(p)

a2u eg

b1g + a1g + eu (s)

b2g

eu b1g a1g

Fig. 24.34 MO diagram for a square planar complex showing both σ as well as Π bonding

Coordination Compounds-II Theories of Bonding

24.31

Likewise, crystal field theory considers the occupation of lower energy t2g set (3dxy, 3dyz and 3dxz) of orbitals. The spin of inner-orbital and outer-orbital octahedral complexes of valence bond theory is also consistent with those of low-spin and high-spin octahedral complexes of crystal field theory respectively. However, valence bond theory considers the promotion of an electron during bond formation, while this idea is not used in crystal field theory. The most important difference between the two theories is that crystal field theory considers the bonding between positively charged metal ion and the ligands bearing the partial negative charge as purely electrostatic whereas the valence bond theory considers the partial donation of ligand electrons to metal orbitals leading to covalent bonding.

2. Valence Bond Theory and Molecular Orbital Theory Molecular orbital theory considers the overlapping of six ‘s’ bonding atomic orbitals (s3d(x2 – y2), s3dz2, s4s, s4px, s4py, s4pz) with the six s orbitals of the ligands, while 3dxy, 3dyz, 3dxz, remains nonbonding as in valence bond theory treatment. However, molecular orbital theory considers the formation of six s-bonding orbitals and six s* antibonding molecular orbitals and provides an explanation for the spectral bonds of the metal complexes; this is not possible in valence bond theory treatment.

3. Molecular Orbital Theory and Crystal Field Theory It is clear from the energy-level diagram of the complexes that the splitting of the atomic orbitals of the central metal ion is consistent to that of the crystal field splitting diagram. The crystal field splitting parameter also corresponds to the energy gap between the highest occupied molecular orbital and the lowest occupied molecular orbital. However, there are some basic differences between the two theories. These basic differences are as follows. (a) Crystal field theory considers the metal ligand interaction as purely electrostatic, whereas the molecular orbital theory considers this interaction by means of overlap between metal and ligand atomic orbitals leading to the formation of molecular orbitals. (b) Molecular orbital theory considers the participation of all metal ion orbitals and all (s and p) ligand orbitals; whereas crystal field theory ignores the participation of s and p-orbitals of the metal ion and does not consider any of the ligand orbitals. (c) According to molecular orbital theory, the splitting of the atomic orbitals is due to covalent bond formation, whereas crystal field theory considers the splitting of 3d metal orbitals due to electrostatic field exerted by the ligands on the central metal ion. (d) Molecular orbital theory considers s as well as p-bonding in the complexes and also gives an explanation for the charge transfer bond. This aspect has not been considered in crystal field theory. The coordination compounds can be studied with the help of colour change of the solution, solubility change of a compound, conductance measurements (more the number of ions formed, more is the conductance), cryoscopic methods, magnetic moment studies, dipole moment studies and spectral studies. Werner gave his theory in terms of primary and secondary valencies. Primary valency (oxidation state) of an atom is satisfied by the negative ions, while the secondary valency (coordination number) is satisfied either by the negative ions or by the neutral molecules. According to Sidgwick’s concept, the central atom accepts the same number of electrons, as possessed by the next noble gas, from the ligand. The total number of electrons now possessed by the cental metal atom is termed its effective atomic number.

24.32 Inorganic Chemistry

The geometry and kind of bonding and magnetic properties were explained by valence bond theory developed by Pauling. According to this theory, the empty s, p and d atomic orbitals of the central atom undergo hybridisation followed by overlapping with the fully filled orbitals of the ligands to form coordinate bonds. The inner orbital or low-spin complex is formed by using (n – 1) d-orbitals, while the outer orbital or high-spin complex is formed by using nd orbital. The complexes containing unpaired electrons are paramagnetic, while those with all paired electrons are diamagnetic. Crystal field theory considers the attraction between the central metal atom and its ligands (considered as point charges) as purely electrostatic. Due to interaction of ligand with the central atom in specific geometries, the degeneracy of ‘d’ orbitals splits up. As a result in case of an octahedral complex, two sets of ‘d’ orbitals are formed, i.e. t2g orbitals (lower energy) and eg orbitals (higher energy). The splitting is known as crystal field splitting and the corresponding difference in energy is known as crystal-field splitting energy. The electrons are filled in these orbitals as per rules and if the crystal- field splitting energy exceeds the energy of electrostatic repulsions between the electrons, these are paired up (from d4 onwards) and form low-spin complex and otherwise high-spin complex is formed. The splitting in a tetrahedral complex is just reverse to that in an octahedral complex and only high-spin complexes are formed. The ligand field theory considers the overlap of the central metal atom with the suitable atomic orbitals of the ligand after appropriate linear combination based on the principles of the group theory. As a result, bonding molecular orbitals and antibonding molecular orbitals are formed. The orbitals nearer in energy to the energy of the ligand orbitals are considered to possess more ligand character, while those near to the energy of the central metal orbitals are considered to possess more metal orbital character. The molecular orbital diagram is thus unsymmetrical due to more electronegative ligands with considerably lower energy than the metal atom orbitals. The electrons occupied in the highest occupied molecular orbital can be excited into the lowest unoccupied molecular orbital to produce charge-transfer bonds.

The compound PtCl4·4NH3 precipitates four Cl– ions on the addition of Ag+ ions. Write down the formula and draw its structure on the basis of Werner’s coordination theory.

EXAMPLE 1

The coordination number of Pt is 4 and since four Cl– ions are precipitated, these ions are outside the coordination sphere, while the NH3 molecules are inside the coordination sphere. Thus, the formula of the compound can be written as [Pt(NH3)4]Cl4 [Pt(NH3)4]Cl4 $ [Pt(NH3)4]4+ + 4Cl–

Cl H3N Cl H3N

4Ag+ + 4Cl– $ 4AgCl. The structure of [Pt(NH3)4]Cl4 can be drawn as shown in the figure.

EXAMPLE 2 Calculate the EAN of Cr in (a) Cr(CO)6 and (b) [Cr(NH3)6]3+. Atomic number of Cr = 24 (a) Oxidation number of Cr in Cr(CO)6 = 0 No. of electrons in Cr(0) = 24 No. of electrons donated by six CO ligands = 6 × 2 = 12

Pt Cl

NH3 Cl NH3

Coordination Compounds-II Theories of Bonding

24.33

EAN of Cr = Number of electrons in Cr(0) + Number of electrons donated by six CO ligands = 24 + 12 = 36 (b) Oxidation number of Cr in [Cr(NH3)6]3+ = +3 No. of electrons in Cr2+ = 24 – 3 = 21 No. of electrons donated by six NH3 ligands = 6 × 2 = 12 EAN of Cr = 21 + 12 = 33

EXAMPLE 3 Discuss the type of hybridisation and magnetic behaviour of [V(H2O)6]3+ ion using VBT. [V(H2O)6]3+ contains V3+ ion with 3d2 configuration. H2O is a weak ligand, and there is no change in the number of unpaired electrons in V3+. Hence, it contains three vacant 3d-orbitals, one vacant 4s orbital and three vacant 4p orbitals and undergoes d2sp3 hybridisation to form an inner orbital complex with octahedral geometry as shown in the figure. 3+

OH2

V3+ ion 3d

4s

4p

OH2

H2O V

V3+ ion in [V(H2O)6]3+ ion

OH2

H2O OH2 OH2

OH2

OH2 OH2 OH2

OH2

Here, number of unpaired electrons, n = 2 Thus, m = n(n + 2) = 2(2 + 2) = 2.828 BM and the complex is paramagnetic.

EXAMPLE 4

Calculate the CFSE and magnetic moment for the following complexes using CFT. (a) [Fe(CN)6]4– (b) [Mn(H2O)6]3+ (a) [Fe(CN)6]4– contains Fe2+ ion with 3d6 configuration and CN– ion is a strong ligand, so the complex will be a low-spin complex with all paired electrons so that µ = o Fe2+(3d6) $ t62g e0g; CFSE = – 4 × 6Dq + 3P = –24 Dq + 3P (b) [Mn(H2O)6]3+ contains Mn3+ ion with d4 configuration. Since H2O is weak ligand, so the complex will be high-spin complex with four unpaired electrons. Mn(H2O)63+ $ t32g eg1; CFSE = –(4 × 3 + 6 × 1)Dq– = – 6Dq Since

n=4

m = n(n + 2) BM = 4(4 + 2) = 24 = 4.90 BM

EXAMPLE 5

Write down the distribution of electrons in the molecular orbitals of [Fe(CN)6] 3– ion with the help of MOT. Also determine its magnetic moment.

The molecular electronic configuration of an octahedral complex can be written as (ss) (sx) (sy) (sz) (sx2–y2) (sz2) (3dxy) (3dyz) (3dxz) (sx2–y2)* (sz2)* [Fe(CN)6]3– contains Fe3+ ion with 3d5 configuration and total (5 + 12) = 17 electrons.

24.34 Inorganic Chemistry

The distribution of electrons can be represented as (ss)2 (sx)2 (sy)2 (sz)2 (sx2–y2)2 (sz2)2 (3dxy)2 (3dyz)2 (3dxz)1 Hence, there is one unpaired electron, n = 1 m = n(n + 2) = 2 = 1.414 BM

QUESTIONS Q.1 Discuss the main points of Werner theory. What is the significance? Q.2 Discuss in brief the crystal field theory and represent the splitting of ‘d ’ orbitals in octahedral and tetrahedral fields. Q.3 Discuss the salient features of valence bond theory. What are all its applications? Q.4 Explain the following: (a) [Fe(H2O)6]3+ is paramagnetic and coloured. (b) (Ni(CN)4)2– is square planar, but [Ni(Cl)4]2– is tetrahedral. (c) [Co(NH3)6]3+ is more stable than [Co(NH3)6]2+, while [Co(H2O)6]2+ is more stable than [Co(H2O)6]3+. (d) Octahedral complexes are less stable than the square planar complexes. (e) [Co(CN)6]3– is a low-spin complex, but [CoF6]3– is high-spin complex. Q.5 Predict the colour and magnetic behaviour of the following complexes. [Sc(H2O)6]3+, [Ti(H2O)6]3+, [Ni(CO)4], [Cr(NH3)6]3+ Q.6 How does crystal field theory differ from valence bond theory? Q.7 Discuss Sidgwick’s concept of effective atomic number with the help of suitable example. Q.8 What are the limitations of valence bond theory? Q.9 Calculate CFSE for the following complexes: [Fe(CN)6]4–, [Cu(NH3)4]2+, [Co(CN)6]3–, [NiCl4]2– Q.10 Discuss the crystal field splitting in a square planar complex. Q.11 Discuss the factors affecting the magnitude of crystal field splitting. Q.12 Discuss the salient features of ligand field theory. How does it differ from crystal field theory? Q.13 Discuss the molecular orbital treatment for an octahedral complex with the help of a suitable example. Q.14 Compare the molecular orbital diagrams of π-acceptor ligands and π-donor ligands. Q.15 Discuss the molecular orbital diagram of a tetrahedral complex with the help of a suitable example. Q.16 Draw molecular orbital diagram of a square planar complex showing only π-bonding. Q.17 How does ligand field theory describe the origin of charge transfer spectra in coordination complexes.

MULTIPLE-CHOICE QUESTIONS 1. The number of ions produced on dissolution of hexamine cobalt (III) chloride is (a) 1 (b) 2 (c) 3 (d) 4 2. The effective atomic number of 36 corresponds to the complex (a) Fe(CO)5 (b) [Fe(CN)6]3– (c) [Cr(NH3)6]3+ (d) [Ag(NH3)2]+

Coordination Compounds-II Theories of Bonding

24.35

3. The geometry and magnetic behaviour of [MnCl4]2– is (a) tetrahedral and paramagnetic (b) tetrahedra and diamagnetic (c) square planar and paramagnetic (d) square planar and diamagnetic. 4. The example of an inner orbital complex is (a) [Co(NO2)6]4– (b) [CoF6]3– (c) [Co(NH3)6]3+ (d) none of these 2– 5. The CFSE corresponding to [MnCl4] is (a) –4Dq (b) – 6Dq (c) –8Dq (d) 0Dq 6. The symmetries of atomic orbital in the valence shell of the central metal ion in an octahedral complex corresponds to (a) t2g, eg (b) t1u (c) both ‘a’ and ‘b’ (d) none of these 3– 7. The highest occupied molecular orbital in [CoF6] is (a) eg (b) t1g (c) a*1g (d) eg*

chapter

25

Coordination Compounds III: Quantitative Basis of Crystal Field Theory

After studying this chapter, the student will be able to learn about

25.1

INTRODUCTION

The crystal field theory is based on the assumption that a central metal ion surrounded by the coordination sphere of the anion or ligands is under the effect of electrostatic forces of attraction between the cation and a set of negative point charges (anions or ligands) and the repulsive forces between the d-electrons of the central metal ion and the negative point charges. The qualitative aspect of crystal-field splitting has already been discussed in chapter 24 and the quantitative aspect will be discussed in this chapter. The orbital eigen function for a single electron in a central field can be represented as follows: y(r, q, f) =

R(r)

Q(q) F(f)

radial wave function

angular wave function

or

n,l,ml

= Rn,l × Ylml

Where Rn,l is the radial part and Ylml is the angular part. The radial part is unaffected by the octahedral field i.e. it does not lead to splitting of energy. Hence, we will not consider this part. The spherical harmonic Ylml is given by the expression Ylml = Qml l( ) × (2 )–½ eiml

25.2

Inorganic Chemistry

Table 25.1 lists the values of Ylml for d-orbitals. Table 25.1 The values of spherical harmonic Ylml n

ml

0

0

2

0

2

2

Ylml

n

1

Ylml

ml

4

0

9 (35 cos4q - 30 cos2q + 3) 256p

( 3cos2θ − 1)

4

±1

45 sin q cos q (7 cos2q - 3) e ± if 64p

±1

15 sin θ cosθ e ± iφ 8π

4

±2

45 sin 2q (7 cos2q - 1) e ± i 2f 128p

±2

15 sin 2q e + i 2f 32p

4

±3

315 sin 3 θ cosθ e ± i 3φ 64π

4

±4

315 sin 4q e ± i 4f 512p

4 5 16π

Ψ1 Ψ2

Ψ1 H11 − E H 21

Ψ2 H12 H 22 − E

Ψ3 H13 H 23

Ψ4 H14 H 24

Ψ5 H15 H 25

Ψ3 Ψ4 Ψ5

H31 H 41 H 51

H32 H 42 H 52

H33 − E H 43 H 53

H34 H 44 − E H 54

H35 H 45 H 55 − E

=0

For a d1 electron, l = 2 and ml = +2, 1 and 0. Since a single-electron wave function n,l,ml can be simply written as (ml), the wave functions corresponding to the degenerate set of d orbitals 1, 2, 3, 4, 5, are written as (+2), (+1), (0), (–1), and (–2), respectively.

25.2

DETERMINATION OF OCTAHEDRAL CRYSTAL FIELD POTENTIAL

We have to consider the changes taking place in the energy of d-orbitals of the metal ion under the effect of octahedral field, hence the energy of the free metal ion, Eo can be taken arbitrarily as zero. Thus, H11 reduces to H11 =

ÚY

* 1

H ¢ Y1 dt 1 =

ÚY

And H12 reduces to H12 = Ú Y1* H ¢ Y 2 dt 1 dt 2 =

* 1

Voct Y1 dt 1

ÚY

* 1

Voct dt 1 , dt 2

Coordination Compounds III: Quantitative Basis of Crystal Field Theory

25.3

The secular determinant is also rewritten as follows:

+2 +1 +2 H +¢2, +2 - E ¢ H +¢2, +1

0

-1

-2

H +¢2,0

H +¢2, -1

H +¢2, -2

+1 H +¢1, + 2

H +¢1, +1 - E ¢

H +¢1,0

H +¢1, -1

H +¢1, -2

0

H 0¢, + 2

H 0¢, +1

H 0¢,0 - E ¢

H 0¢, -1

H 0¢, -2

-1 H -¢1, +2

H -¢1, +1

H -¢1,0

H -¢1, -1 - E ¢

H -¢1, -2

-2 H -¢2, + 2

H -¢2, +1

H -¢2,0

H -¢2, -1

H -¢2, -2 - E ¢

=0

Now we have to solve the secular equation and obtain the matrix elements Hm , m = l

l

Ú (m ) H ¢ (m ¢) dt = Ú (m ) V (m ¢) dt * l

l

* l

oct

where ml may or may not be equal to ml. We will first obtain the value of Voct and then use it for the determination of the matrix elements. If at any particular instant, the d-electron is present at any point j (x, y, z) and the ligands are placed at the point i, where i = 1, 2, 3, ..... 6 along the Cartesian axes (Fig. 25.1), the electrostatic potential Voct (x, y, z) experienced by the d-electron can be expressed as

Ze2 i =1 rij where rij is the distance of the central metal ion from the ligand. The term 1 can be conveniently expressed as the standard expansion: rij +n 4p r rij 6

Voct(x, y, z) =

∑

l

z x

i=5 (0, 0, a)

i=3 (a, 0, 0) o

i=4 (0, –a, 0, )

y

i=2 (0, a, 0)

i=1 i=6 (–a, 0, 0, ) (0, 0, –a)

Fig. 25.1

Position coordinates of ligand points in an octahedral complex

*

where Ynmi l and Ynmj l are the spherical harmonics as functions of the angles and , and r> and r< are the greater and lesser of the distances of the point i and j from the chosen origin respectively. If the metal ligand distance is considered simply as a, the equation can be written as

1 =Â rij n = 0

+n

4p r n ml ml* Y Y n n n +1 j i ml =- n 2 n + 1 a

Â

where, r> = a and r< = r Coordinate positions of all the ligands and all the angles has been shown in Table 25.2. Figure 25.2 represents the internuclear distance rij. Thus, the octahedral potential r = (x, y, z) close to the origin can be expressed as 2

1 1 1 1 1  1 Ze + + + +  = Ze2  +  r1 j r2 j r3 j r4 j r5 j r6 j  i =1 rij 6

Voct (x, y, z) =

∑

We can solve this equation for different values of n and ml, where ml ranges from –n to +n and the value of n ranges from f to . The above equation can be written in the form of ml and n as follows: +n

* * * * * * rn 4p . n+1 ÈÎYnjml Ynm1l + Ynjml Ynm2l + Ynjml Ynm3l + Ynjml Ynm4l + Ynjml Ynm5l + Ynjml Ynm6l ˘˚ n = 0 ml =- n 2 n + 1 a

Voct (x, y, z) = Ze2 Â

Â

25.4

Inorganic Chemistry

Table 25.2

Position coordinates and values of angle and

Ligand points i=1

Position coordinates (–a, 0, 0)

/2

i=2

(0, a, 0)

/2

/2 0

i=3

(a, 0, 0)

/2

/2

i=4

(0, –a, 0)

i=5 i=6

(0, 0, a) (0, 0, –a)

/2 0

i

z

rij

x

j

ri

i

–

i

rj

qj qi fi

y

fj

Fig. 25.2 Representation of angle and involved in the standard

— —

expansion of

1 rij

1. For n = 0 and ml = 0 The spherical harmonic Y00 is independent of all the angles and thus the value is same for all the points and is equal to

1 4

; thus Voct can be determined as follows.

4/ π/  6  6 Ze2 = a  4 π  a 2. For n = 2 and ml = 0 Voct = Ze2 ×

The spherical harmonic Y 02i = Y 0* 2i = Voct = Ze2 ×

Voct = Ze2 ×

5 ( 3 cos2 θi − 1) 16π

4p r2 5 È(3 cos2 q1 - 1) + (3 cos2 q 2 - 1) + (3 cos2 q 3 - 1) ¥ 2 +1 Y20j 2 ¥ 2 +1 a 16p Î + (3 cos2 q 4 - 1) + (3 cos2 q 5 - 1) + (3 cos2 q 6 - 1)˚˘ 4p r 2 5 ¥ ¥ Y20j 5 a3 16p

ÈÊ ˆ ˆ Ê ˆ Ê 2 p 2 p 2 p ÍÁË 3 cos 2 - 1˜¯ + ÁË 3 cos 2 - 1˜¯ + ÁË 3 cos 2 - 1˜¯ Î

˘ p ˆ Ê + Á 3 cos2 - 1˜ + (3 cos2 (0) - 1) + (3 cos2 p - 1)˙ = 0 Ë 2 ¯ ˚ 3. For n = 2 and ml = +1 The spherical harmonic Y2+i1 = Y2+i1* =

Voct = Ze2 ×

Voct = Ze2 ×

15 sin q i cos q i eifi 8p

4π r2 15 sin θ1 cos θ1 eiφ1 + sin θ 2 cos θ 2 eiφ2 + sin θ3 cos θ3 eiφ3 ⋅ 2 +1 ⋅ Y21j 2× 2 +1 a 8π  + sin θ 4 cos θ 4 eiφ4 + sin θ 5 cos θ 5 eiφ5 + sin θ6 cos θ6 eiφ6  -p 4p r 2 1 15 È p p iÊÁ ˆ˜ p p p p ◊ 3 ◊ Y2 j sin cos e Ë 2 ¯ + sin cos ei(0) + sin cos ei (p / 2 ) Í 5 a 8p Î 2 2 2 2 2 2 p p i(p ) ˘ + sin cos e + sin 0 cos(0) (0) + sin p cos p ¥ (0)˙ = 0 ˚ 2 2

Coordination Compounds III: Quantitative Basis of Crystal Field Theory

4. For n = 2 and ml = –1 –1 The spherical harmonic function Y 2i = Y2i–1* = Thus, Voct = 0, as Ql, + ml = Ql,– ml 5. For n = 2 and ml = 2 The spherical harmonic function Y2i+2 = Y +2* 2i = Voct = Ze2 ×

25.5

15 sin θi cos θi e −iφi 8π 15 sin 2 θi ei 2φi 32π

4π r2 15 sin 2 θ1 ei 2φ1 + sin 2 θ 2 ei 2φ2 + sin 2 θ3 ei 2φ3 ⋅ 2 +1 ⋅ Y22j × 2× 2 +1 a 32π  + sin 2 θ 4 ei 2φ4 + sin 2 θ 5 ei 2φ5 + sin 2 θ6 ei 2φ6 

= Ze2 ×

4p r 2 2 Y2 j 5 a3

Ê -p ˆ p 15 È p i 2Á ˜ p p i(2) Ísin 2 e Ë 2 ¯ + sin 2 ei (2 )(0 ) + sin 2 e 2 32p Î 2 2 2 p ˘ + sin 2 ei 2(p ) + sin 2 (0)(0) + sin 2 p (0)˙ ˚ 2

{

Voct = Ze2 ×

4π r 2 2 Y2 j × 5 a3

15 e −iπ + e0 + eiπ + ei 2π + 0 + 0  32π 

Voct = Ze2 ×

4p r 2 2 ◊ Y2 j 5 a3

15 [(cos p - i sin p ) + (1) + (cos p + i sin p ) + (cos 2p + i sinn 2p )] 32p + 0 + 0]

∵sin 2

}

π =1 2

{∵ e±if = cos f ± i sin f }

4π r 2 2 15 ⋅ 3 Y2 j [ −1 − 0 + 1 − 1 + 0 + 1 + 0 + 0] = 0 5 a 32π 6. For n = 2 and ml = –2 15 –2 = Y2i–2* = The spherical harmonic function Y2i sin 2 θi e −i 2φi 32π Thus Voct = 0, as Ql, + ml = Ql,– ml Voct = Ze2 ×

7. For n = 4 and ml = 0 The spherical harmonic function Y40i = Y40i* = Voct = Ze2 ×

4p r4 ¥ 4 +1 Y40j ¥ 2 ¥ 4 +1 a

9 ( 35 cos4 θi − 30 cos2 θi + 3) 256π

(

) (

9 È 35 cos4 q - 30 cos2 q + 3 + 35 cos4 q - 30 coos2 1 1 2 256p Î

(

) ) ( + (35 cos4 q 5 - 30 cos2 q 5 + 3) + (35 cos4 q 6 - 30 cos2 q 6 + 3)˘˚

q 2 + 3) + 35 cos4 q 3 - 30 cos2 q 3 + 3 + 35 cos4 q 4 - 30 cos2 q 4 + 3

Voct = Ze2 ×

4p r 4 0 ◊ Y4 j ¥ 9 a5

9 256p

È Ê 4p 4 p 4 p 4 p 4 4 ˆ Í35 ÁË cos 2 + cos 2 + cos 2 + cos 2 + cos (0) + coss p ˜¯ Î

˘ p p p p Ê ˆ - 30 Á cos2 + cos2 + cos2 + cos2 + cos2 (0) + cos2p ˜ + 18˙ Ë ¯ 2 2 2 2 ˚

25.6

Inorganic Chemistry

Voct = Ze2 ×

4π r 4 0 9 Y4 j × [35 ( 0 + 0 + 0 + 0 + 1 + 1) − 30 ( 0 + 0 + 0 + 0 + 1 + 1) + 18] 9 a5 256π

Voct = Ze2 ×

4π r 4 0 9 Y4 j × [70 − 60 + 18] 9 a5 256π 9 r4 Ze2 5 Y40j 256π a

Voct = 28 × 4π ×

9 8. For n = 4 ml = ±1

The spherical harmonic function Y ±1* = Y ±1 = 4i 4i Since sin

i

cos

i=

45 sin θi cos θi ( 7 cos2 θi − 3 ) e ± iφi 64π

0 for all combinations of i, hence Voct = 0

9. For n = 4, ml = ± 2 The spherical harmonic function Y ±2* = Y ±2 = 4i 4i

45 sin 2 θi ( 7 cos2 θi − 1) e ± i 2φi 128π

Similar to the earlier calculations, Voct can be shown = 0 10. For n = 4, ml = ± 3

315 sin 3 q i cos q i e ± i 3fi 64p since sin3 i cos i = 0 for all combinations of i, hence Voct = 0

The spherical harmonic function

Y4±i3 = Y4±i3* =

11. For n = 4, ml = + 4 The spherical harmonic function Voct = Ze2 ×

Y4+i4 = Y4+i4* =

315 sin 4 θi e + i 4φi 512π

4π r4 315 sin 4 θ1 ei 4φ1 + sin 4 θ 2 ei 4φ2 + sin 4 θ3 ei 4φ3 ⋅ 4 +1 ⋅ Y4+j4 2× 4 +1 a 512π  + sin 4 θ 4 ei 4φ4 + sin 4 θ 5 ei 4φ5 + sin 4 θ6 ei 4φ6  È 4 p i 4ÊÁË -2p ˆ˜¯ p p e + sin 4 ei ( 4 )( 0 ) + sin 4 ei 4 (p / 2 ) Ísin Î 2 2 2 p ˘ + sin 4 ei 4 (p ) + sin 4 (0)(0) + sin 4 (p )(0)˙ ˚ 2

Voct = Ze2 ×

4p r 4 +4 315 ◊ Y4 j 9 a5 512p

Voct = Ze2 ×

4p r 4 +4 315 È1 ¥ e - i 2p + 1 ¥ e0 + 1 + ei 2p + 1 ¥ ei 4p + 0 + 0 ˘˚ ◊ ◊ Y4 j 9 a5 512p Î

Voct = Ze2 ×

4p r 4 +4 315 [cos 2p – i sin 2p + 1 + cos 2p + i sin 2p + cos 4p + i sin 4p + 0 + 0] Y4 9 a 5 j 512p

Voct = Ze2 ×

4π r 4 +4 315 Y4 j [1 + 1 + 1 + 1 + i(0) + 0 + 0] 9 a5 512π

Voct = 4 ×

4π 9

315 r4 Ze2 5 Y4+j4 512π a

Coordination Compounds III: Quantitative Basis of Crystal Field Theory

25.7

12. For n = 4, ml = –4 The spherical harmonic function Y4-i4 = Y4-i4* = V

ct

= 4×

4π 9

315 sin 4 q i e - i 4fi 512p

315 r4 Ze2 5 Y4−j4 512π a

We have not determined V ct for odd values of n, because for d orbital wave functions with l = 2, the parts of the elements of the determinant containing odd values of n vanish. We are also not considering the case for n > 4 as the expansions with n > 2l becomes reductant. Now we can express the overall value of V ct as the sum of all the determined terms as follows:

6 Ze2 4p 9 r4 4p Ze2 5 Y40j + 4 ¥ + 28 ¥ ¥ a 9 256p 9 a

Voct =

=

315 r2 4p Ze2 5 Y4+j4 + 4 ¥ 512p 9 a

315 r2 Ze2 5 Y4-j4 512p a

4 6 Ze2 784 ¥ 16p 2 ¥ 9 Ze2 r 4 0 256p 2 ¥ 315 256 p 2 ¥ 315 Ze2 r 4 2 r +4 Y Ze Y + + ¥ + ¥ 5 4j 4j a 81 ¥ 256p 512p ¥ 81 512p ¥ 81 a5 a5 a

Voct =

 6 Ze2 49 Ze2 r 4  0 5 + 2π × Y + Y4+j4 + Y4−j4  6 18 a 5  4 j 14 

(

)

This equation can be expressed in terms of two functions as follows: V

ct

= (Voct)1 + (Voct)2

 6 Ze2 49 Ze2 r 4  0 5 and ( Voct )2 = 2π Y4+4 + Y4−4 )  ( Y4 + 5 6 18 a 14   * We have to determine the elements of the 5 × 5 determinant Q m ,m ¢ = Ú (ml ) Voct (ml¢) dt l l where (V ct)1 =

We can obtain the matrix elements related to V1 and V2 separately. 2 (a) Matrix element related to (V ct)1, (m )* Ê 6 Ze ˆ ( m ¢) dt = 6 Ze (m )* ( m ¢) dt Ú l Á ˜ l Ú l l

Ë a ¯ When the integral is extended over the definite volume,

Ú (m ) (m¢) dt = 0 for m π m¢ Ú (m ) (m¢) dt = 1 for m π m ¢ * l

and

a

* l

l

l

* l

* l

l

l

2   Thus (V ct)1 contributes uniformly  equal to 6 Ze  to the change in the energy of five d-orbitals a   but will not split it.

(b) The matrix element related to (V ct)2 will cause splitting of the enegy of the five d-orbitals. Thus, we have to determine the integral related to (V ct)2, i.e.

Ê 49 Ze2r 4 * * ( m ) ( V ) ( m ) d t ( m ) = ¢ p 2 oct 2 l l l Ú Ú Á 18 a5 Ë Considering the d-function, (ml) = Rn,l Ylml or

È 0 ˘ˆ 5 +4 Y4 j + Y4-j4 ˙˜ (ml¢)dt ÍY4 j + 14 Î ˚¯

(ml) = Rn,2 Y2ml

(

)

25.8

Inorganic Chemistry

We can rewrite the integral as follows:

ˆ Ê 49 Ze2 ˆ Ê * 4 * V m ¢ ( m ) d = t R r Rn,2 r 2 dr ˜ ¥ ( ) ( ) 2 p oct l l Ú ÁË 2 5 ˜ Á Ú n ,2 ¯ a ¯Ë0 18 Êp ÁÚ Ë0

2p

ÚY

ml* 2

(Y40 ) +

0

ˆ 5 +4 Y4 j + Y4-j4 Y2ml¢ sin q dq df ˜ 14 ¯

(

)

s Now considering the general property of the radial functions for average value of rn,l,

∞

∫ Rn,l r *

0

s

s Rn,1 r2dr = , we can rewrite the equation as

p Ê 49 Ze2 4 ˆ Ê * V m ¢ ( m ) d = t ( ) ( ) 2 p r l oct l Ú ÁË ˜ ÁÚ 2 a5 2 ¯ Ë 0 18

2p

ÚY

ml* 2

0

Ê 49 ˆ È Ze2 or (ml )* (Voct ) ( ml¢) dt = 2p 5 r24 ˜ ¥ Í Ú Ú ÁË 2 ¯ Î0 18 a p

5 14

ˆ 5 +4 Y4 j + Y4-j4 Y2ml¢ sinq dq df ˜ 14 ¯

(

)

2p

ÚY

ml* 2

m¢ Y40 Y2 l sin q dq df +

0

p

2p

Ú

m m¢ Ú Y2 l Y44 Y2 l sin q dq df +

0

(Y40 ) +

*

0

5 14

p

2p

Ú ÚY

ml* 2

0

0

˘ m¢ Y4-4 Y2 l sin q dq df ˙ ˚

Again considering the spherical harmonics expression as

1

Ylml = Q lml (q ) ¥

eiml f

2p

And considering the properties of the spherical harmonics that the integral 2p

Ú Yl

ml1

1

¥ Yl

ml2

¥ Yl

1

ml3

0 only when ml1 + ml2 + ml3 = 0.

df

1

0 2π

which in our case,

∫ Y2

m*

m * Y44 Y2 l dφ ≠ 0 only when ( ml* + 4 + ml′ ) = 0 , i.e. ml + m´l = –4 ′

0

For all other values of m*l and ml,

2p

ml*

Ú Y2

m¢

Y44 Y2 l df = 0

0

This is because the involved integral part 2p

Úe

i ( ml 1 + ml 2 + ml 3 )f

df = 0 when (ml + ml + ml ) 1 2 3

0

and 2p

Similarly,

0

= 2 when (ml1 + ml2 + ml3) = 0 ml*

Ú Y2

(

m¢

)

* Y40 Y2 l df π 0 only when ml* + 0 + ml¢ = 0 i.e, ml = - ml¢

0

This means that the secular determinant can be reduced with vanishing the parts where these rules are not followed and we are finally left with the reduced 5 × 5 determinant as shown ahead. Remembering that we have considered only the (V ct)2 contribution towards the matrix element Hm ,m , i.e. H ml , ml¢ = l

l

Ú (m ) (V ) (m ¢) dt * l

oct 2

l

Coordination Compounds III: Quantitative Basis of Crystal Field Theory

25.9

Using the integral values as r

0 r

r

0 |1| sin i di = – ; s Q |1| 2 Q4 Q 2

18 7

s Q02 Q04 Q02 sin i di =

0 |2| sin i di = s Q|2| 2 Q4 Q 2

0

0

2 14

r

|4| |2| sin i di = ; s Q |2| 2 Q4 Q 2 0

8 7 35 7

we can obtain matrix elements as follows: For m*l = 0 and ml = 0 * Ú (0) (Voct )2 d = (0)

Ze2 r24 a5

= 6 Dq

For ml* = ± 1 and ml* = ± 1;

* Ú (±1) (Voct )2 (±1) dt =

-2 Ê Ze2 r24 ˆ = - 4 Dq 3 ÁË a 5 ˜¯

For ml* = ± 2 and ml* = ± 2;

Ú (±2) (V )

(±2) dt =

1 Ê Ze r24 ˆ = Dq 6 ÁË a 5 ˜¯

For ml* = ± 2 and ml* = ∓ 2;

Ú (±2) (V )

(±2) dt =

5 Ê Ze r24 ˆ = 5Dq 6 ÁË a 5 ˜¯

*

oct 2

*

oct 2

Here, we have to introduce a parameter Dq (in CGS units), called the crystal-field splitting parameter;

 Ze2 2  , where D = 35 5 and q r24 105 a 4  D depends upon the ligands and q represents the properties of the metal ion. Thus, Dq varies from one complex to another. Now we can rewrite the determinant as follows: Dq =

1  Ze2 r24  6  a 5

(+2) (+2) Dq - E ¢ (+1) (0 ) (-1)

0 0 0

(-2)

5Dq

(+1) 0

(-1) 0

(0 ) 0

-4 Dq - E ¢ 0 0 0 -6 Dq - E ¢ 0 0 0 -4 Dq - E ¢ 0

0

0

(-2) 5Dq 0 0 0

=0

Dq - E ¢

This determinant is easily reduced to the subdeterminants: (a) For (+1) and (–1) States

(+1) (+1) −4 Dq − E ′ (−1)

0

(−1) 0 −4 Dq − E ′

This gives C+1 (–4Dq – E ) + C–1 (0) = 0

& E = –4Dq [For (+1) state]

and

& E = –4Dq [For (–1) state]

C+1 (0) + C–1 (–4Dq – E ) = 0

25.10 Inorganic Chemistry

This means that the wave function (+1) and the wave function (–1) both have same energy and remain degenerate under the effect of Voct. Also, for the normalised wave functions, (C+1)2 + (C–1)2 = 0 or C+1 = 1 and C-1 = -1

2

2

Thus, the acceptable linear combinations of two degenerate wave functions can be represented as

1

[(+1) + (-1)] and

2 d xz orbital

1

[(+1) - (-1)]¸Ô

2 d yz orbital

˝ E ¢ = -4 Dq Ô ˛

(b) For (+2) and (–2) States

(+2) (-2) (+2) Dq - E ¢ 5Dq =0 (-2) 5Dq Dq - E ¢ This gives C+2 (Dq – E ) + C–2 (5Dq) C+2 (5Dq) + C–2 (Dq – E ) = 0 The solution of this sub determinant gives (Dq – E )2 – (5Dq)2 = 0 or (Dq – E + 5Dq ) (Dq – E – 5Dq ) = 0 We can solve this equation as E + 6Dq = 0 or

E = + 6Dq

and and

–E – 4Dq = 0 E = –4Dq

Using E = + 6Dq we can solve the first equation as C+2 (Dq – 6Dq) + C–2 (5Dq) = 0 or

C+2 (5Dq) + C–2 (5Dq) = 0

or C+2 + C–2 = 0 Considering the normalisation of wave functions, (C+2)2 + (C–2)2 = 1 Thus

C+2 =

1 2

and C–2 = –

1 2

1

It means that the wave function

[(+2) + (–2)] is associated with energy, E = 6Dq 2 Now we can use E = –4Dq to solve the same equation as C+2 (Dq + 4Dq) + C–2 (5Dq) = 0 or

C+2 (5Dq) + C–2 (5Dq) = 0 or C+2 + C–2 = 0

This means the wave function

1

[(+2) (–2) ] is associated with energy E = – 4Dq 2 These wave functions can be represented as 1 [(+2) + (–2)] $ E = + 6Dq 2

Coordination Compounds III: Quantitative Basis of Crystal Field Theory

1

6Dq

[(+2) + (–2)] $ E = – 4Dq

2

Energy E

(c) For (0) States (0) (0)

x 6Ze2

25.11

D0 = 10Dq

–4Dq

a

Fig. 25.3 Splitting of energy of five d-orbital

6 Dq – E = 0

This gives E = 6Dq meaning that the wave function (0) has energy E = 6Dq This discussion shows that the degenerate set of five d-orbitals split up into two levels, one consisting of higher energy eg set of two degenerate orbitals of energy equal to 6 Dq and another consisting of lower energy t2g set of three degenerate orbitals of energy equal to –4Dq as shown in Fig 25.3.

25.3

DETERMINATION OF TETRAGONAL CRYSTAL FIELD POTENTIAL

Consider the case of d1 metal ion under the effect of tetragonal crystal field, i.e. tetragonally surrounded z by six ligands considered as point charges. In the tetragonal (0, 0, b) environment, four ligands are present at equal distances from i=5 x the metal (say a along x and y direction) while the two ligands are present at comparatively larger distances from the metal i = 3 (a, 0, 0) b (say b along z direction) as shown in Fig. 25.4. a The values of and will remain the same and we will y a a i=2 work on the same principle. This means that we have to obtain i=4 a (0, a, 0) (0, –a, 0) the value of tetragonal electrostatic potential, Vtetg expressed as 4

Vtetg =

n

 Ze   2

i =1

n=0

ml = - n

4p r n ml ml* Yn Yn 2 n + 1 a n +1 j i

i=1 (–a, 0, 0)

b (0, 0, –b)

Fig. 25.4 Position coordinates of ligand 4p r n ml ml* Y Yni points in tetragonal complex n +1 n j 2 n + 1 b i=5 n = 0 ml = - n This equation has been written by considering the metal–ligand distance equal to a for the ligand point charges i = 1 to 4 and the metal–ligand distance equal to b for the ligand point charges for i = 5 and 6. This equation can be expanded and solved for the different values of n and ml as shown ahead. n * * * * 4p rn Vtetg(x,y,z) = Ze2 Â Â ◊ n +1 ÈYnmj l Ynm1l + Ynmj l Ynm2l + Ynmj l Ynm3l + Ynmj l Ynm4l ˘ Î ˚ n = 0 ml = - n 2 n + 1 a 6

n

+ Â Ze2 Â

Â

+ Ze2 Â

Â

n=0

n

ml = - n

* * 4p rn ◊ n +1 ÈYnmj l Ynm5l + Ynmj l Ynm6l ˘ Î ˚ 2n + 1 b

1. For n = 0 and ml = 0 The spherical harmonic Y00 is independent of all the angles and thus the value is same for all the point and is equal to

1

4 Vtetg = Ze2 .

; thus, Vtetg can be determined as follows:

1 1 1 1 1 1 1 ˘ 4p r0 È 1 ¥ + ¥ + ¥ + ¥ . 0 +1 Í ˙ 2(0) + 1 a Î 4p 4p 4p 4p 4p 4p 4p 4p ˚ + Ze 2

4p a (0 )+1

.

1 1 1 ˘ r0 È 1 . . + ˙ 0 +1 Í 4p 4p ˚ b Î 4p 4p

25.12 Inorganic Chemistry

Vtetg = 4 Ze

2

2

a

Ze2 b

2. For n = 2 and ml = 0 The spherical harmonic Y 02i = Y 0* 2i = Vtetg = Ze2 ¥

4p 5 r2 È(3 cos2 q1 - 1) + (3 cos2 q 2 - 1) + (3 cos2 q 3 - 1) + (3 cos2 q 4 - 1)˘˚ ¥ 2 +1 Y20j 2 ¥ 2 +1 a 16p Î

+ Ze2 ¥

4p r2 5 È(3 cos2 q 5 - 1) + (3 cos2 q 6 - 1)˘˚ ¥ 2 +1 Y20j 2 ¥ 2 +1 b 16p Î

Vtetg = Ze2 ¥

4p r 2 5 0 ÈÊ p ˆ˘ p ˆ Ê p ˆ Ê p ˆ Ê ¥ 3 Y Á 3 cos2 - 1˜ + Á 3 cos2 - 1˜ + Á 3 cos2 - 1˜ + Á 3 cos2 - 1˜ ˙ 5 a 16p 2 j ÍÎË 2 ¯ Ë 2 ¯˚ 2 ¯ Ë 2 ¯ Ë

+ Ze2 ¥

4p r 2 ¥ 5 b3 2

Vtetg = Ze2 ¥ 4p ¥ r . 3

5

Vtetg =

5 ( 3 cos2 θi − 1) 16π

a

5 0 Y È(3 cos2 (0) - 1) + (3 cos2 p - 1)˘˚ 16p 2 j Î 5 0 4p r 2 5 Y2 j ( -1 - 1 - 1 - 1) + Ze2 ¥ ¥ 3. Y20j (2 + 2 ) 16p 5 b 16p

16π 2 5 1 1 r . Ze2Y20j  3 − 3  5 16π a  b

3. For n = 2 and ml = ±1 The spherical harmonic Y2+i1 = Y2+i1* =

Vtetg =

Ze2 ¥

15 sin θi sin θi eiφi 8π

4p 15 È r2 ¥ 2 +1 . Y21j sin q1 cos q1 eif1 + sin q 2 cos q 2 eif2 + siin q 3 cos q 3 eif3 2 ¥ 2 +1 a 8p Î + sin q 4 cos q 4 eif4 ˘˚

+ Ze2 ¥

4p r2 15 Èsin q 5 cos q 5 eif5 + sin q 6 cos q 6 eif6 ˘˚ ¥ 2 +1 . Y21j 2 ¥ 2 +1 b 8p Î Êpˆ

Ê -p ˆ 4p r 2 1 15 È p p p i (0 ) p p i ÁË 2 ˜¯ p iÁ ˜ . 3 . Y2 j Ísin cos e Ë 2 ¯ + sin cos e + sin cos e 5 a 8p Î 2 2 2 2 2 2 p p ip ˘ + sin cos e ˙ ˚ 2 2 2 15 r 1 2 4p + Ze . ¥ .Y j [sin(0) cos(0)(0) + sin p cos p (0)] = 0 5 b3 2 8p 4. For n = 2 and ml = –1 The spherical harmonic function Y −1 = Y −1* = 15 sin θ cos θ e −iφ1 2i 2i i i 8π

Vtetg = Ze2 .

Coordination Compounds III: Quantitative Basis of Crystal Field Theory

25.13

Thus, Vtetg = 0, as Ql,+ ml = Ql, – ml 5. For n = 2 and ml = 2; The spherical harmonic function Y2+i2 = Y2+i2* = Vtetg = Ze2 .

4p 15 r2 Èsin 2 q1 ei 2f1 + sin 2 q 2 ei 2f2 + sin 2 q 3 ei 2f3 + sin 2 q 4 ei 2f4 ˘˚ . 2 +1 .Y2+i2 2 ¥ 2 +1 a 32p Î

+ Ze2 . Vtetg = Ze2 .

4p r2 15 Èsin 2 q 5 ei 2f5 + sin 2 q 6 ei 2f6 ˘˚ . 2 +1 .Y2+i2 2 ¥ 2 +1 a 32p Î

4p r 2 2 15 . .Y 5 a 3 2 j 32p

+ Ze2 .

Êpˆ Ê -p ˆ È ˘ p i ( 2 )Á ˜ p p i ( 2 )Á ˜ p Ísin 2 e Ë 2 ¯ + sin 2 ei (2 )(0) + sin 2 e Ë 2 ¯ + sin 2 ei ( 2 )(p ) ˙ Î ˚ 2 2 2 2

4p r 2 2 15 Èsin 2 (0)(0) + sin 2 p (0)˘˚ . .Y2 j 5 a3 32p Î 2

Vtetg = Ze2 . 4p . r .Y22j 3

5 a

2

Vtetg = Ze2 . 4p . r .Y22j 3

5 a

2

Vtetg = Ze2 . 4p . r .Y22j 3

5 a

15 - ip [e + e0 + eip + ei 2p ] + 0 32p 15 [(cos p - i sin p ) + 1 + (cos p + i sin p ) + (cos 2p + i sin 2p )] 32p 15 [-1 - 0 + 1 - 1 + 0 + 1] = 0 32p

6. For n = 2 and ml = –2 The spherical harmonic function Y2-i2 = Y2-i2* = Thus, Vtetg = 0, as Ql,+ ml = Ql, – ml 7. For n = 4 and ml = 0 The spherical harmonic function Y40i = Y40i* = 2 Vtetg = Ze ◊

15 sin 2 q i ei 2fi 32p

15 sin 2 q i e - i 2fi 32p 9 ( 35 cos4 θi − 30 cos2 θi + 3) 256π

4p 9 r4 È (35 cos4q - 30 cos2q + 3) + (35 cos4q - 30 cos2q + 3) ◊ 4 +1 ◊ Y40j ◊ 1 1 2 2 2 ¥ 4 +1 a 256p Î

(

)

+ (35 cos4q 3 - 30 cos2q 3 + 3) + (35 cos4q 4 - 30 cos2q 4 + 3) ˘˚ + Ze2 ◊

4p r4 9 È (35 cos4q - 30 cos2q + 3) + (35 cos4q - 30 cos2q + 3) ˘ ◊ 4 +1 ◊ Y40j ◊ 5 5 6 6 ˚ 2 ¥ 4 +1 b 256p Î

(

4

Vtetg = Ze2 . 4p . r .Y 0 . 5 4j

9 a

9 256p

)

ÈÏÊ p p p ˆ ˆ Ê 4 p - 30 cos2 + 3˜ + Á 35 cos4 - 30 cos2 + 3˜ ÍÌÁË 35 cos Ë ¯ ¯ 2 2 2 2 ÎÓ

p p p p ˆ ¸˘ Ê ˆ Ê + Á 35 cos4 - 30 cos2 + 3˜ + Á 35 cos4 - 30 cos2 + 3˜ ˝˙ ¯ ˛˚ ¯ Ë 2 2 2 2 4π r 4 0 9 Ë 4 2 + Ze2 . . 5 .Y4 j . + ( 35 cos04 − 30 cos20 + 3 )  35 cos 30 cos 3 − + ( ) 0 0  9 a 256π 

{

}

25.14 Inorganic Chemistry

4

Vtetg = Ze2 . 4p . r .Y40j . 5

9 a

Vtetg = 16π r 4

9

9 4p r 4 9 (12 ) + Ze2 ◊ ◊ 5 Y40j (16 ) 256p 9 b 256p

9 4   3 ⋅ Ze2Y40j  5 + 5  256π a b  

8. For n = 4, ml = ±1

45 sin θi cos θi ( 7 cos2 θi − 3 ) e ± iφi 64π Since sinqi cosqi = 0 for all combinations of qi, hence, Vtetg = 0 The spherical harmonic function Y4±i1* = Y4±i1 =

9. For n = 4, ml = ±2

45 sin 2 θi ( 7 cos2 θi − 1) e ± i 2φi 128π Similar to the earlier calculations, Vtetg can be shown = 0 The spherical harmonic function

Y4±i2* = Y4±i2 =

10. For n = 4, ml = ±3 The spherical harmonic function Since sin3qi cos qi

Y4±i3* = Y4±i3 =

for all combinations of qi, hence Vtetg = 0

11. For n = 4, ml = +4 The spherical harmonic function Y4+i4 = Y4+i4* = Vtetg = Ze2 ⋅

315 sin 3 θi cos θi e ± i 3φi 64π

315 sin 4 θi e + i 4φi 512π

4π 315 r4 sin 4 θ1ei 4φ1 + sin 4 θ 2 ei 4φ2 + sin 4 θ3 ei 4φ3 + sin 4 θ 4 ei 4φ4  ⋅ 4 +1 ⋅ Y4+j4 2× 4 +1 a 512π 

4π 315 r4 siin 4 θ 5 ei 4φ5 + sin 4 θ6 ei 4φ6  ⋅ 4 +1 ⋅ Y4+j4 2× 4 +1 b 512π  4 Ê -p ˆ Êpˆ ˘ i ( 4 )Á ˜ Vtetg = Ze2 ◊ 4p ◊ r ◊ Y +4 315 È 4 p i ( 4 )ÁË 2 ˜¯ p Ë 2¯ 4 p 4 p i ( 4 )( 0 ) Í 4 j + e e sin sin + sin e + sin 4 ei ( 4 )(p ) ˙ 9 a5 512p Î ˚ 2 2 2 2 + Ze2 ⋅

4p r 4 +4 315 Èsin 4 (0)(0) + sin 4 (p )(0)˘˚ ◊ ◊Y 9 b5 4 j 512p Î 4p r 4 +4 315 Vtetg = Ze2 ◊ È1 ¥ e - i 2p + 1 ¥ e0 + 1 ¥ ei 2p + 1 ¥ ei 4p ˘˚ + 0 ◊ ◊ Y4 j 9 a5 512p Î + Ze2 ◊

Vtetg = Ze2 ⋅

4π r 4 +4 315 ⋅ ⋅ Y4 j [ cos 2π − i sin 2π + 1 + cos 2π + i sin 2π + cos 4π + i sin 4π ] 9 a5 512π

Vtetg = Ze2 ⋅

4π r 4 +4 315 ⋅ ⋅ Y4 j [1 + 1 + 1 + 1] 9 a5 512π

Vtetg =

16π 4 r 9

315 2 +4 1 Ze Y4 j ⋅ 5 512π a

Coordination Compounds III: Quantitative Basis of Crystal Field Theory

25.15

12. For n = 4, ml = –4 The spherical harmonic function

Y4−i4 = Y4−i4* =

315 sin 4 θi e −i 4φi 512π

Vtetg can be determined as solved above to obtain

16π 4 315 1 r Ze2 5 Y4−j4 9 512π a As discussed in case of Voct, we have not considered odd values of n and n > 4. Now we can express the overall value of Vtetg as the sum of all the determined terms as follows: Vtetg =

Vtetg =

4 Ze2 2 Ze2 16p 2 5 1 ˘ 16p 4 9 4ˆ È1 Ê 3 r r + + . Ze2 Y20j Í 2 - 3 ˙ + . Ze2Y40j Á 5 + 5 ˜ Ë a b 5 16p 9 256p a ˚ a b ¯ Îb

16p 4 315 1 16p 4 315 1 r Ze2 Y4+j4 5 + r Ze2 5 Y4-j4 9 512p 9 512 p a a This equation can be expressed in terms of two function as follows: +

Vtetg = (Vtetg)1 + (Vtetg)2 where

(Vtetg)1=

and

(Vtetg)2 =

4 Ze2 a

2 Ze2 b

16p 2 5 9 1 ˘ 16p 4 4ˆ È1 Ê 3 r r ◊ Ze2 Y20j Í 3 - 3 ˙ + ◊ Ze2 Y40j Á 5 + 5 ˜ Ëa 5 16p 9 256p a ˚ b ¯ Îb +

16p 4 315 1 16 315 1 r Ze2Y4+j4 5 + p r 4 Ze2 5 Y4-j4 9 512p 9 512p a a 4 Ze2

2 Ze2

(Vtetg)1 does not contain any coordinates of j and hence it contributes the term to the a b change in the energy of five d-orbitals but no splitting is resulted. On the other hand, (Vtetg)2 causes splitting of energy of the d-orbitals of the d1 metal ion. However, these calculations are beyond the scope of this book and will not be discussed.

25.4

DETERMINATION OF SQUARE PLANAR CRYSTAL FIELD POTENTIAL

Consider the d1 metal ion in a square planar crystal field surrounded by only four ligands only in X and Y directions. It can be considered as if the two ligands along the Z axes are removed so that the metal–ligand distance, b = or b1 = 0. Thus, we can derive the expression for square planar crystal-field potential from the expression for tetragonal crystal field potential as shown below: Taking

1 b

= 0, Vtetg reduces to Vsp as

Vsp = 4 Ze2

a

-

16p r 2 5 16p 4 9 3 r ◊ Ze2Y20j + ◊ Ze2 ◊ 5 Y40j + 16p r 4 315 Ze2 Y 4 + 1 3 4j 5 a 16p 9 256p a 9 512p a5 +

16p 4 315 1 r Ze2 5 Y4-j4 9 512p a

25.16 Inorganic Chemistry

Vsp =

4 Ze2 16π r 2 5 16π 4 9 Ze2  0 35 −4  35 +4 2 0 Ze Y r − ⋅ + ⋅ Y4 j + Y4 j  j 2 3Y + a 5 a 3 16π 9 256π a 5  4 j 2π 2π 

4 Ze2 does not cause splitting of d-orbitals and simply the energy a of five d-orbitals are raised by equal extent, while the remaining term causes splitting of energy of the five d-degenerate orbitals. Similar to the earlier discussion, the term

25.5

DETERMINATION OF TETRAHEDRAL CRYSTAL FIELD POTENTIAL

Consider a d1 metal ion placed in a tetrahedral crystal field. It can be considered as the ligands are placed at the alternate lattice points of a cube of length, l and the metal ion at the centre of the cube taken as origin of the Cartesian coordinates as shown in Fig. z x 25.5. Consider the metal–ligand distance as a and ligand 2 (–l/2, l/2, l/2) point charges i = 1, 2, 3 and 4. The position coordinates can be determined as C follows: B

In right-angled triangle ABC; BC2 + AB2 = AC2 2

2

Ê lˆ Ê lˆ 2 ÁË ˜¯ + ÁË ˜¯ = AC 2 2 fi AC 2 = fi AC =

2

2

q3 o q1

A

1 (–l/2, l/2, l/2)

q2

l

y 4 (–l/2, l/2, l/2)

2

l l l + = 4 4 2 l

3 (-l/2, l/2, l/2)

Position coordinates of ligand points in a tetrahedral complex

Fig. 25.5

2 In right-angled triangle AOC,

2 Ê lˆ Ê l ˆ fi a2 = Á ˜ + Á Ë 2 ¯ Ë 2 ˜¯

OA2 = OC 2 + AC 2 fi

a2 =

l 2 l 2 3l 2 + = 4 2 4

fi a=

3l 2

2

or l =

2a 3

fi

l a = 2 3

l l l , ; 2 2 2  −l l −l   −l −l l   l −l −l   , ,  ;  , ,  and  , ,  and in terms of metal–ligand distance of ligand points are 2 2 2  2 2 2 2 2 2  a a a a a a − −  −a a −a       a −a −a  , , , , , , , ,  respectively.  ;  ;   and   3 3 3  3 3 3  3 3 3  3 3 3

Therefore, the coordinates of lattice points 1, 2, 3 and 4 in terms of cube length are  ,

The value of associated angles

cos

and

can be determined as

Ê lˆ Ê a ˆ Ê lˆ Ê a ˆ ÁË ˜¯ ÁË ˜¯ ÁË ˜¯ ÁË ˜¯ 1 1 3 3 2 = = ; cosq 2 = 2 = = a a a a 3 3

Coordination Compounds III: Quantitative Basis of Crystal Field Theory

l   −2 = a

cos sin

sin

 a  −   3  = − 1 ; cos θ = 4 a 3

sin

sin

l −   2 = a

25.17

 a  −   3  = −1 a 3

2 3

π π 5π π 7π 5π 3π ; φ2 = π + = ; φ3 = 2π − = ; φ4 = 2π − = 4 4 4 4 4 4 4

f=

Now we can express the tetrahedral crystal field potential as 4

Vtet =

 i =1

Ze2 =Â rij n=0

+4

 ml =- n

* 4p r4 ◊ n +1 ÈYnmj l Ynmi l ˘ Î ˚ 2n + 1 a

Working on similar terms, this equation can be expanded and solved for all different values of n and ml as shown ahead. 1. For n = 0 and ml = 0 The spherical harmonic Y 00 is independent of all the angles and thus the value is same for all the points and is equal to Vtet =

4 Ze2 a

1 4

; thus Vtet can be determined as follows:

2. Forn n = 2 and ml = 0

5 3 cos2 q i - 1 16p

(

The spherical harmonic Y20i = Y20i* =

)

Vtet = Ze2 ×

4π 5 r2 ( 3 cos2 θ − 1) + ( 3 cos2 θ − 1) + ( 3 cos2 θ − 1) + ( 3 cos2 θ − 1)  × 2 +1 Y20j 3 4 1 2  2 × 2 +1 a 16π 

Vtet = Ze2 ⋅

4π r 2 0 5    1 2    1 2    −1 2    −1 2  ⋅ 3 ⋅ Y2 j  3 − 1 + 3   − 1 + 3   − 1  − 1 + 3  5 a 16π    3     3     3      3 

or Vtet = 0 3. For n = 2 and ml = +1 The spherical harmonic Y2+i1 = Y2+i1* = Vtet = Ze2 ⋅

15 sin θi cos θi eiφi 8π

4π 15 r2 sin θ1 cos θ1eiφ1 + sin θ 2 cos θ 2 eiφ2 + siin θ3 cos θ3 eiφ3 ⋅ 2 +1 ⋅ Y21j 2× 2 +1 a 8π  + sin θ 4 cos θ 4 eiφ4 

Vtet = Ze2 ◊

5p 7p 3p 4p r 2 1 15 È 2 1 i ÊÁË p ˆ˜¯ 2 1 i4 2 Ê -1 ˆ i 4 2 Ê -1 ˆ i 4 ˘ ◊ 3 ◊ Y2 j Í e 4 + e + ◊ ◊ ◊Á ˜ e + ◊Á ˜ e ˙ 5 a 8p ÍÎ 3 3 3 3 3 Ë 3¯ 3 Ë 3¯ ˙˚

Vtet = Ze2 ⋅

5π 7π 3π  4π r 2 1 15 2  i π i i i ⋅ 3 ⋅ Y2 j ⋅ 4 +e 4 −e 4 −e 4   e  5 a 8π 3 

25.18 Inorganic Chemistry

2 Vtet = Ze2 ◊ 4p ◊ r ◊ Y 1 15 ◊ 2 5 a 3 2 j 8p 3

Vtet = Ze2 ⋅

p 5p 5p 7p 3p 3p ˘ 7p È p ÍÎcos 4 + i sin 4 + cos 4 + i sin 4 - cos 4 - i sin 4 - cos 4 - i sin 4 ˙˚

1 1 1 1 1 1 1  4π r 2 1 15 2  1 +i − −i − +i + −i ⋅ ⋅ Y2 j ⋅  5 a3 8π 3  2 2 2 2 2 2 2 2

Vtet = 0 4. For n = 2 and ml = –1 The spherical harmonic function Y2−i1 = Y2−i1* = Thus, Vtet = 0, as Ql,+ ml = Ql, – ml

15 sin θi cos θi e −iφi 8π

5. For n = 2 and ml = 2 The spherical harmonic function Y2+i2 = Y2+i2* =

15 sin 2 q i ei 2fi 32p

Vtet = Ze2 ⋅

4π 15 r2 sin 2 θ1 ⋅ ei 2φ1 + sin 2 θ 2 ⋅ ei 2φ2 + sinn 2 θ3 ⋅ ei 2φ3 + sin 2 θ 4 ⋅ ei 2φ4  ⋅ 2 +1 ⋅ Y2+j2 ⋅ 2× 2 +1 a 32π 

Vtet = Ze2 ⋅

2 2 2  5π   7π  π  4π r 2 +2 15  2  i 2 4   2  i 2 4   2  i 2 4   2 2 i 2 3π    4   ⋅ 3 ⋅ Y2 j ⋅ e e + ⋅ e ⋅ + ⋅ +       ⋅e 5 a 32π  3  3 3      3   

4π r 2 +2 15 2  π 7π 7π π 5π 5π ⋅ ⋅ Y2 j ⋅ ⋅ cos + i sin + cos + i sin + i sin + cos 5 a3 32π 3  2 2 2 2 2 2 3π 3π  + cos + i sin  2 2  2 4 π 15 2 r Vtet = Ze2 ⋅ ⋅ ⋅ Y2+j2 ⋅ ⋅ [ 0 + i + 0 + i + 0 + (−i ) + 0 + (−i )] 5 a3 32π 3 Vtet = 0 Vtet = Ze2 ⋅

6. For n = 2 and ml = –2 The spherical harmonic function Y4−i2 = Y4−i2* = Thus, Vteth = 0, as Ql,+ ml = Ql, – ml

15 sin 2 θi e −i 2φi 32π

7. For n = 4 and ml = 0 The spherical harmonic function Y40i = Y40i* = Vtet = Ze2 ◊

9 ( 35 cos4 θi − 30 cos2 θi + 3) 256π

4p 9 r4 È(35 cos4q1 - 30 cos2q1 + 3) + (35 cos4q 2 - 30 cos2q 2 + 3) ◊ 4 +1 ◊ Y40j ◊ 2 ¥ 4 +1 a 256p Î + (35 cos4q 3 - 30 cos2q 3 + 3) + (35 cos4q 4 - 30 cos2q 4 + 3)˘˚

4 4 Vtet = Ze2 ◊ p ◊ r ◊ Y40j ◊ 5

9 a

9 256p

2 2 ÈÏÔ Ê 1 ˆ 4 ¸Ô ÏÔ Ê 1 ˆ 4 ¸Ô Ê 1 ˆ Ê 1 ˆ ÍÌ35 Á ˜ - 30 Á ˜ + 3˝ + Ì35 Á ˜ - 30 Á ˜ + 3˝ Ë 3¯ Ë 3¯ ÍÎÓÔ Ë 3 ¯ ˛Ô ÓÔ Ë 3 ¯ ˛Ô

4 2 4 2 ¸Ô˘ ÏÔ ¸Ô ÏÔ -1 -1 -1 -1 + Ì35 ÊÁ ˆ˜ - 30 ÊÁ ˆ˜ + 3˝ + Ì35 ÊÁ ˆ˜ - 30 ÊÁ ˆ˜ + 3˝˙ Ë 3¯ Ë 3¯ ˛Ô˙˚ ÓÔ Ë 3 ¯ ˛Ô ÓÔ Ë 3 ¯

Coordination Compounds III: Quantitative Basis of Crystal Field Theory

Vtet = Ze2 ⋅

25.19

4π r 4 0 9   1 1  4  35 × − 30 × + 3   ⋅ ⋅ Y4 j ⋅ 9 a5 256π   9 3 

4 Vtet = Ze2 ⋅ 4π ⋅ r ⋅ Y40j ⋅ 5

9 a

4 Vtet = −   28 × 9

9  −112  256π  9 

9 4π r 4 ⋅ ⋅ ⋅ Ze2Y40j 256π 9 a 5

8. For n = 4, ml = ±1

45 sin cos i 64π Since sin i cos i = 0 for all combinations of i, hence Vtet = 0 The spherical harmonic function Y4±i1* = Y4±i1 =

i

(7 cos2

i

–3) e±i

i

9. For n = 4, ml = ±2

45 sin2 i (7 cos2 i –1) e±i2 128π Similar to the earlier calculations, Vtet can be shown = 0 The spherical harmonic function Y4±i2* = Y4±i2 =

i

10. For n = 4, ml = ±3

315 sin3 i cos i e±i3 64p Since sin3 i cos i = 0 for all combinations of i, hence Vtet = 0 The spherical harmonic function Y4±i3 = Y4±i3* =

i

11. For n = 4, and ml = +4 The spherical harmonic function Y4±i4 = Y4±i4* =

Vtet = Ze2 ◊ Vtet = Ze2 ◊

315 sin4 e±i4 i 512p

i

4p 315 È 4 r4 ◊ 4 +1 ◊ Y4+j4 ◊ sin q1 ei 4f1 + sin 4 q 2 ei 4f2 + sinn 4 q 3 ei 4f3 + sin 4 q 4 ei 4f4 ˘˚ 2 ¥ 4 +1 a 512p Î

p 5p 7p 3p 4 4 4 4 4p r 4 +4 315 ÈÊ 2 ˆ i ¥ 4 ¥ 4 Ê 2 ˆ i ¥ 4 ¥ 4 Ê 2 ˆ i ¥ 4 ¥ 4 Ê 2 ˆ i ¥ 4 ¥ 4 ˘ ◊ 5 ◊ Y4 j ◊ ÍÁ ˙ + ◊ e + ◊ e ◊ + ◊ e e ÁË ˜ ÁË ˜ ÁË ˜ 9 a 512p ÍÎË 3 ˜¯ ˙˚ 3¯ 3¯ 3¯

Vtet = Ze2 ⋅

4π r 4 +4 315 4 iπ ⋅ 5 ⋅ Y4 j ⋅ ⋅ [ e + ei 5π + ei 7π + ei 3π ] 9 a 512π 9

Vtet = Ze2 ⋅

4π r 4 +4 315 4 ⋅ 5 ⋅ Y4 j ⋅ ⋅ [ cos π + i sin π + cos 5π + i sin 5π + cos 7π + i sin 7π 9 a 512π 9 + cos 3π + i sin 3π ]

Vteth = Ze2 ⋅ Vteth =

4π r 4 +4 315 4 ⋅ 5 ⋅ Y4 j ⋅ ⋅ [ −1 − 1 − 1 − 1] 9 a 512π 9

−4 315 4π r 4 ×4× ⋅ ⋅ Ze2 .Y4+j4 9 512π 9 a 5

25.20 Inorganic Chemistry

12. For n = 4, ml = –4 The spherical harmonic function Y4-i4 = Y4-i4* =

Ê 4ˆ 9

Vtet = - Á ˜ ¥ 4 Ë ¯

315 sin 4 q i e - i 4fi 512p

315 4p r 4 Ze2 Y4-j4 ◊ ◊ 512p 9 a 5

Now we can express the overall value of Vtet as the sum of all the determined terms as follows: Vtet =

4 Ze2 Ê -4 ˆ 9 4p r 4 2 0 Ê -4 ˆ 315 4p r 4 ¥ ◊ Ze Y4 j + Á ˜ ¥ 4 ¥ + Á ˜ ¥ 28 ¥ ◊ ◊ Ze2 ◊ Y4+j4 5 Ë 9¯ Ë 9¯ a 256p 9 a 512p 9 a 5 315 4p r 4 2 -4 Ê -4 ˆ ◊ +Á ˜ ¥ 4 Ze Y4 j Ë 9¯ 512p 9 a 5

Vtet =

 4 Ze2  −4   49 Ze2 r 4  0 5 +    2π Y4+j4 + Y4−j4  5 Y4 j + a 18  9  a  14 

(

)

Vtet can be resolved in two parts: Vtet = (Vtet)1 + (Vtet)2 2 È where (Vtet)1 = 4e and (Vtet ) = ÊÁ -4 ˆ˜ Í 2p 2

Ë 9 ¯Î

a

˘ 49 Ze2 r 4 È 0 5 +4 Y4 j + Y4 j + Y4-j4 ˙ Í 5 18 a Î 14 ˚

(

)

4 Ze2 to the energy term but does not cause any splitting of d-orbitals, whereas a  −4  (Vtet)2 results in splitting of the energy levels. It is clear from the expression that (Vtet)2 =   (Voct)2  9   −4  It means that Vtet =   Voct  9  −4 Thus, the tetrahedral field of ligands results the th splitting of the metal d-orbitals as caused by 9 Voct. The negative sign of the relation signifies that the order of the d-orbital splitting is opposite to that (Vtet)1 term contributes

resulted in an octahedral ligand environment.

25.6

DETERMINATION OF CUBIC CRYSTAL FIELD POTENTIAL

Consider a d1 ion in a cubic crystal field represented by eight ligands at the eight corners of a cube containing metal ion at its centre, as if two concentric tetrahedra are present as shown in Fig. 25.6.

l

O

Thus, the cubic crystal field potential can be represented as Vcube = 2 V = 2 Ê -4 V ˆ = - 8 V ÁË tet oct oct ˜ ¯

9

9

Fig. 25.6

Representation of ligands in a cubic crystal field potential

Coordination Compounds III: Quantitative Basis of Crystal Field Theory

25.7

25.21

STRUCTURAL AND THERMODYNAMIC EFFECTS OF SPLITTING OF ORBITALS

We have discussed that in presence of electrostatic crystal-field potential of ligands, splitting of d-orbitals take place and affects the structure and the thermodynamic properties of the complexed metal ion, known as crystal-field effects as discussed below.

25.7.1 Effect on Ionic Radii If we consider the crystalline compounds of transition metal ions of a particular series with same charge and same set of ligands in the same geometric environment, we expect Ionic that the ionic radii of the metal ions should radii decrease smoothly as the effective nuclear charge Zeff goes on increasing. However, this is possible only in case of spherically d 0 d1 d 2 d 3 d4 d5 d 6 d7 d 8 d 9 d10 symmetrical environment, i.e. distribution of Ca2+ Sc2+ Ti2+ V2+ Cr2+ Mn2+ Fe2+ Co2+ Ni2+ Cu2+ Zn2+ the electron charge of the metal ions equally in Fig. 25.7 Variation of ionic radii of trivalent metal ions in all directions. However, we observe that as we octahedral environment move from Sc2+ to Zn2+, an irregular decrease in radii is observed. However, if we consider the ionic radii of the divalent cations of alkaline earth metals, a smooth decrease in ionic radii with increase in Zeff is observed. Now if we consider the divalent metal ions of first transition series along with Ca2+ (main group element just preceeding the Sc2+ ion) (Fig. 25.7), It is observed that from Ca2+ to Ti2+, there is a smooth regular decrease in ionic radii and later on irregularity is observed. We can explain these observations as under. 1. In case of Ca2+ ion (d0) due to absence of any d-electrons and any crystal field effects, no splitting of d-orbitals take place. Its electronic configuration can be shown as t02g eg0 for the sake of comparison. 2. In case of Mn2+ (d5) and Zn2+ (d10), the electronic configurations are t32g eg2 and t62g eg4 respectively. In these cases the d-electron density is spherically distributed around the metal ion as all the d orbitals are equally occupied and hence there is no splitting of d-orbitals under the effect of the negative charge of the ligands. That is why a smooth dotted line has been drawn for the three points for the radii of Ca2+, Mn2+ and Zn2+ indicating the smooth decrease of ionic radii of three metal ions with increase in Zeff or the increase of d-electrons. 3. In the plot, points for radii of other ions are lying below the dotted curve. This is due to the reason that in these ions, there is unequal distribution of d-electrons and under the effect of negative chage of the ligands, splitting of d-orbitals takes place. However, the ionic radii of these metal ions decrease irregularly, (indicated by the solid lines), as discussed below. (a) In case of Ti2+ (d2) with t2g configuration, the d-electrons are concentrated in the region lying between the x, y and z axis rather than along the directions of the ligands. As a result, there is minimum repulsion between these d-electrons and the negative charge of the ligands, or in other words, the ligands can be drawn more closer to Ti2+ ion than as compared to the case if there would have been a spherical distribution of d-electrons around the nucleus of Ti2+ ion. As in this case, the negative charge of ligands would have felt more repulsion from the negative charge of the d electrons of the metal ion. Hence it results in the smaller octahedral ionic radius of Ti2+ ion than those expected from the consideration of only Zeff. (b) In case of V2+ (d3) with t32g configuration, the 3d electrons are again concentrated in between the metal–ligand directions instead of the concentration along the metal–ligand directions. Hence, the

25.22 Inorganic Chemistry

ligands feel further less hindrance and greater attraction towards the metal ion resulting in shorter metal–ligand distance. Hence, the octahedral ionic radius of V2+ ion decreases than that of Ti2+ ion. (c) In case of high-spin Cr2+ ion (d4) with t32g eg1 configuration, there is unequal distribution of d-electrons due to presence of three electrons in the t2g orbitals and one electron in the eg orbital. As the t2g orbitals lie in between the metal ligand bond directions and the eg orbitals lie along the directions of metal-ligand bonds, the negative electron charge of the ligands feel repulsion from the negative electron charge of the eg electron. Hence, the ligands cannot approach the metal ion more closely as in case of V2+ resulting in increase of octahedral radius of Cr2+ ion. Since, more of the d-electron charge is concentrated in between the metal ligand directions, the approach of ligand towards the metal ion is still easier as compared to the spherical symmetrical distribution of d-electrons around the metal ion. i.e. the value of the ionic radii is still lesser than expected on the basics of Zeff alone. (d) From Fe2+ to Ni2+ ions in high spin complexes with t42g eg2, t52g eg2 and t62g eg2 configurations respectively, the more of the d electron charge goes on concentrating in between the directions of the metal–ligand bond and results in greater attraction of the ligands towards the metal ion, the octahedral ionic radius goes on decreasing from Fe2+ to Ni2+ ions and these values are even lesser than expected due to the effect of Zeff only. (e) In case of Cu2+ ion (d9) with t62g eg3 electronic configuration, an extra electron is present in eg orbital than compared to that of Ni2+ ion (t62g eg2). It means comparatively more electron density is concentrated along the metal–ligand directions resulting in more repulsion between the d-electron charge and the negative charge of the ligands. Hence, the octahedral ionic radius of Cu2+ is larger as compared to that of Ni2+. However, the value is still lesser than expected under the effect of only Zeff. 4. The plot does not show the point for Sc2+ as it is not stable and its octahedral ionic radius cannot be determined with complete certainity. It is clear from the above discussion that the octahedral ionic radius of a metal ion in a high-spin complex of first transition series is lesser than expected under the effect of Zeff only due to ease of approach of ligands towards the metal ion under the effect of splitting of d-orbitals which differentiate the d electron density in between the metal–ligand axes and along the metal–ligand axes.

Variation of Ionic Radii of Lanthanide Ions under the Effect of Crystal Field If we consider the lanthanide ions with 4f electrons increasing from 4f 0 to 4f14, a regular decrease in ionic radii with increase in Zeff is observed (Fig. 25.8). This is in contrary to the observation in case of transition metal ions. This is due to the reason that the splitting of 4f orbitals is negligible when compared to that of d orbitals. In case of lanthanide ions, the 4f electrons are much shielded from the effect of the negative charge of the ligands by the intervening electrons. As a result, the repulsion between 4f electrons and the negative charge of the ligands is negligibly small resulting in negligible splitting of 4f orbitals and the 4f electrons remain practically spherically symmetrical even under the affect of ligands. Thus under the similar crystal environment, the ionic radii of trivalent lanthanide ions decrease regularly with increase in number of 4f electrons.

Ionic radii

1 2 3 4 5 6 7 8 9 10 11 12 13 14 Number of f-electrons

Fig. 25.8

Variation of ionic radii of Ln3+ with increase in number of 4f electrons

Coordination Compounds III: Quantitative Basis of Crystal Field Theory

25.23

25.7.2 Effect on Lattice Energy We have discussed that the splitting of d-orbitals of the metal ions under the effect of crystal-field potential impacts extra stability to the octahedral complexes and this stabilisation is discussed in terms of Crystal-Field Stabilisation Energy (CFSE). As a result, the lattice energies of these compounds have been found higher than those expected. If we compare the lattice energies of bivalent metal fluorides of the first transition series along with Ca2+ ion, an irregular increase of lattice energy is observed with the increase in Zeff. This can be explained as follows. 1. In case of CaF2 (d0), MnF2 (d5) and ZnF2 (d10), due to equal distribution of d electrons around the nuclei, there is no splitting of d-orbitals and the ionic radii decrease regularly under the effect of Zeff. Since lattice energy is inversely proportional to the interionic distance under the effect of same anions (ligands), a regular increase in lattice energies is observed. 2. In case of all other bivalent metal fluorides, the value of lattice energies are higher than those expected under the effect of Zeff only. (a) In the given plot (Fig. 25.9), the dotted line shows the hypothetical variation of the lattice energies of bivalent metal fluorides under the effect of only Zeff and in the absence of any crystal-field effects. On the other hand, the solid line represents the observed lattice energies of these ions in presence of crystal field effects. Thus, AB represents the lattice energy of CaF2, CD represents the lattice energy of Mn2+ and EF represents the lattice energy of Zn2+ ion. Let us consider the case of Ni2+ ion. In the given plot, GH represents the observed lattice energy and IH represents the hypothetical lattice energy of Ni2+ ion. It is clear that due to crystal-field effect, GI represents the extra stabilisation energy or CFSE. The lattice energy of Mn2+, CD = 2782.4 kJ mol–l = JH = KF The lattice energy of Zn2+,

EF = 2986.1 kJ mol–l EK = EF – KF = 2986.1 – 2782.4 = 203.7 kJ mol–1

As

ECK are similar triangles hence, IJ = EK

ICJ and

Therefore, IJ =

CJ

3 (203.7) = 122.2 kJ mol–1 5

fi

CK

IJ EK = 3 5

fi IJ =

3 EK 5

G E

Lattice energy(kJ/mol)

I L M

K

J

C

N A

B d0

d1

d2

d3

P

D

d4

d5

H d6

d7

Ca2+ Sc2+ Ti2+ V2+ Cr2+ Mn2+ Fe2+ Co2+

F

d8

d9

d10

Ni2+

Cu2+

Zn2+

Fig. 25.9 Variation of lattice energies of bivalent metal ions in octahedral environment

25.24 Inorganic Chemistry the hypothetical lattice energy of Ni2+, IH is equal to IJ + JH IH = 122.2 + 2782.4 = 2904.6 kJ mol–1 Thus, we can obtain the CFSE for NiF2 as the difference between the observed lattice energy and the hypothetical lattice energy as GI = GH – IH or CFSE, GI = 3063.5 – 2904.6 = 158.9 kJ mol–1. We have already determined the CFSE for NiF2 (in octahedral ligand environment) equal to 12Dq. Thus, 12Dq is equal to the calculated CFSE for NiF2.

or

Since

= 10Dq = (10Dq/12Dq) (158.9) = 132.4 kJ mol–1

Similarly, for other metal ions, the CFSE can be calculated by using the observed lattice energies from the table and the following relationships.

2 OC 5 3 For VF2, the hypothetical lattice energy = AB + OC 5 4 For CrF2, the hypothetical lattice energy = AB + OC 5 1 For FeF2, the hypothetical lattice energy = CD + EK 5 2 For CoF2, the hypothetical lattice energy = CD + EK 5 4 For CuF2, the hypothetical lattice energy = CD + EK 5 In case of the bivalent metal complexes of other halides, similar curves are obtained but the deflection seen at the point of Mn2+ ion is greater. This is probably due to the reason that the larger halide ions are more polarisable. Since the polarising powers of Ca2+, Mn2+ and Zn2+ differ appreciably, the contribution of the covalent character introduced in their complexes are also different. As F – is the smallest and least polarisable, the extent of induced covalent character is minimum and increases with the increase in size of the halide ion as seen in Table 25.3.

(i) For TiF2, the hypothetical lattice energy = AB + (ii) (iii) (iv) (v) (vi) (b)

Table 25.3 Halide F–

Lattice energies of halides of Ca2+ and bivalent transition metal ions of first transition series (in kJ mol–1) Ca2+

Sc2+

2607

Ti2+

V2+

Cr2+

Mn2+

Fe2+

Co2+

Ni2+

Cu2+

Zn2+

2757

2770

2887

2782

2929

2979

3063

3067

2987

–

2234

2477

2548

2569

2502

2611

2690

2753

2786

2711

Br–

2133

2393

2489

2506

2435

2540

2619

2699

2740

2644

–

2054

2309

2397

2397

2360

2477

2531

2619

2669

2581

Cl

I

—

25.7.3 Effect of Heat of Hydration Heat of hydration is defined as the heat evolved when one mole of the metal ions in gaseous state reacts with the six moles of H2O molecules. M2+(g) + 6H2O $ [M(H2O)6]2+ (aq) + Hhyd. (where the Hhyd. its heat of hydration) In case of alkaline earth metal cations, as the ionic radii decreases from Ba2+ to Mg2+, resulting in greater attraction of the ligands towards the metal ion, the heat of hydration increases regularly from Ba2+ to Mg2+.

Coordination Compounds III: Quantitative Basis of Crystal Field Theory

25.25

However, this regular increase is not observed in case of bivalent transition metal ions. We have already discussed the effect of splitting of d-orbitals under the effect of crystal field potential of ligands on the ionic radii of these ions. Since heat of hydration is directly related to the ease of approach of the H2O molecules (ligands) towards the M2+ ion and hence upon the attraction between the M2+ ions and the ligands, as the octahedral ionic radius of the M2+ ion varies, corresponding variation in heat of hydration is observed. Thus, we see a regular increase in heat of hydration for Ca2+ to Mn2+ to Zn2+ (due to spherical symmetrical distribution of d-electrons and regular decrease of ionic radius), corresponding to zero CFSE, while for other M2+ ions as the electrostatic attraction of ligands increases, Ca2+ (t02g eg0) < Ti2+ (t22g eg0) < V2+ (t32g eg0) and ionic radius of M2+ decreases, an increase in Hhyd. is observed. Similar is the case from Mn2+ to Ni2+, as the ionic radius decreases from Mn2+ (t32g eg2) > Fe2+ (t42g eg2) > Co2+ (t52g eg2) > Ni2+ (t62g eg2), Hhyd. increases. On the other hand, the ionic radius increases from V2+ (t32g eg0) < Cr2+ (t32g eg1) < Mn2+ (t32g eg2) and similarly from Ni2+ (t62g eg2) < Cu2+ (t62g eg3) < Zn2+ (t32g eg4) and Hhyd. decreases. Just as in case of lattice energies, we can determine the hypothetical hydration energy for the metal ions and then use it to determine the CFSE of the metal complexes, e.g. the dotted line in the plot (figure 25.10) represents the points for the hypothetical hydration energies for the metal ions whereas the solid line represents the observed hydration energies of these ions. Suppose we want to determine the hypothetical heat of hydration of Ti2+ ion with observed value equal to 2732 kJ mol–1. The observed heat of hydration values for Ca2+ and Mn2+ ions are 2469 kJ mol–1 and 2736 kJ mol–1 respecively i.e. ED = AB = 2469 kJ mol–1 and FG = 2736 kJ mol–1

2 2 FH = (FG – AB) 5 5 2 = × (2736 – 2469) = 107 kJ mol–1 5 The hypothetical hydration energy for Ti2+ ion, CD = CE + ED in FHA,

CE =

= 107 + 2469 = 2576 kJ mol–1 or CFSE of [Ti(H2O)6] 2+, CE = CD – ED = 2736 – 2576 = 160 kJ mol–1 In case of trivalent ions, due to high polarisation power of M3+ ions, the extent of covalent character in 3+ M – OH2 bond increases and gives more prominent bend at Fe2+ than Mn 2+. Hence, these are not used for the calculation of CFSEs.

2+

M Ca2+ Ti2+ V2+ Cr2+ Mn2+ Fe2+ Co2+ Ni2+ Cu2+ Zn2+

Heat of hydration from some bivalent metal ions

O

–1

Hhyd. (kJ)mol 2469 2732 2778 2795 2736 2845 2916 2996 3000 2933

L

I

∆Hhyd.(kJ/mol)

Table 25.4

F

J

K

M

N

C A

D

B d0

H

E

d1

d2

G d3

d4

d5

Ca2+ Sc2+ Ti2+ V2+ Cr2+ Mn2+

d 6 d7 d 8 d 9 d10 Fe2+ Co2+ Ni2+ Cu2+ Zn2+

Fig. 25.10 Variation of hydration energy for bivalent ions

25.26 Inorganic Chemistry

25.7.4 Effect on the Geometry of the Coordination Complexes The magnitude of d-orbitals splitting greatly affects the geometry of the coordination complexes. In general, greater CFSE results in greater stability of a particular configuration and the metal ion preferably exists in that configuration. Consider the CFSE values for the high-spin octahedral and tetrahedral complexes as given in Table 25.5.

4 of Dqoct for a particular ligand. This means that we can 9 convert the Dqtet into the corresponding Dqoct as given in the table 25.5. We can find the following points: 1. In case of d3 and d8 configurations of high spin complexes, the state in octahedral environment is much more stabilized than in case of tetrahedral environment. Hence, the metal ions preferably adopt octahedral geometry and the change in geometry from octahedral to tetrahedral would result in loss of stabilization by about 9.3Dqoct which is not preferred. 2. In case of d1 and d6 configurations of high-spin complexes, the change in geometry can take place in presence of strongly dominating and favouring factor for the octahedral or tetrahedral geometry as the loss of stabilisation is very small, about 1.3Dqoct. 3. In case of d2 and d7 configurations of high-spin complexes, the difference of stabilisation between the two geometries is just equal to 0.67Dqoct. Hence, any factor stabilising the tetrahedral geometry would favour the complex in a tetrahedral geometry. 4. The complexes with d0, d5 and d10 configurations, there is equal probability for the occurrence of the tetrahedral and octahedral geometry and depending upon the other factors either of the geometries can be preferred. Now we will discuss some particular cases. (a) Co3+ and Cr3+ preferably exist in octahedral geometry under the effect of strong field ligands. (b) Ti4+, V5+ and Mo6+ with d0 configuration have no particular preference for the octahedral or tetrahedral geometry. Similarly, Mn2+, Fe3+, Zn2+, Cd2+, Hg2+, Cu+, Ag+, Au+, with d10 configuration show both octahedral as well as the tetrahedral complexes. We have already discussed that Dqtet is

Crystal field stabilisation energies of metal ions in high-spin and low-spin octahedral and tetrahedral complexes

Table 25.5 dn dº

CFSE for high spin octahedral complex

CFSE for low spin octahedral complex

0

CFSE for tetrahedral complex 0

1

d

4Dqoct

6Dqtet (= 2.7Dqoct)

d2

6Dqoct

12Dqtet (= 5.33Dqoct)

d3

12Dqoct

6Dqtet (= 2.7Dqoct)

4

d

6Dqoct

16Dqoct

4Dqtet (= 1.78Dqoct)

d5

0

20Dqoct

0Dqtet

4Dqoct

24Dqoct

6Dqtet (= 2.7Dqoct)

d

6Dqoct

18Dqoct

12Dqtet (= 5.33Dqoct)

d8

12Dqoct

6Dqtet

9

6Dqoct

4Dqtet (= 1.78Dqoct)

10

0

0

d

6 7

d

d

Coordination Compounds III: Quantitative Basis of Crystal Field Theory

25.27

(c) Ni2+ with d8 configuration exists mainly as octahedral high-spin complexes, but the low-spin complexes are known in square planar configurations. (iv) All Pt2+ complexes are low-spin square planar complexes due to higher CFSE associated with the geometry.

25.7.5 Effect on Crystal Structure of Spinels Spinels are metallic oxides with the general formula AB2O4, where A is a divalent metallic cation and B is a trivalent metallic cation of same or different element. These are named after the name of the mineral spinel MgAl2O4. Due to large electronegativity of oxygen, all oxide spinels show ionic bonding and are composed of FCC array of O2– anions. The cations A and B occupy the tetrahedral and the octahedral voids in two different ways giving rise to two different classes of spinels, namely normal spinel and inverse spinels.

1. Normal Spinels In case of normal spinels with the general formula, A2+B23+O4, the divalent cations occupy the eight tetrahedral sites while the trivalent cations occupy all the sixteen octahedral sites.

2. Inverse Spinels

Table 25.6 Structure of some spinels with variation

In case of inverse spinels with the general formula of d electrons A3+ B2+ B3+O4, half of the trivalent cations occupy the 2+ A B3+ Structure tetrahedral sites, while the octahedral sites are occupied 0 5 10 1–4 6–9 Normal spinel , d , d d or d d by all the bivalent cations and half of the remaining 1–4 6–9 0 5 10 Inverse spinel or d d , d , d d trivalent cations. In the above general formulae, the tetrahedral sites are represented as A while the octahedral sites are represented by B. Since oxide ions are weak-field ligands, the spinels are high-spin complexes due to weak crystal-field potential. It can be recalled from our earlier knowledge that the size of a cation and thus, the radius ratio decides the site occupancy by the cation in the lattice. However, it has been observed that Mn3O4 and Co3O4 exist in normal spinel structure while Fe3O4 exists in inverse spinel structure. This change in spinel structure cannot be justified on the basis of the effect of size only, but can be explained on the basis of crystal field stabilisation energy. As a general rule, if the B3+ ion has more CFSE for octahedral site than that of A2+ ion, the normal spinel structure is preferred. On the other hand, if A2+ ion has more CFSE for octahedral geometry than the B3+ ion, an inverse spinel structure is preferred. The above generalisations can be explained by the consideration of the CFSE of different cations at the tetrahedral (A) and octahedral sites (B). We will use the following assumptions. 1. The CFSE of a cation at the tetrahedral site is

4 times greater than that of the cation at the octahedral 9

4 Voct 9 2. The CFSE of M3+ ion is 1.5 times the CFSE of the M2+ ion. 3. There is no significant change in the lattice energy with change in the distribution of the cations at the tetrahedral and octahedral sites, but it changes CFSE values significantly. We have already mentioned that O2– is a weak field ligand and lead to generation of high-spin configurations. We can use this data for the calculation of CFSE for the normal and inverse spinel structures of Mn3O4, Fe3O4 and Co3O4 as follows: Thus, it is clear from the above calculations that due to greater CFSE for normal spinel structures in case of Mn3O4 and Co3O4 and for inverse spinel structure in case of Fe3O4, the above-said pattern is observed. Some other cases of normal spinel structures are MgAl2O4 (no CFSE because of nontransition metal ions) and FeCr2O4. On the other hand, the example for inverse spinel structures are NiFe2O4 and CoFe2O4. site as Vtet =

25.28 Inorganic Chemistry Table 25.7 CFSE for bivalent and trivalent ions CFSE (Dq)

High spin Configuration d0, d5

B2+

Octahedral (B) B3+ 0

0

4Dq

1.5 × 4Dq = 6Dq

d2, d7

6Dq

1.5 × 6Dq = 9Dq

d3, d8

12Dq

1.5 × 12Dq = 18Dq

d4, d9

6Dq

1.5 × 6Dq = 9Dq

4 9 4 9 4 9 4 9

d5, d10

0

0

1

d ,d

6

0

Tetrahedral (A) A2+

A3+

0 × 6Dq = 2.67Dq

4Dq

× 12Dq = 5.33Dq

8Dq

× 6Dq = 2.67Dq

4Dq

× 4Dq = 1.78Dq

2.67Dq

0

0

Table 25.8 Calculation of CFSE for Mn3O4, Fe3O4 and Co3O4 Normal spinel structure (A2+ B23+ O4)

25.8

Inverse spinel structure (A3+ B2+ B3+) O4

Structure

Mn3O4

Mn2+ (d5) (Mn3+(d)4)2O4 = 0 + 2 × 9 = 18Dq

Mn3+(d4) Mn2+(d5) Mn3(d4)O4 = 2.67 + 0 + 9 = 11.67Dq

Normal

Fe3O4

Fe2+(d6) (Fe3+d5)2O4 = 2.67 + 0 = 2.76Dq

Fe3+(d5) Fe2+(d6) Fe3(d5)O4 = 0 + 4 = Dq + 0 = 4Dq

Inverse

Co3O4

Co2+ (d7) (Co3+(d6))2O4 = 5.33 = 5.33 + 12 = 17.33Dq

Co3+(d6) Co2+(d7) Co3+(d6)O4 = 4 + 6 +6 = 16Dq

Normal

JAHN–TELLER EFFECT (DISTORTION OF GEOMETRY)

It has been observed that most of the six coordinated complexes have regular or symmetrical octahedral geometry with all the six metal–ligand distances equal. However, there are some complexes in which distortion of the regular geometry takes place resulting in a change in their shape, e.g. in case of CuCl2 crystal, the four chloride ions are at a distance of 230 pm and two Cl– ions are at a distance of 295 pm from the Cu2+ ion, whereas in case of CuF2 crystal, four F – ions are at a distance of 193 pm and two F – ions are at a distance of 227 pm from Cu2+ ion. These octahedral complexes are said to be tetragonally distorted octahedral complexes or simply as tetragonal complexes. On the other hand, in case of low-spin octahedral complexes of Ni2+, Pt2+ and Pd2+, there is strong distortion of the octahedral geometry and conversion into square planar geometry with the removal of two ligands. These distortions have been explained by Jahn and Teller in 1937 in terms of Jahn–Teller effect or Jahn– Teller distortion. This effect states that any nonlinear molecular system with degenerate electronic state will be unstable and will attain stabilisation by distortion from its geometry so as to remove the degeneracy by splitting in its degenerate electronic state. As a result, the system will attain lower symmetry and hence lower energy and more stability.

Coordination Compounds III: Quantitative Basis of Crystal Field Theory

25.29

25.8.1 Concept of Electronically Degenerate State An electronically degenerate state represents more than one electronic arrangements of the same energy or the availaibility of more than one degenerate orbitals to be occupied for an electron, e.g. an octahedral complex of a metal ion with d1 configuration, under the effect of crystal-field potential of ligands, is said to be electronically degenerate as the one of electrons can occupy any of the three t2g orbitals (dxy, dyz, dxz) with the same energy or, in other words, the degenerate orbitals of the t2g set are asymmetrically occupied by the electrons as shown in Fig. 25.11. On the other hand, the complex with metal ions in d3 configuration is said to be electronically nondegenerate as each of the three degenerate t2g orbitals is symmetrically and singly occupied Fig. 2512. According to Jahn–Teller effect, the asymmetrically occupied electronically degenerate states are higher in energy and hence undergo distortion to attain the configuration with lower energy. John– Teller effect can be illustrated with the help of an octahedral complex of Cu2+ ion with d9 configuration surrounded by unidentate ligands. The t2g orbitals in this configuration are completely occupied and hence their electron charge density is uniformly distributed in all the directions. However, this is not so in case of eg orbitals which are occupied by three electrons. The three electrons can occupy the degenerate eg orbitals to have two electronically degenerate states as follows: (a) (dz2)2 (dx2–y2)1 (b) (dz2)1 (dx2 – y2)2

dz2 dx2 – y2 dz2 dx2–y2

dz2 dx2–y2

dz2 dx2–y2 dxz dyz dxy

dxz dyz dxy

Fig. 25.11

dxz dyz dxy

dxz dyz dxy 1

Possible filling of t2g orbitals for d configuration

Fig. 25.12

Symmetrical filling of t2g orbitals for d3 configuration

The electronic state (a) corresponds to the increase of dx2 – y2 dz2 electron charge density in the ‘z’ direction resulting in greater 1 + 2 δ1 + 12 δ1 electrostatic repulsion to the negative charge density of the eg ligands in z directions. It means the ligands along the x – y – 12 δ1 – 12 δ1 directions are more attracted as compared to the ligands along dz2 dx2 – y2 the z-direction. It results in elongation of the metal–ligand bond Possibility 1 Possibility 2 along the z-direction as compared to the metal–ligand bonds along the x and y directions. This splits the degeneracy of the Fig. 25.13 Representation of splitting of eg orbitals due to Jahn–Teller eg orbitals in such a way that the energy of dz2 orbital decreases distortion while that of dx2–y2 increase by the same amount as represented in Fig. 25.13. The electronic state (b) corresponds to the increase of electron charge density along the x–y directions resulting in greater electrostatic repulsions to the negative charge density of the ligands in the x and y directions. It means the ligands along the z-direction are more attracted as compared to the ligands along the x and y directions. It results in elongation of the metal–ligand bonds along the x and y directions as compared to the metal–ligand bonds along the z-direction. This splits the degeneracy of the eg orbitals in such a way that the energy of dz2 orbital increases while that of dx2 – y2 decreases by the same amount as represented in Fig. 25.13.

25.30 Inorganic Chemistry

Both these cases lead to the distortion of the octahedral geometry to the tetrahedral geometry in two equally possible ways: 1. Elongation of the two trans ligands lying on the z axis and compression of the other four ligands along the x and y directions. 2. Elongation of the two trans ligands lying on the x and y axes and compression of the other four ligands along the z-direction. Let us consider the case of an octahedral complex of a metal ion with d1 configuration surrounded by unidentate ligand. This single electron can occupy any of the three degerate t2g orbital resulting in two electronically degenerate states as given below. (a) (dxy1)1 dyz dxz

(b) dxy (dyz)1 dxz or dxy dyz (dxz)1

The electronic state (a) corresponds to the increase of electron dxy dxz, dyz charge density in the x–y plane as compared to that in x–z or y–z 2 3 δ2 plane. It results in greater electrostatic repulsion to the negative + 13 δ2 – 13 δ2 charge density of the ligands in the x–y plane and elongation of – 23 δ2 t2g the metal–ligand bonds along the x and y axes. On the other hand, dxz, dyz dxy the ligands along the z-axis are attracted more and the metal– Possibility 1 Possibility 2 ligand bonds along this direction are compressed. This increases Fig. 25.14 Representation of splitting of the energy of dxz and dyz orbitals as compared to the decrease in eg orbitals due to Jahn–Teller the energy of dxy orbitals as shown in Fig. 25.14. distortion On the other hand, the electronic state (b) corresponds to the increase of electron charge density in the xz or yz plane as compared to the xy plane. As a result, the ligands along the z-direction face more electrostatic repulsion and the metal-ligand bonds along the z-direction elongate while that in x and y direction are compressed. This results in increase in the energy of dxy orbital while decrease in the enegy of dxz and dyz orbitals as shown in Fig. 25.14. Table 25.9 Prediction of distortion in the octahedral complexes Configuration

High-spin complexes Distribution of electrons

d0 d1 d2 d3 d4 d5 d6 d7

t02g symm. t12g unsymm. t22g unsymm. t32g symm. t32g symm. t32g symm. t42g unsymm. t52g unsymm.

eg0 symm. eg0 symm. eg0 symm. eg0 symm. eg1 unsymm. eg2 symm. eg2 symm. eg2 symm.

Low-spin complexes

Prediction — Slight distortion Slight distortion — Strong — Slight Slight

Distribution of electorns t02g symm. t12g unsymm. t22g unsymm. t32g symm. t42g unsymm. t52g unsymm. t62g symm. t62g unsymm.

eg0 symm. eg0 symm. eg0 symm. eg0 symm. eg0 symm. eg0 symm. eg0 symm. eg2 symm.

Prediction — Slight distortion Slight distortion — Slight — Strong Strong

Coordination Compounds III: Quantitative Basis of Crystal Field Theory

Configuration

High-spin complexes Distribution of electrons t62g

eg2

symm. t62g symm. t62g symm.

symm. eg3 unsymm. eg4 symm.

8

d

d9 d10

25.31

Low-spin complexes

Prediction

Distribution of electorns t62g

eg2

symm. t62g symm. t62g unsymm.

symm. eg3 unsymm. eg4 symm.

— Strong Strong

Prediction Strong Strong Strong

Now we can consider the relative enegies of the t2g and eg orbitals with the consideration of tetragonal distortion as shown in Fig. 25.15 corresponding to the elongation of metal–ligand bonds along the z-direction or compression along the z-direction. It is evident from the figure that distortion in eg level is more distinctive than in t2g level, i.e 2 H2O > urea > NH3 > en > ox2 – > NCS – > Cl– > CN– > Br – > I– (smaller

)

(smaller )

Pt4+ < Co3+ < Ph3+ < Ir3+ < Fe3+ < Cr3+ < Ni2+ < V 4+ < Pt 2+ < Mn2+ (larger b)

The approximate value of B and B can be determined using the following empirical relations B = B (1 – hk) × 103 cm–1 and = (1 – hk) where, h and k are empirical Racah parameters characteristic of the ligand and metal ions. Their values have been assigned to various metal ions and ligands as given in Table. 26.8.

26.26 Inorganic Chemistry Table 26.8 Empirical Racah parameters for metal ions and ligands Metal ion V

2+ 3+

Empirical value of k 0.1

ligand F

Empirical value of h

–

0.8

Cr

0.20

H2O

1.0

Mn2+

0.07

en

1.5

3+

Fe

0.24

Cl

–

2.0

3+

0.33

CN

Ni2+

0.12

Br–

2.3

Mn4+

0.50

I–

2.7

Co

–

2.1

4+

0.6

NH3

1.4

Ni4+

0.8

ox2–

1.5

Pt

For example, the value can be determined for [Cr (CN)6]3– as = 1 – (2.0) (0.21) = 0.580 Now, using value of B from table, we can determine the value of B1 for the complex as B = 810 = 469.8 cm–1

26.7

× B = 0.58 ×

TERMS CORRELATION DIAGRAMS UNDER THE EFFECT OF WEAK AND STRONG FIELD EFFECTS

We have already discussed that the energies of the energy terms vary with strength of the ligand field. We can further illustrate this effect with the help of correlation diagram between free atom/ion terms and term of a complex with increasing field effects. As we know that under the effect of weak ligand field, inter-electronic repulsions and hence the energies of the term are expressed in terms of Racah parameters (B and C). On the other hand, under the effect of strong ligand field, the energies of the terms are expressed in term of Dq. The basic characteristics of the correlation diagrams are as follows: 1. The free ion terms are represented at the extreme left while the strong field terms are represented at the extreme right. There is a one-to-one correspondence between these terms. 2. The total number of energy terms remains the same. 3. The energy terms are represented in increasing degeneracy and symmetry and do not cross with change in the ligand field strength order of energy. Thus the energy levels are of the same spin. 4. Non-crossing rule is observed in all the correlation diagrams.

1. Correlation Diagram for d1 and d9 Metal Ion In case of d1 and d9 configuration with only one electron, there is good correlation between the terms of free ion and that of the strong-field term because of the absence of any inter-electronic repulsion as shown in Fig. 26.23. The correlation diagram for d1 octahedral also applies for d9 tetrahedral system and the correlation diagram for d9 octahedral also applies for d1 tetrahedral system. However, for the tetrahedral system, the subscript g is not used as shown in the parentheses in Fig. 26.23.

2. Correlation Diagram for d2 and d8 Metal Ion In case of a d2 configuration, the ground–state term for a free ion is 3F. The only spin allowed transitions are possible for the term 3P, as shown by strong lines in Fig. 26.24 while the transitions to the states of different spin multiplicity have been shown by means of dotted lines.

Coordination Complexes IV: Spectroscopic and Magnetic Properties of Coordination Compounds

5 4 t2g eg

5 4 (t2g e )

2T

2 2g ( T2)

2E 2T

2g

6 t2g eg3

2D 6 (t2g e3)

2E

(2E) g

g

( 2E

g)

Energy

Strong configuration

g

(2E)

1e 1 g ( e)

g

+ 6Dq

Weak ligand field

2T

2g

(2T2)

2T

2g

(2T)

1t

2g

(1t2)

– 4Dq

– 4Dq d9 octahedral and d 1 tetrahedral 3 Strong ligand field

2E

(2T2)

+ 6Dq 2E

(2E)

26.27

d1 octahedral and d 9 tetrahedral Free ion term

Weak ligand field

Strong configuration

3 Strong ligand field

Dq

Fig. 26.23 Correlation diagram for d1 and d9 ion in octahedral and tetrahadral ligand field

Fig. 26.24 Correlation diagram for d2 and d8 ion in octahedral and tetrahedral ligand field

Under the effect of strong ligand field, the possible d2 configurations of different energies in the increasing order of energies are t22g (–8Dq) < t12g e1g (+2Dq) < eg2 (+ 12Dq), Table 26.9.

26.28 Inorganic Chemistry

These strong field configurations give various terms as shown in the table 26.9 along with their configuration degeneracy shown in parentheses. The energy order of various ground-state and strong- field levels can be determined using Hund’s rule, as follows.

Table 26.9

(t2g)2 $ 3T1g < 1T2g < 1Eg < 1A1g

t12g e1g(24)

Orbital energies and terms with configuration degeneracy in parentheses

Electronic Energy configuration t22g(15) 2 × ( – 4Dq) = – 8Dq

(t2g)1 (eg)1 $ 3T1g < 3T2g < 1T1g < 1T2g

eg2(6)

(eg)2 $ 3A2g < 1Eg < 1A1g

Terms 1

A1g + 1Eg + 1T2g + 3T1g (1) (2) (3) (9) = 15 1 –4Dq + 6Dq = +2Dq T2g + 1T1g + 3T2g + 3T1g (3) (3) (9) (9) = 24 2 × ( + 6Dq) = +12Dq 1A1g + 1Eg + 3A2g (1) (2) (3) = 6

It is observed that in the correlation diagram, strength of ligand field does not alter the ground-state term, but strongly alters the energies of other terms. It can further be seen that 3T1g from t22g correlates with 3 T1g from 3F, while that from t12ge1g correlates with 3T1g from 3P. Thus bending of energies is observed for all the terms. This correlation diagram also applies for d8 tetrahedral system. However, the corresponding electronic configuration under extremely strong ligand field are in the order t24e4 < t25e3 < t26e2 as shown in the parentheses (Fig. 26.24). In case of d2 metal ion in tetrahedral ligand-field environment, the energy order of strong field configrations is (e)2 < (e1) (t2)1 < (t2)2. It should be noted that the ordering of weak-field terms remain the same as in octahedral field as 3F < 1D < 3P < 1G < 1S. However, the term splitting gets inverted as discussed earlier. Thus, the energy order of various weak-field terms are as follows: 1

G $ 1A1 < 1E < 1T2 < 1T1 1 3

D $ 1E < 1T2

F $ 3A2 < 3T2 < 3A2

But in case of strong ligand field, the split levels of the strong field configuration are normal and identical to there in octahedral ligand field (Fig. 26.26). The same diagram applies for d8 metal ion in octahedral ligand field.

3. Correlation Diagram for d3 and d7 Metal Ion d3 and d7 configurations yield F term as ground-state and P term as excited-state and shows the splitting patterns as observed for d2 and d8 metal ions. However, in case of d3 octahedral system, a good correlation of free ion ground-state term and the lowest strong-field configuration is observed. But it is not so for d7 octahedral system. Here, the ground-state freeion term correlates with the strong-field configuration which is about 10Dq higher than the ground-state strong-field configuration. This means that as the effect of octahedral ligand field is increased, spin-pairing is resulted. In case of d3 and d7 tetrahedral systems, the term-splitting pattern is inverted to that of the corresponding octahedral systems. Thus, in case of d3 octahedral and d7 tetrahedral system, the ground-state free-ion term is 4T2g while in case of d3 tetrahedral and d7 octahedral system, the ground-state free-ion term is 4T1g. (Fig. 26.25)

4. Correlation Diagram for d4 and d6 Metal Ion In case of d4 and d6 system, the ground-state free-ion term is a quintet (5D), but there is no corresponding strong-field configuration the ground state. As a result, the free ion ground-state terms for d4 octahedral systems correlates with a strong-field configuration which is about 10Dq higher energy than the lowest strong-field configuration, while for d6 octahedral system, the correlation is with the strong-field configuration which is about 20Dq higher in energy than the lowest strong-field configuration. This means that as the field

Coordination Complexes IV: Spectroscopic and Magnetic Properties of Coordination Compounds

2T

1(g)

2T

1(g)

2T

2(g)

2E (g) 2T

1(g)

2A

2A 1(g) 2E (g) 2G

1(g)

2T

2(g)

4T

4T

1(g)

2T

2(g)

2T

t25g e1g (t12 e2) – 8 Dq

26.29

4P

2A

1(g)

1(g)

4T

t22g e1g (t42 e3)

1(g)

4T

2(g)

4T

1(g)

1(g)

– 2 Dq

Energy 4A

2(g)

4T 4T

4F

4T

t26g e1g

(e3)

2E (g)

1(g)

2(g)

4T

2(g)

4A

2(g)

– 18 Dq d7 octahedral and d 3 tetrahedral Strong 3 Strong configuration ligand field

2T

2(g)

1(g)

2T

1(g)

2E

– 12 Dq d3 octahedral and d 7 tetrahedral Weak Strong 3 Strong ligand field configuration ligand field 4A

Weak ligand field

Free ion term Dq

t23g (t32 e4)

1(g)

2(g)

Fig. 26.25 Correlation diagram for d3 and d7 ion in octahedral and tetrahedral ligand field

strength increases, spin pairing is favoured and the crossover leads to a change in the ground state. Thus, at some specific value of Dq, (Dq crossover) the ground-state 5Eg, arising from 5D term changes to 3Tg, arising from 3H term. (Fig. 26.26)

5. Correlation Diagram for d5 Metal Ion The d5 metal ion is a special case, as the ground-state free-ion term is 6S which results into 6A1g. However, at some specific value of Dq crossover, a high-field ground term 2T2g is obtained. This changes the spin multiplicity from six to two and a transition from high-spin to low-spin state in such a way that the groundstate free-ion term correlates with that strong-field configuration which is about 20Dq higher in energy than the ground strong-field configuration (Fig. 26.27). Thus, spin-pairing is much favoured.

26.8

TANABE-SUGANO DIAGRAMS (T-S DIAGRAM)

The correlation diagrams and Orgel diagrams present a qualitative view of transition from weak ligand field to the strong ligand field. These diagrams are not helpful in the correct interpretation of spin-forbidden transitions and inter-electronic repulsions. Tanabe and Sugano put forward a quantitative means of showing the variation in the energies of various terms of dn system with change in ligand- field strength. We have discussed earlier that the inter-electronic repulsion of d electrons are determined in terms of Racah parameters, B and C. Although the values of B and C in a free ion are higher than that of a complex, the corresponding C/B ratio is quite similar (usually C = 4B). Hence, these diagram are plotted for specific C/B ratios (in the range of 4–5).

26.30 Inorganic Chemistry

1I 1T 1(g)

3T

1A

1(g)

3E (g)

1(g)

t23g eg1 (t23 e3)

3T 1(g)

t25g eg1 (t21 e3)

1T 2(g)

3T 2(g)

3T 2(g) 3E

3E (g)

3T 2(g) 3T 1(g)

3T 1(g)

5E (g)

Energy 5E

1(g)

5T 2(g)

5D

1A

– 6 Dq

3H 1(g)

3T 1(g)

6 (e4) t2g

3T 2(g)

1A

1(g)

1E

5T 2(g)

1(g)

1T 2(g) 5E

1(g)

t24g (t22 e4) – 16 Dq

(g) 3T 1(g)

– 24 Dq d6 octahedral and d 4 tetrahedral Strong 3 Strong configuration ligand field

Weak ligand field

Free ion term Dq

Weak ligand field

d4 octahedral and d 6 tetrahedral 3 Strong Strong configuration ligand field

Fig. 26.26 Correlation diagram for d4 and d6 ion in octahedral and tetrahedral ligand field

Fig. 26.27 Correlation diagram for d5 ion in octahedral and tetrahedral ligand field

Coordination Complexes IV: Spectroscopic and Magnetic Properties of Coordination Compounds

26.31

Some basic conventions are as follows: 1. The axes represent dimensionless units of E/B (along the vertical axis) and Dq/B (along the horizontal axis). 2. The ground-state is represented by a horizontal line. Its energy is taken as zero for all values of Dq/B. The free-ion terms are represented along the horizontal line on the left-hand side and the molecular term symbols are on the right-hand side. 3. The spin pairing is represented by a vertical line separating the diagram into two parts. The left side of the diagram represents the weak field and right side of the diagram represents the strong field. Thus, the vertical line represents the spin crossover. It should be noted that these diagrams are general for various dn metal ions (with the same electronic configuration).

26.8.1Applications of Tanable-Sugano Diagrams Interpretation of 10Dq and B Values (a) d2 System [V(H2O)6]3+ The [V(H2O)6]3+ complex ion shows two transitions in aqueous medium (Fig.26.28) which are assigned as follows: 1

Fig. 26.28 Tanabe–Sugano diagram for d2 octahedral ion

= 17,200 cm–1 [3T1g (F) " 3T2g (F)].

= 25,600 cm–1 [3T1g (F) " 3T1g (P)] (i) We have to find the ratio of these frequencies and then correlate the position of this ratio with the height 2

E2 can be interpreted in terms of E1 E2 25, 600 = = 1.49 E1 17, 200

of the 3T1g (P)/3T2g(F). Since

2

, we can write

1

The ratio 1.49 is obtained in the diagram at Dq/B value equal to 2.8. (ii) Now we have to find the corresponding value for E/B on the vertical line which comes out at the values of 25.9 [3T2g (F)] and 38.6 [3T1g (P)]. Thus, E1/B = 25.9

17, 200 cm -1 = 25.9 B or B = 665 cm–1 The value of B for V3+ ion is 860 cm–1 which is quite higher than the observed B value of [V(H2O)6]3+ complex ion, as expected due to nephelauxetic effects. (iii) We can also predict the third electronic transition 3T1g(F) " 3A2g(F) from the corresponding E3/B value for Dq/B value of 2.8. It is clear from the figure that E3/B value is 52. Thus E3 = 52 × B = 52 × 665 cm–1 = 34580 cm–1 This transition falls in uv region and hence is not observed. It also corresponds to Laporte forbidden transition and simultaneous excitation of two electrons. Thus, it is forbidden. or

26.32 Inorganic Chemistry

(iv) We can determine the value of 10Dq as Dq/B = 2.8 or 10Dq = 10 × 2.8 × 660 cm–1 = 18480 cm–1

(b) d3 System [Cr(H2O)6]3+ Complex Ion This complex ion shows three electronic transitions (Fig. 26.29) which are assigned as follows 1

= 17,400 cm–1 [4A2g(F) " 4T2g(F)]

2

= 24,500 cm–1 [4A2g(F) " 4T1g(F)]

3

= 36,500 cm–1 [4A2g(F) " 4T1g(P)]

(i) We can determine E2/E1 ratio as

E2 E1

24, 500 17, 400

1.4

The ratio 1.4 is obtained in the diagram at Dq/B value equal to 2.40. We can draw a vertical line at this value. (ii) The corrosponding value of E/B on this vertical line comes at 34 and 24.

Fig. 26.29 Tanage-Sugano diagram for d3 ion

E1 = 24 cm–1 B 17, 400 17, 400 or = 24 and B = = 725 cm–1 B 24 E2 Similarly, = 34 B Thus,

24, 500 = 34 and B = 725 cm–1 B (iii) We can determine the value of 10Dq as Dq/B = 24 cm–1 or 10Dq = 10 × 2.4 × B = 10 × 2.4 × 725 = 17,400 cm–1 or

(c) d5 System [Mn(H2O)6]2+ Complex Ion This complex ion in aqueous system is a special case as the electronic (Fig. 26.30) spectrum shows all weak spin forbidden transitions as follows: 1

= 18,000 cm–1 [6A1g " 4T1g]

2

= 23,000 cm–1 [6A1g " 4T2g]

3

= 24,500 cm–1 [6A1g " 2T2g]

There is no corresponding excited term of the same multiplicity as that of 6A1g, the ground-state term. We can determine the ratio E2/E1 from low spin portion of the TS diagram as follows: 4

= 24,700 cm–1 [6A1g " 4A1g, 4Eg]

5

= 28,300 cm–1 [6A1g " 4A2g, 2T1g]

6

= 30,000 cm–1 [6A1g " 4Eg]

Fig. 26.30 Tanabe-Sugano diagram for d5 configuration

Coordination Complexes IV: Spectroscopic and Magnetic Properties of Coordination Compounds

26.33

E2 = 23,000/18,600 = 1.2 corresponding to Dq/B E1 value of 1.1 This corresponds to E2/B value at 24. Thus, B comes out to be (23,000 cm–1/24) = 750 cm–1 (quite less than the free ion value of 960 m–1). We can also determine the value of 10Dq as 10 × 1.1 × 750 cm–1 = 9000 cm–1 As the ligand-field strength increases, crossover point is observed. As a result, the ground-state 6A1g of low-spin case ascends rapidly. On the other hand, the 2 T2g state from 2I free-ion term descends rapidly and becomes the ground state.

(d) d6 System [Co(ox)3] 3– Complex Ion This complex ion in aqueous system shows two electronic transitions (Fig. 26.31) which can be assigned as follows: 1

= 16,611 cm–1 [1A1g " 1T1g]

2

= 23,529 cm–1 [1A1g " 1T2g]

E2 23, 529 1.42 which comes out at Dq/B E1 16, 611 value of 3.5. This gives E2/B value at 32 cm–1.

Thus,

16, 611 cm 1 It means B = = 518 cm–1 32 (quite lesser as compared to that of the free ion value of 1400 cm–1. ) We can determine 10Dq value as 10 × 3.5 × 518 cm–1 = 18,153 cm–1

Fig. 26.31 Tanabe–Sugano diagram for d6 configuration

(e) d7 System [Co(H2O)6]2+ Complex Ion In aqueous system, three electronic transitions are observed for aqueous solution of [Co(H2O)6]2+ complex ion (Fig. 26.32). –1 4 4 1 = 8,000 cm [ T1g(F) " T2g (F)] 2

= 19,600 cm–1 [4T1g (F) " 4A2g (F)]

3

= 21,600 cm–1 [4T1g (F) " 4T1g (P)]

E3 21, 600 cm −1 = 2.7 which comes = E1 8, 000 cm −1 out at Dq/B value of 0.95. The corresponding energy gap 8, 000 = 980 cm–1(value is E1/B folds at 8.2. Thus, B = 8.2 again reduced than that of free ion value 1120 cm–1). We can determine 10Dq value as 10 × 0.95 × 980 cm–1 = 9300 cm –1. This gives

Fig. 26.32 Tanabe–Sugano diagram for d7 configuration

26.34 Inorganic Chemistry

(f) d8 System [Ni(H2O)6]2+ Complex Ion The band assignments for the electronic transition for aqueous solution of [Ni(H2O)6]2+ (Fig. 26.33) can be shown as follows. –1 3 3 1 = 8,700 cm [ A2g(F) " T2g (F)] 2

= 14,500 cm–1 [3A2g (F) " 3T1g (F)]

3

= 25,300 cm–1 [3A2g (F) " 3T1g (P)]

This gives

E3 25, 300 cm −1 = = 1.72 which comes out at E2 14, 700 cm −1

Dq/B value of 1.0. The corresponding E3/B comes at 28

25, 300 cm 1 = 900 cm–1. This value is 28 lower than the free ion value of 1,040 cm–1. The 10Dq value comes out as 10 × 1.0 × 900 cm–1 = 9000 cm –1. and gives B =

26.9

CHARGE-TRANSFER TRANSITIONS

Fig. 26.33 Tanabe–Sugano diagram for d8 configuration

We have discussed that colour of transition-metal compounds (complexes) is associated with d-d transitions. However, some transition metal compounds (complexes) show intense colour in solution in spite of absence of d-electrons. For example, the orange colour of TiBr4, yellow colour of CrO42–, orange colour of Cr2O72–, intense red colour of [Fe(SCN)4]– and deep purple of MnO4– are not due to d–d transitions. Similarly, the red colour of HgS and blue copper protein of the biological systems do not account for the d-d transitions. These colours are accounted due to charge-transfer transitions, the transition (transfer) of an electron from an orbital of one atom, called the electron donor, to an orbital of another atom, called the electron acceptor.

26.9.1 Characteristics of Charge-transfer Transitions 1. Charge-transfer transitions are generally more intense than the d-d transitions. Thus, the molar absorption coefficients for charge-transfer transitions are in the order of 103–104L mol–1 cm–1 than the typical molar absorption coefficients for d-d transitions (in the order of 20–1000L mol–1 cm–1). 2. Charge-transfer transitions are spin-allowed and Laporte-allowed. 3. Charge-transfer transitions take place at comparatively shorter wavelength than that of the d–d transitions of the same compound.

26.9.2 Types of Charge-transfer Transitions 1. Ligand to Metal Charge-Transfer Transitions (L " MCT Transitions) These type of transitions take place when transfer of electrons from a molecular orbital of the complex with mainly ligandlike character to those with mainly metal-like character. In these complexes, metals are generally in high oxidation state and ligands possess high-energy lone pairs. In case of an octahedral complex, the acceptor MO’s are either t*2g or e*g and the donor MO’s are t1u ro t2u. For example, in case of [IrBr6]2–, two charge-transfer bands are observed near 300 nm (transitions to e*g level) and near 600

Coordination Complexes IV: Spectroscopic and Magnetic Properties of Coordination Compounds

26.35

nm (transition to t*2g level). However, in case of [IrBr6]3–, only the higher frequency and higher intensity transition to e*g is possible at near 300 nm. This is due to the reason that t*2g are filled and charge-transfer transitions to t*2g is not possible. In case of tetrahedral complexes, the donor MO’s are -orbitals such as a1, t2 or -orbitals such as e, t1, t2. For example, in case of TiX4 (d9), the spectral bonds correspond to the ligand-metal charge transfer bands with transition from t1 molecular orbital to e* (lowest energy transition) and t*2 (highest-energy transition), Fig. 26.34.

Fig. 26.34 Representation of ligand " Metal charge transition (LMCT) in TiX4

It has been observed that the wave number of LMCT increases with increase in the electronegativity of the halogen. Thus, the transition takes place at 19,600 cm–1 in TiI4, at 29,500 cm–1 in TiBr4 and 35,400 cm–1 in TiCl4 (Fig. 26.35). Similarly in case of MnO4– (d0), charge transfer at 17,700 cm–1 [t1 " e] accounts for its deep purple colour. However, the other three possible L " M charge transitions are as follows: 2

= 29,500 cm–1 (t1 " t2*)

= 30,300 cm–1 (t2 " e2*) * –1 4 = 44,400 cm (t2 " t2 ) [CuBr4]2– is more intense in colour with a shift in charge-transfer band towards the lower energy Fig. 26.35 Absorption cross sections of TiX4 IR region. This is due to lesser electronegativity of 2– bromine. However, [CuI4] , with least electronegative I–, is very unstable and leads to reduction of Cu2+ to Cu+ and oxidation of I– to I2. 2Cu2+ + 4I– $ 2CuI + I2 In general, the wave number of the charge-transition band increases with the increase in electronegativity of the ligand, as observed in case of nickel (II) halide complexes (Fig. 26.36). On the other hand, more reducible the metal ion, lower is the energy transition. For example, in case of tetraoxido anions of transition metals, the transition energies increase in the following order: 3

VII

MnO4–

VII