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2, , CHEMISTRY, , ISBN, First Edition, March 2006, Reprinted, October 2006, November 2007, January 2009, December 2009, November 2010, January 2012, November 2012, November 2013, December 2014, May 2016, January 2018, December 2018, , 81-7450-494-X (Part I), 81-7450-535-0 (Part II), , ALL RIGHTS RESERVED, , Phalguna 1927, Kartika 1928, Kartika 1929, Magha 1930, Pausa 1931, Kartika 1932, Pausha 1933, Kartika 1934, Kartika 1935, Pausa 1936, Vaishakha 1938, Magha 1939, Agrahayana 1940, , , , No part of this publication may be reproduced, stored in a retrieval system or, transmitted, in any form or by any means, electronic, mechanical, photocopying,, recording or otherwise without the prior permission of the publisher., , , , This book is sold subject to the condition that it shall not, by way of trade, be lent, resold, hired out or otherwise disposed of without the publisher’s consent, in any form, of binding or cover other than that in which it is published., , , , The correct price of this publication is the price printed on this page. Any revised, price indicated by a rubber stamp or by a sticker or by any other means is incorrect, and should be unacceptable., , OFFICES OF THE PUBLICATION, DIVISION, NCERT, , PD 400T BS, © National Council of Educational, Research and Training, 2006, , NCERT Campus, Sri Aurobindo Marg, New Delhi 110 016, , Phone : 011-26562708, , 108, 100 Feet Road, Hosdakere Halli Extension, Banashankari III Stage, Bengaluru 560 085, , Phone : 080-26725740, , Navjivan Trust Building, P.O. Navjivan, Ahmedabad 380 014, , Phone : 079-27541446, , CWC Campus, Opp. Dhankal Bus Stop, Panihati, Kolkata 700 114, , Phone : 033-25530454, , CWC Complex, Maligaon, Guwahati 781 021, , Phone : 0361-2674869, , Publication Team,  ??.00, , Printed on 80 GSM paper with NCERT, watermark, , Head, Publication, Division, , : M. Siraj Anwar, , Chief Editor, , : Shveta Uppal, , Chief Business, Manager, , : Gautam Ganguly, , Chief Production, Officer, , : Arun Chitkara, , Editor, , : Binoy Banerjee, , Production Assistant, , : Mukesh Gaur, , Cover, Shweta Rao, , Published at the Publication Division, by the Secretary, National Council of, Educational Research and Training, Sri, Aurobindo Marg, New Delhi 110 016 and printed, at Chandra Prabhu Offset Printing Works (P.), Ltd., C-40, Sector-8, Noida - 201 301 (U.P.), , Illustrations, Nidhi Wadhwa, Anil Nayal, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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ANSWERS, , 3, , FOREWORD, The National Curriculum Framework (NCF), 2005 recommends that children’s life at, school must be linked to their life outside the school. This principle marks a departure, from the legacy of bookish learning which continues to shape our system and causes a, gap between the school, home and community. The syllabi and textbooks developed on, the basis of NCF signify an attempt to implement this basic idea. They also attempt to, discourage rote learning and the maintenance of sharp boundaries between different, subject areas. We hope these measures will take us significantly further in the direction of, a child-centred system of education outlined in the National Policy on Education (1986)., The success of this effort depends on the steps that school principals and teachers, will take to encourage children to reflect on their own learning and to pursue, imaginative activities and questions. We must recognise that, given space, time and, freedom, children generate new knowledge by engaging with the information passed, on to them by adults. Treating the prescribed textbook as the sole basis of examination, is one of the key reasons why other resources and sites of learning are ignored., Inculcating creativity and initiative is possible if we perceive and treat children as, participants in learning, not as receivers of a fixed body of knowledge., These aims imply considerable change in school routines and mode of functioning., Flexibility in the daily time-table is as necessary as rigour in implementing the annual, calender so that the required number of teaching days are actually devoted to teaching., The methods used for teaching and evaluation will also determine how effective this, textbook proves for making children’s life at school a happy experience, rather than a, source of stress or boredom. Syllabus designers have tried to address the problem of, curricular burden by restructuring and reorienting knowledge at different stages, with greater consideration for child psychology and the time available for teaching., The textbook attempts to enhance this endeavour by giving higher priority and space, to opportunities for contemplation and wondering, discussion in small groups, and, activities requiring hands-on experience., The National Council of Educational Research and Training (NCERT) appreciates, the hard work done by the textbook development committee responsible for this book., We wish to thank the Chairperson of the advisory group in science and mathematics,, Professor J.V. Narlikar and the Chief Advisor for this book, Professor B. L. Khandelwal, for guiding the work of this committee. Several teachers contributed to the development, of this textbook; we are grateful to their principals for making this possible. We are, indebted to the institutions and organisations which have generously permitted us to, draw upon their resources, material and personnel. As an organisation committed to, systemic reform and continuous improvement in the quality of its products, NCERT, welcomes comments and suggestions which will enable us to undertake further, revision and refinement., Director, National Council of Educational, Research and Training, , New Delhi, 20 December 2005, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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ANSWERS, , 7, , CONTENTS, , Unit 1, , Unit 2, , Unit 3, , Foreword, , iii, , Some Basic Concepts of Chemistry, , 1, , 1.1, , Importance of Chemistry, , 4, , 1.2, , Nature of Matter, , 4, , 1.3, , Properties of Matter and their Measurement, , 6, , 1.4, , Uncertainty in Measurement, , 10, , 1.5, , Laws of Chemical Combinations, , 14, , 1.6, , Dalton’s Atomic Theory, , 16, , 1.7, , Atomic and Molecular Masses, , 16, , 1.8, , Mole Concept and Molar Masses, , 18, , 1.9, , Percentage Composition, , 18, , 1.10 Stoichiometry and Stoichiometric Calculations, , 20, , Structure of Atom, , 29, , 2.1, , Discovery of Sub-atomic Particles, , 30, , 2.2, , Atomic Models, , 32, , 2.3, , Developments Leading to the Bohr’s Model of Atom, , 37, , 2.4, , Bohr’s Model for Hydrogen Atom, , 46, , 2.5, , Towards Quantum Mechanical Model of the Atom, , 49, , 2.6, , Quantum Mechanical Model of Atom, , 53, , Classification of Elements and Periodicity in Properties, , 74, , 3.1, , Why do we Need to Classify Elements ?, , 74, , 3.2, , Genesis of Periodic Classification, , 75, , 3.3, , Modern Periodic Law and the present form of the Periodic Table, , 79, , 3.4, , Nomenclature of Elements with Atomic Numbers > 100, , 79, , 3.5, , Electronic Configurations of Elements and the Periodic Table, , 82, , 3.6, , Electronic Configurations and Types of Elements:, , 83, , s-, p-, d-, f- Blocks, 3.7, , Periodic Trends in Properties of Elements, Download all NCERT books PDFs from www.ncert.online, , 2019-20, , 86

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8, , Unit 4, , Unit 5, , Unit 6, , (viii ), , CHEMISTRY, , Chemical Bonding and Molecular Structure, , 100, , 4.1, , Kössel-Lewis Approach to Chemical Bonding, , 101, , 4.2, , Ionic or Electrovalent Bond, , 106, , 4.3, , Bond Parameters, , 107, , 4.4, , The Valence Shell Electron Pair Repulsion (VSEPR) Theory, , 112, , 4.5, , Valence Bond Theory, , 117, , 4.6, , Hybridisation, , 120, , 4.7, , Molecular Orbital Theory, , 125, , 4.8, , Bonding in Some Homonuclear Diatomic Molecules, , 129, , 4.9, , Hydrogen Bonding, , 131, , States of Matter, , 136, , 5.1, , Intermolecular Forces, , 137, , 5.2, , Thermal Energy, , 139, , 5.3, , Intermolecular Forces vs Thermal Interactions, , 139, , 5.4, , The Gaseous State, , 139, , 5.5, , The Gas Laws, , 140, , 5.6, , Ideal Gas Equation, , 145, , 5.7, , Kinetic Energy and Molecular Speeds, , 147, , 5.8, , Kinetic Molecular Theory of Gases, , 149, , 5.9, , Behaviour of Real Gases: Deviation from Ideal Gas Behaviour, , 150, , 5.10 Liquefaction of Gases, , 152, , 5.11 Liquid State, , 154, , Thermodynamics, , 160, , 6.1, , Thermodynamic Terms, , 161, , 6.2, , Applications, , 164, , 6.3, , Measurement of ∆U and ∆H: Calorimetry, , 169, , 6.4, , Enthalpy Change, ∆rH of a Reaction – Reaction Enthalpy, , 171, , 6.5, , Enthalpies for Different Types of Reactions, , 176, , 6.6, , Spontaneity, , 181, , 6.7, , Gibbs Energy Change and Equilibrium, , 186, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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ANSWERS, , Unit 7, , (i x), , 9, , Equilibrium, , 192, , 7.1, , Equilibrium in Physical Processes, , 193, , 7.2, , Equilibrium in Chemical Processes – Dynamic Equilibrium, , 196, , 7.3, , Law of Chemical Equilibrium and Equilibrium Constant, , 198, , 7.4, , Homogeneous Equilibria, , 201, , 7.5, , Heterogeneous Equilibria, , 203, , 7.6, , Applications of Equilibrium Constants, , 205, , 7.7, , Relationship between Equilibrium Constant K,, , 208, , Reaction Quotient Q and Gibbs Energy G, 7.8, , Factors Affecting Equilibria, , 208, , 7.9, , Ionic Equilibrium in Solution, , 212, , 7.10 Acids, Bases and Salts, , 213, , 7.11 Ionization of Acids and Bases, , 216, , 7.12 Buffer Solutions, , 226, , 7.13 Solubility Equilibria of Sparingly Soluble Salts, , 228, , Appendices, , 239, , Answer to some Selected Questions, , 253, , Index, , 259, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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UNIT 1, , SOME BASIC CONCEPTS OF CHEMISTRY, , After studying this unit, you will be, able to, • appreciate the contribution of, India in the development of, chemistry understand the role of, chemistry in different spheres of, life;, • explain the characteristics of, three states of matter;, • classify different substances into, elements, compounds and, mixtures;, • use scientific notations and, determine significant figures;, • differentiate between precision and, accuracy;, • define SI base units and convert, physical quantities from one, system of units to another;, • explain various laws of chemical, combination;, • appreciate significance of atomic, mass, average atomic mass,, molecular mass and formula, mass;, • describe the terms – mole and, molar mass;, • calculate the mass per cent of, component elements constituting, a compound;, • determine empirical formula and, molecular formula for a compound, from the given experimental data;, and, • perform the stoichiometric, calculations., , Chemistry is the science of molecules and their, transformations. It is the science not so much of the one, hundred elements but of the infinite variety of molecules that, may be built from them., Roald Hoffmann, , Science can be viewed as a continuing human effort to, systematise knowledge for describing and understanding, nature. You have learnt in your previous classes that we come, across diverse substances present in nature and changes in, them in daily life. Curd formation from milk, formation of, vinegar from sugarcane juice on keeping for prolonged time, and rusting of iron are some of the examples of changes which, we come across many times. For the sake of convenience,, science is sub-divided into various disciplines: chemistry,, physics, biology, geology, etc. The branch of science that, studies the preparation, properties, structure and reactions, of material substances is called chemistry., DEVELOPMENT OF CHEMISTRY, Chemistry, as we understand it today, is not a very old, discipline. Chemistry was not studied for its own sake, rather, it came up as a result of search for two interesting things:, i. Philosopher’s stone (Paras) which would convert, all baser metals e.g., iron and copper into gold., ii.‘Elexir of life’ which would grant immortality., People in ancient India, already had the knowledge of many, scientific phenomenon much before the advent of modern, science. They applied that knowledge in various walks of, life. Chemistry developed mainly in the form of Alchemy, and Iatrochemistry during 1300-1600 CE. Modern, chemistry took shape in the 18th century Europe, after a, few centuries of alchemical traditions which were, introduced in Europe by the Arabs., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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2, , CHEMISTRY, , Other cultures – especially the Chinese and, the Indian – had their own alchemical traditions., These included much knowledge of chemical, processes and techniques., In ancient India, chemistry was called, Rasayan Shastra, Rastantra, Ras Kriya or, Rasvidya. It included metallurgy, medicine,, manufacture of cosmetics, glass, dyes, etc., Systematic excavations at Mohenjodaro in, Sindh and Harappa in Punjab prove that the, story of development of chemistry in India is, very old. Archaeological findings show that, baked bricks were used in construction work., It shows the mass production of pottery, which, can be regarded as the earliest chemical process,, in which materials were mixed, moulded and, subjected to heat by using fire to achieve, desirable qualities. Remains of glazed pottery, have been found in Mohenjodaro. Gypsum, cement has been used in the construction work., It contains lime, sand and traces of CaCO3., Harappans made faience, a sort of glass which, was used in ornaments. They melted and forged, a variety of objects from metals, such as lead,, silver, gold and copper. They improved the, hardness of copper for making artefacts by, using tin and arsenic. A number of glass objects, were found in Maski in South India (1000–900, BCE), and Hastinapur and Taxila in North, India (1000–200 BCE). Glass and glazes were, coloured by addition of colouring agents like, metal oxides., Copper metallurgy in India dates back to, the beginning of chalcolithic cultures in the, subcontinent. There are much archeological, evidences to support the view that technologies, for extraction of copper and iron were developed, indigenously., According to Rigveda, tanning of leather, and dying of cotton were practised during, 1000–400 BCE. The golden gloss of the black, polished ware of northen India could not be, replicated and is still a chemical mystery. These, wares indicate the mastery with which kiln, temperatures could be controlled. Kautilya’s, Arthashastra describes the production of salt, from sea., A vast number of statements and material, described in the ancient Vedic literature can, , be shown to agree with modern scientific, findings. Copper utensils, iron, gold, silver, ornaments and terracotta discs and painted, grey pottery have been found in many, archaeological sites in north India. Sushruta, Samhita explains the importance of Alkalies., The Charaka Samhita mentions ancient, indians who knew how to prepare sulphuric, acid, nitric acid and oxides of copper, tin and, zinc; the sulphates of copper, zinc and iron and, the carbonates of lead and iron., Rasopanishada describes the preparation, of gunpowder mixture. Tamil texts also, describe the preparation of fireworks using, sulphur, charcoal, saltpetre (i.e., potassium, nitrate), mercury, camphor, etc., Nagarjuna was a great Indian scientist. He, was a reputed chemist, an alchemist and a, metallurgist. His work Rasratnakar deals with, the formulation of mercury compounds. He has, also discussed methods for the extraction of, metals, like gold, silver, tin and copper. A book,, Rsarnavam, appeared around 800 CE. It, discusses the uses of various furnaces, ovens, and crucibles for different purposes. It, describes methods by which metals could be, identified by flame colour., Chakrapani discovered mercury sulphide., The credit for inventing soap also goes to him., He used mustard oil and some alkalies as, ingredients for making soap. Indians began, making soaps in the 18th century CE. Oil of, Eranda and seeds of Mahua plant and calcium, carbonate were used for making soap., The paintings found on the walls of Ajanta, and Ellora, which look fresh even after ages,, testify to a high level of science achieved in, ancient India. Varähmihir’s Brihat Samhita is, a sort of encyclopaedia, which was composed, in the sixth century CE. It informs about the, preparation of glutinous material to be applied, on walls and roofs of houses and temples. It, was prepared entirely from extracts of various, plants, fruits, seeds and barks, which were, concentrated by boiling, and then, treated with, various resins. It will be interesting to test such, materials scientifically and assess them for use., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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SOME BASIC CONCEPTS OF CHEMISTRY, , 3, , A number of classical texts, like, Atharvaveda (1000 BCE) mention some dye, stuff, the material used were turmeric, madder,, sunflower, orpiment, cochineal and lac. Some, other substances having tinting property were, kamplcica, pattanga and jatuka., Varähmihir’s Brihat Samhita gives, references to perfumes and cosmetics. Recipes, for hair dying were made from plants, like, indigo and minerals like iron power, black iron, or steel and acidic extracts of sour rice gruel., Gandhayukli describes recipes for making, scents, mouth perfumes, bath powders,, incense and talcum power., Paper was known to India in the, 17th century as account of Chinese traveller, I-tsing describes. Excavations at Taxila indicate, that ink was used in India from the fourth, century. Colours of ink were made from chalk,, red lead and minimum., It seems that the process of fermentation, was well-known to Indians. Vedas and, Kautilya’s Arthashastra mention about many, types of liquors. Charaka Samhita also, mentions ingredients, such as barks of plants,, stem, flowers, leaves, woods, cereals, fruits and, sugarcane for making Asavas., The concept that matter is ultimately made, of indivisible building blocks, appeared in, India a few centuries BCE as a part of, philosophical speculations. Acharya Kanda,, born in 600 BCE, originally known by the, name Kashyap, was the first proponent of the, ‘atomic theory’. He formulated the theory of, very small indivisible particles, which he, named ‘Paramãnu’ (comparable to atoms). He, authored the text Vaiseshika Sutras., According to him, all substances are, aggregated form of smaller units called atoms, (Paramãnu), which are eternal, indestructible,, spherical, suprasensible and in motion in the, original state. He explained that this individual, entity cannot be sensed through any human, organ. Kanda added that there are varieties of, atoms that are as different as the different, classes of substances. He said these, (Paramãnu) could form pairs or triplets, among, other combinations and unseen forces cause, , interaction between them. He conceptualised, this theory around 2500 years before John, Dalton (1766-1844)., Charaka Samhita is the oldest Ayurvedic, epic of India. It describes the treatment of, diseases. The concept of reduction of particle, size of metals is clearly discussed in Charaka, Samhita. Extreme reduction of particle size is, termed as nanotechnology. Charaka Samhita, describes the use of bhasma of metals in the, treatment of ailments. Now-a-days, it has been, proved that bhasmas have nanoparticles of, metals., After the decline of alchemy, Iatrochemistry, reached a steady state, but it too declined due, to the introduction and practise of western, medicinal system in the 20th century. During, this period of stagnation, pharmaceutical, industry based on Ayurveda continued to, exist, but it too declined gradually. It took, about 100-150 years for Indians to learn and, adopt new techniques. During this time, foreign, products poured in. As a result, indigenous, traditional techniques gradually declined., Modern science appeared in Indian scene in, the later part of the nineteenth century. By the, mid-nineteenth century, European scientists, started coming to India and modern chemistry, started growing., From the above discussion, you have learnt, that chemistry deals with the composition,, structure, properties and interection of matter, and is of much use to human beings in daily, life. These aspects can be best described and, understood in terms of basic constituents of, matter that are atoms and molecules. That, is why, chemistry is also called the science of, atoms and molecules. Can we see, weigh and, perceive these entities (atoms and molecules)?, Is it possible to count the number of atoms, and molecules in a given mass of matter and, have a quantitative relationship between the, mass and the number of these particles? We, will get the answer of some of these questions, in this Unit. We will further describe how, physical properties of matter can be, quantitatively described using numerical, values with suitable units., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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4, , CHEMISTRY, , 1.1 IMPORTANCE OF CHEMISTRY, Chemistry plays a central role in science and, is often intertwined with other branches of, science., Principles of chemistry are applicable in, diverse areas, such as weather patterns,, functioning of brain and operation of a, computer, production in chemical industries,, manufacturing fertilisers, alkalis, acids, salts,, dyes, polymers, drugs, soaps, detergents,, metals, alloys, etc., including new material., Chemistry contributes in a big way to the, national economy. It also plays an important, role in meeting human needs for food,, healthcare products and other material, aimed at improving the quality of life. This, is exemplified by the large-scale production, of a variety of fertilisers, improved variety of, pesticides and insecticides. Chemistry, provides methods for the isolation of lifesaving drugs from natural sources and, makes possible synthesis of such drugs., Some of these drugs are cisplatin and taxol,, which are effective in cancer therapy. The, drug AZT (Azidothymidine) is used for, helping AIDS patients., Chemistry contributes to a large extent, in the development and growth of a nation., With a better understanding of chemical, principles it has now become possible to, design and synthesise new material having, specific magnetic, electric and optical, properties. This has lead to the production, of superconducting ceramics, conducting, polymers, optical fibres, etc. Chemistry has, helped in establishing industries which, manufacture utility goods, like acids,, alkalies, dyes, polymesr metals, etc. These, industries contribute in a big way to the, e c o no my o f a natio n and g e n e r a t e, employment., In recent years, chemistry has helped, in dealing with some of the pressing aspects, of environmental degradation with a fair, degree of success. Safer alternatives to, environmentally hazardous refrigerants, like, CFCs (chlorofluorocarbons), responsible for, ozone depletion in the stratosphere, have, , been successfully synthesised. However,, many big environmental problems continue, to be matters of grave concern to the, chemists. One such problem is the, management of the Green House gases, like, methane, carbon dioxide, etc. Understanding, of biochemical processes, use of enzymes for, large-scale production of chemicals and, synthesis of new exotic material are some of, the intellectual challenges for the future, generation of chemists. A developing country,, like India, needs talented and creative, chemists for accepting such challenges. To, be a good chemist and to accept such, challanges, one needs to understand the, basic concepts of chemistry, which begin with, the concept of matter. Let us start with the, nature of matter., 1.2 NATURE OF MATTER, You are already familiar with the term matter, from your earlier classes. Anything which has, mass and occupies space is called matter., Everything around us, for example, book, pen,, pencil, water, air, all living beings, etc., are, composed of matter. You know that they have, mass and they occupy space. Let us recall the, characteristics of the states of matter, which, you learnt in your previous classes., 1.2.1 States of Matter, You are aware that matter can exist in three, physical states viz. solid, liquid and gas. The, constituent particles of matter in these three, states can be represented as shown in Fig. 1.1., Particles are held very close to each other, in solids in an orderly fashion and there is not, much freedom of movement. In liquids, the, particles are close to each other but they can, move around. However, in gases, the particles, are far apart as compared to those present in, solid or liquid states and their movement is, easy and fast. Because of such arrangement, of particles, different states of matter exhibit, the following characteristics:, (i) Solids have definite volume and definite, shape., (ii) Liquids have definite volume but do not, have definite shape. They take the shape, of the container in which they are placed., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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SOME BASIC CONCEPTS OF CHEMISTRY, , 5, , Fig. 1.1 Arrangement of particles in solid, liquid, and gaseous state, , (iii) Gases have neither definite volume nor, definite shape. They completely occupy the, space in the container in which they are placed., These three states of matter are, interconvertible by changing the conditions of, temperature and pressure., Solid, liquid, Gas, On heating, a solid usually changes to a, liquid, and the liquid on further heating changes, to gas (or vapour). In the reverse process, a gas, on cooling liquifies to the liquid and the liquid, on further cooling freezes to the solid., 1.2.2. Classification of Matter, In class IX (Chapter 2), you have learnt that, at the macroscopic or bulk level, matter can be, classified as mixture or pure substance. These, can be further sub-divided as shown in Fig. 1.2., When all constituent particles of a, substance are same in chemical nature, it is, said to be a pure substance. A mixture, contains many types of particles., A mixture contains particles of two or more, pure substances which may be present in it in, any ratio. Hence, their composition is variable., Pure sustances forming mixture are called its, components. Many of the substances present, around you are mixtures. For example, sugar, solution in water, air, tea, etc., are all mixtures., A mixture may be homogeneous or, heterogeneous. In a homogeneous mixture,, the components completely mix with each other., This means particles of components of the, mixture are uniformly distributed throughout, , Fig. 1.2 Classification of matter, , the bulk of the mixture and its composition is, uniform throughout. Sugar solution and air, are the examples of homogeneous mixtures., In contrast to this, in a heterogeneous, mixture, the composition is not uniform, throughout and sometimes different, components are visible. For example, mixtures, of salt and sugar, grains and pulses along with, some dirt (often stone pieces), are, heterogeneous mixtures. You can think of, many more examples of mixtures which you, come across in the daily life. It is worthwhile to, mention here that the components of a, mixture can be separated by using physical, methods,, such, as, simple, hand-picking, filtration, crystallisation,, distillation, etc., Pure substances have characteristics, different from mixtures. Constituent particles, of pure substances have fixed composition., Copper, silver, gold, water and glucose are, some examples of pure substances. Glucose, contains carbon, hydrogen and oxygen in a, fixed ratio and its particles are of same, composition. Hence, like all other pure, substances, glucose has a fixed composition., Also, its constituents—carbon, hydrogen and, oxygen—cannot be separated by simple, physical methods., Pure substances can further be, classified into elements and compounds., Particles of an element consist of only one, type of atoms. These particles may exist as, atoms or molecules. You may be familiar, with atoms and molecules from the, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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6, , CHEMISTRY, , previous classes; however, you will be, studying about them in detail in Unit 2., Sodium, copper, silver, hydrogen, oxygen,, etc., are some examples of elements. Their, all atoms are of one type. However, the, atoms of different elements are different in, nature. Some elements, such as sodium or, copper, contain atoms as their constituent, particles, whereas, in some others, the, constituent particles are molecules which, are formed by two or more atoms. For, example, hydrogen, nitrogen and oxygen, gases consist of molecules, in which two, atoms combine to give their respective, molecules. This is illustrated in Fig. 1.3., , Water molecule, (H2O), Fig. 1.4, , Carbon dioxide, molecule (CO2), , A depiction of molecules of water and, carbon dioxide, , carbon atom. Thus, the atoms of different, elements are present in a compound in a fixed, and definite ratio and this ratio is characteristic, of a particular compound. Also, the properties, of a compound are different from those of its, constituent elements. For example, hydrogen, and oxygen are gases, whereas, the compound, formed by their combination i.e., water is a, liquid. It is interesting to note that hydrogen, burns with a pop sound and oxygen is a, supporter of combustion, but water is used, as a fire extinguisher., 1.3, , PROPERTIES OF MATTER AND, THEIR MEASUREMENT, , 1.3.1 Physical and chemical properties, , Fig. 1.3 A representation of atoms and molecules, , When two or more atoms of different, elements combine together in a definite ratio,, the molecule of a compound is obtained., Moreover, the constituents of a compound, cannot be separated into simpler, substances by physical methods. They can, be separated by chemical methods., Examples of some compounds are water,, ammonia, carbon dioxide, sugar, etc. The, molecules of water and carbon dioxide are, represented in Fig. 1.4., Note that a water molecule comprises two, hydrogen atoms and one oxygen atom., Similarly, a molecule of carbon dioxide, contains two oxygen atoms combined with one, , Every substance has unique or characteristic, properties. These properties can be classified, into two categories — physical properties,, such as colour, odour, melting point, boiling, point, density, etc., and chemical properties,, like composition, combustibility, ractivity with, acids and bases, etc., Physical properties can be measured or, observed without changing the identity or the, composition of the substance. The measurement, or observation of chemical properties requires, a chemical change to occur. Measurement of, physical properties does not require occurance, of a chemical change. The examples of chemical, properties are characteristic reactions of different, substances; these include acidity or basicity,, combustibility, etc. Chemists describe, interpret, and predict the behaviour of substances on the, basis of knowledge of their physical and chemical, properties, which are determined by careful, measurement and experimentation. In the, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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SOME BASIC CONCEPTS OF CHEMISTRY, , 7, , following section, we will learn about the, measurement of physical properties., , Maintaining the National, Standards of Measurement, , 1.3.2 Measurement of physical properties, , The s ystem of units, including unit, definitions, keeps on changing with time., Whenever the accuracy of measurement, of a particular unit was enhanced, substantially by adopting new principles,, member nations of metre treaty (signed in, 1875), agreed to change the formal, definition of that unit. Each modern, industrialised country, including India, has, a National Metrology Institute (NMI), which, maintains standards of measurements., This responsibility has been given to the, National Physical Laboratory (NPL),, New Delhi. This laboratory establishes, experiments to realise the base units and, derived units of measurement and, maintains National Standards of, Measurement. These standards are, periodically, inter -compar ed, with, standards maintained at other National, Metrology Institutes in the world, as well, as those, established at the International, Bureau of Standards in Paris., , Quantitative measurement of properties is, reaquired for scientific investigation. Many, properties of matter, such as length, area,, volume, etc., are quantitative in nature. Any, quantitative observation or measurement is, represented by a number followed by units, in which it is measured. For example, length, of a room can be represented as 6 m; here, 6, is the number and m denotes metre, the unit, in which the length is measured., Earlier, two dif ferent systems of, measurement, i.e., the English System and, the Metric System were being used in, different parts of the world. The metric system,, which originated in France in late eighteenth, century, was more convenient as it was based, on the decimal system. Late, need of a common, standard system was felt by the scientific, community. Such a system was established, in 1960 and is discussed in detail below., 1.3.3 The International System of Units (SI), The International System of Units (in French, Le Systeme Inter national d’Unités —, abbreviated as SI) was established by the, 11 th General Conference on Weights and, Measures (CGPM from Conference, Generale des Poids et Measures). The CGPM, is an inter-governmental treaty organisation, created by a diplomatic treaty known as, , Metre Convention, which was signed in Paris, in 1875., The SI system has seven base units and, they are listed in Table 1.1. These units pertain, to the seven fundamental scientific quantities., The other physical quantities, such as speed,, volume, density, etc., can be derived from these, quantities., , Table 1.1 Base Physical Quantities and their Units, Base Physical, Quantity, , Symbol, for, Quantity, , Name of, SI Unit, , Symbol, for SI, Unit, , Length, , l, , metre, , m, , Mass, , m, , kilogram, , kg, , Time, , t, , second, , s, , Electric current, , I, , ampere, , A, , Thermodynamic, temperature, , T, , kelvin, , K, , Amount of substance, Luminous intensity, , n, Iv, , mole, candela, , mol, cd, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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8, , CHEMISTRY, , The definitions of the SI base units are given, in Table 1.2., The SI system allows the use of prefixes to, indicate the multiples or submultiples of a unit., , These prefixes are listed in Table 1.3., Let us now quickly go through some of the, quantities which you will be often using in this, book., , Table 1.2 Definitions of SI Base Units, Unit of length, , metre, , The metre is the length of the path travelled, by light in vacuum during a time interval of, 1/299 792 458 of a second., , Unit of mass, , kilogram, , The kilogram is the unit of mass; it is equal, to the mass of the international prototype, of the kilogram., , Unit of time, , second, , The second is the duration of 9 192 631 770, periods of the radiation corresponding to the, transition between the two hyperfine levels, of the ground state of the caesium-133 atom., , Unit of electric, current, , ampere, , The ampere is that constant current, which, if maintained in two straight parallel, conductors of infinite length of negligible, circular cross-section and placed 1 metre, apart in vacuum, would produce between, these conductors a force equal to 2 × 10–7, newton per metre of length., , Unit of thermodynamic, temperature, , kelvin, , The kelvin, unit of thermodynamic, temperature, is the fraction 1/273.16 of the, thermodynamic temperature of the triple*, point of water., , Unit of amount of substance, , mole, , 1. The mole is the amount of substance of a, system, which contains as many, elementary entities as there are atoms in, 0.012 kilogram of carbon-12; its symbol is, ‘mol’., 2. When the mole is used, the elementary, entities must be specified and these may, be atoms, molecules, ions, electrons, other, particles, or specified groups of such, particles., , Unit of luminous intensity, , candela, , The candela is the luminous intensity, in a, given direction, of a source that emits, mono chromatic radiation of frequency, 540 × 1012 hertz and that has a radiant, intensity in that direction of 1/683 watt per, steradian., , * Triple point of water is 0.01 °C or 279.16K (32.01°F), Download all NCERT books PDFs from www.ncert.online, , 2019-20

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SOME BASIC CONCEPTS OF CHEMISTRY, , 9, , Table 1.3 Prefixes used in the SI System, Multiple, , Prefix, , Symbol, , 10–24, , yocto, , y, , 10–21, , zepto, , z, , –18, , atto, , a, , –15, , femto, , f, , –12, , pico, , p, , –9, , nano, , n, , –6, , 10, , 10, 10, , 10, 10, , micro, , µ, , 10–3, , milli, , m, , 10–2, , centi, , c, , 10, , deci, , d, , 10, , –1, , deca, , da, , 2, , hecto, , h, , 3, , kilo, , k, , 6, , 10, 10, , 10, , mega, , M, , 109, , giga, , G, , 1012, , tera, , T, , 15, , 10, , peta, , P, , 1018, , exa, , E, , 21, , zeta, , Z, , 24, , yotta, , Y, , 10, , 10, , Fig. 1.5, , Analytical balance, , SI system, volume has units of m3. But again,, in chemistry laboratories, smaller volumes are, used. Hence, volume is often denoted in cm3, or dm3 units., , 1.3.4 Mass and Weight, Mass of a substance is the amount of matter, present in it, while weight is the force exerted, by gravity on an object. The mass of a, substance is constant, whereas, its weight, may vary from one place to another due to, change in gravity. You should be careful in, using these terms., , A common unit, litre (L) which is not an SI, unit, is used for measurement of volume of, liquids., 1 L = 1000 mL , 1000 cm3 = 1 dm3, Fig. 1.6 helps to visualise these relations., , The mass of a substance can be determined, accurately in the laboratory by using an, analytical balance (Fig. 1.5)., The SI unit of mass as given in Table 1.1 is, kilogram. However, its fraction named as gram, (1 kg = 1000 g), is used in laboratories due to, the smaller amounts of chemicals used in, chemical reactions., 1.3.5 Volume, Volume is the amont of space occupied by a, substance. It has the units of (length)3. So in, , Fig. 1.6 Different units used to express, volume, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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10, , CHEMISTRY, , In the laboratory, the volume of liquids or, solutions can be measured by graduated, cylinder, burette, pipette, etc. A volumetric, flask is used to prepare a known volume of a, solution. These measuring devices are shown, in Fig. 1.7., , fahrenheit) and K (kelvin). Here, K is the SI, unit. The thermometers based on these scales, are shown in Fig. 1.8. Generally, the, thermometer with celsius scale are calibrated, from 0° to 100°, where these two, temperatures are the freezing point and the, boiling point of water, respectively. The, fahrenheit scale is represented between 32°, to 212°., The temperatures on two scales are related, to each other by the following relationship:, 9, (°C) + 32, 5, The kelvin scale is related to celsius scale, as follows:, °F =, , K = °C + 273.15, , Fig. 1.7, , It is interesting to note that temperature, below 0 °C (i.e., negative values) are possible, in Celsius scale but in Kelvin scale, negative, temperature is not possible., , Some volume measuring devices, , 1.3.6 Density, The two properties — mass and volume, discussed above are related as follows:, , Mass, Volume, Density of a substance is its amount of mass, per unit volume. So, SI units of density can be, obtained as follows:, Density =, , SI unit of density =, , SI unit of mass, SI unit of volume, , kg, or kg m–3, m3, This unit is quite large and a chemist often, expresses density in g cm–3, where mass is, expressed in gram and volume is expressed in, 3, cm . Density of a substance tells us about how, closely its particles are packed. If density is, more, it means particles are more closely, packed., =, , 1.3.7 Temperature, There are three common scales to measure, temperature — °C (degree celsius), °F (degree, , Fig. 1.8, , 1.4, , Thermometers, using, temperature scales, , different, , UNCERTAINTY IN MEASUREMENT, , Many a time in the study of chemistry, one, has to deal with experimental data as well as, theoretical calculations. There are meaningful, ways to handle the numbers conveniently and, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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SOME BASIC CONCEPTS OF CHEMISTRY, , Reference Standard, After defining a unit of measurement such, as the kilogram or the metre, scientists, agreed on reference standards that make, it possible to calibrate all measuring, devices. For getting reliable measurements,, all devices such as metre sticks and, analytical balances have been calibrated by, their manufacturers to give correct, readings. However, each of these devices, is standardised or calibrated against some, reference. The mass standard is the, kilogram since 1889. It has been defined, as the mass of platinum-iridium (Pt-Ir), cylinder that is stored in an airtight jar at, Inter national Bureau of Weights and, Measures in Sevres, France. Pt-Ir was, chosen for this standard because it is, highly resistant to chemical attack and its, mass will not change for an extremely long, time., Scientists are in search of a new, standard for mass. This is being attempted, through accurate determination of, Avogadro constant. Work on this new, standard focuses on ways to measure, accurately the number of atoms in a welldefined mass of sample. One such method,, which uses X-rays to determine the atomic, density of a crystal of ultrapure silicon, has, an accuracy of about 1 part in 106 but has, not yet been adopted to serve as a, standard. There are other methods but, none of them are presently adequate to, replace the Pt-Ir cylinder. No doubt,, changes are expected within this decade., The metre was originally defined as the, length between two marks on a Pt-Ir bar, kept at a temperature of 0°C (273.15 K). In, 1960 the length of the metre was defined, as 1.65076373 × 106 times the wavelength, of light emitted by a krypton laser., Although this was a cumbersome number,, it preserved the length of the metre at its, agreed value. The metre was redefined in, 1983 by CGPM as the length of path, travelled by light in vacuum during a time, interval of 1/299 792 458 of a second., Similar to the length and the mass, there, are reference standards for other physical, quantities., , 11, , present the data realistically with certainty to, the extent possible. These ideas are discussed, below in detail., 1.4.1 Scientific Notation, As chemistry is the study of atoms and, molecules, which have extremely low masses, and are present in extremely large numbers,, a chemist has to deal with numbers as large, as 602, 200,000,000,000,000,000,000 for, the molecules of 2 g of hydrogen gas or as, small as 0.00000000000000000000000166, g mass of a H atom. Similarly, other constants, such as Planck’s constant, speed of light,, charges on particles, etc., involve numbers of, the above magnitude., It may look funny for a moment to write, or count numbers involving so many zeros, but it offers a real challenge to do simple, mathematical operations of addition,, subtraction, multiplication or division with, such numbers. You can write any two, numbers of the above type and try any one of, the operations you like to accept as a, challenge, and then, you will really appreciate, the difficulty in handling such numbers., This problem is solved by using scientific, notation for such numbers, i.e., exponential, notation in which any number can be, represented in the form N × 10n, where n is an, exponent having positive or negative values, and N is a number (called digit term) which, varies between 1.000... and 9.999...., Thus, we can write 232.508 as, 2.32508 ×102 in scientific notation. Note that, while writing it, the decimal had to be moved, to the left by two places and same is the, exponent (2) of 10 in the scientific notation., Similarly, 0.00016 can be written as, 1.6 × 10–4. Here, the decimal has to be moved, four places to the right and (–4) is the exponent, in the scientific notation., While performing mathematical operations, on numbers expressed in scientific notations,, the following points are to be kept in mind., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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12, , CHEMISTRY, , Multiplication and Division, These two operations follow the same rules, which are there for exponential numbers, i.e., , (5.6 × 10 ) × (6.9 × 10 ) = (5.6 × 6.9) (10 ), 5, , 8, , 5+ 8, , = (5.6 × 6.9 ) × 1013, = 38..64 × 1013, = 3.864 × 1014, , (9.8 × 10 ) × (2.5 × 10 ) = (9.8 × 2.5) (10, = ( 9.8 × 2.5) (10, −2, , −2 + ( −6 ), , −6, , = 24.50 × 10, , −2 − 6, , ), , ), , −8, , = 2.450 × 10−7, 2.7 × 10−3, = (2.7 ÷ 5.5 ) 10−3 − 4 = 0.4909 × 10 −7, 5 × 104, 5.5, = 4.909 × 10−8, , (, , ), , Addition and Subtraction, For these two operations, first the numbers are, written in such a way that they have the same, exponent. After that, the coefficients (digit, terms) are added or subtracted as the case, may be., Thus, for adding 6.65 × 104 and 8.95 × 103,, exponent is made same for both the numbers., Thus, we get (6.65 × 104) + (0.895 × 104), Then, these numbers can be added as follows, (6.65 + 0.895) × 104 = 7.545 × 104, Similarly, the subtraction of two numbers can, be done as shown below:, (2.5 × 10–2 ) – (4.8 × 10–3), = (2.5 × 10–2) – (0.48 × 10–2), = (2.5 – 0.48) × 10–2 = 2.02 × 10–2, 1.4.2 Significant Figures, Every experimental measurement has some, amount of uncertainty associated with it, because of limitation of measuring instrument, and the skill of the person making the, measurement. For example, mass of an object, is obtained using a platform balance and it, comes out to be 9.4g. On measuring the mass, of this object on an analytical balance, the, mass obtained is 9.4213g. The mass obtained, , by an analytical balance is slightly higher, than the mass obtained by using a platform, balance. Therefore, digit 4 placed after, decimal in the measurement by platform, balance is uncertain., The uncertainty in the experimental or the, calculated values is indicated by mentioning, the number of significant figures. Significant, figures are meaningful digits which are, known with certainty plus one which is, estimated or uncertain. The uncertainty is, indicated by writing the certain digits and the, last uncertain digit. Thus, if we write a result, as 11.2 mL, we say the 11 is certain and 2 is, uncertain and the uncertainty would be +1, in the last digit. Unless otherwise stated, an, uncertainty of +1 in the last digit is always, understood., There are certain rules for determining the, number of significant figures. These are, stated below:, (1) All non-zero digits are significant. For, example in 285 cm, there are three, significant figures and in 0.25 mL, there, are two significant figures., (2) Zeros preceding to first non-zero digit are, not significant. Such zero indicates the, position of decimal point. Thus, 0.03 has, one significant figure and 0.0052 has two, significant figures., (3) Zeros between two non-zero digits are, significant. Thus, 2.005 has four, significant figures., (4) Zeros at the end or right of a number are, significant, provided they are on the right, side of the decimal point. For example,, 0.200 g has three significant figures. But,, if otherwise, the terminal zeros are not, significant if there is no decimal point. For, example, 100 has only one significant, figure, but 100. has three significant, figures and 100.0 has four significant, figures. Such numbers are better, represented in scientific notation. We can, express the number 100 as 1×102 for one, significant figure, 1.0×10 2 for two, significant figures and 1.00×102 for three, significant figures., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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SOME BASIC CONCEPTS OF CHEMISTRY, , 13, , (5) Counting the numbers of object,, for example, 2 balls or 20 eggs, have, infinite significant figures as these, are exact numbers and can be, represented by writing infinite number of, zeros after placing a decimal i.e.,, 2 = 2.000000 or 20 = 20.000000., In numbers written in scientific notation,, all digits are significant e.g., 4.01×102 has three, –3, significant figures, and 8.256 × 10 has four, significant figures., However, one would always like the results, to be precise and accurate. Precision and, accuracy are often referred to while we talk, about the measurement., Precision refers to the closeness of various, measurements for the same quantity. However,, accuracy is the agreement of a particular value, to the true value of the result. For example, if, the true value for a result is 2.00 g and student, ‘A’ takes two measurements and reports the, results as 1.95 g and 1.93 g. These values are, precise as they are close to each other but are, not accurate. Another student ‘B’ repeats the, experiment and obtains 1.94 g and 2.05 g as, the results for two measurements. These, observations are neither precise nor accurate., When the third student ‘C’ repeats these, measurements and reports 2.01 g and 1.99 g, as the result, these values are both precise and, accurate. This can be more clearly understood, from the data given in Table 1.4., Table 1.4 Data to Illustrate Precision, and Accuracy, Measurements/g, 1, , 2, , Average (g), , Student A, , 1.95, , 1.93, , 1.940, , Student B, , 1.94, , 2.05, , 1.995, , Student C, , 2.01, , 1.99, , 2.000, , Addition and Subtraction of, Significant Figures, The result cannot have more digits to the right, of the decimal point than either of the original, numbers., 12.11, 18.0, 1.012, 31.122, , Here, 18.0 has only one digit after the decimal, point and the result should be reported only, up to one digit after the decimal point, which, is 31.1., Multiplication and Division of, Significant Figures, In these operations, the result must be, reported with no more significant figures as, in the measurement with the few significant, figures., 2.5×1.25 = 3.125, Since 2.5 has two significant figures, the, result should not have more than two, significant figures, thus, it is 3.1., While limiting the result to the required, number of significant figures as done in the, above mathematical operation, one has to, keep in mind the following points for, rounding off the numbers, 1. If the rightmost digit to be removed is more, than 5, the preceding number is increased, by one. For example, 1.386. If we have to, remove 6, we have to round it to 1.39., 2. If the rightmost digit to be removed is less, than 5, the preceding number is not, changed. For example, 4.334 if 4 is to be, removed, then the result is rounded upto, 4.33., 3. If the rightmost digit to be removed is 5,, then the preceding number is not changed, if it is an even number but it is increased, by one if it is an odd number. For example,, if 6.35 is to be rounded by removing 5,, we have to increase 3 to 4 giving 6.4 as, the result. However, if 6.25 is to be, rounded off it is rounded off to 6.2., 1.4.3 Dimensional Analysis, Often while calculating, there is a need to, convert units from one system to the other. The, method used to accomplish this is called factor, label method or unit factor method or, dimensional analysis. This is illustrated, below., Example, A piece of metal is 3 inch (represented by in), long. What is its length in cm?, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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14, , CHEMISTRY, , Solution, , The above is multiplied by the unit factor, , We know that 1 in = 2.54 cm, , 2 × 1000 cm 3 ×, , From this equivalence, we can write, , 1 in, 2.54 cm, =1=, 2.54 cm, 1 in, 1 in, 2.54 cm, Thus, 2.54 cm equals 1 and, 1 in, , Example, How many seconds are there in 2 days?, Solution, Here, we know 1 day = 24 hours (h), , also equals 1. Both of these are called unit, factors. If some number is multiplied by these, unit factors (i.e., 1), it will not be affected, otherwise., , or, , Now, the unit factor by which multiplication, is to be done is that unit factor (, , 1h, 60 min, or 60 min = 1 = 1 h, , It should also be noted in the above, example that units can be handled just like, other numerical part. It can be cancelled,, divided, multiplied, squared, etc. Let us study, one more example., Example, A jug contains 2L of milk. Calculate the volume, of the milk in m3., Solution, Since 1 L = 1000 cm3, and 1m = 100 cm, which gives, , 1m, 100 cm, =1=, 100 cm, 1m, To get m3 from the above unit factors, the, first unit factor is taken and it is cubed., 3, ,  1m , 1m3, 3, ⇒, = (1) = 1, 6, 3,  100 cm , 10 cm, Now 2 L = 2×1000 cm3, , so, for converting 2 days to seconds,, i.e., 2 days – – – – – – = – – – seconds, The unit factors can be multiplied in, series in one step only as follows:, , 2.54 cm, in, 1 in, , the above case) which gives the desired units, i.e., the numerator should have that part which, is required in the desired result., , 1 day, 24 h, =1=, 24 h, 1 day, , then, 1h = 60 min, , Say, the 3 in given above is multiplied by, the unit factor. So,, , 2.54 cm, 3 in = 3 in × 1 in, = 3 × 2.54 cm = 7.62 cm, , 1 m3, 2 m3, =, = 2 × 10 −3 m 3, 106 cm 3, 103, , 2 day ×, , 24 h 60 min, 60 s, ×, ×, 1 day, 1h, 1 min, , = 2 × 24 × 60 × 60 s, = 172800 s, 1.5 LAWS OF CHEMICAL, COMBINATIONS, The combination of elements, to form compounds is, governed by the following five, basic laws., , Antoine Lavoisier, (1743–1794), , 1.5.1 Law of Conservation of Mass, This law was put forth by Antoine Lavoisier, in 1789. He performed careful experimental, studies for combustion reactions and reached, to the conclusion that in all physical and, chemical changes, there is no net change in, mass duting the process. Hence, he reached, to the conclusion that matter can neither be, created nor destroyed. This is called ‘Law of, Conservation of Mass’. This law formed the, basis for several later developments in, chemistry. Infact, this was the result of exact, measurement of masses of reactants and, products, and carefully planned experiments, performed by Lavoisier., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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SOME BASIC CONCEPTS OF CHEMISTRY, , 15, , 1.5.2 Law of Definite Proportions, This law was given by, a, French chemist, Joseph, Proust. He stated that a given, compound always contains, exactly the same proportion, of elements by weight., Proust worked with two, samples of cupric carbonate Joseph Proust, (1754–1826), — one of which was of natural, origin and the other was synthetic. He found, that the composition of elements present in it, was same for both the samples as shown below:, % of, % of, % of, copper carbon oxygen, Natural Sample, , 51.35, , 9.74, , 38.91, , Synthetic Sample, , 51.35, , 9.74, , 38.91, , Thus, he concluded that irrespective of the, source, a given compound always contains, same elements combined together in the same, proportion by mass. The validity of this law, has been confirmed by various experiments., It is sometimes also referred to as Law of, Definite Composition., 1.5.3 Law of Multiple Proportions, This law was proposed by Dalton in 1803., According to this law, if two elements can, combine to form more than one compound, the, masses of one element that combine with a, fixed mass of the other element, are in the, ratio of small whole numbers., For example, hydrogen combines with, oxygen to form two compounds, namely, water, and hydrogen peroxide., Hydrogen + Oxygen → Water, 2g, 16g, 18g, Hydrogen + Oxygen → Hydrogen Peroxide, 2g, 32g, 34g, Here, the masses of oxygen (i.e., 16 g and 32 g),, which combine with a fixed mass of hydrogen, (2g) bear a simple ratio, i.e., 16:32 or 1: 2., 1.5.4 Gay Lussac’s Law of Gaseous, Volumes, This law was given by Gay Lussac in 1808. He, observed that when gases combine or are, , produced in a chemical, reaction they do so in a, simple ratio by volume,, provided all gases are at, the same temperature and, pressure., Thus, 100 mL of hydrogen, Joseph Louis, combine with 50 mL of oxygen, Gay Lussac, to give 100 mL of water, vapour., Hydrogen + Oxygen → Water, 100 mL, 50 mL, 100 mL, Thus, the volumes of hydrogen and oxygen, which combine (i.e., 100 mL and, 50 mL) bear a simple ratio of 2:1., Gay Lussac’s discovery of integer ratio in, volume relationship is actually the law of, definite proportions by volume. The law of, definite proportions, stated earlier, was with, respect to mass. The Gay Lussac’s law was, explained properly by the work of Avogadro, in 1811., 1.5.5 Avogadro’s Law, In 1811, Avogadro proposed that equal, volumes of all gases at the same temperature, and pressure should contain equal number, of molecules. Avogadro made a distinction, between atoms and molecules which is quite, understandable in present times. If we, consider again the reaction of hydrogen and, oxygen to produce water, we see that two, volumes of hydrogen combine with one volume, of oxygen to give two volumes of water without, leaving any unreacted oxygen., Note that in the Fig. 1.9 (Page 16) each box, contains equal number of, molecules. In fact, Avogadro, could explain the above result, by considering the molecules, to be polyatomic. If, hydrogen and oxygen were, considered as diatomic as, recognised now, then the Lorenzo Romano, above results are easily, Amedeo Carlo, Avogadro di, understandable. However,, Quareqa edi, Dalton and others believed at, Carreto, that time that atoms of the, (1776–1856), , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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16, , CHEMISTRY, , Fig. 1.9 Two volumes of hydrogen react with one volume of oxygen to give two volumes of water vapour, , same kind cannot combine and molecules of, oxygen or hydrogen containing two atoms did, not exist. Avogadro’s proposal was published, in the French Journal de Physique. In spite of, being correct, it did not gain much support., After about 50 years, in 1860, the first, international conference on chemistry was held, in Karlsruhe, Germany, to resolve various, ideas. At the meeting, Stanislao Cannizaro, presented a sketch of a course of chemical, philosophy, which emphasised on the, importance of Avogadro’s work., , Dalton’s theory could explain the laws of, chemical combination. However, it could not, explain the laws of gaseous volumes. It could, not provide the reason for combining of, atoms, which was answered later by other, scientists., , 1.6 DALTON’S ATOMIC THEORY, Although the origin of the idea that matter is, composed of small indivisible particles called ‘atomio’ (meaning, indivisible), dates back to the, time of Democritus, a Greek, Philosopher (460–370 BC), it, again started emerging as a, result of several experimental, studies which led to the laws, mentioned above., In, 1808,, Dalton, John Dalton, published ‘A New System of, (1776–1884), Chemical Philosophy’, in, which he proposed the following :, 1. Matter consists of indivisible atoms., 2. All atoms of a given element have identical, properties, including identical mass. Atoms, of different elements differ in mass., 3. Compounds are formed when atoms of, different elements combine in a fixed ratio., 4. Chemical reactions involve reorganisation, of atoms. These are neither created nor, destroyed in a chemical reaction., , 1.7.1 Atomic Mass, , 1.7, , ATOMIC AND MOLECULAR MASSES, , After having some idea about the terms atoms, and molecules, it is appropriate here to, understand what do we mean by atomic and, molecular masses., The atomic mass or the mass of an atom is, actually very-very small because atoms are, extremely small. Today, we have, s o p h i s t i c a t e d t e c h n i q u e s e . g . , mass, spectrometry for determining the atomic, masses fairly accurately. But in the, n i n e t e e n t h c e n t u r y , s c i e n t i s t s cou l d, determine the mass of one atom relative to, another by experimental means, as has been, mentioned earlier. Hydrogen, being the, lightest atom was arbitrarily assigned a mass, of 1 (without any units) and other elements, were assigned masses relative to it. However,, the present system of atomic masses is based, on carbon-12 as the standard and has been, agreed upon in 1961. Here, Carbon-12 is one, of the isotopes of carbon and can be, represented as 12C. In this system, 12C is, assigned a mass of exactly 12 atomic mass, unit (amu) and masses of all other atoms are, given relative to this standard. One atomic, mass unit is defined as a mass exactly equal, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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SOME BASIC CONCEPTS OF CHEMISTRY, , 17, , to one-twelfth of the mass of one carbon - 12, atom., And 1 amu = 1.66056×10–24 g, Mass of an atom of hydrogen, = 1.6736×10–24 g, Thus, in terms of amu, the mass, of hydrogen atom =, = 1.0078 amu, = 1.0080 amu, , 1.7.3 Molecular Mass, Molecular mass is the sum of atomic masses, of the elements present in a molecule. It is, obtained by multiplying the atomic mass of, each element by the number of its atoms and, adding them together. For example, molecular, mass of methane, which contains one carbon, atom and four hydrogen atoms, can be, obtained as follows:, Molecular mass of methane,, , Similarly, the mass of oxygen - 16 (16O), atom would be 15.995 amu., At present, ‘amu’ has been replaced by ‘u’,, which is known as unified mass., When we use atomic masses of elements, in calculations, we actually use average, atomic masses of elements, which are, explained below., , (CH4) = (12.011 u) + 4 (1.008 u), = 16.043 u, Similarly, molecular mass of water (H2O), = 2 × atomic mass of hydrogen + 1 × atomic, mass of oxygen, = 2 (1.008 u) + 16.00 u, = 18.02 u, , 1.7.2 Average Atomic Mass, , 1.7.4 Formula Mass, , Many naturally occurring elements exist as, more than one isotope. When we take into, account the existence of these isotopes and, their relative abundance (per cent occurrence),, the average atomic mass of that element can, be computed. For example, carbon has the, following three isotopes with relative, abundances and masses as shown against, , Some substances, such as sodium chloride,, do not contain discrete molecules as their, constituent units. In such compounds, positive, (sodium ion) and negative (chloride ion) entities, are arranged in a three-dimensional structure,, as shown in Fig. 1.10., , Isotope, , Relative, Abundance, (%), , Atomic, Mass (amu), , 98.892, , 12, , 12, , C, , 13, , C, , 1.108, , 13.00335, , 14, , C, , 2 ×10–10, , 14.00317, , each of them., From the above data, the average atomic, mass of carbon will come out to be:, (0.98892) (12 u) + (0.01108) (13.00335 u) +, (2 × 10–12) (14.00317 u) = 12.011 u, Similarly, average atomic masses for other, elements can be calculated. In the periodic, table of elements, the atomic masses, mentioned for different elements actually, represent their average atomic masses., , Fig. 1.10 Packing of Na+ and Cl– ions, in sodium chloride, , It may be noted that in sodium chloride,, one Na+ ion is surrounded by six Cl– ion and, vice-versa., The formula, such as NaCl, is used to, calculate the formula mass instead of molecular, mass as in the solid state sodium chloride does, not exist as a single entity., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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18, , CHEMISTRY, , Thus, the formula mass of sodium chloride is, atomic mass of sodium + atomic mass of chlorine, = 23.0 u + 35.5 u = 58.5 u, Problem 1.1, Calculate the molecular mass of glucose, (C6H12O6) molecule., Solution, Molecular mass of glucose (C6H12O6), = 6(12.011 u) + 12(1.008 u) +, 6(16.00 u), = (72.066 u) + (12.096 u) +, (96.00 u), = 180.162 u, 1.8, , MOLE CONCEPT AND MOLAR, MASSES, Atoms and molecules are extremely small in, size and their numbers in even a small amount, of any substance is really very large. To handle, such large numbers, a unit of convenient, magnitude is required., Just as we denote one dozen for 12 items,, score for 20 items, gross for 144 items, we, use the idea of mole to count entities at the, microscopic level (i.e., atoms, molecules,, particles, electrons, ions, etc)., In SI system, mole (symbol, mol) was, introduced as seventh base quantity for the, amount of a substance., One mole is the amount of a substance, that contains as many particles or entities, as there are atoms in exactly 12 g (or 0.012, kg) of the 12C isotope. It may be emphasised, that the mole of a substance always, contains the same number of entities, no matter, what the substance may be. In order to, determine this number precisely, the mass of, a carbon–12 atom was determined by a mass, spectrometer and found to be equal to, 1.992648 × 10–23 g. Knowing that one mole of, carbon weighs 12 g, the number of atoms in it, is equal to:, , This number of entities in 1 mol is so, important that it is given a separate name and, symbol. It is known as ‘Avogadro constant’,, or Avogadro number denoted by NA in honour, of Amedeo Avogadro. To appreciate the, largeness of this number, let us write it with, all zeroes without using any powers of ten., 602213670000000000000000, Hence, so many entities (atoms, molecules or, any other particle) constitute one mole of a, particular substance., We can, therefore, say that 1 mol of hydrogen, atoms = 6.022×1023 atoms, 1 mol of water molecules = 6.022×1023 water, molecules, 1 mol of sodium chloride = 6.022 × 1023, formula units of sodium chloride, Having defined the mole, it is easier to know, the mass of one mole of a substance or the, constituent entities. The mass of one mole, of a substance in grams is called its, molar mass. The molar mass in grams is, numerically equal to atomic/molecular/, formula mass in u., Molar mass of water = 18.02 g mol-1, Molar mass of sodium chloride = 58.5 g mol-1, 1.9, , PERCENTAGE COMPOSITION, , So far, we were dealing with the number of, entities present in a given sample. But many a, time, information regarding the percentage of, a particular element present in a compound is, required. Suppose, an unknown or new, compound is given to you, the first question, , 12 g / mol 12 C, 1.992648 × 10 −23 g /12 C atom, , = 6.0221367 × 1023 atoms/mol, , Fig. 1.11 One mole of various substances, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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SOME BASIC CONCEPTS OF CHEMISTRY, , 19, , you would ask is: what is its formula or what, are its constituents and in what ratio are they, present in the given compound? For known, compounds also, such information provides a, check whether the given sample contains the, same percentage of elements as present in a, pure sample. In other words, one can check, the purity of a given sample by analysing this, data., Let us understand it by taking the example, of water (H2O). Since water contains hydrogen, and oxygen, the percentage composition of, both these elements can be calculated as, follows:, Mass % of an element =, , information can be obtained from the, per cent composition data., 1.9.1 Empirical Formula for Molecular, Formula, An empirical formula represents the simplest, whole number ratio of various atoms present in, a compound, whereas, the molecular formula, shows the exact number of different types of, atoms present in a molecule of a compound., If the mass per cent of various elements, present in a compound is known, its empirical, formula can be determined. Molecular formula, can further be obtained if the molar mass is, known. The following example illustrates this, sequence., , mass of that element in the compound × 100, molar mass of the compound, , Molar mass of water = 18.02 g, Mass % of hydrogen =, = 11.18, , 16.00, × 100, 18.02, = 88.79, Let us take one more example. What is the, percentage of carbon, hydrogen and oxygen, in ethanol?, Molecular formula of ethanol is: C2H5OH, Molar mass of ethanol is:, (2×12.01 + 6×1.008 + 16.00) g = 46.068 g, Mass per cent of carbon, Mass % of oxygen, , =, , 24.02 g, = 46.068 g × 100 = 52.14%, Mass per cent of hydrogen, , 6.048 g, = 46.068 g × 100 = 13.13%, , Problem 1.2, A compound contains 4.07% hydrogen,, 24.27% carbon and 71.65% chlorine. Its, molar mass is 98.96 g. What are its, empirical and molecular formulas?, Solution, Step 1. Conversion of mass per cent, to grams, Since we are having mass per cent, it is, convenient to use 100 g of the compound, as the starting material. Thus, in the, 100 g sample of the above compound,, 4.07g hydrogen, 24.27g carbon and, 71.65g chlorine are present., Step 2. Convert into number moles of, each element, Divide the masses obtained above by, respective atomic masses of various, elements. This gives the number of moles, of constituent elements in the compound, Moles of hydrogen =, , Mass per cent of oxygen, , 16.00 g, = 46.068 g × 100 = 34.73%, After understanding the calculation of, per cent of mass, let us now see what, , Moles of carbon =, , 2019-20, , 24.27 g, = 2.021, 12.01 g, , Moles of chlorine =, , Download all NCERT books PDFs from www.ncert.online, , 4.07 g, 1.008 g = 4.04, , 71.65 g, = 2.021, 35.453 g

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20, , CHEMISTRY, , available from the balanced chemical equation, of a given reaction. Let us consider the, combustion of methane. A balanced equation, for this reaction is as given below:, CH4 (g) + 2O2 (g) → CO2 (g) + 2 H2O (g), Here, methane and dioxygen are called, reactants and carbon dioxide and water are, called products. Note that all the reactants and, the products are gases in the above reaction, and this has been indicated by letter (g) in the, brackets next to its formula. Similarly, in case, of solids and liquids, (s) and (l) are written, respectively., The coefficients 2 for O2 and H2O are called, stoichiometric coefficients. Similarly the, coefficient for CH4 and CO2 is one in each case., They represent the number of molecules (and, moles as well) taking part in the reaction or, formed in the reaction., Thus, according to the above chemical, reaction,, , Step 3. Divide each of the mole values, obtained above by the smallest number, amongst them, Since 2.021 is smallest value, division by, it gives a ratio of 2:1:1 for H:C:Cl ., In case the ratios are not whole numbers, then, they may be converted into whole number by, multiplying by the suitable coefficient., Step 4. Write down the empirical formula, by mentioning the numbers after writing, the symbols of respective elements, CH2Cl is, thus, the empirical formula of, the above compound., Step 5. Writing molecular formula, (a) Determine empirical formula mass by, adding the atomic masses of various, atoms present in the empirical formula., For CH2Cl, empirical formula mass is, 12.01 + (2 × 1.008) + 35.453, = 49.48 g, (b) Divide Molar mass by empirical, formula mass, , = 2 = (n), (c) Multiply empirical formula by n, obtained above to get the molecular, formula, Empirical formula = CH2Cl, n = 2. Hence, molecular formula is C2H4Cl2., , •, , One mole of CH4(g) reacts with two moles, of O2(g) to give one mole of CO2(g) and, two moles of H2O(g), , •, , One molecule of CH 4(g) reacts with, 2 molecules of O2(g) to give one molecule, of CO2(g) and 2 molecules of H2O(g), , •, , 22.7 L of CH4(g) reacts with 45.4 L of O2 (g), to give 22.7 L of CO2 (g) and 45.4 L of H2O(g), , 16 g of CH4 (g) reacts with 2×32 g of O2 (g) to, give 44 g of CO2 (g) and 2×18 g of H2O (g)., From these relationships, the given data can, be interconverted as follows:, •, , 1.10 STOICHIOMETRY AND, STOICHIOMETRIC CALCULATIONS, The word ‘stoichiometry’ is derived from two, Greek words — stoicheion (meaning, element), and, metron, (meaning,, measure)., Stoichiometry, thus, deals with the calculation, of masses (sometimes volumes also) of the, reactants and the products involved in a, chemical reaction. Before understanding how, to calculate the amounts of reactants required, or the products produced in a chemical, reaction, let us study what information is, , Mass, = Density, Volume, 1.10.1 Limiting Reagent, Many a time, reactions are carried out with the, amounts of reactants that are different than, the amounts as required by a balanced, chemical reaction. In such situations, one, reactant is in more amount than the amount, required by balanced chemical reaction. The, reactant which is present in the least amount, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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SOME BASIC CONCEPTS OF CHEMISTRY, , 21, , gets consumed after sometime and after that, further reaction does not take place whatever, be the amount of the other reactant. Hence,, the reactant, which gets consumed first, limits, the amount of product formed and is, therefore,, called the limiting reagent., In performing stoichiometric calculations,, this aspect is also to be kept in mind., , important to understand as how the amount, of substance is expressed when it is present in, the solution. The concentration of a solution, or the amount of substance present in its, given volume can be expressed in any of the, following ways., , 1.10.2, , 3. Molarity, , Reactions in Solutions, , A majority of reactions in the laboratories are, carried out in solutions. Therefore, it is, , 1. Mass per cent or weight per cent (w/w %), 2. Mole fraction, 4. Molality, Let us now study each one of them in detail., , Balancing a chemical equation, According to the law of conservation of mass, a balanced chemical equation has the same, number of atoms of each element on both sides of the equation. Many chemical equations can, be balanced by trial and error. Let us take the reactions of a few metals and non-metals with, oxygen to give oxides, (a) balanced equation, 4 Fe(s) + 3O2(g) → 2Fe2O3(s), (b) balanced equation, 2 Mg(s) + O2(g) → 2MgO(s), (c) unbalanced equation, P4(s) + O2 (g) → P4O10(s), Equations (a) and (b) are balanced, since there are same number of metal and oxygen atoms on, each side of the equations. However equation (c) is not balanced. In this equation, phosphorus, atoms are balanced but not the oxygen atoms. To balance it, we must place the coefficient 5 on, the left of oxygen on the left side of the equation to balance the oxygen atoms appearing on the, right side of the equation., balanced equation, P4(s) + 5O2(g) → P4O10(s), Now, let us take combustion of propane, C3H8. This equation can be balanced in steps., Step 1 Write down the correct formulas of reactants and products. Here, propane and oxygen, are reactants, and carbon dioxide and water are products., C3H8(g) + O2(g) → CO2 (g) + H2O(l) unbalanced equation, Step 2 Balance the number of C atoms: Since 3 carbon atoms are in the reactant, therefore,, three CO2 molecules are required on the right side., C3H8 (g) + O2 (g) → 3CO2 (g) + H2O (l), Step 3 Balance the number of H atoms: on the left there are 8 hydrogen atoms in the reactants, however, each molecule of water has two hydrogen atoms, so four molecules of water will be, required for eight hydrogen atoms on the right side., C3H8 (g) +O2 (g) → 3CO2 (g)+4H2O (l), Step 4 Balance the number of O atoms: There are 10 oxygen atoms on the right side (3 × 2 = 6 in, CO2 and 4 × 1= 4 in water). Therefore, five O 2 molecules are needed to supply the required 10, CO2 and 4 × 1= 4 in water). Therefore, five O 2 molecules are needed to supply the required 10, oxygen atoms., C3H8 (g) +5O2 (g) → 3CO2 (g) + 4H2O (l), Step 5 Verify that the number of atoms of each element is balanced in the final equation. The, equation shows three carbon atoms, eight hydrogen atoms, and 10 oxygen atoms on each side., All equations that have correct formulas for all reactants and products can be balanced. Always, remember that subscripts in formulas of reactants and products cannot be changed to balance, an equation., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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SOME BASIC CONCEPTS OF CHEMISTRY, , 23, , 3. Molarity, It is the most widely used unit and is denoted, by M. It is defined as the number of moles of, the solute in 1 litre of the solution. Thus,, , = 3.30×103×17 g NH3 (g), = 56.1×103 g NH3, = 56.1 kg NH3, 1. Mass per cent, It is obtained by using the following relation:, , Molarity (M) =, , No. of moles of solute, Volume of solution in litres, , Suppose, we have 1 M solution of a, substance, say NaOH, and we want to prepare, a 0.2 M solution from it., 1 M NaOH means 1 mol of NaOH present, in 1 litre of the solution. For 0.2 M solution,, we require 0.2 moles of NaOH dissolved in, 1 litre solution., Hence, for making 0.2M solution from 1M, solution, we have to take that volume of 1M NaOH, solution, which contains 0.2 mol of NaOH and, dilute the solution with water to 1 litre., Now, how much volume of concentrated, (1M) NaOH solution be taken, which contains, 0.2 moles of NaOH can be calculated as follows:, If 1 mol is present in 1L or 1000 mL, solution, then, 0.2 mol is present in, , Problem 1.6, A solution is prepared by adding 2 g of a, substance A to 18 g of water. Calculate, the mass per cent of the solute., Solution, , 2. Mole Fraction, It is the ratio of number of moles of a particular, component to the total number of moles of the, solution. If a substance ‘A’ dissolves in, substance ‘B’ and their number of moles are, nA and nB, respectively, then the mole fractions, of A and B are given as:, , 1000 mL, × 0.2 mol solution, 1 mol, = 200 mL solution, Thus, 200 mL of 1M NaOH are taken and, enough water is added to dilute it to make it 1 litre., In fact for such calculations, a general, formula, M1 × V1 = M2 × V2 where M and V are, molarity and volume, respectively, can be used., In this case, M1 is equal to 0.2M; V1 = 1000 mL, and, M 2 = 1.0M; V 2 is to be calculated., Substituting the values in the formula:, 0.2 M × 1000 mL = 1.0 M × V2, , Note that the number of moles of solute, (NaOH) was 0.2 in 200 mL and it has remained, the same, i.e., 0.2 even after dilution ( in 1000, mL) as we have changed just the amount of, solvent (i.e., water) and have not done anything, with respect to NaOH. But keep in mind the, concentration., Download all NCERT books PDFs from www.ncert.online, , 2019-20

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24, , CHEMISTRY, , Problem 1.7, Calculate the molarity of NaOH in the, solution prepared by dissolving its 4 g in, enough water to form 250 mL of the solution., , Problem 1.8, The density of 3 M solution of NaCl is, –1, 1.25 g mL . Calculate the molality of the, solution., , Solution, Since molarity (M), , Solution, M = 3 mol L–1, Mass of NaCl, in 1 L solution = 3 × 58.5 = 175.5 g, Mass of, 1L solution = 1000 × 1.25 = 1250 g, –1, (since density = 1.25 g mL ), Mass of water in solution = 1250 –75.5, = 1074.5 g, Molality =, , =, , Note that molarity of a solution depends, upon temperature because volume of a, solution is temperature dependent., 4. Molality, It is defined as the number of moles of solute, present in 1 kg of solvent. It is denoted by m., Thus, Molality (m) =, , No. of moles of solute, Mass of solvent in kg, , No. of moles of solute, Mass of solvent in kg, 3 mol, 1.0745 kg = 2.79 m, , Often in a chemistry laboratory, a solution, of a desired concentration is prepared by, diluting a solution of known higher, concentration. The solution of higher, concentration is also known as stock, solution. Note that the molality of a solution, does not change with temperature since, mass remains unaffected with temperature., , SUMMARY, Chemistry, as we understand it today is not a very old discipline. People in ancient, India, already had the knowledge of many scientific phenomenon much before the, advent of modern science. They applied the knowledge in various walks of life., The study of chemistry is very important as its domain encompasses every sphere, of life. Chemists study the properties and structure of substances and the changes, undergone by them. All substances contain matter, which can exist in three states, – solid, liquid or gas. The constituent particles are held in different ways in these, states of matter and they exhibit their characteristic properties. Matter can also be, classified into elements, compounds or mixtures. An element contains particles of, only one type, which may be atoms or molecules. The compounds are formed where, atoms of two or more elements combine in a fixed ratio to each other. Mixtures occur, widely and many of the substances present around us are mixtures., When the properties of a substance are studied, measurement is inherent. The, quantification of properties requires a system of measurement and units in which, the quantities are to be expressed. Many systems of measurement exist, of which, Download all NCERT books PDFs from www.ncert.online, , 2019-20

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SOME BASIC CONCEPTS OF CHEMISTRY, , 25, , the English and the Metric Systems are widely used. The scientific community, however,, has agreed to have a uniform and common system throughout the world, which is, abbreviated as SI units (International System of Units)., Since measurements involve recording of data, which are always associated with a, certain amount of uncertainty, the proper handling of data obtained by measuring the, quantities is very important. The measurements of quantities in chemistry are spread, over a wide range of 10–31 to 10+23. Hence, a convenient system of expressing the numbers, in scientific notation is used. The uncertainty is taken care of by specifying the number, of significant figures, in which the observations are reported. The dimensional analysis, helps to express the measured quantities in different systems of units. Hence, it is possible, to interconvert the results from one system of units to another., The combination of different atoms is governed by basic laws of chemical combination, — these being the Law of Conservation of Mass, Law of Definite Proportions, Law of, Multiple Proportions, Gay Lussac’s Law of Gaseous Volumes and Avogadro Law. All, these laws led to the Dalton’s atomic theory, which states that atoms are building, blocks of matter. The atomic mass of an element is expressed relative to 12C isotope of, carbon, which has an exact value of 12u. Usually, the atomic mass used for an element is, the average atomic mass obtained by taking into account the natural abundance of, different isotopes of that element. The molecular mass of a molecule is obtained by, taking sum of the atomic masses of different atoms present in a molecule. The molecular, formula can be calculated by determining the mass per cent of different elements present, in a compound and its molecular mass., The number of atoms, molecules or any other particles present in a given system are, expressed in the terms of Avogadro constant (6.022 × 1023). This is known as 1 mol of, the respective particles or entities., Chemical reactions represent the chemical changes undergone by different elements, and compounds. A balanced chemical equation provides a lot of information. The, coefficients indicate the molar ratios and the respective number of particles taking part, in a particular reaction. The quantitative study of the reactants required or the products, formed is called stoichiometry. Using stoichiometric calculations, the amount of one or, more reactant(s) required to produce a particular amount of product can be determined, and vice-versa. The amount of substance present in a given volume of a solution is, expressed in number of ways, e.g., mass per cent, mole fraction, molarity and molality., , EXERCISES, 1.1, , Calculate the molar mass of the following:, (i) H2O (ii) CO2 (iii) CH4, , 1.2, , Calculate the mass per cent of different elements present in sodium sulphate, (Na2SO4)., , 1.3, , Determine the empirical formula of an oxide of iron, which has 69.9% iron and, 30.1% dioxygen by mass., , 1.4, , Calculate the amount of carbon dioxide that could be produced when, , 1.5, , (i), , 1 mole of carbon is burnt in air., , (ii), , 1 mole of carbon is burnt in 16 g of dioxygen., , (iii), , 2 moles of carbon are burnt in 16 g of dioxygen., , Calculate the mass of sodium acetate (CH3COONa) required to make 500 mL of, 0.375 molar aqueous solution. Molar mass of sodium acetate is 82.0245 g mol–1., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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26, , CHEMISTRY, , 1.6, , Calculate the concentration of nitric acid in moles per litre in a sample which, has a density, 1.41 g mL–1 and the mass per cent of nitric acid in it being 69 %., , 1.7, , How much copper can be obtained from 100 g of copper sulphate (CuSO4)?, , 1.8, , Determine the molecular formula of an oxide of iron, in which the mass per cent, of iron and oxygen are 69.9 and 30.1, respectively., , 1.9, , Calculate the atomic mass (average) of chlorine using the following data:, % Natural Abundance, , Molar Mass, , 35, , 75.77, , 34.9689, , 37, , 24.23, , 36.9659, , Cl, Cl, , 1.10, , In three moles of ethane (C2H6), calculate the following:, (i), , Number of moles of carbon atoms., , (ii), , Number of moles of hydrogen atoms., , (iii), , Number of molecules of ethane., , 1.11, , What is the concentration of sugar (C12H22O11) in mol L–1 if its 20 g are dissolved in, enough water to make a final volume up to 2L?, , 1.12, , If the density of methanol is 0.793 kg L–1, what is its volume needed for making, 2.5 L of its 0.25 M solution?, , 1.13, , Pressure is determined as force per unit area of the surface. The SI unit of, pressure, pascal is as shown below:, 1Pa = 1N m–2, If mass of air at sea level is 1034 g cm–2, calculate the pressure in pascal., , 1.14, , What is the SI unit of mass? How is it defined?, , 1.15, , Match the following prefixes with their multiples:, Prefixes, , Multiples, , (i), , micro, , 106, , (ii), , deca, , 109, , (iii) mega, , 10–6, , (iv) giga, , 10–15, , (v), , 10, , femto, , 1.16, , What do you mean by significant figures?, , 1.17, , A sample of drinking water was found to be severely contaminated with chloroform,, CHCl3, supposed to be carcinogenic in nature. The level of contamination was 15, ppm (by mass)., , 1.18, , 1.19, , (i), , Express this in per cent by mass., , (ii), , Determine the molality of chloroform in the water sample., , Express the following in the scientific notation:, (i), , 0.0048, , (ii), , 234,000, , (iii), , 8008, , (iv), , 500.0, , (v), , 6.0012, , How many significant figures are present in the following?, (i), , 0.0025, , (ii), , 208, , (iii), , 5005, Download all NCERT books PDFs from www.ncert.online, , 2019-20

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SOME BASIC CONCEPTS OF CHEMISTRY, , 1.20, , 1.21, , (iv), , 126,000, , (v), , 500.0, , (vi), , 2.0034, , 27, , Round up the following upto three significant figures:, (i), , 34.216, , (ii), , 10.4107, , (iii), , 0.04597, , (iv), , 2808, , The following data are obtained when dinitrogen and dioxygen react together to, form different compounds:, Mass of dinitrogen, , Mass of dioxygen, , (i), , 14 g, , 16 g, , (ii), , 14 g, , 32 g, , (iii), , 28 g, , 32 g, , (iv), , 28 g, , 80 g, , (a), , Which law of chemical combination is obeyed by the above experimental data?, Give its statement., , (b), , Fill in the blanks in the following conversions:, (i), , 1 km = ...................... mm = ...................... pm, , (ii), , 1 mg = ...................... kg = ...................... ng, , (iii), , 1 mL = ...................... L = ...................... dm3, , 1.22, , If the speed of light is 3.0 × 108 m s–1, calculate the distance covered by light in, 2.00 ns., , 1.23, , In a reaction, A + B2, , , , AB2, , Identify the limiting reagent, if any, in the following reaction mixtures., , 1.24, , (i), , 300 atoms of A + 200 molecules of B, , (ii), , 2 mol A + 3 mol B, , (iii), , 100 atoms of A + 100 molecules of B, , (iv), , 5 mol A + 2.5 mol B, , (v), , 2.5 mol A + 5 mol B, , Dinitrogen and dihydrogen react with each other to produce ammonia according, to the following chemical equation:, N2 (g) + H2 (g), , , , 2NH3 (g), , (i), , Calculate the mass of ammonia produced if 2.00 × 103 g dinitrogen reacts, with 1.00 ×103 g of dihydrogen., , (ii), , Will any of the two reactants remain unreacted?, , (iii), , If yes, which one and what would be its mass?, , 1.25, , How are 0.50 mol Na2CO3 and 0.50 M Na2CO3 different?, , 1.26, , If 10 volumes of dihydrogen gas reacts with five volumes of dioxygen gas, how, many volumes of water vapour would be produced?, , 1.27, , Convert the following into basic units:, (i), , 28.7 pm, , (ii), , 15.15 pm, , (iii), , 25365 mg, Download all NCERT books PDFs from www.ncert.online, , 2019-20

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28, , CHEMISTRY, , 1.28, , Which one of the following will have the largest number of atoms?, (i), , 1 g Au (s), , (ii), , 1 g Na (s), , (iii), , 1 g Li (s), , (iv), , 1 g of Cl2(g), , 1.29, , Calculate the molarity of a solution of ethanol in water, in which the mole fraction of, ethanol is 0.040 (assume the density of water to be one)., , 1.30, , What will be the mass of one, , 1.31, , How many significant figures should be present in the answer of the following, calculations?, , 12, , C atom in g?, , 0.02856 × 298.15 × 0.112, 0.5785, (iii), 0.0125 + 0.7864 + 0.0215, , (i), , 1.32, , Use the data given in the following table to calculate the molar mass of naturally, occuring argon isotopes:, Isotope, , Isotopic molar mass, , 0.337%, , –1, , 0.063%, , 35.96755 g mol, , 38, , 37.96272 g mol, , 40, , –1, , Ar, Ar, , Abundance, , –1, , 36, , Ar, , 1.33, , (ii) 5 × 5.364, , 39.9624 g mol, , 99.600%, , Calculate the number of atoms in each of the following (i) 52 moles of Ar (ii), 52 u of He (iii) 52 g of He., , 1.34 A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it, in oxygen gives 3.38 g carbon dioxide, 0.690 g of water and no other products. A, volume of 10.0 L (measured at STP) of this welding gas is found to weigh 11.6 g., Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular, formula., 1.35 Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the, reaction, CaCO3 (s) + 2 HCl (aq) → CaCl2 (aq) + CO2(g) + H2O(l), What mass of CaCO3 is required to react completely with 25 mL of 0.75 M HCl?, 1.36 Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2) with, aqueous hydrochloric acid according to the reaction, 4 HCl (aq) + MnO2(s) → 2H2O (l) + MnCl2(aq) + Cl2 (g), How many grams of HCl react with 5.0 g of manganese dioxide?, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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29, , STRUCTURE OF ATOM, , UNIT 2, , STRUCTURE OF ATOM, , The rich diversity of chemical behaviour of different elements, can be traced to the differences in the internal structure of, atoms of these elements., , Objectives, After studying this unit you will be, able to, •, , know about the discovery of, electron, proton and neutron and, their characteristics;, , •, , describe Thomson, Rutherford, and Bohr atomic models;, , •, , understand the important, featur es of the quantum, mechanical model of atom;, , •, , understand, natur e, of, electromagnetic radiation and, Planck’s quantum theory;, , •, , explain the photoelectric effect, and describe features of atomic, spectra;, , •, , state the de Broglie relation and, Heisenberg uncertainty principle;, , •, , define an atomic orbital in terms, of quantum numbers;, , •, , state aufbau principle, Pauli, exclusion principle and Hund’s, rule of maximum multiplicity; and, , •, , write the electronic configurations, of atoms., , The existence of atoms has been proposed since the time, of early Indian and Greek philosophers (400 B.C.) who, were of the view that atoms are the fundamental building, blocks of matter. According to them, the continued, subdivisions of matter would ultimately yield atoms which, would not be further divisible. The word ‘atom’ has been, derived from the Greek word ‘a-tomio’ which means, ‘uncut-able’ or ‘non-divisible’. These earlier ideas were, mere speculations and there was no way to test them, experimentally. These ideas remained dormant for a very, long time and were revived again by scientists in the, nineteenth century., The atomic theory of matter was first proposed on a, firm scientific basis by John Dalton, a British school, teacher in 1808. His theory, called Dalton’s atomic, theory, regarded the atom as the ultimate particle of, matter (Unit 1). Dalton’s atomic theory was able to explain, the law of conservation of mass, law of constant, composition and law of multiple proportion very, successfully. However, it failed to explain the results of, many experiments, for example, it was known that, substances like glass or ebonite when rubbed with silk, or fur get electrically charged., In this unit we start with the experimental, observations made by scientists towards the end of, nineteenth and beginning of twentieth century. These, established that atoms are made of sub-atomic particles,, i.e., electrons, protons and neutrons — a concept very, different from that of Dalton., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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30, , CHEMISTRY, , 2.1 DISCOVERY OF SUB-ATOMIC, PARTICLES, An insight into the structure of atom was, obtained from the experiments on electrical, discharge through gases. Before we discuss, these results we need to keep in mind a basic, rule regarding the behaviour of charged, particles : “Like charges repel each other and, unlike charges attract each other”., , Fig. 2.1(a) A cathode ray discharge tube, , 2.1.1 Discovery of Electron, In 1830, Michael Faraday showed that if, electricity is passed through a solution of an, electrolyte, chemical reactions occurred at, the electrodes, which resulted in the, liberation and deposition of matter at the, electrodes. He formulated certain laws which, you will study in class XII. These results, s ugge s t ed the p articu late na t u r e o f, electricity., In mid 1850s many scientists mainly, Faraday began to study electrical discharge, in partially evacuated tubes, known as, cathode ray discharge tubes. It is depicted, in Fig. 2.1. A cathode ray tube is made of, glass containing two thin pieces of metal,, called electrodes, sealed in it. The electrical, discharge through the gases could be, observed only at very low pressures and at, very high voltages. The pressure of different, gases could be adjusted by evacuation of the, glass tubes. When sufficiently high voltage, is applied across the electrodes, current, starts flowing through a stream of particles, moving in the tube from the negative electrode, (cathode) to the positive electrode (anode)., These were called cathode rays or cathode, ray particles. The flow of current from, cathode to anode was further checked by, making a hole in the anode and coating the, tube behind anode with phosphorescent, material zinc sulphide. When these rays, after, passing through anode, strike the zinc, sulphide coating, a bright spot is developed, on the coating [Fig. 2.1(b)]., , Fig. 2.1(b), , A cathode ray discharge tube with, perforated anode, , The results of these experiments are, summarised below., (i), , The cathode rays start from cathode and, move towards the anode., (ii) These rays themselves are not visible but, their behaviour can be observed with the, help of certain kind of materials, (fluorescent or phosphorescent) which, glow when hit by them. Television picture, tubes are cathode ray tubes and, television pictures result due to, fluorescence on the television screen, coated with certain fluorescent or, phosphorescent materials., (iii) In the absence of electrical or magnetic, field, these rays travel in straight lines, (Fig. 2.2)., (iv) In the presence of electrical or magnetic, field, the behaviour of cathode rays are, similar to that expected from negatively, charged particles, suggesting that the, cathode rays consist of negatively, charged particles, called electrons., (v) The characteristics of cathode rays, (electrons) do not depend upon the, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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31, , STRUCTURE OF ATOM, , material of electrodes and the nature of, the gas present in the cathode ray tube., Thus, we can conclude that electrons are, basic constituent of all the atoms., 2.1.2 Charge to Mass Ratio of Electron, In 1897, British physicist J.J. Thomson, measured the ratio of electrical charge (e) to, the mass of electron (me ) by using cathode ray, tube and applying electrical and magnetic field, perpendicular to each other as well as to the, path of electrons (Fig. 2.2). When only electric, field is applied, the electrons deviate from their, path and hit the cathode ray tube at point A, (Fig. 2.2). Similarly when only magnetic field, is applied, electron strikes the cathode ray tube, at point C. By carefully balancing the electrical, and magnetic field strength, it is possible to, bring back the electron to the path which is, followed in the absence of electric or magnetic, field and they hit the screen at point B., Thomson argued that the amount of deviation, of the particles from their path in the presence, of electrical or magnetic field depends upon:, (i), , (ii), , the magnitude of the negative charge on, the particle, greater the magnitude of the, charge on the particle, greater is the, interaction with the electric or magnetic, field and thus greater is the deflection., the mass of the particle — lighter the, particle, greater the deflection., , (iii) the strength of the electrical or magnetic, field — the deflection of electrons from its, original path increases with the increase, in the voltage across the electrodes, or the, strength of the magnetic field., By carrying out accurate measurements on, the amount of deflections observed by the, electrons on the electric field strength or, magnetic field strength, Thomson was able to, determine the value of e/me as:, , e, 11, –1, m e = 1.758820 × 10 C kg, , (2.1), , Where me is the mass of the electron in kg and, e is the magnitude of the charge on the electron, in coulomb (C). Since electrons are negatively, charged, the charge on electron is –e., 2.1.3 Charge on the Electron, R.A. Millikan (1868-1953) devised a method, known as oil drop experiment (1906-14), to, determine the charge on the electrons. He found, the charge on the electron to be, – 1.6 × 10–19 C. The present accepted value of, electrical charge is – 1.602176 × 10–19 C. The, mass of the electron (me) was determined by, combining these results with Thomson’s value, of e/me ratio., , = 9.1094×10–31 kg, , Fig. 2.2 The apparatus to determine the charge to the mass ratio of electron, Download all NCERT books PDFs from www.ncert.online, , 2019-20, , (2.2)

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32, , CHEMISTRY, , 2.1.4 Discovery of Protons and Neutrons, Electrical discharge carried out in the modified, cathode ray tube led to the discovery of canal, rays carrying positively charged particles. The, characteristics of these positively charged, particles are listed below., (i) Unlike cathode rays, mass of positively, charged particles depends upon the, nature of gas present in the cathode ray, tube. These are simply the positively, charged gaseous ions., (ii) The charge to mass ratio of the particles, depends on the gas from which these, originate., (iii) Some of the positively charged particles, carry a multiple of the fundamental unit, of electrical charge., (iv) The behaviour of these particles in the, magnetic or electrical field is opposite to, that observed for electron or cathode, rays., The smallest and lightest positive ion was, obtained from hydrogen and was called, proton. This positively charged particle was, characterised in 1919. Later, a need was felt, for the presence of electrically neutral particle, as one of the constituent of atom. These, particles were discovered by Chadwick (1932), by bombarding a thin sheet of beryllium by, α-particles. When electrically neutral particles, having a mass slightly greater than that of, protons were emitted. He named these, particles as neutrons. The important, properties of all these fundamental particles, are given in Table 2.1., 2.2 ATOMIC MODELS, Observations obtained from the experiments, mentioned in the previous sections have, suggested that Dalton’s indivisible atom is, composed of sub-atomic particles carrying, positive and negative charges. The major, problems before the scientists after the, discovery of sub-atomic particles were:, • to account for the stability of atom,, •, , to compare the behaviour of elements in, terms of both physical and chemical, properties,, , Millikan’s Oil Drop Method, In this method, oil droplets in the form of, mist, produced by the atomiser, were allowed, to enter through a tiny hole in the upper plate, of electrical condenser. The downward motion, of these droplets was viewed through the, telescope, equipped with a micrometer eye, piece. By measuring the rate of fall of these, droplets, Millikan was able to measure the, mass of oil droplets. The air inside the, chamber was ionized by passing a beam of, X-rays through it. The electrical charge on, these oil droplets was acquired by collisions, with gaseous ions. The fall of these charged, oil droplets can be retarded, accelerated or, made stationary depending upon the charge, on the droplets and the polarity and strength, of the voltage applied to the plate. By carefully, measuring the effects of electrical field, strength on the motion of oil droplets,, Millikan concluded that the magnitude of, electrical charge, q, on the droplets is always, an integral multiple of the electrical charge,, e, that is, q = n e, where n = 1, 2, 3... ., , Fig. 2.3 The Millikan oil drop apparatus for, measuring charge ‘e’. In chamber, the, forces acting on oil drop are:, gravitational, electrostatic due to, electrical field and a viscous drag force, when the oil drop is moving., , •, , •, , to explain the formation of different kinds, of molecules by the combination of different, atoms and,, to understand the origin and nature of the, characteristics of electromagnetic radiation, absorbed or emitted by atoms., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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33, , STRUCTURE OF ATOM, , Table 2.1 Properties of Fundamental Particles, Name, , Symbol, , Absolute, , Relative, , charge/C, , charge, , Mass/kg, , Mass/u, , Approx., mass/u, , Electron, , e, , –1.602176×10–19, , –1, , 9.109382×10–31, , 0.00054, , 0, , Proton, , p, , +1.602176×10–19, , +1, , 1.6726216×10–27, , 1.00727, , 1, , Neutron, , n, , 0, , 1.674927×10–27, , 1.00867, , 1, , 0, , Different atomic models were proposed to, explain the distributions of these charged, particles in an atom. Although some of these, models were not able to explain the stability of, atoms, two of these models, one proposed by, J.J. Thomson and the other proposed by, Ernest Rutherford are discussed below., 2.2.1 Thomson Model of Atom, J. J. Thomson, in 1898, proposed that an atom, possesses a spherical shape (radius, approximately 10–10 m) in which the positive, charge is uniformly distributed. The electrons, are embedded into it in such a manner as to, give the most stable electrostatic arrangement, (Fig. 2.4). Many different names are given to, this model, for example, plum pudding, raisin, pudding or watermelon. This model can be, , Fig.2.4 Thomson model of atom, , visualised as a pudding or watermelon of, positive charge with plums or seeds (electrons), embedded into it. An important feature of this, model is that the mass of the atom is assumed, to be uniformly distributed over the atom., Although this model was able to explain the, overall neutrality of the atom, but was not, consistent with the results of later experiments., Thomson was awarded Nobel Prize for physics, in 1906, for his theoretical and experimental, investigations on the conduction of electricity, by gases., , In the later half of the nineteenth century, different kinds of rays were discovered,, besides those mentioned earlier. Wilhalm, Röentgen (1845-1923) in 1895 showed, that when electrons strike a material in, the cathode ray tubes, produce rays, which can cause fluorescence in the, fluorescent materials placed outside the, cathode ray tubes. Since Röentgen did not, know the nature of the radiation, he, named them X-rays and the name is still, carried on. It was noticed that X-rays are, produced effectively when electrons strike, the dense metal anode, called targets., These are not deflected by the electric and, magnetic fields and have a very high, penetrating power through the matter, and that is the reason that these rays are, used to study the interior of the objects., These rays are of very short wavelengths, (∼0.1 nm) and possess electro-magnetic, character (Section 2.3.1)., Henri Becqueral (1852-1908), observed that there are certain elements, which emit radiation on their own and, named, this, phenomenon, as, radioactivity and the elements known, as radioactive elements. This field was, developed by Marie Curie, Piere Curie,, Rutherford and Fredrick Soddy. It was, observed that three kinds of rays i.e., α,, β- and γ-rays are emitted. Rutherford, found that α-rays consists of high energy, particles carrying two units of positive, charge and four unit of atomic mass. He, concluded that α- particles are helium, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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34, , CHEMISTRY, , represented in Fig. 2.5. A stream of high energy, α–particles from a radioactive source was, directed at a thin foil (thickness ∼ 100 nm) of, gold metal. The thin gold foil had a circular, fluorescent zinc sulphide screen around it., Whenever α–particles struck the screen, a tiny, flash of light was produced at that point., , nuclei as when α- particles combined with, two electrons yielded helium gas. β-rays, are negatively charged particles similar to, electrons. The γ-rays are high energy, radiations like X-rays, are neutral in, nature and do not consist of particles. As, regards penetrating power, α-particles are, the least, followed by β-rays (100 times, that of α–particles) and γ-rays (1000 times, of that α-particles)., 2.2.2 Rutherford’s Nuclear Model of Atom, Rutherford and his students (Hans Geiger and, Ernest Marsden) bombarded very thin gold foil, with α–particles. Rutherford’s famous, α –particle scattering experiment is, , The results of scattering experiment were, quite unexpected. According to Thomson, model of atom, the mass of each gold atom in, the foil should have been spread evenly over, the entire atom, and α– particles had enough, energy to pass directly through such a uniform, distribution of mass. It was expected that the, particles would slow down and change, directions only by a small angles as they passed, through the foil. It was observed that:, (i), , most of the α–particles passed through, the gold foil undeflected., , (ii), , a small fraction of the α–particles was, deflected by small angles., , (iii) a very few α–particles (∼1 in 20,000), bounced back, that is, were deflected by, nearly 180°., On the basis of the observations,, Rutherford drew the following conclusions, regarding the structure of atom:, , A. Rutherford’s scattering experiment, , B. Schematic molecular view of the gold foil, Fig. 2.5 Schematic view of Rutherford’s, scattering experiment. When a beam, of alpha (α) particles is “shot” at a thin, gold foil, most of them pass through, without much effect. Some, however,, are deflected., , (i), , Most of the space in the atom is empty as, most of the α–particles passed through, the foil undeflected., , (ii), , A few positively charged α–particles were, deflected. The deflection must be due to, enormous repulsive force showing that, the positive charge of the atom is not, spread throughout the atom as Thomson, had presumed. The positive charge has, to be concentrated in a very small volume, that repelled and deflected the positively, charged α–particles., , (iii) Calculations by Rutherford showed that, the volume occupied by the nucleus is, negligibly small as compared to the total, volume of the atom. The radius of the, atom is about 10–10 m, while that of, nucleus is 10–15 m. One can appreciate, this difference in size by realising that if, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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35, , STRUCTURE OF ATOM, , a cricket ball represents a nucleus, then, the radius of atom would be about 5 km., , The total number of nucleons is termed as, mass number (A) of the atom., , On the basis of above observations and, conclusions, Rutherford proposed the nuclear, model of atom. According to this model:, , mass number (A) = number of protons (Z ), + number of, neutrons (n), (2.4), , (i), , (ii), , The positive charge and most of the mass, of the atom was densely concentrated in, extremely small region. This very small, portion of the atom was called nucleus, by Rutherford., The nucleus is surrounded by electrons, that move around the nucleus with a very, high speed in circular paths called orbits., Thus, Rutherford’s model of atom, resembles the solar system in which the, nucleus plays the role of sun and the, electrons that of revolving planets., , (iii) Electrons and the nucleus are held, together by electrostatic forces of, attraction., 2.2.3 Atomic Number and Mass Number, The presence of positive charge on the, nucleus is due to the protons in the nucleus., As established earlier, the charge on the, proton is equal but opposite to that of, electron. The number of protons present in, the nucleus is equal to atomic number (Z )., For example, the number of protons in the, hydrogen nucleus is 1, in sodium atom it is, 11, therefore their atomic numbers are 1 and, 11 respectively. In order to keep the electrical, neutrality, the number of electrons in an, atom is equal to the number of protons, (atomic number, Z ). For example, number of, electrons in hydrogen atom and sodium atom, are 1 and 11 respectively., Atomic number (Z) = number of protons in, the nucleus of an atom, = number of electrons, in a nuetral atom (2.3), While the positive charge of the nucleus, is due to protons, the mass of the nucleus,, due to protons and neutrons. As discussed, earlier protons and neutrons present in the, nucleus are collectively known as nucleons., , 2.2.4 Isobars and Isotopes, The composition of any atom can be, represented by using the normal element, symbol (X) with super-script on the left hand, side as the atomic mass number (A) and, subscript (Z ) on the left hand side as the atomic, number (i.e., AZ X)., Isobars are the atoms with same mass, number but different atomic number for, 14, 14, example, 6 C and 7 N. On the other hand, atoms, with identical atomic number but different, atomic mass number are known as Isotopes., In other words (according to equation 2.4), it, is evident that difference between the isotopes, is due to the presence of different number of, neutrons present in the nucleus. For example,, considering of hydrogen atom again, 99.985%, of hydrogen atoms contain only one proton., 1, This isotope is called protium (1 H). Rest of the, percentage of hydrogen atom contains two other, isotopes, the one containing 1 proton and 1, 2, neutron is called deuterium (1 D, 0.015%), and the other one possessing 1 proton and 2, 3, neutrons is called tritium ( 1 T ). The latter, isotope is found in trace amounts on the earth., Other examples of commonly occuring, isotopes are: carbon atoms containing 6, 7 and, 13, 14, );, 8 neutrons besides 6 protons ( 12, 6 C, 6 C, 6 C, chlorine atoms containing 18 and 20 neutrons, 35, 37, besides 17 protons ( 17, Cl, 17, Cl )., Lastly an important point to mention, regarding isotopes is that chemical properties, of atoms are controlled by the number of, electrons, which are determined by the, number of protons in the nucleus. Number of, neutrons present in the nucleus have very little, effect on the chemical properties of an element., Therefore, all the isotopes of a given element, show same chemical behaviour., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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36, , CHEMISTRY, , massive sun and the electrons being similar, to the lighter planets. When classical, mechanics* is applied to the solar system, it, shows that the planets describe well-defined, orbits around the sun. The gravitational force, between the planets is given by the expression, , Problem 2.1, Calculate the number of protons,, neutrons and electrons in 80, ., 35 Br, Solution, , Z = 35, A = 80, species, In this case, 80, 35 Br, is neutral, Number of protons = number of electrons, = Z = 35, Number of neutrons = 80 – 35 = 45,, (equation 2.4), ,  m1m 2 ,  G. 2  where m1 and m2 are the masses, r, r, , Problem 2.2, The number of electrons, protons and, neutrons in a species are equal to 18, 16, and 16 respectively. Assign the proper, symbol to the species., Solution, The atomic number is equal to, number of protons = 16. The element is, sulphur (S)., Atomic mass number = number of, protons + number of neutrons, = 16 + 16 = 32, Species is not neutral as the number of, protons is not equal to electrons. It is, anion (negatively charged) with charge, equal to excess electrons = 18 – 16 = 2., 32 2–, Symbol is 16, S ., A, , Note : Before using the notation Z X, find, out whether the species is a neutral atom,, a cation or an anion. If it is a neutral atom,, equation (2.3) is valid, i.e., number of, protons = number of electrons = atomic, number. If the species is an ion, determine, whether the number of protons are larger, (cation, positive ion) or smaller (anion,, negative ion) than the number of electrons., Number of neutrons is always given by, A–Z, whether the species is neutral or ion., 2.2.5 Drawbacks of Rutherford Model, As you have learnt above, Rutherford nuclear, model of an atom is like a small scale solar, system with the nucleus playing the role of the, , is the distance of separation of the masses and, G is the gravitational constant. The theory can, also calculate precisely the planetary orbits and, these are in agreement with the experimental, measurements., The similarity between the solar system, and nuclear model suggests that electrons, should move around the nucleus in well, defined orbits. Further, the coulomb force, (kq1q2/r2 where q1 and q2 are the charges, r is, the distance of separation of the charges and, k is the proportionality constant) between, electron and the nucleus is mathematically, similar to the gravitational force. However,, when a body is moving in an orbit, it, undergoes acceleration even if it is moving with, a constant speed in an orbit because of, changing direction. So an electron in the, nuclear model describing planet like orbits is, under acceleration. According to the, electromagnetic theory of Maxwell, charged, particles when accelerated should emit, electromagnetic radiation (This feature does, not exist for planets since they are uncharged)., Therefore, an electron in an orbit will emit, radiation, the energy carried by radiation, comes from electronic motion. The orbit will, thus continue to shrink. Calculations show, that it should take an electron only 10–8 s to, spiral into the nucleus. But this does not, happen. Thus, the Rutherford model, cannot explain the stability of an atom., If the motion of an electron is described on the, basis of the classical mechanics and, electromagnetic theory, you may ask that, since the motion of electrons in orbits is, leading to the instability of the atom, then, why not consider electrons as stationary, , * Classical, , mechanics is a theoretical science based on Newton’s laws of motion. It specifies the laws of motion of, macroscopic objects., Download all NCERT books PDFs from www.ncert.online, , 2019-20

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37, , STRUCTURE OF ATOM, , around the nucleus. If the electrons were, stationary, electrostatic attraction between, the dense nucleus and the electrons would, pull the electrons toward the nucleus to form, a miniature version of Thomson’s model, of atom., Another serious drawback of the, Rutherford model is that it says nothing about, distribution of the electrons around the, nucleus and the energies of these electrons., 2.3 DEVELOPMENTS LEADING TO THE, BOHR’S MODEL OF ATOM, Historically, results observed from the studies, of interactions of radiations with matter have, provided immense information regarding the, structure of atoms and molecules. Neils Bohr, utilised these results to improve upon the, model proposed by Rutherford. Two, developments played a major role in the, formulation of Bohr’s model of atom. These, were:, (i) Dual character of the electromagnetic, radiation which means that radiations, possess both wave like and particle like, properties, and, (ii) Experimental results regarding atomic, spectra., First, we will discuss about the duel nature, of electromagnetic radiations. Experimental, results regarding atomic spectra will be, discussed in Section 2.4., , in the early 1870’s by James Clerk Maxwell,, which was experimentally confirmed later by, Heinrich Hertz. Here, we will learn some facts, about electromagnetic radiations., James Maxwell (1870) was the first to give, a comprehensive explanation about the, interaction between the charged bodies and, the behaviour of electrical and magnetic fields, on macroscopic level. He suggested that when, electrically charged particle moves under, accelaration, alternating electrical and, magnetic fields are produced and transmitted., These fields are transmitted in the forms of, waves called electromagnetic waves or, electromagnetic radiation., Light is the form of radiation known from, early days and speculation about its nature, dates back to remote ancient times. In earlier, days (Newton) light was supposed to be made, of particles (corpuscules). It was only in the, 19th century when wave nature of light was, established., Maxwell was again the first to reveal that, light waves are associated with oscillating, electric and magnetic character (Fig. 2.6)., , 2.3.1 Wave Nature of Electromagnetic, Radiation, In the mid-nineteenth century, physicists, actively studied absorption and emission of, radiation by heated objects. These are called, thermal radiations. They tried to find out of, what the thermal radiation is made. It is now, a well-known fact that thermal radiations, consist of electromagnetic waves of various, frequencies or wavelengths. It is based on a, number of modern concepts, which were, unknown in the mid-nineteenth century. First, active study of thermal radiation laws occured, in the 1850’s and the theory of electromagnetic, waves and the emission of such waves by, accelerating charged particles was developed, , Fig.2.6 The electric and magnetic field, components of an electromagnetic wave., These components have the same, wavelength, frequency, speed and, amplitude, but they vibrate in two, mutually perpendicular planes., , Although electromagnetic wave motion is, complex in nature, we will consider here only, a few simple properties., (i) The oscillating electric and magnetic fields, produced by oscillating charged particles, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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38, , CHEMISTRY, , are perpendicular to each other and both, are perpendicular to the direction of, propagation of the wave. Simplified, picture of electromagnetic wave is shown, in Fig. 2.6., (ii) Unlike sound waves or waves produced, in water, electromagnetic waves do not, require medium and can move in, vacuum., (iii) It is now well established that there are, many types of electromagnetic radiations,, which differ from one another in, wavelength (or frequency). These, constitute what is called electromagnetic, spectrum (Fig. 2.7). Different regions of, the spectrum are identified by different, names. Some examples are: radio, frequency region around 106 Hz, used for, broadcasting; microwave region around, 1010 Hz used for radar; infrared region, around 10 13 Hz used for heating;, ultraviolet region around 10 16 Hz a, component of sun’s radiation. The small, portion around 10 15 Hz, is what is, ordinarily called visible light. It is only, this part which our eyes can see (or, detect). Special instruments are required, to detect non-visible radiation., , (iv) Different kinds of units are used to, represent electromagnetic radiation., These radiations are characterised by the, properties, namely, frequency ( ν ) and, wavelength (λ)., The SI unit for frequency ( ν ) is hertz, (Hz, s–1), after Heinrich Hertz. It is defined as, the number of waves that pass a given point, in one second., Wavelength should have the units of length, and as you know that the SI units of length is, meter (m). Since electromagnetic radiation, consists of different kinds of waves of much, smaller wavelengths, smaller units are used., Fig.2.7 shows various types of electromagnetic radiations which differ from one, another in wavelengths and frequencies., In vaccum all types of electromagnetic, radiations, regardless of wavelength, travel at, the same speed, i.e., 3.0 × 108 m s–1 (2.997925, × 108 m s–1, to be precise). This is called speed, of light and is given the symbol ‘c‘. The, frequency (ν ), wavelength (λ) and velocity of light, (c) are related by the equation (2.5)., c=ν λ, , (2.5), , ν, (a), , (b), , Fig. 2.7, , (a) The spectrum of electromagnetic radiation. (b) Visible spectrum. The visible region is only, a small part of the entire spectrum., Download all NCERT books PDFs from www.ncert.online, , 2019-20

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39, , STRUCTURE OF ATOM, , The other commonly used quantity, specially in spectroscopy, is the wavenumber, ( ). It is defined as the number of wavelengths, per unit length. Its units are reciprocal of, wavelength unit, i.e., m–1. However commonly, used unit is cm–1 (not SI unit)., , Frequency of red light, , The range of visible spectrum is from, 4.0 × 1014 to 7.5 × 1014 Hz in terms of, frequency units., , Problem 2.3, The Vividh Bharati station of All India, Radio, Delhi, broadcasts on a frequency, of 1,368 kHz (kilo hertz). Calculate the, wavelength of the electromagnetic, radiation emitted by transmitter. Which, part of the electromagnetic spectrum, does it belong to?, , Problem 2.5, Calculate (a) wavenumber and (b), frequency of yellow radiation having, wavelength 5800 Å., Solution, (a) Calculation of wavenumber ( ), λ=5800Å = 5800 × 10–8 cm, = 5800 × 10–10 m, , Solution, The wavelength, λ, is equal to c/ν, where, c is the speed of electromagnetic radiation, in vacuum and ν is the frequency., Substituting the given values, we have, , λ=, , c, v, , (b), , This is a characteristic radiowave, wavelength., Problem 2.4, The wavelength range of the visible, spectrum extends from violet (400 nm) to, red (750 nm). Express these wavelengths, in frequencies (Hz). (1nm = 10 –9 m), Solution, Using equation 2.5, frequency of violet, light, , = 7.50 × 1014 Hz, *, **, , = 4.00 × 1014 Hz, , ν=, , Calculation of the frequency (ν ), , 2.3.2 Particle Nature of Electromagnetic, Radiation: Planck’s Quantum, Theory, Some of the experimental phenomenon such, as diffraction* and interference** can be, explained by the wave nature of the, electromagnetic radiation. However, following, are some of the observations which could not, be explained with the help of even the, electromagentic theory of 19th century, physics (known as classical physics):, (i) the nature of emission of radiation from, hot bodies (black -body radiation), (ii) ejection of electrons from metal surface, when radiation strikes it (photoelectric, effect), (iii) variation of heat capacity of solids as a, function of temperature, , Diffraction is the bending of wave around an obstacle., Interference is the combination of two waves of the same or different frequencies to give a wave whose distribution at, each point in space is the algebraic or vector sum of disturbances at that point resulting from each interfering wave., Download all NCERT books PDFs from www.ncert.online, , 2019-20

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40, , CHEMISTRY, , (iv) Line spectra of atoms with special, reference to hydrogen., These phenomena indicate that the system, can take energy only in discrete amounts. All, possible energies cannot be taken up or, radiated., It is noteworthy that the first concrete, explanation for the phenomenon of the black, body radiation mentioned above was given by, Max Planck in 1900. Let us first try to, understand this phenomenon, which is given, below:, Hot objects emit electromagnetic radiations, over a wide range of wavelengths. At high, temperatures, an appreciable proportion of, radiation is in the visible region of the, spectrum. As the temperature is raised, a, higher proportion of short wavelength (blue, light) is generated. For example, when an iron, rod is heated in a furnace, it first turns to dull, red and then progressively becomes more and, more red as the temperature increases. As this, is heated further, the radiation emitted becomes, white and then becomes blue as the, temperature becomes very high. This means, that red radiation is most intense at a particular, temperature and the blue radiation is more, intense at another temperature. This means, intensities of radiations of different wavelengths, emitted by hot body depend upon its, temperature. By late 1850’s it was known that, objects made of different material and kept at, different temperatures emit different amount of, radiation. Also, when the surface of an object is, irradiated with light (electromagnetic radiation),, a part of radiant energy is generally reflected, as such, a part is absorbed and a part of it is, transmitted. The reason for incomplete, absorption is that ordinary objects are as a rule, imperfect absorbers of radiation. An ideal body,, which emits and absorbs radiations of all, frequencies uniformly, is called a black body, and the radiation emitted by such a body is, called black body radiation. In practice, no, such body exists. Carbon black approximates, fairly closely to black body. A good physical, approximation to a black body is a cavity with, a tiny hole, which has no other opening. Any, ray entering the hole will be reflected by the, cavity walls and will be eventually absorbed by, the walls. A black body is also a perfect radiator, , of radiant energy. Furthermore, a black body, is in thermal equilibrium with its surroundings., It radiates same amount of energy per unit area, as it absorbs from its surrounding in any given, time. The amount of light emitted (intensity of, radiation) from a black body and its spectral, distribution depends only on its temperature., At a given temperature, intensity of radiation, emitted increases with the increase of, wavelength, reaches a maximum value at a, given wavelength and then starts decreasing, with further increase of wavelength, as shown, in Fig. 2.8. Also, as the temperature increases,, maxima of the curve shifts to short wavelength., Several attempts were made to predict the, intensity of radiation as a function of, wavelength., But the results of the above experiment, could not be explained satisfactorily on the, basis of the wave theory of light. Max Planck, , Fig. 2.8, , Wavelength-intensity relationship, , Fig. 2.8(a) Black body, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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41, , STRUCTURE OF ATOM, , arrived at a satisfactory relationship by, making an assumption that absorption and, emmission of radiation arises from oscillator, i.e., atoms in the wall of black body. Their, frequency of oscillation is changed by, interaction with oscilators of electromagnetic, radiation. Planck assumed that radiation, could be sub-divided into discrete chunks of, energy. He suggested that atoms and, molecules could emit or absorb energy only, in discrete quantities and not in a continuous, manner. He gave the name quantum to the, smallest quantity of energy that can be, emitted or absorbed in the form of, electromagnetic radiation. The energy (E ) of a, quantum of radiation is proportional, to its frequency (ν ) and is expressed by, equation (2.6)., (2.6), E = hυ, The proportionality constant, ‘h’ is known, as Planck’s constant and has the value, 6.626×10–34 J s., With this theory, Planck was able to explain, the distribution of intensity in the radiation, from black body as a function of frequency or, wavelength at different temperatures., Quantisation has been compared to, standing on a staircase. A person can stand, on any step of a staircase, but it is not possible, for him/her to stand in between the two steps., The energy can take any one of the values from, the following set, but cannot take on any, values between them., E = 0, hυ, 2hυ, 3hυ....nhυ....., , Fig.2.9, , Equipment for studying the photoelectric, effect. Light of a particular frequency strikes, a clean metal surface inside a vacuum, chamber. Electrons are ejected from the, metal and are counted by a detector that, measures their kinetic energy., , Max Planck, (1858 – 1947), Max Planck, a German physicist,, received his Ph.D in theoretical, physics from the University of, Munich in 1879. In 1888, he was, appointed Director of the Institute, of Theoretical Physics at the, University of Berlin. Planck was awarded the Nobel, Prize in Physics in 1918 for his quantum theory., Planck also made significant contributions in, thermodynamics and other areas of physics., , Photoelectric Effect, In 1887, H. Hertz performed a very interesting, experiment in which electrons (or electric, current) were ejected when certain metals (for, example potassium, rubidium, caesium etc.), were exposed to a beam of light as shown, in Fig.2.9. The phenomenon is called, Photoelectric effect. The results observed in, this experiment were:, (i) The electrons are ejected from the metal, surface as soon as the beam of light strikes, the surface, i.e., there is no time lag, between the striking of light beam and the, ejection of electrons from the metal surface., (ii) The number of electrons ejected is, proportional to the intensity or brightness, of light., (iii) For each metal, there is a characteristic, minimum frequency,ν0 (also known as, threshold frequency) below which, photoelectric effect is not observed. At a, frequency ν >ν 0, the ejected electrons come, out with certain kinetic energy. The kinetic, energies of these electrons increase with, the increase of frequency of the light used., All the above results could not be explained, on the basis of laws of classical physics., According to latter, the energy content of the, beam of light depends upon the brightness of, the light. In other words, number of electrons, ejected and kinetic energy associated with, them should depend on the brightness of light., It has been observed that though the number, of electrons ejected does depend upon the, brightness of light, the kinetic energy of the, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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42, , CHEMISTRY, , Table 2.2 Values of Work Function (W0) for a Few Metals, Metal, W0 /eV, , Li, , Na, , K, , Mg, , Cu, , Ag, , 2.42, , 2.3, , 2.25, , 3.7, , 4.8, , 4.3, , ejected electrons does not. For example, red, light [ν = (4.3 to 4.6) × 1014 Hz] of any brightness, (intensity) may shine on a piece of potassium, metal for hours but no photoelectrons are, ejected. But, as soon as even a very weak yellow, light (ν = 5.1–5.2 × 1014 Hz) shines on the, potassium metal, the photoelectric effect is, observed. The threshold frequency (ν0) for, potassium metal is 5.0×1014 Hz., Einstein (1905) was able to explain the, photoelectric effect using Planck’s quantum, theory of electromagnetic radiation as a, starting point., Albert Einstein, a German, born American physicist, is, regarded by many as one of, the two great physicists the, world has known (the other, is Isaac Newton). His three, research papers (on special, relativity, Brownian motion, Albert Einstein, and the photoelectric effect), (1879-1955), which he published in 1905,, while he was employed as a technical, assistant in a Swiss patent office in Ber ne, have profoundly influenced the development, of physics. He received the Nobel Prize in, Physics in 1921 for his explanation of the, photoelectric effect., , Shining a beam of light on to a metal, surface can, therefore, be viewed as shooting, a beam of particles, the photons. When a, photon of sufficient energy strikes an electron, in the atom of the metal, it transfers its energy, instantaneously to the electron during the, collision and the electron is ejected without, any time lag or delay. Greater the energy, possessed by the photon, greater will be, transfer of energy to the electron and greater, the kinetic energy of the ejected electron. In, other words, kinetic energy of the ejected, electron is proportional to the frequency of the, electromagnetic radiation. Since the striking, photon has energy equal to hν and the, , minimum energy required to eject the electron, is hν0 (also called work function, W0 ; Table 2.2),, then, the, difference, in, energy, (hν – hν0 ) is transferred as the kinetic energy of, the photoelectron. Following the conservation, of energy principle, the kinetic energy of the, ejected electron is given by the equation 2.7., (2.7), where me is the mass of the electron and v is, the velocity associated with the ejected electron., Lastly, a more intense beam of light consists, of larger number of photons, consequently the, number of electrons ejected is also larger as, compared to that in an experiment in which a, beam of weaker intensity of light is employed., Dual Behaviour of Electromagnetic, Radiation, The particle nature of light posed a dilemma, for scientists. On the one hand, it could explain, the black body radiation and photoelectric, effect satisfactorily but on the other hand, it, was not consistent with the known wave, behaviour of light which could account for the, phenomena of interference and diffraction. The, only way to resolve the dilemma was to accept, the idea that light possesses both particle and, wave-like properties, i.e., light has dual, behaviour. Depending on the experiment, we, find that light behaves either as a wave or as a, stream of particles. Whenever radiation, interacts with matter, it displays particle like, properties in contrast to the wavelike, properties (interference and diffraction), which, it exhibits when it propagates. This concept, was totally alien to the way the scientists, thought about matter and radiation and it took, them a long time to become convinced of its, validity. It turns out, as you shall see later,, that some microscopic particles like electrons, also exhibit this wave-particle duality., , Download all NCERT books PDFs from www.ncert.online, C:\Chemistry XI\Unit-2\Unit-2(2)-Lay-3(reprint).pmd, , 27.7.6, 16.10.6 (Reprint), 2019-20

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43, , STRUCTURE OF ATOM, , Solution, The energy (E) of a 300 nm photon is, given by, , Problem 2.6, Calculate energy of one mole of photons, of radiation whose frequency is 5 ×1014, Hz., Solution, Energy (E) of one photon is given by the, expression, , = 6.626 × 10-19 J, The energy of one mole of photons, = 6.626 ×10–19 J × 6.022 ×1023 mol–1, = 3.99 × 105 J mol–1, The minimum energy needed to remove, one mole of electrons from sodium, 5, –1, = (3.99 –1.68) 10 J mol, 5, –1, = 2.31 × 10 J mol, The minimum energy for one electron, , E = hν, h = 6.626 ×10, 14, , –34, , Js, , –1, , ν = 5×10 s (given), E = (6.626 ×10–34 J s) × (5 ×1014 s–1), = 3.313 ×10–19 J, Energy of one mole of photons, = (3.313 ×10–19 J) × (6.022 × 1023 mol–1), = 199.51 kJ mol–1, Problem 2.7, A 100 watt bulb emits monochromatic, light of wavelength 400 nm. Calculate the, number of photons emitted per second, by the bulb., Solution, Power of the bulb = 100 watt, –1, = 100 J s, , This corresponds to the wavelength, , hc, E, 6.626 × 10 −34 J s × 3.0 × 108 m s −1, =, 3.84 × 10 −19 J, , ∴λ=, , Energy of one photon E = hν = hc/λ, , = 517 nm, (This corresponds to green light), Problem 2.9, The threshold frequency ν0 for a metal is, 14 –1, 7.0 ×10 s . Calculate the kinetic energy, of an electron emitted when radiation of, 15 –1, frequency ν =1.0 ×10 s hits the metal., Solution, According to Einstein’s equation, 2, Kinetic energy = ½ mev =h(ν – ν0 ), –34, 15 –1, = (6.626 ×10 J s) (1.0 × 10 s – 7.0, 14 –1, ×10 s ), –34, 14 –1, = (6.626 ×10 J s) (10.0 ×10 s – 7.0, 14 –1, ×10 s ), –34, 14 –1, = (6.626 ×10 J s) × (3.0 ×10 s ), –19, = 1.988 ×10 J, , 6.626 × 10 −34 J s × 3 × 108 m s−1, =, 400 × 10 −9 m, -19, , = 4.969 × 10 J, Number of photons emitted, , 100 J s−1, = 2.012 × 1020 s −1, 4.969 × 10 −19 J, Problem 2.8, When electromagnetic radiation of, wavelength 300 nm falls on the surface, of sodium, electrons are emitted with a, kinetic energy of 1.68 ×105 J mol–1. What, is the minimum energy needed to remove, an electron from sodium? What is the, maximum wavelength that will cause a, photoelectron to be emitted?, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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44, , CHEMISTRY, , 2.3.3 Evidence for the quantized*, Electronic Energy Levels: Atomic, spectra, The speed of light depends upon the nature of, the medium through which it passes. As a, result, the beam of light is deviated or refracted, from its original path as it passes from one, medium to another. It is observed that when a, ray of white light is passed through a prism,, the wave with shorter wavelength bends more, than the one with a longer wavelength. Since, ordinary white light consists of waves with all, the wavelengths in the visible range, a ray of, white light is spread out into a series of, coloured bands called spectrum. The light of, red colour which has longest wavelength is, deviated the least while the violet light, which, has shortest wavelength is deviated the most., The spectrum of white light, that we can see,, ranges from violet at 7.50 × 1014 Hz to red at, 4×10 14 Hz. Such a spectrum is called, continuous spectrum. Continuous because, violet merges into blue, blue into green and so, on. A similar spectrum is produced when a, rainbow forms in the sky. Remember that, visible light is just a small portion of the, electromagnetic radiation (Fig.2.7). When, electromagnetic radiation interacts with matter,, atoms and molecules may absorb energy and, reach to a higher energy state. With higher, energy, these are in an unstable state. For, returning to their normal (more stable, lower, energy states) energy state, the atoms and, molecules emit radiations in various regions, of the electromagnetic spectrum., Emission and Absorption Spectra, The spectrum of radiation emitted by a, substance that has absorbed energy is called, an emission spectrum. Atoms, molecules or, ions that have absorbed radiation are said to, be “excited”. To produce an emission, spectrum, energy is supplied to a sample by, heating it or irradiating it and the wavelength, (or frequency) of the radiation emitted, as the, sample gives up the absorbed energy, is, recorded., An absorption spectrum is like the, photographic negative of an emission, *, , spectrum. A continuum of radiation is passed, through a sample which absorbs radiation of, certain wavelengths. The missing wavelength, which corresponds to the radiation absorbed, by the matter, leave dark spaces in the bright, continuous spectrum., The study of emission or absorption, spectra is referred to as spectroscopy. The, spectrum of the visible light, as discussed, above, was continuous as all wavelengths (red, to violet) of the visible light are represented in, the spectra. The emission spectra of atoms in, the gas phase, on the other hand, do not show, a continuous spread of wavelength from red, to violet, rather they emit light only at specific, wavelengths with dark spaces between them., Such spectra are called line spectra or atomic, spectra because the emitted radiation is, identified by the appearance of bright lines in, the spectra (Fig. 2.10 page 45)., Line emission spectra are of great, interest in the study of electronic structure., Each element has a unique line emission, spectrum. The characteristic lines in atomic, spectra can be used in chemical analysis to, identify unknown atoms in the same way as, fingerprints are used to identify people. The, exact matching of lines of the emission, spectrum of the atoms of a known element with, the lines from an unknown sample quickly, establishes the identity of the latter, German, chemist, Robert Bunsen (1811-1899) was one, of the first investigators to use line spectra to, identify elements., Elements like rubidium (Rb), caesium (Cs), thallium (Tl), indium (In), gallium (Ga) and, scandium (Sc) were discovered when their, minerals were analysed by spectroscopic, methods. The element helium (He) was, discovered in the sun by spectroscopic method., Line Spectrum of Hydrogen, When an electric discharge is passed through, gaseous hydrogen, the H2 molecules dissociate, and the energetically excited hydrogen atoms, produced emit electromagnetic radiation of, discrete frequencies. The hydrogen spectrum, consists of several series of lines named after, their discoverers. Balmer showed in 1885 on, the basis of experimental observations that if, , The restriction of any property to discrete values is called quantization., Download all NCERT books PDFs from www.ncert.online, , 2019-20

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45, , STRUCTURE OF ATOM, , (a), , (b), , Fig. 2.10 (a) Atomic emission. The light emitted by a sample of excited hydrogen atoms (or any other, element) can be passed through a prism and separated into certain discrete wavelengths. Thus an, emission spectrum, which is a photographic recording of the separated wavelengths is called as line, spectrum. Any sample of reasonable size contains an enormous number of atoms. Although a single, atom can be in only one excited state at a time, the collection of atoms contains all possible excited, states. The light emitted as these atoms fall to lower energy states is responsible for the spectrum. (b), Atomic absorption. When white light is passed through unexcited atomic hydrogen and then through, a slit and prism, the transmitted light is lacking in intensity at the same wavelengths as are emitted in, (a) The recorded absorption spectrum is also a line spectrum and the photographic negative of the, emission spectrum., , spectral lines are expressed in terms of, wavenumber ( ), then the visible lines of the, hydrogen spectrum obey the following formula:, (2.8), where n is an integer equal to or greater than, 3 (i.e., n = 3,4,5,....), The series of lines described by this formula, are called the Balmer series. The Balmer series, of lines are the only lines in the hydrogen, spectrum which appear in the visible region, of the electromagnetic spectrum. The Swedish, spectroscopist, Johannes Rydberg, noted that, all series of lines in the hydrogen spectrum, could be described by the following, expression :, , The value 109,677 cm –1 is called the, Rydberg constant for hydrogen. The first five, series of lines that correspond to n1 = 1, 2, 3,, 4, 5 are known as Lyman, Balmer, Paschen,, Bracket and Pfund series, respectively,, Table 2.3 shows these series of transitions in, the hydrogen spectrum. Fig 2.11 (page, 46), shows the Lyman, Balmer and Paschen series, of transitions for hydrogen atom., Of all the elements, hydrogen atom has the, simplest line spectrum. Line spectrum becomes, Table 2.3 The Spectral Lines for Atomic, Hydrogen, , (2.9), where n1=1,2........, n2 = n1 + 1, n1 + 2......, Download all NCERT books PDFs from www.ncert.online, , 2019-20

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46, , CHEMISTRY, , rationalize many points in the atomic structure, and spectra. Bohr’s model for hydrogen atom, is based on the following postulates:, i), The electron in the hydrogen atom can, move around the nucleus in a circular path, of fixed radius and energy. These paths are, called orbits, stationary states or allowed, energy states. These orbits are arranged, concentrically around the nucleus., ii) The energy of an electron in the orbit does, not change with time. However, the, electron will move from a lower stationary, state to a higher stationary state when, required amount of energy is absorbed, by the electron or energy is emitted when, electron moves from higher stationary, state to lower stationary state (equation, 2.16). The energy change does not take, place in a continuous manner., Angular Momentum, Just as linear momentum is the product, of mass (m) and linear velocity (v), angular, momentum is the product of moment of, inertia (I) and angular velocity (ω). For an, electron of mass me, moving in a circular, path of radius r around the nucleus,, angular momentum = I × ω, , Fig. 2.11 T ransitions of the electron in the, hydrogen atom (The diagram shows, the Lyman, Balmer and Paschen series, of transitions), , more and more complex for heavier atom. There, are, however, certain features which are, common to all line spectra, i.e., (i) line spectrum, of element is unique and (ii) there is regularity, in the line spectrum of each element. The, questions which arise are: What are the, reasons for these similarities? Is it something, to do with the electronic structure of atoms?, These are the questions need to be answered., We shall find later that the answers to these, questions provide the key in understanding, electronic structure of these elements., 2.4 BOHR’S MODEL FOR HYDROGEN, ATOM, Neils Bohr (1913) was the first to explain, quantitatively the general features of the, structure of hydrogen atom and its spectrum., He used Planck’s concept of quantisation of, energy. Though the theory is not the modern, quantum mechanics, it can still be used to, , Since I = mer2, and ω = v/r where v is the, linear velocity,, ∴angular momentum = mer2 × v/r = mevr, , iii), , The frequency of radiation absorbed or, emitted when transition occurs between, two stationary states that differ in energy, by ∆E, is given by:, , ∆E E 2 − E1, =, (2.10), h, h, Where E1 and E2 are the energies of the, lower and higher allowed energy states, respectively. This expression is commonly, known as Bohr’s frequency rule., The angular momentum of an electron is, quantised. In a given stationary state it, can be expressed as in equation (2.11), , ν=, , iv), , m e v r = n., , h, n = 1,2,3....., 2π, , Download all NCERT books PDFs from www.ncert.online, , 2019-20, , (2.11)

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47, , STRUCTURE OF ATOM, , Where me is the mass of electron, v is the, velocity and r is the radius of the orbit in which, electron is moving., Thus an electron can move only in those, orbits for which its angular momentum is, integral multiple of h/2π. That means angular, momentum is quantised. Radiation is emitted, or obsorbed only when transition of electron, takes place from one quantised value of angular, momentum to another. Therefore, Maxwell’s, electromagnetic theory does not apply here that, is why only certain fixed orbits are allowed., The details regarding the derivation of, energies of the stationary states used by Bohr,, are quite complicated and will be discussed in, higher classes. However, according to Bohr’s, theory for hydrogen atom:, a), , b), , c), , The stationary states for electron are, numbered n = 1,2,3.......... These integral, numbers (Section 2.6.2) are known as, Principal quantum numbers., The radii of the stationary states are, expressed as:, rn = n2 a0, (2.12), where a0 = 52.9 pm. Thus the radius of, the first stationary state, called the Bohr, orbit, is 52.9 pm. Normally the electron, in the hydrogen atom is found in this orbit, (that is n=1). As n increases the value of r, will increase. In other words the electron, will be present away from the nucleus., The most important property associated, with the electron, is the energy of its, stationary state. It is given by the, expression., ,  1, En = − R H  2 , n , , n = 1,2,3...., , Niels Bohr, (1885–1962), N i e l s B o h r, a D a n i s h, physicist received his Ph.D., from the University of, Copenhagen in 1911. He, then spent a year with J.J., Thomson and Er nest Rutherford in England., In 1913, he returned to Copenhagen where, he remained for the rest of his life. In 1920, he was named Director of the Institute of, theoretical Physics. After first World War,, Bohr worked energetically for peaceful uses, of atomic energy. He received the first Atoms, for Peace award in 1957. Bohr was awarded, the Nobel Prize in Physics in 1922., , Fig. 2.11 depicts the energies of different, stationary states or energy levels of hydrogen, atom. This representation is called an energy, level diagram., When the electron is free from the influence, of nucleus, the energy is taken as zero. The, electron in this situation is associated with the, stationary state of Principal Quantum number, = n = ∞ and is called as ionized hydrogen atom., When the electron is attracted by the nucleus, and is present in orbit n, the energy is emitted, , (2.13), , where RH is called Rydberg constant and its, value is 2.18×10–18 J. The energy of the lowest, state, also called as the ground state, is, E1 = –2.18×10–18 (, , 1, ) = –2.18×10–18 J. The, 12, , energy of the stationary state for n = 2, will, be : E2 = –2.18×10–18J (, , 1, )= –0.545×10–18 J., 22, , What does the negative electronic, energy (En) for hydrogen atom mean?, The energy of the electron in a hydrogen, atom has a negative sign for all possible, orbits (eq. 2.13). What does this negative, sign convey? This negative sign means that, the energy of the electron in the atom is, lower than the energy of a free electron at, rest. A free electron at rest is an electron, that is infinitely far away from the nucleus, and is assigned the energy value of zero., Mathematically, this corresponds to, setting n equal to infinity in the equation, (2.13) so that E∞=0. As the electron gets, closer to the nucleus (as n decreases), En, becomes larger in absolute value and more, and more negative. The most negative, energy value is given by n=1 which, corresponds to the most stable orbit. We, call this the ground state., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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48, , CHEMISTRY, , and its energy is lowered. That is the reason, for the presence of negative sign in equation, (2.13) and depicts its stability relative to the, reference state of zero energy and n = ∞., d) Bohr’s theory can also be applied to the, ions containing only one electron, similar, to that present in hydrogen atom. For, example, He+ Li2+, Be3+ and so on. The, energies of the stationary states associated, with these kinds of ions (also known as, hydrogen like species) are given by the, expression., ,  Z2, E n = − 2.18 × 10 −18  2  J, n , ,  R   R , ∆ E =  − 2H  −  − 2H  (where n and n, i, f,  nf   ni , stand for initial orbit and final orbits),  1,  1, 1, 1, −18, ∆E = R H  2 − 2  = 2.18 × 10 J  2 − 2 ,  n i nf ,  ni nf , , (2,17), The frequency (ν ) associated with the, absorption and emission of the photon can be, evaluated by using equation, (2.18), , ν=, , (2.14), , and radii by the expression, =, , 2, , rn =, , 52.9 (n ), pm, Z, , (2.15), , where Z is the atomic number and has values, 2,3 for the helium and lithium atoms, respectively. From the above equations, it is, evident that the value of energy becomes more, negative and that of radius becomes smaller, with increase of Z . This means that electron, will be tightly bound to the nucleus., e), It is also possible to calculate the velocities, of electrons moving in these orbits., Although the precise equation is not given, here, qualitatively the magnitude of, velocity of electron increases with increase, of positive charge on the nucleus and, decreases with increase of principal, quantum number., 2.4.1 Explanation of Line Spectrum of, Hydrogen, Line spectrum observed in case of hydrogen, atom, as mentioned in section 2.3.3, can be, explained quantitatively using Bohr’s model., According to assumption 2, radiation (energy), is absorbed if the electron moves from the orbit, of smaller Principal quantum number to the, orbit of higher Principal quantum number,, whereas the radiation (energy) is emitted if the, electron moves from higher orbit to lower orbit., The energy gap between the two orbits is given, by equation (2.16), (2.16), ∆E = Ef – Ei, Combining equations (2.13) and (2.16), , ∆ E RH  1, 1, =, − 2, 2, , h, h  ni nf , , 2.18 × 10 −18 J  1, 1, − 2, −34, 2, , 6.626 × 10 J s  n i n f , ,  1, 1, = 3.29 × 1015  2 − 2  Hz,  ni n f , , (2.18), , (2.19), , and in terms of wavenumbers ( ), , ν RH  1 − 1 , ν= =, c hc  n i2 n f2 , =, , (2.20), , 1, 3.29 × 1015 s−1  1, − 2, 8, −s  2, 3 × 10 m s  n i n f , ,  1, 1, = 1.09677 × 107  2 − 2  m −1,  ni nf , , (2.21), , In case of absorption spectrum, nf > ni and, the term in the parenthesis is positive and energy, is absorbed. On the other hand in case of, emission spectrum ni > nf , ∆ E is negative and, energy is released., The expression (2.17) is similar to that used, by Rydberg (2.9) derived empirically using the, experimental data available at that time. Further,, each spectral line, whether in absorption or, emission spectrum, can be associated to the, particular transition in hydrogen atom. In case, of large number of hydrogen atoms, different, possible transitions can be observed and thus, leading to large number of spectral lines. The, brightness or intensity of spectral lines depends, upon the number of photons of same wavelength, or frequency absorbed or emitted., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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49, , STRUCTURE OF ATOM, , Problem 2.10, What are the frequency and wavelength, of a photon emitted during a transition, from n = 5 state to the n = 2 state in the, hydrogen atom?, Solution, Since ni = 5 and nf = 2, this transition gives, rise to a spectral line in the visible region, of the Balmer series. From equation (2.17), , 2.4.2 Limitations of Bohr’s Model, Bohr’s model of the hydrogen atom was no, doubt an improvement over Rutherford’s, nuclear model, as it could account for the, stability and line spectra of hydrogen atom and, hydrogen like ions (for example, He+, Li2+, Be3+,, and so on). However, Bohr’s model was too, simple to account for the following points., i), , 1, 1, ∆E = 2.18 × 10 −18 J  2 − 2 , 2 , 5, −19, = − 4.58 × 10 J, It is an emission energy, The frequency of the photon (taking, energy in terms of magnitude) is given by, , ν=, , ∆E, h, , It fails to account for the finer details, (doublet, that is two closely spaced lines), of the hydrogen atom spectrum observed, by using sophisticated spectroscopic, techniques. This model is also unable to, explain the spectrum of atoms other than, hydrogen, for example, helium atom which, possesses only two electrons. Further,, Bohr’s theory was also unable to explain, the splitting of spectral lines in the presence, of magnetic field (Zeeman effect) or an, electric field (Stark effect)., , ii) It could not explain the ability of atoms to, form molecules by chemical bonds., In other words, taking into account the, points mentioned above, one needs a better, theory which can explain the salient features, of the structure of complex atoms., , = 6.91×1014 Hz, , Problem 2.11, Calculate the energy associated with the, first orbit of He+. What is the radius of this, orbit?, Solution, En = −, , (2.18 × 10 −18 J )Z 2, atom–1, n2, , For He , n = 1, Z = 2, 2, , (2.18 × 10 J)(2 ), = −8.72 × 10 −18 J, 12, The radius of the orbit is given by equation, (2.15), E1 = −, , (0.0529 nm )n 2, Z, Since n = 1, and Z = 2, rn =, , rn =, , In view of the shortcoming of the Bohr’s model,, attempts were made to develop a more suitable, and general model for atoms. Two important, developments which contributed significantly, in the formulation of such a model were :, 1. Dual behaviour of matter,, 2. Heisenberg uncertainty principle., , +, , −18, , 2.5 TOWARDS QUANTUM MECHANICAL, MODEL OF THE ATOM, , (0.0529 nm )12, = 0.02645 nm, 2, , 2.5.1 Dual Behaviour of Matter, The French physicist, de Broglie, in 1924, proposed that matter, like radiation, should, also exhibit dual behaviour i.e., both particle, and wavelike properties. This means that just, as the photon has momentum as well as, wavelength, electrons should also have, momentum as well as wavelength, de Broglie,, from this analogy, gave the following relation, between wavelength (λ) and momentum (p) of, a material particle., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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50, , CHEMISTRY, , Solution, According to de Brogile equation (2.22), , Louis de Broglie (1892 – 1987), Louis de Broglie, a French, physicist, studied history as an, undergraduate in the early, 1910’s. His interest turned to, science as a result of his, assignment, to, radio, communications in World War I., He received his Dr. Sc. from the University of, Paris in 1924. He was professor of theoretical, physics at the University of Paris from 1932 untill, his retirement in 1962. He was awarded the, Nobel Prize in Physics in 1929., , h, h, λ=, =, mv p, , λ=, , = 6.626 × 10–34 m (J = kg m2 s–2), Problem 2.13, The mass of an electron is 9.1×10–31 kg. If, its K.E. is 3.0×10 –25 J, calculate its, wavelength., Solution, Since K. E. = ½ mv2,  2K.E., v =,  m , , (2.22), , where m is the mass of the particle, v its, velocity and p its momentum. de Broglie’s, prediction was confirmed experimentally, when it was found that an electron beam, undergoes diffraction, a phenomenon, characteristic of waves. This fact has been put, to use in making an electron microscope,, which is based on the wavelike behaviour of, electrons just as an ordinary microscope, utilises the wave nature of light. An electron, microscope is a powerful tool in modern, scientific research because it achieves a, magnification of about 15 million times., It needs to be noted that according to de, Broglie, every object in motion has a wave, character. The wavelengths associated with, ordinary objects are so short (because of their, large masses) that their wave properties cannot, be detected. The wavelengths associated with, electrons and other subatomic particles (with, very small mass) can however be detected, experimentally. Results obtained from the, following problems prove these points, qualitatively., Problem 2.12, What will be the wavelength of a ball of, mass 0.1 kg moving with a velocity of 10, m s–1 ?, , h, (6.626 × 10 −34 Js), =, mv (0.1 kg )(10 m s −1 ), , 1/ 2, ,  2 × 3.0 × 10−25 kg m 2 s −2 , =, , 9.1 × 10 −31 kg, , , 1/ 2, , = 812 m s–1, , λ=, , h, 6.626 × 10−34 Js, =, m v (9.1 × 10 −31 kg )(812 m s−1 ), , = 8967 × 10–10 m = 896.7 nm, Problem 2.14, Calculate the mass of a photon with, wavelength 3.6 Å., Solution, λ = 3.6 Å = 3.6 × 10–10 m, Velocity of photon = velocity of light, , = 6.135 × 10–29 kg, 2.5.2 Heisenberg’s Uncertainty Principle, Werner Heisenberg a German physicist in, 1927, stated uncertainty principle which is the, consequence of dual behaviour of matter and, radiation. It states that it is impossible to, determine simultaneously, the exact, position and exact momentum (or velocity), of an electron., Mathematically, it can be given as in, equation (2.23)., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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51, , STRUCTURE OF ATOM, , (2.23), , where ∆x is the uncertainty in position and ∆px, (or ∆vx ) is the uncertainty in momentum (or, velocity) of the particle. If the position of the, electron is known with high degree of accuracy, (∆x is small), then the velocity of the electron, will be uncertain [∆(vx) is large]. On the other, hand, if the velocity of the electron is known, precisely (∆(vx ) is small), then the position of, the, electron, will, be, uncertain, (∆x will be large). Thus, if we carry out some, physical measurements on the electron’s, position or velocity, the outcome will always, depict a fuzzy or blur picture., The uncertainty principle can be best, understood with the help of an example., Suppose you are asked to measure the, thickness of a sheet of paper with an, unmarked metrestick. Obviously, the results, obtained would be extremely inaccurate and, meaningless, In order to obtain any accuracy,, you should use an instrument graduated in, units smaller than the thickness of a sheet of, the paper. Analogously, in order to determine, the position of an electron, we must use a, meterstick calibrated in units of smaller than, the dimensions of electron (keep in mind that, an electron is considered as a point charge and, is therefore, dimensionless). To observe an, electron, we can illuminate it with “light” or, electromagnetic radiation. The “light” used, must have a wavelength smaller than the, dimensions of an electron. The high, , h, , momentum photons of such light  p = , λ, would change the energy of electrons by, collisions. In this process we, no doubt, would, be able to calculate the position of the electron,, but we would know very little about the, velocity of the electron after the collision., Significance of Uncertainty Principle, One of the important implications of the, Heisenberg Uncertainty Principle is that it, rules out existence of definite paths or, trajectories of electrons and other similar, particles. The trajectory of an object is, determined by its location and velocity at, various moments. If we know where a body is, at a particular instant and if we also know its, velocity and the forces acting on it at that, instant, we can tell where the body would be, sometime later. We, therefore, conclude that the, position of an object and its velocity fix its, trajectory. Since for a sub-atomic object such, as an electron, it is not possible simultaneously, to determine the position and velocity at any, given instant to an arbitrary degree of, precision, it is not possible to talk of the, trajectory of an electron., The effect of Heisenberg Uncertainty, Principle is significant only for motion of, microscopic objects and is negligible for, that of macroscopic objects. This can be, seen from the following examples., If uncertainty principle is applied to an, object of mass, say about a milligram (10–6 kg),, then, , Werner Heisenberg (1901 – 1976) Werner Heisenberg (1901 – 1976) received his Ph.D. in, physics from the University of Munich in 1923. He then spent a year working with Max, Born at Gottingen and three years with Niels Bohr in Copenhagen. He was professor of, physics at the University of Leipzig from 1927 to 1941. During World War II, Heisenberg, was in charge of German research on the atomic bomb. After the war he was named, director of Max Planck Institute for physics in Gottingen. He was also accomplished, mountain climber. Heisenberg was awarded the Nobel Prize in Physics in 1932., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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52, , CHEMISTRY, , The value of ∆v∆x obtained is extremely, small and is insignificant. Therefore, one may, say that in dealing with milligram-sized or, heavier, objects,, the, associated, uncertainties are hardly of any real, consequence., In the case of a microscopic object like an, electron on the other hand. ∆v.∆x obtained is, much larger and such uncertainties are of real, consequence. For example, for an electron, whose mass is 9.11×10–31 kg., according to, Heisenberg uncertainty principle, , = 0.579×107 m s–1 (1J = 1 kg m2 s–2), = 5.79×106 m s–1, Problem 2.16, A golf ball has a mass of 40g, and a speed, of 45 m/s. If the speed can be measured, within accuracy of 2%, calculate the, uncertainty in the position., Solution, The uncertainty in the speed is 2%, i.e.,, , Using the equation (2.22), It, therefore, means that if one tries to find, the exact location of the electron, say to an, uncertainty of only 10–8 m, then the uncertainty, ∆v in velocity would be, , which is so large that the classical picture of, electrons moving in Bohr’s orbits (fixed) cannot, hold good. It, therefore, means that the, precise statements of the position and, momentum of electrons have to be, replaced by the statements of probability,, that the electron has at a given position, and momentum. This is what happens in, the quantum mechanical model of atom., Problem 2.15, A microscope using suitable photons is, employed to locate an electron in an atom, within a distance of 0.1 Å. What is the, uncertainty involved in the measurement, of its velocity?, Solution, , ∆x ∆p =, , h, h, or ∆x m ∆v =, 4π, 4π, , = 1.46×10–33 m, This is nearly ~ 1018 times smaller than, the diameter of a typical atomic nucleus., As mentioned earlier for large particles, the, uncertainty principle sets no meaningful, limit to the precision of measurements., Reasons for the Failure of the Bohr Model, One can now understand the reasons for the, failure of the Bohr model. In Bohr model, an, electron is regarded as a charged particle, moving in well defined circular orbits about, the nucleus. The wave character of the electron, is not considered in Bohr model. Further, an, orbit is a clearly defined path and this path, can completely be defined only if both the, position and the velocity of the electron are, known exactly at the same time. This is not, possible according to the Heisenberg, uncertainty principle. Bohr model of the, hydrogen atom, therefore, not only ignores, dual behaviour of matter but also contradicts, Heisenberg uncertainty principle. In view of, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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53, , STRUCTURE OF ATOM, , properties. It specifies the laws of motion that, these objects obey. When quantum mechanics, is applied to macroscopic objects (for which, wave like properties are insignificant) the, results are the same as those from the classical, mechanics., Quantum mechanics was developed, independently in 1926 by Werner Heisenberg, and Erwin Schrödinger. Here, however, we, shall be discussing the quantum mechanics, which is based on the ideas of wave motion., The fundamental equation of quantum, mechanics was developed by Schrödinger and, it won him the Nobel Prize in Physics in 1933., This equation which incorporates waveparticle duality of matter as proposed by de, Broglie is quite complex and knowledge of, higher mathematics is needed to solve it. You, will learn its solutions for different systems in, higher classes., , Erwin Schrödinger, an, Austrian, physicist, received his Ph.D. in, theoretical physics from, the University of Vienna, in 1910. In 1927, Schrödinger succeeded, Max Planck at the, University of Berlin at, Planck’s request. In 1933, Erwin Schrödinger, (1887-1961), Schrödinger left Berlin, because of his opposition to Hitler and Nazi, policies and returned to Austria in 1936. After, the invasion of Austria by Ger many,, Schrödinger was forcibly removed from his, professorship. He then moved to Dublin, Ireland, where he remained for seventeen years., Schrödinger shared the Nobel Prize for Physics, with P.A.M. Dirac in 1933., , these inherent weaknesses in the Bohr model,, there was no point in extending Bohr model, to other atoms. In fact an insight into the, structure of the atom was needed which could, account for wave-particle duality of matter and, be consistent with Heisenberg uncertainty, principle. This came with the advent of, quantum mechanics., 2.6 QUANTUM MECHANICAL MODEL OF, ATOM, Classical mechanics, based on Newton’s laws, of motion, successfully describes the motion, of all macroscopic objects such as a falling, stone, orbiting planets etc., which have, essentially a particle-like behaviour as shown, in the previous section. However it fails when, applied to microscopic objects like electrons,, atoms, molecules etc. This is mainly because, of the fact that classical mechanics ignores the, concept of dual behaviour of matter especially, for sub-atomic particles and the uncertainty, principle. The branch of science that takes into, account this dual behaviour of matter is called, quantum mechanics., Quantum mechanics is a theoretical, science that deals with the study of the motions, of the microscopic objects that have both, observable wave like and particle like, , For a system (such as an atom or a, molecule whose energy does not change with, time) the Schrödinger equation is written as, where is a mathematical operator, called Hamiltonian. Schrödinger gave a recipe, of constructing this operator from the, expression for the total energy of the system., The total energy of the system takes into, account the kinetic energies of all the subatomic particles (electrons, nuclei), attractive, potential between the electrons and nuclei and, repulsive potential among the electrons and, nuclei individually. Solution of this equation, gives E and ψ., Hydrogen Atom and the Schrödinger, Equation, When Schrödinger equation is solved for, hydrogen atom, the solution gives the possible, energy levels the electron can occupy and the, corresponding wave function(s) (ψ) of the, electron associated with each energy level., These quantized energy states and, corresponding wave functions which are, characterized by a set of three quantum, numbers (principal quantum number n,, azimuthal quantum number l and, magnetic quantum number m l ) arise as a, natural consequence in the solution of the, Schrödinger equation. When an electron is in, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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54, , CHEMISTRY, , any energy state, the wave function, corresponding to that energy state contains all, information about the electron. The wave, function is a mathematical function whose, value depends upon the coordinates of the, electron in the atom and does not carry any, physical meaning. Such wave functions of, hydrogen or hydrogen like species with one, electron are called atomic orbitals. Such wave, functions pertaining to one-electron species, are called one-electron systems. The, probability of finding an electron at a point, within an atom is proportional to the |ψ|2 at, that point. The quantum mechanical results, of the hydrogen atom successfully predict all, aspects of the hydrogen atom spectrum, including some phenomena that could not be, explained by the Bohr model., Application of Schrödinger equation to, multi-electron atoms presents a difficulty: the, Schrödinger equation cannot be solved exactly, for a multi-electron atom. This difficulty can, be overcome by using approximate methods., Such calculations with the aid of modern, computers show that orbitals in atoms other, than hydrogen do not differ in any radical way, from the hydrogen orbitals discussed above., The principal difference lies in the consequence, of increased nuclear charge. Because of this, all the orbitals are somewhat contracted., Further, as you shall see later (in subsections, 2.6.3 and 2.6.4), unlike orbitals of hydrogen, or hydrogen like species, whose energies, depend only on the quantum number n, the, energies of the orbitals in multi-electron atoms, depend on quantum numbers n and l., , 2. The existence of quantized electronic, energy levels is a direct result of the wave, like properties of electrons and are, allowed solutions of Schrödinger wave, equation., 3. Both the exact position and exact velocity, of an electron in an atom cannot be, determined simultaneously (Heisenberg, uncertainty principle). The path of an, electron in an atom therefore, can never, be determined or known accurately., That is why, as you shall see later on,, one talks of only probability of finding, the electron at different points in, an atom., 4. An atomic orbital is the wave function, ψ for an electron in an atom., Whenever an electron is described by a, wave function, we say that the electron, occupies that orbital. Since many such, wave functions are possible for an, electron, there are many atomic orbitals, in an atom. These “one electron orbital, wave functions” or orbitals form the, basis of the electronic structure of atoms., In each orbital, the electron has a, definite energy. An orbital cannot, contain more than two electrons. In a, multi-electron atom, the electrons are, filled in various orbitals in the order of, increasing energy. For each electron of, a multi-electron atom, there shall,, therefore, be an orbital wave function, characteristic of the orbital it occupies., All the information about the electron, in an atom is stored in its orbital wave, function ψ and quantum mechanics, makes it possible to extract this, information out of ψ., 5. The probability of finding an electron at, a point within an atom is proportional, to the square of the orbital wave function, 2, 2, i.e., |ψ| at that point. |ψ| is known, as probability density and is always, 2, positive. From the value of |ψ | at, different points within an atom, it is, possible to predict the region around, the nucleus where electron will most, probably be found., , Important Features of the Quantum, Mechanical Model of Atom, Quantum mechanical model of atom is the, picture of the structure of the atom, which, emerges from the application of the, Schrödinger equation to atoms. The, following are the important features of the, quantum-mechanical model of atom:, 1. The energy of electrons in atoms is, quantized (i.e., can only have certain, specific values), for example when, electrons are bound to the nucleus in, atoms., , 2.6.1 Orbitals and Quantum Numbers, A large number of orbitals are possible in an, atom. Qualitatively these orbitals can be, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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55, , STRUCTURE OF ATOM, , distinguished by their size, shape and, orientation. An orbital of smaller size means, there is more chance of finding the electron near, the nucleus. Similarly shape and orientation, mean that there is more probability of finding, the electron along certain directions than, along others. Atomic orbitals are precisely, distinguished by what are known as quantum, numbers. Each orbital is designated by three, quantum numbers labelled as n, l and m l., The principal quantum number ‘n’ is, a positive integer with value of n = 1,2,3......., The principal quantum number determines the, size and to large extent the energy of the, orbital. For hydrogen atom and hydrogen like, species (He+, Li2+, .... etc.) energy and size of, the orbital depends only on ‘n’., The principal quantum number also, identifies the shell. With the increase in the, value of ‘n’, the number of allowed orbital, increases and are given by ‘n 2’ All the, orbitals of a given value of ‘n’ constitute a, single shell of atom and are represented by, the following letters, n = 1 2 3 4 ............, Shell = K L M N ............, Size of an orbital increases with increase of, principal quantum number ‘n’. In other words, the electron will be located away from the, nucleus. Since energy is required in shifting, away the negatively charged electron from the, positively charged nucleus, the energy of the, orbital will increase with increase of n., Azimuthal quantum number. ‘l’ is also, known as orbital angular momentum or, subsidiary quantum number. It defines the, three-dimensional shape of the orbital. For a, given value of n, l can have n values ranging, from 0 to n – 1, that is, for a given value of n,, the possible value of l are : l = 0, 1, 2, .........., (n–1), For example, when n = 1, value of l is only, 0. For n = 2, the possible value of l can be 0, and 1. For n = 3, the possible l values are 0, 1, and 2., Each shell consists of one or more subshells or sub-levels. The number of sub-shells, in a principal shell is equal to the value of n., , For example in the first shell (n = 1), there is, only one sub-shell which corresponds to l = 0., There are two sub-shells (l = 0, 1) in the second, shell (n = 2), three (l = 0, 1, 2) in third shell (n =, 3) and so on. Each sub-shell is assigned an, azimuthal quantum number (l). Sub-shells, corresponding to different values of l are, represented by the following symbols., Value for l : 0 1 2 3 4 5 ............, notation for s p d f g h ............, sub-shell, Table 2.4 shows the permissible values of, ‘l ’ for a given principal quantum number and, the corresponding sub-shell notation., Table 2.4 Subshell Notations, , Magnetic orbital quantum number. ‘ml’, gives information about the spatial, orientation of the orbital with respect to, standard set of co-ordinate axis. For any, sub-shell (defined by ‘l’ value) 2l+1 values, of ml are possible and these values are given, by :, ml = – l, – (l –1), – (l – 2)... 0,1... (l – 2), (l –1), l, Thus for l = 0, the only permitted value of, ml = 0, [2(0)+1 = 1, one s orbital]. For l = 1, ml, can be –1, 0 and +1 [2(1)+1 = 3, three p, orbitals]. For l = 2, ml = –2, –1, 0, +1 and +2,, [2(2)+1 = 5, five d orbitals]. It should be noted, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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56, , CHEMISTRY, , that the values of ml are derived from l and that, the value of l are derived from n., Each orbital in an atom, therefore, is, defined by a set of values for n, l and ml. An, orbital described by the quantum numbers, n = 2, l = 1, ml = 0 is an orbital in the p sub-shell, of the second shell. The following chart gives, the relation between the subshell and the, number of orbitals associated with it., Value of l, , 0, , 1, , 2, , 3, , 4, , 5, , Subshell notation, , s, , p, , d, , f, , g, , h, , number of orbitals, , 1, , 3, , 5, , 7, , 9, , 11, , Electron spin ‘s’ : The three quantum, numbers labelling an atomic orbital can be, used equally well to define its energy, shape, and orientation. But all these quantum, numbers are not enough to explain the line, spectra observed in the case of multi-electron, atoms, that is, some of the lines actually occur, in doublets (two lines closely spaced), triplets, (three lines, closely spaced) etc. This suggests, the presence of a few more energy levels than, predicted by the three quantum numbers., In 1925, George Uhlenbeck and Samuel, Goudsmit proposed the presence of the fourth, quantum number known as the electron, spin quantum number (ms ). An electron, spins around its own axis, much in a similar, way as earth spins around its own axis while, revolving around the sun. In other words, an, , electron has, besides charge and mass,, intrinsic spin angular quantum number. Spin, angular momentum of the electron — a vector, quantity, can have two orientations relative to, the chosen axis. These two orientations are, distinguished by the spin quantum numbers, ms which can take the values of +½ or –½., These are called the two spin states of the, electron and are normally represented by two, arrows, ↑ (spin up) and ↓ (spin down). Two, electrons that have different ms values (one +½, and the other –½) are said to have opposite, spins. An orbital cannot hold more than two, electrons and these two electrons should have, opposite spins., To sum up, the four quantum numbers, provide the following information :, i), , n defines the shell, determines the size of, the orbital and also to a large extent the, energy of the orbital., , ii) There are n subshells in the nth shell. l, identifies the subshell and determines the, shape of the orbital (see section 2.6.2)., There are (2l+1) orbitals of each type in a, subshell, that is, one s orbital (l = 0), three, p orbitals (l = 1) and five d orbitals (l = 2), per subshell. To some extent l also, determines the energy of the orbital in a, multi-electron atom., iii) ml designates the orientation of the orbital., For a given value of l, ml has (2l+1) values,, , Orbit, orbital and its importance, Orbit and orbital are not synonymous. An orbit, as proposed by Bohr, is a circular path, around the nucleus in which an electron moves. A precise description of this path of the, electron is impossible according to Heisenberg uncertainty principle. Bohr orbits, therefore,, have no real meaning and their existence can never be demonstrated experimentally. An, atomic orbital, on the other hand, is a quantum mechanical concept and refers to the one, electron wave function ψ in an atom. It is characterized by three quantum numbers (n, l and, ml) and its value depends upon the coordinates of the electron. ψ has, by itself, no physical, 2, 2, meaning. It is the square of the wave function i.e., |ψ| which has a physical meaning. |ψ|, at any point in an atom gives the value of probability density at that point. Probability density, 2, 2, (|ψ| ) is the probability per unit volume and the product of |ψ| and a small volume (called a, volume element) yields the probability of finding the electron in that volume (the reason for, 2, specifying a small volume element is that |ψ| varies from one region to another in space but, its value can be assumed to be constant within a small volume element). The total probability, of finding the electron in a given volume can then be calculated by the sum of all the products, of | ψ| 2 and the corresponding volume elements. It is thus possible to get the probable, distribution of an electron in an orbital., Download all NCERT books PDFs from www.ncert.online, , 2019-20

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57, , STRUCTURE OF ATOM, , the same as the number of orbitals per, subshell. It means that the number of, orbitals is equal to the number of ways in, which they are oriented., iv) ms refers to orientation of the spin of the, electron., Problem 2.17, What is the total number of orbitals, associated with the principal quantum, number n = 3 ?, Solution, For n = 3, the possible values of l are 0, 1, and 2. Thus there is one 3s orbital, (n = 3, l = 0 and ml = 0); there are three 3p, orbitals (n = 3, l = 1 and ml = –1, 0, +1);, there are five 3d orbitals (n = 3, l = 2 and, ml = –2, –1, 0, +1+, +2)., Therefore, the total number of orbitals is, 1+3+5 = 9, The same value can also be obtained by, using the relation; number of orbitals, = n2, i.e. 32 = 9., , Fig. 2.12 The plots of (a) the orbital wave, function ψ (r ); (b) the variation of, 2, probability density ψ (r) as a function, of distance r of the electron from the, nucleus for 1s and 2s orbitals., , Problem 2.18, Using s, p, d, f notations, describe the, orbital with the following quantum, numbers, (a) n = 2, l = 1, (b) n = 4, l = 0, (c) n = 5,, l = 3, (d) n = 3, l = 2, Solution, a), b), c), d), , n, 2, 4, 5, 3, , l, 1, 0, 3, 2, , orbital, 2p, 4s, 5f, 3d, , 2.6.2 Shapes of Atomic Orbitals, The orbital wave function or ψ for an electron, in an atom has no physical meaning. It is, simply a mathematical function of the, coordinates of the electron. However, for, different orbitals the plots of corresponding, wave functions as a function of r (the distance, from the nucleus) are different. Fig. 2.12(a),, gives such plots for 1s (n = 1, l = 0) and 2s (n =, 2, l = 0) orbitals., , According to the German physicist, Max, Born, the square of the wave function, 2, (i.e.,ψ ) at a point gives the probability density, of the electron at that point. The variation of, ψ 2 as a function of r for 1s and 2s orbitals is, given in Fig. 2.12(b). Here again, you may note, that the curves for 1s and 2s orbitals are, different., It may be noted that for 1s orbital the, probability density is maximum at the nucleus, and it decreases sharply as we move away from, it. On the other hand, for 2s orbital the, probability density first decreases sharply to, zero and again starts increasing. After reaching, a small maxima it decreases again and, approaches zero as the value of r increases, further. The region where this probability, density function reduces to zero is called, nodal surfaces or simply nodes. In general,, it has been found that ns-orbital has (n – 1), nodes, that is, number of nodes increases with, increase of principal quantum number n. In, other words, number of nodes for 2s orbital is, one, two for 3s and so on., These probability density variation can be, visualised in terms of charge cloud diagrams, [Fig. 2.13(a)]. In these diagrams, the density, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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58, , CHEMISTRY, , of the dots in a region represents electron, probability density in that region., Boundary surface diagrams of constant, probability density for different orbitals give a, fairly good representation of the shapes of the, orbitals. In this representation, a boundary, surface or contour surface is drawn in space, for an orbital on which the value of probability, 2, density |ψ| is constant. In principle many, such boundary surfaces may be possible., However, for a given orbital, only that, boundary surface diagram of constant, probability density* is taken to be good, representation of the shape of the orbital which, encloses a region or volume in which the, probability of finding the electron is very high,, say, 90%. The boundary surface diagram for, 1s and 2s orbitals are given in Fig. 2.13(b)., One may ask a question : Why do we not draw, a boundary surface diagram, which bounds a, region in which the probability of finding the, electron is, 100 %? The answer to this question, 2, is that the probability density |ψ| has always, some value, howsoever small it may be, at any, finite distance from the nucleus. It is therefore,, not possible to draw a boundary surface, diagram of a rigid size in which the probability, of finding the electron is 100%. Boundary, surface diagram for a s orbital is actually a, sphere centred on the nucleus. In two, dimensions, this sphere looks like a circle. It, encloses a region in which probability of, finding the electron is about 90%., , Fig. 2.13 (a) Probability density plots of 1s and, 2s atomic orbitals. The density of the, dots represents the probability, density of finding the electron in that, region. (b) Boundary surface diagram, for 1s and 2s orbitals., , Thus, we see that 1s and 2s orbitals are, spherical in shape. In reality all the s-orbitals, are spherically symmetric, that is, the probability, of finding the electron at a given distance is equal, in all the directions. It is also observed that the, Fig. 2.14 Boundary surface diagrams of the, size of the s orbital increases with increase in n,, three 2p orbitals., that is, 4s > 3s > 2s > 1s and the electron is, located further away from the nucleus as the, diagrams are not spherical. Instead each, principal quantum number increases., p orbital consists of two sections called lobes, that are on either side of the plane that passes, Boundary surface diagrams for three 2p, through the nucleus. The probability density, orbitals (l = 1) are shown in Fig. 2.14. In these, function is zero on the plane where the two, diagrams, the nucleus is at the origin. Here,, lobes touch each other. The size, shape and, unlike s-orbitals, the boundary surface, 2, * If probability density |ψ| is constant on a given surface, |ψ| is also constant over the surface. The boundary, 2, surface for |ψ| and |ψ| are identical., Download all NCERT books PDFs from www.ncert.online, , 2019-20

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59, , STRUCTURE OF ATOM, , energy of the three orbitals are identical. They, differ however, in the way the lobes are, oriented. Since the lobes may be considered, to lie along the x, y or z axis, they are given the, designations 2px, 2py, and 2pz. It should be, understood, however, that there is no simple, relation between the values of ml (–1, 0 and, +1) and the x, y and z directions. For our, purpose, it is sufficient to remember that,, because there are three possible values of m l,, , there are, therefore, three p orbitals whose axes, are mutually perpendicular. Like s orbitals, p, orbitals increase in size and energy with, increase in the principal quantum number and, hence the order of the energy and size of, various p orbitals is 4p > 3p > 2p. Further, like, s orbitals, the probability density functions for, p-orbital also pass through value zero, besides, at zero and infinite distance, as the distance, from the nucleus increases. The number of, nodes are given by the n –2, that is number of, radial node is 1 for 3p orbital, two for 4p orbital, and so on., For l = 2, the orbital is known as d-orbital, and the minimum value of principal quantum, number (n) has to be 3. as the value of l cannot, be greater than n–1. There are five ml values (–, 2, –1, 0, +1 and +2) for l = 2 and thus there are, five d orbitals. The boundary surface diagram, of d orbitals are shown in Fig. 2.15., The five d-orbitals are designated as dxy, dyz,, dxz, dx2–y2 and dz2. The shapes of the first four dorbitals are similar to each other, where as that, of the fifth one, dz2, is different from others, but, all five 3d orbitals are equivalent in energy. The, d orbitals for which n is greater than 3 (4d,, 5d...) also have shapes similar to 3d orbital,, but differ in energy and size., Besides the radial nodes (i.e., probability, density function is zero), the probability, density functions for the np and nd orbitals, are zero at the plane (s), passing through the, nucleus (origin). For example, in case of pz, orbital, xy-plane is a nodal plane, in case of dxy, orbital, there are two nodal planes passing, through the origin and bisecting the xy plane, containing z-axis. These are called angular, nodes and number of angular nodes are given, by ‘l’, i.e., one angular node for p orbitals, two, angular nodes for ‘d’ orbitals and so on. The, total number of nodes are given by (n–1),, i.e., sum of l angular nodes and (n – l – 1), radial nodes., 2.6.3 Energies of Orbitals, , Fig. 2.15 Boundary surface diagrams of the five, 3d orbitals., , The energy of an electron in a hydrogen atom, is determined solely by the principal quantum, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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60, , CHEMISTRY, , number. Thus the energy of the orbitals in, hydrogen atom increases as follows :, 1s < 2s = 2p < 3s = 3p = 3d <4s = 4p = 4d, = 4f <, (2.23), and is depicted in Fig. 2.16. Although the, shapes of 2s and 2p orbitals are different, an, electron has the same energy when it is in the, 2s orbital as when it is present in 2p orbital., The orbitals having the same energy are called, degenerate. The 1s orbital in a hydrogen, atom, as said earlier, corresponds to the most, stable condition and is called the ground state, and an electron residing in this orbital is most, strongly held by the nucleus. An electron in, the 2s, 2p or higher orbitals in a hydrogen atom, is in excited state., , Fig. 2.16 Energy level diagrams for the few, electronic shells of (a) hydrogen atom, and (b) multi-electronic atoms. Note that, orbitals for the same value of principal, quantum number, have the same, energies even for different azimuthal, quantum number for hydrogen atom., In case of multi-electron atoms, orbitals, with same principal quantum number, possess different energies for different, azimuthal quantum numbers., , The energy of an electron in a multielectron atom, unlike that of the hydrogen, atom, depends not only on its principal, quantum number (shell), but also on its, azimuthal quantum number (subshell). That, is, for a given principal quantum number, s,, p, d, f ... all have different energies. Within a, given principal quantum number, the energy, of orbitals increases in the order s<p<d<f. For, higher energy levels, these differences are, sufficiently pronounced and straggering of, orbital energy may result, e.g., 4s<3d and, 6s<5d ; 4f<6p. The main reason for having, different energies of the subshells is the mutual, repulsion among the electrons in multielectron atoms. The only electrical interaction, present in hydrogen atom is the attraction, between the negatively charged electron and, the positively charged nucleus. In multielectron atoms, besides the presence of, attraction between the electron and nucleus,, there are repulsion terms between every, electron and other electrons present in the, atom. Thus the stability of an electron in a, multi-electron atom is because total attractive, interactions are more than the repulsive, interactions. In general, the repulsive, interaction of the electrons in the outer shell, with the electrons in the inner shell are more, important. On the other hand, the attractive, interactions of an electron increases with, increase of positive charge (Ze) on the nucleus., Due to the presence of electrons in the inner, shells, the electron in the outer shell will not, experience the full positive charge of the, nucleus (Ze). The effect will be lowered due to, the partial screening of positive charge on the, nucleus by the inner shell electrons. This is, known as the shielding of the outer shell, electrons from the nucleus by the inner, shell electrons, and the net positive charge, experienced by the outer electrons is known, as effective nuclear charge (Zeff e). Despite, the shielding of the outer electrons from the, nucleus by the inner shell electrons, the, attractive force experienced by the outer shell, electrons increases with increase of nuclear, charge. In other words, the energy of, interaction between, the nucleus and electron, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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61, , STRUCTURE OF ATOM, , (that is orbital energy) decreases (that is, more negative) with the increase of atomic, number (Z )., Both the attractive and repulsive, interactions depend upon the shell and shape, of the orbital in which the electron is present., For example electrons present in spherical, shaped, s orbital shields the outer electrons, from the nucleus more effectively as compared, to electrons present in p orbital. Similarly, electrons present in p orbitals shield the outer, electrons from the nucleus more than the, electrons present in d orbitals, even though all, these orbitals are present in the same shell., Further within a shell, due to spherical shape, of s orbital, the s orbital electron spends more, time close to the nucleus in comparison to p, orbital electron which spends more time in the, vicinity of nucleus in comparison to d orbital, electron. In other words, for a given shell, (principal quantum number), the Z eff, experienced by the electron decreases with, increase of azimuthal quantum number (l),, that is, the s orbital electron will be more tightly, bound to the nucleus than p orbital electron, which in turn will be better tightly bound than, the d orbital electron. The energy of electrons, in s orbital will be lower (more negative) than, that of p orbital electron which will have less, energy than that of d orbital electron and so, on. Since the extent of shielding from the, nucleus is different for electrons in different, orbitals, it leads to the splitting of energy levels, within the same shell (or same principal, quantum number), that is, energy of electron, in an orbital, as mentioned earlier, depends, upon the values of n and l. Mathematically,, the dependence of energies of the orbitals on n, and l are quite complicated but one simple rule, is that, the lower the value of (n + l) for an, orbital, the lower is its energy. If two, orbitals have the same value of (n + l), the, orbital with lower value of n will have the, lower energy. The Table 2.5 illustrates the (n, + l ) rule and Fig. 2.16 depicts the energy levels, of multi-electrons atoms. It may be noted that, different subshells of a particular shell have, different energies in case of multi–electrons, atoms. However, in hydrogen atom, these have, , Table 2.5 Arrangement of Orbitals with, Increasing Energy on the Basis of, (n+l ) Rule, , the same energy. Lastly it may be mentioned, here that energies of the orbitals in the, same subshell decrease with increase in, the atomic number (Zeff). For example, energy, of 2s orbital of hydrogen atom is greater than, that of 2s orbital of lithium and that of lithium, is greater than that of sodium and so on, that, is, E2s(H) > E2s(Li) > E2s(Na) > E2s(K)., 2.6.4 Filling of Orbitals in Atom, The filling of electrons into the orbitals of, different atoms takes place according to the, aufbau principle which is based on the Pauli’s, exclusion principle, the Hund’s rule of, maximum multiplicity and the relative, energies of the orbitals., Aufbau Principle, The word ‘aufbau’ in German means ‘building, up’. The building up of orbitals means the, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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62, , CHEMISTRY, , filling up of orbitals with electrons. The, principle states : In the ground state of the, atoms, the orbitals are filled in order of, their increasing energies. In other words,, electrons first occupy the lowest energy orbital, available to them and enter into higher energy, orbitals only after the lower energy orbitals are, filled. As you have learnt above, energy of a, given orbital depends upon effective nuclear, charge and different type of orbitals are affected, to different extent. Thus, there is no single, ordering of energies of orbitals which will be, universally correct for all atoms., However, following order of energies of the, orbitals is extremely useful:, 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 4f,, 5d, 6p, 7s..., The order may be remembered by using, the method given in Fig. 2.17. Starting from, , Fig.2.17 Order of filling of orbitals, , the top, the direction of the arrows gives the, order of filling of orbitals, that is starting from, right top to bottom left. With respect to, placement of outermost valence electrons, it is, remarkably accurate for all atoms. For, example, valence electron in potassium must, choose between 3d and 4s orbitals and as, predicted by this sequence, it is found in 4s, orbital. The above order should be assumed, to be a rough guide to the filling of energy, levels. In many cases, the orbitals are similar, in energy and small changes in atomic, structure may bring about a change in the, order of filling. Even then, the above series is a, useful guide to the building of the electronic, structure of an atom provided that it is, remembered that exceptions may occur., Pauli Exclusion Principle, The number of electrons to be filled in various, orbitals is restricted by the exclusion principle,, given by the Austrian scientist Wolfgang Pauli, (1926). According to this principle : No two, electrons in an atom can have the same, set of four quantum numbers. Pauli, exclusion principle can also be stated as : “Only, two electrons may exist in the same orbital, and these electrons must have opposite, spin.” This means that the two electrons can, have the same value of three quantum numbers, n, l and ml, but must have the opposite spin, quantum number. The restriction imposed by, Pauli’s exclusion principle on the number of, electrons in an orbital helps in calculating the, capacity of electrons to be present in any, subshell. For example, subshell 1s comprises, one orbital and thus the maximum number of, electrons present in 1s subshell can be two, in, p and d subshells, the maximum number of, electrons can be 6 and 10 and so on. This can, be summed up as : the maximum number, of electrons in the shell with principal, quantum number n is equal to 2n2., Hund’s Rule of Maximum Multiplicity, This rule deals with the filling of electrons into, the orbitals belonging to the same subshell, (that is, orbitals of equal energy, called, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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63, , STRUCTURE OF ATOM, , degenerate orbitals). It states : pairing of, electrons in the orbitals belonging to the, same subshell (p, d or f) does not take place, until each orbital belonging to that, subshell has got one electron each i.e., it, is singly occupied., Since there are three p, five d and seven f, orbitals, therefore, the pairing of electrons will, start in the p, d and f orbitals with the entry of, 4th, 6th and 8th electron, respectively. It has, been observed that half filled and fully filled, degenerate set of orbitals acquire extra stability, due to their symmetry (see Section, 2.6.7)., 2.6.5 Electronic Configuration of Atoms, The distribution of electrons into orbitals of an, atom is called its electronic configuration., If one keeps in mind the basic rules which, govern the filling of different atomic orbitals,, the electronic configurations of different atoms, can be written very easily., The electronic configuration of different, atoms can be represented in two ways. For, example :, (i) s a p bd c ...... notation, (ii) Orbital diagram, p, d, In the first notation, the subshell is, represented by the respective letter symbol and, the number of electrons present in the subshell, is depicted, as the super script, like a, b, c, ..., etc. The similar subshell represented for, different shells is differentiated by writing the, principal quantum number before the, respective subshell. In the second notation, each orbital of the subshell is represented by, a box and the electron is represented by an, arrow (↑) a positive spin or an arrow (↓) a, negative spin. The advantage of second notation, over the first is that it represents all the four, quantum numbers., The hydrogen atom has only one electron, which goes in the orbital with the lowest, energy, namely 1s. The electronic, configuration of the hydrogen atom is 1s1, meaning that it has one electron in the 1s, orbital. The second electron in helium (He) can, , also occupy the 1s orbital. Its configuration, is, therefore, 1s2. As mentioned above, the two, electrons differ from each other with opposite, spin, as can be seen from the orbital diagram., , The third electron of lithium (Li) is not, allowed in the 1s orbital because of Pauli, exclusion principle. It, therefore, takes the next, available choice, namely the 2s orbital. The, electronic configuration of Li is 1s22s1. The 2s, orbital can accommodate one more electron., The configuration of beryllium (Be) atom is,, therefore, 1s2 2s2 (see Table 2.6, page 66 for, the electronic configurations of elements)., In the next six elements—boron, (B, 1s22s22p1), carbon (C, 1s22s22p2), nitrogen, (N, 1s22s22p3), oxygen (O, 1s22s22p4), fluorine, (F, 1s22s22p5) and neon (Ne, 1s22s22p6), the 2p, orbitals get progressively filled. This process, is completed with the neon atom. The orbital, picture of these elements can be represented, as follows :, , s, , The electronic configuration of the elements, sodium (Na, 1s 2 2s 2 2p 6 3s 1 ) to argon, (Ar,1s22s22p63s23p6), follow exactly the same, pattern as the elements from lithium to neon, with the difference that the 3s and 3p orbitals, are getting filled now. This process can be, simplified if we represent the total number of, electrons in the first two shells by the name of, element neon (Ne). The electronic configuration, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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64, , CHEMISTRY, , of the elements from sodium to argon can be, written as (Na, [Ne]3s1) to (Ar, [Ne] 3s23p6). The, electrons in the completely filled shells are, known as core electrons and the electrons that, are added to the electronic shell with the, highest principal quantum number are called, valence electrons. For example, the electrons, in Ne are the core electrons and the electrons, from Na to Ar are the valence electrons. In, potassium (K) and calcium (Ca), the 4s orbital,, being lower in energy than the 3d orbitals, is, occupied by one and two electrons respectively., A new pattern is followed beginning with, scandium (Sc). The 3d orbital, being lower in, energy than the 4p orbital, is filled first., Consequently, in the next ten elements,, scandium (Sc), titanium (Ti), vanadium (V),, chromium (Cr), manganese (Mn), iron (Fe),, cobalt (Co), nickel (Ni), copper (Cu) and zinc, (Zn), the five 3d orbitals are progressively, occupied. We may be puzzled by the fact that, chromium and copper have five and ten, electrons in 3d orbitals rather than four and, nine as their position would have indicated with, two-electrons in the 4s orbital. The reason is, that fully filled orbitals and half-filled orbitals, have extra stability (that is, lower energy). Thus, p3, p6, d5, d10,f 7, f14 etc. configurations, which, are either half-filled or fully filled, are more, stable. Chromium and copper therefore adopt, the d 5 and d 10 configuration (Section, 2.6.7)[caution: exceptions do exist], With the saturation of the 3d orbitals, the, filling of the 4p orbital starts at gallium (Ga), and is complete at krypton (Kr). In the next, eighteen elements from rubidium (Rb) to xenon, (Xe), the pattern of filling the 5s, 4d and 5p, orbitals are similar to that of 4s, 3d and 4p, orbitals as discussed above. Then comes the, turn of the 6s orbital. In caesium (Cs) and the, barium (Ba), this orbital contains one and two, electrons, respectively. Then from lanthanum, (La) to mercury (Hg), the filling up of electrons, takes place in 4f and 5d orbitals. After this,, , filling of 6p, then 7s and finally 5f and 6d, orbitals takes place. The elements after, uranium (U) are all short-lived and all of them, are produced artificially. The electronic, configurations of the known elements (as, determined by spectroscopic methods) are, tabulated in Table 2.6 (page 66)., One may ask what is the utility of knowing, the electron configuration? The modern, approach to the chemistry, infact, depends, almost entirely on electronic distribution to, understand and explain chemical behaviour., For example, questions like why two or more, atoms combine to form molecules, why some, elements are metals while others are nonmetals, why elements like helium and argon, are not reactive but elements like the halogens, are reactive, find simple explanation from the, electronic configuration. These questions have, no answer in the Daltonian model of atom. A, detailed understanding of the electronic, structure of atom is, therefore, very essential, for getting an insight into the various aspects, of modern chemical knowledge., 2.6.6 Stability of Completely Filled and, Half Filled Subshells, The ground state electronic configuration of the, atom of an element always corresponds to the, state of the lowest total electronic energy. The, electronic configurations of most of the atoms, follow the basic rules given in Section 2.6.5., However, in certain elements such as Cu, or, Cr, where the two subshells (4s and 3d) differ, slightly in their energies, an electron shifts from, a subshell of lower energy (4s) to a subshell of, higher energy (3d), provided such a shift, results in all orbitals of the subshell of higher, energy getting either completely filled or half, filled. The valence electronic configurations of, Cr and Cu, therefore, are 3d5 4s1 and 3d10 4s1, respectively and not 3d4 4s2 and 3d9 4s2. It has, been found that there is extra stability, associated with these electronic configurations., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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65, , STRUCTURE OF ATOM, , Causes of Stability of Completely Filled and Half-filled Subshells, The completely filled and completely, half-filled subshells are stable due to, the following reasons:, 1. Symmetrical distribution of, electrons: It is well known that symmetry, leads to stability. The completely filled or, half filled subshells have symmetrical, distribution of electrons in them and are, therefore more stable. Electrons in the, same subshell (here 3d) have equal energy, but dif ferent spatial distribution., Consequently, their shielding of oneanother is relatively small and the, electrons are more strongly attracted by, the nucleus., 2. Exchange Energy : The stabilizing, effect arises whenever two or more, electrons with the same spin are present, in the degenerate orbitals of a subshell., These electrons tend to exchange their, positions and the energy released due to, this exchange is called exchange energy., The number of exchanges that can take, place is maximum when the subshell is, either half filled or completely filled (Fig., 2.18). As a result the exchange energy is, maximum and so is the stability., You may note that the exchange, energy is at the basis of Hund’s rule that, electrons which enter orbitals of equal, energy have parallel spins as far as, possible. In other words, the extra, stability of half-filled and completely filled, subshell is due to: (i) relatively small, shielding, (ii) smaller coulombic repulsion, energy, and (iii) larger exchange energy., Details about the exchange energy will be, dealt with in higher classes., , Fig. 2.18 Possible exchange for a d5, configuration, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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66, , CHEMISTRY, , Table 2.6 Electronic Configurations of the Elements, , * Elements with exceptional electronic configurations, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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67, , STRUCTURE OF ATOM, , ** Elements with atomic number 112 and above have been reported but not yet fully authenticated and named., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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68, , CHEMISTRY, , SUMMARY, Atoms are the building blocks of elements. They are the smallest parts of an element, that chemically react. The first atomic theory, proposed by John Dalton in 1808, regarded, atom as the ultimate indivisible particle of matter. Towards the end of the nineteenth, century, it was proved experimentally that atoms are divisible and consist of three, fundamental particles: electrons, protons and neutrons. The discovery of sub-atomic, particles led to the proposal of various atomic models to explain the structure of atom., Thomson in 1898 proposed that an atom consists of uniform sphere of positive electricity, with electrons embedded into it. This model in which mass of the atom is considered to be, evenly spread over the atom was proved wrong by Rutherford’s famous alpha-particle, scattering experiment in 1909. Rutherford concluded that atom is made of a tiny positively, charged nucleus, at its centre with electrons revolving around it in circular orbits., Rutherford model, which resembles the solar system, was no doubt an improvement over, Thomson model but it could not account for the stability of the atom i.e., why the electron, does not fall into the nucleus. Further, it was also silent about the electronic structure of, atoms i.e., about the distribution and relative energies of electrons around the nucleus., The difficulties of the Rutherford model were overcome by Niels Bohr in 1913 in his model, of the hydrogen atom. Bohr postulated that electron moves around the nucleus in circular, orbits. Only certain orbits can exist and each orbit corresponds to a specific energy. Bohr, calculated the energy of electron in various orbits and for each orbit predicted the distance, between the electron and nucleus. Bohr model, though offering a satisfactory model for, explaining the spectra of the hydrogen atom, could not explain the spectra of multi-electron, atoms. The reason for this was soon discovered. In Bohr model, an electron is regarded as, a charged particle moving in a well defined circular orbit about the nucleus. The wave, character of the electron is ignored in Bohr’s theory. An orbit is a clearly defined path and, this path can completely be defined only if both the exact position and the exact velocity of, the electron at the same time are known. This is not possible according to the Heisenberg, uncertainty principle. Bohr model of the hydrogen atom, therefore, not only ignores the, dual behaviour of electron but also contradicts Heisenberg uncertainty principle., Erwin Schrödinger, in 1926, proposed an equation called Schrödinger equation to describe, the electron distributions in space and the allowed energy levels in atoms. This equation, incorporates de Broglie’s concept of wave-particle duality and is consistent with Heisenberg, uncertainty principle. When Schrödinger equation is solved for the electron in a hydrogen, atom, the solution gives the possible energy states the electron can occupy [and the, corresponding wave function(s) (ψ) (which in fact are the mathematical functions) of the, electron associated with each energy state]. These quantized energy states and corresponding, wave functions which are characterized by a set of three quantum numbers (principal, quantum number n, azimuthal quantum number l and magnetic quantum number ml), arise as a natural consequence in the solution of the Schrödinger equation. The restrictions, on the values of these three quantum numbers also come naturally from this solution. The, quantum mechanical model of the hydrogen atom successfully predicts all aspects of the, hydrogen atom spectrum including some phenomena that could not be explained by the, Bohr model., According to the quantum mechanical model of the atom, the electron distribution of an, atom containing a number of electrons is divided into shells. The shells, in turn, are thought, to consist of one or more subshells and subshells are assumed to be composed of one or, more orbitals, which the electrons occupy. While for hydrogen and hydrogen like systems, +, 2+, (such as He , Li etc.) all the orbitals within a given shell have same energy, the energy of, the orbitals in a multi-electron atom depends upon the values of n and l: The lower the, value of (n + l ) for an orbital, the lower is its energy. If two orbitals have the same (n + l ), value, the orbital with lower value of n has the lower energy. In an atom many such orbitals, are possible and electrons are filled in those orbitals in order of increasing energy in, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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69, , STRUCTURE OF ATOM, , accordance with Pauli exclusion principle (no two electrons in an atom can have the, same set of four quantum numbers) and Hund’s rule of maximum multiplicity (pairing, of electrons in the orbitals belonging to the same subshell does not take place until, each orbital belonging to that subshell has got one electron each, i.e., is singly occupied)., This forms the basis of the electronic structure of atoms., , EXERCISES, 2.1, , (i), (ii), , 2.2, , (i), (ii), , Calculate the total number of electrons present in one mole of methane., Find (a) the total number and (b) the total mass of neutrons in 7 mg of 14C., –27, (Assume that mass of a neutron = 1.675 × 10 kg)., (iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3 at, STP., Will the answer change if the temperature and pressure are changed ?, , 2.3, , How many neutrons and protons are there in the following nuclei ?, 13, 6, , C,, , Calculate the number of electrons which will together weigh one gram., Calculate the mass and charge of one mole of electrons., , 16, 8, , O,, , 24, 12, , Mg,, , 56, 26, , Fe,, , 88, 38, , Sr, , 2.4, , Write the complete symbol for the atom with the given atomic number (Z) and, atomic mass (A), (i) Z = 17 , A = 35., (ii) Z = 92 , A = 233., (iii) Z = 4 , A = 9., , 2.5, , Yellow light emitted from a sodium lamp has a wavelength (λ) of 580 nm. Calculate, the frequency (ν) and wavenumber ( ν ) of the yellow light., , 2.6, , Find energy of each of the photons which, 15, (i) correspond to light of frequency 3×10 Hz., (ii) have wavelength of 0.50 Å., , 2.7, , Calculate the wavelength, frequency and wavenumber of a light wave whose period, is 2.0 × 10–10 s., , 2.8, , What is the number of photons of light with a wavelength of 4000 pm that provide, 1J of energy?, , 2.9, , A photon of wavelength 4 × 10 m strikes on metal surface, the work function of, the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic, energy of the emission, and (iii) the velocity of the photoelectron, –19, (1 eV= 1.6020 × 10 J)., , 2.10, , Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the, sodium atom. Calculate the ionisation energy of sodium in kJ mol–1., , 2.11, , A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57µm., Calculate the rate of emission of quanta per second., , 2.12, , Electrons are emitted with zero velocity from a metal surface when it is exposed to, radiation of wavelength 6800 Å. Calculate threshold frequency (ν0 ) and work function, (W0 ) of the metal., , 2.13, , What is the wavelength of light emitted when the electron in a hydrogen atom, undergoes transition from an energy level with n = 4 to an energy level with n = 2?, , –7, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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70, , CHEMISTRY, , 2.14, , How much energy is required to ionise a H atom if the electron occupies n = 5, orbit? Compare your answer with the ionization enthalpy of H atom ( energy required, to remove the electron from n =1 orbit)., , 2.15, , What is the maximum number of emission lines when the excited electron of a, H atom in n = 6 drops to the ground state?, , 2.16, , (i), , The energy associated with the first orbit in the hydrogen atom is, –18, –1, –2.18 × 10 J atom . What is the energy associated with the fifth orbit?, (ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom., , 2.17, , Calculate the wavenumber for the longest wavelength transition in the Balmer, series of atomic hydrogen., , 2.18, , What is the energy in joules, required to shift the electron of the hydrogen atom, from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the, light emitted when the electron returns to the ground state? The ground state, –11, electron energy is –2.18 × 10 ergs., , 2.19, , The electron energy in hydrogen atom is given by En = (–2.18 × 10 )/n J. Calculate, the energy required to remove an electron completely from the n = 2 orbit. What is, the longest wavelength of light in cm that can be used to cause this transition?, , 2.20, , Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 m s–1., , 2.21, , The mass of an electron is 9.1 × 10, wavelength., , 2.22, , Which of the following are isoelectronic species i.e., those having the same number, of electrons?, +, +, 2+, 2+, 2–, Na , K , Mg , Ca , S , Ar., , 2.23, , Write the electronic configurations of the following ions: (a) H – (b) Na+ (c) O2–, –, (d) F, (ii) What are the atomic numbers of elements whose outermost electrons are, 1, 3, 5, represented by (a) 3s (b) 2p and (c) 3p ?, , –18, , –31, , –25, , kg. If its K.E. is 3.0 × 10, , 2, , J, calculate its, , (i), , (iii) Which atoms are indicated by the following configurations ?, 1, 2, 3, 2, 1, (a) [He] 2s (b) [Ne] 3s 3p (c) [Ar] 4s 3d ., 2.24, , What is the lowest value of n that allows g orbitals to exist?, , 2.25, , An electron is in one of the 3d orbitals. Give the possible values of n, l and ml for, this electron., , 2.26, , An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the, number of protons and (ii) the electronic configuration of the element., , 2.27, 2.28, , Give the number of electrons in the species, (i) An atomic orbital has n = 3. What are the possible values of l and ml ?, (ii) List the quantum numbers (ml and l ) of electrons for 3d orbital., (iii) Which of the following orbitals are possible?, 1p, 2s, 2p and 3f, , 2.29, , Using s, p, d notations, describe the orbital with the following quantum numbers., (a) n=1, l=0; (b) n = 3; l=1 (c) n = 4; l =2; (d) n=4; l=3., , 2.30, , Explain, giving reasons, which of the following sets of quantum numbers are not, possible., (a), n = 0,, l = 0,, ml = 0,, ms = + ½, (b), , n = 1,, , l = 0,, , ml = 0,, , ms = – ½, , (c), , n = 1,, , l = 1,, , ml =, , 0,, , ms = + ½, , (d), , n = 2,, , l = 1,, , ml =, , 0,, , ms = – ½, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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71, , STRUCTURE OF ATOM, , (e), , n = 3,, , l = 3,, , ml = –3,, , ms = + ½, , (f), , n = 3,, , l = 1,, , ml =, , ms = + ½, , 0,, , 2.31, , How many electrons in an atom may have the following quantum numbers?, (a) n = 4, ms = – ½, (b) n = 3, l = 0, , 2.32, , Show that the circumference of the Bohr orbit for the hydrogen atom is an integral, multiple of the de Broglie wavelength associated with the electron revolving around, the orbit., , 2.33, , What transition in the hydrogen spectrum would have the same wavelength as the, +, Balmer transition n = 4 to n = 2 of He spectrum ?, , 2.34, , Calculate the energy required for the process, +, , 2+, , –, , He (g)  He (g) + e, –18, –1, The ionization energy for the H atom in the ground state is 2.18 × 10 J atom, 2.35, , If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms, which can be placed side by side in a straight line across length of scale of length, 20 cm long., , 2.36, , 2 ×108 atoms of carbon are arranged side by side. Calculate the radius of carbon, atom if the length of this arrangement is 2.4 cm., , 2.37, , The diameter of zinc atom is 2.6 Å.Calculate (a) radius of zinc atom in pm and (b), number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side, by side lengthwise., , 2.38, , A certain particle carries 2.5 × 10, of electrons present in it., , 2.39, , In Milikan’s experiment, static electric charge on the oil drops has been obtained, by shining X-rays. If the static electric charge on the oil drop is –1.282 × 10–18C,, calculate the number of electrons present on it., , 2.40, , In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum, etc. have been used to be bombarded by the α-particles. If the thin foil of light, atoms like aluminium etc. is used, what difference would be observed from the, above results ?, , 2.41, , Symbols, , 79, 35 Br, , and, , 79, , –16, , C of static electric charge. Calculate the number, , Br can be written, whereas symbols, , 35, 79 Br, , and, , 35, , Br are not, , acceptable. Answer briefly., 2.42, , An element with mass number 81 contains 31.7% more neutrons as compared to, protons. Assign the atomic symbol., , 2.43, , An ion with mass number 37 possesses one unit of negative charge. If the ion, conatins 11.1% more neutrons than the electrons, find the symbol of the ion., , 2.44, , An ion with mass number 56 contains 3 units of positive charge and 30.4% more, neutrons than electrons. Assign the symbol to this ion., , 2.45, , Arrange the following type of radiations in increasing order of frequency: (a) radiation, from microwave oven (b) amber light from traffic signal (c) radiation from FM radio, (d) cosmic rays from outer space and (e) X-rays., , 2.46, , Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of, 24, photons emitted is 5.6 × 10 , calculate the power of this laser., , 2.47, , Neon gas is generally used in the sign boards. If it emits strongly at 616 nm,, calculate (a) the frequency of emission, (b) distance traveled by this radiation in, 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of, energy., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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72, , CHEMISTRY, , 2.48, , In astronomical observations, signals observed from the distant stars are, generally weak. If the photon detector receives a total of 3.15 × 10–18 J from the, radiations of 600 nm, calculate the number of photons received by the detector., , 2.49, , Lifetimes of the molecules in the excited states are often measured by using, pulsed radiation source of duration nearly in the nano second range. If the, radiation source has the duration of 2 ns and the number of photons emitted, 15, during the pulse source is 2.5 × 10 , calculate the energy of the source., , 2.50, , The longest wavelength doublet absorption transition is observed at 589 and, 589.6 nm. Calcualte the frequency of each transition and energy difference, between two excited states., , 2.51, , The work function for caesium atom is 1.9 eV. Calculate (a) the threshold, wavelength and (b) the threshold frequency of the radiation. If the caesium, element is irradiated with a wavelength 500 nm, calculate the kinetic energy, and the velocity of the ejected photoelectron., , 2.52, , Following results are observed when sodium metal is irradiated with different, wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant., λ (nm), 500 450 400, v × 10–5 (cm s–1), 2.55 4.35 5.35, , 2.53, , The ejection of the photoelectron from the silver metal in the photoelectric effect, experiment can be stopped by applying the voltage of 0.35 V when the radiation, 256.7 nm is used. Calculate the work function for silver metal., , 2.54, , If the photon of the wavelength 150 pm strikes an atom and one of tis inner bound, 7, –1, electrons is ejected out with a velocity of 1.5 × 10 m s , calculate the energy with, which it is bound to the nucleus., , 2.55, , Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n, and can be represeted as v = 3.29 × 1015 (Hz) [ 1/32 – 1/n2], Calculate the value of n if the transition is observed at 1285 nm. Find the region of, the spectrum., , 2.56, , Calculate the wavelength for the emission transition if it starts from the orbit having, radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition, belongs and the region of the spectrum., , 2.57, , Dual behaviour of matter proposed by de Broglie led to the discovery of electron, microscope often used for the highly magnified images of biological molecules and, 6, other type of material. If the velocity of the electron in this microscope is 1.6 × 10, –1, ms , calculate de Broglie wavelength associated with this electron., , 2.58, , Similar to electron diffraction, neutron diffraction microscope is also used for the, determination of the structure of molecules. If the wavelength used here is 800 pm,, calculate the characteristic velocity associated with the neutron., , 2.59, , If the velocity of the electron in Bohr’s first orbit is 2.19 × 10 ms , calculate the, de Broglie wavelength associated with it., , 2.60, , The velocity associated with a proton moving in a potential difference of 1000 V is, 5, –1, 4.37 × 10 ms . If the hockey ball of mass 0.1 kg is moving with this velocity,, calcualte the wavelength associated with this velocity., , 2.61, , If the position of the electron is measured within an accuracy of + 0.002 nm, calculate, the uncertainty in the momentum of the electron. Suppose the momentum of the, electron is h/4πm × 0.05 nm, is there any problem in defining this value., , 2.62, , The quantum numbers of six electrons are given below. Arrange them in order of, increasing energies. If any of these combination(s) has/have the same energy lists:, 1. n = 4, l = 2, ml = –2 , ms = –1/2, 2. n = 3, l = 2, ml = 1 , ms = +1/2, , 6, , Download all NCERT books PDFs from www.ncert.online, , 2019-20, , –1

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73, , STRUCTURE OF ATOM, , 3., 4., 5., 6., , n, n, n, n, , = 4, l = 1, ml = 0 , ms = +1/2, = 3, l = 2, ml = –2 , ms = –1/2, = 3, l = 1, ml = –1 , ms = +1/2, = 4, l = 1, ml = 0 , ms = +1/2, , 2.63, , The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital,, 6 electrons in 3p orbital and 5 electron in 4p orbital. Which of these electron, experiences the lowest effective nuclear charge ?, , 2.64, , Among the following pairs of orbitals which orbital will experience the larger effective, nuclear charge? (i) 2s and 3s, (ii) 4d and 4f, (iii) 3d and 3p., , 2.65, , The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will, experience more effective nuclear charge from the nucleus ?, , 2.66, , Indicate the number of unpaired electrons in : (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr., , 2.67, , (a) How many subshells are associated with n = 4 ? (b) How many electrons will, be present in the subshells having ms value of –1/2 for n = 4 ?, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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74, , CHEMISTRY, , UNIT 3, , CLASSIFICATION OF ELEMENTS AND, PERIODICITY IN PROPERTIES, , After studying this Unit, you will be, able to, , • appreciate how the concept of, , •, •, , •, •, •, •, •, •, , grouping elements in accordance to, their properties led to the, development of Periodic Table., understand the Periodic Law;, understand the significance of, atomic number and electronic, configuration as the basis for, periodic classification;, name, the, elements, with, Z >100 according to IUPAC, nomenclature;, classify elements into s, p, d, f, blocks and learn their main, characteristics;, recognise the periodic trends in, physical and chemical properties of, elements;, compare the reactivity of elements, and correlate it with their, occurrence in nature;, explain the relationship between, ionization enthalpy and metallic, character;, use, scientific, vocabulary, appropriately to communicate ideas, related to certain important, properties of atoms e.g., atomic/, ionic radii, ionization enthalpy,, electron, gain, enthalpy,, electronegativity, valence of, elements., , The Periodic Table is arguably the most important concept in, chemistry, both in principle and in practice. It is the everyday, support for students, it suggests new avenues of research to, professionals, and it provides a succinct organization of the, whole of chemistry. It is a remarkable demonstration of the, fact that the chemical elements are not a random cluster of, entities but instead display trends and lie together in families., An awareness of the Periodic Table is essential to anyone who, wishes to disentangle the world and see how it is built up, from the fundamental building blocks of the chemistry, the, chemical elements., Glenn T. Seaborg, , In this Unit, we will study the historical development of the, Periodic Table as it stands today and the Modern Periodic, Law. We will also learn how the periodic classification, follows as a logical consequence of the electronic, configuration of atoms. Finally, we shall examine some of, the periodic trends in the physical and chemical properties, of the elements., 3.1 WHY DO WE NEED TO CLASSIFY ELEMENTS ?, We know by now that the elements are the basic units of all, types of matter. In 1800, only 31 elements were known. By, 1865, the number of identified elements had more than, doubled to 63. At present 114 elements are known. Of, them, the recently discovered elements are man-made., Efforts to synthesise new elements are continuing. With, such a large number of elements it is very difficult to study, individually the chemistry of all these elements and their, innumerable compounds individually. To ease out this, problem, scientists searched for a systematic way to, organise their knowledge by classifying the elements. Not, only that it would rationalize known chemical facts about, elements, but even predict new ones for undertaking further, study., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , 3.2 GENESIS OF PERIODIC, CLASSIFICATION, Classification of elements into groups and, development of Periodic Law and Periodic, Table are the consequences of systematising, the knowledge gained by a number of scientists, through their observations and experiments., The German chemist, Johann Dobereiner in, early 1800’s was the first to consider the idea, of trends among properties of elements. By, 1829 he noted a similarity among the physical, and chemical properties of several groups of, three elements (Triads). In each case, he, noticed that the middle element of each of the, Triads had an atomic weight about half way, between the atomic weights of the other two, (Table 3.1). Also the properties of the middle, element were in between those of the other, , 75, , chemist, John Alexander Newlands in 1865, profounded the Law of Octaves. He arranged, the elements in increasing order of their atomic, weights and noted that every eighth element, had properties similar to the first element, (Table 3.2). The relationship was just like every, eighth note that resembles the first in octaves, of music. Newlands’s Law of Octaves seemed, to be true only for elements up to calcium., Although his idea was not widely accepted at, that time, he, for his work, was later awarded, Davy Medal in 1887 by the Royal Society,, London., The Periodic Law, as we know it today owes, its development to the Russian chemist, Dmitri, Mendeleev (1834-1907) and the German, chemist, Lothar Meyer (1830-1895). Working, independently, both the chemists in 1869, , Table 3.1 Dobereiner’s Triads, Element, , Atomic, weight, , Element, , Atomic, weight, , Element, , Atomic, weight, , Li, , 7, , Ca, , 40, , Cl, , 35.5, , Na, , 23, , Sr, , 88, , Br, , 80, , K, , 39, , Ba, , 137, , I, , 127, , two members. Since Dobereiner’s relationship,, referred to as the Law of Triads, seemed to, work only for a few elements, it was dismissed, as coincidence. The next reported attempt to, classify elements was made by a French, geologist, A.E.B. de Chancourtois in 1862. He, arranged the then known elements in order of, increasing atomic weights and made a, cylindrical table of elements to display the, periodic recurrence of properties. This also did, not attract much attention. The English, , proposed that on arranging elements in the, increasing order of their atomic weights,, similarities appear in physical and chemical, properties at regular intervals. Lothar Meyer, plotted the physical properties such as atomic, volume, melting point and boiling point, against atomic weight and obtained a, periodically repeated pattern. Unlike, Newlands, Lothar Meyer observed a change in, length of that repeating pattern. By 1868,, Lothar Meyer had developed a table of the, , Table 3.2 Newlands’ Octaves, Element, , Li, , Be, , B, , C, , N, , O, , F, , At. wt., , 7, , 9, , 11, , 12, , 14, , 16, , 19, , Element, , Na, , Mg, , Al, , Si, , P, , S, , Cl, , At. wt., , 23, , 24, , 27, , 29, , 31, , 32, , 35.5, , Element, , K, , Ca, , At. wt., , 39, , 40, Download all NCERT books PDFs from www.ncert.online, , 2019-20

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76, , CHEMISTRY, , elements that closely resembles the Modern, Periodic Table. However, his work was not, published until after the work of Dmitri, Mendeleev, the scientist who is generally, credited with the development of the Modern, Periodic Table., While Dobereiner initiated the study of, periodic relationship, it was Mendeleev who, was responsible for publishing the Periodic, Law for the first time. It states as follows :, The properties of the elements are a, periodic function of their atomic, weights., Mendeleev arranged elements in horizontal, rows and vertical columns of a table in order, of their increasing atomic weights in such a, way that the elements with similar properties, occupied the same vertical column or group., Mendeleev’s system of classifying elements was, more elaborate than that of Lothar Meyer’s., He fully recognized the significance of, periodicity and used broader range of physical, and chemical properties to classify the, elements. In particular, Mendeleev relied on, the similarities in the empirical formulas and, properties of the compounds formed by the, elements. He realized that some of the elements, did not fit in with his scheme of classification, if the order of atomic weight was strictly, followed. He ignored the order of atomic, Table 3.3, , weights, thinking that the atomic, measurements might be incorrect, and placed, the elements with similar properties together., For example, iodine with lower atomic weight, than that of tellurium (Group VI) was placed, in Group VII along with fluorine, chlorine,, bromine because of similarities in properties, (Fig. 3.1). At the same time, keeping his, primary aim of arranging the elements of, similar properties in the same group, he, proposed that some of the elements were still, undiscovered and, therefore, left several gaps, in the table. For example, both gallium and, germanium were unknown at the time, Mendeleev published his Periodic Table. He left, the gap under aluminium and a gap under, silicon, and called these elements EkaAluminium and Eka-Silicon. Mendeleev, predicted not only the existence of gallium and, germanium, but also described some of their, general physical properties. These elements, were discovered later. Some of the properties, predicted by Mendeleev for these elements and, those found experimentally are listed in, Table 3.3., The boldness of Mendeleev’s quantitative, predictions and their eventual success made, him and his Periodic Table famous., Mendeleev’s Periodic Table published in 1905, is shown in Fig. 3.1., , Mendeleev’s Predictions for the Elements Eka-aluminium (Gallium) and, Eka-silicon (Germanium), , Property, , Eka-aluminium, (predicted), , Gallium, (found), , Eka-silicon, (predicted), , Germanium, (found), , Atomic weight, , 68, , 70, , 72, , 72.6, , Density / (g/cm3), , 5.9, , 5.94, , 5.5, , 5.36, , Melting point /K, , Low, , 302.93, , High, , 1231, , Formula of oxide, , E2O3, , Ga2O3, , EO2, , GeO2, , Formula of chloride, , ECl3, , GaCl3, , ECl4, , GeCl4, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , PERIODIC SYSTEM OF THE ELEMENTS IN GROUPS AND SERIES, , Fig. 3.1 Mendeleev’s Periodic Table published earlier, 77, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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78, , CHEMISTRY, , 3.3 MODERN PERIODIC LAW AND THE, PRESENT FORM OF THE PERIODIC, TABLE, We must bear in mind that when Mendeleev, developed his Periodic Table, chemists knew, nothing about the internal structure of atom., However, the beginning of the 20th century, witnessed profound developments in theories, about sub-atomic particles. In 1913, the, English physicist, Henry Moseley observed, regularities in the characteristic X-ray spectra, of the elements. A plot of ν (where ν is, frequency of X-rays emitted) against atomic, number (Z ) gave a straight line and not the, plot of, , ν vs atomic mass. He thereby showed, , that the atomic number is a more fundamental, property of an element than its atomic mass., Mendeleev’s Periodic Law was, therefore,, accordingly modified. This is known as the, Modern Periodic Law and can be stated as :, The physical and chemical properties, of the elements are periodic functions, of their atomic numbers., The Periodic Law revealed important, analogies among the 94 naturally occurring, elements (neptunium and plutonium like, actinium and protoactinium are also found in, pitch blende – an ore of uranium). It stimulated, renewed interest in Inorganic Chemistry and, has carried into the present with the creation, of artificially produced short-lived elements., You may recall that the atomic number is, equal to the nuclear charge (i.e., number of, protons) or the number of electrons in a neutral, atom. It is then easy to visualize the significance, of quantum numbers and electronic, configurations in periodicity of elements. In, fact, it is now recognized that the Periodic Law, is essentially the consequence of the periodic, variation in electronic configurations, which, indeed determine the physical and chemical, properties of elements and their compounds., , *, , Numerous forms of Periodic Table have, been devised from time to time. Some forms, emphasise chemical reactions and valence,, whereas others stress the electronic, configuration of elements. A modern version,, the so-called “long form” of the Periodic Table, of the elements (Fig. 3.2), is the most convenient, and widely used. The horizontal rows (which, Mendeleev called series) are called periods and, the vertical columns, groups. Elements having, similar outer electronic configurations in their, atoms are arranged in vertical columns,, referred to as groups or families. According, to the recommendation of International Union, of Pure and Applied Chemistry (IUPAC), the, groups are numbered from 1 to 18 replacing, the older notation of groups IA … VIIA, VIII, IB, … VIIB and 0., There are altogether seven periods. The, period number corresponds to the highest, principal quantum number (n) of the elements, in the period. The first period contains 2, elements. The subsequent periods consists of, 8, 8, 18, 18 and 32 elements, respectively. The, seventh period is incomplete and like the sixth, period would have a theoretical maximum (on, the basis of quantum numbers) of 32 elements., In this form of the Periodic Table, 14 elements, of both sixth and seventh periods (lanthanoids, and actinoids, respectively) are placed in, separate panels at the bottom*., 3.4 NOMENCLATURE OF ELEMENTS WITH, ATOMIC NUMBERS > 100, The naming of the new elements had been, traditionally the privilege of the discoverer (or, discoverers) and the suggested name was, ratified by the IUPAC. In recent years this has, led to some controversy. The new elements with, very high atomic numbers are so unstable that, only minute quantities, sometimes only a few, atoms of them are obtained. Their synthesis, and characterisation, therefore, require highly, , Glenn T. Seaborg’s work in the middle of the 20th century starting with the discovery of plutonium in 1940, followed by, those of all the transuranium elements from 94 to 102 led to reconfiguration of the periodic table placing the actinoids, below the lanthanoids. In 1951, Seaborg was awarded the Nobel Prize in chemistry for his work. Element 106 has been, named Seaborgium (Sg) in his honour., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , Download all NCERT books PDFs from www.ncert.online, , 2019-20, , 79, , Fig. 3.2 Long form of the Periodic Table of the Elements with their atomic numbers and ground state outer, electronic configurations. The groups are numbered 1-18 in accordance with the 1984 IUPAC, recommendations. This notation replaces the old numbering scheme of IA–VIIA, VIII, IB–VIIB and 0 for, the elements.

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CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , Thus, the new element first gets a, temporary name, with symbol consisting of, three letters. Later permanent name and, symbol are given by a vote of IUPAC, representatives from each country. The, permanent name might reflect the country (or, state of the country) in which the element was, discovered, or pay tribute to a notable, scientist. As of now, elements with atomic, numbers up to 118 have been discovered., Official names of all elements have been, announced by IUPAC., Problem 3.1, What would be the IUPAC name and, symbol for the element with atomic, number 120?, Solution, From Table 3.4, the roots for 1, 2 and 0, are un, bi and nil, respectively. Hence, the, symbol and the name respectively are Ubn, and unbinilium., 3.5 ELECTRONIC CONFIGURATIONS OF, ELEMENTS AND THE PERIODIC, TABLE, In the preceding unit we have learnt that an, electron in an atom is characterised by a set of, four quantum numbers, and the principal, quantum number (n ) defines the main energy, level known as shell. We have also studied, about the filling of electrons into different, subshells, also referred to as orbitals (s, p, d,, f ) in an atom. The distribution of electrons into, orbitals of an atom is called its electronic, configuration. An element’s location in the, Periodic Table reflects the quantum numbers, of the last orbital filled. In this section we will, observe a direct connection between the, electronic configurations of the elements and, the long form of the Periodic Table., (a) Electronic Configurations in Periods, The period indicates the value of n for the, outermost or valence shell. In other words,, successive period in the Periodic Table is, associated with the filling of the next higher, principal energy level (n = 1, n = 2, etc.). It can, , 81, , be readily seen that the number of elements in, each period is twice the number of atomic, orbitals available in the energy level that is, being filled. The first period (n = 1) starts with, the filling of the lowest level (1s) and therefore, has two elements — hydrogen (ls1) and helium, (ls2) when the first shell (K) is completed. The, second period (n = 2) starts with lithium and, the third electron enters the 2s orbital. The next, element, beryllium has four electrons and has, 2, 2, the electronic configuration 1s 2s . Starting, from the next element boron, the 2p orbitals, are filled with electrons when the L shell is, 2, 6, completed at neon (2s 2p ). Thus there are, 8 elements in the second period. The third, period (n = 3) begins at sodium, and the added, electron enters a 3s orbital. Successive filling, of 3s and 3p orbitals gives rise to the third, period of 8 elements from sodium to argon. The, fourth period (n = 4) starts at potassium, and, the added electrons fill up the 4s orbital. Now, you may note that before the 4p orbital is filled,, filling up of 3d orbitals becomes energetically, favourable and we come across the so called, 3d transition series of elements. This starts, from scandium (Z = 21) which has the electronic, 1, 2, configuration 3d 4s . The 3d orbitals are filled, at zinc (Z=30) with electronic configuration, 10, 2, 3d 4s . The fourth period ends at krypton, with the filling up of the 4p orbitals. Altogether, we have 18 elements in this fourth period. The, fifth period (n = 5) beginning with rubidium is, similar to the fourth period and contains the, 4d transition series starting at yttrium, (Z = 39). This period ends at xenon with the, filling up of the 5p orbitals. The sixth period, (n = 6) contains 32 elements and successive, electrons enter 6s, 4f, 5d and 6p orbitals, in, the order — filling up of the 4f orbitals begins, with cerium (Z = 58) and ends at lutetium, (Z = 71) to give the 4f-inner transition series, which is called the lanthanoid series. The, seventh period (n = 7) is similar to the sixth, period with the successive filling up of the 7s,, 5f, 6d and 7p orbitals and includes most of, the man-made radioactive elements. This, period will end at the element with atomic, number 118 which would belong to the noble, gas family. Filling up of the 5f orbitals after, actinium (Z = 89) gives the 5f-inner transition, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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82, , CHEMISTRY, , series known as the actinoid series. The 4fand 5f-inner transition series of elements are, placed separately in the Periodic Table to, maintain its structure and to preserve the, principle of classification by keeping elements, with similar properties in a single column., Problem 3.2, How would you justify the presence of 18, elements in the 5th period of the Periodic, Table?, Solution, When n = 5, l = 0, 1, 2, 3. The order in, which the energy of the available orbitals, 4d, 5s and 5p increases is 5s < 4d < 5p., The total number of orbitals available are, 9. The maximum number of electrons that, can be accommodated is 18; and therefore, 18 elements are there in the 5th period., (b) Groupwise Electronic Configurations, Elements in the same vertical column or group, have similar valence shell electronic, configurations, the same number of electrons, in the outer orbitals, and similar properties., For example, the Group 1 elements (alkali, 1, metals) all have ns valence shell electronic, configuration as shown below., Atomic number, , Symbol, , theoretical foundation for the periodic, classification. The elements in a vertical column, of the Periodic Table constitute a group or, family and exhibit similar chemical behaviour., This similarity arises because these elements, have the same number and same distribution, of electrons in their outermost orbitals. We can, classify the elements into four blocks viz.,, s-block, p-block, d-block and f-block, depending on the type of atomic orbitals that, are being filled with electrons. This is illustrated, in Fig. 3.3. We notice two exceptions to this, categorisation. Strictly, helium belongs to the, s-block but its positioning in the p-block along, with other group 18 elements is justified, because it has a completely filled valence shell, (1s 2) and as a result, exhibits properties, characteristic of other noble gases. The other, exception is hydrogen. It has only one, s-electron and hence can be placed in group 1, (alkali metals). It can also gain an electron to, achieve a noble gas arrangement and hence it, can behave similar to a group 17 (halogen, family) elements. Because it is a special case,, we shall place hydrogen separately at the top, of the Periodic Table as shown in Fig. 3.2 and, Fig. 3.3. We will briefly discuss the salient, features of the four types of elements marked in, , Electronic configuration, , 3, , Li, , 1s22s1 (or) [He]2s1, , 11, , Na, , 1s22s22p63s1 (or) [Ne]3s1, , 19, , K, , 37, , Rb, , 1s22s22p63s23p63d104s24p65s1 (or) [Kr]5s1, , 55, , Cs, , 1s22s22p63s23p63d104s24p64d105s25p66s1 (or) [Xe]6s1, , 87, , Fr, , [Rn]7s1, , 1s22s22p63s23p64s1 (or) [Ar]4s1, , Thus it can be seen that the properties of, an element have periodic dependence upon its, atomic number and not on relative atomic, mass., 3.6, ELECTRONIC CONFIGURATIONS, AND TYPES OF ELEMENTS:, s-, p-, d-, f- BLOCKS, The aufbau (build up) principle and the, electronic configuration of atoms provide a, , the Periodic Table. More about these elements, will be discussed later. During the description, of their features certain terminology has been, used which has been classified in section 3.7., 3.6.1 The s-Block Elements, The elements of Group 1 (alkali metals) and, Group 2 (alkaline earth metals) which have ns1, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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Download all NCERT books PDFs from www.ncert.online, , 2019-20, , Ts, , Og, , 83, , Fig. 3.3 The types of elements in the Periodic Table based on the orbitals that, are being filled. Also shown is the broad division of elements into METALS, (, ) , NON-METALS (, ) and METALLOIDS (, )., , Mc, , CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , Nh

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84, , CHEMISTRY, , and ns2 outermost electronic configuration, belong to the s-Block Elements. They are all, reactive metals with low ionization enthalpies., They lose the outermost electron(s) readily to, form 1+ ion (in the case of alkali metals) or 2+, ion (in the case of alkaline earth metals). The, metallic character and the reactivity increase, as we go down the group. Because of high, reactivity they are never found pure in nature., The compounds of the s-block elements, with, the exception of those of lithium and beryllium, are predominantly ionic., 3.6.2 The p-Block Elements, The p-Block Elements comprise those, belonging to Group 13 to 18 and these, together with the s-Block Elements are called, the Representative Elements or Main Group, Elements. The outermost electronic, configuration varies from ns2np1 to ns2np6 in, each period. At the end of each period is a noble, gas element with a closed valence shell ns2np6, configuration. All the orbitals in the valence, shell of the noble gases are completely filled, by electrons and it is very difficult to alter this, stable arrangement by the addition or removal, of electrons. The noble gases thus exhibit very, low chemical reactivity. Preceding the noble gas, family are two chemically important groups of, non-metals. They are the halogens (Group 17), and the chalcogens (Group 16). These two, groups of elements have highly negative, electron gain enthalpies and readily add one, or two electrons respectively to attain the stable, noble gas configuration. The non-metallic, character increases as we move from left to right, across a period and metallic character increases, as we go down the group., , used as catalysts. However, Zn, Cd and Hg, which have the electronic configuration,, (n-1) d10ns2 do not show most of the properties, of transition elements. In a way, transition, metals form a bridge between the chemically, active metals of s-block elements and the less, active elements of Groups 13 and 14 and thus, take their familiar name “Transition, Elements”., 3.6.4 The f-Block Elements, (Inner-Transition Elements), The two rows of elements at the bottom of the, Periodic Table, called the Lanthanoids,, Ce(Z = 58) – Lu(Z = 71) and Actinoids,, Th(Z = 90) – Lr (Z = 103) are characterised by, the outer electronic configuration (n-2)f1-14, (n-1)d0–1ns2. The last electron added to each, element is filled in f- orbital. These two series, of elements are hence called the InnerTransition Elements (f-Block Elements)., They are all metals. Within each series, the, properties of the elements are quite similar. The, chemistry of the early actinoids is more, complicated than the corresponding, lanthanoids, due to the large number of, oxidation states possible for these actinoid, elements. Actinoid elements are radioactive., Many of the actinoid elements have been made, only in nanogram quantities or even less by, nuclear reactions and their chemistry is not, fully studied. The elements after uranium are, called Transuranium Elements., , 3.6.3 The d-Block Elements (Transition, Elements), These are the elements of Group 3 to 12 in the, centre of the Periodic Table. These are, characterised by the filling of inner d orbitals, by electrons and are therefore referred to as, d-Block Elements. These elements have the, general outer electronic configuration, (n-1)d1-10ns0-2 . They are all metals. They mostly, form coloured ions, exhibit variable valence, (oxidation states), paramagnetism and oftenly, , Problem 3.3, The elements Z = 117 and 120 have not, yet been discovered. In which family /, group would you place these elements, and also give the electronic configuration, in each case., Solution, We see from Fig. 3.2, that element with Z, = 117, would belong to the halogen family, (Group 17) and the electronic, configuration, would, be, [Rn], 5f146d107s27p5. The element with Z = 120,, will be placed in Group 2 (alkaline earth, metals), and will have the electronic, configuration [Uuo]8s2., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , 3.6.5 Metals, Non-metals and Metalloids, In addition to displaying the classification of, elements into s-, p-, d-, and f-blocks, Fig. 3.3, shows another broad classification of elements, based on their properties. The elements can, be divided into Metals and Non-Metals. Metals, comprise more than 78% of all known elements, and appear on the left side of the Periodic, Table. Metals are usually solids at room, temperature [mercury is an exception; gallium, and caesium also have very low melting points, (303K and 302K, respectively)]. Metals usually, have high melting and boiling points. They are, good conductors of heat and electricity. They, are malleable (can be flattened into thin sheets, by hammering) and ductile (can be drawn into, wires). In contrast, non-metals are located at, the top right hand side of the Periodic Table., In fact, in a horizontal row, the property of, elements change from metallic on the left to, non-metallic on the right. Non-metals are, usually solids or gases at room temperature, with low melting and boiling points (boron and, carbon are exceptions). They are poor, conductors of heat and electricity. Most nonmetallic solids are brittle and are neither, malleable nor ductile. The elements become, more metallic as we go down a group; the nonmetallic character increases as one goes from, left to right across the Periodic Table. The, change from metallic to non-metallic character, is not abrupt as shown by the thick zig-zag, line in Fig. 3.3. The elements (e.g., silicon,, germanium, arsenic, antimony and tellurium), bordering this line and running diagonally, across the Periodic Table show properties that, are characteristic of both metals and nonmetals. These elements are called Semi-metals, or Metalloids., Problem 3.4, Considering the atomic number and, position in the periodic table, arrange the, following elements in the increasing order, of metallic character : Si, Be, Mg, Na, P., Solution, Metallic character increases down a group, and decreases along a period as we move, , 85, , from left to right. Hence the order of, increasing metallic character is: P < Si <, Be < Mg < Na., 3.7 PERIODIC TRENDS IN PROPERTIES, OF ELEMENTS, There are many observable patterns in the, physical and chemical properties of elements, as we descend in a group or move across a, period in the Periodic Table. For example,, within a period, chemical reactivity tends to be, high in Group 1 metals, lower in elements, towards the middle of the table, and increases, to a maximum in the Group 17 non-metals., Likewise within a group of representative, metals (say alkali metals) reactivity increases, on moving down the group, whereas within a, group of non-metals (say halogens), reactivity, decreases down the group. But why do the, properties of elements follow these trends? And, how can we explain periodicity? To answer, these questions, we must look into the theories, of atomic structure and properties of the atom., In this section we shall discuss the periodic, trends in certain physical and chemical, properties and try to explain them in terms of, number of electrons and energy levels., 3.7.1 Trends in Physical Properties, There are numerous physical properties of, elements such as melting and boiling points,, heats of fusion and vaporization, energy of, atomization, etc. which show periodic, variations. However, we shall discuss the, periodic trends with respect to atomic and ionic, radii, ionization enthalpy, electron gain, enthalpy and electronegativity., (a) Atomic Radius, You can very well imagine that finding the size, of an atom is a lot more complicated than, measuring the radius of a ball. Do you know, why? Firstly, because the size of an atom, (~ 1.2 Å i.e., 1.2 × 10–10 m in radius) is very, small. Secondly, since the electron cloud, surrounding the atom does not have a sharp, boundary, the determination of the atomic size, cannot be precise. In other words, there is no, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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86, , CHEMISTRY, , practical way by which the size of an individual, atom can be measured. However, an estimate, of the atomic size can be made by knowing the, distance between the atoms in the combined, state. One practical approach to estimate the, size of an atom of a non-metallic element is to, measure the distance between two atoms when, they are bound together by a single bond in a, covalent molecule and from this value, the, “Covalent Radius” of the element can be, calculated. For example, the bond distance in, the chlorine molecule (Cl 2) is 198 pm and half, this distance (99 pm), is taken as the atomic, radius of chlorine. For metals, we define the, term “Metallic Radius” which is taken as half, the internuclear distance separating the metal, cores in the metallic crystal. For example, the, distance between two adjacent copper atoms, in solid copper is 256 pm; hence the metallic, radius of copper is assigned a value of 128 pm., For simplicity, in this book, we use the term, Atomic Radius to refer to both covalent or, metallic radius depending on whether the, element is a non-metal or a metal. Atomic radii, can be measured by X-ray or other, spectroscopic methods., The atomic radii of a few elements are listed, in Table 3.6 . Two trends are obvious. We can, , explain these trends in terms of nuclear charge, and energy level. The atomic size generally, decreases across a period as illustrated in, Fig. 3.4(a) for the elements of the second period., It is because within the period the outer, electrons are in the same valence shell and the, effective nuclear charge increases as the atomic, number increases resulting in the increased, attraction of electrons to the nucleus. Within a, family or vertical column of the periodic table,, the atomic radius increases regularly with, atomic number as illustrated in Fig. 3.4(b). For, alkali metals and halogens, as we descend the, groups, the principal quantum number (n), increases and the valence electrons are farther, from the nucleus. This happens because the, inner energy levels are filled with electrons,, which serve to shield the outer electrons from, the pull of the nucleus. Consequently the size, of the atom increases as reflected in the atomic, radii., Note that the atomic radii of noble gases, are not considered here. Being monoatomic,, their (non-bonded radii) values are very large., In fact radii of noble gases should be compared, not with the covalent radii but with the van der, Waals radii of other elements., , Table 3.6(a) Atomic Radii/pm Across the Periods, Atom (Period II), , Li, , Be, , B, , C, , N, , O, , F, , Atomic radius, , 152, , 111, , 88, , 77, , 74, , 66, , 64, , Atom (Period III), , Na, , Mg, , Al, , Si, , P, , S, , Cl, , Atomic radius, , 186, , 160, , 143, , 117, , 110, , 104, , 99, , Table 3.6(b) Atomic Radii/pm Down a Family, Atom, (Group I), , Atomic, Radius, , Atom, (Group 17), , Atomic, Radius, , Li, , 152, , F, , 64, , Na, , 186, , Cl, , 99, , K, , 231, , Br, , 114, , Rb, , 244, , I, , 133, , Cs, , 262, , At, , 140, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , Fig. 3.4 (a), , 87, , Fig. 3.4 (b) Variation of atomic radius with, atomic number for alkali metals, and halogens, , Variation of atomic radius with, atomic number across the second, period, , (b) Ionic Radius, The removal of an electron from an atom results, in the formation of a cation, whereas gain of, an electron leads to an anion. The ionic radii, can be estimated by measuring the distances, between cations and anions in ionic crystals., In general, the ionic radii of elements exhibit, the same trend as the atomic radii. A cation is, smaller than its parent atom because it has, fewer electrons while its nuclear charge remains, the same. The size of an anion will be larger, than that of the parent atom because the, addition of one or more electrons would result, in increased repulsion among the electrons, and a decrease in effective nuclear charge. For, –, example, the ionic radius of fluoride ion (F ) is, 136 pm whereas the atomic radius of fluorine, is only 64 pm. On the other hand, the atomic, radius of sodium is 186 pm compared to the, +, ionic radius of 95 pm for Na ., When we find some atoms and ions which, contain the same number of electrons, we call, them isoelectronic species*. For example,, 2–, –, +, 2+, O , F , Na and Mg have the same number of, electrons (10). Their radii would be different, because of their different nuclear charges. The, cation with the greater positive charge will have, a smaller radius because of the greater, , attraction of the electrons to the nucleus. Anion, with the greater negative charge will have the, larger radius. In this case, the net repulsion of, the electrons will outweigh the nuclear charge, and the ion will expand in size., Problem 3.5, Which of the following species will have, the largest and the smallest size?, Mg, Mg2+, Al, Al3+., Solution, Atomic radii decrease across a period., Cations are smaller than their parent, atoms. Among isoelectronic species, the, one with the larger positive nuclear charge, will have a smaller radius., Hence the largest species is Mg; the, smallest one is Al3+., (c) Ionization Enthalpy, A quantitative measure of the tendency of an, element to lose electron is given by its, Ionization Enthalpy. It represents the energy, required to remove an electron from an isolated, gaseous atom (X) in its ground state. In other, words, the first ionization enthalpy for an, , * Two or more species with same number of atoms, same number of valence electrons and same structure,, regardless of the nature of elements involved., Download all NCERT books PDFs from www.ncert.online, , 2019-20

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88, , CHEMISTRY, , element X is the enthalpy change (∆i H) for the, reaction depicted in equation 3.1., +, , –, , X(g) → X (g) + e, , (3.1), , The ionization enthalpy is expressed in, units of kJ mol–1. We can define the second, ionization enthalpy as the energy required to, remove the second most loosely bound, electron; it is the energy required to carry out, the reaction shown in equation 3.2., +, , 2+, , X (g) → X (g) + e, , –, , (3.2), , Energy is always required to remove, electrons from an atom and hence ionization, enthalpies are always positive. The second, ionization enthalpy will be higher than the first, ionization enthalpy because it is more difficult, to remove an electron from a positively charged, ion than from a neutral atom. In the same way, the third ionization enthalpy will be higher than, the second and so on. The term “ionization, enthalpy”, if not qualified, is taken as the first, ionization enthalpy., The first ionization enthalpies of elements, having atomic numbers up to 60 are plotted, in Fig. 3.5. The periodicity of the graph is quite, striking. You will find maxima at the noble gases, which have closed electron shells and very, stable electron configurations. On the other, hand, minima occur at the alkali metals and, their low ionization enthalpies can be correlated, , Fig. 3.5 Variation of first ionization enthalpies, (∆i H) with atomic number for elements, with Z = 1 to 60, , with their high reactivity. In addition, you will, notice two trends the first ionization enthalpy, generally increases as we go across a period, and decreases as we descend in a group. These, trends are illustrated in Figs. 3.6(a) and 3.6(b), respectively for the elements of the second, period and the first group of the periodic table., You will appreciate that the ionization enthalpy, and atomic radius are closely related, properties. To understand these trends, we, have to consider two factors : (i) the attraction, of electrons towards the nucleus, and (ii) the, repulsion of electrons from each other. The, effective nuclear charge experienced by a, valence electron in an atom will be less than, , 3.6 (b), 3.6 (a), Fig. 3.6(a) First ionization enthalpies (∆ i H ) of elements of the second period as a, function of atomic number (Z) and Fig. 3.6(b) ∆ i H of alkali metals as a function of Z., Download all NCERT books PDFs from www.ncert.online, , 2019-20

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CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , the actual charge on the nucleus because of, “shielding” or “screening” of the valence, electron from the nucleus by the intervening, core electrons. For example, the 2s electron in, lithium is shielded from the nucleus by the, inner core of 1s electrons. As a result, the, valence electron experiences a net positive, charge which is less than the actual charge of, +3. In general, shielding is effective when the, orbitals in the inner shells are completely filled., This situation occurs in the case of alkali metals, which have single outermost ns-electron, preceded by a noble gas electronic, configuration., When we move from lithium to fluorine, across the second period, successive electrons, are added to orbitals in the same principal, quantum level and the shielding of the nuclear, charge by the inner core of electrons does not, increase very much to compensate for the, increased attraction of the electron to the, nucleus. Thus, across a period, increasing, nuclear charge outweighs the shielding., Consequently, the outermost electrons are held, more and more tightly and the ionization, enthalpy increases across a period. As we go, down a group, the outermost electron being, increasingly farther from the nucleus, there is, an increased shielding of the nuclear charge, by the electrons in the inner levels. In this case,, increase in shielding outweighs the increasing, nuclear charge and the removal of the, outermost electron requires less energy down, a group., From Fig. 3.6(a), you will also notice that, the first ionization enthalpy of boron (Z = 5) is, slightly less than that of beryllium (Z = 4) even, though the former has a greater nuclear charge., When we consider the same principal quantum, level, an s-electron is attracted to the nucleus, more than a p-electron. In beryllium, the, electron removed during the ionization is an, s-electron whereas the electron removed during, ionization of boron is a p-electron. The, penetration of a 2s-electron to the nucleus is, more than that of a 2p-electron; hence the 2p, electron of boron is more shielded from the, nucleus by the inner core of electrons than the, 2s electrons of beryllium. Therefore, it is easier, , 89, , to remove the 2p-electron from boron compared, to the removal of a 2s- electron from beryllium., Thus, boron has a smaller first ionization, enthalpy than beryllium. Another “anomaly”, is the smaller first ionization enthalpy of oxygen, compared to nitrogen. This arises because in, the nitrogen atom, three 2p-electrons reside in, different atomic orbitals (Hund’s rule) whereas, in the oxygen atom, two of the four 2p-electrons, must occupy the same 2p-orbital resulting in, an increased electron-electron repulsion., Consequently, it is easier to remove the fourth, 2p-electron from oxygen than it is, to remove, one of the three 2p-electrons from nitrogen., Problem 3.6, The first ionization enthalpy (∆i H ) values, of the third period elements, Na, Mg and, Si are respectively 496, 737 and 786 kJ, mol–1. Predict whether the first ∆i H value, for Al will be more close to 575 or 760 kJ, mol–1 ? Justify your answer., Solution, –1, It will be more close to 575 kJ mol . The, value for Al should be lower than that of, Mg because of effective shielding of 3p, electrons from the nucleus by, 3s-electrons., (d) Electron Gain Enthalpy, When an electron is added to a neutral gaseous, atom (X) to convert it into a negative ion, the, enthalpy change accompanying the process is, defined as the Electron Gain Enthalpy (∆egH)., Electron gain enthalpy provides a measure of, the ease with which an atom adds an electron, to form anion as represented by equation 3.3., –, , X(g) + e – → X (g), (3.3), Depending on the element, the process of, adding an electron to the atom can be either, endothermic or exothermic. For many elements, energy is released when an electron is added, to the atom and the electron gain enthalpy is, negative. For example, group 17 elements (the, halogens) have very high negative electron gain, enthalpies because they can attain stable noble, gas electronic configurations by picking up an, electron. On the other hand, noble gases have, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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90, , CHEMISTRY, , Table 3.7 Electron Gain Enthalpies* / (kJ mol–1) of Some Main Group Elements, Group 1, , ∆ eg H, , Group 16, , ∆ eg H, , Group 17, , H, , – 73, , Li, , – 60, , O, , – 141, , F, , Na, , – 53, , S, , – 200, , K, , – 48, , Se, , Rb, , – 47, , Cs, , – 46, , Group 0, , ∆ eg H, , He, , + 48, , – 328, , Ne, , + 116, , Cl, , – 349, , Ar, , + 96, , – 195, , Br, , – 325, , Kr, , + 96, , Te, , – 190, , I, , – 295, , Xe, , + 77, , Po, , – 174, , At, , – 270, , Rn, , + 68, , large positive electron gain enthalpies because, the electron has to enter the next higher, principal quantum level leading to a very, unstable electronic configuration. It may be, noted that electron gain enthalpies have large, negative values toward the upper right of the, periodic table preceding the noble gases., The variation in electron gain enthalpies of, elements is less systematic than for ionization, enthalpies. As a general rule, electron gain, enthalpy becomes more negative with increase, in the atomic number across a period. The, effective nuclear charge increases from left to, right across a period and consequently it will, be easier to add an electron to a smaller atom, since the added electron on an average would, be closer to the positively charged nucleus. We, should also expect electron gain enthalpy to, become less negative as we go down a group, because the size of the atom increases and the, added electron would be farther from the, nucleus. This is generally the case (Table 3.7)., However, electron gain enthalpy of O or F is, less negative than that of the succeeding, element. This is because when an electron is, added to O or F, the added electron goes to the, smaller n = 2 quantum level and suffers, significant repulsion from the other electrons, present in this level. For the n = 3 quantum, level (S or Cl), the added electron occupies a, larger region of space and the electron-electron, repulsion is much less., *, , ∆ eg H, , Problem 3.7, Which of the following will have the most, negative electron gain enthalpy and which, the least negative?, P, S, Cl, F., Explain your answer., Solution, Electron gain enthalpy generally becomes, more negative across a period as we move, from left to right. Within a group, electron, gain enthalpy becomes less negative down, a group. However, adding an electron to, the 2p-orbital leads to greater repulsion, than adding an electron to the larger, 3p-orbital. Hence the element with most, negative electron gain enthalpy is chlorine;, the one with the least negative electron, gain enthalpy is phosphorus., (e) Electronegativity, A qualitative measure of the ability of an atom, in a chemical compound to attract shared, electrons to itself is called electronegativity., Unlike ionization enthalpy and electron gain, enthalpy, it is not a measureable quantity., However, a number of numerical scales of, electronegativity of elements viz., Pauling scale,, Mulliken-Jaffe scale, Allred-Rochow scale have, been developed. The one which is the most, , In many books, the negative of the enthalpy change for the process depicted in equation 3.3 is defined as the, ELECTRON AFFINITY (Ae ) of the atom under consideration. If energy is released when an electron is added to an atom,, the electron affinity is taken as positive, contrary to thermodynamic convention. If energy has to be supplied to add an, electron to an atom, then the electron affinity of the atom is assigned a negative sign. However, electron affinity is, defined as absolute zero and, therefore at any other temperature (T) heat capacities of the reactants and the products, have to be taken into account in ∆egH = –Ae – 5/2 RT., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , 91, , widely used is the Pauling scale. Linus Pauling,, electrons and the nucleus increases as the, an American scientist, in 1922 assigned, atomic radius decreases in a period. The, arbitrarily a value of 4.0 to fluorine, the element, electronegativity also increases. On the same, considered to have the greatest ability to attract, account electronegativity values decrease with, electrons. Approximate values for the, the increase in atomic radii down a group. The, electronegativity of a few elements are given in, trend is similar to that of ionization enthalpy., Table 3.8(a), Knowing the relationship between, The electronegativity of any given element, electronegativity and atomic radius, can you, is not constant; it varies depending on the, now visualise the relationship between, element to which it is bound. Though it is not, electronegativity and non-metallic properties?, a measurable quantity, it does provide a means, of predicting the nature of force, that holds a pair of atoms together, – a relationship that you will, explore later., Electronegativity generally, increases across a period from left, to right (say from lithium to, fluorine) and decrease down a group, (say from fluorine to astatine) in, the periodic table. How can these, trends be explained? Can the, electronegativity be related to, atomic radii, which tend to, decrease across each period from, left to right, but increase down, each group ? The attraction, between the outer (or valence), Fig. 3.7 The periodic trends of elements in the periodic table, Table 3.8(a) Electronegativity Values (on Pauling scale) Across the Periods, Atom (Period II), , Li, , Be, , B, , C, , N, , O, , F, , Electronegativity, , 1.0, , 1.5, , 2.0, , 2.5, , 3.0, , 3.5, , 4.0, , Atom (Period III), , Na, , Mg, , Al, , Si, , P, , S, , Cl, , Electronegativity, , 0.9, , 1.2, , 1.5, , 1.8, , 2.1, , 2.5, , 3.0, , Table 3.8(b) Electronegativity Values (on Pauling scale) Down a Family, Atom, (Group I), , Electronegativity, Value, , Atom, (Group 17), , Electronegativity, Value, , Li, , 1.0, , F, , 4.0, , Na, , 0.9, , Cl, , 3.0, , K, , 0.8, , Br, , 2.8, , Rb, , 0.8, , I, , 2.5, , Cs, , 0.7, , At, , 2.2, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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92, , CHEMISTRY, , Non-metallic elements have strong tendency, to gain electrons. Therefore, electronegativity, is directly related to that non-metallic, properties of elements. It can be further, extended to say that the electronegativity is, inversely related to the metallic properties of, elements. Thus, the increase in, electronegativities across a period is, accompanied by an increase in non-metallic, properties (or decrease in metallic properties), of elements. Similarly, the decrease in, electronegativity down a group is accompanied, by a decrease in non-metallic properties (or, increase in metallic properties) of elements., All these periodic trends are summarised, in figure 3.7., 3.7.2 Periodic Trends in Chemical, Properties, Most of the trends in chemical properties of, elements, such as diagonal relationships, inert, pair effect, effects of lanthanoid contraction etc., will be dealt with along the discussion of each, group in later units. In this section we shall, study the periodicity of the valence state shown, by elements and the anomalous properties of, the second period elements (from lithium to, fluorine)., (a) Periodicity of Valence or Oxidation, States, The valence is the most characteristic property, of the elements and can be understood in terms, of their electronic configurations. The valence, of representative elements is usually (though, not necessarily) equal to the number of, electrons in the outermost orbitals and / or, equal to eight minus the number of outermost, electrons as shown below., Nowadays the term oxidation state is, frequently used for valence. Consider the two, oxygen containing compounds: OF2 and Na2O., The order of electronegativity of the three, elements involved in these compounds is F >, O > Na. Each of the atoms of fluorine, with outer, , electronic configuration 2s22p5, shares one, electron with oxygen in the OF2 molecule. Being, highest electronegative element, fluorine is, given oxidation state –1. Since there are two, fluorine atoms in this molecule, oxygen with, 2, outer electronic configuration 2s 2p4 shares, two electrons with fluorine atoms and thereby, exhibits oxidation state +2. In Na2O, oxygen, being more electronegative accepts two, electrons, one from each of the two sodium, atoms and, thus, shows oxidation state –2. On, the other hand sodium with electronic, configuration 3s1 loses one electron to oxygen, and is given oxidation state +1. Thus, the, oxidation state of an element in a particular, compound can be defined as the charge, acquired by its atom on the basis of, electronegative consideration from other atoms, in the molecule., Problem 3.8, Using the Periodic Table, predict the, formulas of compounds which might be, formed by the following pairs of elements;, (a) silicon and bromine (b) aluminium and, sulphur., Solution, (a) Silicon is group 14 element with a, valence of 4; bromine belongs to the, halogen family with a valence of 1., Hence the formula of the compound, formed would be SiBr4., (b) Aluminium belongs to group 13 with, a valence of 3; sulphur belongs to, group 16 elements with a valence of, 2. Hence, the formula of the compound, formed would be Al2S3., Some periodic trends observed in the, valence of elements (hydrides and oxides) are, shown in Table 3.9. Other such periodic trends, which occur in the chemical behaviour of the, elements are discussed elsewhere in this book., , Group, , 1, , 2, , 13, , 14, , 15, , 16, , 17, , 18, , Number of valence, electron, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , Valence, , 1, , 2, , 3, , 4, , 3,5, , 2,6, , 1,7, , 0,8, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , 93, , Table 3.9 Periodic Trends in Valence of Elements as shown by the Formulas, of Their Compounds, Group, , 1, , Formula, , LiH, , of hydride, , NaH, , 2, , CaH2, , 13, , 14, , B2H6, , CH4, , NH3, , H2O, , HF, , AlH3, , SiH4, , PH3, , H2S, , HCl, , GeH4, , AsH3, , H2Se, , HBr, , SnH4, , SbH3, , H2Te, , HI, , KH, , 15, , 16, , 17, , Formula, , Li2O, , MgO, , B2O3, , CO2, , N2O3, N2O5, , of oxide, , Na2O, , CaO, , Al2O3, , SiO2, , P4O6, P4O10, , SO3, , Cl2 O7, , K2O, , SrO, , Ga2O3, , GeO2, , As2O3, As2O5, , SeO3, , –, , BaO, , In2O3, , SnO2, , Sb2O3, Sb2O5, , TeO3, , –, , PbO2, , Bi2O3, , There are many elements which exhibit variable, valence. This is particularly characteristic of, transition elements and actinoids, which we, shall study later., (b) Anomalous Properties of Second Period, Elements, The first element of each of the groups 1, (lithium) and 2 (beryllium) and groups 13-17, (boron to fluorine) differs in many respects from, the other members of their respective group., For example, lithium unlike other alkali metals,, and beryllium unlike other alkaline earth, metals, form compounds with pronounced, covalent character; the other members of these, groups predominantly form ionic compounds., In fact the behaviour of lithium and beryllium, is more similar with the second element of the, Property, , −, , [BF4 ], , +, , Ionic radius M / pm, , Li, , Be, , B, , 152, , 111, , 88, , Na, , Mg, , Al, , 186, , 160, , Li, , Be, , 76, , 31, , Na, , Mg, , 102, , 72, , –, , following group i.e., magnesium and, aluminium, respectively. This sort of similarity, is commonly referred to as diagonal, relationship in the periodic properties., What are the reasons for the different, chemical behaviour of the first member of a, group of elements in the s- and p-blocks, compared to that of the subsequent members, in the same group? The anomalous behaviour, is attributed to their small size, large charge/, radius ratio and high electronegativity of the, elements. In addition, the first member of group, has only four valence orbitals (2s and 2p), available for bonding, whereas the second, member of the groups have nine valence, orbitals (3s, 3p, 3d). As a consequence of this,, the maximum covalency of the first member of, each group is 4 (e.g., boron can only form, , Element, , Metallic radius M/ pm, , –, , –, , 143, , , whereas the other members, , of the groups can expand their, valence shell to accommodate, more than four pairs of electrons, e.g., aluminium, , 6, , forms)., , Furthermore, the first member of, p-block elements displays greater, ability to form pπ – pπ multiple bonds, to itself (e.g., C = C, C ≡ C, N = N,, N ≡ Ν) and to other second period, elements (e.g., C = O, C = N, C ≡ N,, N = O) compared to subsequent, members of the same group., , Download all NCERT books PDFs from www.ncert.online, , 2019-20, , 3−, , [ AlF ]

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94, , CHEMISTRY, , Problem 3.9, Are the oxidation state and covalency of, 2+, Al in [AlCl(H2O)5] same ?, Solution, No. The oxidation state of Al is +3 and the, covalency is 6., 3.7.3 Periodic Trends and Chemical, Reactivity, We have observed the periodic trends in certain, fundamental properties such as atomic and, ionic radii, ionization enthalpy, electron gain, enthalpy and valence. We know by now that, the periodicity is related to electronic, configuration. That is, all chemical and, physical properties are a manifestation of the, electronic configuration of elements. We shall, now try to explore relationships between these, fundamental properties of elements with their, chemical reactivity., The atomic and ionic radii, as we know,, generally decrease in a period from left to right., As a consequence, the ionization enthalpies, generally increase (with some exceptions as, outlined in section 3.7.1(a)) and electron gain, enthalpies become more negative across a, period. In other words, the ionization enthalpy, of the extreme left element in a period is the, least and the electron gain enthalpy of the, element on the extreme right is the highest, negative (note : noble gases having completely, filled shells have rather positive electron gain, enthalpy values). This results into high, chemical reactivity at the two extremes and the, lowest in the centre. Thus, the maximum, chemical reactivity at the extreme left (among, alkali metals) is exhibited by the loss of an, electron leading to the formation of a cation, and at the extreme right (among halogens), shown by the gain of an electron forming an, anion. This property can be related with the, reducing and oxidizing behaviour of the, , elements which you will learn later. However,, here it can be directly related to the metallic, and non-metallic character of elements. Thus,, the metallic character of an element, which is, highest at the extremely left decreases and the, non-metallic character increases while moving, from left to right across the period. The, chemical reactivity of an element can be best, shown by its reactions with oxygen and, halogens. Here, we shall consider the reaction, of the elements with oxygen only. Elements on, two extremes of a period easily combine with, oxygen to form oxides. The normal oxide, formed by the element on extreme left is the, most basic (e.g., Na2O), whereas that formed, by the element on extreme right is the most, acidic (e.g., Cl2O7). Oxides of elements in the, centre are amphoteric (e.g., Al2O3, As2O3) or, neutral (e.g., CO, NO, N2O). Amphoteric oxides, behave as acidic with bases and as basic with, acids, whereas neutral oxides have no acidic, or basic properties., Problem 3.10, Show by a chemical reaction with water, that Na2O is a basic oxide and Cl2O7 is an, acidic oxide., Solution, Na2O with water forms a strong base, whereas Cl2O7 forms strong acid., Na2O + H2O → 2NaOH, Cl2O7 + H2O → 2HClO4, Their basic or acidic nature can be, qualitatively tested with litmus paper., Among transition metals (3d series), the change, in atomic radii is much smaller as compared, to those of representative elements across the, period. The change in atomic radii is still, smaller among inner -transition metals, (4f series). The ionization enthalpies are, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , intermediate between those of s- and p-blocks., As a consequence, they are less electropositive, than group 1 and 2 metals., In a group, the increase in atomic and ionic, radii with increase in atomic number generally, results in a gradual decrease in ionization, enthalpies and a regular decrease (with, exception in some third period elements as, shown in section 3.7.1(d)) in electron gain, , 95, , enthalpies in the case of main group elements., Thus, the metallic character increases down, the group and non-metallic character, decreases. This trend can be related with their, reducing and oxidizing property which you, will learn later. In the case of transition, elements, however, a reverse trend is observed., This can be explained in terms of atomic size, and ionization enthalpy., , SUMMARY, In this Unit, you have studied the development of the Periodic Law and the Periodic, Table. Mendeleev’s Periodic Table was based on atomic masses. Modern Periodic Table, arranges the elements in the order of their atomic numbers in seven horizontal rows, (periods) and eighteen vertical columns (groups or families). Atomic numbers in a period, are consecutive, whereas in a group they increase in a pattern. Elements of the same, group have similar valence shell electronic configuration and, therefore, exhibit similar, chemical properties. However, the elements of the same period have incrementally, increasing number of electrons from left to right, and, therefore, have different valencies., Four types of elements can be recognized in the periodic table on the basis of their, electronic configurations. These are s-block, p-block, d-block and f-block elements., Hydrogen with one electron in the 1s orbital occupies a unique position in the periodic, table. Metals comprise more than seventy eight per cent of the known elements. Nonmetals, which are located at the top of the periodic table, are less than twenty in number., Elements which lie at the border line between metals and non-metals (e.g., Si, Ge, As), are called metalloids or semi-metals. Metallic character increases with increasing atomic, number in a group whereas decreases from left to right in a period. The physical and, chemical properties of elements vary periodically with their atomic numbers., Periodic trends are observed in atomic sizes, ionization enthalpies, electron, gain enthalpies, electronegativity and valence. The atomic radii decrease while going, from left to right in a period and increase with atomic number in a group. Ionization, enthalpies generally increase across a period and decrease down a group. Electronegativity, also shows a similar trend. Electron gain enthalpies, in general, become more negative, across a period and less negative down a group. There is some periodicity in valence, for, example, among representative elements, the valence is either equal to the number of, electrons in the outermost orbitals or eight minus this number. Chemical reactivity is, highest at the two extremes of a period and is lowest in the centre. The reactivity on the, left extreme of a period is because of the ease of electron loss (or low ionization enthalpy)., Highly reactive elements do not occur in nature in free state; they usually occur in the, combined form. Oxides formed of the elements on the left are basic and of the elements, on the right are acidic in nature. Oxides of elements in the centre are amphoteric or, neutral., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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96, , CHEMISTRY, , EXERCISES, 3.1, , What is the basic theme of organisation in the periodic table?, , 3.2, , Which important property did Mendeleev use to classify the elements in his periodic, table and did he stick to that?, , 3.3, , What is the basic difference in approach between the Mendeleev’s Periodic Law, and the Modern Periodic Law?, , 3.4, , On the basis of quantum numbers, justify that the sixth period of the periodic, table should have 32 elements., , 3.5, , In terms of period and group where would you locate the element with Z =114?, , 3.6, , Write the atomic number of the element present in the third period and seventeenth, group of the periodic table., , 3.7, , Which element do you think would have been named by, (i), , Lawrence Berkeley Laboratory, , (ii) Seaborg’s group?, 3.8, , Why do elements in the same group have similar physical and chemical properties?, , 3.9, , What does atomic radius and ionic radius really mean to you?, , 3.10, , How do atomic radius vary in a period and in a group? How do you explain the, variation?, , 3.11, , What do you understand by isoelectronic species? Name a species that will be, isoelectronic with each of the following atoms or ions., (i) F, , 3.12, , –, , (ii), , Ar, , (iii) Mg, , 2+, , (iv), , +, , Rb, , Consider the following species :, 3–, , 2–, , –, , +, , 2+, , N , O , F , Na , Mg, , 3+, , and Al, , (a) What is common in them?, (b) Arrange them in the order of increasing ionic radii., 3.13, , Explain why cation are smaller and anions larger in radii than their parent atoms?, , 3.14, , What is the significance of the terms — ‘isolated gaseous atom’ and ‘ground state’, while defining the ionization enthalpy and electron gain enthalpy?, Hint : Requirements for comparison purposes., , 3.15, , Energy of an electron in the ground state of the hydrogen atom is, –2.18×10–18J. Calculate the ionization enthalpy of atomic hydrogen in terms of, J mol–1., Hint: Apply the idea of mole concept to derive the answer., , 3.16, , Among the second period elements the actual ionization enthalpies are in the, order Li < B < Be < C < O < N < F < Ne., Explain why, (i), , Be has higher ∆i H than B, , (ii) O has lower ∆i H than N and F?, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , 3.17, , How would you explain the fact that the first ionization enthalpy of sodium is, lower than that of magnesium but its second ionization enthalpy is higher than, that of magnesium?, , 3.18, , What are the various factors due to which the ionization enthalpy of the main, group elements tends to decrease down a group?, , 3.19, , The first ionization enthalpy values (in kJ mol–1) of group 13 elements are :, B Al, , Ga, , In, , Tl, , 801, , 577, , 579, , 558, , 589, , How would you explain this deviation from the general trend ?, 3.20, , Which of the following pairs of elements would have a more negative electron gain, enthalpy?, (i) O or F (ii) F or Cl, , 3.21, , Would you expect the second electron gain enthalpy of O as positive, more negative, or less negative than the first? Justify your answer., , 3.22, , What is the basic difference between the terms electron gain enthalpy and, electronegativity?, , 3.23, , How would you react to the statement that the electronegativity of N on Pauling, scale is 3.0 in all the nitrogen compounds?, , 3.24, , Describe the theory associated with the radius of an atom as it, (a) gains an electron, (b) loses an electron, , 3.25, , Would you expect the first ionization enthalpies for two isotopes of the same element, to be the same or different? Justify your answer., , 3.26, , What are the major differences between metals and non-metals?, , 3.27, , Use the periodic table to answer the following questions., (a) Identify an element with five electrons in the outer subshell., (b) Identify an element that would tend to lose two electrons., (c), , Identify an element that would tend to gain two electrons., , (d) Identify the group having metal, non-metal, liquid as well as gas at the room, temperature., 3.28, , The increasing order of reactivity among group 1 elements is Li < Na < K < Rb <Cs, whereas that among group 17 elements is F > CI > Br > I. Explain., , 3.29, , Write the general outer electronic configuration of s-, p-, d- and f- block elements., , 3.30, , Assign the position of the element having outer electronic configuration, (i) ns 2np4 for n=3 (ii) (n-1)d2ns2 for n=4, and (iii) (n-2) f 7 (n-1)d1ns2 for n=6, in the, periodic table., , Download all NCERT books PDFs from www.ncert.online, , 2019-20, , 97

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98, , CHEMISTRY, , 3.31, , The first (∆iH1) and the second (∆iH2) ionization enthalpies (in kJ mol–1), and the (∆egH) electron gain enthalpy (in kJ mol–1) of a few elements are, given below:, Elements, , ∆H1, , ∆H2, , ∆egH, , I, , 520, , 7300, , –60, , II, , 419, , 3051, , –48, , III, , 1681, , 3374, , –328, , IV, , 1008, , 1846, , –295, , V, , 2372, , 5251, , +48, , VI, , 738, , 1451, , –40, , Which of the above elements is likely to be :, (a) the least reactive element., (b) the most reactive metal., (c), , the most reactive non-metal., , (d) the least reactive non-metal., , 3.32, , 3.33, , (e), , the metal which can form a stable binary halide of the formula, MX2(X=halogen)., , (f), , the metal which can form a predominantly stable covalent halide, of the formula MX (X=halogen)?, , Predict the formulas of the stable binary compounds that would be, formed by the combination of the following pairs of elements., (a) Lithium and oxygen, , (b) Magnesium and nitrogen, , (c), , Aluminium and iodine, , (d) Silicon and oxygen, , (e), , Phosphorus and fluorine, , (f) Element 71 and fluorine, , In the modern periodic table, the period indicates the value of :, (a) atomic number, (b) atomic mass, (c), , principal quantum number, , (d) azimuthal quantum number., 3.34, , Which of the following statements related to the modern periodic table, is incorrect?, (a) The p-block has 6 columns, because a maximum of 6 electrons, can occupy all the orbitals in a p-shell., (b) The d-block has 8 columns, because a maximum of 8 electrons, can occupy all the orbitals in a d-subshell., (c), , Each block contains a number of columns equal to the number of, electrons that can occupy that subshell., , (d) The block indicates value of azimuthal quantum number (l) for the, last subshell that received electrons in building up the electronic, configuration., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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CLASSIFICATION OF ELEMENTS AND PERIODICITY IN PROPERTIES, , 3.35, , Anything that influences the valence electrons will affect the chemistry of the, element. Which one of the following factors does not affect the valence shell?, (a) Valence principal quantum number (n), (b) Nuclear charge (Z ), (c), , Nuclear mass, , (d) Number of core electrons., 3.36, , –, , The size of isoelectronic species — F , Ne and Na+ is affected by, (a) nuclear charge (Z ), (b) valence principal quantum number (n), (c), , electron-electron interaction in the outer orbitals, , (d) none of the factors because their size is the same., 3.37, , Which one of the following statements is incorrect in relation to ionization, enthalpy?, (a) Ionization enthalpy increases for each successive electron., (b) The greatest increase in ionization enthalpy is experienced on removal of, electron from core noble gas configuration., (c), , End of valence electrons is marked by a big jump in ionization enthalpy., , (d) Removal of electron from orbitals bearing lower n value is easier than from, orbital having higher n value., 3.38, , 3.39, , 3.40, , Considering the elements B, Al, Mg, and K, the correct order of their metallic, character is :, (a) B > Al > Mg > K, , (b) Al > Mg > B > K, , (c) Mg > Al > K > B, , (d) K > Mg > Al > B, , Considering the elements B, C, N, F, and Si, the correct order of their non-metallic, character is :, (a) B > C > Si > N > F, , (b) Si > C > B > N > F, , (c) F > N > C > B > Si, , (d) F > N > C > Si > B, , Considering the elements F, Cl, O and N, the correct order of their chemical reactivity, in terms of oxidizing property is :, (a) F > Cl > O > N, , (b) F > O > Cl > N, , (c) Cl > F > O > N, , (d) O > F > N > Cl, , Download all NCERT books PDFs from www.ncert.online, , 2019-20, , 99

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100, , CHEMISTRY, , UNIT 4, , CHEMICAL BONDING AND, MOLECULAR STRUCTURE, , After studying this Unit, you will be, able to, , Scientists are constantly discovering new compounds, orderly, arranging the facts about them, trying to explain with the, existing knowledge, organising to modify the earlier views or, evolve theories for explaining the newly observed facts., , • understand, , K Ö ssel-Lewis, approach to chemical bonding;, , • explain the octet rule and its, limitations,, draw, Lewis, structures of simple molecules;, , • explain the formation of different, types of bonds;, , • describe the VSEPR theory and, predict the geometry of simple, molecules;, , • explain the valence bond, approach for the formation of, covalent bonds;, , • predict the directional properties, of covalent bonds;, , • explain the different types of, hybridisation involving s, p and, d orbitals and draw shapes of, simple covalent molecules;, , • describe the molecular orbital, theory of homonuclear diatomic, molecules;, , • explain the concept of hydrogen, bond., , Matter is made up of one or different type of elements., Under normal conditions no other element exists as an, independent atom in nature, except noble gases. However,, a group of atoms is found to exist together as one species, having characteristic properties. Such a group of atoms is, called a molecule. Obviously there must be some force, which holds these constituent atoms together in the, molecules. The attractive force which holds various, constituents (atoms, ions, etc.) together in different, chemical species is called a chemical bond. Since the, formation of chemical compounds takes place as a result, of combination of atoms of various elements in different, ways, it raises many questions. Why do atoms combine?, Why are only certain combinations possible? Why do some, atoms combine while certain others do not? Why do, molecules possess definite shapes? To answer such, questions different theories and concepts have been put, forward from time to time. These are Kössel-Lewis, approach, Valence Shell Electron Pair Repulsion (VSEPR), Theory, Valence Bond (VB) Theory and Molecular Orbital, (MO) Theory. The evolution of various theories of valence, and the interpretation of the nature of chemical bonds have, closely been related to the developments in the, understanding of the structure of atom, the electronic, configuration of elements and the periodic table. Every, system tends to be more stable and bonding is nature’s, way of lowering the energy of the system to attain stability., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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101, , CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 4.1 KÖSSEL-LEWIS APPROACH TO, CHEMICAL BONDING, In order to explain the formation of chemical, bond in terms of electrons, a number of, attempts were made, but it was only in 1916, when Kössel and Lewis succeeded, independently in giving a satisfactory, explanation. They were the first to provide, some logical explanation of valence which was, based on the inertness of noble gases., Lewis pictured the atom in terms of a, positively charged ‘Kernel’ (the nucleus plus, the inner electrons) and the outer shell that, could accommodate a maximum of eight, electrons. He, further assumed that these, eight electrons occupy the corners of a cube, which surround the ‘Kernel’. Thus the single, outer shell electron of sodium would occupy, one corner of the cube, while in the case of a, noble gas all the eight corners would be, occupied. This octet of electrons, represents, a particularly stable electronic arrangement., Lewis postulated that atoms achieve the, stable octet when they are linked by, chemical bonds. In the case of sodium and, chlorine, this can happen by the transfer of, an electron from sodium to chlorine thereby, –, giving the Na+ and Cl ions. In the case of, other molecules like Cl2, H2, F2, etc., the bond, is formed by the sharing of a pair of electrons, between the atoms. In the process each atom, attains a stable outer octet of electrons., Lewis Symbols: In the for mation of a, molecule, only the outer shell electrons take, part in chemical combination and they are, known as valence electrons. The inner shell, electrons are well protected and are generally, not involved in the combination process., G.N. Lewis, an American chemist introduced, simple notations to represent valence, electrons in an atom. These notations are, called Lewis symbols. For example, the Lewis, symbols for the elements of second period are, as under:, , Significance of Lewis Symbols : The, number of dots around the symbol represents, , the number of valence electrons. This number, of valence electrons helps to calculate the, common or group valence of the element. The, group valence of the elements is generally, either equal to the number of dots in Lewis, symbols or 8 minus the number of dots or, valence electrons., Kössel, in relation to chemical bonding,, drew attention to the following facts:, • In the periodic table, the highly, electronegative halogens and the highly, electropositive alkali metals are separated, by the noble gases;, • The formation of a negative ion from a, halogen atom and a positive ion from an, alkali metal atom is associated with the, gain and loss of an electron by the, respective atoms;, • The negative and positive ions thus, formed attain stable noble gas electronic, configurations. The noble gases (with the, exception of helium which has a duplet, of electrons) have a particularly stable, outer shell configuration of eight (octet), electrons, ns2np6., • The negative and positive ions are, stabilized by electrostatic attraction., For example, the formation of NaCl from, sodium and chlorine, according to the above, scheme, can be explained as:, –, Na, →, Na+ + e, [Ne] 3s1, [Ne], –, →, Cl–, Cl + e, [Ne] 3s2 3p5, [Ne] 3s2 3p6 or [Ar], –, –, →, NaCl or Na+Cl, Na+ + Cl, Similarly the formation of CaF2 may be, shown as:, –, Ca, →, Ca2+ + 2e, [Ar]4s2, [Ar], –, –, →, F, F +e, [He] 2s2 2p5, [He] 2s2 2p6 or [Ne], –, –, →, CaF2 or Ca2+(F )2, Ca2+ + 2F, The bond formed, as a result of the, electrostatic attraction between the, positive and negative ions was termed as, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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102, , CHEMISTRY, , the electrovalent bond. The electrovalence, is thus equal to the number of unit, charge(s) on the ion. Thus, calcium is, assigned a positive electrovalence of two,, while chlorine a negative electrovalence of, one., Kössel’s postulations provide the basis for, the modern concepts regarding ion-formation, by electron transfer and the formation of ionic, crystalline compounds. His views have proved, to be of great value in the understanding and, systematisation of the ionic compounds. At, the same time he did recognise the fact that, a large number of compounds did not fit into, these concepts., 4.1.1 Octet Rule, Kössel and Lewis in 1916 developed an, important theory of chemical combination, between atoms known as electronic theory, of chemical bonding. According to this,, atoms can combine either by transfer of, valence electrons from one atom to another, (gaining or losing) or by sharing of valence, electrons in order to have an octet in their, valence shells. This is known as octet rule., 4.1.2 Covalent Bond, Langmuir (1919) refined the Lewis, postulations by abandoning the idea of the, stationary cubical arrangement of the octet,, and by introducing the term covalent bond., The Lewis-Langmuir theory can be, understood by considering the formation of, the chlorine molecule,Cl2. The Cl atom with, electronic configuration, [Ne]3s 2 3p5, is one, electron short of the argon configuration., The formation of the Cl2 molecule can be, understood in terms of the sharing of a pair, of electrons between the two chlorine atoms,, each chlorine atom contributing one electron, to the shared pair. In the process both, , chlorine atoms attain the outer shell octet of, the nearest noble gas (i.e., argon)., The dots represent electrons. Such, structures are referred to as Lewis dot, structures., The Lewis dot structures can be written, for other molecules also, in which the, combining atoms may be identical or, different. The important conditions being that:, • Each bond is formed as a result of sharing, of an electron pair between the atoms., • Each combining atom contributes at least, one electron to the shared pair., • The combining atoms attain the outershell noble gas configurations as a result, of the sharing of electrons., • Thus in water and carbon tetrachloride, molecules, formation of covalent bonds, can be represented as:, , Thus, when two atoms share one, electron pair they are said to be joined by, a single covalent bond. In many compounds, we have multiple bonds between atoms. The, formation of multiple bonds envisages, sharing of more than one electron pair, between two atoms. If two atoms share two, pairs of electrons, the covalent bond, between them is called a double bond. For, example, in the carbon dioxide molecule, we, have two double bonds between the carbon, and oxygen atoms. Similarly in ethene, molecule the two carbon atoms are joined by, a double bond., , or Cl – Cl, Covalent bond between two Cl atoms, , Double bonds in CO2 molecule, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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103, , CHEMICAL BONDING AND MOLECULAR STRUCTURE, , C2H4 molecule, , When combining atoms share three, electron pairs as in the case of two, nitrogen atoms in the N2 molecule and the, two carbon atoms in the ethyne molecule,, a triple bond is formed., , •, , •, , N2 molecule, , •, C2H2 molecule, , 4.1.3 Lewis Representation of Simple, Molecules (the Lewis Structures), The Lewis dot structures provide a picture, of bonding in molecules and ions in terms, of the shared pairs of electrons and the, octet rule. While such a picture may not, explain the bonding and behaviour of a, molecule completely, it does help in, understanding the formation and properties, of a molecule to a large extent. Writing of, Lewis dot structures of molecules is,, therefor e, very useful. The Lewis dot, structures can be written by adopting the, following steps:, • The total number of electrons required for, writing the structures are obtained by, adding the valence electrons of the, combining atoms. For example, in the CH4, molecule there are eight valence electrons, available for bonding (4 from carbon and, 4 from the four hydrogen atoms)., • For anions, each negative charge would, mean addition of one electron. For, cations, each positive charge would result, , in subtraction of one electron from the total, number of valence electrons. For example,, 2–, for the CO3 ion, the two negative charges, indicate that there are two additional, electrons than those provided by the, +, neutral atoms. For NH 4 ion, one positive, charge indicates the loss of one electron, from the group of neutral atoms., Knowing the chemical symbols of the, combining atoms and having knowledge, of the skeletal structure of the compound, (known or guessed intelligently), it is easy, to distribute the total number of electrons, as bonding shared pairs between the, atoms in proportion to the total bonds., In general the least electronegative atom, occupies the central position in the, molecule/ion. For example in the NF3 and, 2–, CO3 , nitrogen and carbon are the central, atoms whereas fluorine and oxygen, occupy the terminal positions., After accounting for the shared pairs of, electrons for single bonds, the remaining, electron pairs are either utilized for multiple, bonding or remain as the lone pairs. The, basic requirement being that each bonded, atom gets an octet of electrons., Lewis representations of a few molecules/, ions are given in Table 4.1., , Table 4.1 The Lewis Representation of Some, Molecules, , * Each H atom attains the configuration of helium (a duplet, of electrons), , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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104, , CHEMISTRY, , each of the oxygen atoms completing the, octets on oxygen atoms. This, however,, does not complete the octet on nitrogen, if the remaining two electrons constitute, lone pair on it., , Problem 4.1, Write the Lewis dot structure of CO, molecule., Solution, Step 1. Count the total number of, valence electrons of carbon and oxygen, atoms. The outer (valence) shell, configurations of carbon and oxygen, atoms are: 2s 2 2p 2 and 2s 2 2p 4 ,, respectively. The valence electrons, available are 4 + 6 =10., Step 2. The skeletal structure of CO is, written as: C O, Step 3. Draw a single bond (one shared, electron pair) between C and O and, complete the octet on O, the remaining, two electrons are the lone pair on C., , This does not complete the octet on, carbon and hence we have to resort to, multiple bonding (in this case a triple, bond) between C and O atoms. This, satisfies the octet rule condition for both, atoms., , Problem 4.2, Write the Lewis structure of the nitrite, –, ion, NO2 ., Solution, Step 1. Count the total number of, valence electrons of the nitrogen atom,, the oxygen atoms and the additional one, negative charge (equal to one electron)., N(2s2 2p3), O (2s2 2p4), 5 + (2 × 6) +1 = 18 electrons, , Hence we have to resort to multiple, bonding between nitrogen and one of the, oxygen atoms (in this case a double, bond). This leads to the following Lewis, dot structures., , 4.1.4 Formal Charge, Lewis dot structures, in general, do not, represent the actual shapes of the molecules., In case of polyatomic ions, the net charge is, possessed by the ion as a whole and not by a, particular atom. It is, however, feasible to, assign a formal charge on each atom. The, formal charge of an atom in a polyatomic, molecule or ion may be defined as the, difference between the number of valence, electrons of that atom in an isolated or free, state and the number of electrons assigned, to that atom in the Lewis structure. It is, expressed as :, Formal charge (F.C.), on an atom in a Lewis, structure, , =, , total number of non, total number of valence, — bonding (lone pair), electrons in the free, electrons, atom, , –, , Step 2. The skeletal structure of NO2 is, written as : O N O, , total number of, — (1/2) bonding(shared), electrons, , Step 3. Draw a single bond (one shared, electron pair) between the nitrogen and, Download all NCERT books PDFs from www.ncert.online, , 2019-20

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105, , CHEMICAL BONDING AND MOLECULAR STRUCTURE, , The counting is based on the assumption, that the atom in the molecule owns one, electron of each shared pair and both the, electrons of a lone pair., Let us consider the ozone molecule (O3)., The Lewis structure of O3 may be drawn as :, , The atoms have been numbered as 1, 2, and 3. The formal charge on:, •, , The central O atom marked 1, , •, , 1, (6) = +1, 2, The end O atom marked 2, =6–2–, , 1, (4) = 0, 2, The end O atom marked 3, , =6–4–, •, , 1, (2) = –1, 2, Hence, we represent O3 along with the, formal charges as follows:, , 4.1.5 Limitations of the Octet Rule, The octet rule, though useful, is not universal., It is quite useful for understanding the, structures of most of the organic compounds, and it applies mainly to the second period, elements of the periodic table. There are three, types of exceptions to the octet rule., The incomplete octet of the central atom, In some compounds, the number of electrons, surrounding the central atom is less than, eight. This is especially the case with elements, having less than four valence electrons., Examples are LiCl, BeH2 and BCl3., , Li, Be and B have 1,2 and 3 valence electrons, only. Some other such compounds are AlCl3, and BF3., Odd-electron molecules, In molecules with an odd number of electrons, like nitric oxide, NO and nitrogen dioxide,, NO2, the octet rule is not satisfied for all the, atoms, , =6–6–, , We must understand that formal charges, do not indicate real charge separation within, the molecule. Indicating the charges on the, atoms in the Lewis structure only helps in, keeping track of the valence electrons in the, molecule. For mal charges help in the, selection of the lowest energy structure from, a number of possible Lewis structures for a, given species. Generally the lowest energy, structure is the one with the smallest, formal charges on the atoms. The formal, charge is a factor based on a pure covalent, view of bonding in which electron pairs, are shared equally by neighbouring atoms., , The expanded octet, Elements in and beyond the third period of, the periodic table have, apart from 3s and 3p, orbitals, 3d orbitals also available for bonding., In a number of compounds of these elements, there are more than eight valence electrons, around the central atom. This is termed as, the expanded octet. Obviously the octet rule, does not apply in such cases., Some of the examples of such compounds, are: PF 5 , SF 6 , H 2 SO 4 and a number of, coordination compounds., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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106, , CHEMISTRY, , Interestingly, sulphur also forms many, compounds in which the octet rule is obeyed., In sulphur dichloride, the S atom has an octet, of electrons around it., , Other drawbacks of the octet theory, • It is clear that octet rule is based upon, the chemical inertness of noble gases., However, some noble gases (for example, xenon and krypton) also combine with, oxygen and fluorine to form a number of, compounds like XeF2, KrF2, XeOF2 etc.,, • This theory does not account for the shape, of molecules., • It does not explain the relative stability of, the molecules being totally silent about, the energy of a molecule., 4.2 IONIC OR ELECTROVALENT BOND, From the Kössel and Lewis treatment of the, formation of an ionic bond, it follows that the, for mation of ionic compounds would, primarily depend upon:, • The ease of formation of the positive and, negative ions from the respective neutral, atoms;, • The arrangement of the positive and, negative ions in the solid, that is, the, lattice of the crystalline compound., The formation of a positive ion involves, ionization, i.e., removal of electron(s) from, the neutral atom and that of the negative ion, involves the addition of electron(s) to the, neutral atom., –, M(g), → M+(g) + e ;, Ionization enthalpy, –, –, X(g) + e → X (g) ;, Electron gain enthalpy, –, M+(g) + X (g) → MX(s), The electron gain enthalpy, ∆eg H, is the, enthalpy change (Unit 3), when a gas phase atom, in its ground state gains an electron. The, electron gain process may be exothermic or, endothermic. The ionization, on the other hand,, is always endothermic. Electron affinity, is the, negative of the energy change accompanying, electron gain., , Obviously ionic bonds will be formed, more easily between elements with, comparatively low ionization enthalpies, and elements with comparatively high, negative value of electron gain enthalpy., Most ionic compounds have cations, derived from metallic elements and anions, from non-metallic elements. The, +, ammonium ion, NH4 (made up of two nonmetallic elements) is an exception. It forms, the cation of a number of ionic compounds., Ionic compounds in the crystalline state, consist of orderly three-dimensional, arrangements of cations and anions held, together by coulombic interaction energies., These compounds crystallise in different, crystal structures determined by the size, of the ions, their packing arrangements and, other factors. The crystal structure of, sodium chloride, NaCl (rock salt), for, example is shown below., , Rock salt structure, In ionic solids, the sum of the electron, gain enthalpy and the ionization enthalpy, may be positive but still the crystal, structure gets stabilized due to the energy, released in the formation of the crystal, lattice. For example: the ionization, +, enthalpy for Na (g) formation from Na(g), is 495.8 kJ mol –1 ; while the electron gain, –, enthalpy for the change Cl(g) + e →, –, –1, Cl (g) is, – 348.7 kJ mol only. The sum, of the two, 147.1 kJ mol -1 is more than, compensated for by the enthalpy of lattice, for mation of NaCl(s) (–788 kJ mol –1 )., Therefore, the energy released in the, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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107, , CHEMICAL BONDING AND MOLECULAR STRUCTURE, , processes is more than the energy absorbed., Thus a qualitative measure of the, stability of an ionic compound is, provided by its enthalpy of lattice, formation and not simply by achieving, octet of electrons around the ionic species, in gaseous state., Since lattice enthalpy plays a key role, in the formation of ionic compounds, it is, important that we learn more about it., 4.2.1 Lattice Enthalpy, The Lattice Enthalpy of an ionic solid is, defined as the energy required to, completely separate one mole of a solid, ionic compound into gaseous constituent, ions. For example, the lattice enthalpy of NaCl, is 788 kJ mol–1. This means that 788 kJ of, energy is required to separate one mole of, solid NaCl into one mole of Na+ (g) and one, mole of Cl– (g) to an infinite distance., This process involves both the attractive, forces between ions of opposite charges and, the repulsive forces between ions of like, charge. The solid crystal being thr eedimensional; it is not possible to calculate, lattice enthalpy directly from the interaction, of forces of attraction and repulsion only., Factors associated with the crystal geometry, have to be included., , Fig. 4.1 The bond length in a covalent, molecule AB., R = rA + rB (R is the bond length and rA and rB, are the covalent radii of atoms A and B, respectively), , covalent bond in the same molecule. The van, der Waals radius represents the overall size, of the atom which includes its valence shell, in a nonbonded situation. Further, the van, der Waals radius is half of the distance, between two similar atoms in separate, molecules in a solid. Covalent and van der, Waals radii of chlorine are depicted in Fig.4.2, 8, , pm, , =, , w, , r vd, 0, , 18, 36, , 0, , pm, , pm, , 4.3.1 Bond Length, Bond length is defined as the equilibrium, distance between the nuclei of two bonded, atoms in a molecule. Bond lengths are, measured by spectroscopic, X-ray diffraction, and electron-diffraction techniques about, which you will learn in higher classes. Each, atom of the bonded pair contributes to the, bond length (Fig. 4.1). In the case of a covalent, bond, the contribution from each atom is, called the covalent radius of that atom., The covalent radius is measured, approximately as the radius of an atom’s, core which is in contact with the core of, an adjacent atom in a bonded situation., The covalent radius is half of the distance, between two similar atoms joined by a, ,, , 19, , rc = 99 pm, , 4.3 BOND PARAMETERS, , Fig. 4.2 Covalent and van der Waals radii in a, chlorine molecule. The inner circles, correspond to the size of the chlorine atom, (r vdw and r c are van der Waals and, covalent radii respectively)., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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108, , CHEMISTRY, , Some typical average bond lengths for, single, double and triple bonds are shown in, Table 4.2. Bond lengths for some common, molecules are given in Table 4.3., The covalent radii of some common, elements are listed in Table 4.4., 4.3.2 Bond Angle, It is defined as the angle between the orbitals, containing bonding electron pairs around the, central atom in a molecule/complex ion. Bond, angle is expressed in degree which can be, experimentally determined by spectroscopic, methods. It gives some idea regarding the, distribution of orbitals around the central, atom in a molecule/complex ion and hence it, helps us in determining its shape. For, example H–O–H bond angle in water can be, represented as under :, , Table 4.2 Average Bond Lengths for Some, Single, Double and Triple Bonds, Bond Type, , Covalent Bond Length, (pm), , O–H, C–H, N–O, C–O, C–N, C–C, C=O, N=O, C=C, C=N, C≡N, C≡C, , 96, 107, 136, 143, 143, 154, 121, 122, 133, 138, 116, 120, , Table 4.3 Bond Lengths in Some Common, Molecules, Molecule, H2 (H – H), F2 (F – F), Cl2 (Cl – Cl), Br2 (Br – Br), I2 (I – I), N2 (N ≡ N), O2 (O = O), HF (H – F), HCl (H – Cl), HBr (H – Br), HI (H – I), , 4.3.3 Bond Enthalpy, It is defined as the amount of energy required, to break one mole of bonds of a particular, type between two atoms in a gaseous state., The unit of bond enthalpy is kJ mol–1. For, example, the H – H bond enthalpy in hydrogen, molecule is 435.8 kJ mol–1., , H2(g) → H(g) + H(g); ∆aH = 435.8 kJ mol–1, Similarly the bond enthalpy for molecules, containing multiple bonds, for example O2 and, N2 will be as under :, O2 (O = O) (g) → O(g) + O(g);, , ∆aH = 498 kJ mol–1, N2 (N ≡ N) (g) → N(g) + N(g);, , ∆ aH = 946.0 kJ mol–1, It is important that larger the bond, dissociation enthalpy, stronger will be the, bond in the molecule. For a heteronuclear, diatomic molecules like HCl, we have, , HCl (g) → H(g) + Cl (g); ∆aH = 431.0 kJ mol–1, In case of polyatomic molecules, the, measurement of bond strength is more, complicated. For example in case of H 2O, molecule, the enthalpy needed to break the, two O – H bonds is not the same., , Bond Length, (pm), 74, 144, 199, 228, 267, 109, 121, 92, 127, 141, 160, , Table 4.4 Covalent Radii, *rcov/(pm), , * The values cited are for single bonds, except where, otherwise indicated in parenthesis. (See also Unit 3 for, periodic trends)., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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109, , CHEMICAL BONDING AND MOLECULAR STRUCTURE, , , H2O(g) → H(g) + OH(g); ∆aH1 = 502 kJ mol–1, , , –1, , OH(g) → H(g) + O(g); ∆aH2 = 427 kJ mol, , equally represented by the structures I and II, shown below:, , , , The difference in the ∆aH value shows that, the second O – H bond undergoes some change, because of changed chemical environment., This is the reason for some difference in energy, of the same O – H bond in different molecules, like C2H5OH (ethanol) and water. Therefore in, polyatomic molecules the term mean or, average bond enthalpy is used. It is obtained, by dividing total bond dissociation enthalpy, by the number of bonds broken as explained, below in case of water molecule,, Average bond enthalpy =, , 502 + 427, 2, , = 464.5 kJ mol, , Fig. 4.3 Resonance in the O3 molecule, –1, , (structures I and II represent the two canonical, forms while the structure III is the resonance, hybrid), , 4.3.4 Bond Order, In the Lewis description of covalent bond,, the Bond Order is given by the number of, bonds between the two atoms in a, molecule. The bond order, for example in H2, (with a single shared electron pair), in O2, (with two shared electron pairs) and in N2, (with three shared electron pairs) is 1,2,3, respectively. Similarly in CO (three shared, electron pairs between C and O) the bond, order is 3. For N2, bond order is 3 and its, ∆a H  is 946 kJ mol–1; being one of the, highest for a diatomic molecule., Isoelectronic molecules and ions have, identical bond orders; for example, F2 and, 2–, O2 have bond order 1. N2, CO and NO+, have bond order 3., A general correlation useful for, understanding the stablities of molecules, is that: with increase in bond order, bond, enthalpy increases and bond length, decreases., 4.3.5 Resonance Structures, It is often observed that a single Lewis, structure is inadequate for the representation, of a molecule in confor mity with its, experimentally determined parameters. For, example, the ozone, O 3 molecule can be, , In both structures we have a O–O single, bond and a O=O double bond. The normal, O–O and O=O bond lengths are 148 pm and, 121 pm respectively. Experimentally, determined oxygen-oxygen bond lengths in, the O3 molecule are same (128 pm). Thus the, oxygen-oxygen bonds in the O3 molecule are, intermediate between a double and a single, bond. Obviously, this cannot be represented, by either of the two Lewis structures shown, above., The concept of resonance was introduced, to deal with the type of difficulty experienced, in the depiction of accurate structures of, molecules like O3. According to the concept, of resonance, whenever a single Lewis, structure cannot describe a molecule, accurately, a number of structures with, similar energy, positions of nuclei, bonding, and non-bonding pairs of electrons are taken, as the canonical structures of the hybrid, which describes the molecule accurately., Thus for O3, the two structures shown above, constitute the canonical structures or, resonance structures and their hybrid i.e., the, III structure represents the structure of O3, more accurately. This is also called resonance, hybrid. Resonance is represented by a double, headed arrow., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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110, , CHEMISTRY, , Some of the other examples of resonance, structures are provided by the carbonate ion, and the carbon dioxide molecule., Problem 4.3, 2–, Explain the structure of CO3 ion in terms, of resonance., Solution, The single Lewis structure based on the, presence of two single bonds and one, double bond between carbon and oxygen, atoms is inadequate to represent the, molecule accurately as it represents, unequal bonds. According to the, experimental findings, all carbon to, oxygen bonds in CO32– are equivalent., Therefore the carbonate ion is best, described as a resonance hybrid of the, canonical forms I, II, and III shown below., , Fig. 4.5, , In general, it may be stated that, • Resonance stabilizes the molecule as the, energy of the resonance hybrid is less, than the energy of any single cannonical, structure; and,, • Resonance averages the bond, characteristics as a whole., Thus the energy of the O 3 resonance, hybrid is lower than either of the two, cannonical froms I and II (Fig 4.3)., Many misconceptions are associated, with resonance and the same need to be, dispelled. You should remember that :, • The cannonical forms have no real, existence., • The molecule does not exist for a, certain fraction of time in one, cannonical for m and for other, fractions of time in other cannonical, forms., • There is no such equilibrium between, the cannonical forms as we have, between tautomeric forms (keto and, enol) in tautomerism., • The molecule as such has a single, structure which is the resonance, hybrid of the cannonical forms and, which cannot as such be depicted by, a single Lewis structure., , Fig.4.4 Resonance in CO32–, I, II and, III represent the three, canonical forms., , Problem 4.4, Explain the structure of CO2 molecule., Solution, The experimentally determined carbon, to oxygen bond length in CO 2 is, 115 pm. The lengths of a nor mal, carbon to oxygen double bond (C=O), and carbon to oxygen triple bond (C≡O), are 121 pm and 110 pm respectively., The carbon-oxygen bond lengths in, CO2 (115 pm) lie between the values, for C=O and C≡O. Obviously, a single, Lewis structure cannot depict this, position and it becomes necessary to, write more than one Lewis structures, and to consider that the structure of, CO2 is best described as a hybrid of, the canonical or resonance forms I, II, and III., , Resonance in CO2 molecule, I, II, and III represent the three, canonical forms., , 4.3.6 Polarity of Bonds, The existence of a hundred percent ionic or, covalent bond represents an ideal situation., In reality no bond or a compound is either, completely covalent or ionic. Even in case of, covalent bond between two hydrogen atoms,, there is some ionic character., When covalent bond is formed between, two similar atoms, for example in H2, O2, Cl2,, N2 or F2, the shared pair of electrons is equally, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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111, , CHEMICAL BONDING AND MOLECULAR STRUCTURE, , attracted by the two atoms. As a result electron, pair is situated exactly between the two, identical nuclei. The bond so formed is called, nonpolar covalent bond. Contrary to this in, case of a heteronuclear molecule like HF, the, shared electron pair between the two atoms, gets displaced more towards fluorine since the, electronegativity of fluorine (Unit 3) is far, greater than that of hydrogen. The resultant, covalent bond is a polar covalent bond., As a result of polarisation, the molecule, possesses the dipole moment (depicted, below) which can be defined as the product, of the magnitude of the charge and the, distance between the centres of positive and, negative charge. It is usually designated by a, Greek letter ‘µ’. Mathematically, it is expressed, as follows :, Dipole moment (µ) = charge (Q) × distance of, separation (r), Dipole moment is usually expressed in, Debye units (D). The conversion factor is, 1 D = 3.33564 × 10–30 C m, where C is coulomb and m is meter., Further dipole moment is a vector quantity, and by convention it is depicted by a small, arrow with tail on the negative centre and head, pointing towards the positive centre. But in, chemistry presence of dipole moment is, ) put, represented by the crossed arrow (, on Lewis structure of the molecule. The cross, is on positive end and arrow head is on negative, end. For example the dipole moment of HF may, be represented as :, H, , F, , In case of polyatomic molecules the dipole, moment not only depend upon the individual, dipole moments of bonds known as bond, dipoles but also on the spatial arrangement of, various bonds in the molecule. In such case,, the dipole moment of a molecule is the vector, sum of the dipole moments of various bonds., For example in H2O molecule, which has a bent, structure, the two O–H bonds are oriented at, an angle of 104.50. Net dipole moment of 6.17, × 10–30 C m (1D = 3.33564 × 10–30 C m) is the, resultant of the dipole moments of two O–H, bonds., , Net Dipole moment, µ = 1.85 D, –30, –30, = 1.85 × 3.33564 × 10 C m = 6.17 ×10 C m, The dipole moment in case of BeF2 is zero., This is because the two equal bond dipoles, point in opposite directions and cancel the, effect of each other., , In tetra-atomic molecule, for example in, BF3, the dipole moment is zero although the, o, B – F bonds are oriented at an angle of 120 to, one another, the three bond moments give a, net sum of zero as the resultant of any two is, equal and opposite to the third., , This arrow symbolises the direction of the, shift of electron density in the molecule. Note, that the direction of crossed arrow is opposite, to the conventional direction of dipole moment, vector., Peter Debye, the Dutch chemist, received Nobel prize in 1936 for, his work on X-ray diffraction and, dipole moments. The magnitude, of the dipole moment is given in, Debye units in order to honour him., , Let us study an interesting case of NH3, and NF3 molecule. Both the molecules have, pyramidal shape with a lone pair of electrons, on nitrogen atom. Although fluorine is more, electronegative than nitrogen, the resultant, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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112, , CHEMISTRY, , dipole moment of NH3 ( 4.90 × 10–30 C m) is, greater than that of NF3 (0.8 × 10–30 C m). This, is because, in case of NH3 the orbital dipole, due to lone pair is in the same direction as the, resultant dipole moment of the N – H bonds,, whereas in NF3 the orbital dipole is in the, direction opposite to the resultant dipole, moment of the three N–F bonds. The orbital, dipole because of lone pair decreases the effect, of the resultant N – F bond moments, which, results in the low dipole moment of NF3 as, represented below :, , Dipole moments of some molecules are, shown in Table 4.5., Just as all the covalent bonds have, some partial ionic character, the ionic, bonds also have partial covalent, character. The partial covalent character, of ionic bonds was discussed by Fajans, , in terms of the following rules:, • The smaller the size of the cation and the, larger the size of the anion, the greater the, covalent character of an ionic bond., • The greater the charge on the cation, the, greater the covalent character of the ionic bond., • For cations of the same size and charge,, the one, with electronic configuration, (n-1)dnnso, typical of transition metals, is, more polarising than the one with a noble, gas configuration, ns2 np6, typical of alkali, and alkaline earth metal cations., The cation polarises the anion, pulling the, electronic charge toward itself and thereby, increasing the electronic charge between, the two. This is precisely what happens in, a covalent bond, i.e., buildup of electron, charge density between the nuclei. The, polarising power of the cation, the, polarisability of the anion and the extent, of distortion (polarisation) of anion are the, factors, which determine the per cent, covalent character of the ionic bond., 4.4 THE VALENCE SHELL ELECTRON, PAIR REPULSION (VSEPR) THEORY, As already explained, Lewis concept is unable, to explain the shapes of molecules. This theory, provides a simple procedure to predict the, shapes of covalent molecules. Sidgwick, , Table 4.5 Dipole Moments of Selected Molecules, Type of, Molecule, , Example, , Dipole, Moment, µ(D), , Molecule (AB), , HF, HCl, HBr, HI, H2, , 1.78, 1.07, 0.79, 0.38, 0, , linear, linear, linear, linear, linear, , Molecule (AB2), , H2O, H2 S, CO2, , 1.85, 0.95, 0, , bent, bent, linear, , Molecule (AB3), , NH3, NF3, BF3, , 1.47, 0.23, 0, , trigonal-pyramidal, trigonal-pyramidal, trigonal-planar, , Molecule (AB4), , CH4, CHCl3, CCl4, , 0, 1.04, 0, , tetrahedral, tetrahedral, tetrahedral, , Download all NCERT books PDFs from www.ncert.online, , 2019-20, , Geometry

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113, , CHEMICAL BONDING AND MOLECULAR STRUCTURE, , and Powell in 1940, proposed a simple theory, based on the repulsive interactions of the, electron pairs in the valence shell of the atoms., It was further developed and redefined by, Nyholm and Gillespie (1957)., The main postulates of VSEPR theory are, as follows:, •, , The shape of a molecule depends upon, the number of valence shell electron pairs, (bonded or nonbonded) around the central, atom., , •, , Pairs of electrons in the valence shell repel, one another since their electron clouds are, negatively charged., , •, , These pairs of electrons tend to occupy, such positions in space that minimise, repulsion and thus maximise distance, between them., , •, , The valence shell is taken as a sphere with, the electron pairs localising on the, spherical surface at maximum distance, from one another., , •, , A multiple bond is treated as if it is a single, electron pair and the two or three electron, pairs of a multiple bond are treated as a, single super pair., , •, , Where two or more resonance structures, can represent a molecule, the VSEPR, model is applicable to any such structure., , result in deviations from idealised shapes and, alterations in bond angles in molecules., For the prediction of geometrical shapes of, molecules with the help of VSEPR theory, it is, convenient to divide molecules into two, categories as (i) molecules in which the, central atom has no lone pair and (ii), molecules in which the central atom has, one or more lone pairs., Table 4.6 (page114) shows the, arrangement of electron pairs about a central, atom A (without any lone pairs) and, geometries of some molecules/ions of the type, AB. Table 4.7 (page 115) shows shapes of some, simple molecules and ions in which the central, atom has one or more lone pairs. Table 4.8, (page 116) explains the reasons for the, distortions in the geometry of the molecule., As depicted in Table 4.6, in the, compounds of AB2, AB3, AB4, AB5 and AB6,, the arrangement of electron pairs and the B, atoms around the central atom A are : linear,, trigonal planar, tetrahedral, trigonalbipyramidal and octahedral, respectively., Such arrangement can be seen in the, molecules like BF3 (AB3), CH4 (AB4) and PCl5, (AB5) as depicted below by their ball and, stick models., , The repulsive interaction of electron pairs, decrease in the order:, Lone pair (lp) – Lone pair (lp) > Lone pair (lp), – Bond pair (bp) > Bond pair (bp) –, Bond pair (bp), , Fig. 4.6 The shapes of molecules in which, central atom has no lone pair, , Nyholm and Gillespie (1957) refined the, VSEPR model by explaining the important, difference between the lone pairs and bonding, pairs of electrons. While the lone pairs are, localised on the central atom, each bonded pair, is shared between two atoms. As a result, the, lone pair electrons in a molecule occupy more, space as compared to the bonding pairs of, electrons. This results in greater repulsion, between lone pairs of electrons as compared, to the lone pair - bond pair and bond pair bond pair repulsions. These repulsion effects, , The VSEPR Theory is able to predict, geometry of a large number of molecules,, especially the compounds of p-block elements, accurately. It is also quite successful in, determining the geometry quite-accurately, even when the energy difference between, possible structures is very small. The, theoretical basis of the VSEPR theory, regarding the effects of electron pair repulsions, on molecular shapes is not clear and, continues to be a subject of doubt and, discussion., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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114, , CHEMISTRY, , Table 4.6 Geometry of Molecules in which the Central Atom has No Lone Pair of Electrons, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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115, , CHEMICAL BONDING AND MOLECULAR STRUCTURE, , Table 4.7, , Shape (geometry) of Some Simple Molecules/Ions with Central Ions having One or, More Lone Pairs of Electrons(E)., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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116, , CHEMISTRY, , Table 4.8 Shapes of Molecules containing Bond Pair and Lone Pair, Molecule, type, AB2E, , No. of, lone, pairs, , 4, , 1, , Bent, , Theoretically the shape, should have been triangular, planar but actually it is found, to be bent or v-shaped. The, reason being the lone pairbond pair repulsion is much, more as compared to the, bond pair-bond pair repulsion. So the angle is reduced, to 119.5° from 120°., , Trigonal, pyramidal, , Had there been a bp in place, of lp the shape would have, been tetrahedral but one, lone pair is present and due, to the repulsion between, lp-bp (which is more than, bp-bp repulsion) the angle, between bond pairs is, reduced to 107° from 109.5°., , AB3E, , 3, , 1, , AB2E2, , 2, , 2, , AB4E, , 4, , 1, , Arrangement, of electrons, , Reason for the, shape acquired, , No. of, bonding, pairs, , Shape, , Bent, , The shape should have been, tetrahedral if there were all bp, but two lp are present so the, shape is distorted tetrahedral, or angular. The reason is, lp-lp repulsion is more than, lp-bp repulsion which is more, than bp-bp repulsion. Thus,, the angle is reduced to 104.5°, from 109.5°., , See- In (a) the lp is present at axial, saw position so there are three, lp—bp repulsions at 90°. In(b), the lp is in an equatorial, position, and there are two, lp—bp repulsions. Hence,, arrangement (b) is more, stable. The shape shown in (b), is described as a distorted, tetrahedron, a folded square or, (More stable), a see-saw., Download all NCERT books PDFs from www.ncert.online, , 2019-20

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117, , CHEMICAL BONDING AND MOLECULAR STRUCTURE, , Molecule, type, , AB3E2, , No. of, bonding, pairs, 3, , No. of, lone, pairs, , Arrangement, of electrons, , Shape, , T-shape, , 2, , 4.5 VALENCE BOND THEORY, As we know that Lewis approach helps in, writing the structure of molecules but it fails, to explain the formation of chemical bond. It, also does not give any reason for the difference, in bond dissociation enthalpies and bond, lengths in molecules like H2 (435.8 kJ mol-1,, 74 pm) and F 2 (155 kJ mol -1 , 144 pm),, although in both the cases a single covalent, bond is formed by the sharing of an electron, pair between the respective atoms. It also gives, no idea about the shapes of polyatomic, molecules., Similarly the VSEPR theory gives the, geometry of simple molecules but, theoretically, it does not explain them and also, it has limited applications. To overcome these, limitations the two important theories based, on quantum mechanical principles are, introduced. These are valence bond (VB) theory, and molecular orbital (MO) theory., Valence bond theory was introduced by, Heitler and London (1927) and developed, further by Pauling and others. A discussion, of the valence bond theory is based on the, , Reason for the, shape acquired, , In (a) the lp are at, equatorial position so, there are less lp-bp, repulsions, as, compared to others in, which the lp are at, axial positions. So, structure (a) is most, stable. (T -shaped)., , knowledge of atomic orbitals, electronic, configurations of elements (Units 2), the, overlap criteria of atomic orbitals, the, hybridization of atomic orbitals and the, principles of variation and superposition. A, rigorous treatment of the VB theory in terms, of these aspects is beyond the scope of this, book. Therefore, for the sake of convenience,, valence bond theory has been discussed in, terms of qualitative and non-mathematical, treatment only. To start with, let us consider, the formation of hydrogen molecule which is, the simplest of all molecules., Consider two hydrogen atoms A and B, approaching each other having nuclei NA and, N B and electrons present in them are, represented by eA and eB. When the two atoms, are at large distance from each other, there is, no interaction between them. As these two, atoms approach each other, new attractive and, repulsive forces begin to operate., Attractive forces arise between:, (i) nucleus of one atom and its own electron, that is NA – eA and NB– eB., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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118, , CHEMISTRY, , (ii) nucleus of one atom and electron of other, atom i.e., NA– eB, NB– eA., Similarly repulsive forces arise between, (i) electrons of two atoms like e A – e B ,, (ii) nuclei of two atoms NA – NB., Attractive forces tend to bring the two, atoms close to each other whereas repulsive, forces tend to push them apart (Fig. 4.7)., , hydrogen atoms are said to be bonded together, to form a stable molecule having the bond, length of 74 pm., Since the energy gets released when the, bond is formed between two hydrogen atoms,, the hydrogen molecule is more stable than that, of isolated hydrogen atoms. The energy so, released is called as bond enthalpy, which is, corresponding to minimum in the curve, depicted in Fig. 4.8. Conversely, 435.8 kJ of, energy is required to dissociate one mole of, H2 molecule., H2(g) + 435.8 kJ mol–1 → H(g) + H(g), , Fig. 4.8 The potential energy curve for the, formation of H2 molecule as a function of, internuclear distance of the H atoms. The, minimum in the curve corresponds to the, most stable state of H2., , 4.5.1 Orbital Overlap Concept, , Fig. 4.7, , Forces of attraction and repulsion during, the formation of H2 molecule., , Experimentally it has been found that the, magnitude of new attractive force is more than, the new repulsive forces. As a result, two, atoms approach each other and potential, energy decreases. Ultimately a stage is, reached where the net force of attraction, balances the force of repulsion and system, acquires minimum energy. At this stage two, , In the formation of hydrogen molecule, there, is a minimum energy state when two hydrogen, atoms are so near that their atomic orbitals, undergo partial interpenetration. This partial, merging of atomic orbitals is called overlapping, of atomic orbitals which results in the pairing, of electrons. The extent of overlap decides the, strength of a covalent bond. In general, greater, the overlap the stronger is the bond formed, between two atoms. Therefore, according to, orbital overlap concept, the formation of a, covalent bond between two atoms results by, pairing of electrons present in the valence shell, having opposite spins., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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119, , CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 4.5.2 Directional Properties of Bonds, As we have already seen, the covalent bond is, formed by overlapping of atomic orbitals. The, molecule of hydrogen is formed due to the, overlap of 1s-orbitals of two H atoms., In case of polyatomic molecules like CH4,, NH3 and H2O, the geometry of the molecules is, also important in addition to the bond, formation. For example why is it so that CH4, molecule has tetrahedral shape and HCH bond, angles are 109.5°? Why is the shape of NH3, molecule pyramidal ?, The valence bond theory explains the, shape, the formation and directional properties, of bonds in polyatomic molecules like CH4, NH3, and H 2 O, etc. in terms of overlap and, hybridisation of atomic orbitals., 4.5.3 Overlapping of Atomic Orbitals, When orbitals of two atoms come close to form, bond, their overlap may be positive, negative, or zero depending upon the sign (phase) and, direction of orientation of amplitude of orbital, wave function in space (Fig. 4.9). Positive and, negative sign on boundary surface diagrams, in the Fig. 4.9 show the sign (phase) of orbital, wave function and are not related to charge., Orbitals forming bond should have same sign, (phase) and orientation in space. This is called, positive overlap. Various overlaps of s and p, orbitals are depicted in Fig. 4.9., The criterion of overlap, as the main factor, for the formation of covalent bonds applies, uniformly to the homonuclear/heteronuclear, diatomic molecules and polyatomic molecules., We know that the shapes of CH4, NH3, and H2O, molecules are tetrahedral, pyramidal and bent, respectively. It would be therefore interesting, to use VB theory to find out if these geometrical, shapes can be explained in terms of the orbital, overlaps., Let us first consider the CH4 (methane), molecule. The electronic configuration of, carbon in its ground state is [He]2s2 2p2 which, in the excited state becomes [He] 2s1 2px1 2py1, 2pz1. The energy required for this excitation is, compensated by the release of energy due to, overlap between the orbitals of carbon and the, , Fig.4.9, , Positive, negative and zero overlaps of, s and p atomic orbitals, , hydrogen.The four atomic orbitals of carbon,, each with an unpaired electron can overlap, with the 1s orbitals of the four H atoms which, are also singly occupied. This will result in the, formation of four C-H bonds. It will, however,, be observed that while the three p orbitals of, carbon are at 90° to one another, the HCH, angle for these will also be 90 °. That is three, C-H bonds will be oriented at 90° to one, another. The 2s orbital of carbon and the 1s, orbital of H are spherically symmetrical and, they can overlap in any direction. Therefore, the direction of the fourth C-H bond cannot, be ascertained. This description does not fit, in with the tetrahedral HCH angles of 109.5°., Clearly, it follows that simple atomic orbital, overlap does not account for the directional, characteristics of bonds in CH4. Using similar, procedure and arguments, it can be seen that in the, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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120, , CHEMISTRY, , case of NH3 and H2O molecules, the HNH and, HOH angles should be 90 ° . This is in, disagreement with the actual bond angles of, 107 ° and 104.5 ° in the NH 3 and H 2 O, molecules respectively., 4.5.4 Types of Overlapping and Nature of, Covalent Bonds, The covalent bond may be classified into two, types depending upon the types of, overlapping:, (i) Sigma(σ) bond, and (ii) pi(π) bond, (i) Sigma(σ, σ) bond : This type of covalent bond, is formed by the end to end (head-on), overlap of bonding orbitals along the, internuclear axis. This is called as head, on overlap or axial overlap. This can be, formed by any one of the following types, of combinations of atomic orbitals., • s-s overlapping : In this case, there is, overlap of two half filled s-orbitals along, the internuclear axis as shown below :, , charged clouds above and below the plane, of the participating atoms., , 4.5.5 Strength of Sigma and pi Bonds, Basically the strength of a bond depends upon, the extent of overlapping. In case of sigma bond,, the overlapping of orbitals takes place to a, larger extent. Hence, it is stronger as compared, to the pi bond where the extent of overlapping, occurs to a smaller extent. Further, it is, important to note that in the formation of, multiple bonds between two atoms of a, molecule, pi bond(s) is formed in addition to a, sigma bond., 4.6 HYBRIDISATION, , •, , s-p overlapping: This type of overlap, occurs between half filled s-orbitals of one, atom and half filled p-orbitals of another, atom., , •, , p–p overlapping : This type of overlap, takes place between half filled p-orbitals, of the two approaching atoms., , In order to explain the characteristic, geometrical shapes of polyatomic molecules, like CH4, NH3 and H2O etc., Pauling introduced, the concept of hybridisation. According to him, the atomic orbitals combine to form new set of, equivalent orbitals known as hybrid orbitals., Unlike pure orbitals, the hybrid orbitals are, used in bond formation. The phenomenon is, known as hybridisation which can be defined, as the process of intermixing of the orbitals of, slightly different energies so as to redistribute, their energies, resulting in the formation of new, set of orbitals of equivalent energies and shape., For example when one 2s and three 2p-orbitals, of carbon hybridise, there is the formation of, four new sp 3 hybrid orbitals., Salient features of hybridisation: The main, features of hybridisation are as under :, , (ii) pi(π ) bond : In the formation of π bond, the atomic orbitals overlap in such a way, that their axes remain parallel to each other, and perpendicular to the internuclear axis., The orbitals formed due to sidewise, overlapping consists of two saucer type, , 1. The number of hybrid orbitals is equal to, the number of the atomic orbitals that get, hybridised., 2. The hybridised orbitals are always, equivalent in energy and shape., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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121, , CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 3. The hybrid orbitals are more effective in, forming stable bonds than the pure atomic, orbitals., 4. These hybrid orbitals are directed in space, in some preferred direction to have, minimum repulsion between electron, pairs and thus a stable arrangement., Therefore, the type of hybridisation, indicates the geometry of the molecules., Important conditions for hybridisation, (i) The orbitals present in the valence shell, of the atom are hybridised., (ii) The orbitals undergoing hybridisation, should have almost equal energy., (iii) Promotion of electron is not essential, condition prior to hybridisation., (iv) It is not necessary that only half filled, orbitals participate in hybridisation. In, some cases, even filled orbitals of valence, shell take part in hybridisation., , vacant 2p orbital to account for its bivalency., One 2s and one 2p-orbital gets hybridised to, form two sp hybridised orbitals. These two, sp hybrid orbitals are oriented in opposite, direction forming an angle of 180°. Each of, the sp hybridised orbital overlaps with the, 2p-orbital of chlorine axially and form two BeCl sigma bonds. This is shown in Fig. 4.10., , 4.6.1 Types of Hybridisation, There are various types of hybridisation, involving s, p and d orbitals. The different, types of hybridisation are as under:, (I) sp hybridisation: This type of, hybridisation involves the mixing of one s and, one p orbital resulting in the formation of two, equivalent sp hybrid orbitals. The suitable, orbitals for sp hybridisation are s and pz, if, the hybrid orbitals are to lie along the z-axis., Each sp hybrid orbitals has 50% s-character, and 50% p-character. Such a molecule in, which the central atom is sp-hybridised and, linked directly to two other central atoms, possesses linear geometry. This type of, hybridisation is also known as diagonal, hybridisation., The two sp hybrids point in the opposite, direction along the z-axis with projecting, positive lobes and very small negative lobes,, which provides more effective overlapping, resulting in the formation of stronger bonds., Example of molecule having sp, hybridisation, BeCl 2 : The ground state electronic, configuration of Be is 1s22s2. In the exited state, one of the 2s-electrons is promoted to, , Fig.4.10, , Be, , (a) Formation of sp hybrids from s and, p orbitals; (b) Formation of the linear, BeCl2 molecule, , (II) sp2 hybridisation : In this hybridisation, there is involvement of one s and two, p-orbitals in order to form three equivalent sp2, hybridised orbitals. For example, in BCl3, molecule, the ground state electronic, configuration of central boron atom is, 1s22s22p1. In the excited state, one of the 2s, electrons is promoted to vacant 2p orbital as, , Fig.4.11 Formation of sp2 hybrids and the BCl3, molecule, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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122, , CHEMISTRY, , a result boron has three unpaired electrons., These three orbitals (one 2s and two 2p), hybridise to form three sp2 hybrid orbitals. The, three hybrid orbitals so formed are oriented in, a trigonal planar arrangement and overlap with, 2p orbitals of chlorine to form three B-Cl, bonds. Therefore, in BCl3 (Fig. 4.11), the, geometry is trigonal planar with ClBCl bond, angle of 120°., (III) sp 3 hybridisation: This type of, hybridisation can be explained by taking the, example of CH4 molecule in which there is, mixing of one s-orbital and three p-orbitals of, the valence shell to form four sp3 hybrid orbital, of equivalent energies and shape. There is 25%, s-character and 75% p-character in each sp3, hybrid orbital. The four sp3 hybrid orbitals so, formed are directed towards the four corners, of the tetrahedron. The angle between sp3, hybrid orbital is 109.5° as shown in Fig. 4.12., , ground state is 2S 2 2 p1x 2 p1y 2 p1z having three, unpaired electrons in the sp3 hybrid orbitals, and a lone pair of electrons is present in the, fourth one. These three hybrid orbitals overlap, with 1s orbitals of hydrogen atoms to form, three N–H sigma bonds. We know that the force, of repulsion between a lone pair and a bond, pair is more than the force of repulsion, between two bond pairs of electrons. The, molecule thus gets distorted and the bond, angle is reduced to 107° from 109.5°. The, geometry of such a molecule will be pyramidal, as shown in Fig. 4.13., , Fig.4.13 Formation of NH3 molecule, , In case of H2O molecule, the four oxygen, orbitals (one 2s and three 2p) undergo sp3, hybridisation forming four sp3 hybrid orbitals, out of which two contain one electron each and, the other two contain a pair of electrons. These, four sp3 hybrid orbitals acquire a tetrahedral, geometry, with two corners occupied by, hydrogen atoms while the other two by the lone, pairs. The bond angle in this case is reduced, to 104.5° from 109.5° (Fig. 4.14) and the, molecule thus acquires a V-shape or, angular geometry., , σ, σ, , σ, , σ, , Fig.4.12 For mation of sp 3 hybrids by the, combination of s , px , py and pz atomic, orbitals of carbon and the formation of, CH4 molecule, , The structure of NH3 and H2O molecules, can also be explained with the help of sp3, hybridisation. In NH3, the valence shell (outer), electronic configuration of nitrogen in the, , Fig.4.14 Formation of H2O molecule, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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123, , CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 4.6.2 Other Examples of sp3, sp2 and sp, Hybridisation, sp3 Hybridisation in C2H 6 molecule: In, ethane molecule both the carbon atoms, assume sp3 hybrid state. One of the four sp3, hybrid orbitals of carbon atom overlaps axially, with similar orbitals of other atom to form, sp3-sp3 sigma bond while the other three, hybrid orbitals of each carbon atom are used, in forming sp3–s sigma bonds with hydrogen, atoms as discussed in section 4.6.1(iii)., Therefore in ethane C–C bond length is 154, pm and each C–H bond length is 109 pm., sp2 Hybridisation in C2H4: In the formation, of ethene molecule, one of the sp2 hybrid, orbitals of carbon atom overlaps axially with, sp2 hybridised orbital of another carbon atom, to form C–C sigma bond. While the other two, , sp2 hybrid orbitals of each carbon atom are, used for making sp2–s sigma bond with two, hydrogen atoms. The unhybridised orbital (2px, or 2py) of one carbon atom overlaps sidewise, with the similar orbital of the other carbon, atom to form weak π bond, which consists of, two equal electron clouds distributed above, and below the plane of carbon and hydrogen, atoms., Thus, in ethene molecule, the carboncarbon bond consists of one sp2–sp2 sigma, bond and one pi (π ) bond between p orbitals, which are not used in the hybridisation and, are perpendicular to the plane of molecule;, the bond length 134 pm. The C–H bond is, sp2–s sigma with bond length 108 pm. The H–, C–H bond angle is 117.6° while the H–C–C, angle is 121°. The formation of sigma and pi, bonds in ethene is shown in Fig. 4.15., , Fig. 4.15 Formation of sigma and pi bonds in ethene, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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124, , CHEMISTRY, , sp Hybridisation in C2H2 : In the formation, of ethyne molecule, both the carbon atoms, undergo sp-hybridisation having two, unhybridised orbital i.e., 2py and 2px., One sp hybrid orbital of one carbon atom, overlaps axially with sp hybrid orbital of the, other carbon atom to form C–C sigma bond,, while the other hybridised orbital of each, carbon atom overlaps axially with the half, filled s orbital of hydrogen atoms forming σ, bonds. Each of the two unhybridised p orbitals, of both the carbon atoms overlaps sidewise to, form two π bonds between the carbon atoms., So the triple bond between the two carbon, atoms is made up of one sigma and two pi, bonds as shown in Fig. 4.16., , 4.6.3 Hybridisation of Elements involving, d Orbitals, The elements present in the third period, contain d orbitals in addition to s and p, orbitals. The energy of the 3d orbitals are, comparable to the energy of the 3s and 3p, orbitals. The energy of 3d orbitals are also, comparable to those of 4s and 4p orbitals. As, a consequence the hybridisation involving, either 3s, 3p and 3d or 3d, 4s and 4p is, possible. However, since the difference in, energies of 3p and 4s orbitals is significant, no, hybridisation involving 3p, 3d and 4s orbitals, is possible., The important hybridisation schemes, involving s, p and d orbitals are summarised, below:, , Shape of, molecules/, ions, , Hybridisation, type, , Atomic, orbitals, , Examples, , Square, planar, , dsp2, , d+s+p(2), , [Ni(CN)4]2–,, [Pt(Cl)4]2–, , Trigonal, bipyramidal, , sp3d, , s+p(3)+d, , PF5, PCl5, , Square, pyramidal, , sp3d2, , s+p(3)+d(2), , BrF5, , Octahedral, , sp3d2, d2sp3, , s+p(3)+d(2), d(2)+s+p(3), , SF6, [CrF6]3–, [Co(NH3)6]3+, , (i) Formation of PCl5 (sp3d hybridisation):, The ground state and the excited state outer, electronic configurations of phosphorus (Z=15), are represented below., , Fig.4.16 Formation of sigma and pi bonds in, ethyne, , sp3d hybrid orbitals filled by electron pairs, donated by five Cl atoms., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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125, , CHEMICAL BONDING AND MOLECULAR STRUCTURE, , Now the five orbitals (i.e., one s, three p and, one d orbitals) are available for hybridisation, to yield a set of five sp3d hybrid orbitals which, are directed towards the five corners of a, trigonal bipyramidal as depicted in the Fig., 4.17., , hybrid orbitals overlap with singly occupied, orbitals of fluorine atoms to form six S–F sigma, bonds. Thus SF 6 molecule has a regular, octahedral geometry as shown in Fig. 4.18., , sp3d2 hybridisation, , Fig. 4.17 Trigonal bipyramidal geometry of PCl5, molecule, , It should be noted that all the bond angles, in trigonal bipyramidal geometry are not, equivalent. In PCl5 the five sp3d orbitals of, phosphorus overlap with the singly occupied, p orbitals of chlorine atoms to form five P–Cl, sigma bonds. Three P–Cl bond lie in one plane, and make an angle of 120° with each other;, these bonds are termed as equatorial bonds., The remaining two P–Cl bonds–one lying, above and the other lying below the equatorial, plane, make an angle of 90° with the plane., These bonds are called axial bonds. As the axial, bond pairs suffer more repulsive interaction, from the equatorial bond pairs, therefore axial, bonds have been found to be slightly longer, and hence slightly weaker than the equatorial, bonds; which makes PCl5 molecule more, reactive., (ii) Formation of SF6 (sp3d2 hybridisation):, In SF6 the central sulphur atom has the, ground state outer electronic configuration, 3s23p4. In the exited state the available six, orbitals i.e., one s, three p and two d are singly, occupied by electrons. These orbitals hybridise, to form six new sp3d2 hybrid orbitals, which, are projected towards the six corners of a, regular octahedron in SF6. These six sp3d2, , Fig. 4.18 Octahedral geometry of SF6 molecule, , 4.7 MOLECULAR ORBITAL THEORY, Molecular orbital (MO) theory was developed, by F. Hund and R.S. Mulliken in 1932. The, salient features of this theory are :, (i), The electrons in a molecule are present, in the various molecular orbitals as the, electrons of atoms are present in the, various atomic orbitals., (ii) The atomic orbitals of comparable, energies and proper symmetry combine, to form molecular orbitals., (iii) While an electron in an atomic orbital is, influenced by one nucleus, in a, molecular orbital it is influenced by two, or more nuclei depending upon the, number of atoms in the molecule. Thus,, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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126, , CHEMISTRY, , an atomic orbital is monocentric while a, molecular orbital is polycentric., (iv), , (v), , (vi), , The number of molecular orbital formed, is equal to the number of combining, atomic orbitals. When two atomic, orbitals combine, two molecular orbitals, are formed. One is known as bonding, molecular orbital while the other is, called antibonding molecular orbital., The bonding molecular orbital has lower, energy and hence greater stability than, the corresponding antibonding, molecular orbital., Just as the electron probability, distribution around a nucleus in an, atom is given by an atomic orbital, the, electron probability distribution around, a group of nuclei in a molecule is given, by a molecular orbital., , Mathematically, the formation of molecular, orbitals may be described by the linear, combination of atomic orbitals that can take, place by addition and by subtraction of wave, functions of individual atomic orbitals as, shown below :, , ψMO = ψA + ψB, Therefore, the two molecular orbitals, σ and σ* are formed as :, σ = ψA + ψB, σ* = ψA – ψB, The molecular orbital σ formed by the, addition of atomic orbitals is called the, bonding molecular orbital while the, molecular orbital σ* formed by the subtraction, of atomic orbital is called antibonding, molecular orbital as depicted in Fig. 4.19., , (vii) The molecular orbitals like atomic, orbitals are filled in accordance with the, aufbau principle obeying the Pauli’s, exclusion principle and the Hund’s rule., 4.7.1 Formation of Molecular Orbitals, Linear Combination of Atomic, Orbitals (LCAO), According to wave mechanics, the atomic, orbitals can be expressed by wave functions, (ψ ’s) which represent the amplitude of the, electron waves. These are obtained from the, solution of Schrödinger wave equation., However, since it cannot be solved for any, system containing more than one electron,, molecular orbitals which are one electron wave, functions for molecules are difficult to obtain, directly from the solution of Schrödinger wave, equation. To overcome this problem, an, approximate method known as linear, combination of atomic orbitals (LCAO) has, been adopted., Let us apply this method to the, homonuclear diatomic hydrogen molecule., Consider the hydrogen molecule consisting, of two atoms A and B. Each hydrogen atom in, the ground state has one electron in 1s orbital., The atomic orbitals of these atoms may be, represented by the wave functions ψA and ψB., , σ* = ψA – ψB, , ψB, , ψA, σ = ψA + ψB, , Fig.4.19 For mation of bonding (σ) and, antibonding (σ*) molecular orbitals by the, linear combination of atomic orbitals ψA, and ψB centered on two atoms A and B, respectively., , Qualitatively, the formation of molecular, orbitals can be understood in terms of the, constructive or destructive interference of the, electron waves of the combining atoms. In the, formation of bonding molecular orbital, the, two electron waves of the bonding atoms, reinforce each other due to constructive, interference while in the formation of, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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127, , CHEMICAL BONDING AND MOLECULAR STRUCTURE, , antibonding molecular orbital, the electron, waves cancel each other due to destructive, interference. As a result, the electron density, in a bonding molecular orbital is located, between the nuclei of the bonded atoms, because of which the repulsion between the, nuclei is very less while in case of an, antibonding molecular orbital, most of the, electron density is located away from the space, between the nuclei. Infact, there is a nodal, plane (on which the electron density is zero), between the nuclei and hence the repulsion, between the nuclei is high. Electrons placed in, a bonding molecular orbital tend to hold the, nuclei together and stabilise a molecule., Therefore, a bonding molecular orbital always, possesses lower energy than either of the, atomic orbitals that have combined to form it., In contrast, the electrons placed in the, antibonding molecular orbital destabilise the, molecule. This is because the mutual repulsion, of the electrons in this orbital is more than the, attraction between the electrons and the nuclei,, which causes a net increase in energy., It may be noted that the energy of the, antibonding orbital is raised above the energy, of the parent atomic orbitals that have, combined and the energy of the bonding orbital, has been lowered than the parent orbitals. The, total energy of two molecular orbitals, however,, remains the same as that of two original atomic, orbitals., 4.7.2 Conditions for the Combination of, Atomic Orbitals, The linear combination of atomic orbitals to, form molecular orbitals takes place only if the, following conditions are satisfied:, 1. The combining atomic orbitals must, have the same or nearly the same energy., This means that 1s orbital can combine with, another 1s orbital but not with 2s orbital, because the energy of 2s orbital is appreciably, higher than that of 1s orbital. This is not true, if the atoms are very different., 2. The combining atomic orbitals must, have the same symmetry about the, , molecular axis. By convention z-axis is, taken as the molecular axis. It is important, to note that atomic orbitals having same or, nearly the same energy will not combine if, they do not have the same symmetry. For, example, 2pz orbital of one atom can combine, with 2p z orbital of the other atom but not, with the 2px or 2py orbitals because of their, different symmetries., 3. The combining atomic orbitals must, overlap to the maximum extent. Greater, the extent of overlap, the greater will be the, electron-density between the nuclei of a, molecular orbital., 4.7.3 Types of Molecular Orbitals, Molecular orbitals of diatomic molecules are, designated as σ (sigma), π (pi), δ (delta), etc., In this nomenclature, the sigma (σ, σ), molecular orbitals are symmetrical around, the bond-axis while pi (π) molecular orbitals, are not symmetrical. For example, the linear, combination of 1s orbitals centered on two, nuclei produces two molecular orbitals which, are symmetrical around the bond-axis. Such, molecular orbitals are of the σ type and are, designated as σ1s and σ*1s [Fig. 4.20(a),page, 124]. If internuclear axis is taken to be in, the z-direction, it can be seen that a linear, combination of 2pz- orbitals of two atoms, also produces two sigma molecular orbitals, designated as σ2pz and σ*2pz. [Fig. 4.20(b)], Molecular orbitals obtained from 2px and, 2py orbitals are not symmetrical around the, bond axis because of the presence of positive, lobes above and negative lobes below the, molecular plane. Such molecular orbitals, are, labelled as π and π * [Fig. 4.20(c)]. A π bonding, MO has larger electron density above and, below the inter -nuclear axis. The π *, antibonding MO has a node between the nuclei., 4.7.4 Energy Level Diagram for Molecular, Orbitals, We have seen that 1s atomic orbitals on two, atoms form two molecular orbitals designated, as σ1s and σ*1s. In the same manner, the 2s, and 2p atomic orbitals (eight atomic orbitals, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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129, , CHEMICAL BONDING AND MOLECULAR STRUCTURE, , energies of various molecular orbitals for O2, and F2 is given below:, σ1s < σ * 1s < σ 2s < σ * 2s < σ 2 p z < ( π 2 p x = π 2 p y ), < ( π * 2 p x = π * 2 p y ) < σ * 2 pz, , However, this sequence of energy levels of, molecular orbitals is not correct for the, remaining molecules Li2, Be2, B2, C2, N2. For, instance, it has been observed experimentally, that for molecules such as B2, C2, N2, etc. the, increasing order of energies of various, molecular orbitals is, σ1s < σ * 1s < σ 2s < σ * 2s < ( π 2 p x = π 2 p y ), , The rules discussed above regarding the, stability of the molecule can be restated in, terms of bond order as follows: A positive bond, order (i.e., Nb > Na) means a stable molecule, while a negative (i.e., Nb<Na) or zero (i.e.,, N b = N a ) bond order means an unstable, molecule., Nature of the bond, Integral bond order values of 1, 2 or 3, correspond to single, double or triple bonds, respectively as studied in the classical, concept., Bond-length, , < σ 2 pz < ( π * 2 p x = π * 2 p y ) < σ * 2 p z, , The important characteristic feature of this, order is that the energy of σ 2p z, molecular orbital is higher than that, of π 2px and π 2py molecular orbitals., 4.7.5 Electronic Configuration and, Molecular Behaviour, The distribution of electrons among various, molecular orbitals is called the electronic, configuration of the molecule. From the, electronic configuration of the molecule, it is, possible to get important information about, the molecule as discussed below., Stability of Molecules: If Nb is the number, of electrons occupying bonding orbitals and, Na the number occupying the antibonding, orbitals, then, (i) the molecule is stable if Nb is greater than, Na, and, (ii) the molecule is unstable if N b is less, than Na., In (i) more bonding orbitals are occupied, and so the bonding influence is stronger and a, stable molecule results. In (ii) the antibonding, influence is stronger and therefore the molecule, is unstable., , The bond order between two atoms in a, molecule may be taken as an approximate, measure of the bond length. The bond length, decreases as bond order increases., Magnetic nature, If all the molecular orbitals in a molecule are, doubly occupied, the substance is, diamagnetic (repelled by magnetic field)., However if one or more molecular orbitals are, singly occupied it is paramagnetic (attracted, by magnetic field), e.g., O2 molecule., 4.8 BONDING IN SOME HOMONUCLEAR, DIATOMIC MOLECULES, In this section we shall discuss bonding in, some homonuclear diatomic molecules., 1. Hydrogen molecule (H2 ): It is formed by, the combination of two hydrogen atoms. Each, hydrogen atom has one electron in 1s orbital., Therefore, in all there are two electrons in, hydrogen molecule which are present in σ1s, molecular orbital. So electronic configuration, of hydrogen molecule is, H2 : (σ1s)2, The bond order of H2 molecule can be, calculated as given below:, , Bond order, , Bond order =, , Bond order (b.o.) is defined as one half the, difference between the number of electrons, present in the bonding and the antibonding, orbitals i.e.,, Bond order (b.o.) = ½ (Nb–Na), , N b − Na 2 − 0, =, =1, 2, 2, , This means that the two hydrogen atoms, are bonded together by a single covalent bond., The bond dissociation energy of hydrogen, molecule has been found to be 438 kJ mol–1, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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130, , CHEMISTRY, , and bond length equal to 74 pm. Since no, unpaired electron is present in hydrogen, molecule, therefore, it is diamagnetic., 2. Helium molecule (He2 ): The electronic, configuration of helium atom is 1s2. Each helium, atom contains 2 electrons, therefore, in He2, molecule there would be 4 electrons. These, electrons will be accommodated in σ1s and σ*1s, molecular orbitals leading to electronic, configuration:, He2 : (σ1s)2 (σ*1s)2, Bond order of He2 is ½(2 – 2) = 0, He2 molecule is therefore unstable and, does not exist., Similarly, it can be shown that Be 2, molecule (σ1s)2 (σ*1s)2 (σ2s)2 (σ*2s)2 also does, not exist., 3. Lithium molecule (Li2 ): The electronic, configuration of lithium is 1s2, 2s1 . There are, six electrons in Li2. The electronic configuration, of Li2 molecule, therefore, is, Li2 : (σ1s)2 (σ*1s)2 (σ2s)2, The above configuration is also written as, KK(σ2s)2 where KK represents the closed K shell, structure (σ1s)2 (σ*1s)2., From the electronic configuration of Li2, molecule it is clear that there are four electrons, present in bonding molecular orbitals and two, electrons present in antibonding molecular, orbitals. Its bond order, therefore, is ½ (4 – 2), = 1. It means that Li2 molecule is stable and, since it has no unpaired electrons it should be, diamagnetic. Indeed diamagnetic Li 2, molecules are known to exist in the vapour, phase., 4. Carbon molecule (C2 ): The electronic, configuration of carbon is 1s2 2s2 2p2. There, are twelve electrons in C2. The electronic, configuration of C2 molecule, therefore, is, C2 : (σ1s ) (σ * 1s ) (σ * 2s ) (π 2 p 2x = π 2 py2 ), 2, , 2, , 2, , 2, 2, 2, 2, or KK (σ2s) (σ * 2s ) (π 2p x = π 2p y ), , The bond order of C2 is ½ (8 – 4) = 2 and C2, should be diamagnetic. Diamagnetic C 2, , molecules have indeed been detected in vapour, phase. It is important to note that double bond, in C2 consists of both pi bonds because of the, presence of four electrons in two pi molecular, orbitals. In most of the other molecules a, double bond is made up of a sigma bond and, a pi bond. In a similar fashion the bonding in, N2 molecule can be discussed., 5. Oxygen molecule (O2 ): The electronic, configuration of oxygen atom is 1s2 2s2 2p4., Each oxygen atom has 8 electrons, hence, in, O 2 molecule there are 16 electrons. The, electronic configuration of O 2 molecule,, therefore, is, , O2 : ( σ1s) 2 ( σ*1s) 2 ( σ*2s) 2 ( σ*2s) 2 ( σ2pz ) 2, , (π 2 p, , 2, x, , ≡ π 2 py 2, , ) (π * 2 p, , 1, x, , ≡ π * 2 p y1, , ), , KK (σ2s )2 (σ * 2s )2 (σ2p z )2, O2 : , , 2, 2, 1, 1,  π 2p x ≡ π 2p y , π * 2p x ≡ π * 2p y, , (, , )(, , , , , , ), , From the electronic configuration of O 2, molecule it is clear that ten electrons are present, in bonding molecular orbitals and six electrons, are present in antibonding molecular orbitals., Its bond order, therefore, is, , Bond order =, , 1, 1, [ N b − N a ] = [10 − 6 ] = 2, 2, 2, , So in oxygen molecule, atoms are held, by a double bond. Moreover, it may be noted, that it contains two unpaired electrons in, π * 2p x and π * 2p y molecular orbitals,, therefore, O 2 molecule should be, paramagnetic, a prediction that, corresponds to experimental observation., In this way, the theory successfully explains, the paramagnetic nature of oxygen., Similarly, the electronic configurations of, other homonuclear diatomic molecules of the, second row of the periodic table can be written., In Fig.4.21 are given the molecular orbital, occupancy and molecular properties for B2, through Ne2. The sequence of MOs and their, electron population are shown. The bond, energy, bond length, bond order, magnetic, properties and valence electron configuration, appear below the orbital diagrams., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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131, , CHEMICAL BONDING AND MOLECULAR STRUCTURE, , Fig. 4.21 MO occupancy and molecular properties for B2 through Ne2., , 4.9 HYDROGEN BONDING, Nitrogen, oxygen and fluorine are the higly, electronegative elements. When they are, attached to a hydrogen atom to form covalent, bond, the electrons of the covalent bond are, shifted towards the more electronegative atom., This partially positively charged hydrogen, atom forms a bond with the other more, electronegative atom. This bond is known as, hydrogen bond and is weaker than the, covalent bond. For example, in HF molecule,, the hydrogen bond exists between hydrogen, atom of one molecule and fluorine atom of, another molecule as depicted below :, , − − −H δ+ – F δ− − − − H δ+ – F δ− − − − H δ+ – F δ−, Here, hydrogen bond acts as a bridge between, two atoms which holds one atom by covalent, bond and the other by hydrogen bond., , Hydrogen bond is represented by a dotted line, (– – –) while a solid line represents the covalent, bond. Thus, hydrogen bond can be defined, as the attractive force which binds, hydrogen atom of one molecule with the, electronegative atom (F, O or N) of another, molecule., 4.9.1 Cause of Formation of Hydrogen, Bond, When hydrogen is bonded to strongly, electronegative element ‘X’, the electron pair, shared between the two atoms moves far away, from hydrogen atom. As a result the hydrogen, atom becomes highly electropositive with, respect to the other atom ‘X’. Since there is, displacement of electrons towards X, the, hydrogen acquires fractional positive charge, (δ +) while ‘X’ attain fractional negative charge, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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132, , CHEMISTRY, , (δ–). This results in the formation of a polar, molecule having electrostatic force of attraction, which can be represented as :, , Hδ + − X δ − − − − H δ + − X δ − − − − Hδ + − X δ −, The magnitude of H-bonding depends on, the physical state of the compound. It is, maximum in the solid state and minimum in, the gaseous state. Thus, the hydrogen bonds, have strong influence on the structure and, properties of the compounds., 4.9.2 Types of H-Bonds, There are two types of H-bonds, (i), Intermolecular hydrogen bond, (ii), Intramolecular hydrogen bond, (1) Intermolecular hydrogen bond : It is, formed between two different molecules of the, same or different compounds. For example, H-, , bond in case of HF molecule, alcohol or water, molecules, etc., (2) Intramolecular hydrogen bond : It is, formed when hydrogen atom is in between the, two highly electronegative (F, O, N) atoms, present within the same molecule. For example,, in o-nitrophenol the hydrogen is in between, the two oxygen atoms., , Fig. 4.22 Intramolecular hydrogen bonding in, o-nitrophenol molecule, , SUMMARY, Kössel’s first insight into the mechanism of formation of electropositive and electronegative, ions related the process to the attainment of noble gas configurations by the respective, ions. Electrostatic attraction between ions is the cause for their stability. This gives the, concept of electrovalency., The first description of covalent bonding was provided by Lewis in terms of the sharing, of electron pairs between atoms and he related the process to the attainment of noble gas, configurations by reacting atoms as a result of sharing of electrons. The Lewis dot symbols, show the number of valence electrons of the atoms of a given element and Lewis dot, structures show pictorial representations of bonding in molecules., An ionic compound is pictured as a three-dimensional aggregation of positive and, negative ions in an ordered arrangement called the crystal lattice. In a crystalline solid, there is a charge balance between the positive and negative ions. The crystal lattice is, stabilized by the enthalpy of lattice formation., While a single covalent bond is formed by sharing of an electron pair between two, atoms, multiple bonds result from the sharing of two or three electron pairs. Some bonded, atoms have additional pairs of electrons not involved in bonding. These are called lonepairs of electrons. A Lewis dot structure shows the arrangement of bonded pairs and lone, pairs around each atom in a molecule. Important parameters, associated with chemical, bonds, like: bond length, bond angle, bond enthalpy, bond order and bond polarity, have significant effect on the properties of compounds., A number of molecules and polyatomic ions cannot be described accurately by a single, Lewis structure and a number of descriptions (representations) based on the same skeletal, structure are written and these taken together represent the molecule or ion. This is a very, important and extremely useful concept called resonance. The contributing structures or, canonical forms taken together constitute the resonance hybrid which represents the, molecule or ion., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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133, , CHEMICAL BONDING AND MOLECULAR STRUCTURE, , The VSEPR model used for predicting the geometrical shapes of molecules is based on, the assumption that electron pairs repel each other and, therefore, tend to remain as far, apart as possible. According to this model, molecular geometry is determined by repulsions, between lone pairs and lone pairs ; lone pairs and bonding pairs and bonding pairs and, bonding pairs. The order of these repulsions being : lp-lp > lp-bp > bp-bp, The valence bond (VB) approach to covalent bonding is basically concerned with the, energetics of covalent bond formation about which the Lewis and VSEPR models are silent., Basically the VB theory discusses bond formation in terms of overlap of orbitals. For, example the formation of the H2 molecule from two hydrogen atoms involves the overlap of, the 1s orbitals of the two H atoms which are singly occupied. It is seen that the potential, energy of the system gets lowered as the two H atoms come near to each other. At the, equilibrium inter-nuclear distance (bond distance) the energy touches a minimum. Any, attempt to bring the nuclei still closer results in a sudden increase in energy and consequent, destabilization of the molecule. Because of orbital overlap the electron density between the, nuclei increases which helps in bringing them closer. It is however seen that the actual, bond enthalpy and bond length values are not obtained by overlap alone and other variables, have to be taken into account., For explaining the characteristic shapes of polyatomic molecules Pauling introduced, the concept of hybridisation of atomic orbitals. sp,sp2, sp3 hybridizations of atomic orbitals, of Be, B,C, N and O are used to explain the formation and geometrical shapes of molecules, like BeCl2, BCl3, CH4, NH3 and H2O. They also explain the formation of multiple bonds in, molecules like C2H2 and C2H4., The molecular orbital (MO) theory describes bonding in terms of the combination, and arrangment of atomic orbitals to form molecular orbitals that are associated with the, molecule as a whole. The number of molecular orbitals are always equal to the number of, atomic orbitals from which they are formed. Bonding molecular orbitals increase electron, density between the nuclei and are lower in energy than the individual atomic orbitals., Antibonding molecular orbitals have a region of zero electron density between the nuclei, and have more energy than the individual atomic orbitals., The electronic configuration of the molecules is written by filling electrons in the, molecular orbitals in the order of increasing energy levels. As in the case of atoms, the, Pauli exclusion principle and Hund’s rule are applicable for the filling of molecular orbitals., Molecules are said to be stable if the number of elctrons in bonding molecular orbitals is, greater than that in antibonding molecular orbitals., Hydrogen bond is formed when a hydrogen atom finds itself between two highly, electronegative atoms such as F, O and N. It may be intermolecular (existing between two, or more molecules of the same or different substances) or intramolecular (present within, the same molecule). Hydrogen bonds have a powerful effect on the structure and properties, of many compounds., , EXERCISES, 4.1, , Explain the formation of a chemical bond., , 4.2, , Write Lewis dot symbols for atoms of the following elements : Mg, Na, B, O, N, Br., , 4.3, , Write Lewis symbols for the following atoms and ions:, S and S2–; Al and Al3+; H and H–, , 4.4, , Draw the Lewis structures for the following molecules and ions :, H2S, SiCl4, BeF2, CO23 − , HCOOH, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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134, , CHEMISTRY, , 4.5, , Define octet rule. Write its significance and limitations., , 4.6, , Write the favourable factors for the formation of ionic bond., , 4.7, , Discuss the shape of the following molecules using the VSEPR model:, BeCl2, BCl3, SiCl4, AsF5, H2S, PH3, , 4.8, , Although geometries of NH3 and H2O molecules are distorted tetrahedral, bond, angle in water is less than that of ammonia. Discuss., , 4.9, , How do you express the bond strength in terms of bond order ?, , 4.10, , Define the bond length., , 4.11, , ion., Explain the important aspects of resonance with reference to the CO 2−, 3, , 4.12, , H3PO3 can be represented by structures 1 and 2 shown below. Can these two, structures be taken as the canonical forms of the resonance hybrid representing, H3PO3 ? If not, give reasons for the same., , 4.13, , Write the resonance structures for SO3, NO2 and NO3− ., , 4.14, , Use Lewis symbols to show electron transfer between the following atoms to form, cations and anions : (a) K and S (b) Ca and O (c) Al and N., , 4.15, , Although both CO2 and H2O are triatomic molecules, the shape of H2O molecule, is bent while that of CO2 is linear. Explain this on the basis of dipole moment., , 4.16, , Write the significance/applications of dipole moment., , 4.17, , Define electronegativity. How does it differ from electron gain enthalpy ?, , 4.18, , Explain with the help of suitable example polar covalent bond., , 4.19, , Arrange the bonds in order of increasing ionic character in the molecules: LiF,, K2O, N2, SO2 and ClF3., , 4.20, , The skeletal structure of CH3COOH as shown below is correct, but some of the, bonds are shown incorrectly. Write the correct Lewis structure for acetic acid., , 4.21, , Apart from tetrahedral geometry, another possible geometry for CH4 is square, planar with the four H atoms at the corners of the square and the C atom at its, centre. Explain why CH4 is not square planar ?, , 4.22, , Explain why BeH2 molecule has a zero dipole moment although the Be–H bonds, are polar., , 4.23, , Which out of NH3 and NF3 has higher dipole moment and why ?, , 4.24, , What is meant by hybridisation of atomic orbitals? Describe the shapes of sp,, sp2, sp3 hybrid orbitals., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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135, , CHEMICAL BONDING AND MOLECULAR STRUCTURE, , 4.25, , Describe the change in hybridisation (if any) of the Al atom in the following, reaction., AlCl3 + Cl − → AlCl4−, , 4.26, , Is there any change in the hybridisation of B and N atoms as a result of the, following reaction?, BF3 + NH3 → F3B.NH 3, , 4.27, , Draw diagrams showing the formation of a double bond and a triple bond between, carbon atoms in C2H4 and C2H2 molecules., , 4.28, , What is the total number of sigma and pi bonds in the following molecules?, (a) C2H2 (b) C2H4, , 4.29, , Considering x-axis as the internuclear axis which out of the following will not, form a sigma bond and why? (a) 1s and 1s (b) 1s and 2px ; (c) 2py and 2py, (d) 1s and 2s., , 4.30, , Which hybrid orbitals are used by carbon atoms in the following molecules?, , 4.31, , What do you understand by bond pairs and lone pairs of electrons? Illustrate by, giving one exmaple of each type., , 4.32, , Distinguish between a sigma and a pi bond., , 4.33, , Explain the formation of H2 molecule on the basis of valence bond theory., , 4.34, , Write the important conditions required for the linear combination of atomic orbitals, to form molecular orbitals., , 4.35, , Use molecular orbital theory to explain why the Be2 molecule does not exist., , 4.36, , Compare the relative stability of the following species and indicate their magnetic, properties;, , CH3–CH3; (b) CH3–CH=CH2; (c) CH3-CH2-OH; (d) CH3-CHO (e) CH3COOH, , O2 ,O2+ ,O 2− (superoxide), O22 − (peroxide), , 4.37, , Write the significance of a plus and a minus sign shown in representing the, orbitals., , 4.38, , Describe the hybridisation in case of PCl5. Why are the axial bonds longer as, compared to equatorial bonds?, , 4.39, , Define hydrogen bond. Is it weaker or stronger than the van der Waals forces?, , 4.40, , What is meant by the term bond order? Calculate the bond order of : N2, O2, O2, and O2–., , +, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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136, , CHEMISTRY, , UNIT 5, , STATES OF MATTER, , The snowflake falls, yet lays not long, Its feath’ry grasp on Mother Earth, Ere Sun returns it to the vapors Whence it came,, Or to waters tumbling down the rocky slope., After studying this unit you will be, able to, , Rod O’ Connor, , • explain the existence of different, states of matter in ter ms of, balance between intermolecular, forces and ther mal energy of, particles;, , • explain the laws gover ning, behaviour of ideal gases;, , • apply gas laws in various real life, situations;, , • explain the behaviour of real, gases;, , • describe the conditions required, for liquifaction of gases;, , • realise that there is continuity in, gaseous and liquid state;, , • differentiate between gaseous, state and vapours; and, , • explain properties of liquids in, ter ms, of, attractions., , inter molecular, , INTRODUCTION, In previous units we have learnt about the properties, related to single particle of matter, such as atomic size,, ionization enthalpy, electronic charge density, molecular, shape and polarity, etc. Most of the observable, characteristics of chemical systems with which we are, familiar represent bulk properties of matter, i.e., the, properties associated with a collection of a large number, of atoms, ions or molecules. For example, an individual, molecule of a liquid does not boil but the bulk boils., Collection of water molecules have wetting properties;, individual molecules do not wet. Water can exist as ice,, which is a solid; it can exist as liquid; or it can exist in, the gaseous state as water vapour or steam. Physical, properties of ice, water and steam are very different. In, all the three states of water chemical composition of water, remains the same i.e., H2O. Characteristics of the three, states of water depend on the energies of molecules and, on the manner in which water molecules aggregate. Same, is true for other substances also., Chemical properties of a substance do not change with, the change of its physical state; but rate of chemical, reactions do depend upon the physical state. Many times, in calculations while dealing with data of experiments we, require knowledge of the state of matter. Therefore, it, becomes necessary for a chemist to know the physical, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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STATES OF MATTER, , 137, , laws which govern the behaviour of matter in, different states. In this unit, we will learn, more about these three physical states of, matter particularly liquid and gaseous states., To begin with, it is necessary to understand, the nature of intermolecular forces, molecular, interactions and effect of thermal energy on, the motion of particles because a balance, between these determines the state of a, substance., 5.1 INTERMOLECULAR FORCES, Intermolecular forces are the forces of, attraction and repulsion between interacting, particles (atoms and molecules). This term, does not include the electrostatic forces that, exist between the two oppositely charged ions, and the forces that hold atoms of a molecule, together i.e., covalent bonds., Attractive intermolecular forces are known, as van der Waals forces, in honour of Dutch, scientist Johannes van der Waals (18371923), who explained the deviation of real, gases from the ideal behaviour through these, forces. We will learn about this later in this, unit. van der Waals forces vary considerably, in magnitude and include dispersion forces, or London forces, dipole-dipole forces, and, dipole-induced dipole forces. A particularly, strong type of dipole-dipole interaction is, hydrogen bonding. Only a few elements can, participate in hydrogen bond formation,, therefore it is treated as a separate, category. We have already learnt about this, interaction in Unit 4., At this point, it is important to note that, attractive forces between an ion and a dipole, are known as ion-dipole forces and these are, not van der Waals forces. We will now learn, about different types of van der Waals forces., 5.1.1 Dispersion Forces or London Forces, Atoms and nonpolar molecules are electrically, symmetrical and have no dipole moment, because their electronic charge cloud is, symmetrically distributed. But a dipole may, develop momentarily even in such atoms and, molecules. This can be understood as follows., Suppose we have two atoms ‘A’ and ‘B’ in the, close vicinity of each other (Fig. 5.1a). It may, , so happen that momentarily electronic charge, distribution in one of the atoms, say ‘A’,, becomes unsymmetrical i.e., the charge cloud, is more on one side than the other (Fig. 5.1 b, and c). This results in the development of, instantaneous dipole on the atom ‘A’ for a very, short time. This instantaneous or transient, dipole distorts the electron density of the, other atom ‘B’, which is close to it and as a, consequence a dipole is induced in the, atom ‘B’., The temporary dipoles of atom ‘A’ and ‘B’, attract each other. Similarly temporary dipoles, are induced in molecules also. This force of, attraction was first proposed by the German, physicist Fritz London, and for this reason, force of attraction between two temporary, , Fig. 5.1 Dispersion forces or London forces, between atoms., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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138, , CHEMISTRY, , dipoles is known as London force. Another, name for this force is dispersion force. These, forces are always attractive and interaction, energy is inversely proportional to the sixth, power of the distance between two interacting, particles (i.e., 1/r 6 where r is the distance, between two particles). These forces are, important only at short distances (~500 pm), and their magnitude depends on the, polarisability of the particle., 5.1.2 Dipole - Dipole Forces, Dipole-dipole forces act between the molecules, possessing permanent dipole. Ends of the, dipoles possess “partial charges” and these, charges are shown by Greek letter delta (δ)., Partial charges are always less than the unit, electronic charge (1.6×10–19 C). The polar, molecules interact with neighbouring, molecules. Fig 5.2 (a) shows electron cloud, distribution in the dipole of hydrogen chloride, and Fig. 5.2 (b) shows dipole-dipole interaction, between two HCl molecules. This interaction, is stronger than the London forces but is, weaker than ion-ion interaction because only, partial charges are involved. The attractive, force decreases with the increase of distance, between the dipoles. As in the above case here, also, the interaction energy is inversely, proportional to distance between polar, molecules. Dipole-dipole interaction energy, between stationary polar molecules (as in, solids) is proportional to 1/r 3 and that, between rotating polar molecules is, , proportional to 1/r 6, where r is the distance, between polar molecules. Besides dipoledipole interaction, polar molecules can, interact by London forces also. Thus, cumulative ef fect is that the total of, intermolecular forces in polar molecules, increase., 5.1.3 Dipole–Induced Dipole Forces, This type of attractive forces operate between, the polar molecules having permanent dipole, and the molecules lacking permanent dipole., Permanent dipole of the polar molecule, induces dipole on the electrically neutral, molecule by deforming its electronic cloud, (Fig. 5.3). Thus an induced dipole is developed, in the other molecule. In this case also, interaction energy is proportional to 1/r 6, where r is the distance between two, molecules. Induced dipole moment depends, upon the dipole moment present in the, permanent dipole and the polarisability of the, electrically neutral molecule. We have already, learnt in Unit 4 that molecules of larger size, can be easily polarized. High polarisability, increases the strength of attractive, interactions., , Fig. 5.3 Dipole - induced dipole interaction, between permanent dipole and induced, dipole, , In this case also cumulative effect of, dispersion forces and dipole-induced dipole, interactions exists., Fig. 5.2 (a) Distribution of electron cloud in HCl –, a polar molecule, (b) Dipole-dipole, interaction between two HCl molecules, , 5.1.4 Hydrogen bond, As already mentioned in section (5.1); this is, special case of dipole-dipole interaction. We, have already learnt about this in Unit 4. This, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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STATES OF MATTER, , 139, , is found in the molecules in which highly polar, N–H, O–H or H–F bonds are present. Although, hydrogen bonding is regarded as being limited, to N, O and F; but species such as Cl may, also participate in hydrogen bonding. Energy, of hydrogen bond varies between 10 to 100, kJ mol–1. This is quite a significant amount of, energy; therefore, hydrogen bonds are, powerful force in determining the structure and, properties of many compounds, for example, proteins and nucleic acids. Strength of the, hydrogen bond is determined by the coulombic, interaction between the lone-pair electrons of, the electronegative atom of one molecule and, the hydrogen atom of other molecule., Following diagram shows the formation of, hydrogen bond., δ+, , δ−, , δ+, , thermal energy of the molecules tends to keep, them apart. Three states of matter are the result, of balance between intermolecular forces and, the thermal energy of the molecules., When molecular interactions are very, weak, molecules do not cling together to make, liquid or solid unless thermal energy is, reduced by lowering the temperature. Gases, do not liquify on compression only, although, molecules come very close to each other and, intermolecular forces operate to the maximum., However, when thermal energy of molecules, is reduced by lowering the temperature; the, gases can be very easily liquified., Predominance of thermal energy and the, molecular interaction energy of a substance, in three states is depicted as follows :, , δ−, , H − F ⋅⋅ ⋅ H − F, , Intermolecular forces discussed so far are, all attractive. Molecules also exert repulsive, forces on one another. When two molecules, are brought into close contact with each other,, the repulsion between the electron clouds and, that between the nuclei of two molecules comes, into play. Magnitude of the repulsion rises very, rapidly as the distance separating the, molecules decreases. This is the reason that, liquids and solids are hard to compress. In, these states molecules are already in close, contact; therefore they resist further, compression; as that would result in the, increase of repulsive interactions., , We have already learnt the cause for the, existence of the three states of matter. Now, we will learn more about gaseous and liquid, states and the laws which govern the, behaviour of matter in these states. We shall, deal with the solid state in class XII., , 5.2 THERMAL ENERGY, Thermal energy is the energy of a body arising, from motion of its atoms or molecules. It is, directly proportional to the temperature of the, substance. It is the measure of average, kinetic energy of the particles of the matter, and is thus responsible for movement of, particles. This movement of particles is called, thermal motion., , 5.4 THE GASEOUS STATE, This is the simplest state of matter., Throughout our life we remain immersed in, the ocean of air which is a mixture of gases., We spend our life in the lowermost layer of, the atmosphere called troposphere, which is, held to the surface of the earth by gravitational, force. The thin layer of atmosphere is vital to, our life. It shields us from harmful radiations, and contains substances like dioxygen,, dinitrogen, carbon dioxide, water vapour, etc., , 5.3 INTERMOLECULAR FORCES vs, THERMAL INTERACTIONS, We have already learnt that intermolecular, forces tend to keep the molecules together but, , Let us now focus our attention on the, behaviour of substances which exist in the, gaseous state under normal conditions of, temperature and pressure. A look at the, periodic table shows that only eleven elements, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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140, , CHEMISTRY, , centuries on the physical properties of gases., The first reliable measurement on properties, of gases was made by Anglo-Irish scientist, Robert Boyle in 1662. The law which he, formulated is known as Boyle’s Law. Later, on attempts to fly in air with the help of hot, air balloons motivated Jaccques Charles and, Joseph Lewis Gay Lussac to discover, additional gas laws. Contribution from, Avogadro and others provided lot of, information about gaseous state., , Fig. 5.4 Eleven elements that exist as gases, , exist as gases under normal conditions, (Fig 5.4)., The gaseous state is characterized by the, following physical properties., • Gases are highly compressible., • Gases exert pressure equally in all, directions., • Gases have much lower density than the, solids and liquids., • The volume and the shape of gases are, not fixed. These assume volume and shape, of the container., • Gases mix evenly and completely in all, proportions without any mechanical aid., Simplicity of gases is due to the fact that, the forces of interaction between their, molecules are negligible. Their behaviour is, governed by same general laws, which were, discovered as a result of their experimental, studies. These laws are relationships between, measurable properties of gases. Some of these, properties like pressure, volume, temperature, and mass are very important because, relationships between these variables describe, state of the gas. Interdependence of these, variables leads to the formulation of gas laws., In the next section we will learn about gas, laws., , 5.5.1 Boyle’s Law (Pressure - Volume, Relationship), On the basis of his experiments, Robert Boyle, reached to the conclusion that at constant, temperature, the pressure of a fixed, amount (i.e., number of moles n) of gas, varies inversely with its volume. This is, known as Boyle’s law. Mathematically, it can, be written as, , 1, ( at constant T and n), V, , p ∝, , ⇒ p = k1, , 1, V, , (5.1), , (5.2), , where k1 is the proportionality constant. The, value of constant k 1 depends upon the, amount of the gas, temperature of the gas, and the units in which p and V are expressed., On rearranging equation (5.2) we obtain, pV = k1, , (5.3), , It means that at constant temperature,, product of pressure and volume of a fixed, amount of gas is constant., If a fixed amount of gas at constant, temperature T occupying volume V 1 at, pressure p1 undergoes expansion, so that, volume becomes V2 and pressure becomes p2,, then according to Boyle’s law :, , 5.5 THE GAS LAWS, The gas laws which we will study now are the, result of research carried on for several, , p V = p V = constant, 1, , ⇒, , 1, , 2, , 2, , p1 V2, =, p2, V1, , Download all NCERT books PDFs from www.ncert.online, , 2019-20, , (5.4), , (5.5)

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STATES OF MATTER, , 141, , Figure 5.5 shows two conventional ways, of graphically presenting Boyle’s law., Fig. 5.5 (a) is the graph of equation (5.3) at, different temperatures. The value of k1 for each, curve is different because for a given mass of, gas, it varies only with temperature. Each, curve corresponds to a different constant, temperature and is known as an isotherm, (constant temperature plot). Higher curves, correspond to higher temperature. It should, be noted that volume of the gas doubles if, pressure is halved. Table 5.1 gives effect of, pressure on volume of 0.09 mol of CO2 at, 300 K., Fig 5.5 (b) represents the graph between p, Fig. 5.5(a) Graph of pressure, p vs. Volume, V of, a gas at different temperatures., , and, , 1, . It is a straight line passing through, V, , origin. However at high pressures, gases, deviate from Boyle’s law and under such, conditions a straight line is not obtained in the, graph., Experiments of Boyle, in a quantitative, manner prove that gases are highly, compressible because when a given mass of a, gas is compressed, the same number of, molecules occupy a smaller space. This means, that gases become denser at high pressure. A, relationship can be obtained between density, and pressure of a gas by using Boyle’s law:, By definition, density ‘d’ is related to the, mass ‘m’ and the volume ‘V’ by the relation, , m, . If we put value of V in this equation, V, from Boyle’s law equation, we obtain the, relationship., d =, , Fig. 5.5 (b) Graph of pressure of a gas, p vs., , 1, V, , Table 5.1 Effect of Pressure on the Volume of 0.09 mol CO2 Gas at 300 K., Pressure/104 Pa, , Volume/10–3 m3, , (1/V )/m–3, , pV/102 Pa m3, , 2.0, , 112.0, , 8.90, , 22.40, , 2.5, , 89.2, , 11.2, , 22.30, , 3.5, , 64.2, , 15.6, , 22.47, , 4.0, , 56.3, , 17.7, , 22.50, , 6.0, , 37.4, , 26.7, , 22.44, , 8.0, , 28.1, , 35.6, , 22.48, , 10.0, , 22.4, , 44.6, , 22.40, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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142, , CHEMISTRY, ,  273.15 + t , ⇒ Vt = V0 ,  273.15 , , m , d =   p = k′ p,  k1 , This shows that at a constant temperature,, pressure is directly proportional to the density, of a fixed mass of the gas., Problem 5.1, A balloon is filled with hydrogen at room, temperature. It will burst if pressure, exceeds 0.2 bar. If at 1 bar pressure the, gas occupies 2.27 L volume, upto what, volume can the balloon be expanded ?, Solution, According to Boyle’s Law p1V1 = p2V2, If p1 is 1 bar, V1 will be 2.27 L, If p2 = 0.2 bar, then V2 =, , ⇒ V2 =, , p1V1, p2, , 1 bar × 2.27 L, =11.35 L, 0.2 bar, , At this stage, we define a new scale of, temperature such that t °C on new scale is given, by T = 273.15 + t and 0 °C will be given by, T0 = 273.15. This new temperature scale is, called the Kelvin temperature scale or, Absolute temperature scale., Thus 0°C on the celsius scale is equal to, 273.15 K at the absolute scale. Note that, degree sign is not used while writing the, temperature in absolute temperature scale,, i.e., Kelvin scale. Kelvin scale of temperature, is also called Thermodynamic scale of, temperature and is used in all scientific, works., Thus we add 273 (more precisely 273.15), to the celsius temperature to obtain, temperature at Kelvin scale., If we write Tt = 273.15 + t and T0 = 273.15, in the equation (5.6) we obtain the, relationship, , Since balloon bursts at 0.2 bar pressure,, the volume of balloon should be less than, 11.35 L., , T , Vt = V0  t ,  T0 , , 5.5.2 Charles’ Law (Temperature - Volume, Relationship), Charles and Gay Lussac performed several, experiments on gases independently to, improve upon hot air balloon technology., Their investigations showed that for a fixed, mass of a gas at constant pressure, volume, of a gas increases on increasing temperature, and decreases on cooling. They found that, for each degree rise in temperature, volume, 1, of the original, 273.15, volume of the gas at 0 °C. Thus if volumes of, the gas at 0 °C and at t °C are V0 and Vt, respectively, then, , of a gas increases by, , t, V0, 273.15, t, , , ⇒ Vt = V0 1 +,  273.15 , , Vt = V0 +, , (5.6), , ⇒, , Vt, T, = t, V0, T0, , (5.7), , Thus we can write a general equation as, follows., , V2, T, = 2, V1, T1, ⇒, , (5.8), , V1 V2, =, T1 T2, , V, = constant = k 2, (5.9), T, (5.10), Thus V = k2 T, The value of constant k2 is determined by, the pressure of the gas, its amount and the, units in which volume V is expressed., ⇒, , Equation (5.10) is the mathematical, expression for Charles’ law, which states that, pressure remaining constant, the volume, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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STATES OF MATTER, , 143, , of a fixed mass of a gas is directly, proportional to its absolute temperature., Charles found that for all gases, at any given, pressure, graph of volume vs temperature (in, celsius) is a straight line and on extending to, zero volume, each line intercepts the, temperature axis at – 273.15 °C. Slopes of, lines obtained at different pressure are, different but at zero volume all the lines meet, the temperature axis at – 273.15 °C (Fig. 5.6)., , with 2 L air. What will be the volume of, the balloon when the ship reaches Indian, ocean, where temperature is 26.1°C ?, Solution, V1 = 2 L, T1 = (23.4 + 273) K, , T2 = 26.1 + 273, = 299.1 K, , = 296.4 K, From Charles law, , V1 V2, =, T1 T2, V1T2, T1, 2 L × 299.1 K, ⇒ V2 =, 296.4 K, ⇒ V2 =, , = 2L × 1.009, = 2.018L, 5.5.3 Gay Lussac’s Law, Temperature Relationship), , Fig. 5.6 Volume vs Temperature ( °C) graph, , Each line of the volume vs temperature, graph is called isobar., Observations of Charles can be interpreted, if we put the value of t in equation (5.6) as, – 273.15 °C. We can see that the volume of, the gas at – 273.15 °C will be zero. This means, that gas will not exist. In fact all the gases get, liquified before this temperature is reached., The lowest hypothetical or imaginary, temperature at which gases are supposed to, occupy zero volume is called Absolute zero., All gases obey Charles’ law at very low, pressures and high temperatures., Problem 5.2, On a ship sailing in Pacific Ocean where, temperature is 23.4°C, a balloon is filled, , (Pressure-, , Pressure in well inflated tyres of automobiles, is almost constant, but on a hot summer day, this increases considerably and tyre may, burst if pressure is not adjusted properly., During winters, on a cold morning one may, find the pressure in the tyres of a vehicle, decreased considerably. The mathematical, relationship between pressure and, temperature was given by Joseph Gay Lussac, and is known as Gay Lussac’s law. It states, that at constant volume, pressure of a fixed, amount of a gas varies directly with the, temperature. Mathematically,, , p ∝T, p, ⇒, = constant = k 3, T, This relationship can be derived from, Boyle’s law and Charles’ Law. Pressure vs, temperature (Kelvin) graph at constant molar, volume is shown in Fig. 5.7. Each line of this, graph is called isochore., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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144, , CHEMISTRY, , we came across while discussing definition of, a ‘mole’ (Unit 1)., Since volume of a gas is directly, proportional to the number of moles; one mole, of each gas at standard temperature and, pressure (STP)* will have same volume., Standard temperature and pressure means, 273.15 K (0°C) temperature and 1 bar (i.e.,, exactly 10 5 pascal) pressure. These, values approximate freezing temperature, of water and atmospheric pressure at sea, level. At STP molar volume of an ideal gas, or a combination of ideal gases is, 22.71098 L mol–1., Molar volume of some gases is given in, (Table 5.2)., , Fig. 5.7 Pressure vs temperature (K) graph, (Isochores) of a gas., , 5.5.4 Avogadro Law (Volume - Amount, Relationship), In 1811 Italian scientist Amedeo Avogadro, tried to combine conclusions of Dalton’s atomic, theory and Gay Lussac’s law of combining, volumes (Unit 1) which is now known as, Avogadro law. It states that equal volumes, of all gases under the same conditions of, temperature and pressure contain equal, number of molecules. This means that as, long as the temperature and pressure remain, constant, the volume depends upon number, of molecules of the gas or in other words, amount of the gas. Mathematically we can write, , V ∝n, , Table 5.2 Molar volume in litres per mole of, some gases at 273.15 K and 1 bar, (STP)., Argon, , 22.37, , Carbon dioxide, , 22.54, , Dinitrogen, , 22.69, , Dioxygen, , 22.69, , Dihydrogen, , 22.72, , Ideal gas, , 22.71, , Number of moles of a gas can be calculated, as follows, , m, (5.12), M, Where m = mass of the gas under, investigation and M = molar mass, Thus,, n=, , where n is the number of moles, , of the gas., , ⇒ V = k4 n, , V = k4, (5.11), , The number of molecules in one mole of a, gas has been determined to be 6.022 ×1023 and, is known as Avogadro constant. You, will find that this is the same number which, , m, M, , (5.13), , Equation (5.13) can be rearranged as, follows :, M = k4, , m, = k4d, M, , (5.14), , * The previous standard is still often used, and applies to all chemistry data more than decade old. In this definition STP, , denotes the same temperature of 0°C (273.15 K), but a slightly higher pressure of 1 atm (101.325 kPa). One mole of any gas, of a combination of gases occupies 22.413996 L of volume at STP., Standard ambient temperature and pressure (SATP), conditions are also used in some scientific works. SATP conditions, means 298.15 K and 1 bar (i.e., exactly 105 Pa). At SATP (1 bar and 298.15 K), the molar volume of an ideal gas is, 24.789 L mol –1., Download all NCERT books PDFs from www.ncert.online, , 2019-20

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STATES OF MATTER, , 145, , Here ‘d’ is the density of the gas. We can conclude, from equation (5.14) that the density of a gas is, directly proportional to its molar mass., , that at constant temperature and pressure, n moles of any gas will have the same volume, , A gas that follows Boyle’s law, Charles’ law, and Avogadro law strictly is called an ideal, gas. Such a gas is hypothetical. It is assumed, that intermolecular forces are not present, between the molecules of an ideal gas. Real, gases follow these laws only under certain, specific conditions when forces of interaction, are practically negligible. In all other situations, these deviate from ideal behaviour. You will, learn about the deviations later in this unit., , because V =, , n RT, and n,R,T and p are, p, , constant. This equation will be applicable to, any gas, under those conditions when, behaviour of the gas approaches ideal, behaviour. Volume of one mole of an ideal, gas under STP conditions (273.15 K and 1, bar pressure) is 22.710981 L mol–1. Value, of R for one mole of an ideal gas can be, c a l c u l a t e d u n d e r t h e s e c o n di ti on s, as follows :, , 5.6 IDEAL GAS EQUATION, The three laws which we have learnt till now, can be combined together in a single equation, which is known as ideal gas equation., , 1, Boyle’s Law, p, At constant p and n; V ∝ T Charles’ Law, At constant p and T ; V ∝ n Avogadro Law, Thus,, , = 8.314 Pa m3 K, , nT, p, , ⇒ V =R, , (5.15), , nT, p, , (5.16), , where R is proportionality constant. On, rearranging the equation (5.16) we obtain, pV = n RT, (5.17), , ⇒ R=, , pV, nT, , (5.18), , –1, , –1, , –1, , mol, , –1, , –1, , = 8.314 J K mol, , At STP conditions used earlier, (0 °C and 1 atm pressure), value of R is, 8.20578 × 10–2 L atm K–1 mol–1., Ideal gas equation is a relation between, four variables and it describes the state of, any gas, therefore, it is also called equation, of state., Let us now go back to the ideal gas, equation. This is the relationship for the, simultaneous variation of the variables. If, temperature, volume and pressure of a fixed, amount of gas vary from T1, V1 and p1 to T2,, V2 and p2 then we can write, , p1V1, = nR and, T1, , R is called gas constant. It is same for all gases., Therefore it is also called Universal Gas, Constant. Equation (5.17) is called ideal gas, equation., Equation (5.18) shows that the value of, R depends upon units in which p, V and T, are measured. If three variables in this, e quat io n are k no w n, fo u rth c a n b e, calculated. From this equation we can see, , mol, , = 8.314 × 10–2 bar L K, , At constant T and n; V ∝, , V ∝, , –1, , ⇒, , p1V1, p V, = 2 2, T1, T2, , p 2V2, = nR, T2, (5.19), , Equation (5.19) is a very useful equation., If out of six, values of five variables are known,, the value of unknown variable can be, calculated from the equation (5.19). This, equation is also known as Combined gas law., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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146, , CHEMISTRY, , by the mixture of non-reactive gases is, equal to the sum of the partial pressures, of individual gases i.e., the pressures which, these gases would exert if they were enclosed, separately in the same volume and under the, same conditions of temperature. In a mixture, of gases, the pressure exerted by the individual, gas is called partial pressure. Mathematically,, , Problem 5.3, At 25°C and 760 mm of Hg pressure a, gas occupies 600 mL volume. What will, be its pressure at a height where, temperature is 10°C and volume of the, gas is 640 mL., Solution, , pTotal = p1+p2+p3+......(at constant T, V) (5.23), , p1 = 760 mm Hg, V1= 600 mL, T1 = 25 + 273 = 298 K, V2 = 640 mL and T2 = 10 + 273 = 283 K, According to Combined gas law, , where pTotal is the total pressure exerted by, the mixture of gases and p1, p2 , p3 etc. are, partial pressures of gases., Gases are generally collected over water, and therefore are moist. Pressure of dry gas, can be calculated by subtracting vapour, pressure of water from the total pressure of, the moist gas which contains water vapours, also. Pressure exerted by saturated water, vapour is called aqueous tension. Aqueous, tension of water at different temperatures is, given in Table 5.3., , p1V1 p 2V2, =, T1, T2, ⇒ p2 =, , p1 V1 T2, T1 V2, , ⇒ p2 =, , (760 mm Hg ) × (600 mL ) × (283 K ), (640 mL ) × (298 K ), , = 676.6 mm Hg, , 5.6.1 Density and Molar Mass of a, Gaseous Substance, Ideal gas equation can be rearranged as follows:, , n, p, =, V, RT, Replacing n by, , m, p, =, MV, RT, , Table 5.3 Aqueous Tension of Water (Vapour, Pressure) as a Function of, Temperature, , (5.20), , On rearranging equation (5.21) we get the, relationship for calculating molar mass of a gas., , d RT, p, , (5.24), , m, , we get, M, , d, p, =, M R T (where d is the density) (5.21), , M=, , pDry gas = pTotal – Aqueous tension, , (5.22), , Partial pressure in terms of mole fraction, Suppose at the temperature T, three gases,, enclosed in the volume V, exert partial, pressure p1, p2 and p3 respectively, then,, , p1 =, , 5.6.2 Dalton’s Law of Partial Pressures, The law was formulated by John Dalton in, 1801. It states that the total pressure exerted, , n1RT, V, , p2 =, , n 2 RT, V, , Download all NCERT books PDFs from www.ncert.online, , 2019-20, , (5.25), , (5.26)

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STATES OF MATTER, , p3 =, , 147, , n 3 RT, V, , Number of moles of neon, , (5.27), , =, , where n1 n2 and n3 are number of moles of these, gases. Thus, expression for total pressure will be, pTotal = p1 + p2 + p3, , = n1, , = 8.375 mol, Mole fraction of dioxygen, , RT, RT, RT, + n2, + n3, V, V, V, , = (n1 + n2 + n3), , RT, V, , 2.21, 2.21 + 8.375, 2.21, =, 10.585, = 0.21, =, , (5.28), , On dividing p1 by ptotal we get, , Mole fraction of neon =, ,  RTV, , p1, n1, =, p total  n1 +n 2 +n 3  RTV, , =, , 167.5 g, 20 g mol −1, , 8.375, 2.21 + 8.375, , = 0.79, Alternatively,, mole fraction of neon = 1– 0.21 = 0.79, , n1, n, = 1 = x1, n1 +n 2 +n 3 n, , where n = n1+n2+n3, x1 is called mole fraction of first gas., Thus, p1 = x1 ptotal, Similarly for other two gases we can write, p2 = x2 ptotal and p3 = x3 ptotal, Thus a general equation can be written as, pi = xi ptotal, , Partial pressure, of a gas, , = mole fraction ×, total pressure, , ⇒ Partial pressure, of oxygen, , = 0.21 × (25 bar), = 5.25 bar, , Partial pressure, of neon, , = 0.79 × (25 bar), = 19.75 bar, , (5.29), , where pi and xi are partial pressure and mole, fraction of ith gas respectively. If total pressure, of a mixture of gases is known, the equation, (5.29) can be used to find out pressure exerted, by individual gases., Problem 5.4, A neon-dioxygen mixture contains, 70.6 g dioxygen and 167.5 g neon. If, pressure of the mixture of gases in the, cylinder is 25 bar. What is the partial, pressure of dioxygen and neon in the, mixture ?, Number of moles of dioxygen, , =, , 70.6 g, 32 g mol −1, , 5.7 KINETIC ENERGY AND MOLECULAR, SPEEDS, Molecules of gases remain in continuous, motion. While moving they collide with each, other and with the walls of the container. This, results in change of their speed and, redistribution of energy. So the speed and, energy of all the molecules of the gas at any, instant are not the same. Thus, we can obtain, only average value of speed of molecules. If, there are n number of molecules in a sample, and their individual speeds are u1, u2,…….un,, then average speed of molecules uav can be, calculated as follows:, , uav =, , = 2.21 mol, , u 1 + u 2 + .........un, n, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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148, , CHEMISTRY, , Maxwell and Boltzmann have shown that, actual distribution of molecular speeds, depends on temperature and molecular mass, of a gas. Maxwell derived a formula for, calculating the number of molecules, possessing a particular speed. Fig. 5.8 shows, schematic plot of number of molecules vs., molecular speed at two different temperatures, T 1 and T 2 (T 2 is higher than T 1 ). The, distribution of speeds shown in the plot is, called Maxwell-Boltzmann distribution, of speeds., , the molecular speed distribution curve of, chlorine and nitrogen given in Fig. 5.9., Though at a particular temperature the, i n d i v i d u a l s p e e d o f m o l e c u l e s keeps, changing, the distribution of speeds remains, same., , Fig. 5.9: Distribution of molecular speeds for chlorine, and nitrogen at 300 K, , We know that kinetic energy of a particle is, given by the expression:, 1, mu2, 2, Therefore, if we want to know average, Kinetic Energy =, , Fig. 5.8: Maxwell-Boltzmann distribution of speeds, , translational kinetic energy,, The graph shows that number of molecules, possessing very high and very low speed is very, small. The maximum in the curve represents, speed possessed by maximum number of, molecules. This speed is called most, probable speed, ump. This is very close to the, average speed of the molecu les. On, increasing the temperature most probable, speed increases. Also, speed distribution, curve broadens at higher temperature., Broadening of the curve shows that number, of molecules moving at higher speed, increases. Speed distribution also depends, upon mass of molecules. At the same, temperature, gas molecules with heavier, mass have slower speed than lighter gas, molecules. For example, at the same, temperature lighter nitrogen molecules move, faster than heavier chlorine molecules., Hence, at any given temperature, nitrogen, molecules have higher value of most probable, speed than the chlorine molecules. Look at, , 1, mu2 , for the, 2, , movement of a gas particle in a straight line,, we require the value of mean of square of, speeds, u2 , of all molecules. This is, represented as follows:, u12 +u22 +..........un2, n, The mean square speed is the direct, measure of the average kinetic energy of gas, molecules. If we take the square root of the, mean of the square of speeds then we get a, value of speed which is different from most, probable speed and average speed. This speed, is called root mean square speed and is given, by the expression as follows:, u2 =, , urms, , =, , u2, , Root mean square speed, average speed, and the most probable speed have following, relationship:, urms > uav > ump, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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STATES OF MATTER, , 149, , The ratio between the three speeds is given, below :, ump: uav : urms : : 1 : 1.128 : 1.224, 5.8 KINETIC MOLECULAR THEORY OF, GASES, So far we have learnt the laws (e.g., Boyle’s law,, Charles’ law etc.) which are concise statements, of experimental facts observed in the laboratory, by the scientists. Conducting careful, experiments is an important aspect of scientific, method and it tells us how the particular, system is behaving under different conditions., However, once the experimental facts are, established, a scientist is curious to know why, the system is behaving in that way. For, example, gas laws help us to predict that, pressure increases when we compress gases, but we would like to know what happens at, molecular level when a gas is compressed ? A, theory is constructed to answer such, questions. A theory is a model (i.e., a mental, picture) that enables us to better understand, our observations. The theory that attempts to, elucidate the behaviour of gases is known as, kinetic molecular theory., Assumptions or postulates of the kineticmolecular theory of gases are given below., These postulates are related to atoms and, molecules which cannot be seen, hence it is, said to provide a microscopic model of gases., • Gases consist of large number of identical, particles (atoms or molecules) that are so, small and so far apart on the average that, the actual volume of the molecules is, negligible in comparison to the empty space, between them. They are considered as point, masses. This assumption explains the, great compressibility of gases., • There is no force of attraction between the, particles of a gas at ordinary temperature and, pressure. The support for this assumption, comes from the fact that gases expand and, occupy all the space available to them., • Particles of a gas are always in constant and, random motion. If the particles were at rest, and occupied fixed positions, then a gas would, have had a fixed shape which is not observed., , •, , •, , •, , •, , Particles of a gas move in all possible, directions in straight lines. During their, random motion, they collide with each, other and with the walls of the container., Pressure is exerted by the gas as a result, of collision of the particles with the walls of, the container., Collisions of gas molecules are perfectly, elastic. This means that total energy of, molecules before and after the collision, remains same. There may be exchange of, energy between colliding molecules, their, individual energies may change, but the, sum of their energies remains constant. If, there were loss of kinetic energy, the motion, of molecules will stop and gases will settle, down. This is contrary to what is actually, observed., At any particular time, different particles, in the gas have different speeds and hence, different kinetic energies. This assumption, is reasonable because as the particles, collide, we expect their speed to change., Even if initial speed of all the particles was, same, the molecular collisions will disrupt, this uniformity. Consequently, the particles, must have different speeds, which go on, changing constantly. It is possible to show, that though the individual speeds are, changing, the distribution of speeds, remains constant at a particular, temperature., If a molecule has variable speed, then it, must have a variable kinetic energy. Under, these circumstances, we can talk only, about average kinetic energy. In kinetic, theory, it is assumed that average kinetic, energy of the gas molecules is directly, proportional to the absolute temperature., It is seen that on heating a gas at constant, volume, the pressure increases. On heating, the gas, kinetic energy of the particles, increases and these strike the walls of the, container more frequently, thus, exerting, more pressure., , Kinetic theory of gases allows us to derive, theoretically, all the gas laws studied in the, previous sections. Calculations and predictions, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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150, , CHEMISTRY, , 5.9 BEHAVIOUR OF REAL GASES:, DEVIATION FROM IDEAL GAS, BEHAVIOUR, Our theoritical model of gases corresponds, very well with the experimental observations., Difficulty arises when we try to test how far, the relation pV = nRT reproduce actual, pressure-volume-temperature relationship, of gases. To test this point we plot pV vs p plot, of gases because at constant temperature, pV, will be constant (Boyle’s law) and pV vs p, graph at all pressures will be a straight line, parallel to x-axis. Fig. 5.10 shows such a plot, constructed from actual data for several gases, at 273 K., , negative deviation from ideal behaviour, the pV, value decreases with increase in pressure and, reaches to a minimum value characteristic of, a gas. After that pV value starts increasing. The, curve then crosses the line for ideal gas and, after that shows positive deviation, continuously. It is thus, found that real gases, do not follow ideal gas equation perfectly under, all conditions., Deviation from ideal behaviour also, becomes apparent when pressure vs volume, plot is drawn. The pressure vs volume plot of, experimental data (real gas) and that, theoretically calculated from Boyle’s law (ideal, gas) should coincide. Fig 5.11 shows these, plots. It is apparent that at very high pressure, the measured volume is more than the, calculated volume. At low pressures, measured, and calculated volumes approach each other., , Fig. 5.10 Plot of pV vs p for real gas and, ideal gas, , Fig. 5.11 Plot of pressure vs volume for real gas, and ideal gas, , It can be seen easily that at constant, temperature pV vs p plot for real gases is not a, straight line. There is a significant deviation, from ideal behaviour. Two types of curves are, seen. In the curves for dihydrogen and helium,, as the pressure increases the value of pV also, increases. The second type of plot is seen in, the case of other gases like carbon monoxide, and methane. In these plots first there is a, , It is found that real gases do not follow,, Boyle’s law, Charles law and Avogadro law, perfectly under all conditions. Now two, questions arise., , based on kinetic theory of gases agree very well, with the experimental observations and thus, establish the correctness of this model., , (i) Why do gases deviate from the ideal, behaviour?, (ii) What are the conditions under which gases, deviate from ideality?, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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STATES OF MATTER, , 151, , We get the answer of the first question if we, look into postulates of kinetic theory once, again. We find that two assumptions of the, kinetic theory do not hold good. These are, (a) There is no force of attraction between the, molecules of a gas., (b) Volume of the molecules of a gas is, negligibly small in comparison to the space, occupied by the gas., If assumption (a) is correct, the gas will, never liquify. However, we know that gases do, liquify when cooled and compressed. Also,, liquids formed are very difficult to compress., This means that forces of repulsion are, powerful enough and prevent squashing of, molecules in tiny volume. If assumption (b) is, correct, the pressure vs volume graph of, experimental data (real gas) and that, theoritically calculated from Boyles law (ideal, gas) should coincide., Real gases show deviations from ideal gas, law because molecules interact with each other., At high pressures molecules of gases are very, close to each other. Molecular interactions start, operating. At high pressure, molecules do not, strike the walls of the container with full impact, because these are dragged back by other, molecules due to molecular attractive forces., This affects the pressure exerted by the, molecules on the walls of the container. Thus,, the pressure exerted by the gas is lower than, the pressure exerted by the ideal gas., , pideal = preal, , +, , observed, pressure, , an 2, V2, , because instead of moving in volume V, these, are now restricted to volume (V–nb) where nb, is approximately the total volume occupied by, the molecules themselves. Here, b is a constant., Having taken into account the corrections for, pressure and volume, we can rewrite equation, (5.17) as, , , an 2 , p, +, (V − nb) = nRT, , V 2 , , Equation (5.31) is known as van der Waals, equation. In this equation n is number of moles, of the gas. Constants a and b are called van, der Waals constants and their value depends, on the characteristic of a gas. Value of ‘a’ is, measure of magnitude of intermolecular, attractive forces within the gas and is, independent of temperature and pressure., Also, at very low temperature,, intermolecular forces become significant. As, the molecules travel with low average speed,, these can be captured by one another due to, attractive forces. Real gases show ideal, behaviour when conditions of temperature and, pressure are such that the intermolecular, forces are practically negligible. The real gases, show ideal behaviour when pressure, approaches zero., The deviation from ideal behaviour can be, measured in terms of compressibility factor, Z, which is the ratio of product pV and nRT., Mathematically, , (5.30), , correction, term, , Here, a is a constant., Repulsive forces also become significant., Repulsive interactions are short-range, interactions and are significant when, molecules are almost in contact. This is the, situation at high pressure. The repulsive forces, cause the molecules to behave as small but, impenetrable spheres. The volume occupied, by the molecules also becomes significant, , (5.31), , Z =, , pV, n RT, , (5.32), , For ideal gas Z = 1 at all temperatures and, pressures because pV = n RT. The graph of Z, vs p will be a straight line parallel to pressure, axis (Fig. 5.12, page 152). For gases which, deviate from ideality, value of Z deviates from, unity. At very low pressures all gases shown, have Z ≈1 and behave as ideal gas. At high, pressure all the gases have Z > 1. These are, more difficult to compress. At intermediate, pressures, most gases have Z < 1. Thus gases, show ideal behaviour when the volume, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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152, , CHEMISTRY, , Videal =, , nRT, n RT, . On putting this value of, p, p, , in equation (5.33) we have Z =, , Fig. 5.12 Variation of compressibility factor for, some gases, , occupied is large so that the volume of the, molecules can be neglected in comparison, to it. In other words, the behaviour of the gas, becomes more ideal when pressure is very low., Upto what pressure a gas will follow the ideal, gas law, depends upon nature of the gas and, its temperature. The temperature at which a real, gas obeys ideal gas law over an appreciable, range of pressure is called Boyle temperature, or Boyle point. Boyle point of a gas depends, upon its nature. Above their Boyle point, real, gases show positive deviations from ideality and, Z values are greater than one. The forces of, attraction between the molecules are very feeble., Below Boyle temperature real gases first show, decrease in Z value with increasing pressure,, which reaches a minimum value. On further, increase in pressure, the value of Z increases, continuously. Above explanation shows that at, low pressure and high temperature gases show, ideal behaviour. These conditions are different, for different gases., More insight is obtained in the significance, of Z if we note the following derivation, , Z =, , pVreal, n RT, , (5.33), , If the gas shows ideal behaviour then, , Vreal, Videal, , (5.34), , From equation (5.34) we can see that, compressibility factor is the ratio of actual, molar volume of a gas to the molar volume of, it, if it were an ideal gas at that temperature, and pressure., In the following sections we will see that it, is not possible to distinguish between gaseous, state and liquid state and that liquids may be, considered as continuation of gas phase into a, region of small volumes and very high, molecular attraction. We will also see how we, can use isotherms of gases for predicting the, conditions for liquifaction of gases., 5.10, , LIQUIFACTION OF GASES, , First complete data on pressure-volumetemperature relations of a substance in both, gaseous and liquid state was obtained by, Thomas Andrews on Carbon dioxide. He plotted, isotherms of carbon dioxide at various, temperatures (Fig. 5.13). Later on it was found, that real gases behave in the same manner as, carbon dioxide. Andrews noticed that at high, temperatures isotherms look like that of an, ideal gas and the gas cannot be liquified even at, very high pressure. As the temperature is, lowered, shape of the curve changes and data, show considerable deviation from ideal, behaviour. At 30.98°C carbon dioxide remains, gas upto 73 atmospheric pressure. (Point E in, Fig. 5.13). At 73 atmospheric pressure, liquid, carbon dioxide appears for the first time. The, temperature 30.98 ° C is called critical, temperature (TC) of carbon dioxide. This is the, highest temperature at which liquid carbon, dioxide is observed. Above this temperature it, is gas. Volume of one mole of the gas at critical, temperature is called critical volume (VC) and, pressure at this temperature is called critical, pressure (pC). The critical temperature, pressure, and volume are called critical constants. Further, increase in pressure simply compresses the, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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STATES OF MATTER, , 153, , Fig. 5.13 Isotherms of carbon dioxide at various, temperatures, , liquid carbon dioxide and the curve represents, the compressibility of the liquid. The steep line, represents the isotherm of liquid. Even a slight, compression results in steep rise in pressure, indicating very low compressibility of the liquid., Below 30.98 °C, the behaviour of the gas on, compression is quite different. At 21.5 °C,, carbon dioxide remains as a gas only upto, point B. At point B, liquid of a particular volume, appears. Further compression does not change, the pressure. Liquid and gaseous carbon, dioxide coexist and further application of, pressure results in the condensation of more, gas until the point C is reached. At point C, all, the gas has been condensed and further, application of pressure merely compresses the, liquid as shown by steep line. A slight, compression from volume V2 to V3 results in, steep rise in pressure from p2 to p3 (Fig. 5.13)., Below 30.98 °C (critical temperature) each, curve shows the similar trend. Only length of, the horizontal line increases at lower, temperatures. At critical point horizontal, portion of the isotherm merges into one point., , Thus we see that a point like A in the Fig. 5.13, represents gaseous state. A point like D, represents liquid state and a point under the, dome shaped area represents existence of liquid, and gaseous carbon dioxide in equilibrium. All, the gases upon compression at constant, temperature (isothermal compression) show the, same behaviour as shown by carbon dioxide., Also above discussion shows that gases should, be cooled below their critical temperature for, liquification. Critical temperature of a gas is, highest temperature at which liquifaction of the, gas first occurs. Liquifaction of so called, permanent gases (i.e., gases which show, continuous positive deviation in Z value), requires cooling as well as considerable, compression. Compression brings the, molecules in close vicinity and cooling slows, down the movement of molecules therefore,, intermolecular interactions may hold the closely, and slowly moving molecules together and the, gas liquifies., It is possible to change a gas into liquid or, a liquid into gas by a process in which always, a single phase is present. For example in, Fig. 5.13 we can move from point A to F, vertically by increasing the temperature, then, we can reach the point G by compressing the, gas at the constant temperature along this, isotherm (isotherm at 31.1°C). The pressure, will increase. Now we can move vertically down, towards D by lowering the temperature. As, soon as we cross the point H on the critical, isotherm we get liquid. We end up with liquid, but in this series of changes we do not pass, through two-phase region. If process is carried, out at the critical temperature, substance, always remains in one phase., Thus there is continuity between the, gaseous and liquid state. The term fluid is used, for either a liquid or a gas to recognise this, continuity. Thus a liquid can be viewed as a, very dense gas. Liquid and gas can be, distinguished only when the fluid is below its, critical temperature and its pressure and, volume lie under the dome, since in that, situation liquid and gas are in equilibrium and, a surface separating the two phases is visible., In the absence of this surface there is no, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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154, , CHEMISTRY, , fundamental way of distinguishing between, two states. At critical temperature, liquid, passes into gaseous state imperceptibly and, continuously; the surface separating two, phases disappears (Section 5.11.1). A gas, below the critical temperature can be liquified, by applying pressure, and is called vapour of, the substance. Carbon dioxide gas below its, critical temperature is called carbon dioxide, vapour. Critical constants for some common, substances are given in Table 5.4., Table 5.4 Critical Constants, Substances, , for, , Some, , between them and under normal conditions, liquids are denser than gases., Molecules of liquids are held together by, attractive intermolecular forces. Liquids have, definite volume because molecules do not, separate from each other. However, molecules, of liquids can move past one another freely,, therefore, liquids can flow, can be poured and, can assume the shape of the container in which, these are stored. In the following sections we, will look into some of the physical properties, of the liquids such as vapour pressure, surface, tension and viscosity., 5.11.1 Vapour Pressure, If an evacuated container is partially filled with, a liquid, a portion of liquid evaporates to fill, the remaining volume of the container with, vapour. Initially the liquid evaporates and, pressure exerted by vapours on the walls of, the container (vapour pressure) increases. After, some time it becomes constant, an equilibrium, is established between liquid phase and, vapour phase. Vapour pressure at this stage, is known as equilibrium vapour pressure or, saturated vapour pressure.. Since process of, vapourisation is temperature dependent; the, temperature must be mentioned while, reporting the vapour pressure of a liquid., , Problem 5.5, Gases possess characteristic critical, temperature which depends upon the, magnitude of intermolecular forces, between the gas particles. Critical, temperatures of ammonia and carbon, dioxide are 405.5 K and 304.10 K, respectively. Which of these gases will, liquify first when you start cooling from, 500 K to their critical temperature ?, Solution, Ammonia will liquify first because its, critical temperature will be reached first., Liquifaction of CO 2 will require more, cooling., 5.11 LIQUID STATE, Intermolecular forces are stronger in liquid, state than in gaseous state. Molecules in liquids, are so close that there is very little empty space, , When a liquid is heated in an open vessel,, the liquid vapourises from the surface. At the, temperature at which vapour pressure of the, liquid becomes equal to the external pressure,, vapourisation can occur throughout the bulk, of the liquid and vapours expand freely into, the surroundings. The condition of free, vapourisation throughout the liquid is called, boiling. The temperature at which vapour, pressure of liquid is equal to the external, pressure is called boiling temperature at that, pressure. Vapour pressure of some common, liquids at various temperatures is given in, (Fig. 5.14, page 155). At 1 atm pressure boiling, temperature is called normal boiling point., If pressure is 1 bar then the boiling point is, called standard boiling point of the liquid., Standard boiling point of the liquid is slightly, lower than the normal boiling point because, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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STATES OF MATTER, , 155, , more and more molecules go to vapour phase, and density of vapours rises. At the same time, liquid becomes less dense. It expands because, molecules move apart. When density of liquid, and vapours becomes the same; the clear, boundary between liquid and vapours, disappears. This temperature is called critical, temperature about which we have already, discussed in section 5.10., , Fig. 5.14 Vapour pressure vs temperature curve, of some common liquids., , 1 bar pressure is slightly less than 1 atm, pressure. The normal boiling point of water is, 100 °C (373 K), its standard boiling point is, 99.6 °C (372.6 K)., At high altitudes atmospheric pressure is, low. Therefore liquids at high altitudes boil at, lower temperatures in comparison to that at, sea level. Since water boils at low temperature, on hills, the pressure cooker is used for, cooking food. In hospitals surgical instruments, are sterilized in autoclaves in which boiling, point of water is increased by increasing the, pressure above the atmospheric pressure by, using a weight covering the vent., Boiling does not occur when liquid is, heated in a closed vessel. On heating, continuously vapour pressure increases. At, first a clear boundary is visible between liquid, and vapour phase because liquid is more dense, than vapour. As the temperature increases, , 5.11.2 Surface Tension, It is well known fact that liquids assume the, shape of the container. Why is it then small, drops of mercury form spherical bead instead, of spreading on the surface. Why do particles, of soil at the bottom of river remain separated, but they stick together when taken out ? Why, does a liquid rise (or fall) in a thin capillary as, soon as the capillary touches the surface of, the liquid ? All these phenomena are caused, due to the characteristic property of liquids,, called surface tension. A molecule in the bulk, of liquid experiences equal intermolecular, forces from all sides. The molecule, therefore, does not experience any net force. But for the, molecule on the surface of liquid, net attractive, force is towards the interior of the liquid (Fig., 5.15), due to the molecules below it. Since there, are no molecules above it., Liquids tend to minimize their surface area., The molecules on the surface experience a net, downward force and have more energy than, , Fig. 5.15 Forces acting on a molecule on liquid, surface and on a molecule inside the, liquid, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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156, , CHEMISTRY, , the molecules in the bulk, which do not, experience any net force. Therefore, liquids tend, to have minimum number of molecules at their, surface. If surface of the liquid is increased by, pulling a molecule from the bulk, attractive, forces will have to be overcome. This will, require expenditure of energy. The energy, required to increase the surface area of the, liquid by one unit is defined as surface energy., Its dimensions are J m–2. Surface tension is, defined as the force acting per unit length, perpendicular to the line drawn on the surface, of liquid. It is denoted by Greek letter γ, (Gamma). It has dimensions of kg s–2 and in SI, unit it is expressed as N m–1. The lowest energy, state of the liquid will be when surface area is, minimum. Spherical shape satisfies this, condition, that is why mercury drops are, spherical in shape. This is the reason that sharp, glass edges are heated for making them, smooth. On heating, the glass melts and the, surface of the liquid tends to take the rounded, shape at the edges, which makes the edges, smooth. This is called fire polishing of glass., Liquid tends to rise (or fall) in the capillary, because of surface tension. Liquids wet the, things because they spread across their surfaces, as thin film. Moist soil grains are pulled together, because surface area of thin film of water is, reduced. It is surface tension which gives, stretching property to the surface of a liquid., On flat surface, droplets are slightly flattened, by the effect of gravity; but in the gravity free, environments drops are perfectly spherical., The magnitude of surface tension of a liquid, depends on the attractive forces between the, molecules. When the attractive forces are large,, the surface tension is large. Increase in, temperature increases the kinetic energy of the, molecules and effectiveness of intermolecular, attraction decreases, so surface tension, decreases as the temperature is raised., , between layers of fluid as they slip past one, another while liquid flows. Strong, intermolecular forces between molecules hold, them together and resist movement of layers, past one another., When a liquid flows over a fixed surface,, the layer of molecules in the immediate contact, of surface is stationary. The velocity of upper, layers increases as the distance of layers from, the fixed layer increases. This type of flow in, which there is a regular gradation of velocity, in passing from one layer to the next is called, laminar flow. If we choose any layer in the, flowing liquid (Fig.5.16), the layer above it, accelerates its flow and the layer below this, , Fig. 5.16 Gradation of velocity in the laminar, flow, , retards its flow., If the velocity of the layer at a distance dz, is changed by a value du then velocity gradient, , du, A force is required, dz, to maintain the flow of layers. This force is, proportional to the area of contact of layers, and velocity gradient i.e., , is given by the amount ., , F ∝ A (A is the area of contact), , du, du, (where,, is velocity gradient;, dz, dz, the change in velocity with distance), , 5.11.3 Viscosity, , F ∝ A., , F ∝ A., , It is one of the characteristic properties of, liquids. Viscosity is a measure of resistance to, flow which arises due to the internal friction, , du, dz, , ⇒ F = ηA, , du, dz, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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STATES OF MATTER, , 157, , ‘ η ’ is proportionality constant and is called, coefficient of viscosity. Viscosity coefficient, is the force when velocity gradient is unity and, the area of contact is unit area. Thus ‘ ’ is, measure of viscosity. SI unit of viscosity, coefficient is 1 newton second per square metre, (N s m–2) = pascal second (Pa s = 1kg m–1s–1). In, cgs system the unit of coefficient of viscosity is, poise (named after great scientist Jean Louise, Poiseuille)., 1 poise = 1 g cm–1s–1 = 10–1kg m–1s–1, , Greater the viscosity, the more slowly the, liquid flows. Hydrogen bonding and van der, Waals forces are strong enough to cause high, viscosity. Glass is an extremely viscous liquid., It is so viscous that many of its properties, resemble solids., Viscosity of liquids decreases as the, temperature rises because at high temperature, molecules have high kinetic energy and can, overcome the intermolecular forces to slip past, one another between the layers., , SUMMARY, Intermolecular forces operate between the particles of matter. These forces differ from, pure electrostatic forces that exist between two oppositely charged ions. Also, these do, not include forces that hold atoms of a covalent molecule together through covalent, bond. Competition between thermal energy and intermolecular interactions determines, the state of matter. “Bulk” properties of matter such as behaviour of gases, characteristics, of solids and liquids and change of state depend upon energy of constituent particles, and the type of interaction between them. Chemical properties of a substance do not, change with change of state, but the reactivity depends upon the physical state., , η, , Forces of interaction between gas molecules are negligible and are almost independent, of their chemical nature. Interdependence of some observable properties namely, pressure, volume, temperature and mass leads to different gas laws obtained from, experimental studies on gases. Boyle’s law states that under isothermal condition,, pressure of a fixed amount of a gas is inversely proportional to its volume. Charles’ law, is a relationship between volume and absolute temperature under isobaric condition. It, states that volume of a fixed amount of gas is directly proportional to its absolute, temperature (V ∝ T ) . If state of a gas is represented by p1, V1 and T1 and it changes to, state at p2, V2 and T2, then relationship between these two states is given by combined, gas law according to which, , p1V1, T1, , =, , p 2V2, T2, , . Any one of the variables of this gas can be, , found out if other five variables are known. Avogadro law states that equal volumes of, all gases under same conditions of temperature and pressure contain equal number of, molecules. Dalton’s law of partial pressure states that total pressure exerted by a, mixture of non-reacting gases is equal to the sum of partial pressures exerted by them., Thus p = p1+p2+p3+ ... . Relationship between pressure, volume, temperature and number, of moles of a gas describes its state and is called equation of state of the gas. Equation, of state for ideal gas is pV=nRT, where R is a gas constant and its value depends upon, units chosen for pressure, volume and temperature., At high pressure and low temperature intermolecular forces start operating strongly, between the molecules of gases because they come close to each other. Under suitable, temperature and pressure conditions gases can be liquified. Liquids may be considered, as continuation of gas phase into a region of small volume and very strong molecular, attractions. Some properties of liquids e.g., surface tension and viscosity are due to, strong intermolecular attractive forces., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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158, , CHEMISTRY, , EXERCISES, 5.1, , What will be the minimum pressure required to compress 500 dm3 of air at 1 bar to, 200 dm3 at 30°C?, , 5.2, , A vessel of 120 mL capacity contains a certain amount of gas at 35 °C and 1.2 bar, pressure. The gas is transferred to another vessel of volume 180 mL at 35 °C. What, would be its pressure?, , 5.3, , Using the equation of state pV=nRT; show that at a given temperature density of a, gas is proportional to gas pressure p., , 5.4, , At 0°C, the density of a certain oxide of a gas at 2 bar is same as that of dinitrogen, at 5 bar. What is the molecular mass of the oxide?, , 5.5, , Pressure of 1 g of an ideal gas A at 27 °C is found to be 2 bar. When 2 g of another, ideal gas B is introduced in the same flask at same temperature the pressure, becomes 3 bar. Find a relationship between their molecular masses., , 5.6, , The drain cleaner, Drainex contains small bits of aluminum which react with caustic, soda to produce dihydrogen. What volume of dihydrogen at 20 °C and one bar will, be released when 0.15g of aluminum reacts?, , 5.7, , What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of, carbon dioxide contained in a 9 dm3 flask at 27 °C ?, , 5.8, , What will be the pressure of the gaseous mixture when 0.5 L of H2 at 0.8 bar and, 2.0 L of dioxygen at 0.7 bar are introduced in a 1L vessel at 27°C?, , 5.9, , Density of a gas is found to be 5.46 g/dm3 at 27 °C at 2 bar pressure. What will be, its density at STP?, , 5.10, , 34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 0.1 bar pressure., What is the molar mass of phosphorus?, , 5.11, , A student forgot to add the reaction mixture to the round bottomed flask at 27 °C, but instead he/she placed the flask on the flame. After a lapse of time, he realized, his mistake, and using a pyrometer he found the temperature of the flask was 477, °C. What fraction of air would have been expelled out?, , 5.12, , Calculate the temperature of 4.0 mol of a gas occupying 5 dm 3 at 3.32 bar., (R = 0.083 bar dm3 K–1 mol–1)., , 5.13, , Calculate the total number of electrons present in 1.4 g of dinitrogen gas., , 5.14, , How much time would it take to distribute one Avogadro number of wheat grains, if, 1010 grains are distributed each second ?, , 5.15, , Calculate the total pressure in a mixture of 8 g of dioxygen and 4 g of dihydrogen, confined in a vessel of 1 dm3 at 27°C. R = 0.083 bar dm3 K–1 mol–1., , 5.16, , Pay load is defined as the difference between the mass of displaced air and the, mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass, 100 kg is filled with helium at 1.66 bar at 27°C. (Density of air = 1.2 kg m–3 and, R = 0.083 bar dm3 K–1 mol–1)., , 5.17, , Calculate the volume occupied by 8.8 g of CO 2 at 31.1°C and 1 bar pressure., R = 0.083 bar L K–1 mol–1., , 5.18, , 2.9 g of a gas at 95 °C occupied the same volume as 0.184 g of dihydrogen at 17 °C,, at the same pressure. What is the molar mass of the gas?, , 5.19, , A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight, of dihydrogen. Calculate the partial pressure of dihydrogen., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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STATES OF MATTER, , 159, , 5.20, , What would be the SI unit for the quantity pV 2T 2/n ?, , 5.21, , In terms of Charles’ law explain why –273 °C is the lowest possible temperature., , 5.22, , Critical temperature for carbon dioxide and methane are 31.1 °C and –81.9 °C, respectively. Which of these has stronger intermolecular forces and why?, , 5.23, , Explain the physical significance of van der Waals parameters., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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160, , CHEMISTRY, , UNIT 6, , THERMODYNAMICS, , It is the only physical theory of universal content concerning, which I am convinced that, within the framework of the, applicability of its basic concepts, it will never be overthrown., After studying this Unit, you will, be able to, •, •, •, •, , •, , •, •, •, •, •, •, •, •, •, , •, •, , explain the terms : system and, surroundings;, discriminate between close,, open and isolated systems;, explain internal energy, work, and heat;, state, first, law, of, thermodynamics and express, it mathematically;, calculate energy changes as, work and heat contributions, in chemical systems;, explain state functions: U, H., correlate ∆U and ∆H;, measure experimentally ∆U, and ∆H;, define standard states for ∆H;, calculate enthalpy changes for, various types of reactions;, state and apply Hess’s law of, constant heat summation;, differentiate between extensive, and intensive properties;, define spontaneous and nonspontaneous processes;, explain entropy as a, thermodynamic state function, and apply it for spontaneity;, explain Gibbs energy change, (∆G); and, establish relationship between, ∆G and spontaneity, ∆G and, equilibrium constant., , Albert Einstein, , Chemical energy stored by molecules can be released as heat, during chemical reactions when a fuel like methane, cooking, gas or coal burns in air. The chemical energy may also be, used to do mechanical work when a fuel burns in an engine, or to provide electrical energy through a galvanic cell like, dry cell. Thus, various forms of energy are interrelated and, under certain conditions, these may be transformed from, one form into another. The study of these energy, transformations forms the subject matter of thermodynamics., The laws of thermodynamics deal with energy changes of, macroscopic systems involving a large number of molecules, rather than microscopic systems containing a few molecules., Thermodynamics is not concerned about how and at what, rate these energy transformations are carried out, but is, based on initial and final states of a system undergoing the, change. Laws of thermodynamics apply only when a system, is in equilibrium or moves from one equilibrium state to, another equilibrium state. Macroscopic properties like, pressure and temperature do not change with time for a, system in equilibrium state. In this unit, we would like to, answer some of the important questions through, thermodynamics, like:, How do we determine the energy changes involved in a, chemical reaction/process? Will it occur or not?, What drives a chemical reaction/process?, To what extent do the chemical reactions proceed?, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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THERMODYNAMICS, , 161, , 6.1 THERMODYNAMIC TERMS, We are interested in chemical reactions and the, energy changes accompanying them. For this, we need to know certain thermodynamic, terms. These are discussed below., 6.1.1 The System and the Surroundings, A system in thermodynamics refers to that, part of universe in which observations are, made and remaining universe constitutes the, surroundings. The surroundings include, everything other than the system. System and, the surroundings together constitute the, universe ., The universe = The system + The surroundings, However, the entire universe other than, the system is not affected by the changes, taking place in the system. Therefore, for, all practical purposes, the surroundings, are that portion of the remaining universe, which can interact with the system., Usually, the region of space in the, neighbourhood of the system constitutes, its surroundings., For example, if we are studying the, reaction between two substances A and B, kept in a beaker, the beaker containing the, reaction mixture is the system and the room, where the beaker is kept is the surroundings, (Fig. 6.1)., , be real or imaginary. The wall that separates, the system from the surroundings is called, boundary. This is designed to allow us to, control and keep track of all movements of, matter and energy in or out of the system., 6.1.2 Types of the System, We, further classify the systems according to, the movements of matter and energy in or out, of the system., 1. Open System, In an open system, there is exchange of energy, and matter between system and surroundings, [Fig. 6.2 (a)]. The presence of reactants in an, open beaker is an example of an open system*., Here the boundary is an imaginary surface, enclosing the beaker and reactants., 2. Closed System, In a closed system, there is no exchange of, matter, but exchange of energy is possible, between system and the surroundings, [Fig. 6.2 (b)]. The presence of reactants in a, closed vessel made of conducting material e.g.,, copper or steel is an example of a closed, system., , Fig. 6.1 System and the surroundings, , Note that the system may be defined by, physical boundaries, like beaker or test tube,, or the system may simply be defined by a set, of Cartesian coordinates specifying a, particular volume in space. It is necessary to, think of the system as separated from the, surroundings by some sort of wall which may, , *, , Fig. 6.2 Open, closed and isolated systems., , We could have chosen only the reactants as system then walls of the beakers will act as boundary., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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162, , CHEMISTRY, , 3. Isolated System, In an isolated system, there is no exchange of, energy or matter between the system and the, surroundings [Fig. 6.2 (c)]. The presence of, reactants in a thermos flask or any other closed, insulated vessel is an example of an isolated, system., 6.1.3 The State of the System, The system must be described in order to make, any useful calculations by specifying, quantitatively each of the properties such as, its pressure (p), volume (V), and temperature, (T ) as well as the composition of the system., We need to describe the system by specifying, it before and after the change. You would recall, from your Physics course that the state of a, system in mechanics is completely specified at, a given instant of time, by the position and, velocity of each mass point of the system. In, thermodynamics, a different and much simpler, concept of the state of a system is introduced., It does not need detailed knowledge of motion, of each particle because, we deal with average, measurable properties of the system. We specify, the state of the system by state functions or, state variables., The state of a thermodynamic system is, described by its measurable or macroscopic, (bulk) properties. We can describe the state of, a gas by quoting its pressure (p), volume (V),, temperature (T ), amount (n) etc. Variables like, p, V, T are called state variables or state, functions because their values depend only, on the state of the system and not on how it is, reached. In order to completely define the state, of a system it is not necessary to define all the, properties of the system; as only a certain, number of properties can be varied, independently. This number depends on the, nature of the system. Once these minimum, number of macroscopic properties are fixed,, others automatically have definite values., The state of the surroundings can never, be completely specified; fortunately it is not, necessary to do so., 6.1.4 The Internal Energy as a State, Function, When we talk about our chemical system, losing or gaining energy, we need to introduce, a quantity which represents the total energy, , of the system. It may be chemical, electrical,, mechanical or any other type of energy you, may think of, the sum of all these is the energy, of the system. In thermodynamics, we call it, the internal energy, U of the system, which may, change, when, • heat passes into or out of the system,, • work is done on or by the system,, • matter enters or leaves the system., These systems are classified accordingly as, you have already studied in section 6.1.2., (a) Work, Let us first examine a change in internal, energy by doing work. We take a system, containing some quantity of water in a thermos, flask or in an insulated beaker. This would not, allow exchange of heat between the system, and surroundings through its boundary and, we call this type of system as adiabatic. The, manner in which the state of such a system, may be changed will be called adiabatic, process. Adiabatic process is a process in, which there is no transfer of heat between the, system and surroundings. Here, the wall, separating the system and the surroundings, is called the adiabatic wall (Fig 6.3)., Let us bring the change in the internal, energy of the system by doing some work on, , Fig. 6.3 An adiabatic system which does not, permit the transfer of heat through its, boundary., , it. Let us call the initial state of the system as, state A and its temperature as TA. Let the, internal energy of the system in state A be, called UA. We can change the state of the system, in two different ways., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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THERMODYNAMICS, , 163, , One way: We do some mechanical work, say, 1 kJ, by rotating a set of small paddles and, thereby churning water. Let the new state be, called B state and its temperature, as TB. It is, found that T B > T A and the change in, temperature, ∆T = T B–TA. Let the internal, energy of the system in state B be UB and the, change in internal energy, ∆U =UB– UA., Second way: We now do an equal amount (i.e.,, 1kJ) electrical work with the help of an, immersion rod and note down the temperature, change. We find that the change in temperature, is same as in the earlier case, say, TB – TA., In fact, the experiments in the above, manner were done by J. P. Joule between, 1840–50 and he was able to show that a given, amount of work done on the system, no matter, how it was done (irrespective of path) produced, the same change of state, as measured by the, change in the temperature of the system., So, it seems appropriate to define a, quantity, the internal energy U, whose value, is characteristic of the state of a system,, whereby the adiabatic work, wad required to, bring about a change of state is equal to the, difference between the value of U in one state, and that in another state, ∆U i.e.,, ∆U = U 2 − U 1 = w ad, Therefore, internal energy, U, of the system, is a state function., By conventions of IUPAC in chemical, thermodynamics. The positive sign expresses, that wad is positive when work is done on the, system and the internal energy of system, increases. Similarly, if the work is done by the, system,wad will be negative because internal, energy of the system decreases., Can you name some other familiar state, functions? Some of other familiar state, functions are V, p, and T. For example, if we, bring a change in temperature of the system, from 25°C to 35°C, the change in temperature, is 35°C–25°C = +10°C, whether we go straight, up to 35°C or we cool the system for a few, degrees, then take the system to the final, temperature. Thus, T is a state function and, the change in temperature is independent of, , *, , the route taken. Volume of water in a pond,, for example, is a state function, because, change in volume of its water is independent, of the route by which water is filled in the, pond, either by rain or by tubewell or by both., (b) Heat, We can also change the internal energy of a, system by transfer of heat from the, surroundings to the system or vice-versa, without expenditure of work. This exchange, of energy, which is a result of temperature, difference is called heat, q. Let us consider, bringing about the same change in temperature, (the same initial and final states as before in, section 6.1.4 (a) by transfer of heat through, thermally conducting walls instead of, adiabatic walls (Fig. 6.4)., , Fig. 6.4, , A system which allows heat transfer, through its boundary., , We take water at temperature, TA in a, container having thermally conducting walls,, say made up of copper and enclose it in a huge, heat reservoir at temperature, TB. The heat, absorbed by the system (water), q can be, measured in terms of temperature difference ,, TB – TA. In this case change in internal energy,, ∆U= q, when no work is done at constant, volume., By conventions of IUPAC in chemical, thermodynamics. The q is positive, when, heat is transferred from the surroundings to, the system and the internal energy of the, system increases and q is negative when, heat is transferred from system to the, surroundings resulting in decrease of the, internal energy of the system.., , Earlier negative sign was assigned when the work is done on the system and positive sign when the work is done by the, system. This is still followed in physics books, although IUPAC has recommended the use of new sign convention., Download all NCERT books PDFs from www.ncert.online, , 2019-20

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164, , CHEMISTRY, , (c) The general case, Let us consider the general case in which a, change of state is brought about both by, doing work and by transfer of heat. We write, change in internal energy for this case as:, ∆U = q + w, , (6.1), , For a given change in state, q and w can, vary depending on how the change is carried, out. However, q +w = ∆U will depend only on, initial and final state. It will be independent of, the way the change is carried out. If there is, no transfer of energy as heat or as work, (isolated system) i.e., if w = 0 and q = 0, then, ∆ U = 0., The equation 6.1 i.e., ∆U = q + w is, mathematical statement of the first law of, thermodynamics, which states that, The energy of an isolated system is, constant., It is commonly stated as the law of, conservation of energy i.e., energy can neither, be created nor be destroyed., Note: There is considerable difference between, the character of the thermodynamic property, energy and that of a mechanical property such, as volume. We can specify an unambiguous, (absolute) value for volume of a system in a, particular state, but not the absolute value of, the internal energy. However, we can measure, only the changes in the internal energy, ∆U of, the system., Problem 6.1, Express the change in internal energy of, a system when, (i) No heat is absorbed by the system, from the surroundings, but work (w), is done on the system. What type of, wall does the system have ?, (ii) No work is done on the system, but, q amount of heat is taken out from, the system and given to the, surroundings. What type of wall does, the system have?, (iii) w amount of work is done by the, system and q amount of heat is, supplied to the system. What type of, system would it be?, , Solution, (i) ∆ U = w ad, wall is adiabatic, (ii) ∆ U = – q, thermally conducting walls, (iii) ∆ U = q – w, closed system., 6.2 APPLICATIONS, Many chemical reactions involve the generation, of gases capable of doing mechanical work or, the generation of heat. It is important for us to, quantify these changes and relate them to the, changes in the internal energy. Let us see how!, 6.2.1 Work, First of all, let us concentrate on the nature of, work a system can do. We will consider only, mechanical work i.e., pressure-volume work., For understanding pressure-volume, work, let us consider a cylinder which, contains one mole of an ideal gas fitted with a, frictionless piston. Total volume of the gas is, Vi and pressure of the gas inside is p. If, external pressure is pex which is greater than, p, piston is moved inward till the pressure, inside becomes equal to pex. Let this change, , Fig. 6.5(a) Work done on an ideal gas in a, cylinder when it is compressed by a, constant external pressure, p ex, (in single step) is equal to the shaded, area., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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THERMODYNAMICS, , 165, , be achieved in a single step and the final, volume be Vf . During this compression,, suppose piston moves a distance, l and is, cross-sectional area of the piston is A, [Fig. 6.5(a)]., then, volume change = l × A = ∆V = (Vf – Vi ), We also know, pressure =, , force, area, , Vf, , Therefore, force on the piston = pex . A, If w is the work done on the system by, movement of the piston then, w = force × distance = pex . A .l, = pex . (–∆V) = – pex ∆V = – pex (Vf – Vi ), , If the pressure is not constant but changes, during the process such that it is always, infinitesimally greater than the pressure of the, gas, then, at each stage of compression, the, volume decreases by an infinitesimal amount,, dV. In such a case we can calculate the work, done on the gas by the relation, , (6.2), , The negative sign of this expression is, required to obtain conventional sign for w,, which will be positive. It indicates that in case, of compression work is done on the system., Here (Vf – Vi ) will be negative and negative, multiplied by negative will be positive. Hence, the sign obtained for the work will be positive., If the pressure is not constant at every, stage of compression, but changes in number, of finite steps, work done on the gas will be, summed over all the steps and will be equal, to −  p ∆V [Fig. 6.5 (b)], , Fig. 6.5 (b) pV-plot when pressure is not constant, and changes in finite steps during, compression from initial volume, Vi to, final volume, Vf . Work done on the gas, is represented by the shaded area., , w=−, , ∫p, , ex, , dV, , ( 6.3), , Vi, , Here, pex at each stage is equal to (pin + dp) in, case of compression [Fig. 6.5(c)]. In an, expansion process under similar conditions,, the external pressure is always less than the, pressure of the system i.e., pex = (pin– dp). In, general case we can write, pex = (pin + dp). Such, processes are called reversible processes., A process or change is said to be, reversible, if a change is brought out in, such a way that the process could, at any, moment, be reversed by an infinitesimal, change. A reversible process proceeds, infinitely slowly by a series of equilibrium, states such that system and the, surroundings are always in near, equilibrium with each other. Processes, , Fig. 6.5 (c) pV-plot when pressure is not constant, and changes in infinite steps, (reversible conditions) during, compression from initial volume, Vi to, final volume, Vf . Work done on the gas, is represented by the shaded area., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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166, , CHEMISTRY, , other than reversible processes are known, as irreversible processes., In chemistry, we face problems that can, be solved if we relate the work term to the, internal pressure of the system. We can, relate work to internal pressure of the system, under reversible conditions by writing, equation 6.3 as follows:, Vf, , w rev = −, , ∫p, , Vf, ex, , dV = −, , Vi, , ∫ (p, , in, , ± dp ) dV, , Vi, , Since dp × dV is very small we can write, , Isothermal and free expansion of an, ideal gas, For isothermal (T = constant) expansion of an, ideal gas into vacuum ; w = 0 since pex = 0., Also, Joule determined experimentally that, q = 0; therefore, ∆U = 0, Equation 6.1, ∆ U = q + w can be, expressed for isothermal irreversible and, reversible changes as follows:, 1., For isothermal irreversible change, q = – w = pex (Vf – Vi ), 2., For isothermal reversible change, , Vf, , Vf, , w rev = − ∫ pin dV, , q = – w = nRT ln V, i, , (6.4), , Vi, , Now, the pressure of the gas (pin which we, can write as p now) can be expressed in terms, of its volume through gas equation. For n mol, of an ideal gas i.e., pV =nRT, , = 2.303 nRT log, 3., , nRT, V, Therefore, at constant temperature (isothermal, process),, ⇒p =, , Vf, , w rev = − ∫ nRT, Vi, , Vf, dV, = −nRT ln, V, Vi, , = – 2.303 nRT log, , Vf, , (6.5), , Vi, , Free expansion: Expansion of a gas in, vacuum (pex = 0) is called free expansion. No, work is done during free expansion of an ideal, gas whether the process is reversible or, irreversible (equation 6.2 and 6.3)., Now, we can write equation 6.1 in number, of ways depending on the type of processes., Let us substitute w = – pex∆V (eq. 6.2) in, equation 6.1, and we get, ∆U = q − pex ∆V, , If a process is carried out at constant volume, (∆V = 0), then, ∆U = qV, the subscript V in qV denotes that heat is, supplied at constant volume., , Vi, , For adiabatic change, q = 0,, ∆U = wad, Problem 6.2, Two litres of an ideal gas at a pressure of, 10 atm expands isothermally at 25 °C into, a vacuum until its total volume is 10 litres., How much heat is absorbed and how much, work is done in the expansion ?, Solution, We have q = – w = pex (10 – 2) = 0(8) = 0, No work is done; no heat is absorbed., Problem 6.3, Consider the same expansion, but this, time against a constant external pressure, of 1 atm., Solution, We have q = – w = pex (8) = 8 litre-atm, Problem 6.4, Consider the expansion given in problem, 6.2, for 1 mol of an ideal gas conducted, reversibly., Solution, , Vf, Vs 10, = 2.303 × 1 × 0.8206 × 298 × log, , We have q = – w = 2.303 nRT log, , Download all NCERT books PDFs from www.ncert.online, , 2019-20, , Vf, , 2

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THERMODYNAMICS, , 167, , = 2.303 x 0.8206 x 298 x log 5, = 2.303 x 0.8206 x 298 x 0.6990, = 393.66 L atm, 6.2.2 Enthalpy, H, (a) A Useful New State Function, We know that the heat absorbed at constant, volume is equal to change in the internal, energy i.e., ∆U = q . But most of chemical, V, reactions are carried out not at constant, volume, but in flasks or test tubes under, constant atmospheric pressure. We need to, define another state function which may be, suitable under these conditions., We may write equation (6.1) as, ∆U = qp – p∆V at constant pressure, where qp, is heat absorbed by the system and –p ∆V, represent expansion work done by the system., Let us represent the initial state by, subscript 1 and final state by 2, We can rewrite the above equation as, U2–U1 = qp – p (V2 – V1), On rearranging, we get, , Remember ∆H = qp, heat absorbed by the, system at constant pressure., ∆H is negative for exothermic reactions, which evolve heat during the reaction and, ∆H is positive for endothermic reactions, which absorb heat from the surroundings., At constant volume (∆V = 0), ∆U = qV,, therefore equation 6.8 becomes, ∆H = ∆U = q, V, , The difference between ∆H and ∆U is not, usually significant for systems consisting of, only solids and / or liquids. Solids and liquids, do not suffer any significant volume changes, upon heating. The difference, however,, becomes significant when gases are involved., Let us consider a reaction involving gases. If, VA is the total volume of the gaseous reactants,, VB is the total volume of the gaseous products,, nA is the number of moles of gaseous reactants, and nB is the number of moles of gaseous, products, all at constant pressure and, temperature, then using the ideal gas law, we, write,, pVA = nART, , (6.6), qp = (U2 + pV2) – (U1 + pV1), Now we can define another thermodynamic, function, the enthalpy H [Greek word, enthalpien, to warm or heat content] as :, H = U + pV, (6.7), so, equation (6.6) becomes, qp= H2 – H1 = ∆H, Although q is a path dependent function,, H is a state function because it depends on U,, p and V, all of which are state functions., Therefore, ∆H is independent of path. Hence,, qp is also independent of path., For finite changes at constant pressure, we, can write equation 6.7 as, ∆H = ∆U + ∆pV, Since p is constant, we can write, ∆H = ∆U + p∆V, (6.8), It is important to note that when heat is, absorbed by the system at constant pressure,, we are actually measuring changes in the, enthalpy., , and, , pVB = nBRT, , Thus, pVB – pVA = nBRT – nART = (nB–nA)RT, or, , p (VB – VA) = (nB – nA) RT, , or, , p ∆V = ∆ngRT, , (6.9), , Here, ∆ng refers to the number of moles of, gaseous products minus the number of moles, of gaseous reactants., Substituting the value of p∆V from, equation 6.9 in equation 6.8, we get, (6.10), ∆H = ∆U + ∆ngRT, The equation 6.10 is useful for calculating, ∆H from ∆U and vice versa., Problem 6.5, If water vapour is assumed to be a perfect, gas, molar enthalpy change for, vapourisation of 1 mol of water at 1bar, and 100°C is 41kJ mol–1. Calculate the, internal energy change, when, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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168, , CHEMISTRY, , 1 mol of water is vapourised at 1 bar, pressure and 100°C., Solution, (i) The change H2O (l ) → H2O ( g ), , ∆H = ∆U + ∆n g RT, , Fig. 6.6(a) A gas at volume V and temperature T, , or ∆U = ∆H – ∆n g RT , substituting the, values, we get, , ∆U = 41.00 kJ mol −1 − 1, × 8.3 J mol −1K −1 × 373 K, = 41.00 kJ mol-1 – 3.096 kJ mol-1, = 37.904 kJ mol–1, (b) Extensive and Intensive Properties, In thermodynamics, a distinction is made, between extensive properties and intensive, properties. An extensive property is a, property whose value depends on the, quantity or size of matter present in the, system. For example, mass, volume,, internal energy, enthalpy, heat capacity, etc., are extensive properties., Those properties which do not depend, on the quantity or size of matter present are, known as intensive properties. For, example temperature, density, pressure etc., are intensive properties. A molar property,, χm, is the value of an extensive property χ of, the system for 1 mol of the substance. If n, χ, is, is the amount of matter, χ m =, n, independent of the amount of matter. Other, examples are molar volume, Vm and molar, heat capacity, Cm. Let us understand the, distinction between extensive and intensive, properties by considering a gas enclosed in, a container of volume V and at temperature, T [Fig. 6.6(a)]. Let us make a partition such, that volume is halved, each part [Fig. 6.6, (b)] now has one half of the original volume,, V, , but the temperature will still remain the, 2, , same i.e., T. It is clear that volume is an, extensive property and temperature is an, intensive property., , Fig. 6.6 (b) Partition, each part having half the, volume of the gas, , (c) Heat Capacity, In this sub-section, let us see how to measure, heat transferred to a system. This heat appears, as a rise in temperature of the system in case, of heat absorbed by the system., The increase of temperature is proportional, to the heat transferred, , q = coeff × ∆T, The magnitude of the coefficient depends, on the size, composition and nature of the, system. We can also write it as q = C ∆T, The coefficient, C is called the heat capacity., Thus, we can measure the heat supplied, by monitoring the temperature rise, provided, we know the heat capacity., When C is large, a given amount of heat, results in only a small temperature rise. Water, has a large heat capacity i.e., a lot of energy is, needed to raise its temperature., C is directly proportional to amount of, substance. The molar heat capacity of a, , C, substance, Cm =   , is the heat capacity for, n, one mole of the substance and is the quantity, of heat needed to raise the temperature of, one mole by one degree celsius (or one, kelvin). Specific heat, also called specific heat, capacity is the quantity of heat required to, raise the temperature of one unit mass of a, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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THERMODYNAMICS, , 169, , substance by one degree celsius (or one, kelvin). For finding out the heat, q, required, to raise the temperatures of a sample, we, multiply the specific heat of the substance,, c, by the mass m, and temperatures change,, ∆T as, , q = c × m × ∆T = C ∆T, , (6.11), , (d) The Relationship between Cp and CV, for an Ideal Gas, At constant volume, the heat capacity, C is, denoted by CV and at constant pressure, this, is denoted by Cp . Let us find the relationship, between the two., We can write equation for heat, q, , i) at constant volume, qV, ii) at constant pressure, qp, (a) ∆U Measurements, For chemical reactions, heat absorbed at, constant volume, is measured in a bomb, calorimeter (Fig. 6.7). Here, a steel vessel (the, bomb) is immersed in a water bath. The whole, device is called calorimeter. The steel vessel is, immersed in water bath to ensure that no heat, is lost to the surroundings. A combustible, substance is burnt in pure dioxygen supplied, , at constant volume as qV = CV ∆T = ∆U, at constant pressure as qp = C p ∆T = ∆H, The difference between Cp and CV can be, derived for an ideal gas as:, For a mole of an ideal gas, ∆H = ∆U + ∆(pV ), = ∆U + ∆(RT ), = ∆U + R∆T, , ∴ ∆H = ∆U + R ∆T, , (6.12), , On putting the values of ∆H and ∆U,, we have, , C p ∆T = CV ∆T + R∆T, C p = CV + R, Cp – CV = R, , Fig. 6.7 Bomb calorimeter, , (6.13), , 6.3 MEASUREMENT OF ∆ U AND ∆ H:, CALORIMETRY, We can measure energy changes associated, with chemical or physical processes by an, experimental technique called calorimetry. In, calorimetry, the process is carried out in a, vessel called calorimeter, which is immersed, in a known volume of a liquid. Knowing the, heat capacity of the liquid in which calorimeter, is immersed and the heat capacity of, calorimeter, it is possible to determine the heat, evolved in the process by measuring, temperature changes. Measurements are, made under two different conditions:, , in the steel bomb. Heat evolved during the, reaction is transferred to the water around the, bomb and its temperature is monitored. Since, the bomb calorimeter is sealed, its volume does, not change i.e., the energy changes associated, with reactions are measured at constant, volume. Under these conditions, no work is, done as the reaction is carried out at constant, volume in the bomb calorimeter. Even for, reactions involving gases, there is no work, done as ∆V = 0. Temperature change of the, calorimeter produced by the completed, reaction is then converted to qV, by using the, known heat capacity of the calorimeter with, the help of equation 6.11., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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170, , CHEMISTRY, , (b) ∆H Measurements, Measurement of heat change at constant pressure, (generally under atmospheric pressure) can be, done in a calorimeter shown in Fig. 6.8. We know, that ∆Η = qp (at constant p) and, therefore, heat, absorbed or evolved, qp at constant pressure is, also called the heat of reaction or enthalpy of, reaction, ∆rH., In an exothermic reaction, heat is evolved,, and system loses heat to the surroundings., Therefore, qp will be negative and ∆rH will also, be negative. Similarly in an endothermic, reaction, heat is absorbed, qp is positive and, ∆rH will be positive., , of the bomb calorimeter is 20.7kJ/K,, what is the enthalpy change for the above, reaction at 298 K and 1 atm?, Solution, Suppose q is the quantity of heat from the, reaction mixture and CV is the heat capacity, of the calorimeter, then the quantity of heat, absorbed by the calorimeter., q = CV × ∆T, Quantity of heat from the reaction will, have the same magnitude but opposite, sign because the heat lost by the system, (reaction mixture) is equal to the heat, gained by the calorimeter., q = − CV × ∆T = − 20.7 kJ/K × (299 − 298) K, = − 20.7 kJ, , (Here, negative sign indicates the, exothermic nature of the reaction), Thus, ∆U for the combustion of the 1g of, graphite = – 20.7 kJK–1, For combustion of 1 mol of graphite,, , =, , 12.0 g mol −1 × ( −20.7 kJ ), 1g, , = – 2.48 ×102 kJ mol–1 ,, Since ∆ ng = 0,, ∆ H = ∆ U = – 2.48 ×102 kJ mol–1, , Fig. 6.8 Calorimeter for measuring heat changes, at constant pressure (atmospheric, pressure)., , Problem 6.6, 1g of graphite is burnt in a bomb, calorimeter in excess of oxygen at 298 K, and 1 atmospheric pressure according to, the equation, C (graphite) + O2 (g) → CO2 (g), During the reaction, temperature rises, from 298 K to 299 K. If the heat capacity, , 6.4 ENTHALPY CHANGE, ∆ r H OF A, REACTION – REACTION ENTHALPY, In a chemical reaction, reactants are converted, into products and is represented by,, Reactants → Products, The enthalpy change accompanying a, reaction is called the reaction enthalpy. The, enthalpy change of a chemical reaction, is given, by the symbol ∆rH, ∆rH = (sum of enthalpies of products) – (sum, of enthalpies of reactants), , =, , ∑a H, i, , i, , products, , − ∑ bi H reactants (6.14), , Here symbol, , i, , ∑, , (sigma) is used for, , summation and ai and bi are the stoichiometric, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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THERMODYNAMICS, , 171, , coefficients of the products and reactants, respectively in the balanced chemical, equation. For example, for the reaction, CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l), ∆r H = ∑ a i H Pr oducts − ∑ bi H reac tan ts, i, , i, , = [Hm (CO2 ,g) + 2Hm (H2O, l)]– [Hm (CH4 , g), , ethanol at 298 K is pure liquid ethanol at, 1 bar; standard state of solid iron at 500 K is, pure iron at 1 bar. Usually data are taken, at 298 K., Standard conditions are denoted by adding, the superscript to the symbol ∆H, e.g., ∆H, , + 2Hm (O2, g)], where Hm is the molar enthalpy., , (b) Enthalpy Changes during Phase, Transformations, , Enthalpy change is a very useful quantity., Knowledge of this quantity is required when, one needs to plan the heating or cooling, required to maintain an industrial chemical, reaction at constant temperature. It is also, required to calculate temperature dependence, of equilibrium constant., , Phase transformations also involve energy, changes. Ice, for example, requires heat for, melting. Normally this melting takes place at, constant pressure (atmospheric pressure) and, during phase change, temperature remains, constant (at 273 K)., , , H2O(s) → H2O(l); ∆fusH = 6.00 kJ moI–1, , (a) Standard Enthalpy of Reactions, , , , Enthalpy of a reaction depends on the, conditions under which a reaction is carried, out. It is, therefore, necessary that we must, specify some standard conditions. The, standard enthalpy of reaction is the, enthalpy change for a reaction when all, the participating substances are in their, standard states., The standard state of a substance at a, specified temperature is its pure form at, 1 bar. For example, the standard state of liquid, , Here ∆fusH is enthalpy of fusion in standard, state. If water freezes, then process is reversed, and equal amount of heat is given off to the, surroundings., The enthalpy change that accompanies, melting of one mole of a solid substance, in standard state is called standard, enthalpy of fusion or molar enthalpy of, fusion, ∆fusH ., Melting of a solid is endothermic, so all, enthalpies of fusion are positive. Water requires, , Table 6.1 Standard Enthalpy Changes of Fusion and Vaporisation, , (Tf and Tb are melting and boiling points, respectively), Download all NCERT books PDFs from www.ncert.online, , 2019-20

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172, , CHEMISTRY, , heat for evaporation. At constant temperature, of its boiling point Tb and at constant pressure:, , , H2O(l) → H2O(g); ∆vapH = + 40.79 kJ moI–1 ∆vapH, is the standard enthalpy of vaporisation., , , , Amount of heat required to vaporize, one mole of a liquid at constant, temperature and under standard pressure, (1bar) is called its standard enthalpy of, vaporization or molar enthalpy of, vaporization, ∆vapH ., Sublimation is direct conversion of a solid, into its vapour. Solid CO2 or ‘dry ice’ sublimes, , at 195K with ∆ sub H =25.2 kJ mol – 1 ;, naphthalene sublimes slowly and for this, ∆sub H = 73.0 kJ mol–1 ., Standard enthalpy of sublimation,, , ∆subH is the change in enthalpy when one, mole of a solid substance sublimes at a, constant temperature and under standard, pressure (1bar)., The magnitude of the enthalpy change, depends on the strength of the intermolecular, interactions in the substance undergoing the, phase transfomations. For example, the strong, hydrogen bonds between water molecules hold, them tightly in liquid phase. For an organic, liquid, such as acetone, the intermolecular, dipole-dipole interactions are significantly, weaker. Thus, it requires less heat to vaporise, 1 mol of acetone than it does to vaporize 1 mol, of water. Table 6.1 gives values of standard, enthalpy changes of fusion and vaporisation, for some substances., Problem 6.7, A swimmer coming out from a pool is, covered with a film of water weighing, about 18g. How much heat must be, supplied to evaporate this water at, 298 K ? Calculate the internal energy of, vaporisation at 298K., ∆vap H  for water, at 298K= 44.01kJ mol–1, , Solution, We can represent the process of, evaporation as, vaporisation, → H2 O(g), H 2O(1) , 1mol, 1mol, , No. of moles in 18 g H2O(l) is, , =, , 18g, = 1mol, 18 g mol −1, , Heat supplied to evaporate18g water at, , 298 K, = n × ∆vap H, = (1 mol) × (44.01 kJ mol–1), = 44.01 kJ, (assuming steam behaving as an ideal, gas)., ∆vapU = ∆vap H  − p ∆V = ∆vap H  − ∆n g RT, , ∆vap H V − ∆n g RT = 44.01 kJ, − (1)(8.314 JK −1 mol −1 )(298K )(10 −3 kJ J −1 ), ∆vapU V = 44.01 kJ − 2.48 kJ, = 41.53 kJ, Problem 6.8, Assuming the water vapour to be a perfect, gas, calculate the internal energy change, when 1 mol of water at 100°C and 1 bar, pressure is converted to ice at 0°C. Given, the enthalpy of fusion of ice is 6.00 kJ mol1, heat capacity of water is 4.2 J/g°C, The change take place as follows:, Step - 1, 1 mol H2O (l, 100°C)  1, mol (l, 0°C) Enthalpy, change ∆H1, Step - 2, , 1 mol H2O (l, 0°C)  1 mol, H2O( S, 0°C) Enthalpy, change ∆H2, Total enthalpy change will be ∆H = ∆H1 + ∆H2, ∆H1 = - (18 x 4.2 x 100) J mol-1, = - 7560 J mol-1 = - 7.56 k J mol-1, ∆H2 = - 6.00 kJ mol-1, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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THERMODYNAMICS, , 173, , , Table 6.2 Standard Molar Enthalpies of Formation (∆, ∆f H ) at 298K of a, Few Selected Substances, , Therefore,, ∆H = - 7.56 kJ mol-1 + (-6.00 kJ mol-1), = -13.56 kJ mol-1, There is negligible change in the volume, during the change form liquid to solid, state., Therefore, p∆v = ∆ng RT = 0, ∆H = ∆U = - 13.56kJ mol-1, (c) Standard Enthalpy of Formation, The standard enthalpy change for the, formation of one mole of a compound from, its elements in their most stable states of, , aggregation (also known as reference, states) is called Standard Molar Enthalpy, of Formation. Its symbol is ∆f H , where, the subscript ‘ f ’ indicates that one mole of, the compound in question has been formed, in its standard state from its elements in their, most stable states of aggregation. The reference, state of an element is its most stable state of, aggregation at 25°C and 1 bar pressure., For example, the reference state of dihydrogen, is H2 gas and those of dioxygen, carbon and, sulphur are O 2 gas, C graphite and S rhombic, respectively. Some reactions with standard, molar enthalpies of formation are as follows., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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174, , CHEMISTRY, , H2(g) + ½O2 (g) → H2O(1);, , ∆f H = –74.81 kJ mol–1, , Here, we can make use of standard enthalpy, of formation and calculate the enthalpy, change for the reaction. The following general, equation can be used for the enthalpy change, calculation., , 2C (graphite, s)+3H2 (g)+ ½O2(g) → C2H5OH(1);, , ∆rH= ∑ ai ∆f H(products) – ∑ bi ∆f H(reactants), , , , ∆f H = –285.8 kJ mol, , –1, , C (graphite, s) + 2H2(g) → Ch4 (g);, , ∆f H = – 277.7kJ mol–1, It is important to understand that a, , standard molar enthalpy of formation, ∆f H ,, , is just a special case of ∆rH , where one mole, of a compound is formed from its constituent, elements, as in the above three equations,, where 1 mol of each, water, methane and, ethanol is formed. In contrast, the enthalpy, change for an exothermic reaction:, CaO(s) + CO2(g) → CaCo3(s);, ∆rH= – 178.3kJ mol–1, is not an enthalpy of formation of calcium, carbonate, since calcium carbonate has been, formed from other compounds, and not from, its constituent elements. Also, for the reaction, given below, enthalpy change is not standard, , enthalpy of formation, ∆fH for HBr(g)., H2(g) + Br2(l) → 2HBr(g);, ∆r H= – 178.3kJ mol–1, Here two moles, instead of one mole of the, product is formed from the elements, i.e., ., ∆r H= 2∆f H, Therefore, by dividing all coefficients in the, balanced equation by 2, expression for, enthalpy of formation of HBr (g) is written as, ½H2(g) + ½Br2(1) → HBr(g);, ∆f H= – 36.4 kJ mol–1, Standard enthalpies of formation of some, common substances are given in Table 6.2., By convention, standard enthalpy for, formation, ∆f H , of an element in reference, state, i.e., its most stable state of aggregation, is taken as zero., Suppose, you are a chemical engineer and, want to know how much heat is required to, decompose calcium carbonate to lime and, carbon dioxide, with all the substances in their, standard state., CaCO3(s) → CaO(s) + CO2(g); ∆r H = ?, , i, , i, , (6.15), where a and b represent the coefficients of the, products and reactants in the balanced, equation. Let us apply the above equation for, decomposition of calcium carbonate. Here,, coefficients ‘a’ and ‘b’ are 1 each. Therefore,, ∆rH= ∆f H= [CaO(s)]+ ∆f H[CO2(g)], – ∆f H= [CaCO3(s)], =1 (–635.1 kJ mol–1) + 1(–393.5 kJ mol–1), –1(–1206.9 kJ mol–1), = 178.3 kJ mol–1, Thus, the decomposition of CaCO3 (s) is an, endothermic process and you have to heat it, for getting the desired products., (d) Thermochemical Equations, A balanced chemical equation together with, the value of its ∆rH is called a thermochemical, equation. We specify the physical state, (alongwith allotropic state) of the substance in, an equation. For example:, C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l);, ∆rH= – 1367 kJ mol–1, The above equation describes the, combustion of liquid ethanol at constant, temperature and pressure. The negative sign, of enthalpy change indicates that this is an, exothermic reaction., It would be necessary to remember the, following conventions regarding thermochemical equations., 1. The coefficients in a balanced thermochemical equation refer to the number of, moles (never molecules) of reactants and, products involved in the reaction., , , 2. The numerical value of ∆rH refers to the, number of moles of substances specified, by an equation. Standard enthalpy change, , ∆rH will have units as kJ mol–1., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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THERMODYNAMICS, , 175, , To illustrate the concept, let us consider, the calculation of heat of reaction for the, following reaction :, , Fe2 O3 (s ) + 3H2 ( g ) → 2Fe ( s) + 3H2 O ( l ) ,, From the Table (6.2) of standard enthalpy of, formation (∆f H ), we find :, ∆f H  (H2O,l) = –285.83 kJ mol–1;, ∆f H  (Fe2O3,s) = – 824.2 kJ mol–1;, Also ∆f H  (Fe, s) = 0 and, ∆f H  (H2, g) = 0 as per convention, Then,, ∆f H 1 = 3(–285.83 kJ mol–1), – 1(– 824.2 kJ mol–1), = (–857.5 + 824.2) kJ mol–1, = –33.3 kJ mol–1, Note that the coefficients used in these, calculations are pure numbers, which are, equal to the respective stoichiometric, , is, coefficients. The unit for ∆ r H, –1, kJ mol , which means per mole of reaction., Once we balance the chemical equation in a, particular way, as above, this defines the mole, of reaction. If we had balanced the equation, differently, for example,, , 1, 3, 3, Fe2 O3 (s ) + H2 ( g ) → Fe ( s) + H 2O ( l ), 2, 2, 2, then this amount of reaction would be one, , mole of reaction and ∆rH would be, ∆f H 2 =, –, , 3, (–285.83 kJ mol–1), 2, , (e) Hess’s Law of Constant Heat, Summation, We know that enthalpy is a state function,, therefore the change in enthalpy is, independent of the path between initial state, (reactants) and final state (products). In other, words, enthalpy change for a reaction is the, same whether it occurs in one step or in a, series of steps. This may be stated as follows, in the form of Hess’s Law., If a reaction takes place in several steps, then its standard reaction enthalpy is the, sum of the standard enthalpies of the, intermediate reactions into which the, overall reaction may be divided at the same, temperature., Let us understand the importance of this, law with the help of an example., Consider the enthalpy change for the, reaction, , 1, O (g) → CO (g); ∆r H = ?, 2 2, Although CO(g) is the major product, some, CO2 gas is always produced in this reaction., Therefore, we cannot measure enthalpy change, for the above reaction directly. However, if we, can find some other reactions involving related, species, it is possible to calculate the enthalpy, change for the above reaction., C (graphite,s) +, , Let us consider the following reactions:, C (graphite,s) + O2 (g) → CO2 (g);, ∆r H = – 393.5 kJ mol–1 (i), , 1, (–824.2 kJ mol–1), 2, , CO (g) +, , = (– 428.7 + 412.1) kJ mol–1, = –16.6 kJ mol–1 = ½ ∆r H 1, It shows that enthalpy is an extensive quantity., 3. When a chemical equation is reversed, the, , value of ∆rH is reversed in sign. For, example, N2(g) + 3H2 (g) → 2NH3 (g);, ∆r H = – 91.8 kJ. mol–1, 2NH3(g) → N2(g) + 3H2 (g);, ∆r H = + 91.8 kJ mol–1, , 1, O (g) → CO2 (g), 2 2, ∆r H = – 283.0 kJ mol–1 (ii), , We can combine the above two reactions, in such a way so as to obtain the desired, reaction. To get one mole of CO(g) on the right,, we reverse equation (ii). In this, heat is, absorbed instead of being released, so we, , change sign of ∆rH value, CO2 (g) → CO (g) +, , ∆r H = + 283.0 kJ mol–1 (iii), , Download all NCERT books PDFs from www.ncert.online, , 2019-20, , 1, O (g);, 2 2

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176, , CHEMISTRY, , Adding equation (i) and (iii), we get the, desired equation,, , C (graphite, s) +, for which ∆r H, , , , 1, O2 ( g ) → CO ( g ) ;, 2, , C6 H12 O6 ( g ) + 6O2 ( g ) → 6CO2 ( g ) + 6H 2 O(1);, ∆C H  = – 2802.0 kJ mol–1, , = (– 393.5 + 283.0), , = – 110.5 kJ mol–1, In general, if enthalpy of an overall reaction, A→B along one route is ∆rH and ∆rH1, ∆rH2,, ∆rH3..... representing enthalpies of reactions, leading to same product, B along another, route,then we have, ∆rH = ∆rH1 + ∆rH2 + ∆rH3 ..., , Similarly, combustion of glucose gives out, 2802.0 kJ/mol of heat, for which the overall, equation is :, , Our body also generates energy from food, by the same overall process as combustion,, although the final products are produced after, a series of complex bio-chemical reactions, involving enzymes., , (6.16), , It can be represented as:, A, , ∆rH, , B, ∆rH3, , ∆H1, , C, , CO2(g) and H2 O(l) are –393.5 kJ mol–1, and – 285.83 kJ mol–1 respectively., , D, ∆rH2, , 6.5 ENTHALPIES FOR DIFFERENT TYPES, OF REACTIONS, It is convenient to give name to enthalpies, specifying the types of reactions., (a), , Standard Enthalpy of Combustion, , (symbol : ∆cH ), Combustion reactions are exothermic in, nature. These are important in industry,, rocketry, and other walks of life. Standard, enthalpy of combustion is defined as the, enthalpy change per mole (or per unit amount), of a substance, when it undergoes combustion, and all the reactants and products being in, their standard states at the specified, temperature., Cooking gas in cylinders contains mostly, butane (C4H10). During complete combustion, of one mole of butane, 2658 kJ of heat is, released. We can write the thermochemical, reactions for this as:, , C4 H10 ( g ) +, , Problem 6.9, The combustion of one mole of benzene, takes place at 298 K and 1 atm. After, combustion, CO2(g) and H2O (1) are, produced and 3267.0 kJ of heat is, liberated. Calculate the standard, , enthalpy of formation, ∆f H of benzene., Standard enthalpies of formation of, , 13, O2 ( g ) → 4CO2 ( g ) + 5H2 O(1);, 2, ∆C H  = – 2658.0 kJ mol–1, , Solution, The formation reaction of benezene is, given by :, , 6C (graphite ) + 3H2 ( g ) → C6 H6 ( l ) ;, ∆f H  = ? ... (i), The enthalpy of combustion of 1 mol of, benzene is :, , C 6 H6 ( l ) +, , 15, O2 → 6CO2 ( g ) + 3H2 O ( l ) ;, 2, ∆C H  = – 3267 kJ mol–1... (ii), , The enthalpy of formation of 1 mol of, CO2(g) :, , C ( graphite ) + O2 ( g ) → CO2 ( g ) ;, ∆f H  = – 393.5 kJ mol–1... (iii), The enthalpy of formation of 1 mol of, H2O(l) is :, , H2 ( g ) +, , 1, O 2 ( g ) → H2 O ( l ) ;, 2, ∆C H  = – 285.83 kJ mol–1... (iv), , multiplying eqn. (iii) by 6 and eqn. (iv), by 3 we get:, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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THERMODYNAMICS, , 177, , 6C ( graphite ) + 6O2 ( g ) → 6CO2 (g ) ;, ∆f H  = – 2361 kJ mol–1, , 3 H2 ( g ) +, , 3, O2 ( g ) → 3H 2 O (1) ;, 2, ∆f H  = – 857.49 kJ mol–1, , Summing up the above two equations :, 6C ( graphite ) + 3H2 ( g ) +, , 15, O2 ( g ) → 6CO2 ( g ), 2, + 3 H2 O ( l ) ;, , ∆f H  = – 3218.49 kJ mol–1... (v), Reversing equation (ii);, , 15, 6CO2 ( g ) + 3H2 O ( l ) → C6 H6 ( l ) +, O2 ;, 2, ∆f H  = – 3267.0 kJ mol–1... (vi), Adding equations (v) and (vi), we get, , 6C ( graphite ) + 3H2 (g ) → C6 H6 ( l ) ;, ∆f H  = – 48.51 kJ mol–1... (iv), (b) Enthalpy of Atomization, , (symbol: ∆aH ), Consider the following example of atomization, of dihydrogen, , , –1, , H2(g) → 2H(g); ∆aH = 435.0 kJ mol, You can see that H atoms are formed by, breaking H–H bonds in dihydrogen. The, enthalpy change in this process is known as, enthalpy of atomization, ∆ aH . It is the, enthalpy change on breaking one mole of, bonds completely to obtain atoms in the gas, phase., In case of diatomic molecules, like, dihydrogen (given above), the enthalpy of, atomization is also the bond dissociation, enthalpy. The other examples of enthalpy of, atomization can be, , , CH4(g) → C(g) + 4H(g); ∆aH = 1665 kJ mol, , –1, , Note that the products are only atoms of C, and H in gaseous phase. Now see the following, reaction:, , , Na(s) → Na(g) ; ∆aH = 108.4 kJ mol, , –1, , In this case, the enthalpy of atomization is, same as the enthalpy of sublimation., , (c) Bond Enthalpy (symbol: ∆bondH ), Chemical reactions involve the breaking and, making of chemical bonds. Energy is required, to break a bond and energy is released when, a bond is formed. It is possible to relate heat, of reaction to changes in energy associated, with breaking and making of chemical bonds., With reference to the enthalpy changes, associated with chemical bonds, two different, terms are used in thermodynamics., , (i), , Bond dissociation enthalpy, , (ii), , Mean bond enthalpy, , Let us discuss these terms with reference, to diatomic and polyatomic molecules., Diatomic Molecules: Consider the following, process in which the bonds in one mole of, dihydrogen gas (H2) are broken:, , –1, H2(g) → 2H(g) ; ∆H–HH = 435.0 kJ mol, The enthalpy change involved in this process, is the bond dissociation enthalpy of H–H bond., The bond dissociation enthalpy is the change, in enthalpy when one mole of covalent bonds, of a gaseous covalent compound is broken to, form products in the gas phase., Note that it is the same as the enthalpy of, atomization of dihydrogen. This is true for all, diatomic molecules. For example:, , , –1, , Cl2(g) → 2Cl(g) ; ∆Cl–ClH = 242 kJ mol, , , –1, , O2(g) → 2O(g) ; ∆O=OH = 428 kJ mol, In the case of polyatomic molecules, bond, dissociation enthalpy is different for different, bonds within the same molecule., Polyatomic Molecules: Let us now consider, a polyatomic molecule like methane, CH4. The, overall thermochemical equation for its, atomization reaction is given below:, , CH4 ( g ) → C(g ) + 4H( g );, ∆a H  = 1665 kJ mol–1, In methane, all the four C – H bonds are, identical in bond length and energy. However,, the energies required to break the individual, C – H bonds in each successive step differ :, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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178, , CHEMISTRY, , CH4(g) → CH3(g)+H(g);∆bond H= +427 kJ mol–1, , , –1, , CH3(g) → CH2(g)+H(g);∆bond H = +439 kJ mol, , CH2(g) → CH(g)+H(g);∆bond H= +452 kJ mol–1, CH(g) → C(g)+H(g);∆bond H= +347 kJ mol–1, Therefore,, CH4(g) → C(g)+4H(g);∆a H= 1665 kJ mol–1, In such cases we use mean bond enthalpy, of C – H bond., , For example in CH4, ∆C–HH is calculated as:, , ∆C–HH = ¼ (∆a H) = ¼ (1665 kJ mol–1), = 416 kJ mol–1, We find that mean C–H bond enthalpy in, methane is 416 kJ/mol. It has been found that, mean C–H bond enthalpies differ slightly from, compound, to, compound,, as, in, CH3CH2Cl,CH3NO2, etc, but it does not differ, in a great deal*. Using Hess’s law, bond, enthalpies can be calculated. Bond enthalpy, values of some single and multiple bonds are, , given in Table 6.3. The reaction enthalpies are, very important quantities as these arise from, the changes that accompany the breaking of, old bonds and formation of the new bonds., We can predict enthalpy of a reaction in gas, phase, if we know different bond enthalpies., The standard enthalpy of reaction, ∆rH is, related to bond enthalpies of the reactants and, products in gas phase reactions as:, , ∑ bond enthalpiesreactants, − ∑ bond enthalpies products, , ∆r H  =, , , (6.17)**, This relationship is particularly more, useful when the required values of ∆f H are, not available. The net enthalpy change of a, reaction is the amount of energy required to, break all the bonds in the reactant molecules, minus the amount of energy required to break, all the bonds in the product molecules., Remember that this relationship is, , Table 6.3(a) Some Mean Single Bond Enthalpies in kJ mol, H, , C, , 435.8 414, 347, , N, , O, , F, , Si, , P, , 389, 293, 159, , 464, 351, 201, 138, , 569, 439, 272, 184, 155, , 293, 289, 368, 540, 176, , 318, 264, 209, 351, 490, 213, 213, , S, , Cl, , 339, 259, 327, 226, 230, 213, , 431, 330, 201, 205, 255, 360, 331, 251, 243, , –1, , Br, 368, 276, 243, 197, 289, 272, 213, 218, 192, , Table 6.3(b) Some Mean Multiple Bond Enthalpies in kJ mol, N=N, , 418, , C=C, , 611, , N≡N, , 946, , C≡ C, , 837, , C=N, , 615, , C=O, , 741, , C≡N, , 891, , C≡O, , 1070, , at 298 K, , –1, , I, 297, 238, 201, 213, 213, 209, 180, 151, , H, C, N, O, F, Si, P, S, Cl, Br, I, , at 298 K, , O=O, , 498, , * Note that symbol used for bond dissociation enthalpy and mean bond enthalpy is the same.,  ), which is the enthalpy change when one mole of a particular type of, ** If we use enthalpy of bond formation, (∆f H bond, bond is formed from gaseous atom, then ∆f H =, , ∑ ∆f H  bonds of products – ∑ ∆f H  bonds of reactants, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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THERMODYNAMICS, , 179, , approximate and is valid when all substances, (reactants and products) in the reaction are in, gaseous state., (d) Lattice Enthalpy, The lattice enthalpy of an ionic compound is, the enthalpy change which occurs when one, mole of an ionic compound dissociates into its, ions in gaseous state., , Na + Cl − ( s) → Na + ( g ) + Cl − ( g ) ;, , ∆latticeH = +788 kJ mol–1, Since it is impossible to determine lattice, enthalpies directly by experiment, we use an, indirect method where we construct an, enthalpy diagram called a Born-Haber Cycle, (Fig. 6.9)., Let us now calculate the lattice enthalpy, of Na+Cl–(s) by following steps given below :, 1. Na(s) → Na( g ) , sublimation of sodium, , metal, ∆subH = 108.4 kJ mol–1, , 2. Na( g ) → Na + ( g ) + e −1 ( g ) , the ionization of, sodium atoms, ionization enthalpy, ∆iH= 496 kJ mol–1, 3., , 1, Cl2 ( g ) → Cl( g ) , the dissociation of, 2, chlorine, the reaction enthalpy is half the, bond dissociation enthalpy., 1, , –1, ∆bondH = 121 kJ mol, 2, , 4. Cl( g ) + e −1 ( g ) → Cl( g ) electron gained by, chlorine atoms. The electron gain enthalpy,, ∆egH = – 348.6 kJ mol –1., You have learnt about ionization enthalpy, and electron gain enthalpy in Unit 3. In, fact, these terms have been taken from, thermodynamics. Earlier terms, ionization, energy and electron affinity were in, practice in place of the above terms (see, the box for justification)., Ionization Energy and Electron Affinity, Ionization energy and electron affinity are, defined at absolute zero. At any other, temperature, heat capacities for the, reactants and the products have to be, taken into account. Enthalpies of reactions, for, +, , M (g), , M(g) →, M(g) + e, , –, , →, , +e, , –, , (for ionization), , –, , M (g) (for electron gain), , at temperature, T is, T, , , , , ∆rH (T ) = ∆rH (0) +, , ∫∆C, r, , , P, , dT, , 0, , The value of C p for each species in the, above reaction is 5/2 R (CV = 3/2R), So, ∆rCp = + 5/2 R (for ionization), , ∆rCp = – 5/2 R (for electron gain), Therefore,, , ∆rH (ionization enthalpy), ∆rH, , Fig. 6.9 Enthalpy diagram for lattice enthalpy, of NaCl, ,  , , = E0 (ionization energy) + 5/2 RT, (electron gain enthalpy), = – A( electron affinity) – 5/2 RT, , +, −, +, −, 5. Na ( g ) + Cl ( g ) → Na Cl ( s ), The sequence of steps is shown in Fig. 6.9,, and is known as a Born-Haber cycle. The, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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180, , CHEMISTRY, , importance of the cycle is that, the sum of, the enthalpy changes round a cycle is, zero. A p p l y i n g H e s s ’ s l a w , w e g e t ,, ∆latticeH= 411.2 + 108.4 + 121 + 496 – 348.6, , , ∆latticeH = + 788kJ, for NaCl(s) Na+(g) + Cl–(g), Internal energy is smaller by 2RT ( because, ∆ng = 2) and is equal to + 783 kJ mol –1., Now we use the value of lattice enthalpy to, calculate enthalpy of solution from the, expression:, , , , , , , ∆solH = ∆latticeH + ∆hydH, For one mole of NaCl(s),, lattice enthalpy = + 788 kJ mol–1, , and ∆hydH = – 784 kJ mol –1( from the, literature), , ∆sol H = + 788 kJ mol –1 – 784 kJ mol –1, = + 4 kJ mol–1, The dissolution of NaCl(s) is accompanied, by very little heat change., (e) Enthalpy of Solution (symbol : ∆solH), Enthalpy of solution of a substance is the, enthalpy change when one mole of it dissolves, in a specified amount of solvent. The enthalpy, of solution at infinite dilution is the enthalpy, change observed on dissolving the substance, in an infinite amount of solvent when the, interactions between the ions (or solute, molecules) are negligible., When an ionic compound dissolves in a, solvent, the ions leave their ordered positions on, the crystal lattice. These are now more free in, solution. But solvation of these ions (hydration, in case solvent is water) also occurs at the same, time. This is shown diagrammatically, for an, ionic compound, AB (s), , , , The enthalpy of solution of AB(s), ∆solH , in, water is, therefore, determined by the selective, , values of the lattice enthalpy,∆latticeH and, , , enthalpy of hydration of ions, ∆hydH as, , , , , , , ∆sol H = ∆latticeH + ∆hydH, , For most of the ionic compounds, ∆sol H is, positive and the dissociation process is, endothermic. Therefore the solubility of most, salts in water increases with rise of, temperature. If the lattice enthalpy is very, high, the dissolution of the compound may not, take place at all. Why do many fluorides tend, to be less soluble than the corresponding, chlorides? Estimates of the magnitudes of, enthalpy changes may be made by using tables, of bond energies (enthalpies) and lattice, energies (enthalpies)., (f) Enthalpy of Dilution, It is known that enthalpy of solution is the, enthalpy change associated with the addition, of a specified amount of solute to the specified, amount of solvent at a constant temperature, and pressure. This argument can be applied, to any solvent with slight modification., Enthalpy change for dissolving one mole of, gaseous hydrogen chloride in 10 mol of water, can be represented by the following equation., For convenience we will use the symbol aq. for, water, HCl(g) + 10 aq. → HCl.10 aq., ∆H = –69.01 kJ / mol, Let us consider the following set of enthalpy, changes:, (S-1) HCl(g) + 25 aq. → HCl.25 aq., ∆H = –72.03 kJ / mol, (S-2) HCl(g) + 40 aq. → HCl.40 aq., ∆H = –72.79 kJ / mol, (S-3) HCl(g) + ∞ aq. → HCl. ∞ aq., ∆H = –74.85 kJ / mol, The values of ∆H show general dependence, of the enthalpy of solution on amount of solvent., As more and more solvent is used, the enthalpy, of solution approaches a limiting value, i.e, the, value in infinitely dilute solution. For, hydrochloric acid this value of ∆H is given, above in equation (S-3)., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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THERMODYNAMICS, , 181, , If we subtract the first equation (equation, S-1) from the second equation (equation S-2), in the above set of equations, we obtain–, HCl.25 aq. + 15 aq. → HCl.40 aq., ∆H = [ –72.79 – (–72.03)] kJ / mol, = – 0.76 kJ / mol, This value (–0.76kJ/mol) of ∆H is enthalpy, of dilution. It is the heat withdrawn from the, surroundings when additional solvent is, added to the solution. The enthalpy of dilution, of a solution is dependent on the original, concentration of the solution and the amount, of solvent added., 6.6 SPONTANEITY, The first law of thermodynamics tells us about, the relationship between the heat absorbed, and the work performed on or by a system. It, puts no restrictions on the direction of heat, flow. However, the flow of heat is unidirectional, from higher temperature to lower, temperature. In fact, all naturally occurring, processes whether chemical or physical will, tend to proceed spontaneously in one, direction only. For example, a gas expanding, to fill the available volume, burning carbon, in dioxygen giving carbon dioxide., But heat will not flow from colder body to, warmer body on its own, the gas in a container, will not spontaneously contract into one, corner or carbon dioxide will not form carbon, and dioxygen spontaneously. These and many, other spontaneously occurring changes show, unidirectional change. We may ask ‘what is the, driving force of spontaneously occurring, changes ? What determines the direction of a, spontaneous change ? In this section, we shall, establish some criterion for these processes, whether these will take place or not., Let us first understand what do we mean, by spontaneous reaction or change ? You may, think by your common observation that, spontaneous reaction is one which occurs, immediately when contact is made between, the reactants. Take the case of combination of, hydrogen and oxygen. These gases may be, mixed at room temperature and left for many, years without observing any perceptible, change. Although the reaction is taking place, , between them, it is at an extremely slow rate., It is still called spontaneous reaction. So, spontaneity means ‘having the potential to, proceed without the assistance of external, agency’. However, it does not tell about the, rate of the reaction or process. Another aspect, of spontaneous reaction or process, as we see, is that these cannot reverse their direction on, their own. We may summarise it as follows:, A spontaneous process is an, irreversible process and may only be, reversed by some external agency., (a) Is Decrease in Enthalpy a Criterion, for Spontaneity ?, If we examine the phenomenon like flow of, water down hill or fall of a stone on to the, ground, we find that there is a net decrease in, potential energy in the direction of change. By, analogy, we may be tempted to state that a, chemical reaction is spontaneous in a given, direction, because decrease in energy has, taken place, as in the case of exothermic, reactions. For example:, , 3, 1, N2(g) + H2(g) = NH3(g) ;, 2, 2, , ∆ r H = – 46.1 kJ mol–1, 1, 1, H2(g) + Cl2(g) = HCl (g) ;, 2, 2, ∆r H = – 92.32 kJ mol–1, 1, O (g) → H2O(l) ;, 2 2, , ∆r H = –285.8 kJ mol–1, The decrease in enthalpy in passing from, reactants to products may be shown for any, exothermic reaction on an enthalpy diagram, as shown in Fig. 6.10(a)., Thus, the postulate that driving force for a, chemical reaction may be due to decrease in, energy sounds ‘reasonable’ as the basis of, evidence so far !, Now let us examine the following reactions:, 1, N (g) + O2(g) → NO2(g);, 2 2, , ∆r H = +33.2 kJ mol–1, , H2(g) +, , C(graphite, s) + 2 S(l) → CS2(l);, , ∆r H = +128.5 kJ mol–1, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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182, , Fig. 6.10 (a), , CHEMISTRY, , Enthalpy diagram for exothermic, reactions, , These reactions though endothermic, are, spontaneous. The increase in enthalpy may, be represented on an enthalpy diagram as, shown in Fig. 6.10(b)., , Fig. 6.11 Diffusion of two gases, , Fig. 6.10 (b), , Enthalpy diagram for endothermic, reactions, , Therefore, it becomes obvious that while, decrease in enthalpy may be a contributory, factor for spontaneity, but it is not true for all, cases., (b) Entropy and Spontaneity, Then, what drives the spontaneous process in, a given direction ? Let us examine such a case, in which ∆H = 0 i.e., there is no change in, enthalpy, but still the process is spontaneous., Let us consider diffusion of two gases into, each other in a closed container which is, isolated from the surroundings as shown in, Fig. 6.11., The two gases, say, gas A and gas B are, represented by black dots and white dots, , respectively and separated by a movable, partition [Fig. 6.11 (a)]. When the partition is, withdrawn [Fig.6.11( b)], the gases begin to, diffuse into each other and after a period of, time, diffusion will be complete., Let us examine the process. Before, partition, if we were to pick up the gas, molecules from left container, we would be, sure that these will be molecules of gas A and, similarly if we were to pick up the gas, molecules from right container, we would be, sure that these will be molecules of gas B. But,, if we were to pick up molecules from container, when partition is removed, we are not sure, whether the molecules picked are of gas A or, gas B. We say that the system has become less, predictable or more chaotic., We may now formulate another postulate:, in an isolated system, there is always a, tendency for the systems’ energy to become, more disordered or chaotic and this could be, a criterion for spontaneous change !, At this point, we introduce another, thermodynamic function, entropy denoted as, S. The above mentioned disorder is the, manifestation of entropy. To form a mental, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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THERMODYNAMICS, , 183, , picture, one can think of entropy as a measure, of the degree of randomness or disorder in the, system. The greater the disorder in an isolated, system, the higher is the entropy. As far as a, chemical reaction is concerned, this entropy, change can be attributed to rearrangement of, atoms or ions from one pattern in the reactants, to another (in the products). If the structure, of the products is very much disordered than, that of the reactants, there will be a resultant, increase in entropy. The change in entropy, accompanying a chemical reaction may be, estimated qualitatively by a consideration of, the structures of the species taking part in the, reaction. Decrease of regularity in structure, would mean increase in entropy. For a given, substance, the crystalline solid state is the, state of lowest entropy (most ordered), The, gaseous state is state of highest entropy., Now let us try to quantify entropy. One way, to calculate the degree of disorder or chaotic, distribution of energy among molecules would, be through statistical method which is beyond, the scope of this treatment. Other way would, be to relate this process to the heat involved in, a process which would make entropy a, thermodynamic concept. Entropy, like any, other thermodynamic property such as, internal energy U and enthalpy H is a state, function and ∆S is independent of path., Whenever heat is added to the system, it, increases molecular motions causing, increased randomness in the system. Thus, heat (q) has randomising influence on the, system. Can we then equate ∆S with q ? Wait !, Experience suggests us that the distribution, of heat also depends on the temperature at, which heat is added to the system. A system, at higher temperature has greater randomness, in it than one at lower temperature. Thus,, temperature is the measure of average, chaotic motion of particles in the system., Heat added to a system at lower temperature, causes greater randomness than when the, same quantity of heat is added to it at higher, temperature. This suggests that the entropy, change is inversely proportional to the, temperature. ∆S is related with q and T for a, reversible reaction as :, , ∆S =, , qrev, T, , (6.18), , The total entropy change ( ∆Stotal) for the, system and surroundings of a spontaneous, process is given by, , ∆Stotal = ∆Ssystem + ∆Ssurr > 0, , (6.19), , When a system is in equilibrium, the, entropy is maximum, and the change in, entropy, ∆S = 0., We can say that entropy for a spontaneous, process increases till it reaches maximum and, at equilibrium the change in entropy is zero., Since entropy is a state property, we can, calculate the change in entropy of a reversible, process by, ∆Ssys =, , qsys ,rev, T, , We find that both for reversible and, irreversible expansion for an ideal gas, under, isothermal conditions, ∆U = 0, but ∆Stotal i.e.,, , ∆Ssys + ∆Ssurr is not zero for irreversible, process. Thus, ∆U does not discriminate, between reversible and irreversible process,, whereas ∆S does., Problem 6.10, Predict in which of the following, entropy, increases/decreases :, (i) A liquid crystallizes into a solid., (ii) Temperature of a crystalline solid is, raised from 0 K to 115 K., , ( iii ), , 2NaHCO3 ( s) → Na 2 CO3 ( s) +, CO2 (g ) + H2 O (g ), , (iv), , H2 ( g ) → 2 H ( g ), , Solution, (i) After freezing, the molecules attain an, ordered state and therefore, entropy, decreases., (ii) At 0 K, the contituent particles are, static and entropy is minimum. If, temperature is raised to 115 K, these, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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184, , CHEMISTRY, , begin to move and oscillate about their, equilibrium positions in the lattice and, system becomes more disordered,, therefore entropy increases., (iii) Reactant, NaHCO3 is a solid and it, has low entropy. Among products, there are one solid and two gases., Therefore, the products represent a, condition of higher entropy., (iv) Here one molecule gives two atoms, i.e., number of particles increases, leading to more disordered state., Two moles of H atoms have higher, entropy than one mole of dihydrogen, molecule., Problem 6.11, For oxidation of iron,, , 4Fe ( s) + 3O 2 ( g ) → 2Fe2 O 3 ( s), entropy change is – 549.4 JK–1mol–1at, 298 K. Inspite of negative entropy change, of this reaction, why is the reaction, spontaneous?, , , this, reaction, is, (∆ r H for, 3, –1, –1648 × 10 J mol ), Solution, One decides the spontaneity of a reaction, by considering, , (, , = 4980.6 JK–1 mol–1, This shows that the above reaction is, spontaneous., (c) Gibbs Energy and Spontaneity, We have seen that for a system, it is the total, entropy change, ∆S total which decides the, spontaneity of the process. But most of the, chemical reactions fall into the category of, either closed systems or open systems., Therefore, for most of the chemical reactions, there are changes in both enthalpy and, entropy. It is clear from the discussion in, previous sections that neither decrease in, enthalpy nor increase in entropy alone can, determine the direction of spontaneous change, for these systems., For this purpose, we define a new, thermodynamic function the Gibbs energy or, Gibbs function, G, as, G = H – TS, (6.20), Gibbs function, G is an extensive property, and a state function., The change in Gibbs energy for the system,, ∆Gsys can be written as, ∆Gsys = ∆H sys − T ∆Ssys − Ssys ∆T, , At constant temperature, ∆T = 0, , ), , ∴ ∆G sys = ∆H sys − T ∆Ssys, , ∆Stotal ∆Ssys + ∆Ssurr . For calculating, ∆S surr, we have to consider the heat, absorbed by the surroundings which is, , equal to – ∆rH . At temperature T, entropy, change of the surroundings is, , =−, , ( −1648 ×10, , 3, , J mol −1, , ), , 298 K, , = 5530 JK–1mol–1, Thus, total entropy change for this, reaction, ∆r Stotal = 5530 JK –1mol –1 +, , ( −549.4 JK, , –1, , mol –1, , ), , Usually the subscript ‘system’ is dropped, and we simply write this equation as, ∆G = ∆H − T ∆S, , (6.21), , Thus, Gibbs energy change = enthalpy, change – temperature × entropy change, and, is referred to as the Gibbs equation, one of the, most important equations in chemistry. Here,, we have considered both terms together for, spontaneity: energy (in terms of ∆H) and, entropy (∆S, a measure of disorder) as, indicated earlier. Dimensionally if we analyse,, we find that ∆G has units of energy because,, both ∆H and the T∆S are energy terms, since, T∆S = (K) (J/K) = J., Now let us consider how ∆G is related to, reaction spontaneity., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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THERMODYNAMICS, , 185, , We know,, ∆Stotal = ∆Ssys + ∆Ssurr, If the system is in thermal equilibrium with, the surrounding, then the temperature of the, surrounding is same as that of the system., Also, increase in enthalpy of the surrounding, is equal to decrease in the enthalpy of the, system., Therefore,, entropy, change, of, surroundings,, , ∆Ssurr =, , ∆H sys, ∆H surr, =−, T, T, ,  ∆H sys , ∆Stotal = ∆Ssys +  −, T , , Rearranging the above equation:, T∆Stotal = T∆Ssys – ∆Hsys, For spontaneous process, ∆Stotal > 0 , so, T∆Ssys – ∆Hsys > Ο, , (, , ), , ⇒ − ∆H sys − T ∆Ssys > 0, Using equation 6.21, the above equation can, be written as, –∆G > O, , ∆G = ∆H − T ∆S < 0, , (6.22), , ∆Hsys is the enthalpy change of a reaction,, T∆Ssys is the energy which is not available to, do useful work. So ∆G is the net energy, available to do useful work and is thus a, measure of the ‘free energy’. For this reason, it, is also known as the free energy of the reaction., ∆G gives a criteria for spontaneity at, constant pressure and temperature., (i) If ∆G is negative (< 0), the process is, spontaneous., (ii) If ∆G is positive (> 0), the process is non, spontaneous., Note : If a reaction has a positive enthalpy, change and positive entropy change, it can be, spontaneous when T∆S is large enough to, outweigh ∆H. This can happen in two ways;, (a) The positive entropy change of the system, , can be ‘small’ in which case T must be large., (b) The positive entropy change of the system, can be ’large’, in which case T may be small., The former is one of the reasons why reactions, are often carried out at high temperature., Table 6.4 summarises the effect of temperature, on spontaneity of reactions., (d) Entropy and Second Law of, Thermodynamics, We know that for an isolated system the, change in energy remains constant. Therefore,, increase in entropy in such systems is the, natural direction of a spontaneous change., This, in fact is the second law of, thermodynamics. Like first law of, thermodynamics, second law can also be, stated in several ways. The second law of, thermodynamics explains why spontaneous, exothermic reactions are so common. In, exothermic reactions heat released by the, reaction increases the disorder of the, surroundings and overall entropy change is, positive which makes the reaction, spontaneous., (e) Absolute Entropy and Third Law of, Thermodynamics, Molecules of a substance may move in a, straight line in any direction, they may spin, like a top and the bonds in the molecules may, stretch and compress. These motions of the, molecule are called translational, rotational, and vibrational motion respectively. When, temperature of the system rises, these motions, become more vigorous and entropy increases., On the other hand when temperature is, lowered, the entropy decreases. The entropy, of any pure crystalline substance, approaches zero as the temperature, approaches absolute zero. This is called, third law of thermodynamics. This is so, because there is perfect order in a crystal at, absolute zero. The statement is confined to, pure crystalline solids because theoretical, arguments and practical evidences have, shown that entropy of solutions and super, cooled liquids is not zero at 0 K. The, importance of the third law lies in the fact that, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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186, , CHEMISTRY, , it permits the calculation of absolute values, of entropy of pure substance from thermal, data alone. For a pure substance, this can be, q, done by summing rev increments from 0 K, T, to 298 K. Standard entropies can be used to, calculate standard entropy changes by a, Hess’s law type of calculation., 6.7 GIBBS ENERGY, EQUILIBRIUM, , CHANGE, , AND, , We have seen how a knowledge of the sign and, magnitude of the free energy change of a, chemical reaction allows:, (i) Prediction of the spontaneity of the, chemical reaction., (ii) Prediction of the useful work that could, be extracted from it., So far we have considered free energy, changes in irreversible reactions. Let us now, examine the free energy changes in reversible, reactions., ‘Reversible’ under strict thermodynamic, sense is a special way of carrying out a, process such that system is at all times in, perfect equilibrium with its surroundings., When applied to a chemical reaction, the, term ‘reversible’ indicates that a given, reaction can proceed in either direction, simultaneously, so that a dynamic, equilibrium is set up. This means that the, reactions in both the directions should, , proceed with a decrease in free energy, which, seems impossible. It is possible only if at, equilibrium the free energy of the system is, minimum. If it is not, the system would, spontaneously change to configuration of, lower free energy., So, the criterion for equilibrium, A + B  C + D ; is, ∆rG = 0, Gibbs energy for a reaction in which all, reactants and products are in standard state,, , ∆rG is related to the equilibrium constant of, the reaction as follows:, , 0 = ∆rG + RT ln K, , or ∆rG = – RT ln K, , (6.23), or ∆rG = – 2.303 RT log K, We also know that, (6.24), For strongly endothermic reactions, the, , value of ∆rH may be large and positive. In, such a case, value of K will be much smaller, than 1 and the reaction is unlikely to form, much product. In case of exothermic, reactions, ∆rH is large and negative, and ∆rG, is likely to be large and negative too. In such, cases, K will be much larger than 1. We may, expect strongly exothermic reactions to have, a large K, and hence can go to near, , , completion. ∆rG also depends upon ∆rS , if, the changes in the entropy of reaction is also, taken into account, the value of K or extent, of chemical reaction will also be affected,, , Table 6.4 Effect of Temperature on Spontaneity of Reactions, , ∆rH, , *, , , , ∆rS, , , , ∆rG, , , , Description*, , –, , +, , –, , Reaction spontaneous at all temperatures, , –, , –, , – (at low T ), , Reaction spontaneous at low temperature, , –, , –, , + (at high T ), , Reaction nonspontaneous at high temperature, , +, , +, , + (at low T ), , Reaction nonspontaneous at low temperature, , +, , +, , – (at high T ), , Reaction spontaneous at high temperature, , +, , –, , + (at all T ), , Reaction nonspontaneous at all temperatures, , The term low temperature and high temperature are relative. For a particular reaction, high temperature could even, mean room temperature., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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THERMODYNAMICS, , 187, , , , depending upon whether ∆rS is positive or, negative., Using equation (6.24),, , (i) It is possible to obtain an estimate of ∆G, , , from the measurement of ∆H and ∆S ,, and then calculate K at any temperature, for economic yields of the products., (ii) If K is measured directly in the, , laboratory, value of ∆G at any other, temperature can be calculated., Using equation (6.24),, , ( –13.6 × 10, =, 2.303 (8.314 JK, , 3, , –1, , mol, , –1, , ), , ) (298 K ), , = 2.38, Hence K = antilog 2.38 = 2.4 × 102., Problem 6.14, At 60°C, dinitrogen tetroxide is 50, per cent dissociated. Calculate the, standard free energy change at this, temperature and at one atmosphere., Solution, N2O4(g), , Problem 6.12, Calculate ∆rG  for conversion of oxygen, to ozone, 3/2 O2(g) → O3(g) at 298 K. if Kp, for this conversion is 2.47 × 10 –29., Solution, We know ∆rG  = – 2.303 RT log Kp and, R = 8.314 JK–1 mol–1, , Therefore, ∆rG =, – 2.303 (8.314 J K–1 mol–1), × (298 K) (log 2.47 × 10 –29), = 163000 J mol–1, = 163 kJ mol–1., Problem 6.13, Find out the value of equilibrium constant, for the following reaction at 298 K., , J mol –1, , 2NO2(g), , If N 2O4 is 50% dissociated, the mole, fraction of both the substances is given, by, xN, , 2 O4, , pN, , =, , 2 O4, , 1 − 0.5, 2 × 0.5, : x NO2 =, 1 + 0.5, 1 + 0.5, , =, , 0.5, × 1 atm, p NO =, 2, 1.5, 1, × 1 atm., 1.5, , The equilibrium constant Kp is given by, 2, , Kp =, , (p ), NO2, , p N 2O 4, , =, , 1.5, (1.5)2 (0.5 ), , = 1.33 atm., Since, , , , Standard Gibbs energy change, ∆rG at, the given temperature is –13.6 kJ mol–1., , , , ∆rG = –RT ln Kp, , , ∆rG = (– 8.314 JK–1 mol–1) × (333 K), , Solution, , × (2.303) × (0.1239), We know, log K =, , –1, , = – 763.8 kJ mol, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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188, , CHEMISTRY, , SUMMARY, Thermodynamics deals with energy changes in chemical or physical processes and enables us, to study these changes quantitatively and to make useful predictions. For these purposes, we, divide the universe into the system and the surroundings. Chemical or physical processes, lead to evolution or absorption of heat (q), part of which may be converted into work (w). These, quantities are related through the first law of thermodynamics via ∆U = q + w. ∆U, change in, internal energy, depends on initial and final states only and is a state function, whereas q and, w depend on the path and are not the state functions. We follow sign conventions of q and w by, giving the positive sign to these quantities when these are added to the system. We can measure, the transfer of heat from one system to another which causes the change in temperature. The, magnitude of rise in temperature depends on the heat capacity (C) of a substance. Therefore,, heat absorbed or evolved is q = C∆T. Work can be measured by w = –pex∆V, in case of expansion, of gases. Under reversible process, we can put pex = p for infinitesimal changes in the volume, making wrev = – p dV. In this condition, we can use gas equation, pV = nRT., At constant volume, w = 0, then ∆U = qV , heat transfer at constant volume. But in study of, chemical reactions, we usually have constant pressure. We define another state function, enthalpy. Enthalpy change, ∆H = ∆U + ∆ngRT, can be found directly from the heat changes at, constant pressure, ∆H = qp., There are varieties of enthalpy changes. Changes of phase such as melting, vaporization, and sublimation usually occur at constant temperature and can be characterized by enthalpy, changes which are always positive. Enthalpy of formation, combustion and other enthalpy, changes can be calculated using Hess’s law. Enthalpy change for chemical reactions can be, determined by, ∆r H =, , ∑ (a ∆, i, , f, , f, , ), , (, , H products − ∑ bi ∆ f H reactions, i, , ), , and in gaseous state by, , , ∆rH =, , Σ bond enthalpies of the reactants – Σ bond enthalpies of the products, , First law of thermodynamics does not guide us about the direction of chemical reactions, i.e., what is the driving force of a chemical reaction. For isolated systems,, ∆U = 0. We define another state function, S, entropy for this purpose. Entropy is a measure of, disorder or randomness. For a spontaneous change, total entropy change is positive. Therefore,, for an isolated system, ∆U = 0, ∆S > 0, so entropy change distinguishes a spontaneous change,, while energy change does not. Entropy changes can be measured by the equation, q rev, q rev, for a reversible process., is independent of path., T, T, Chemical reactions are generally carried at constant pressure, so we define another state, function Gibbs energy, G, which is related to entropy and enthalpy changes of the system by, the equation:, , ∆S =, , ∆rG = ∆rH – T ∆rS, For a spontaneous change, ∆Gsys < 0 and at equilibrium, ∆Gsys = 0., Standard Gibbs energy change is related to equilibrium constant by, , , ∆rG = – RT ln K., K can be calculated from this equation, if we know ∆ rG, , , , which can be found from, , . Temperature is an important factor in the equation. Many reactions which, are non-spontaneous at low temperature, are made spontaneous at high temperature for systems, having positive entropy of reaction., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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THERMODYNAMICS, , 189, , EXERCISES, 6.1, , Choose the correct answer. A thermodynamic state function is a quantity, (i), used to determine heat changes, (ii), , whose value is independent of path, , (iii) used to determine pressure volume work, (iv), 6.2, , whose value depends on temperature only., , For the process to occur under adiabatic conditions, the correct condition is:, (i), ∆T = 0, (ii), , ∆p = 0, , (iii) q = 0, (iv), 6.3, , w=0, , The enthalpies of all elements in their standard states are:, (i), unity, (ii), , zero, , (iii) < 0, (iv), 6.4, , different for each element, , , , , ∆U of combustion of methane is – X kJ mol–1. The value of ∆H is, (i), = ∆U , (ii), , > ∆U , , (iii) < ∆U, (iv), 6.5, , , , =0, , The enthalpy of combustion of methane, graphite and dihydrogen at 298 K, are, –890.3 kJ mol–1 –393.5 kJ mol–1, and –285.8 kJ mol–1 respectively. Enthalpy, of formation of CH4(g) will be, (ii), –52.27 kJ mol–1, (i), –74.8 kJ mol–1, (iii) +74.8 kJ mol–1, , 6.6, , (iv), , +52.26 kJ mol–1., , A reaction, A + B → C + D + q is found to have a positive entropy change. The, reaction will be, (i), possible at high temperature, (ii), , possible only at low temperature, , (iii) not possible at any temperature, (v), , possible at any temperature, , 6.7, , In a process, 701 J of heat is absorbed by a system and 394 J of work is done, by the system. What is the change in internal energy for the process?, , 6.8, , The reaction of cyanamide, NH2CN (s), with dioxygen was carried out in a bomb, calorimeter, and ∆U was found to be –742.7 kJ mol–1 at 298 K. Calculate enthalpy, change for the reaction at 298 K., NH2CN(g) +, , 3, O (g) → N2(g) + CO2(g) + H2O(l), 2 2, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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THERMODYNAMICS, , 191, , , , 6.20, , The equilibrium constant for a reaction is 10. What will be the value of ∆G  ?, R = 8.314 JK–1 mol–1, T = 300 K., , 6.21, , Comment on the thermodynamic stability of NO(g), given, , 1, 1, , N (g) +, O (g) → NO(g) ; ∆rH = 90 kJ mol–1, 2 2, 2 2, NO(g) +, 6.22, , 1, , O (g) → NO2(g) : ∆rH = –74 kJ mol–1, 2 2, , Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed, , under standard conditions. ∆f H = –286 kJ mol–1., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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192, , CHEMISTRY, , UNIT 7, , EQUILIBRIUM, , After studying this unit you will be, able to, • identify dynamic nature of, equilibrium involved in physical, and chemical processes;, • state the law of equilibrium;, • explain characteristics of, equilibria involved in physical, and chemical processes;, • write, expressions, for, equilibrium constants;, • establish a relationship between, Kp and K c ;, • explain various factors that, affect the equilibrium state of a, reaction;, • classify substances as acids or, bases according to Arrhenius,, Bronsted-Lowry and Lewis, concepts;, • classify acids and bases as weak, or strong in terms of their, ionization constants;, • explain the dependence of, degree of ionization on, concentration of the electrolyte, and that of the common ion;, • describe, pH, scale, for, representing hydrogen ion, concentration;, • explain ionisation of water and, its duel role as acid and base;, • describe ionic product (Kw ) and, pKw for water;, • appreciate use of buffer, solutions;, • calculate solubility product, constant., , Chemical equilibria are important in numerous biological, and environmental processes. For example, equilibria, involving O2 molecules and the protein hemoglobin play a, crucial role in the transport and delivery of O2 from our, lungs to our muscles. Similar equilibria involving CO, molecules and hemoglobin account for the toxicity of CO., When a liquid evaporates in a closed container,, molecules with relatively higher kinetic energy escape the, liquid surface into the vapour phase and number of liquid, molecules from the vapour phase strike the liquid surface, and are retained in the liquid phase. It gives rise to a constant, vapour pressure because of an equilibrium in which the, number of molecules leaving the liquid equals the number, returning to liquid from the vapour. We say that the system, has reached equilibrium state at this stage. However, this, is not static equilibrium and there is a lot of activity at the, boundary between the liquid and the vapour. Thus, at, equilibrium, the rate of evaporation is equal to the rate of, condensation. It may be represented by, H2O (l)  H2O (vap), The double half arrows indicate that the processes in, both the directions are going on simultaneously. The mixture, of reactants and products in the equilibrium state is called, an equilibrium mixture., Equilibrium can be established for both physical, processes and chemical reactions. The reaction may be fast, or slow depending on the experimental conditions and the, nature of the reactants. When the reactants in a closed vessel, at a particular temperature react to give products, the, concentrations of the reactants keep on decreasing, while, those of products keep on increasing for some time after, which there is no change in the concentrations of either of, the reactants or products. This stage of the system is the, dynamic equilibrium and the rates of the forward and, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 193, , reverse reactions become equal. It is due to, this dynamic equilibrium stage that there is, no change in the concentrations of various, species in the reaction mixture. Based on the, extent to which the reactions proceed to reach, the state of chemical equilibrium, these may, be classified in three groups., (i) The reactions that proceed nearly to, completion and only negligible, concentrations of the reactants are left. In, some cases, it may not be even possible to, detect these experimentally., (ii) The reactions in which only small amounts, of products are formed and most of the, reactants remain unchanged at, equilibrium stage., (iii) The reactions in which the concentrations, of the reactants and products are, comparable, when the system is in, equilibrium., The extent of a reaction in equilibrium, varies with the experimental conditions such, as concentrations of reactants, temperature,, etc. Optimisation of the operational conditions, is very important in industry and laboratory, so that equilibrium is favorable in the, direction of the desired product. Some, important aspects of equilibrium involving, physical and chemical processes are dealt in, this unit along with the equilibrium involving, ions in aqueous solutions which is called as, ionic equilibrium., 7.1 EQUILIBRIUM, IN, PHYSICAL, PROCESSES, The characteristics of system at equilibrium, are better understood if we examine some, physical processes. The most familiar, examples are phase transformation, processes, e.g.,, solid, liquid, solid, , liquid, gas, gas, , 7.1.1 Solid-Liquid Equilibrium, Ice and water kept in a perfectly insulated, thermos flask (no exchange of heat between, its contents and the surroundings) at 273K, , and the atmospheric pressure are in, equilibrium state and the system shows, interesting characteristic features. We observe, that the mass of ice and water do not change, with time and the temperature remains, constant. However, the equilibrium is not, static. The intense activity can be noticed at, the boundary between ice and water., Molecules from the liquid water collide against, ice and adhere to it and some molecules of ice, escape into liquid phase. There is no change, of mass of ice and water, as the rates of transfer, of molecules from ice into water and of reverse, transfer from water into ice are equal at, atmospheric pressure and 273 K., It is obvious that ice and water are in, equilibrium only at particular temperature, and pressure. For any pure substance at, atmospheric pressure, the temperature at, which the solid and liquid phases are at, equilibrium is called the normal melting point, or normal freezing point of the substance., The system here is in dynamic equilibrium and, we can infer the following:, (i) Both the opposing processes occur, simultaneously., (ii) Both the processes occur at the same rate, so that the amount of ice and water, remains constant., 7.1.2 Liquid-Vapour Equilibrium, This equilibrium can be better understood if, we consider the example of a transparent box, carrying a U-tube with mercury (manometer)., Drying agent like anhydrous calcium chloride, (or phosphorus penta-oxide) is placed for a, few hours in the box. After removing the, drying agent by tilting the box on one side, a, watch glass (or petri dish) containing water is, quickly placed inside the box. It will be, observed that the mercury level in the right, limb of the manometer slowly increases and, finally attains a constant value, that is, the, pressure inside the box increases and reaches, a constant value. Also the volume of water in, the watch glass decreases (Fig. 7.1). Initially, there was no water vapour (or very less) inside, the box. As water evaporated the pressure in, the box increased due to addition of water, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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194, , CHEMISTRY, , Fig.7.1 Measuring equilibrium vapour pressure of water at a constant temperature, , molecules into the gaseous phase inside the, box. The rate of evaporation is constant., However, the rate of increase in pressure, decreases with time due to condensation of, vapour into water. Finally it leads to an, equilibrium condition when there is no net, evaporation. This implies that the number of, water molecules from the gaseous state into, the liquid state also increases till the, equilibrium is attained i.e.,, rate of evaporation= rate of condensation, H2O (vap), H2O(l), At equilibrium the pressure exerted by the, water molecules at a given temperature, remains constant and is called the equilibrium, vapour pressure of water (or just vapour, pressure of water); vapour pressure of water, increases with temperature. If the above, experiment is repeated with methyl alcohol,, acetone and ether, it is observed that different, liquids have different equilibrium vapour, pressures at the same temperature, and the, liquid which has a higher vapour pressure is, more volatile and has a lower boiling point., If we expose three watch glasses, containing separately 1mL each of acetone,, ethyl alcohol, and water to atmosphere and, repeat the experiment with different volumes, of the liquids in a warmer room, it is observed, that in all such cases the liquid eventually, disappears and the time taken for complete, evaporation depends on (i) the nature of the, liquid, (ii) the amount of the liquid and (iii) the, temperature. When the watch glass is open to, the atmosphere, the rate of evaporation, remains constant but the molecules are, , dispersed into large volume of the room. As a, consequence the rate of condensation from, vapour to liquid state is much less than the, rate of evaporation. These are open systems, and it is not possible to reach equilibrium in, an open system., Water and water vapour are in equilibrium, position at atmospheric pressure (1.013 bar), and at 100°C in a closed vessel. The boiling, point of water is 100°C at 1.013 bar pressure., For any pure liquid at one atmospheric, pressure (1.013 bar), the temperature at, which the liquid and vapours are at, equilibrium is called normal boiling point of, the liquid. Boiling point of the liquid depends, on the atmospheric pressure. It depends on, the altitude of the place; at high altitude the, boiling point decreases., 7.1.3 Solid – Vapour Equilibrium, Let us now consider the systems where solids, sublime to vapour phase. If we place solid iodine, in a closed vessel, after sometime the vessel gets, filled up with violet vapour and the intensity of, colour increases with time. After certain time the, intensity of colour becomes constant and at this, stage equilibrium is attained. Hence solid iodine, sublimes to give iodine vapour and the iodine, vapour condenses to give solid iodine. The, equilibrium can be represented as,, I2(solid), I2 (vapour), Other examples showing this kind of, equilibrium are,, Camphor (solid)  Camphor (vapour), NH4Cl (solid)  NH4Cl (vapour), , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 195, , 7.1.4 Equilibrium Involving Dissolution of, Solid or Gases in Liquids, Solids in liquids, We know from our experience that we can, dissolve only a limited amount of salt or sugar, in a given amount of water at room, temperature. If we make a thick sugar syrup, solution by dissolving sugar at a higher, temperature, sugar crystals separate out if we, cool the syrup to the room temperature. We, call it a saturated solution when no more of, solute can be dissolved in it at a given, temperature. The concentration of the solute, in a saturated solution depends upon the, temperature. In a saturated solution, a, dynamic equilibrium exits between the solute, molecules in the solid state and in the solution:, Sugar (solution), , Sugar (solid), and, , the rate of dissolution of sugar = rate of, crystallisation of sugar., Equality of the two rates and dynamic, nature of equilibrium has been confirmed with, the help of radioactive sugar. If we drop some, radioactive sugar into saturated solution of, non-radioactive sugar, then after some time, radioactivity is observed both in the solution, and in the solid sugar. Initially there were no, radioactive sugar molecules in the solution, but due to dynamic nature of equilibrium,, there is exchange between the radioactive and, non-radioactive sugar molecules between the, two phases. The ratio of the radioactive to nonradioactive molecules in the solution increases, till it attains a constant value., Gases in liquids, When a soda water bottle is opened, some of, the carbon dioxide gas dissolved in it fizzes, out rapidly. The phenomenon arises due to, difference in solubility of carbon dioxide at, different pressures. There is equilibrium, between the molecules in the gaseous state, and the molecules dissolved in the liquid, under pressure i.e.,, CO2 (gas), , pressure of the gas above the solvent. This, amount decreases with increase of, temperature. The soda water bottle is sealed, under pressure of gas when its solubility in, water is high. As soon as the bottle is opened,, some of the dissolved carbon dioxide gas, escapes to reach a new equilibrium condition, required for the lower pressure, namely its, partial pressure in the atmosphere. This is how, the soda water in bottle when left open to the, air for some time, turns ‘flat’. It can be, generalised that:, liquid equilibrium, there is, (i) For solid, only one temperature (melting point) at, 1 atm (1.013 bar) at which the two phases, can coexist. If there is no exchange of heat, with the surroundings, the mass of the two, phases remains constant., (ii) For liquid, vapour equilibrium, the, vapour pressure is constant at a given, temperature., (iii) For dissolution of solids in liquids, the, solubility is constant at a given, temperature., (iv) For dissolution of gases in liquids, the, concentration of a gas in liquid is, proportional, to, the, pressure, (concentration) of the gas over the liquid., These observations are summarised in, Table 7.1, Table 7.1, , Some Features, Equilibria, , Process, Liquid, H2O (l), Solid, H2O (s), Solute(s), Sugar(s), , Vapour, H2O (g), Liquid, H2O (l), , Gas(g), , Gas (aq), , CO2(g), , CO2(aq), , Download all NCERT books PDFs from www.ncert.online, , 2019-20, , Physical, , Conclusion, pH2Oconstant at given, temperature, Melting point is fixed at, constant pressure, , Solute Concentration of solute, (solution) in solution is constant, Sugar, at a given temperature, (solution), , CO2 (in solution), , This equilibrium is governed by Henry’s, law, which states that the mass of a gas, dissolved in a given mass of a solvent at, any temperature is proportional to the, , of, , [gas(aq)]/[gas(g)] is, constant at a given, temperature, [CO 2 (aq)]/[CO 2 (g)] is, constant at a given, temperature

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196, , CHEMISTRY, , 7.1.5 General Characteristics of Equilibria, Involving Physical Processes, For the physical processes discussed above,, following characteristics are common to the, system at equilibrium:, (i) Equilibrium is possible only in a closed, system at a given temperature., (ii) Both the opposing processes occur at the, same rate and there is a dynamic but, stable condition., (iii) All measurable properties of the system, remain constant., (iv) When equilibrium is attained for a physical, process, it is characterised by constant, value of one of its parameters at a given, temperature. Table 7.1 lists such, quantities., (v) The magnitude of such quantities at any, stage indicates the extent to which the, physical process has proceeded before, reaching equilibrium., 7.2 EQUILIBRIUM IN CHEMICAL, PROCESSES – DYNAMIC, EQUILIBRIUM, Analogous to the physical systems chemical, reactions also attain a state of equilibrium., These reactions can occur both in forward, and backward directions. When the rates of, the forward and reverse reactions become, equal, the concentrations of the reactants, and the products remain constant. This is, the stage of chemical equilibrium. This, equilibrium is dynamic in nature as it, consists of a forward reaction in which the, reactants give product(s) and reverse, reaction in which product(s) gives the, original reactants., For a better comprehension, let us, consider a general case of a reversible reaction,, A+B, , C+D, , With passage of time, there is, accumulation of the products C and D and, depletion of the reactants A and B (Fig. 7.2)., This leads to a decrease in the rate of forward, reaction and an increase in he rate of the, reverse reaction,, , Fig. 7.2 Attainment of chemical equilibrium., , Eventually, the two reactions occur at the, same rate and the system reaches a state of, equilibrium., Similarly, the reaction can reach the state of, equilibrium even if we start with only C and D;, that is, no A and B being present initially, as the, equilibrium can be reached from either direction., The dynamic nature of chemical, equilibrium can be demonstrated in the, synthesis of ammonia by Haber’s process. In, a series of experiments, Haber started with, known amounts of dinitrogen and dihydrogen, maintained at high temperature and pressure, and at regular intervals determined the, amount of ammonia present. He was, successful in determining also the, concentration of unreacted dihydrogen and, dinitrogen. Fig. 7.4 (page 191) shows that after, a certain time the composition of the mixture, remains the same even though some of the, reactants are still present. This constancy in, composition indicates that the reaction has, reached equilibrium. In order to understand, the dynamic nature of the reaction, synthesis, of ammonia is carried out with exactly the, same starting conditions (of partial pressure, and temperature) but using D2 (deuterium), in place of H2. The reaction mixtures starting, either with H2 or D2 reach equilibrium with, the same composition, except that D2 and ND3, are present instead of H2 and NH3. After, equilibrium is attained, these two mixtures, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 197, , Dynamic Equilibrium – A Student’s Activity, Equilibrium whether in a physical or in a chemical system, is always of dynamic, nature. This can be demonstrated by the use of radioactive isotopes. This is not feasible, in a school laboratory. However this concept can be easily comprehended by performing, the following activity. The activity can be performed in a group of 5 or 6 students., Take two 100mL measuring cylinders (marked as 1 and 2) and two glass tubes, each of 30 cm length. Diameter of the tubes may be same or different in the range of, 3-5mm. Fill nearly half of the measuring cylinder -1 with coloured water (for this, purpose add a crystal of potassium permanganate to water) and keep second cylinder, (number 2) empty., Put one tube in cylinder 1 and second in cylinder 2. Immerse one tube in cylinder, 1, close its upper tip with a finger and transfer the coloured water contained in its, lower portion to cylinder 2. Using second tube, kept in 2 nd cylinder, transfer the coloured, water in a similar manner from cylinder 2 to cylinder 1. In this way keep on transferring, coloured water using the two glass tubes from cylinder 1 to 2 and from 2 to 1 till you, notice that the level of coloured water in both the cylinders becomes constant., If you continue intertransferring coloured solution between the cylinders, there will, not be any further change in the levels of coloured water in two cylinders. If we take, analogy of ‘level’ of coloured water with ‘concentration’ of reactants and products in the, two cylinders, we can say the process of transfer, which continues even after the constancy, of level, is indicative of dynamic nature of the process. If we repeat the experiment taking, two tubes of different diameters we find that at equilibrium the level of coloured water in, two cylinders is different. How far diameters are responsible for change in levels in two, cylinders? Empty cylinder (2) is an indicator of no product in it at the beginning., , Fig.7.3, , Demonstrating dynamic nature of equilibrium. (a) initial stage (b) final stage after the, equilibrium is attained., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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198, , CHEMISTRY, , Fig 7.4 Depiction of equilibrium for the reaction, , N 2 ( g ) + 3H2 ( g )  2NH3 ( g ), (H 2, N2, NH3 and D2 , N2, ND3) are mixed, together and left for a while. Later, when this, mixture is analysed, it is found that the, concentration of ammonia is just the same as, before. However, when this mixture is, analysed by a mass spectrometer, it is found, that ammonia and all deuterium containing, forms of ammonia (NH3, NH2D, NHD2 and ND3), and dihydrogen and its deutrated forms, (H2, HD and D2) are present. Thus one can, conclude that scrambling of H and D atoms, in the molecules must result from a, continuation of the forward and reverse, reactions in the mixture. If the reaction had, simply stopped when they reached, equilibrium, then there would have been no, mixing of isotopes in this way., Use of isotope (deuterium) in the formation, of ammonia clearly indicates that chemical, reactions reach a state of dynamic, equilibrium in which the rates of forward, and reverse reactions are equal and there, is no net change in composition., Equilibrium can be attained from both, sides, whether we start reaction by taking,, H2(g) and N2(g) and get NH3(g) or by taking, NH3(g) and decomposing it into N2(g) and, H2(g)., 2NH3(g), N2(g) + 3H2(g), , 2NH3(g), N2(g) + 3H2(g), Similarly let us consider the reaction,, 2HI(g). If we start with equal, H2(g) + I2(g), initial concentration of H2 and I2, the reaction, proceeds in the forward direction and the, concentration of H2 and I2 decreases while that, of HI increases, until all of these become, constant at equilibrium (Fig. 7.5). We can also, start with HI alone and make the reaction to, proceed in the reverse direction; the, concentration of HI will decrease and, concentration of H2 and I2 will increase until, they all become constant when equilibrium is, reached (Fig.7.5). If total number of H and I, atoms are same in a given volume, the same, equilibrium mixture is obtained whether we, start it from pure reactants or pure product., , Fig.7.5 Chemical equilibrium in the reaction, H2(g) + I2(g)  2HI(g) can be attained, from either direction, , 7.3 LAW OF CHEMICAL EQUILIBRIUM, AND EQUILIBRIUM CONSTANT, A mixture of reactants and products in the, equilibrium state is called an equilibrium, mixture. In this section we shall address a, number of important questions about the, composition of equilibrium mixtures: What is, the relationship between the concentrations of, reactants and products in an equilibrium, mixture? How can we determine equilibrium, concentrations from initial concentrations?, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 199, , What factors can be exploited to alter the, composition of an equilibrium mixture? The, last question in particular is important when, choosing conditions for synthesis of industrial, chemicals such as H2, NH3, CaO etc., To answer these questions, let us consider, a general reversible reaction:, C+D, A+B, where A and B are the reactants, C and D are, the products in the balanced chemical, equation. On the basis of experimental studies, of many reversible reactions, the Norwegian, chemists Cato Maximillian Guldberg and Peter, Waage proposed in 1864 that the, concentrations in an equilibrium mixture are, related by the following equilibrium, equation,, , Kc =, , [C ][D], [ A ][B ], , (7.1) where Kc is the equilibrium constant, and the expression on the right side is called, the equilibrium constant expression., The equilibrium equation is also known as, the law of mass action because in the early, days of chemistry, concentration was called, “active mass”. In order to appreciate their work, better, let us consider reaction between, gaseous H2 and I2 carried out in a sealed vessel, at 731K., H2(g) + I2(g)   2HI(g), 1 mol 1 mol, 2 mol, , Six sets of experiments with varying initial, conditions were performed, starting with only, gaseous H2 and I2 in a sealed reaction vessel, in first four experiments (1, 2, 3 and 4) and, only HI in other two experiments (5 and 6)., Experiment 1, 2, 3 and 4 were performed, taking different concentrations of H2 and / or, I2, and with time it was observed that intensity, of the purple colour remained constant and, equilibrium was attained. Similarly, for, experiments 5 and 6, the equilibrium was, attained from the opposite direction., Data obtained from all six sets of, experiments are given in Table 7.2., It is evident from the experiments 1, 2, 3, and 4 that number of moles of dihydrogen, reacted = number of moles of iodine reacted =, ½ (number of moles of HI formed). Also,, experiments 5 and 6 indicate that,, [H2(g)]eq = [I2(g)]eq, Knowing the above facts, in order to, establish, a, relationship, between, concentrations of the reactants and products,, several combinations can be tried. Let us, consider the simple expression,, [HI(g)]eq / [H2(g)]eq [I2(g)]eq, It can be seen from Table 7.3 that if we, put the equilibrium concentrations of the, reactants and products, the above expression, , Table 7.2 Initial and Equilibrium Concentrations of H2, I2 and HI, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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200, , CHEMISTRY, , Table 7.3, , Expression, Involving, the, Equilibrium Concentration of, Reactants, 2HI(g), H2(g) + I2(g), , The equilibrium constant for a general, reaction,, aA + bB, cC + dD, is expressed as,, c, , d, , a, , b, , (7.4), Kc = [C] [D] / [A] [B], where [A], [B], [C] and [D] are the equilibrium, concentrations of the reactants and products., Equilibrium constant for the reaction,, 4NH3(g) + 5O2(g), as, , 4NO(g) + 6H2O(g) is written, , 4, , 6, , 4, , 5, , Kc = [NO] [H2O] / [NH3] [O2], is far from constant. However, if we consider, the expression,, [HI(g)]2eq / [H2(g)]eq [I2(g)]eq, we find that this expression gives constant, value (as shown in Table 7.3) in all the six, cases. It can be seen that in this expression, the power of the concentration for reactants, and products are actually the stoichiometric, coefficients in the equation for the chemical, reaction. Thus, for the reaction H2(g) + I2(g), 2HI(g), following equation 7.1, the equilibrium, constant Kc is written as,, 2, , Kc = [HI(g)]eq / [H2(g)]eq [I2(g)]eq, , (7.2), , Generally the subscript ‘eq’ (used for, equilibrium) is omitted from the concentration, terms. It is taken for granted that the, concentrations in the expression for Kc are, equilibrium values. We, therefore, write,, Kc = [HI(g)]2 / [H2(g)] [I2(g)], , (7.3), , The subscript ‘c’ indicates that K c is, expressed in concentrations of mol L–1., At a given temperature, the product of, concentrations of the reaction products, raised to the respective stoichiometric, coefficient in the balanced chemical, equation divided by the product of, concentrations of the reactants raised to, their individual stoichiometric coefficients, has a constant value. This is known as, the Equilibrium Law or Law of Chemical, Equilibrium., , Molar concentration of different species is, indicated by enclosing these in square bracket, and, as mentioned above, it is implied that these, are equilibrium concentrations. While writing, expression for equilibrium constant, symbol for, phases (s, l, g) are generally ignored., Let us write equilibrium constant for the, reaction, H2(g) + I2(g), 2HI(g), (7.5), 2, as, Kc = [HI] / [H2] [I2] = x, (7.6), The equilibrium constant for the reverse, reaction, 2HI(g), H2(g) + I2(g), at the same, temperature is,, K′c = [H2] [I2] / [HI]2 = 1/ x = 1 / Kc, Thus, K′c = 1 / Kc, , (7.7), (7.8), , Equilibrium constant for the reverse, reaction is the inverse of the equilibrium, constant for the reaction in the forward, direction., If we change the stoichiometric coefficients, in a chemical equation by multiplying, throughout by a factor then we must make, sure that the expression for equilibrium, constant also reflects that change. For, example, if the reaction (7.5) is written as,, ½ H2(g) + ½ I2(g), HI(g), (7.9), the equilibrium constant for the above reaction, is given by, 1/2, , K″c = [HI] / [H2], , 1/2, , [I2], , 2, , = {[HI] / [H2][I2]}, 1/2, , 1/2, , 1/2, , = x = Kc, (7.10), On multiplying the equation (7.5) by n, we get, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 201, , nH2(g) + nI2(g) D, 2nHI(g), (7.11), Therefore, equilibrium constant for the, n, reaction is equal to Kc . These findings are, summarised in Table 7.4. It should be noted, that because the equilibrium constants Kc and, K ′c have different numerical values, it is, important to specify the form of the balanced, chemical equation when quoting the value of, an equilibrium constant., , 800K. What will be Kc for the reaction, 2NO(g), N2(g) + O2(g), Solution, For the reaction equilibrium constant,, Kc can be written as,, , Kc =, , Table 7.4 Relations between Equilibrium, Constants for a General Reaction, and its Multiples., Chemical equation, c C + dD, , Kc, , cC+dD, , aA+bB, , K′c =(1/Kc ), , na A + nb B, , ncC + ndD, , n, , K′″c = (Kc ), , Problem 7.1, The following concentrations were, obtained for the formation of NH3 from N2, and H 2 at equilibrium at 500K., [N2] = 1.5 × 10–2M. [H2] = 3.0 ×10–2 M and, [NH3] = 1.2 ×10–2M. Calculate equilibrium, constant., Solution, The equilibrium constant for the reaction,, 2NH3(g) can be written, N2(g) + 3H2(g), as,, 2, , Kc =, ,  NH3 ( g ), 3, N 2 (g )  H2 ( g ), −2 2, , =, , (1.2 × 10 ), (1.5 × 10 ) (3.0 × 10 ), −2, , =, , −2 3, , = 0.106 × 104 = 1.06 × 103, Problem 7.2, At equilibrium, the concentrations of, N2=3.0 × 10 –3M, O2 = 4.2 × 10–3M and, NO= 2.8 × 10–3M in a sealed vessel at, , 2, , (2.8 × 10 M ), (3.0 × 10 M) (4.2 × 10, -3, , Equilibrium, constant, , aA+bB, , [NO]2, [ N 2 ][O2 ], −3, , −3, , M, , ), , = 0.622, 7.4 HOMOGENEOUS EQUILIBRIA, In a homogeneous system, all the reactants, and products are in the same phase. For, example, in the gaseous reaction,, 2NH3(g), reactants and, N 2(g) + 3H 2(g), products are in the homogeneous phase., Similarly, for the reactions,, CH3COOH (aq), CH3COOC2H5 (aq) + H2O (l), + C2H5OH (aq), –, , and, Fe3+ (aq) + SCN (aq), , Fe(SCN)2+ (aq), , all the reactants and products are in, homogeneous solution phase. We shall now, consider equilibrium constant for some, homogeneous reactions., 7.4.1 Equilibrium Constant in Gaseous, Systems, So far we have expressed equilibrium constant, of the reactions in terms of molar, concentration of the reactants and products,, and used symbol, Kc for it. For reactions, involving gases, however, it is usually more, convenient to express the equilibrium, constant in terms of partial pressure., The ideal gas equation is written as,, pV = n RT, , n, RT, V, Here, p is the pressure in Pa, n is the number, of moles of the gas, V is the volume in m3 and, T is the temperature in Kelvin, ⇒ p=, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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202, , CHEMISTRY, , Therefore,, n/V is concentration expressed in mol/m3, If concentration c, is in mol/L or mol/dm3,, and p is in bar then, p = cRT,, We can also write p = [gas]RT., Here, R= 0.0831 bar litre/mol K, At constant temperature, the pressure of, the gas is proportional to its concentration i.e.,, p ∝ [gas], For reaction in equilibrium, 2HI(g), H2(g) + I2(g), We can write either, , 2, , −2, , or K p = K c ( RT ), aA + bB, , Kp, , or K c =, , H2, , (7.12), , Further, since p HI =  HI (g ) RT, pI2 =  I2 ( g ) RT, pH2 =  H2 ( g ) RT, Therefore,, , Kp =, , 2, , H2, , 2, ,  HI ( g ) [RT ], =,  H2 ( g ) RT . I2 ( g ) RT, , ( p )( p ), I2, , HI ( g ), = Kc, H2 (g ) I2 ( g ), , c, , d, , c +d ), , a, , b, , a +b ), , A, , B, , [C ]c [D]d (RT )(c +d )−(a +b ), [ A ]a [B ]b, ∆n, , = K c (RT ), , (7.15), , Table 7.5 Equilibrium Constants, Kp for a, Few Selected Reactions, , 2, , =, , d, , D, b, , where ∆n = (number of moles of gaseous, products) – (number of moles of gaseous, reactants) in the balanced chemical equation., It is necessary that while calculating the value, of Kp, pressure should be expressed in bar, because standard state for pressure is 1 bar., We know from Unit 1 that :, –2, 1pascal, Pa=1Nm , and 1bar = 105 Pa, Kp values for a few selected reactions at, different temperatures are given in Table 7.5, , I2, , ( pHI )2, , c, , C, a, , c, d, C ] [ D], [, ∆n, =, RT ), a, b (, [ A ] [B ], , 2, , ( p )( p ), , cC + dD, , ( p )( p ) = [C] [D] (RT )(, =, ( p )( p ) [ A ] [B] (RT )(, =, ,  HI ( g ), H 2 ( g ) I2 ( g ), , ( p HI ), , (7.14), , Similarly, for a general reaction, , 2, , Kc =, , −2, ,  NH3 ( g ) [RT ], −2, = K c ( RT ), = , 3,  N 2 ( g )  H 2 ( g ), , (7.13), , In this example, K p = K c i.e., both, equilibrium constants are equal. However, this, is not always the case. For example in reaction, 2NH3(g), N2(g) + 3H2(g), 2, , Kp, , (p ), =, ( p )( p ), NH 3, , 3, , N2, , H2, , 2, , Problem 7.3, , 2, ,  NH3 ( g ) [RT ], =, 3, 3,  N 2 ( g ) RT .  H 2 ( g ) ( RT ), , PCl5, PCl3 and Cl2 are at equilibrium at, 500 K and having concentration 1.59M, PCl3, 1.59M Cl2 and 1.41 M PCl5., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 203, , Calculate Kc for the reaction,, , the value 0.194 should be neglected, because it will give concentration of the, reactant which is more than initial, concentration., Hence the equilibrium concentrations are,, [CO2] = [H2-] = x = 0.067 M, [CO] = [H2O] = 0.1 – 0.067 = 0.033 M, , PCl3 + Cl2, PCl5, Solution, The equilibrium constant Kc for the above, reaction can be written as,, 2, , Kc, , [PCl ] [Cl ] = (1.59), =, (1.41), [PCl ], 3, , 2, , = 1.79, , 5, , Problem 7.5, For the equilibrium,, , Problem 7.4, The value of Kc = 4.24 at 800K for the reaction,, CO2 (g) + H2 (g), CO (g) + H2O (g), Calculate equilibrium concentrations of, CO2, H2, CO and H2O at 800 K, if only CO, and H 2 O are present initially at, concentrations of 0.10M each., Solution, For the reaction,, CO2 (g) + H2 (g), CO (g) + H2O (g), Initial concentration:, 0.1M, 0.1M, 0, 0, Let x mole per litre of each of the product, be formed., At equilibrium:, (0.1-x) M (0.1-x) M, xM, xM, where x is the amount of CO2 and H2 at, equilibrium., Hence, equilibrium constant can be, written as,, Kc = x2/(0.1-x)2 = 4.24, x2 = 4.24(0.01 + x2-0.2x), x2 = 0.0424 + 4.24x2-0.848x, 3.24x2 – 0.848x + 0.0424 = 0, a = 3.24, b = – 0.848, c = 0.0424, (for quadratic equation ax2 + bx + c = 0,, , (− b ±, x=, , b2 − 4ac, , ), , 2a, x = 0.848±√(0.848)2– 4(3.24)(0.0424)/, (3.24×2), x = (0.848 ± 0.4118)/ 6.48, x1 = (0.848 – 0.4118)/6.48 = 0.067, x2 = (0.848 + 0.4118)/6.48 = 0.194, , 2NO(g) + Cl2(g), 2NOCl(g), the value of the equilibrium constant, Kc, is 3.75 × 10–6 at 1069 K. Calculate the Kp, for the reaction at this temperature?, Solution, We know that,, ∆n, Kp = Kc(RT), For the above reaction,, ∆n = (2+1) – 2 = 1, Kp = 3.75 ×10–6 (0.0831 × 1069), Kp = 0.033, 7.5 HETEROGENEOUS EQUILIBRIA, Equilibrium in a system having more than one, phase is called heterogeneous equilibrium., The equilibrium between water vapour and, liquid water in a closed container is an, example of heterogeneous equilibrium., H2O(g), H2O(l), In this example, there is a gas phase and a, liquid phase. In the same way, equilibrium, between a solid and its saturated solution,, Ca2+ (aq) + 2OH–(aq), Ca(OH)2 (s) + (aq), is a heterogeneous equilibrium., Heterogeneous equilibria often involve pure, solids or liquids. We can simplify equilibrium, expressions for the heterogeneous equilibria, involving a pure liquid or a pure solid, as the, molar concentration of a pure solid or liquid, is constant (i.e., independent of the amount, present). In other words if a substance ‘X’ is, involved, then [X(s)] and [X(l)] are constant,, whatever the amount of ‘X’ is taken. Contrary, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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204, , CHEMISTRY, , to this, [X(g)] and [X(aq)] will vary as the, amount of X in a given volume varies. Let us, take thermal dissociation of calcium carbonate, which is an interesting and important example, of heterogeneous chemical equilibrium., CaCO3 (s), , CaO (s) + CO2 (g), , (7.16), , On the basis of the stoichiometric equation,, we can write,, , Kc =, , Since [CaCO3(s)] and [CaO(s)] are both, constant, therefore modified equilibrium, constant for the thermal decomposition of, calcium carbonate will be, , 2, , Kp = PCO2 = 2 × 105 Pa/105 Pa = 2.00, Similarly, in the equilibrium between, nickel, carbon monoxide and nickel carbonyl, (used in the purification of nickel),, , CaO ( s) CO 2 ( g ), CaCO 3 ( s), , K´c = [CO2(g)], or Kp = pCO, , This shows that at a particular, temperature, there is a constant concentration, or pressure of CO2 in equilibrium with CaO(s), and CaCO3(s). Experimentally it has been, found that at 1100 K, the pressure of CO2 in, equilibrium with CaCO3(s) and CaO(s), is, 2.0 ×105 Pa. Therefore, equilibrium constant, at 1100K for the above reaction is:, , (7.17), (7.18), , Units of Equilibrium Constant, The value of equilibrium constant Kc can, be calculated by substituting the, concentration terms in mol/L and for K p, partial pressure is substituted in Pa, kPa,, bar or atm. This results in units of, equilibrium constant based on molarity or, pressure, unless the exponents of both the, numerator and denominator are same., For the reactions,, H2(g) + I2(g), 2HI, Kc and Kp have no unit., , Ni(CO)4 (g),, Ni (s) + 4 CO (g), the equilibrium constant is written as, ,  Ni (CO)4 , Kc = , [CO]4, It must be remembered that for the, existence of heterogeneous equilibrium pure, solids or liquids must also be present, (however small the amount may be) at, equilibrium, but their concentrations or, partial pressures do not appear in the, expression of the equilibrium constant. In the, reaction,, Ag2O(s) + 2HNO3(aq), 2AgNO3(aq) +H2O(l), , Kc =, , [ AgNO ], [HNO ], , 2, , 3, , 2, , 3, , 2NO2 (g), Kc has unit mol/L and, N2O4(g), Kp has unit bar, , Problem 7.6, , Equilibrium constants can also be, expressed as dimensionless quantities if, the standard state of reactants and, products are specified. For a pure gas, the, standard state is 1bar. Therefore a pressure, of 4 bar in standard state can be expressed, as 4 bar/1 bar = 4, which is a, dimensionless number. Standard state (c0), for a solute is 1 molar solution and all, concentrations can be measured with, respect to it. The numerical value of, equilibrium constant depends on the, standard state chosen. Thus, in this, system both K p and Kc are dimensionless, quantities but have different numerical, values due to different standard states., , The value of Kp for the reaction,, CO2 (g) + C (s), 2CO (g), is 3.0 at 1000 K. If initially PCO = 0.48 bar, 2, and PCO = 0 bar and pure graphite is, present, calculate the equilibrium partial, pressures of CO and CO2., Solution, For the reaction,, let ‘x’ be the decrease in pressure of CO2,, then, 2CO(g), CO2(g) + C(s), Initial, pressure: 0.48 bar, 0, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 205, , At equilibrium:, (0.48 – x)bar, , Kp =, , 5. The equilibrium constant K for a reaction, is related to the equilibrium constant of the, corresponding reaction, whose equation is, obtained by multiplying or dividing the, equation for the original reaction by a small, integer., Let us consider applications of equilibrium, constant to:, • predict the extent of a reaction on the basis, of its magnitude,, • predict the direction of the reaction, and, • calculate equilibrium concentrations., , 2x bar, , pC2 O, pC O2, , Kp = (2x)2/(0.48 – x) = 3, 4x2 = 3(0.48 – x), 4x2 = 1.44 – x, 4x2 + 3x – 1.44 = 0, a = 4, b = 3, c = –1.44, , (−b ±, x=, , b2 − 4 ac, , ), , 2a, = [–3 ± √(3)2– 4(4)(–1.44)]/2 × 4, = (–3 ± 5.66)/8, = (–3 + 5.66)/ 8 (as value of x cannot be, negative hence we neglect that value), x = 2.66/8 = 0.33, The equilibrium partial pressures are,, pCO = 2x = 2 × 0.33 = 0.66 bar, 2, , pCO = 0.48 – x = 0.48 – 0.33 = 0.15 bar, 2, , 7.6 APPLICATIONS OF EQUILIBRIUM, CONSTANTS, Before considering the applications of, equilibrium constants, let us summarise the, important features of equilibrium constants as, follows:, 1. Expression for equilibrium constant is, applicable only when concentrations of the, reactants and products have attained, constant value at equilibrium state., 2. The value of equilibrium constant is, independent of initial concentrations of the, reactants and products., 3. Equilibrium constant is temperature, dependent having one unique value for a, particular reaction represented by a, balanced equation at a given temperature., 4. The equilibrium constant for the reverse, reaction is equal to the inverse of the, equilibrium constant for the forward, reaction., , 7.6.1 Predicting the Extent of a Reaction, The numerical value of the equilibrium, constant for a reaction indicates the extent of, the reaction. But it is important to note that, an equilibrium constant does not give any, information about the rate at which the, equilibrium is reached. The magnitude of Kc, or K p is directly proportional to the, concentrations of products (as these appear, in the numerator of equilibrium constant, expression) and inversely proportional to the, concentrations of the reactants (these appear, in the denominator). This implies that a high, value of K is suggestive of a high concentration, of products and vice-versa., We can make the following generalisations, concerning the composition of, equilibrium mixtures:, • If Kc > 103, products predominate over, reactants, i.e., if Kc is very large, the reaction, proceeds nearly to completion. Consider, the following examples:, (a) The reaction of H2 with O2 at 500 K has a, very large equilibrium c o n s t a n t ,, Kc = 2.4 × 1047., (b) H2(g) + Cl2(g), Kc = 4.0 × 1031., , 2HCl(g) at 300K has, , (c) H 2(g) + Br 2(g), Kc = 5.4 × 1018, , 2HBr (g) at 300 K,, , •, , If Kc < 10–3, reactants predominate over, products, i.e., if Kc is very small, the reaction, proceeds rarely. Consider the following, examples:, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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206, , CHEMISTRY, , (a) The decomposition of H2O into H2 and O2, at 500 K has a very small equilibrium, constant, Kc = 4.1 × 10 –48, (b) N2(g) + O2(g), 2NO(g),, at 298 K has Kc = 4.8 ×10 – 31., • If K c is in the range of 10 – 3 to 10 3 ,, appreciable concentrations of both, reactants and products are present., Consider the following examples:, (a) For reaction of H2 with I2 to give HI,, Kc = 57.0 at 700K., (b) Also, gas phase decomposition of N2O4 to, NO2 is another reaction with a value, of Kc = 4.64 × 10 –3 at 25°C which is neither, too small nor too large. Hence,, equilibrium mixtures contain appreciable, concentrations of both N2O4 and NO2., These generarlisations are illustrated in, Fig. 7.6, , Fig.7.6 Dependence of extent of reaction on Kc, , 7.6.2 Predicting the Direction of the, Reaction, The equilibrium constant helps in predicting, the direction in which a given reaction will, proceed at any stage. For this purpose, we, calculate the reaction quotient Q. The, reaction quotient, Q (Q c with molar, concentrations and QP with partial pressures), is defined in the same way as the equilibrium, constant Kc except that the concentrations in, Qc are not necessarily equilibrium values., For a general reaction:, cC+dD, (7.19), aA+bB, c, d, a, b, (7.20), Qc = [C] [D] / [A] [B], , If Qc = Kc, the reaction mixture is already, at equilibrium., Consider the gaseous reaction of H 2, with I2,, H2(g) + I2(g), 2HI(g); Kc = 57.0 at 700 K., Suppose we have molar concentrations, [H2]t=0.10M, [I2]t = 0.20 M and [HI]t = 0.40 M., (the subscript t on the concentration symbols, means that the concentrations were measured, at some arbitrary time t, not necessarily at, equilibrium)., Thus, the reaction quotient, Qc at this stage, of the reaction is given by,, Qc = [HI]t2 / [H2]t [I2]t = (0.40)2/ (0.10)×(0.20), = 8.0, Now, in this case, Qc (8.0) does not equal, Kc (57.0), so the mixture of H2(g), I2(g) and HI(g), is not at equilibrium; that is, more H2(g) and, I2(g) will react to form more HI(g) and their, concentrations will decrease till Qc = Kc., The reaction quotient, Q c is useful in, predicting the direction of reaction by, comparing the values of Qc and Kc., Thus, we can make the following, generalisations concerning the direction of the, reaction (Fig. 7.7) :, , Fig. 7.7 Predicting the direction of the reaction, , •, •, •, , Then,, If Qc > Kc, the reaction will proceed in the, direction of reactants (reverse reaction)., If Qc < Kc, the reaction will proceed in the, direction of the products (forward reaction)., , If Qc < Kc, net reaction goes from left to right, If Qc > Kc, net reaction goes from right to, left., If Qc = Kc, no net reaction occurs., Problem 7.7, The value of Kc for the reaction, 2A, B + C is 2 × 10–3. At a given time,, the composition of reaction mixture is, [A] = [B] = [C] = 3 × 10–4 M. In which, direction the reaction will proceed?, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 207, , Solution, For the reaction the reaction quotient Qc, is given by,, Qc = [B][C]/ [A]2, as [A] = [B] = [C] = 3 × 10 –4M, Qc = (3 ×10 –4)(3 × 10 –4) / (3 ×10 –4)2 = 1, as Qc > Kc so the reaction will proceed in, the reverse direction., , The total pressure at equilbrium was, found to be 9.15 bar. Calculate Kc, Kp and, partial pressure at equilibrium., , 7.6.3 Calculating Equilibrium, Concentrations, In case of a problem in which we know the, initial concentrations but do not know any of, the equilibrium concentrations, the following, three steps shall be followed:, Step 1. Write the balanced equation for the, reaction., Step 2. Under the balanced equation, make a, table that lists for each substance involved in, the reaction:, (a) the initial concentration,, (b) the change in concentration on going to, equilibrium, and, (c) the equilibrium concentration., In constructing the table, define x as the, concentration (mol/L) of one of the substances, that reacts on going to equilibrium, then use, the stoichiometry of the reaction to determine, the concentrations of the other substances in, terms of x., Step 3. Substitute the equilibrium, concentrations into the equilibrium equation, for the reaction and solve for x. If you are to, solve a quadratic equation choose the, mathematical solution that makes chemical, sense., Step 4. Calculate the equilibrium, concentrations from the calculated value of x., Step 5. Check your results by substituting, them into the equilibrium equation., Problem 7.8, 13.8g of N2O4 was placed in a 1L reaction, vessel at 400K and allowed to attain, equilibrium, N 2O4 (g), , 2NO2 (g), , Solution, We know pV = nRT, Total volume (V ) = 1 L, Molecular mass of N2O4 = 92 g, Number of moles = 13.8g/92 g = 0.15, of the gas (n), Gas constant (R) = 0.083 bar L mol–1K–1, Temperature (T ) = 400 K, pV = nRT, p × 1L = 0.15 mol × 0.083 bar L mol–1K–1, × 400 K, p = 4.98 bar, N 2O4, 2NO2, Initial pressure: 4.98 bar, 0, At equilibrium: (4.98 – x) bar 2x bar, Hence,, ptotal at equilibrium = pN O + pNO, 2 4, 2, 9.15 = (4.98 – x) + 2x, 9.15 = 4.98 + x, x = 9.15 – 4.98 = 4.17 bar, Partial pressures at equilibrium are,, pN O = 4.98 – 4.17 = 0.81bar, 2 4, , pNO = 2x = 2 × 4.17 = 8.34 bar, 2, , (, , K p = p NO2, Kp, , 2, , ), , / p N 2O4, , = (8.34)2/0.81 = 85.87, ∆n, = Kc(RT), , 85.87 = Kc(0.083 × 400)1, Kc = 2.586 = 2.6, Problem 7.9, 3.00 mol of PCl5 kept in 1L closed reaction, vessel was allowed to attain equilibrium, at 380K. Calculate composition of the, mixture at equilibrium. Kc= 1.80, Solution, PCl3 + Cl2, PCl5, Initial, concentration: 3.0, 0, 0, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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208, , CHEMISTRY, , , K = e–∆G /RT, (7.23), Hence, using the equation (7.23), the, reaction spontaneity can be interpreted in, terms of the value of ∆G., , , • If ∆G < 0, then –∆G /RT is positive, and, >1, making K >1, which implies, a spontaneous reaction or the reaction, which proceeds in the forward direction to, such an extent that the products are, present predominantly., , , • If ∆G > 0, then –∆G /RT is negative, and, , Let x mol per litre of PCl5 be dissociated,, At equilibrium:, (3-x), x, x, Kc = [PCl3][Cl2]/[PCl5], 1.8 = x2/ (3 – x), x2 + 1.8x – 5.4 = 0, x = [–1.8 ± √(1.8)2 – 4(–5.4)]/2, x = [–1.8 ± √3.24 + 21.6]/2, x = [–1.8 ± 4.98]/2, x = [–1.8 + 4.98]/2 = 1.59, [PCl5] = 3.0 – x = 3 –1.59 = 1.41 M, [PCl3] = [Cl2] = x = 1.59 M, , < 1, that is , K < 1, which implies, , 7.7 RELATIONSHIP BETWEEN, EQUILIBRIUM CONSTANT K,, REACTION QUOTIENT Q AND, GIBBS ENERGY G, The value of Kc for a reaction does not depend, on the rate of the reaction. However, as you, have studied in Unit 6, it is directly related, to the thermodynamics of the reaction and, in particular, to the change in Gibbs energy,, ∆G. If,, • ∆G is negative, then the reaction is, spontaneous and proceeds in the forward, direction., • ∆G is positive, then reaction is considered, non-spontaneous. Instead, as reverse, reaction would have a negative ∆G, the, products of the forward reaction shall be, converted to the reactants., • ∆G is 0, reaction has achieved equilibrium;, at this point, there is no longer any free, energy left to drive the reaction., A mathematical expression of this, thermodynamic view of equilibrium can be, described by the following equation:, , (7.21), ∆G = ∆G + RT lnQ, , where, G is standard Gibbs energy., At equilibrium, when ∆G = 0 and Q = Kc,, the equation (7.21) becomes,, , ∆G = ∆G + RT ln K = 0, , ∆G = – RT lnK, (7.22), , lnK = – ∆G / RT, Taking antilog of both sides, we get,, , a non-spontaneous reaction or a reaction, which proceeds in the forward direction to, such a small degree that only a very minute, quantity of product is formed., Problem 7.10, , The value of ∆G for the phosphorylation, of glucose in glycolysis is 13.8 kJ/mol., Find the value of Kc at 298 K., Solution, , ∆G = 13.8 kJ/mol = 13.8 × 103J/mol, , Also, ∆G = – RT lnKc, Hence, ln Kc = –13.8 × 103J/mol, (8.314 J mol –1K –1 × 298 K), ln Kc = – 5.569, Kc = e–5.569, Kc = 3.81 × 10 –3, Problem 7.11, Hydrolysis of sucrose gives,, Glucose + Fructose, Sucrose + H2O, Equilibrium constant Kc for the reaction, , is 2 ×1013 at 300K. Calculate ∆G at, 300K., Solution, , ∆G = – RT lnKc, , ∆G = – 8.314J mol–1K–1×, 300K × ln(2×1013), , 4, ∆G = – 7.64 ×10 J mol–1, 7.8 FACTORS AFFECTING EQUILIBRIA, One of the principal goals of chemical synthesis, is to maximise the conversion of the reactants, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 209, , to products while minimizing the expenditure, of energy. This implies maximum yield of, products at mild temperature and pressure, conditions. If it does not happen, then the, experimental conditions need to be adjusted., For example, in the Haber process for the, synthesis of ammonia from N2 and H2, the, choice of experimental conditions is of real, economic importance. Annual world, production of ammonia is about hundred, million tones, primarily for use as fertilizers., Equilibrium constant, Kc is independent of, initial concentrations. But if a system at, equilibrium is subjected to a change in the, concentration of one or more of the reacting, substances, then the system is no longer at, equilibrium; and net reaction takes place in, some direction until the system returns to, equilibrium once again. Similarly, a change in, temperature or pressure of the system may, also alter the equilibrium. In order to decide, what course the reaction adopts and make a, qualitative prediction about the effect of a, change in conditions on equilibrium we use, Le Chatelier’s principle. It states that a, change in any of the factors that, determine the equilibrium conditions of a, system will cause the system to change, in such a manner so as to reduce or to, counteract the effect of the change. This, is applicable to all physical and chemical, equilibria., , “When the concentration of any of the, reactants or products in a reaction at, equilibrium is changed, the composition, of the equilibrium mixture changes so as, to minimize the effect of concentration, changes”., Let us take the reaction,, H2(g) + I2(g), 2HI(g), If H2 is added to the reaction mixture at, equilibrium, then the equilibrium of the, reaction is disturbed. In order to restore it, the, reaction proceeds in a direction wherein H2 is, consumed, i.e., more of H2 and I2 react to form, HI and finally the equilibrium shifts in right, (forward) direction (Fig.7.8). This is in, accordance with the Le Chatelier’s principle, which implies that in case of addition of a, reactant/product, a new equilibrium will be, set up in which the concentration of the, reactant/product should be less than what it, was after the addition but more than what it, was in the original mixture., , We shall now be discussing factors which, can influence the equilibrium., 7.8.1 Effect of Concentration Change, In general, when equilibrium is disturbed by, the addition/removal of any reactant/, products, Le Chatelier’s principle predicts that:, • The concentration stress of an added, reactant/product is relieved by net reaction, in the direction that consumes the added, substance., • The concentration stress of a removed, reactant/product is relieved by net reaction, in the direction that replenishes the, removed substance., or in other words,, , Fig. 7.8, , Effect of addition of H2 on change of, concentration for the reactants and, products, in, the, reaction,, H2(g) + I2 (g), 2HI(g), , The same point can be explained in terms, of the reaction quotient, Qc,, 2, Qc = [HI] / [H2][I2], , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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210, , CHEMISTRY, , Addition of hydrogen at equilibrium results, in value of Qc being less than Kc . Thus, in order, to attain equilibrium again reaction moves in, the forward direction. Similarly, we can say, that removal of a product also boosts the, forward reaction and increases the, concentration of the products and this has, great commercial application in cases of, reactions, where the product is a gas or a, volatile substance. In case of manufacture of, ammonia, ammonia is liquified and removed, from the reaction mixture so that reaction, keeps moving in forward direction. Similarly,, in the large scale production of CaO (used as, important building material) from CaCO3,, constant removal of CO2 from the kiln drives, the reaction to completion. It should be, remembered that continuous removal of a, product maintains Qc at a value less than Kc, and reaction continues to move in the forward, direction., Effect of Concentration – An experiment, This can be demonstrated by the following, reaction:, –, , 2+, , [Fe(SCN)] (aq), Fe3+(aq)+ SCN (aq), yellow, colourless, deep red, , (7.24), , (7.25), A reddish colour appears on adding two, drops of 0.002 M potassium thiocynate, solution to 1 mL of 0.2 M iron(III) nitrate, solution due to the formation of [Fe(SCN)]2+., The intensity of the red colour becomes, constant on attaining equilibrium. This, equilibrium can be shifted in either forward, or reverse directions depending on our choice, of adding a reactant or a product. The, equilibrium can be shifted in the opposite, direction by adding reagents that remove Fe3+, –, or SCN ions. For example, oxalic acid, (H2C2O4), reacts with Fe3+ ions to form the, stable complex ion [Fe(C 2 O 4) 3 ] 3 – , thus, decreasing the concentration of free Fe3+(aq)., In accordance with the Le Chatelier’s principle,, the concentration stress of removed Fe3+ is, relieved by dissociation of [Fe(SCN)] 2+ to, , replenish the Fe 3+ ions. Because the, concentration of [Fe(SCN)]2+ decreases, the, intensity of red colour decreases., Addition of aq. HgCl2 also decreases red, –, colour because Hg2+ reacts with SCN ions to, 2–, form stable complex ion [Hg(SCN)4] . Removal, –, of free SCN (aq) shifts the equilibrium in, equation (7.24) from right to left to replenish, –, SCN ions. Addition of potassium thiocyanate, on the other hand increases the colour, intensity of the solution as it shift the, equilibrium to right., 7.8.2 Effect of Pressure Change, A pressure change obtained by changing the, volume can affect the yield of products in case, of a gaseous reaction where the total number, of moles of gaseous reactants and total, number of moles of gaseous products are, different. In applying Le Chatelier’s principle, to a heterogeneous equilibrium the effect of, pressure changes on solids and liquids can, be ignored because the volume (and, concentration) of a solution/liquid is nearly, independent of pressure., Consider the reaction,, CH4(g) + H2O(g), CO(g) + 3H2(g), Here, 4 mol of gaseous reactants (CO + 3H2), become 2 mol of gaseous products (CH4 +, H2O). Suppose equilibrium mixture (for above, reaction) kept in a cylinder fitted with a piston, at constant temperature is compressed to one, half of its original volume. Then, total pressure, will, be, doubled, (according, to, pV = constant). The partial pressure and, therefore, concentration of reactants and, products have changed and the mixture is no, longer at equilibrium. The direction in which, the reaction goes to re-establish equilibrium, can be predicted by applying the Le Chatelier’s, principle. Since pressure has doubled, the, equilibrium now shifts in the forward, direction, a direction in which the number of, moles of the gas or pressure decreases (we, know pressure is proportional to moles of the, gas). This can also be understood by using, reaction quotient, Qc. Let [CO], [H2], [CH4] and, [H 2 O] be the molar concentrations at, equilibrium for methanation reaction. When, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 211, , volume of the reaction mixture is halved, the, partial pressure and the concentration are, doubled. We obtain the reaction quotient by, replacing each equilibrium concentration by, double its value., , CH4 ( g )  H2 O ( g ), Qc = , 3, CO ( g )  H2 (g ), As Qc < Kc , the reaction proceeds in the, forward direction., In reaction C(s) + CO2(g), 2CO(g), when, pressure is increased, the reaction goes in the, reverse direction because the number of moles, of gas increases in the forward direction., 7.8.3 Effect of Inert Gas Addition, , , , If the volume is kept constant and an inert gas, such as argon is added which does not take, part in the reaction, the equilibrium remains, undisturbed. It is because the addition of an, inert gas at constant volume does not change, the partial pressures or the molar, concentrations of the substance involved in the, reaction. The reaction quotient changes only, if the added gas is a reactant or product, involved in the reaction., 7.8.4 Effect of Temperature Change, Whenever an equilibrium is disturbed by a, change in the concentration, pressure or, volume, the composition of the equilibrium, mixture changes because the reaction, quotient, Qc no longer equals the equilibrium, constant, Kc. However, when a change in, temperature occurs, the value of equilibrium, constant, Kc is changed., , Production of ammonia according to the, reaction,, N2(g) + 3H2(g), 2NH3(g) ;, ∆H= – 92.38 kJ mol–1, is an exothermic process. According to, Le Chatelier’s principle, raising the, temperature shifts the equilibrium to left and, decreases the equilibrium concentration of, ammonia. In other words, low temperature is, favourable for high yield of ammonia, but, practically very low temperatures slow down, the reaction and thus a catalyst is used., Effect of Temperature – An experiment, Effect of temperature on equilibrium can be, demonstrated by taking NO2 gas (brown in, colour) which dimerises into N 2 O 4 gas, (colourless)., 2NO2(g), N2O4(g); ∆H = –57.2 kJ mol–1, NO 2 gas prepared by addition of Cu, turnings to conc. HNO3 is collected in two, 5 mL test tubes (ensuring same intensity of, colour of gas in each tube) and stopper sealed, with araldite. Three 250 mL beakers 1, 2 and, 3 containing freezing mixture, water at room, temperature and hot water (363 K ),, respectively, are taken (Fig. 7.9). Both the test, tubes are placed in beaker 2 for 8-10 minutes., After this one is placed in beaker 1 and the, other in beaker 3. The effect of temperature, on direction of reaction is depicted very well, in this experiment. At low temperatures in, beaker 1, the forward reaction of formation of, N2O4 is preferred, as reaction is exothermic, and, thus, intensity of brown colour due to NO2, decreases. While in beaker 3, high, temperature favours the reverse reaction of, , In general, the temperature dependence of, the equilibrium constant depends on the sign, of ∆H for the reaction., •, , The equilibrium constant for an exothermic, reaction (negative ∆H) decreases as the, temperature increases., , •, , The equilibrium constant for an, endothermic reaction (positive ∆H), increases as the temperature increases., , Temperature changes affect the, equilibrium constant and rates of reactions., , Fig. 7.9 Effect of temperature on equilibrium for, N2O4 (g), the reaction, 2NO2 (g), , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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212, , CHEMISTRY, , formation of NO2 and thus, the brown colour, intensifies., Effect of temperature can also be seen in, an endothermic reaction,, 3+, , –, , 2–, , [Co(H2O) 6] (aq) + 4Cl (aq), pink, , [CoCl4] (aq) +, 6H2O(l), , colourless, , blue, , At room temperature, the equilibrium, mixture is blue due to [CoCl4]2–. When cooled, in a freezing mixture, the colour of the mixture, turns pink due to [Co(H2O)6]3+., 7.8.5 Effect of a Catalyst, A catalyst increases the rate of the chemical, reaction by making available a new low energy, pathway for the conversion of reactants to, products. It increases the rate of forward and, reverse reactions that pass through the same, transition state and does not affect, equilibrium. Catalyst lowers the activation, energy for the forward and reverse reactions, by exactly the same amount. Catalyst does not, affect the equilibrium composition of a, reaction mixture. It does not appear in the, balanced chemical equation or in the, equilibrium constant expression., Let us consider the formation of NH3 from, dinitrogen and dihydrogen which is highly, exothermic reaction and proceeds with, decrease in total number of moles formed as, compared to the reactants. Equilibrium, constant decreases with increase in, temperature. At low temperature rate, decreases and it takes long time to reach at, equilibrium, whereas high temperatures give, satisfactory rates but poor yields., German chemist, Fritz Haber discovered, that a catalyst consisting of iron catalyse the, reaction to occur at a satisfactory rate at, temperatures, where the equilibrium, concentration of NH3 is reasonably favourable., Since the number of moles formed in the, reaction is less than those of reactants, the, yield of NH3 can be improved by increasing, the pressure., Optimum conditions of temperature and, pressure for the synthesis of NH 3 using, catalyst are around 500 °C and 200 atm., , Similarly, in manufacture of sulphuric, acid by contact process,, 2SO2(g) + O2(g), , 2SO3(g); Kc = 1.7 × 1026, , though the value of K is suggestive of reaction, going to completion, but practically the oxidation, of SO2 to SO3 is very slow. Thus, platinum or, divanadium penta-oxide (V 2O5) is used as, catalyst to increase the rate of the reaction., Note: If a reaction has an exceedingly small, K, a catalyst would be of little help., 7.9 IONIC EQUILIBRIUM IN SOLUTION, Under the effect of change of concentration on, the direction of equilibrium, you have, incidently come across with the following, equilibrium which involves ions:, Fe3+(aq) + SCN–(aq), , [Fe(SCN)]2+(aq), , There are numerous equilibria that involve, ions only. In the following sections we will, study the equilibria involving ions. It is well, known that the aqueous solution of sugar, does not conduct electricity. However, when, common salt (sodium chloride) is added to, water it conducts electricity. Also, the, conductance of electricity increases with an, increase in concentration of common salt., Michael Faraday classified the substances into, two categories based on their ability to conduct, electricity. One category of substances, conduct electricity in their aqueous solutions, and are called electrolytes while the other do, not and are thus, referred to as nonelectrolytes. Faraday further classified, electrolytes into strong and weak electrolytes., Strong electrolytes on dissolution in water are, ionized almost completely, while the weak, electrolytes are only partially dissociated., For example, an aqueous solution of, sodium chloride is comprised entirely of, sodium ions and chloride ions, while that, of acetic acid mainly contains unionized, acetic acid molecules and only some acetate, ions and hydronium ions. This is because, there is almost 100% ionization in case of, sodium chloride as compared to less, than 5% ionization of acetic acid which is, a weak electrolyte. It should be noted, that in weak electrolytes, equilibrium is, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 213, , established between ions and the unionized, molecules. This type of equilibrium involving, ions in aqueous solution is called ionic, equilibrium. Acids, bases and salts come, under the category of electrolytes and may act, as either strong or weak electrolytes., 7.10, , ACIDS, BASES AND SALTS, , Acids, bases and salts find widespread, occurrence in nature. Hydrochloric acid, present in the gastric juice is secreted by the, lining of our stomach in a significant amount, of 1.2-1.5 L/day and is essential for digestive, processes. Acetic acid is known to be the main, constituent of vinegar. Lemon and orange, juices contain citric and ascorbic acids, and, tartaric acid is found in tamarind paste. As, most of the acids taste sour, the word “acid”, has been derived from a latin word “acidus”, meaning sour. Acids are known to turn blue, litmus paper into red and liberate dihydrogen, on reacting with some metals. Similarly, bases, are known to turn red litmus paper blue, taste, bitter and feel soapy. A common example of a, base is washing soda used for washing, purposes. When acids and bases are mixed in, the right proportion they react with each other, to give salts. Some commonly known, examples of salts are sodium chloride, barium, sulphate, sodium nitrate. Sodium chloride, (common salt ) is an important component of, our diet and is formed by reaction between, hydrochloric acid and sodium hydroxide. It, , exists in solid state as a cluster of positively, charged sodium ions and negatively charged, chloride ions which are held together due to, electrostatic interactions between oppositely, charged species (Fig.7.10). The electrostatic, forces between two charges are inversely, proportional to dielectric constant of the, medium. Water, a universal solvent, possesses, a very high dielectric constant of 80. Thus,, when sodium chloride is dissolved in water,, the electrostatic interactions are reduced by a, factor of 80 and this facilitates the ions to move, freely in the solution. Also, they are wellseparated due to hydration with water, molecules., , Fig.7.10 Dissolution of sodium chloride in water., Na+ and Cl – ions are stablised by their, hydration with polar water molecules., , Comparing, the ionization of hydrochloric, acid with that of acetic acid in water we find, that though both of them are polar covalent, , Faraday was born near London into a family of very limited means. At the age of 14 he, was an apprentice to a kind bookbinder who allowed Faraday to read the books he, was binding. Through a fortunate chance he became laboratory assistant to Davy, and, during 1813-4, Faraday accompanied him to the Continent. During this trip he gained, much from the experience of coming into contact with many of the leading scientists of, the time. In 1825, he succeeded Davy as Director of the Royal Institution laboratories,, and in 1833 he also became the first Fullerian Professor of Chemistry. Faraday’s first Michael Faraday, (1791–1867), important work was on analytical chemistry. After 1821 much of his work was on, electricity and magnetism and different electromagnetic phenomena. His ideas have led to the establishment, of modern field theory. He discovered his two laws of electrolysis in 1834. Faraday was a very modest, and kind hearted person. He declined all honours and avoided scientific controversies. He preferred to, work alone and never had any assistant. He disseminated science in a variety of ways including his, Friday evening discourses, which he founded at the Royal Institution. He has been very famous for his, Christmas lecture on the ‘Chemical History of a Candle’. He published nearly 450 scientific papers., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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214, , CHEMISTRY, , molecules, former is completely ionized into, its constituent ions, while the latter is only, partially ionized (< 5%). The extent to which, ionization occurs depends upon the strength, of the bond and the extent of solvation of ions, produced. The terms dissociation and, ionization have earlier been used with different, meaning. Dissociation refers to the process of, separation of ions in water already existing as, such in the solid state of the solute, as in, sodium chloride. On the other hand, ionization, corresponds to a process in which a neutral, molecule splits into charged ions in the, solution. Here, we shall not distinguish, between the two and use the two terms, interchangeably., , Hydronium and Hydroxyl Ions, Hydrogen ion by itself is a bare proton with, very small size (~10 –15 m radius) and, intense electric field, binds itself with the, water molecule at one of the two available, +, lone pairs on it giving H3O . This species, has been detected in many compounds, +, –, (e.g., H3O Cl ) in the solid state. In aqueous, solution the hydronium ion is further, +, +, hydrated to give species like H5O2 , H7O3 and, +, H9O4 . Similarly the hydroxyl ion is hydrated, –, –, to give several ionic species like H3O2 , H5O3, –, and H7O4 etc., , 7.10.1 Arrhenius Concept of Acids and, Bases, According to Arrhenius theory, acids are, substances that dissociates in water to give, hydrogen ions H + (aq) and bases are, substances that produce hydroxyl ions, –, OH (aq). The ionization of an acid HX (aq) can, be represented by the following equations:, +, , –, , HX (aq) → H (aq) + X (aq), or, +, –, HX(aq) + H2O(l) → H3O (aq) + X (aq), A bare proton, H+ is very reactive and, cannot exist freely in aqueous solutions. Thus,, it bonds to the oxygen atom of a solvent water, molecule to give trigonal pyramidal, +, +, hydronium ion, H3O {[H (H2O)] } (see box)., +, +, In this chapter we shall use H (aq) and H3O (aq), interchangeably to mean the same i.e., a, hydrated proton., , +, , H9O4, , 7.10.2 The Brönsted-Lowry Acids and, Bases, The Danish chemist, Johannes Brönsted and, the English chemist, Thomas M. Lowry gave a, more general definition of acids and bases., According to Brönsted-Lowry theory, acid is, a substance that is capable of donating a, hydrogen ion H+ and bases are substances, capable of accepting a hydrogen ion, H +. In, short, acids are proton donors and bases are, proton acceptors., Consider the example of dissolution of NH3, in H2O represented by the following equation:, , Similarly, a base molecule like MOH, ionizes in aqueous solution according to the, equation:, +, , –, , MOH(aq) → M (aq) + OH (aq), The hydroxyl ion also exists in the hydrated, form in the aqueous solution. Arrhenius, concept of acid and base, however, suffers, from the limitation of being applicable only to, aqueous solutions and also, does not account, for the basicity of substances like, ammonia, which do not possess a hydroxyl group., , The basic solution is formed due to the, presence of hydroxyl ions. In this reaction,, water molecule acts as proton donor and, ammonia molecule acts as proton acceptor, and are thus, called Lowry-Brönsted acid and, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 215, , Arrhenius was born near Uppsala, Sweden. He presented his thesis, on the conductivities, of electrolyte solutions, to the University of Uppsala in 1884. For the next five years he, travelled extensively and visited a number of research centers in Europe. In 1895 he was, appointed professor of physics at the newly formed University of Stockholm, serving its, rector from 1897 to 1902. From 1905 until his death he was Director of physical chemistry, at the Nobel Institute in Stockholm. He continued to work for many years on electrolytic, solutions. In 1899 he discussed the temperature dependence of reaction rates on the, basis of an equation, now usually known as Arrhenius equation., He worked in a variety of fields, and made important contributions to, immunochemistry, cosmology, the origin of life, and the causes of ice age. He was the, Svante Arrhenius, first to discuss the ‘green house effect’ calling by that name. He received Nobel Prize in, (1859-1927), Chemistry in 1903 for his theory of electrolytic dissociation and its use in the development, of chemistry., , base, respectively. In the reverse reaction, H +, is transferred from NH4+ to OH – . In this case,, –, NH4+ acts as a Bronsted acid while OH acted, as a Brönsted base. The acid-base pair that, differs only by one proton is called a conjugate, –, acid-base pair. Therefore, OH is called the, +, conjugate base of an acid H2O and NH4 is, called conjugate acid of the base NH 3. If, Brönsted acid is a strong acid then its, conjugate base is a weak base and viceversa. It may be noted that conjugate acid, has one extra proton and each conjugate base, has one less proton., Consider the example of ionization of, hydrochloric acid in water. HCl(aq) acts as an, acid by donating a proton to H2O molecule, which acts as a base., , ammonia it acts as an acid by donating a, proton., , It can be seen in the above equation, that, water acts as a base because it accepts the, proton. The species H3O+ is produced when, water accepts a proton from HCl. Therefore,, –, Cl is a conjugate base of HCl and HCl is the, –, conjugate acid of base Cl . Similarly, H2O is a, +, +, conjugate base of an acid H3O and H3O is a, conjugate acid of base H2O., It is interesting to observe the dual role of, water as an acid and a base. In case of reaction, with HCl water acts as a base while in case of, , Problem 7.12, What will be the conjugate bases for the, following Brönsted acids: HF, H2SO4 and, –, HCO3 ?, Solution, The conjugate bases should have one, proton less in each case and therefore the, –, corresponding conjugate bases are: F ,, –, HSO4 and CO32– respectively., Problem 7.13, Write the conjugate acids for the following, –, –, Brönsted bases: NH2 , NH3 and HCOO ., Solution, The conjugate acid should have one extra, proton in each case and therefore the, corresponding conjugate acids are: NH3,, +, NH4 and HCOOH respectively., Problem 7.14, –, –, The species: H2O, HCO3 , HSO4 and NH3, can act both as Bronsted acids and bases., For each case give the corresponding, conjugate acid and conjugate base., Solution, The answer is given in the following Table:, Species Conjugate, Conjugate, acid, base, –, +, H3O, OH, H2O, –, HCO3, H2CO3, CO32–, –, HSO4, H2SO4, SO42–, –, NH3, NH4+, NH2, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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216, , CHEMISTRY, , 7.10.3 Lewis Acids and Bases, G.N. Lewis in 1923 defined an acid as a, species which accepts electron pair and base, which donates an electron pair. As far as, bases are concerned, there is not much, difference between Brönsted-Lowry and Lewis, concepts, as the base provides a lone pair in, both the cases. However, in Lewis concept, many acids do not have proton. A typical, example is reaction of electron deficient species, BF3 with NH3., BF3 does not have a proton but still acts, as an acid and reacts with NH3 by accepting, its lone pair of electrons. The reaction can be, represented by,, BF3 + :NH3 → BF3:NH3, Electron deficient species like AlCl3, Co3+,, Mg , etc. can act as Lewis acids while species, –, like H2O, NH3, OH etc. which can donate a pair, of electrons, can act as Lewis bases., 2+, , Problem 7.15, Classify the following species into Lewis, acids and Lewis bases and show how, these act as such:, –, –, +, (a) HO (b)F, (c) H, (d) BCl3, Solution, (a) Hydroxyl ion is a Lewis base as it can, –, donate an electron lone pair (:OH )., (b) Flouride ion acts as a Lewis base as, it can donate any one of its four, electron lone pairs., (c) A proton is a Lewis acid as it can, accept a lone pair of electrons from, bases like hydroxyl ion and fluoride, ion., (d) BCl3 acts as a Lewis acid as it can, accept a lone pair of electrons from, species like ammonia or amine, molecules., 7.11 IONIZATION OF ACIDS AND BASES, Arrhenius concept of acids and bases becomes, useful in case of ionization of acids and bases, as mostly ionizations in chemical and, biological systems occur in aqueous medium., Strong acids like perchloric acid (HClO4),, , hydrochloric acid (HCl), hydrobromic acid, (HBr), hyrdoiodic acid (HI), nitric acid (HNO3), and sulphuric acid (H2SO4) are termed strong, because they are almost completely, dissociated into their constituent ions in an, aqueous medium, thereby acting as proton, (H +) donors. Similarly, strong bases like, lithium hydroxide (LiOH), sodium hydroxide, (NaOH), potassium hydroxide (KOH), caesium, hydroxide (CsOH) and barium hydroxide, Ba(OH)2 are almost completely dissociated into, ions in an aqueous medium giving hydroxyl, ions, OH – . According to Arrhenius concept, they are strong acids and bases as they are, +, able to completely dissociate and produce H3O, –, and OH ions respectively in the medium., Alternatively, the strength of an acid or base, may also be gauged in terms of BrönstedLowry concept of acids and bases, wherein a, strong acid means a good proton donor and a, strong base implies a good proton acceptor., Consider, the acid-base dissociation, equilibrium of a weak acid HA,, HA(aq) + H2O(l)   H3O+(aq) + A–(aq), conjugate conjugate, acid, base, acid, base, In section 7.10.2 we saw that acid (or base), dissociation equilibrium is dynamic involving, a transfer of proton in forward and reverse, directions. Now, the question arises that if the, equilibrium is dynamic then with passage of, time which direction is favoured? What is the, driving force behind it? In order to answer, these questions we shall deal into the issue of, comparing the strengths of the two acids (or, bases) involved in the dissociation equilibrium., Consider the two acids HA and H3O+ present, in the above mentioned acid-dissociation, equilibrium. We have to see which amongst, them is a stronger proton donor. Whichever, exceeds in its tendency of donating a proton, over the other shall be termed as the stronger, acid and the equilibrium will shift in the, direction of weaker acid. Say, if HA is a, stronger acid than H3O+, then HA will donate, protons and not H3O+, and the solution will, mainly contain A – and H 3 O + ions. The, equilibrium moves in the direction of, formation of weaker acid and weaker base, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 217, , because the stronger acid donates a proton, to the stronger base., It follows that as a strong acid dissociates, completely in water, the resulting base formed, would be very weak i.e., strong acids have, very weak conjugate bases. Strong acids like, perchloric acid (HClO4), hydrochloric acid, (HCl), hydrobromic acid (HBr), hydroiodic acid, (HI), nitric acid (HNO3) and sulphuric acid, –, (H2SO4) will give conjugate base ions ClO4 , Cl,, – –, –, –, Br , I , NO3 and HSO4 , which are much weaker, bases than H2O. Similarly a very strong base, would give a very weak conjugate acid. On the, other hand, a weak acid say HA is only partially, dissociated in aqueous medium and thus, the, solution mainly contains undissociated HA, molecules. Typical weak acids are nitrous acid, (HNO2), hydrofluoric acid (HF) and acetic acid, (CH3COOH). It should be noted that the weak, acids have very strong conjugate bases. For, 2–, –, –, example, NH2 , O and H are very good proton, acceptors and thus, much stronger bases than, H2O., Certain water soluble organic compounds, like phenolphthalein and bromothymol blue, behave as weak acids and exhibit different, colours in their acid (HIn) and conjugate base, –, (In ) forms., +, , –, , HIn(aq) + H2O(l), H3O (aq) + In (aq), acid, conjugate conjugate, indicator, acid, base, colour A, colourB, Such compounds are useful as indicators, +, in acid-base titrations, and finding out H ion, concentration., 7.11.1 The Ionization Constant of Water, and its Ionic Product, Some substances like water are unique in their, ability of acting both as an acid and a base., We have seen this in case of water in section, 7.10.2. In presence of an acid, HA it accepts a, proton and acts as the base while in the, –, presence of a base, B it acts as an acid by, donating a proton. In pure water, one H2O, molecule donates proton and acts as an acid, and another water molecules accepts a proton, and acts as a base at the same time. The, following equilibrium exists:, , H3O+(aq) + OH–(aq), conjugate, conjugate, acid, base, , H2O(l) + H2O(l), acid, base, , The dissociation constant is represented by,, –, , (7.26), K = [H3O+] [OH ] / [H2O], The concentration of water is omitted from, the denominator as water is a pure liquid and, its concentration remains constant. [H2O] is, incorporated within the equilibrium constant, to give a new constant, Kw, which is called the, ionic product of water., +, , –, , Kw = [H ][OH ], , (7.27), +, , The concentration of H has been found, out experimentally as 1.0 × 10–7 M at 298 K., And, as dissociation of water produces equal, –, number of H+ and OH ions, the concentration, –, +, of hydroxyl ions, [OH ] = [H ] = 1.0 × 10 –7 M., Thus, the value of Kw at 298K,, +, , –, , Kw = [H3O ][OH ] = (1 × 10–7)2 = 1 × 10–14 M 2, (7.28), The value of Kw is temperature dependent, as it is an equilibrium constant., The density of pure water is 1000 g / L, and its molar mass is 18.0 g /mol. From this, the molarity of pure water can be given as,, [H2O] = (1000 g /L)(1 mol/18.0 g) = 55.55 M., Therefore, the ratio of dissociated water to that, of undissociated water can be given as:, –9, , –9, , 10 –7 / (55.55) = 1.8 × 10 or ~ 2 in 10 (thus,, equilibrium lies mainly towards undissociated, water), We can distinguish acidic, neutral and, basic aqueous solutions by the relative values, of the H3O+ and OH– concentrations:, –, Acidic: [H3O+] > [OH ], –, , Neutral: [H3O+] = [OH ], –, , Basic : [H3O+] < [OH ], 7.11.2 The pH Scale, Hydronium ion concentration in molarity is, more conveniently expressed on a logarithmic, scale known as the pH scale. The pH of a, solution is defined as the negative logarithm, , ( ), , to base 10 of the activity a H+, , Download all NCERT books PDFs from www.ncert.online, , 2019-20, , of hydrogen

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218, , CHEMISTRY, , ion. In dilute solutions (< 0.01 M), activity of, +, hydrogen ion (H ) is equal in magnitude to, +, molarity represented by [H ]. It should, be noted that activity has no units and is, defined as:, +, , = [H ] / mol L–1, From the definition of pH, the following, can be written,, +, , pH = – log aH+ = – log {[H ] / mol L–1}, Thus, an acidic solution of HCl (10–2 M), will have a pH = 2. Similarly, a basic solution, –, –4, +, of NaOH having [OH ] =10 M and [H3O ] =, –10, M will have a pH = 10. At 25 °C, pure, 10, water has a concentration of hydrogen ions,, +, –7, [H ] = 10 M. Hence, the pH of pure water is, given as:, , change in pH by just one unit also means, change in [H+] by a factor of 10. Similarly, when, the hydrogen ion concentration, [H+] changes, by a factor of 100, the value of pH changes by, 2 units. Now you can realise why the change, in pH with temperature is often ignored., Measurement of pH of a solution is very, essential as its value should be known when, dealing with biological and cosmetic, applications. The pH of a solution can be found, roughly with the help of pH paper that has, different colour in solutions of different pH., Now-a-days pH paper is available with four, strips on it. The different strips have different, colours (Fig. 7.11) at the same pH. The pH in, the range of 1-14 can be determined with an, accuracy of ~0.5 using pH paper., , pH = –log(10–7) = 7, Acidic solutions possess a concentration, +, of hydrogen ions, [H ] > 10–7 M, while basic, solutions possess a concentration of hydrogen, +, ions, [H ] < 10–7 M. thus, we can summarise, that, Acidic solution has pH < 7, Basic solution has pH > 7, Neutral solution has pH = 7, Now again, consider the equation (7.28) at, 298 K, –, +, Kw = [H3O ] [OH ] = 10–14, Taking negative logarithm on both sides, of equation, we obtain, , Fig.7.11 pH-paper with four strips that may, have different colours at the same pH, , For greater accuracy pH meters are used., pH meter is a device that measures the, pH-dependent electrical potential of the test, solution within 0.001 precision. pH meters of, the size of a writing pen are now available in, the market. The pH of some very common, substances are given in Table 7.5 (page 212)., , –, , –log Kw = – log {[H3O+] [OH ]}, , The concentration of hydrogen ion in a, sample of soft drink is 3.8 × 10–3M. what, is its pH ?, , –, , = – log [H3O+] – log [OH ], = – log 10 –14, pKw = pH +, , pOH = 14, , Problem 7.16, , (7.29), , Note that although Kw may change with, temperature the variations in pH with, temperature are so small that we often, ignore it., pK w is a very important quantity for, aqueous solutions and controls the relative, concentrations of hydrogen and hydroxyl ions, as their product is a constant. It should be, noted that as the pH scale is logarithmic, a, , Solution, pH = – log[3.8 × 10 –3], = – {log[3.8] + log[10 –3]}, = – {(0.58) + (– 3.0)} = – { – 2.42} = 2.42, Therefore, the pH of the soft drink is 2.42, and it can be inferred that it is acidic., Problem 7.17, Calculate pH of a 1.0 × 10, of HCl., , Download all NCERT books PDFs from www.ncert.online, , 2019-20, , –8, , M solution

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EQUILIBRIUM, , 219, , Table 7.5 The pH of Some Common Substances, Name of the Fluid, Saturated solution of NaOH, 0.1 M NaOH solution, Lime water, Milk of magnesia, Egg white, sea water, Human blood, Milk, Human Saliva, , pH, , Name of the Fluid, , ~15, 13, 10.5, 10, 7.8, 7.4, 6.8, 6.4, , Black Coffee, Tomato juice, Soft drinks and vinegar, Lemon juice, Gastric juice, 1M HCl solution, Concentrated HCl, , –, , 2H2O (l), H3O (aq) + OH (aq), –, Kw = [OH ][H3O+], = 10–14, –, +, Let, x = [OH ] = [H3O ] from H2O. The H3O+, concentration is generated (i) from, the ionization of HCl dissolved i.e.,, +, , –, , H3O (aq) + Cl (aq),, HCl(aq) + H2O(l), and (ii) from ionization of H2O. In these, very dilute solutions, both sources of, H3O+ must be considered:, –8, [H3O+] = 10 + x, –8, –14, Kw = (10 + x)(x) = 10, –8, or x2 + 10 x – 10–14 = 0, –, –8, [OH ] = x = 9.5 × 10, So, pOH = 7.02 and pH = 6.98, 7.11.3 Ionization Constants of Weak Acids, Consider a weak acid HX that is partially, ionized in the aqueous solution. The, equilibrium can be expressed by:, +, , 5.0, ~4.2, ~3.0, ~2.2, ~1.2, ~0, ~–1.0, , constant for the above discussed aciddissociation equilibrium:, Ka = c2α2 / c(1-α) = cα2 / 1-α, Ka is called the dissociation or ionization, constant of acid HX. It can be represented, alternatively in terms of molar concentration, as follows,, +, –, Ka = [H ][X ] / [HX], (7.30), At a given temperature T, K a is a, measure of the strength of the acid HX, i.e., larger the value of Ka, the stronger is, the acid. Ka is a dimensionless quantity, with the understanding that the standard, state concentration of all species is 1M., The values of the ionization constants of, some selected weak acids are given in, Table 7.6., , Solution, +, , pH, , Table 7.6 The Ionization Constants of Some, Selected Weak Acids (at 298K), Acid, , –, , H3O (aq) + X (aq), HX(aq) + H2O(l), Initial, concentration (M), c, 0, 0, Let α be the extent of ionization, Change (M), -cα, +cα, +cα, Equilibrium concentration (M), c-cα, cα, cα, Here, c = initial concentration of the, undissociated acid, HX at time, t = 0. α = extent, up to which HX is ionized into ions. Using, these notations, we can derive the equilibrium, , Ionization Constant,, Ka, –4, , Hydrofluoric Acid (HF), , 3.5 × 10, , Nitrous Acid (HNO2), , 4.5 × 10 –4, , Formic Acid (HCOOH), , 1.8 × 10, , –4, , Niacin (C5H4NCOOH), , 1.5 × 10, , –5, , Acetic Acid (CH3COOH), , 1.74 × 10, , Benzoic Acid (C6H5COOH), , 6.5 × 10 –5, , Hypochlorous Acid (HCIO), , 3.0 × 10, , –8, , Hydrocyanic Acid (HCN), , 4.9 × 10, , –10, , Phenol (C6H5OH), , 1.3 × 10, , –10, , –5, , The pH scale for the hydrogen ion, concentration has been so useful that besides, pKw, it has been extended to other species and, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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220, , CHEMISTRY, , quantities. Thus, we have:, (7.31), pKa = –log (Ka), Knowing the ionization constant, Ka of an, acid and its initial concentration, c, it is, possible to calculate the equilibrium, concentration of all species and also the degree, of ionization of the acid and the pH of the, solution., A general step-wise approach can be, adopted to evaluate the pH of the weak, electrolyte as follows:, Step 1. The species present before dissociation, are, identified, as, Brönsted-Lowry, acids / bases., Step 2. Balanced equations for all possible, reactions i.e., with a species acting both as, acid as well as base are written., Step 3. The reaction with the higher Ka is, identified as the primary reaction whilst the, other is a subsidiary reaction., Step 4. Enlist in a tabular form the following, values for each of the species in the primary, reaction, (a) Initial concentration, c., (b) Change in concentration on proceeding to, equilibrium in terms of α, degree of, ionization., (c) Equilibrium concentration., Step 5. Substitute equilibrium concentrations, into equilibrium constant equation for, principal reaction and solve for α., Step 6. Calculate the concentration of species, in principal reaction., Step 7. Calculate pH = – log[H3O+], The above mentioned methodology has, been elucidated in the following examples., Problem 7.18, The ionization constant of HF is, 3.2 × 10 –4 . Calculate the degree of, dissociation of HF in its 0.02 M solution., Calculate the concentration of all species, +, –, present (H3O , F and HF) in the solution, and its pH., , Solution, The following proton transfer reactions are, possible:, 1) HF + H2O, , H 3 O+ + F, , –, , Ka = 3.2 × 10–4, –, 2) H2O + H2O, H3O + OH, Kw = 1.0 × 10–14, As Ka >> Kw, [1] is the principle reaction., –, HF + H2O, H3O+ + F, Initial, concentration (M), 0.02, 0, 0, Change (M), –0.02α, +0.02α +0.02α, Equilibrium, concentration (M), 0.02 – 0.02 α, 0.02 α 0.02α, Substituting equilibrium concentrations, in the equilibrium reaction for principal, reaction gives:, Ka = (0.02α)2 / (0.02 – 0.02α), = 0.02 α2 / (1 –α) = 3.2 × 10–4, We obtain the following quadratic, equation:, α2 + 1.6 × 10–2α – 1.6 × 10–2 = 0, The quadratic equation in α can be solved, and the two values of the roots are:, α = + 0.12 and – 0.12, The negative root is not acceptable and, hence,, α = 0.12, This means that the degree of ionization,, α = 0.12, then equilibrium concentrations, –, +, of other species viz., HF, F and H3O are, given by:, –, [H3O+] = [F ] = cα = 0.02 × 0.12, = 2.4 × 10–3 M, [HF] = c(1 – α) = 0.02 (1 – 0.12), = 17.6 × 10-3 M, pH = – log[H+] = –log(2.4 × 10–3) = 2.62, +, , Problem 7.19, The pH of 0.1M monobasic acid is 4.50., +, Calculate the concentration of species H ,, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 221, , –, , A and HA at equilibrium. Also, determine, the value of Ka and pKa of the monobasic, acid., Solution, , Percent dissociation, = {[HOCl]dissociated / [HOCl]initial }× 100, = 1.41 × 10–3 × 102/ 0.08 = 1.76 %., pH = –log(1.41 × 10–3) = 2.85., , +, , pH = – log [H ], Therefore, [H+] = 10 –pH = 10 –4.50, = 3.16 × 10, +, , –, , [H ] = [A ] = 3.16 × 10, Thus,, , 7.11.4 Ionization of Weak Bases, –5, , –5, , Ka = [H+][A-] / [HA], , [HA]eqlbm = 0.1 – (3.16 × 10-5) 0.1, Ka = (3.16 × 10–5)2 / 0.1 = 1.0 × 10–8, pKa = – log(10–8) = 8, Alternatively, “Percent dissociation” is, another useful method for measure of, strength of a weak acid and is given as:, Percent dissociation, = [HA]dissociated/[HA]initial × 100%, , (7.32), , Problem 7.20, Calculate the pH of 0.08M solution of, hypochlorous acid, HOCl. The ionization, constant of the acid is 2.5 × 10 –5 ., Determine the percent dissociation of, HOCl., Solution, HOCl(aq) + H2O (l), H3O+(aq) + ClO–(aq), Initial concentration (M), 0.08, 0, 0, Change to reach, equilibrium concentration, (M), –x, +x, +x, equilibrium concentartion (M), 0.08 – x, x, x, Ka = {[H3O+][ClO–] / [HOCl]}, = x2 / (0.08 –x), As x << 0.08, therefore 0.08 – x  0.08, x2 / 0.08 = 2.5 × 10–5, x2 = 2.0 × 10–6, thus, x = 1.41 × 10–3, [H+] = 1.41 × 10–3 M., Therefore,, , The ionization of base MOH can be represented, by equation:, –, M+(aq) + OH (aq), MOH(aq), In a weak base there is partial ionization, –, of MOH into M+ and OH , the case is similar to, that of acid-dissociation equilibrium. The, equilibrium constant for base ionization is, called base ionization constant and is, represented by Kb. It can be expressed in terms, of concentration in molarity of various species, in equilibrium by the following equation:, –, (7.33), Kb = [M+][OH ] / [MOH], Alternatively, if c = initial concentration of, base and α = degree of ionization of base i.e., the extent to which the base ionizes. When, equilibrium is reached, the equilibrium, constant can be written as:, Kb = (cα)2 / c (1-α) = cα2 / (1-α), The values of the ionization constants of, some selected weak bases, Kb are given in, Table 7.7., Table 7.7, , The Values of the Ionization, Constant of Some Weak Bases at, 298 K, , Base, , Kb, –4, , Dimethylamine, (CH3)2NH, , 5.4 × 10, , Triethylamine, (C2H5)3N, , 6.45 × 10, , Ammonia, NH3 or NH4OH, , 1.77 × 10, , –5, , Quinine, (A plant product), , 1.10 × 10, , –6, , Pyridine, C5H5N, , 1.77 × 10, , –5, , –9, , Aniline, C6H5NH2, , 4.27 × 10, , Urea, CO (NH2)2, , 1.3 × 10, , –10, , –14, , Many organic compounds like amines are, weak bases. Amines are derivatives of, ammonia in which one or more hydrogen, atoms are replaced by another group. For, example, methylamine, codeine, quinine and, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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222, , CHEMISTRY, , nicotine all behave as very weak bases due to, their very small Kb. Ammonia produces OH–, in aqueous solution:, +, –, NH4 (aq) + OH (aq), NH3(aq) + H2O(l), The pH scale for the hydrogen ion, concentration has been extended to get:, (7.34), pKb = –log (Kb), , Kb = 10–4.75 = 1.77 × 10–5 M, , [H+] = antilog (–pH), = antilog (–9.7) = 1.67 ×10–10, [OH–] = Kw / [H+] = 1 × 10–14 / 1.67 × 10–10, = 5.98 × 10–5, The concentration of the corresponding, hydrazinium ion is also the same as that, of hydroxyl ion. The concentration of both, these ions is very small so the, concentration of the undissociated base, can be taken equal to 0.004M., Thus,, +, , –, , Kb = [NH2NH3 ][OH ] / [NH2NH2], = (5.98 × 10–5)2 / 0.004 = 8.96 × 10–7, , Calculate the pH of the solution in which, 0.2M NH4Cl and 0.1M NH3 are present. The, pKb of ammonia solution is 4.75., Solution, +, , The ionization constant of NH3,, Kb = antilog (–pKb) i.e., , +, , OH, , OH, , –, , 0.10, , 0.20, , 0, , +x, , +x, , 0.20 + x, , x, , Change to reach, equilibrium (M), –x, At equilibrium (M), 0.10 – x, +, , –, , Kb = [NH4 ][OH ] / [NH3], = (0.20 + x)(x) / (0.1 – x) = 1.77 × 10–5, As K b is small, we can neglect x in, comparison to 0.1M and 0.2M. Thus,, –, , [OH ] = x = 0.88 × 10–5, Therefore, [H+] = 1.12 × 10–9, pH = – log[H+] = 8.95., 7.11.5 Relation between K a and K b, As seen earlier in this chapter, K a and K b, represent the strength of an acid and a base,, respectively. In case of a conjugate acid-base, pair, they are related in a simple manner so, that if one is known, the other can be deduced., +, Considering the example of NH4 and NH3, we see,, NH4+(aq) + H2O(l), H3O+(aq) + NH3(aq), Ka = [H3O+][ NH3] / [NH4+] = 5.6 × 10–10, –, NH3(aq) + H2O(l), NH4+(aq) + OH (aq), –, Kb =[ NH4+][ OH ] / NH3 = 1.8 × 10–5, –, Net: 2 H2O(l), H3O+(aq) + OH (aq), –, –14, Kw = [H3O+][ OH ] = 1.0 × 10 M, +, , Problem 7.22, , NH4, , +, , Where, K a represents the strength of NH4 as, an acid and K b represents the strength of NH3, as a base., , pKb = –logKb = –log(8.96 × 10–7) = 6.04., , NH3 + H2O, , NH4, , Initial concentration (M), , Problem 7.21, The pH of 0.004M hydrazine solution is, 9.7. Calculate its ionization constant Kb, and pKb., Solution, –, NH2NH3+ + OH, NH2NH2 + H2O, From the pH we can calculate the, hydrogen ion concentration. Knowing, hydrogen ion concentration and the ionic, product of water we can calculate the, concentration of hydroxyl ions. Thus we, have:, , +, , NH3 + H2O, , –, , It can be seen from the net reaction that, the equilibrium constant is equal to the, product of equilibrium constants K a and K b, for the reactions added. Thus,, +, , +, , K a × K b = {[H3O+][ NH3] / [NH4 ]} × {[NH4 ], –, [ OH ] / [NH3]}, –, , = [H3O+][ OH ] = Kw, = (5.6x10–10) × (1.8 × 10–5) = 1.0 × 10–14 M, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 223, , This can be extended to make a, generalisation. The equilibrium constant for, a net reaction obtained after adding two, (or more) reactions equals the product of, the equilibrium constants for individual, reactions:, K NET = K1 × K2 × ……, , The value of α is small, therefore the, quadratic equation can be simplified by, neglecting α in comparison to 1 in the, denominator on right hand side of the, equation,, Thus,, Kb = c α2 or α = √ (1.77 × 10–5 / 0.05), , Alternatively, the above expression, K w = K a × K b , can also be obtained by, considering the base-dissociation equilibrium, reaction:, , = 0.018., –, , [OH ] = c α = 0.05 × 0.018 = 9.4 × 10–4M., –, , [H+] = Kw / [OH ] = 10–14 / (9.4 × 10–4), , –, , BH+(aq) + OH (aq), , = 1.06 × 10–11, , –, , K b = [BH ][OH ] / [B], , pH = –log(1.06 × 10–11) = 10.97., , As the concentration of water remains, constant it has been omitted from the, denominator and incorporated within the, dissociation constant. Then multiplying and, dividing the above expression by [H+], we get:, –, , +, , Now, using the relation for conjugate, acid-base pair,, Ka × Kb = Kw, using the value of K b of NH 3 from, Table 7.7., , +, , Kb = [BH ][OH ][H ] / [B][H ], –, , +, , +, , –, , Kb = 0.05 α2 / (1 – α), , Knowing one, the other can be obtained. It, should be noted that a strong acid will have, a weak conjugate base and vice-versa., , +, , OH, , –, , (7.36), , B(aq) + H2O(l), , +, , [OH ] = c α = 0.05 α, , Similarly, in case of a conjugate acid-base, pair,, , +, , NH4, , We use equation (7.33) to calculate, hydroxyl ion concentration,, , (3.35), , K a × Kb = K w, , +, , NH3 + H2O, , +, , ={[ OH ][H ]}{[BH ] / [B][H ]}, = K w / Ka, or K a × K b = K w, It may be noted that if we take negative, logarithm of both sides of the equation, then, pK values of the conjugate acid and base are, related to each other by the equation:, pK a + pK b = pK w = 14 (at 298K), Problem 7.23, Determine the degree of ionization and pH, of a 0.05M of ammonia solution. The, ionization constant of ammonia can be, taken from Table 7.7. Also, calculate the, ionization constant of the conjugate acid, of ammonia., Solution, The ionization of NH 3 in water is, represented by equation:, , We can determine the concentration of, +, conjugate acid NH4, Ka = Kw / Kb = 10–14 / 1.77 × 10–5, = 5.64 × 10–10., 7.11.6 Di- and Polybasic Acids and Di- and, Polyacidic Bases, Some of the acids like oxalic acid, sulphuric, acid and phosphoric acids have more than one, ionizable proton per molecule of the acid., Such acids are known as polybasic or, polyprotic acids., The ionization reactions for example for a, dibasic acid H 2X are represented by the, equations:, +, –, H2X(aq), H (aq) + HX (aq), –, +, 2–, H (aq) + X (aq), HX (aq), And the corresponding equilibrium, constants are given below:, +, –, Ka = {[H ][HX ]} / [H2X] and, 1, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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224, , CHEMISTRY, , Ka2 = {[H+][X2-]} / [HX-], Here, Ka1and Ka2are called the first and second, ionization constants respectively of the acid H2, X. Similarly, for tribasic acids like H3PO4 we, have three ionization constants. The values of, the ionization constants for some common, polyprotic acids are given in Table 7.8., Table 7.8 The Ionization Constants of Some, Common Polyprotic Acids (298K), , It can be seen that higher order ionization, constants (Ka2, Ka3) are smaller than the lower, order ionization constant (Ka1) of a polyprotic, acid. The reason for this is that it is more, difficult to remove a positively charged proton, from a negative ion due to electrostatic forces., This can be seen in the case of removing a, proton from the uncharged H 2 CO 3 as, –, compared from a negatively charged HCO3 ., Similarly, it is more difficult to remove a proton, 2–, from a doubly charged HPO 4 anion as, –, compared to H2PO4 ., Polyprotic acid solutions contain a mixture, 2–, –, of acids like H2A, HA and A in case of a, diprotic acid. H2A being a strong acid, the, primary reaction involves the dissociation of, H2 A, and H3O+ in the solution comes mainly, from the first dissociation step., 7.11.7 Factors Affecting Acid Strength, Having discussed quantitatively the strengths, of acids and bases, we come to a stage where, we can calculate the pH of a given acid, solution. But, the curiosity rises about why, should some acids be stronger than others?, What factors are responsible for making them, stronger? The answer lies in its being a, complex phenomenon. But, broadly speaking, we can say that the extent of dissociation of, an acid depends on the strength and polarity, of the H-A bond., , In general, when strength of H-A bond, decreases, that is, the energy required to break, the bond decreases, HA becomes a stronger, acid. Also, when the H-A bond becomes more, polar i.e., the electronegativity difference, between the atoms H and A increases and, there is marked charge separation, cleavage, of the bond becomes easier thereby increasing, the acidity., But it should be noted that while, comparing elements in the same group of the, periodic table, H-A bond strength is a more, important factor in determining acidity than, its polar nature. As the size of A increases, down the group, H-A bond strength decreases, and so the acid strength increases. For, example,, Size increases, HF << HCl << HBr << HI, Acid strength increases, Similarly, H2S is stronger acid than H2O., But, when we discuss elements in the same, row of the periodic table, H-A bond polarity, becomes the deciding factor for determining, the acid strength. As the electronegativity of A, increases, the strength of the acid also, increases. For example,, Electronegativity of A increases, CH4 < NH3 < H2O < HF, Acid strength increases, 7.11.8 Common Ion Effect in the, Ionization of Acids and Bases, Consider an example of acetic acid dissociation, equilibrium represented as:, –, H+(aq) + CH3COO (aq), CH3COOH(aq), –, H+ (aq) + Ac (aq), or HAc(aq), –, Ka = [H+][Ac ] / [HAc], Addition of acetate ions to an acetic acid, solution results in decreasing the, concentration of hydrogen ions, [H+]. Also, if, H+ ions are added from an external source then, the equilibrium moves in the direction of, undissociated acetic acid i.e., in a direction of, reducing the concentration of hydrogen ions,, +, [H ]. This phenomenon is an example of, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 225, , common ion effect. It can be defined as a, shift in equilibrium on adding a substance, that provides more of an ionic species already, present in the dissociation equilibrium. Thus,, we can say that common ion effect is a, phenomenon based on the Le Chatelier’s, principle discussed in section 7.8., In order to evaluate the pH of the solution, resulting on addition of 0.05M acetate ion to, 0.05M acetic acid solution, we shall consider, the acetic acid dissociation equilibrium once, again,, –, H+(aq) + Ac (aq), HAc(aq), Initial concentration (M), 0.05, 0, 0.05, Let x be the extent of ionization of acetic, acid., Change in concentration (M), –x, +x, +x, Equilibrium concentration (M), 0.05-x, x, 0.05+x, Therefore,, –, K a= [H+][Ac ]/[H Ac] = {(0.05+x)(x)}/(0.05-x), As Ka is small for a very weak acid, x<<0.05., , Thus, x = 1.33 × 10–3 = [OH–], Therefore,[H+] = Kw / [OH–] = 10–14 /, (1.33 × 10–3) = 7.51 × 10–12, pH = –log(7.5 × 10–12) = 11.12, On addition of 25 mL of 0.1M HCl, solution (i.e., 2.5 mmol of HCl) to 50 mL, of 0.1M ammonia solution (i.e., 5 mmol, of NH3), 2.5 mmol of ammonia molecules, are neutralized. The resulting 75 mL, solution contains the remaining, unneutralized 2.5 mmol of NH3 molecules, +, and 2.5 mmol of NH4 ., NH3, , Problem 7.24, Calculate the pH of a 0.10M ammonia, solution. Calculate the pH after 50.0 mL, of this solution is treated with 25.0 mL of, 0.10M HCl. The dissociation constant of, ammonia, K b = 1.77 × 10–5, Solution, H2O, , →, , NH4+ + OH–, , Kb = [NH4+][OH–] / [NH3] = 1.77 × 10–5, Before neutralization,, –, [NH4+] = [OH ] = x, [NH3] = 0.10 – x  0.10, x2 / 0.10 = 1.77 × 10–5, , → NH4+ + Cl–, , The final 75 mL solution after, neutralisation already contains, 2.5 m mol NH4+ ions (i.e. 0.033M), thus, +, total concentration of NH4 ions is given as:, +, [NH4 ] = 0.033 + y, As y is small, [NH4OH]  0.033 M and, [NH4+]  0.033M., We know,, Kb = [NH4+][OH–] / [NH4OH], = y(0.033)/(0.033) = 1.77 × 10–5 M, Thus, y = 1.77 × 10–5 = [OH–], [H+] = 10–14 / 1.77 × 10–5 = 0.56 × 10–9, Hence, pH = 9.24, , Thus,, 1.8 × 10–5 = (x) (0.05 + x) / (0.05 – x), –5, = x(0.05) / (0.05) = x = [H+] = 1.8 × 10 M, pH = – log(1.8 × 10–5) = 4.74, , +, , HCl, , 2.5, 2.5, 0, 0, At equilibrium, 0, 0, 2.5, 2.5, The resulting 75 mL of solution contains, +, 2.5 mmol of NH4 ions (i.e., 0.033 M) and, 2.5 mmol (i.e., 0.033 M ) of uneutralised, NH3 molecules. This NH3 exists in the, following equilibrium:, +, –, NH4OH   NH4 + OH, 0.033M – y, y, y, –, +, where, y = [OH ] = [NH4 ], , Hence, (0.05 + x) ≈ (0.05 – x) ≈ 0.05, , NH3, , +, , 7.11.9 Hydrolysis of Salts and the pH of, their Solutions, Salts formed by the reactions between acids, and bases in definite proportions, undergo, ionization in water. The cations/anions formed, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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226, , CHEMISTRY, , on ionization of salts either exist as hydrated, ions in aqueous solutions or interact with, water to reform corresponding acids/bases, depending upon the nature of salts. The later, process of interaction between water and, cations/anions or both of salts is called, hydrolysis. The pH of the solution gets affected, +, +, by this interaction. The cations (e.g., Na , K ,, 2+, 2+, Ca , Ba , etc.) of strong bases and anions, –, –, –, –, (e.g., Cl , Br , NO3, ClO4 etc.) of strong acids, simply get hydrated but do not hydrolyse, and, therefore the solutions of salts formed from, strong acids and bases are neutral i.e., their, pH is 7. However, the other category of salts, do undergo hydrolysis., We now consider the hydrolysis of the salts, of the following types :, (i) salts of weak acid and strong base e.g.,, CH3COONa., (ii) salts of strong acid and weak base e.g.,, NH4Cl, and, (iii) salts of weak acid and weak base, e.g.,, CH3COONH4., , increased of H+ ion concentration in solution, making the solution acidic. Thus, the pH of, NH4Cl solution in water is less than 7., Consider the hydrolysis of CH3COONH4 salt, formed from weak acid and weak base. The, ions formed undergo hydrolysis as follow:, –, +, CH3COOH +, CH3COO + NH4 + H2O, NH4OH, CH3COOH and NH4OH, also remain into, partially dissociated form:, –, CH3COOH, CH3COO + H+, +, –, NH4OH, NH4 + OH, –, H2O, H+ + OH, Without going into detailed calculation, it, can be said that degree of hydrolysis is, independent of concentration of solution, and, pH of such solutions is determined by their pK, values:, pH = 7 + ½ (pKa – pKb), (7.38), The pH of solution can be greater than 7,, if the difference is positive and it will be less, than 7, if the difference is negative., , In the first case, CH3COONa being a salt of, weak acid, CH3COOH and strong base, NaOH, gets completely ionised in aqueous solution., –, , Problem 7.25, The pK a of acetic acid and pK b of, ammonium hydroxide are 4.76 and 4.75, respectively. Calculate the pH of, ammonium acetate solution., , +, , CH3COONa(aq) → CH3COO (aq)+ Na (aq), Acetate ion thus formed undergoes, hydrolysis in water to give acetic acid and OH–, ions, –, , CH3COO (aq)+H2O(l), , pH = 7 + ½ [pKa – pKb], , CH3COOH(aq)+OH (aq), , Acetic acid being a weak acid, (Ka = 1.8 × 10–5) remains mainly unionised in, –, solution. This results in increase of OH ion, concentration in solution making it alkaline., The pH of such a solution is more than 7., Similarly, NH4Cl formed from weak base,, NH 4OH and strong acid, HCl, in water, dissociates completely., +, , Solution, , –, , –, , NH4Cl(aq) → NH 4(aq) +Cl (aq), Ammonium ions undergo hydrolysis with, water to form NH4OH and H+ ions, +, NH 4 (aq) + H2O (1), NH4OH(aq) + H+(aq), Ammonium hydroxide is a weak base, (Kb = 1.77 × 10–5) and therefore remains almost, unionised in solution. This results in, , = 7 + ½ [4.76 – 4.75], = 7 + ½ [0.01] = 7 + 0.005 = 7.005, 7.12 BUFFER SOLUTIONS, Many body fluids e.g., blood or urine have, definite pH and any deviation in their pH, indicates malfunctioning of the body. The, control of pH is also very important in many, chemical and biochemical processes. Many, medical and cosmetic formulations require, that these be kept and administered at a, particular pH. The solutions which resist, change in pH on dilution or with the, addition of small amounts of acid or alkali, are called Buffer Solutions. Buffer solutions, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 227, , of known pH can be prepared from the, knowledge of pK a of the acid or pK of base, b, and by controlling the ratio of the salt and acid, or salt and base. A mixture of acetic acid and, sodium acetate acts as buffer solution around, pH 4.75 and a mixture of ammonium chloride, and ammonium hydroxide acts as a buffer, around pH 9.25. You will learn more about, buffer solutions in higher classes., 7.12.1 Designing Buffer Solution, Knowledge of pK a, pK b and equilibrium, constant help us to prepare the buffer solution, of known pH. Let us see how we can do this., Preparation of Acidic Buffer, To prepare a buffer of acidic pH we use weak, acid and its salt formed with strong base. We, develop the equation relating the pH, the, equilibrium constant, Ka of weak acid and ratio, of concentration of weak acid and its conjugate, base. For the general case where the weak acid, HA ionises in water,, HA + H2O  H3O+ + A–, For which we can write the expression, , conjugate base (anion) of the acid and the acid, present in the mixture. Since acid is a weak, acid, it ionises to a very little extent and, concentration of [HA] is negligibly different from, concentration of acid taken to form buffer. Also,, most of the conjugate base, [A—], comes from, the ionisation of salt of the acid. Therefore, the, concentration of conjugate base will be, negligibly different from the concentration of, salt. Thus, equation (7.40) takes the form:, , [Salt], [Acid], In the equation (7.39), if the concentration, of [A—] is equal to the concentration of [HA],, then pH = pKa because value of log 1 is zero., Thus if we take molar concentration of acid and, salt (conjugate base) same, the pH of the buffer, solution will be equal to the pKa of the acid. So, for preparing the buffer solution of the required, pH we select that acid whose pKa is close to the, required pH. For acetic acid pKa value is 4.76,, therefore pH of the buffer solution formed by, acetic acid and sodium acetate taken in equal, molar concentration will be around 4.76., A similar analysis of a buffer made with a, weak base and its conjugate acid leads to the, result,, pH=pK a + log, , Rearranging the expression we have,, , pOH= pK b +log, , Taking logarithm on both the sides and, rearranging the terms we get —, , Or, , (7.41), pH of the buffer solution can be calculated, by using the equation pH + pOH =14., We know that pH + pOH = pK w and, pKa + pKb = pKw. On putting these values in, equation (7.41) it takes the form as follows:, pK w - pH = pK w − pK a + log, , (7.39), , (7.40), The expression (7.40) is known as, Henderson–Hasselbalch equation. The, is the ratio of concentration of, , [Conjugate acid,BH + ], [Base,B], , or, pH = pK a + log, , quantity, , [Conjugate acid,BH+ ], [Base,B], , [Conjugate acid,BH+ ], [Base,B], , (7.42), , If molar concentration of base and its, conjugate acid (cation) is same then pH of the, buffer solution will be same as pKa for the base., pKa value for ammonia is 9.25; therefore a, buffer of pH close to 9.25 can be obtained by, taking ammonia solution and ammonium, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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228, , CHEMISTRY, , chloride solution of same molar concentration., For a buffer solution formed by ammonium, chloride and ammonium hydroxide, equation, (7.42) becomes:, , [Conjugate acid,BH + ], pH = 9.25 + log, [Base,B], pH of the buffer solution is not affected by, dilution because ratio under the logarithmic, term remains unchanged., 7.13 SOLUBILITY, EQUILIBRIA, OF, SPARINGLY SOLUBLE SALTS, We have already known that the solubility of, ionic solids in water varies a great deal. Some, of these (like calcium chloride) are so soluble, that they are hygroscopic in nature and even, absorb water vapour from atmosphere. Others, (such as lithium fluoride) have so little solubility, that they are commonly termed as insoluble., The solubility depends on a number of factors, important amongst which are the lattice, enthalpy of the salt and the solvation enthalpy, of the ions in a solution. For a salt to dissolve, in a solvent the strong forces of attraction, between its ions (lattice enthalpy) must be, overcome by the ion-solvent interactions. The, solvation enthalpy of ions is referred to in terms, of solvation which is always negative i.e. energy, is released in the process of solvation. The, amount of solvation enthalpy depends on the, nature of the solvent. In case of a non-polar, (covalent) solvent, solvation enthalpy is small, and hence, not sufficient to overcome lattice, enthalpy of the salt. Consequently, the salt does, not dissolve in non-polar solvent. As a general, rule , for a salt to be able to dissolve in a, particular solvent its solvation enthalpy must, be greater than its lattice enthalpy so that the, latter may be overcome by former. Each salt has, its characteristic solubility which depends on, temperature. We classify salts on the basis of, their solubility in the following three categories., Category I, , Soluble, , Solubility > 0.1M, , Category II, , Slightly, Soluble, , 0.01M<Solubility< 0.1M, , Category III, , Sparingly, Soluble, , Solubility < 0.01M, , We shall now consider the equilibrium, between the sparingly soluble ionic salt and, its saturated aqueous solution., 7.13.1 Solubility Product Constant, Let us now have a solid like barium sulphate, in contact with its saturated aqueous solution., The equilibrium between the undisolved solid, and the ions in a saturated solution can be, represented by the equation:, BaSO4(s) , , Ba2+(aq) + SO42–(aq),, , The equilibrium constant is given by the, equation:, K = {[Ba2+][SO42–]} / [BaSO4], For a pure solid substance the, concentration remains constant and we can, write, (7.43), Ksp = K[BaSO4] = [Ba2+][SO42–], We call Ksp the solubility product constant, or simply solubility product. The experimental, value of Ksp in above equation at 298K is, –10, 1.1 × 10 . This means that for solid barium, sulphate in equilibrium with its saturated, solution, the product of the concentrations of, barium and sulphate ions is equal to its, solubility, product, constant., The, concentrations of the two ions will be equal to, the molar solubility of the barium sulphate. If, molar solubility is S, then, 1.1 × 10–10 = (S)(S) = S2, or, S = 1.05 × 10–5., Thus, molar solubility of barium sulphate, will be equal to 1.05 × 10–5 mol L–1., A salt may give on dissociation two or more, than two anions and cations carrying different, charges. For example, consider a salt like, zirconium phosphate of molecular formula, 3–, 4+, (Zr )3(PO4 )4. It dissociates into 3 zirconium, cations of charge +4 and 4 phosphate anions, of charge –3. If the molar solubility of, zirconium phosphate is S, then it can be seen, from the stoichiometry of the compound that, [Zr4+] = 3S and [PO43–] = 4S, and Ksp = (3S)3 (4S)4 = 6912 (S)7, or S = {Ksp / (33 × 44)}1/7 = (Ksp / 6912)1/7, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 229, , A solid salt of the general formula, q−, M X y with molar solubility S in equilibrium, with its saturated solution may be represented, by the equation:, p+, q–, xM (aq) + yX (aq), M x X y (s), (where x × p+ = y × q–), And its solubility product constant is given, by:, p+ x, q– y, x, y, (7.44), Ksp = [M ] [X ] = (xS) (yS), = xx . yy . S(x + y), S(x + y) = Ksp / xx . yy, S = (Ksp / xx . yy)1 / x + y, (7.45), The term Ksp in equation is given by Qsp, (section 7.6.2) when the concentration of one, or more species is not the concentration under, equilibrium. Obviously under equilibrium, conditions Ksp = Qsp but otherwise it gives the, direction of the processes of precipitation or, dissolution. The solubility product constants, of a number of common salts at 298K are given, in Table 7.9., p+, x, , Table 7.9 The Solubility Product Constants,, Ksp of Some Common Ionic Salts at, 298K., , Problem 7.26, Calculate the solubility of A2 X3 in pure, water, assuming that neither kind of ion, reacts with water. The solubility product, of A 2X 3 , Ksp = 1.1 × 10–23., Solution, A2X3 → 2A3+ + 3X2–, Ksp = [A3+]2 [X2–]3 = 1.1 × 10–23, If S = solubility of A 2X 3 , then, [A3+] = 2S; [X2–] = 3S, therefore, Ksp = (2S)2(3S)3 = 108S5, = 1.1 × 10–23, thus, S5 = 1 × 10–25, S = 1.0 × 10–5 mol/L., Problem 7.27, The values of Ksp of two sparingly soluble, salts Ni(OH)2 and AgCN are 2.0 × 10–15, and 6 × 0–17 respectively. Which salt is, more soluble? Explain., Solution, AgCN, , –, , Ag+ + CN, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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230, , CHEMISTRY, , Ksp = [Ag+][CN–] = 6 × 10–17, Ni(OH)2, Ni2+ + 2OH–, Ksp = [Ni2+][OH–]2 = 2 × 10–15, Let [Ag+] = S1, then [CN-] = S1, Let [Ni2+] = S2, then [OH–] = 2S2, S12 = 6 × 10–17 , S1 = 7.8 × 10–9, (S2)(2S2)2 = 2 × 10–15, S2 = 0.58 × 10–4, Ni(OH)2 is more soluble than AgCN., , Dissolution of S mol/L of Ni(OH)2 provides, –, S mol/L of Ni2+ and 2S mol/L of OH , but, –, the total concentration of OH = (0.10 +, 2S) mol/L because the solution already, contains 0.10 mol/L of OH– from NaOH., –, , 7.13.2 Common Ion Effect on Solubility, of Ionic Salts, It is expected from Le Chatelier’s principle that, if we increase the concentration of any one of, the ions, it should combine with the ion of its, opposite charge and some of the salt will be, precipitated till once again Ksp = Qsp. Similarly,, if the concentration of one of the ions is, decreased, more salt will dissolve to increase, the concentration of both the ions till once, again Ksp = Qsp. This is applicable even to, soluble salts like sodium chloride except that, due to higher concentrations of the ions, we, use their activities instead of their molarities, in the expression for Qsp. Thus if we take a, saturated solution of sodium chloride and, pass HCl gas through it, then sodium chloride, is precipitated due to increased concentration, (activity) of chloride ion available from the, dissociation of HCl. Sodium chloride thus, obtained is of very high purity and we can get, rid of impurities like sodium and magnesium, sulphates. The common ion effect is also used, for almost complete precipitation of a particular, ion as its sparingly soluble salt, with very low, value of solubility product for gravimetric, estimation. Thus we can precipitate silver ion, as silver chloride, ferric ion as its hydroxide, (or hydrated ferric oxide) and barium ion as, its sulphate for quantitative estimations., Problem 7.28, Calculate the molar solubility of Ni(OH)2, in 0.10 M NaOH. The ionic product of, Ni(OH)2 is 2.0 × 10–15., Solution, Let the solubility of Ni(OH)2 be equal to S., , Ksp = 2.0 × 10–15 = [Ni2+] [OH ]2, = (S) (0.10 + 2S)2, As Ksp is small, 2S << 0.10,, thus, (0.10 + 2S) ≈ 0.10, Hence,, 2.0 × 10–15 = S (0.10)2, S = 2.0 × 10–13 M = [Ni2+], The solubility of salts of weak acids like, phosphates increases at lower pH. This is, because at lower pH the concentration of the, anion decreases due to its protonation. This, in turn increase the solubility of the salt so, that Ksp = Qsp. We have to satisfy two equilibria, simultaneously i.e.,, Ksp = [M+] [X–],, , –, , [X ] / [HX] = Ka / [H+], Taking inverse of both side and adding 1, we get, , [HX ] + 1 =,  X − , , H + , +1, Ka, , [HX ] + H − ,  X − , , H+  + K a, =  , Ka, , Now, again taking inverse, we get, –, –, [X ] / {[X ] + [HX]} = f = Ka / (Ka + [H+]) and it, can be seen that ‘f’ decreases as pH decreases. If, S is the solubility of the salt at a given pH then, Ksp = [S] [f S] = S2 {Ka / (Ka + [H+])} and, (7.46), S = {Ksp ([H+] + Ka ) / Ka }1/2, Thus solubility S increases with increase, in [H+] or decrease in pH., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 231, , SUMMARY, When the number of molecules leaving the liquid to vapour equals the number of, molecules returning to the liquid from vapour, equilibrium is said to be attained and is, dynamic in nature. Equilibrium can be established for both physical and chemical, processes and at this stage rate of forward and reverse reactions are equal. Equilibrium, constant, Kc is expressed as the concentration of products divided by reactants, each, term raised to the stoichiometric coefficient., For reaction, a A + b B, c C +d D, c, d, a, b, Kc = [C] [D] /[A] [B], Equilibrium constant has constant value at a fixed temperature and at this stage, all the macroscopic properties such as concentration, pressure, etc. become constant., For a gaseous reaction equilibrium constant is expressed as Kp and is written by replacing, concentration terms by partial pressures in Kc expression. The direction of reaction can, be predicted by reaction quotient Qc which is equal to Kc at equilibrium. Le Chatelier’s, principle states that the change in any factor such as temperature, pressure,, concentration, etc. will cause the equilibrium to shift in such a direction so as to reduce, or counteract the effect of the change. It can be used to study the effect of various, factors such as temperature, concentration, pressure, catalyst and inert gases on the, direction of equilibrium and to control the yield of products by controlling these factors., Catalyst does not effect the equilibrium composition of a reaction mixture but increases, the rate of chemical reaction by making available a new lower energy pathway for, conversion of reactants to products and vice-versa., All substances that conduct electricity in aqueous solutions are called electrolytes., Acids, bases and salts are electrolytes and the conduction of electricity by their aqueous, solutions is due to anions and cations produced by the dissociation or ionization of, electrolytes in aqueous solution. The strong electrolytes are completely dissociated. In, weak electrolytes there is equilibrium between the ions and the unionized electrolyte, molecules. According to Arrhenius, acids give hydrogen ions while bases produce, hydroxyl ions in their aqueous solutions. Brönsted-Lowry on the other hand, defined, an acid as a proton donor and a base as a proton acceptor. When a Brönsted-Lowry, acid reacts with a base, it produces its conjugate base and a conjugate acid corresponding, to the base with which it reacts. Thus a conjugate pair of acid-base differs only by one, proton. Lewis further generalised the definition of an acid as an electron pair acceptor, and a base as an electron pair donor. The expressions for ionization (equilibrium), constants of weak acids (Ka) and weak bases (Kb) are developed using Arrhenius definition., The degree of ionization and its dependence on concentration and common ion are, discussed. The pH scale (pH = -log[H+]) for the hydrogen ion concentration (activity) has, –, been introduced and extended to other quantities (pOH = – log[OH ]) ; pKa = –log[Ka] ;, pKb = –log[Kb]; and pKw = –log[Kw] etc.). The ionization of water has been considered and, we note that the equation: pH + pOH = pK w is always satisfied. The salts of strong acid, and weak base, weak acid and strong base, and weak acid and weak base undergo, hydrolysis in aqueous solution.The definition of buffer solutions, and their importance, are discussed briefly. The solubility equilibrium of sparingly soluble salts is discussed, and the equilibrium constant is introduced as solubility product constant (Ksp). Its, relationship with solubility of the salt is established. The conditions of precipitation of, the salt from their solutions or their dissolution in water are worked out. The role of, common ion and the solubility of sparingly soluble salts is also discussed., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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232, , CHEMISTRY, , SUGGESTED ACTIVITIES FOR STUDENTS REGARDING THIS UNIT, (a) The student may use pH paper in determining the pH of fresh juices of various, vegetables and fruits, soft drinks, body fluids and also that of water samples available., (b) The pH paper may also be used to determine the pH of different salt solutions and, from that he/she may determine if these are formed from strong/weak acids and, bases., (c) They may prepare some buffer solutions by mixing the solutions of sodium acetate, and acetic acid and determine their pH using pH paper., (d) They may be provided with different indicators to observe their colours in solutions, of varying pH., (e) They may perform some acid-base titrations using indicators., (f) They may observe common ion effect on the solubility of sparingly soluble salts., (g) If pH meter is available in their school, they may measure the pH with it and compare, the results obtained with that of the pH paper., , EXERCISES, 7.1, , A liquid is in equilibrium with its vapour in a sealed container at a fixed, temperature. The volume of the container is suddenly increased., , a), , What is the initial effect of the change on vapour pressure?, , b), , How do rates of evaporation and condensation change initially?, , c), , What happens when equilibrium is restored finally and what will be the final, vapour pressure?, , 7.2, , What is Kc for the following equilibrium when the equilibrium concentration of, each substance is: [SO2]= 0.60M, [O2] = 0.82M and [SO3] = 1.90M ?, 2SO2(g) + O2(g), , 7.3, , 2SO3(g), , At a certain temperature and total pressure of 105Pa, iodine vapour contains 40%, by volume of I atoms, I2 (g), 2I (g), Calculate Kp for the equilibrium., , 7.4, , 7.5, , 7.6, , Write the expression for the equilibrium constant, Kc for each of the following, reactions:, 2NO (g) + Cl2 (g), , (i), , 2NOCl (g), , (ii), , 2Cu(NO3)2 (s), , (iii), , CH3COOC2H5(aq) + H2O(l), , (iv), , Fe3+ (aq) + 3OH (aq), , (v), , I2 (s) + 5F2, , 2CuO (s) + 4NO2 (g) + O2 (g), –, , CH3COOH (aq) + C2H5OH (aq), Fe(OH)3 (s), , 2IF5, , Find out the value of Kc for each of the following equilibria from the value of Kp:, (i), , 2NOCl (g), , 2NO (g) + Cl2 (g); Kp= 1.8 × 10–2 at 500 K, , (ii), , CaCO3 (s), , CaO(s) + CO2(g); Kp= 167 at 1073 K, , For the following equilibrium, Kc= 6.3 × 1014 at 1000 K, , NO (g) + O3 (g), , NO2 (g) + O2 (g), , Both the forward and reverse reactions in the equilibrium are elementary, bimolecular reactions. What is Kc, for the reverse reaction?, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 233, , 7.7, , Explain why pure liquids and solids can be ignored while writing the equilibrium, constant expression?, , 7.8, , Reaction between N2 and O2– takes place as follows:, 2N2 (g) + O2 (g), , 2N2O (g), , If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction, vessel and allowed to form N2O at a temperature for which Kc= 2.0 × 10 –37,, determine the composition of equilibrium mixture., 7.9, , Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given, below:, 2NOBr (g), , 2NO (g) + Br2 (g), , When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at, constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate, equilibrium amount of NO and Br2 ., 7.10, , At 450K, Kp= 2.0 × 1010/bar for the given reaction at equilibrium., 2SO2(g) + O2(g), , 2SO3 (g), , What is Kc at this temperature ?, 7.11, , A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the, partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium ?, , 7.12, , A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into, a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant,, Kc for the reaction N2 (g) + 3H2 (g), 2NH3 (g) is 1.7 × 102. Is the reaction mixture, at equilibrium? If not, what is the direction of the net reaction?, , 7.13, , The equilibrium constant expression for a gas reaction is,, , H2 (g) + I2 (g), , 2HI (g), , Kc =, , 4, 5, [NH3 ] [O2 ], 6, [ NO]4 [H2O], , Write the balanced chemical equation corresponding to this expression., 7.14, , One mole of H2O and one mole of CO are taken in 10 L vessel and heated to, 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the, equation,, H2O (g) + CO (g), , H2 (g) + CO2 (g), , Calculate the equilibrium constant for the reaction., 7.15, , At 700 K, equilibrium constant for the reaction:, H2 (g) + I2 (g), , 2HI (g), , –1, , is 54.8. If 0.5 mol L of HI(g) is present at equilibrium at 700 K, what are the, concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and, allowed it to reach equilibrium at 700 K?, 7.16, , What is the equilibrium concentration of each of the substances in the, equilibrium when the initial concentration of ICl was 0.78 M ?, , 7.17, , Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium, concentration of C2H6 when it is placed in a flask at 4.0 atm pressure and allowed, to come to equilibrium?, , 2ICl (g), , I2 (g) + Cl2 (g);, , Kc = 0.14, , C2H6 (g), , C2H4 (g) + H2 (g), , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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234, , CHEMISTRY, , 7.18, , Ethyl acetate is formed by the reaction between ethanol and acetic acid and the, equilibrium is represented as:, CH3COOH (l) + C2H5OH (l), , CH3COOC2H5 (l) + H2O (l), , (i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note:, water is not in excess and is not a solvent in this reaction), (ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol,, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate, the equilibrium constant., (iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining, it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium, been reached?, 7.19, , A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After, equilibrium was attained, concentration of PCl 5 was found to be, 0.5 × 10–1 mol L–1. If value of Kc is 8.3 × 10–3, what are the concentrations of PCl3, and Cl2 at equilibrium?, PCl5 (g), PCl3 (g) + Cl2(g), , 7.20, , One of the reaction that takes place in producing steel from iron ore is the, reduction of iron(II) oxide by carbon monoxide to give iron metal and CO2., FeO (s) + CO (g), , Fe (s) + CO2 (g);, , Kp = 0.265 atm at 1050K, , What are the equilibrium partial pressures of CO and CO2 at 1050 K if the, initial partial pressures are: pCO = 1.4 atm and, 7.21, , = 0.80 atm?, , Equilibrium constant, Kc for the reaction, N2 (g) + 3H2 (g), , 2NH3 (g) at 500 K is 0.061, , At a particular time, the analysis shows that composition of the reaction mixture, is 3.0 mol L–1 N2, 2.0 mol L–1 H2 and 0.5 mol L–1 NH3. Is the reaction at equilibrium?, If not in which direction does the reaction tend to proceed to reach equilibrium?, 7.22, , Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches, the equilibrium:, Br2 (g) + Cl2 (g), 2BrCl (g), for which Kc= 32 at 500 K. If initially pure BrCl is present at a concentration of, 3.3 × 10–3 mol L–1, what is its molar concentration in the mixture at equilibrium?, , 7.23, , At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO 2 in equilibrium, with soild carbon has 90.55% CO by mass, C (s) + CO2 (g), , 2CO (g), , Calculate Kc for this reaction at the above temperature., 7.24, , , , Calculate a) ∆G and b) the equilibrium constant for the formation of NO2 from, NO and O2 at 298K, NO (g) + ½ O2 (g), , NO2 (g), , where, , , ∆fG (NO2) = 52.0 kJ/mol, , , ∆fG (NO) = 87.0 kJ/mol, , , ∆fG (O2) = 0 kJ/mol, 7.25, , Does the number of moles of reaction products increase, decrease or remain, same when each of the following equilibria is subjected to a decrease in pressure, by increasing the volume?, , (a), , PCl5 (g), , PCl3 (g) + Cl2 (g), , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 235, , (b), , CaO (s) + CO2 (g), , CaCO3 (s), , (c), , 3Fe (s) + 4H2O (g), , Fe3O4 (s) + 4H2 (g), , 7.26, , Which of the following reactions will get affected by increasing the pressure?, Also, mention whether change will cause the reaction to go into forward or, backward direction., , (i), , COCl2 (g), , (ii), , CH4 (g) + 2S2 (g), , (iii), , CO2 (g) + C (s), , (iv), , 2H2 (g) + CO (g), , (v), , CaCO3 (s), , (vi), , 4 NH3 (g) + 5O2 (g), , 7.27, , The equilibrium constant for the following reaction is 1.6 ×105 at 1024K, , CO (g) + Cl2 (g), CS2 (g) + 2H2S (g), 2CO (g), CH3OH (g), CaO (s) + CO2 (g), , H2(g) + Br2(g), , 4NO (g) + 6H2O(g), 2HBr(g), , Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a, sealed container at 1024K., 7.28, , Dihydrogen gas is obtained from natural gas by partial oxidation with steam as, per following endothermic reaction:, CH4 (g) + H2O (g), , CO (g) + 3H2 (g), , (a) Write as expression for Kp for the above reaction., (b) How will the values of Kp and composition of equilibrium mixture be affected, by, (i), , increasing the pressure, , (ii) increasing the temperature, (iii) using a catalyst ?, 7.29, , Describe the effect of :, a), b), c), d), on, , addition of H2, addition of CH3OH, removal of CO, removal of CH3OH, the equilibrium of the reaction:, 2H2(g) + CO (g), , 7.30, , CH3OH (g), , At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride,, PCl5 is 8.3 ×10-3. If decomposition is depicted as,, PCl5 (g), , , , ∆rH = 124.0 kJ mol–1, , PCl3 (g) + Cl2 (g), , a), , write an expression for Kc for the reaction., , b), , what is the value of Kc for the reverse reaction at the same temperature ?, , c), , what would be the effect on Kc if (i) more PCl5 is added (ii) pressure is increased, (iii) the temperature is increased ?, , 7.31, , Dihydrogen gas used in Haber’s process is produced by reacting methane from, natural gas with high temperature steam. The first stage of two stage reaction, involves the formation of CO and H2. In second stage, CO formed in first stage is, reacted with more steam in water gas shift reaction,, CO (g) + H2O (g), , CO2 (g) + H2 (g), , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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236, , CHEMISTRY, , If a reaction vessel at 400 °C is charged with an equimolar mixture of CO and, steam such that pco = pH O = 4.0 bar, what will be the partial pressure of H2 at, 2, equilibrium? Kp= 10.1 at 400°C, 7.32, , Predict which of the following reaction will have appreciable concentration of, reactants and products:, 2Cl (g) Kc = 5 ×10–39, , a) Cl2 (g), , b) Cl2 (g) + 2NO (g), , 2NOCl (g), , Kc = 3.7 × 108, , c) Cl2 (g) + 2NO2 (g), , 2NO2Cl (g), , Kc = 1.8, , 7.33, , 2O3 (g) is 2.0 ×10–50 at 25°C. If the, The value of Kc for the reaction 3O2 (g), equilibrium concentration of O 2 in air at 25°C is 1.6 ×10 –2, what is the, concentration of O3?, , 7.34, , The reaction, CO(g) + 3H2(g), , CH4(g) + H2O(g), , is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol, of H2 and 0.02 mol of H2O and an unknown amount of CH4 in the flask. Determine, the concentration of CH4 in the mixture. The equilibrium constant, Kc for the, reaction at the given temperature is 3.90., 7.35, , What is meant by the conjugate acid-base pair? Find the conjugate acid/base, for the following species:, –, , –, , –, , 2–, , HNO2, CN , HClO4, F , OH , CO3 , and S, , 2–, +, , +, , 7.36, , Which of the followings are Lewis acids? H2O, BF3, H , and NH4, , 7.37, , What will be the conjugate bases for the Brönsted acids: HF, H2SO4 and HCO 3?, , 7.38, , Write the conjugate acids for the following Brönsted bases: NH2 , NH3 and HCOO ., , 7.39, , The species: H2O, HCO3, HSO4 and NH3 can act both as Brönsted acids and, bases. For each case give the corresponding conjugate acid and base., , 7.40, , Classify the following species into Lewis acids and Lewis bases and show how, –, –, these act as Lewis acid/base: (a) OH (b) F (c) H+ (d) BCl3 ., , 7.41, , The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10–3 M. what, is its pH?, , 7.42, , The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen, ion in it., , 7.43, , The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10 –4,, 1.8 × 10–4 and 4.8 × 10–9 respectively. Calculate the ionization constants of the, corresponding conjugate base., , 7.44, , The ionization constant of phenol is 1.0 × 10–10. What is the concentration of, phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization, if the solution is also 0.01M in sodium phenolate?, , 7.45, , The first ionization constant of H2S is 9.1 × 10–8. Calculate the concentration of, –, HS ion in its 0.1M solution. How will this concentration be affected if the solution, is 0.1M in HC l also ? If the second dissociation constant of H 2 S is, 1.2 × 10–13, calculate the concentration of S2– under both conditions., , 7.46, , The ionization constant of acetic acid is 1.74 × 10–5. Calculate the degree of, dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of, acetate ion in the solution and its pH., , 7.47, , It has been found that the pH of a 0.01M solution of an organic acid is 4.15., Calculate the concentration of the anion, the ionization constant of the acid, and its pKa., , –, , –, , –, , –, , –, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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EQUILIBRIUM, , 7.48, , 237, , Assuming complete dissociation, calculate the pH of the following solutions:, (a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH, , 7.49, , Calculate the pH of the following solutions:, a), , 2 g of TlOH dissolved in water to give 2 litre of solution., , b), , 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of solution., , c), , 0.3 g of NaOH dissolved in water to give 200 mL of solution., , d), , 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution., , 7.50, , The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate, the pH of the solution and the pKa of bromoacetic acid., , 7.51, , The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate its ionization, constant and pKb., , 7.52, , What is the pH of 0.001M aniline solution ? The ionization constant of aniline, can be taken from Table 7.7. Calculate the degree of ionization of aniline in the, solution. Also calculate the ionization constant of the conjugate acid of aniline., , 7.53, , Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74., How is the degree of dissociation affected when its solution also contains, (a) 0.01M (b) 0.1M in HCl ?, , 7.54, , The ionization constant of dimethylamine is 5.4 × 10–4. Calculate its degree of, ionization in its 0.02M solution. What percentage of dimethylamine is ionized if, the solution is also 0.1M in NaOH?, , 7.55, , Calculate the hydrogen ion concentration in the following biological fluids whose, pH are given below:, (a) Human muscle-fluid, 6.83, , (b), , Human stomach fluid, 1.2, , (c) Human blood, 7.38, , (d), , Human saliva, 6.4., , 7.56, , The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8,, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion, concentration in each., , 7.57, , If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K., Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What, is its pH?, , 7.58, , The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the, concentrations of strontium and hydroxyl ions and the pH of the solution., , 7.59, , The ionization constant of propanoic acid is 1.32 × 10–5. Calculate the degree of, ionization of the acid in its 0.05M solution and also its pH. What will be its, degree of ionization if the solution is 0.01M in HCl also?, , 7.60, , The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization, constant of the acid and its degree of ionization in the solution., , 7.61, , The ionization constant of nitrous acid is 4.5 × 10–4. Calculate the pH of 0.04 M, sodium nitrite solution and also its degree of hydrolysis., , 7.62, , A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the, ionization constant of pyridine., , 7.63, , Predict if the solutions of the following salts are neutral, acidic or basic:, NaCl, KBr, NaCN, NH4NO3, NaNO2 and KF, , 7.64, , The ionization constant of chloroacetic acid is 1.35 × 10–3. What will be the pH of, 0.1M acid and its 0.1M sodium salt solution?, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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238, , CHEMISTRY, , 7.65, , Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at, this temperature?, , 7.66, , Calculate the pH of the resultant mixtures:, a), , 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl, , b), , 10 mL of 0.01M H2SO4 + 10 mL of 0.01M Ca(OH)2, , c), , 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH, , 7.67, , Determine the solubilities of silver chromate, barium chromate, ferric hydroxide,, lead chloride and mercurous iodide at 298K from their solubility product, constants given in Table 7.9. Determine also the molarities of individual ions., , 7.68, , The solubility product constant of Ag 2CrO 4 and AgBr are 1.1 × 10 –12 and, 5.0 × 10–13 respectively. Calculate the ratio of the molarities of their saturated, solutions., , 7.69, , Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are, mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate, Ksp = 7.4 × 10–8 )., , 7.70, , The ionization constant of benzoic acid is 6.46 × 10–5 and Ksp for silver benzoate, is 2.5 × 10–13. How many times is silver benzoate more soluble in a buffer of pH, 3.19 compared to its solubility in pure water?, , 7.71, , What is the maximum concentration of equimolar solutions of ferrous sulphate, and sodium sulphide so that when mixed in equal volumes, there is no, precipitation of iron sulphide? (For iron sulphide, Ksp = 6.3 × 10–18)., , 7.72, , What is the minimum volume of water required to dissolve 1g of calcium sulphate, at 298 K? (For calcium sulphate, Ksp is 9.1 × 10–6)., , 7.73, , The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen, sulphide is 1.0 × 10–19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of, the following: FeSO4, MnCl 2, ZnCl2 and CdCl2. in which of these solutions, precipitation will take place?, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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APPENDICES, , 239, , Appendix I, , Definitions of the SI Base Units, , Metre (m): The metre is the length of path travelled by light in vacuum during a time, interval of 1/299 792 458 of a second (17th CGPM, 1983)., Kilogram (kg): The kilogram is the unit of mass; it is equal to the mass of the, international prototype of the kilogram (3rd CGPM, 1901)., Second (s): The second is the duration of 9192631770 periods of the radiation, corresponding to the transition between the two hyperfine levels of the ground state, of the caesium-133 atom (13th CGPM, 1967)., Ampere (A): The ampere is that constant current which, if maintained in two straight, parallel conductors of infinite length, of negligible circular cross-section, and placed, 1 metre apart in vacuum, would produce between these conductors a force equal to, 2 × 10-7 Newton per metre of length (9th CGPM, 1948)., Kelvin (K): The kelvin, unit of thermodynamic temperature, is the fraction 1/273.16, of the thermodynamic temperature of the triple point of water (13th CGPM, 1967)., Mole (mol): The mole is the amount of substance of a system which contains as, many elementary entities as there are atoms in 0.012 kilogram of carbon-12. When, the mole is used, the elementary entities must be specified and may be atoms,, molecules, ions, electrons, other particles, or specified groups of such particles, (14th CGPM, 1971)., Candela (cd): The candela is the luminous intensity, in a given direction, of a source, that emits monochromatic radiation of frequency 540 × 1012 hertz and that has a, radiant intensity in that direction of (1/683) watt per steradian (16th CGPM, 1979)., (The symbols listed here are internationally agreed and should not be changed in, other languages or scripts)., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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240, , CHEMISTRY, , Appendix II, , Elements, their Atomic Number and Molar Mass, , Element, , Actinium, Aluminium, Americium, Antimony, Argon, Arsenic, Astatine, Barium, Berkelium, Beryllium, Bismuth, Bohrium, Boron, Bromine, Cadmium, Caesium, Calcium, Californium, Carbon, Cerium, Chlorine, Chromium, Cobalt, Copper, Curium, Dubnium, Dysprosium, Einsteinium, Erbium, Europium, Fermium, Fluorine, Francium, Gadolinium, Gallium, Germanium, Gold, Hafnium, Hassium, Helium, Holmium, Hydrogen, Indium, Iodine, Iridium, Iron, Krypton, Lanthanum, Lawrencium, Lead, Lithium, Lutetium, Magnesium, Manganese, Meitneium, Mendelevium, , Symbol, , Atomic, Number, , Molar, mass/, (g mol–1), , Element, , Ac, Al, Am, Sb, Ar, As, At, Ba, Bk, Be, Bi, Bh, B, Br, Cd, Cs, Ca, Cf, C, Ce, Cl, Cr, Co, Cu, Cm, Db, Dy, Es, Er, Eu, Fm, F, Fr, Gd, Ga, Ge, Au, Hf, Hs, He, Ho, H, In, I, Ir, Fe, Kr, La, Lr, Pb, Li, Lu, Mg, Mn, Mt, Md, , 89, 13, 95, 51, 18, 33, 85, 56, 97, 4, 83, 107, 5, 35, 48, 55, 20, 98, 6, 58, 17, 24, 27, 29, 96, 105, 66, 99, 68, 63, 100, 9, 87, 64, 31, 32, 79, 72, 108, 2, 67, 1, 49, 53, 77, 26, 36, 57, 103, 82, 3, 71, 12, 25, 109, 101, , 227.03, 26.98, (243), 121.75, 39.95, 74.92, 210, 137.34, (247), 9.01, 208.98, (264), 10.81, 79.91, 112.40, 132.91, 40.08, 251.08, 12.01, 140.12, 35.45, 52.00, 58.93, 63.54, 247.07, (263), 162.50, (252), 167.26, 151.96, (257.10), 19.00, (223), 157.25, 69.72, 72.61, 196.97, 178.49, (269), 4.00, 164.93, 1.0079, 114.82, 126.90, 192.2, 55.85, 83.80, 138.91, (262.1), 207.19, 6.94, 174.96, 24.31, 54.94, (268), 258.10, , Mercury, Molybdenum, Neodymium, Neon, Neptunium, Nickel, Niobium, Nitrogen, Nobelium, Osmium, Oxygen, Palladium, Phosphorus, Platinum, Plutonium, Polonium, Potassium, Praseodymium, Promethium, Protactinium, Radium, Radon, Rhenium, Rhodium, Rubidium, Ruthenium, Rutherfordium, Samarium, Scandium, Seaborgium, Selenium, Silicon, Silver, Sodium, Strontium, Sulphur, Tantalum, Technetium, Tellurium, Terbium, Thallium, Thorium, Thulium, Tin, Titanium, Tungsten, Ununbium, Ununnilium, Unununium, Uranium, Vanadium, Xenon, Ytterbium, Yttrium, Zinc, Zirconium, , Symbol, , Atomic, Number, , Hg, Mo, Nd, Ne, Np, Ni, Nb, N, No, Os, O, Pd, P, Pt, Pu, Po, K, Pr, Pm, Pa, Ra, Rn, Re, Rh, Rb, Ru, Rf, Sm, Sc, Sg, Se, Si, Ag, Na, Sr, S, Ta, Tc, Te, Tb, Tl, Th, Tm, Sn, Ti, W, Uub, Uun, Uuu, U, V, Xe, Yb, Y, Zn, Zr, , 80, 42, 60, 10, 93, 28, 41, 7, 102, 76, 8, 46, 15, 78, 94, 84, 19, 59, 61, 91, 88, 86, 75, 45, 37, 44, 104, 62, 21, 106, 34, 14, 47, 11, 38, 16, 73, 43, 52, 65, 81, 90, 69, 50, 22, 74, 112, 110, 111, 92, 23, 54, 70, 39, 30, 40, , The value given in parenthesis is the molar mass of the isotope of largest known half-life., , Download all NCERT books PDFs from www.ncert.online, , 2019-20, , Molar, mass/, (g mol–1), 200.59, 95.94, 144.24, 20.18, (237.05), 58.71, 92.91, 14.0067, (259), 190.2, 16.00, 106.4, 30.97, 195.09, (244), 210, 39.10, 140.91, (145), 231.04, (226), (222), 186.2, 102.91, 85.47, 101.07, (261), 150.35, 44.96, (266), 78.96, 28.08, 107.87, 22.99, 87.62, 32.06, 180.95, (98.91), 127.60, 158.92, 204.37, 232.04, 168.93, 118.69, 47.88, 183.85, (277), (269), (272), 238.03, 50.94, 131.30, 173.04, 88.91, 65.37, 91.22

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APPENDICES, , 241, , Appendix III, A. Specific and Molar Heat Capacities for Some Substances at 298 K and, one Atmospheric Pressure, Substance, , Specific Heat Capacity, (J/g), , Molar Heat Capacity, (J/mol), , 0.720, 4.184, 2.06, 0.797, 0.360, 4.70, 2.46, 2.42, 2.06, 0.861, 0.5980, 0.817, 1.03, 0.477, 0.473, 0.460, 0.385, 0.902, 0.128, 0.720, , 20.8, 75.4, 35.1, 29.1, 29.1, 79.9, 113.16, 152.52, 37.08, 132.59, 72.35, 39.2, 20.7, 33.8, 75.6, 25.1, 24.7, 24.35, 25.2, 8.65, , air, water (liquid), ammonia (gas), hydrogen chloride, hydrogen bromide, ammonia (liquid), ethyl alcohol (liquid), ethylene glycol (liquid), water (solid), carbon tetrachloride (liquid), chlorofluorocarbon (CCl F ), 2 2, ozone, neon, chlorine, bromine, iron, copper, aluminium, gold, graphite, B., , Molar Heat Capacities for Some Gases (J/mol), , Gas, Monatomic*, helium, argon, iodine, mercury, Diatomic†, hydrogen, oxygen, nitrogen, hydrogen chloride, carbon monoxide, Triatomic†, nitrous oxide, carbon dioxide, Polyatomic†, ethane, , Cp, , Cv, , Cp - Cv, , C p / Cv, , 20.9, 20.8, 20.9, 20.8, , 12.8, 12.5, 12.6, 12.5, , 8.28, 8.33, 8.37, 8.33, , 1.63, 1.66, 1.66, 1.66, , 28.6, 29.1, 29.0, 29.6, 29.0, , 20.2, 20.8, 20.7, 21.0, 21.0, , 8.33, 8.33, 8.30, 8.60, 8.00, , 1.41, 1.39, 1.40, 1.39, 1.41, , 39.0, 37.5, , 30.5, 29.0, , 8.50, 8.50, , 1.28, 1.29, , 53.2, , 44.6, , 8.60, , 1.19, , *Translational kinetic energy only., †Translational, vibrational and rotational energy., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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242, , CHEMISTRY, , Appendix IV, , Physical Constants, , Quantity, , Symbol Traditional Units, , Acceleration of gravity, Atomic mass unit (1/12, the mass of 12C atom), Avogadro constant, Bohr radius, , g, amu, or u, NA, ao, , Boltzmann constant, Charge-to-mass, ratio of electron, Electronic charge, , k, e/m, e, , Electron rest mass, , me, , Faraday constant, , F, , Gas constant, , R, , 980.6 cm/s, 1.6606 × 10-24 g, , 9.806 m/s, 1.6606 × 10-27 kg, , 6.022 ×1023, particles/mol, 0.52918 , 5.2918 × 10-9 cm, 1.3807 × 10-16 erg/K, 1.758820 ×l08 coulomb/g, , 6.022 × 1023, particles/mol, 5.2918 × 10-11 m, , 1.602176 × 10-19 coulomb, 4.8033 × 10-19 esu, 9.109382 ×10-28 g, 0.00054859 u, 96,487 coulombs/eq, 23.06 kcal/volt. eq, , 1.60219 × 10-19 C, , 0.8206, , 1.987, , L atm, mol K, cal, , mol K, , Molar volume (STP), , Vm, , 22.710981 L/mol, , Neutron rest mass, , mn, , Planck constant, Proton rest mass, , h, mp, , Rydberg constant, , R∞, , 1.674927 × 10-24 g, 1.008665 u, 6.6262 × 10-27 ergs, 1.6726216 ×10-24 g, 1.007277 u, 3.289 × 1015 cycles/s, 2.1799 × 10-11 erg, 2.9979 ×1010 cm/s, (186,281 miles/second), , Speed of light, (in a vacuum), , c, , SI Units, , 1.3807 × 10-23 J/K, 1.7588 × 1011 C/kg, , 9.10952 ×10-31 kg, 96,487 C/mol e96,487 J/V.mol e-, , 8.3145, , 3, , mol K, , 8.3145 J/mol.K, 22.710981 × 10-3 m3/mol, 22.710981 dm3/mol, 1.67495 × 10-27 kg, 6.6262 × 10-34 J s, 1.6726 ×10-27 kg, 1.0974 × 107 m-1, 2.1799 × 10-18 J, 2.9979 × 108 m/s, , π = 3.1416, , 2.303 R = 4.576 cal/mol K = 19.15 J/mol K, , e = 2.71828, , 2.303 RT (at 25°C) = 1364 cal/mol = 5709 J/mol, , ln X = 2.303 log X, , Download all NCERT books PDFs from www.ncert.online, , 2019-20, , kPa dm

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APPENDICES, , 243, , Appendix V, , Some Useful Conversion Factors, , Common Unit of Mass and Weight, 1 pound = 453.59 grams, , Common Units of Length, 1 inch = 2.54 centimetres (exactly), , 1 pound = 453.59 grams = 0.45359 kilogram, 1 kilogram = 1000 grams = 2.205 pounds, 1 gram = 10 decigrams = 100 centigrams, = 1000 milligrams, 1 gram = 6.022 × 1023 atomic mass units or u, 1 atomic mass unit = 1.6606 × 10–24 gram, 1 metric tonne = 1000 kilograms, = 2205 pounds, , 1 mile = 5280 feet = 1.609 kilometres, 1 yard = 36 inches = 0.9144 metre, 1 metre = 100 centimetres = 39.37 inches, = 3.281 feet, = 1.094 yards, 1 kilometre = 1000 metres = 1094 yards, = 0.6215 mile, 1 Angstrom = 1.0 × 10–8 centimetre, = 0.10 nanometre, = 1.0 × 10–10 metre, = 3.937 × 10–9 inch, , Common Unit of Volume, 1 quart = 0.9463 litre, 1 litre = 1.056 quarts, , Common Units of Force* and Pressure, , 1 litre = 1 cubic decimetre = 1000 cubic, centimetres = 0.001 cubic metre, 1 millilitre = 1 cubic centimetre = 0.001 litre, = 1.056 × 10-3 quart, 1 cubic foot = 28.316 litres = 29.902 quarts, = 7.475 gallons, , 1 atmosphere = 760 millimetres of mercury, = 1.013 × 105 pascals, = 14.70 pounds per square inch, 1 bar = 105 pascals, 1 torr = 1 millimetre of mercury, 1 pascal = 1 kg/ms2 = 1 N/m2, , Common Units of Energy, 1 joule = 1 × 107 ergs, , Temperature, SI Base Unit: Kelvin (K), , 1 thermochemical calorie**, = 4.184 joules, = 4.184 × 107 ergs, = 4.129 × 10–2 litre-atmospheres, = 2.612 × 1019 electron volts, 1 ergs = 1 × 10–7 joule = 2.3901 × 10–8 calorie, 1 electron volt = 1.6022 × 10–19 joule, = 1.6022 × 10–12 erg, = 96.487 kJ/mol†, 1 litre-atmosphere = 24.217 calories, = 101.32 joules, = 1.0132 ×109 ergs, 1 British thermal unit = 1055.06 joules, = 1.05506 ×1010 ergs, = 252.2 calories, , K = -273.15°C, K = °C + 273.15, °F = 1.8(°C) + 32, , °C =, , °F − 32, 1.8, , * Force: 1 newton (N) = 1 kg m/s2, i.e.,the force that, when applied for 1 second, gives a, 1-kilogram mass a velocity of 1 metre per second., ** The amount of heat required to raise the temperature of one gram of water from 14.50C to, 15.50C., † Note that the other units are per particle and must be multiplied by 6.022 ×1023 to be strictly, comparable., , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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244, , CHEMISTRY, , Appendix VI, , Thermodynamic Data at 298 K, INORGANIC SUBSTANCES, Substance, , Enthalpy of formation,, ∆fH/ (kJ mol–1), , Gibbs Energy of formation,, ∆fG/ (kJ mol–1), , Entropy,*, S/(J K–1 mol–1), , Aluminium, Al(s), Al3+(aq), Al2O3(s), Al(OH)3(s), AlCl3(s), , 0, – 524.7, –1675.7, –1276, –704.2, , 0, –481.2, –1582.3, —, –628.8, , 28.33, –321.7, 50.92, —, 110.67, , Antimony, SbH3(g), SbCl3(g), SbCl5(g), , 145.11, –313.8, –394.34, , 147.75, –301.2, –334.29, , 232.78, 337.80, 401.94, , 0, –169.0, –888.14, , 0, –168.6, –648.41, , 0, –537.64, –553.5, –1216.3, –1214.78, , 0, –560.77, –525.1, –1137.6, –1088.59, , 62.8, 9.6, 70.42, 112.1, –47.3, , 0, –1272.8, –1137.0, , 0, –1193.7, –1120.3, , 5.86, 53.97, 254.12, , Arsenic, As(s), gray, As2S3(s), AsO43–(aq), , 35.1, 163.6, –162.8, , Barium, Ba(s), Ba2+(aq), BaO(s), BaCO3(s), BaCO3(aq), Boron, B(s), B2O3(s), BF3(g), Bromine, Br2(l), Br2(g), Br(g), Br–(aq), HBr(g), BrF3(g), , 0, 30.91, 111.88, –121.55, –36.40, –255.60, , 0, 3.11, 82.40, –103.96, –53.45, –229.43, , 152.23, 245.46, 175.02, 82.4, 198.70, 292.53, , 0, 178.2, –542.83, , 0, 144.3, –553.58, , 41.42, 154.88, –53.1, , Calcium, Ca(s), Ca(g), Ca2+(aq), , (continued), Download all NCERT books PDFs from www.ncert.online, , 2019-20

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APPENDICES, , Substance, , 245, , Enthalpy of formation,, ∆fH/ (kJ mol–1), , Gibbs Energy of formation,, ∆fG/ (kJ mol–1), , Entropy,*, S/(J K–1 mol–1), , Calcium (continued), CaO(s), Ca(OH)2(s), Ca(OH)2(aq), CaCO3(s), calcite, CaCO3(s), aragonite, CaCO3(aq), CaF2(s), CaF2(aq), CaCl2(s), CaCl2(aq), CaBr2(s), CaC2(s), CaS(s), CaSO4(s), CaSO4(aq), , –635.09, –986.09, –1002.82, –1206.92, –1207.1, –1219.97, –1219.6, –1208.09, –795.8, –877.1, –682.8, –59.8, –482.4, –1434.11, –1452.10, , –604.03, –898.49, –868.07, –1128.8, –1127.8, –1081.39, –1167.3, –1111.15, –748.1, –816.0, –663.6, –64.9, –477.4, –1321.79, –1298.10, , 39.75, 83.39, –74.5, 92.9, 88.7, –110.0, 68.87, –80.8, 104.6, 59.8, 130, 69.96, 56.5, 106.7, –33.1, , Carbon**, C(s), graphite, C(s), diamond, C(g), CO(g), CO2(g), CO32–(aq), CCl4(l), CS2(l), HCN(g), HCN(l), , 0, 1.895, 716.68, –110.53, –393.51, –677.14, –135.44, 89.70, 135.1, 108.87, , 0, 2.900, 671.26, –137.17, –394.36, –527.81, –65.21, 65.27, 124.7, 124.97, , 5.740, 2.377, 158.10, 197.67, 213.74, –56.9, 216.40, 151.34, 201.78, 112.84, , 0, –696.2, –537.2, , 0, –672.0, –503.8, , 0, 121.68, –167.16, –92.31, –167.16, , 0, 105.68, –131.23, –95.30, –131.23, , 223.07, 165.20, 56.5, 186.91, 56.5, , 0, 71.67, 64.77, –168.6, –157.3, –771.36, –2279.7, , 0, 49.98, 65.49, –146.0, –129.7, –661.8, –1879.7, , 33.15, 40.6, –99.6, 93.14, 42.63, 109, 300.4, , Cerium, Ce(s), Ce3+(aq), Ce4+(aq), , 72.0, –205, –301, , Chlorine, Cl2(g), Cl(g), Cl–(aq), HCl(g), HCl(aq), Copper, Cu(s), Cu+ (aq), Cu2+(aq), Cu2O(aq), CuO(s), CuSO4(s), CuSO4.5H2O(s), , ** For organic compounds, a separate table is provided in continuation., Download all NCERT books PDFs from www.ncert.online, , 2019-20, , (continued)

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246, , Substance, , CHEMISTRY, , Enthalpy of formation,, ∆fH/ (kJ mol–1), , Gibbs Energy of formation,, ∆fG/ (kJ mol–1), , Entropy,*, S/(J K–1 mol–1), , Deuterium, D2(g), D2O(g), D2 O(l), , 0, –249.20, –294.60, , 0, –234.54, –243.44, , 144.96, 198.34, 75.94, , Fluorine, F2(g), F–(aq), HF(g), HF(aq), , 0, –332.63, –271.1, –332.63, , 0, –278.79, –273.2, –278.79, , 202.78, –13.8, 173.78, –13.8, , Hydrogen (see also Deuterium), H2(g), 0, H(g), 217.97, H+ (aq), 0, H2 O(l), –285.83, H2O(g), –241.82, –187.78, H2O 2(l), H2O2(aq), –191.17, , 0, 203.25, 0, –237.13, –228.57, –120.35, –134.03, , 130.68, 114.71, 0, 69.91, 188.83, 109.6, 143.9, , 0, 19.33, –51.57, 1.70, , 116.14, 260.69, 111.3, 206.59, , 0, –78.90, –4.7, –1015.4, –742.2, –100.4, 6.9, –166.9, , 27.28, –137.7, –315.9, 146.4, 87.40, 60.29, —, 52.93, , 0, –1.7, –277.4, –919.94, –278.7, –244.8, , 0, –24.43, –217.33, –813.14, –261.92, –232.34, , 64.81, 10.5, 68.6, 148.57, 161.5, 175.3, , 0, 147.70, –466.85, –601.70, –1095.8, –524.3, , 0, 113.10, –454.8, –569.43, –1012.1, –503.8, , 32.68, 148.65, –138.1, 26.94, 65.7, 117.2, , Iodine, I2(s), I2(g), I–(aq), HI(g), Iron, Fe(s), Fe2+(aq), Fe3+(aq), Fe3O4(s), magnetite, Fe2O3(s), haematite, FeS(s,α), FeS(aq), FeS2(s), , 0, 62.44, –55.19, 26.48, , 0, –89.1, –48.5, –1118.4, –824.2, –100.0, —, –178.2, , Lead, Pb(s), Pb2+(aq), PbO2(s), PbSO4(s), PbBr2(s), PbBr2(aq), Magnesium, Mg(s), Mg(g), Mg2+(aq), MgO(s), MgCO3(s), MgBr2(s), , (continued), Download all NCERT books PDFs from www.ncert.online, , 2019-20

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APPENDICES, , Substance, , 247, , Enthalpy of formation,, ∆fH/ (kJ mol–1), , Gibbs Energy of formation,, ∆fG/ (kJ mol–1), , Entropy,*, S/(J K–1 mol–1), , Mercury, Hg(1), Hg(g), HgO(s), Hg2Cl2(s), , 0, 61.32, –90.83, –265.22, , 0, 31.82, –58.54, –210.75, , 76.02, 174.96, 70.29, 192.5, , 0, 90.25, 82.05, 33.18, 9.16, –174.10, –207.36, –205.0, –46.11, –80.29, –132.51, –114.2, 294.1, 50.63, –365.56, –314.43, –295.31, , 0, 86.55, 104.20, 51.31, 97.89, –80.71, –111.25, –108.74, –16.45, –26.50, –79.31, —, 328.1, 149.34, –183.87, –202.87, –88.75, , 191.61, 210.76, 219.85, 240.06, 304.29, 155.60, 146.4, 146.4, 192.45, 111.3, 113.4, —, 238.97, 121.21, 151.08, 94.6, 186.2, , 0, 142.7, –229.99, , 0, 163.2, –157.24, , 205.14, 238.93, –10.75, , 0, 58.91, 5.4, –2984.0, –964.8, –1266.9, –1277.4, –319.7, –287.0, –374.9, , 0, 24.44, 13.4, –2697.0, —, —, –1018.7, –272.3, –267.8, –305.0, , 41.09, 279.98, 210.23, 228.86, —, —, —, 217.18, 311.78, 364.6, , 0, 89.24, –252.38, –424.76, –482.37, –567.27, , 0, 60.59, –283.27, –379.08, –440.50, –537.75, , 64.18, 160.34, 102.5, 78.9, 91.6, 66.57, , Nitrogen, N2(g), NO(g), N2O(g), NO2(g), N2O4(g), HNO3(1), HNO3(aq), NO3– (aq), NH3(g), NH3(aq), NH+4 (aq), NH2OH(s), HN3(g), N2H4(1), NH4NO3(s), NH4Cl(s), NH4ClO4(s), Oxygen, O2(g), O3(g), OH–(aq), Phosphorus, P(s), white, P4(g), PH3(g), P4O10(s), H3PO3(aq), H3PO4(1), H3PO4(aq), PCl3(1), PCl3(g), PCl5(g), Potassium, K(s), K(g), K+ (aq), KOH(s), KOH(aq), KF(s), , (continued), Download all NCERT books PDFs from www.ncert.online, , 2019-20

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248, , Substance, , CHEMISTRY, , Enthalpy of formation,, ∆fH/ (kJ mol–1), , Gibbs Energy of formation,, ∆fG/ (kJ mol–1), , Entropy,*, S/(J K–1 mol–1), , Potassium (continued), KCl(s), KBr(s), KI(s), KClO3(s), KClO4(s), K2S(s), K2S(aq), , –436.75, –393.80, –327.90, –397.73, –432.75, –380.7, –471.5, , –409.14, –380.66, –324.89, –296.25, –303.09, –364.0, –480.7, , 82.59, 95.90, 106.32, 143.1, 151.0, 105, 190.4, , Si(s), SiO2(s,α), , 0, –910.94, , 0, –856.64, , 18.83, 41.84, , Silver, Ag(s), Ag+(aq), Ag2O(s), AgBr(s), AgBr(aq), AgCl(s), AgCl(aq), AgI(s), AgI(aq), AgNO3(s), , 0, 105.58, –31.05, –100.37, –15.98, –127.07, –61.58, –61.84, 50.38, –124.39, , 0, 77.11, –11.20, –96.90, –26.86, –109.79, –54.12, –66.19, 25.52, –33.41, , 42.55, 72.68, 121.3, 107.1, 155.2, 96.2, 129.3, 115.5, 184.1, 140.92, , 0, 107.32, –240.12, –425.61, –470.11, –411.15, –407.3, –361.06, –287.78, –947.7, –1130.9, , 0, 76.76, –261.91, –379.49, –419.15, –384.14, –393.1, –348.98, –286.06, –851.9, –1047.7, , 51.21, 153.71, 59.0, 64.46, 48.1, 72.13, 115.5, 86.82, 98.53, 102.1, 136.0, , 0, 0.33, 33.1, –296.83, –395.72, –813.99, –909.27, –909.27, –20.63, –39.7, –1209, , 0, 0.1, 85.8, –300.19, –371.06, –690.00, –744.53, –744.53, –33.56, –27.83, –1105.3, , 31.80, 32.6, –14.6, 248.22, 256.76, 156.90, 20.1, 20.1, 205.79, 121, 291.82, , Silicon, , Sodium, Na(s), Na(g), Na+(aq), NaOH(s), NaOH(aq), NaCl(s), NaCl(aq), NaBr(s), NaI(s), NaHCO3(s), Na2CO3(s), Sulphur, S(s), rhombic, S(s), monoclinic, S 2–(aq), SO2(g), SO3(g), H2SO4(l), H2SO4(aq), SO42–(aq), H2S(g), H2S(aq), SF6(g), , (continued), Download all NCERT books PDFs from www.ncert.online, , 2019-20

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APPENDICES, , Substance, , 249, , Enthalpy of formation,, ∆fH/ (kJ mol–1), , Gibbs Energy of formation,, ∆fG/ (kJ mol–1), , Entropy,*, S/(J K–1 mol–1), , Tin, Sn(s), white, Sn(s), gray, SnO(s), SnO2(s), , 0, –2.09, –285.8, –580.7, , 0, 0.13, –256.9, –519.6, , 51.55, 44.14, 56.5, 52.3, , 0, –153.89, –348.28, +130.73, , 0, –147.06, –318.30, +95.14, , 41.63, –112.1, 43.64, 160.93, , Zinc, Zn(s), Zn2+(aq), ZnO(s), Zn(g), , *The entropies of individual ions in solution are determined by setting the entropy of H+ in water equal to, 0 and then defining the entropies of all other ions relative to this value; hence a negative entropy is one, that is lower than the entropy of H+ in water., , ORGANIC COMPOUNDS, Substance, , Enthalpy of, Enthalpy of, Gibbs Energy of, combustion,, formation,, formation,, Entropy,, ∆cH/ (kJ mol–1) ∆fH/ (kJ mol–1) ∆fG/ (kJ mol–1) S/(J K–1 mol–1), , Hydrocarbons, CH4(g), methane, C2H2(g), ethyne (acetylene), C2H4(g), ethene(ethylene), C2H6(g), ethane, C3H6(g), propene (propylene), C3H6(g), cyclopropane, C3H8(g), propane, C4H10(g), butane, C5H12(g), pentane, C6H6(l), benzene, C6H6(g), C7H8(l), toluene, C7H8(g), C6H12(l), cyclohexane, C6H12(g),, C8H18(l), octane, , –890, –1300, –1411, –1560, –2058, –2091, –2220, –2878, –3537, –3268, –3302, –3910, –3953, –3920, –3953, –5471, , –74.81, 226.73, 52.26, –84.68, 20.42, 53.30, –103.85, –126.15, –146.44, 49.0, —, 12.0, —, –156.4, —, –249.9, , –50.72, 209.20, 68.15, –32.82, 62.78, 104.45, –23.49, –17.03, –8.20, 124.3, —, 113.8, —, 26.7, —, 6.4, , 186.26, 200.94, 219.56, 229.60, 266.6, 237.4, 270.2, 310.1, 349, 173.3, —, 221.0, —, 204.4, —, 358, , –726, –764, –1368, –1409, –3054, , –238.86, –200.66, –277.69, –235.10, –164.6, , –166.27, –161.96, –174.78, –168.49, –50.42, , 126.8, 239.81, 160.7, 282.70, 144.0, , Alcohols and phenols, CH3OH(l), methanol, CH3OH(g), C2H5OH(l), ethanol, C2H5OH(g), C6H5OH(s), phenol, , (continued), Download all NCERT books PDFs from www.ncert.online, , 2019-20

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ANSWERS, , 253, , Answer to Some Selected Problems, , UNIT 1, 1.17, , ~ 15 × 10–4 g , 1.25 × 10–4 m, , 1.18, , (i) 4.8 × 10–3, , (ii) 2.34 × 105, , (iii) 8.008 × 103, , (iv) 5.000 × 102, , (i) 2, , (ii) 3, , (iii) 4, , (iv) 3, , (v) 4, , (vi) 5, , (i) 34.2, , (ii) 10.4, , (v) 6.0012, 1.19, 1.20, 1.21, , (iii) 0.0460, , (iv) 2810, 6, , (b) (i) Ans : (10 mm, 1015 pm), , (a) law of multiple proportion, , (ii) Ans : (10–6 kg, 106 ng), (iii) Ans : (10–3 L, 10–3 dm3), 1.22, , 6.00 × 10–1 m =0.600 m, , 1.23, , (i) B is limiting, , (ii) A is limiting, , (iii) Stoichiometric mixture –No, , (iv) B is limiting, , (v) A is limiting, 1.24, , (i) 2.43 × 103 g, , (ii) Yes, , (iii) Hydrogen will remain unreacted; 5.72 × 10 2g, 1.26, , Ten volumes, , 1.27, , (i), , 2.87 × 10–11m, , (ii), , 1.515 × 10–11 m, , (iii) 2.5365 × 10–2kg, , (ii), , 4, , (iii) 4, , (ii), , 13 atoms, , (iii) 7.8286 × 1024 atoms, , –23, , 1.30, , 1.99265 × 10, , 1.31, , (i), , g, , 3, –1, , 1.32, , 39.948 g mol, , 1.33, , (i), , 1.34, , Empirical formula CH, molar mass 26.0 g mol –1, molecular formula C2H2, , 1.35, , 0.94 g CaCO3, , 1.36, , 8.40 g HCl, , 3.131 × 1025 atoms, , UNIT 2, 2.1, , (i) 1.099 × 1027 electrons (ii) 5.48 × 10–7 kg, 9.65 × 104C, , 2.2, , (i) 6.022 × 1024electrons, (ii) (a) 2.4088 × 1021 neutrons(b) 4.0347 × 10–6 kg, (iii) (a) 1.2044 × 1022 protons (b) 2.015 × 10–5 kg, , 2.3, , 7,6: 8,8: 12,12: 30,26: 50, 38, , 2.4, , (i) Cl, , 2.5, , 5.17 × 1014 s–1, 1.72 × 106m–1, , 2.6, , (i) 1.988 × 10–18 J, , (ii) U, , (iii) Be, , (ii) 3.98 × 10–15 J, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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254, , CHEMISTRY, , 2.7, , 6.0 × 10–2 m, 5.0 × 109 s–1 and 16.66 m–1, , 2.8, , 2.012 × 1016 photons, , 2.9, , (i) 4.97 × 10–19 J (3.10 eV); (ii) 0.97 eV, , (iii) 5.84 × 105 m s–1, , mol–1, , 2.10, , 494 kJ, , 2.11, , 7.18 × 1019s–1, , 2.12, , 4.41 × 1014s–1, 2.91 × 10–19J, , 2.13, , 486 nm, , 2.14, , 8.72 × 10–20J, , 2.15, , 15 emission lines, , 2.16, , (i) 8.72 × 10–20J, , 2.17, , 1.523 × 106 m–1, , 2.18, , 2.08 × 10–11 ergs, 950 , , 2.19, , 3647, , 2.20, , 3.55 × 10–11m, , 2.21, , 8967, , 2.22, , Na+, Mg2+, Ca2+; Ar, S2– and K+, , 2.23, , (i) (a) 1s2, , 2.24, , n=5, , 2.25, , n = 3; l = 2; ml = –2, –1, 0, +1, +2 (any one value), , 2.26, , (i) 29 protons, , 2.27, , 1, 2, 15, , 2.28, , (i), , (ii) 1.3225 nm, , (b) 1s2 2s2 2p6;, , l, , ml, , 0, , 0, , 1, , –1,0,+1, , 2, , –2,–1,0,+1,+2, , (c) 1s22s22p6, , (d) 1s22s22p6, , (ii) l = 2; m1=–2, –1,0,+1,+2, (iii) 2s, 2p, 2.29, , (a) 1s,, , 2.30, , (a), (c) and (e) are not possible, , 2.31, , (a) 16 electrons, , 2.33, , n = 2 to n = 1, , 2.34, , 8.72 × 10–18J per atom, , 2.35, , 1.33 × 109, , 2.36, , 0.06 nm, , 2.37, , (a) 1.3 × 102 pm, , 2.38, , 1560, , 2.39, , 8, , 2.40, , More number of K–particles will pass as the nucleus of the lighter atoms is small,, smaller number of K–particles will be deflected as a number of positve charges is, less than on the lighter nuclei., , 2.41, , For a given element the number of prontons is the same for the isotopes, whereas, the mass number can be different for the given atomic number., , 2.42, , 81, 35, , (b) 3p, (c) 4d and (d) 4f, (b) 2 electrons, , (b) 6.15 × 107 pm, , Br, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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ANSWERS, , 255, , 2.43, , 37, 17, , Cl −1, , 2.44, , 56, 26, , Fe 3 +, , 2.45, , Cosmic rays > X–rays > amber colour > microwave > FM, , 2.46, , 3.3 × 106 J, , 2.47, , (a) 4.87 × 1014 s–1, , (b) 9.0 × 109 m, , (c) 32.27 × 10–20 J, , (d) 6.2 × 1018 quanta, 2.48, , 10, , 2.49, , 8.28 × 10–10 J, , 2.50, , 3.45 × 10–22 J, , 2.51, , (a) Threshold wave length (b) Threshold frequency of radiation, 652.46 nm, 4.598 ×1014 s–1, (c) Kinetic energy of ejected photoelectron, 9.29 ×10–20 J, Velocity of photoelectron 4.516 × 105 ms–1, , 2.52, , 530.9 nm, , 2.53, , 4.48 eV, , 2.54, , 7.6 × 103 eV, , 2.55, , infrared, 5, , 2.56, , 434 nm, , 2.57, , 455 pm, , 2.58, , 494.5 ms–1, , 2.59, , 332 pm, , 2.60, , 1.516 × 10–38 m, , 2.61, , Cannot be defined as the actual magnitude is smaller than uncertainity., , 2.62, , (v) < (ii) = (iv) < (vi) = (iii) < (i), , 2.63, , 4p, , 2.64, , (i) 2s, , 2.65, , Si, , 2.66, 2.67, , 16, , (ii) 4d, , (iii) 3p, , (a) 3, , (b) 2, , (c) 6, , (d) 4, , (e) zero, , UNIT 5, 5.1, , 2.5 bar, , 5.2, , 0.8 bar, , 5.4, , 70 g/mol, , 5.5, , MB = 4MA, , 5.6, , 203.2 mL, , 5.7, , 8.314 × 104 Pa, , 5.8, , 1.8 bar, , 5.9, , 3g/dm3, , 5.10, , 1249.8 g mol–1, , 5.11, , 3/5, , 5.12, , 50 K, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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256, , CHEMISTRY, , 5.13, , 4.2154 × 1023 electrons, , 5.14, , 1.90956 × 106 year, , 5.15, , 56.025 bar, , 5.16, , 3811.1 kg, , 5.17, , 5.05 L, , 5.18, , 40 g mol–1, , 5.19, , 0.8 bar, , UNIT 6, 6.1, , (ii), , 6.2, , (iii), , 6.3, , (ii), , 6.4, , (iii), , 6.5, , (i), , 6.6, , (iv), , 6.7, , q = + 701 J, w = – 394 J, since work is done by the system, ∆U = 307 J, , 6.8, , –743.939 kJ, , 6.9, 6.10, , 1.067 kJ, ∆H = –7.151 kJ mol–1, , 6.11, , – 314.8 kJ, , 6.12, , ∆rH = –778 kJ, , 6.13, , – 46.2 kJ mol–1, , 6.14, , – 239 kJ mol–1, , 6.15, , 326 kJ mol–1, , 6.16, , ∆S > 0, , 6.17, , 2000 K, , 6.18, , ∆H is negative (bond energy is released) and ∆S is negative (There is less, randomness among the molecules than among the atoms), , 6.19, , 0.164 kJ, the reaction is not spontaneous., , 6.20, , –5.744 kJ mol–1, , 6.21, , NO(g) is unstable, but NO2(g) is formed., , 6.22, , qsurr = + 286 kJ mol–1, ∆Ssurr = 959.73 J K–1, , UNIT 7, 7.2, , 12.229, , 7.3, , 2.67 x 104, , 7.5, , (i) 4.33 × 10–4 (ii) 1.90, , 7.6, , 1.59 × 10–15, , 7.8, , [N2] = 0.0482 molL–1, [O2] = 0.0933 molL–1, [N2O] = 6.6 × 10–21 molL–1, Download all NCERT books PDFs from www.ncert.online, , 2019-20

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ANSWERS, , 257, , 7.9, , 0.0352mol of NO and 0.0178mol of Br2, , 7.10, , 7.47 × 1011 M–1, , 7.11, , 4.0, , 7.12, , Qc = 2.379 × 103. No, reaction is not at equilibrium., , 7.14, , 0.44, , 7.15, , 0.068 molL–1 each of H2 and I2, , 7.16, , [I2] = [Cl2] = 0.167 M, [ICl] = 0.446 M, , 7.17, , [C2H6]eq = 3.62 atm, , 7.18, , (i) [CH3COOC2H5][H2O] / [CH3COOH][C2H5OH], (ii) 3.92 (iii) value of Qc is less than Kc therefore equilibrium is not attained., , 7.19, , 0.02molL–1 for both., , 7.20, , [PCO] = 1.739atm, [PCO2] = 0.461atm., , 7.21, , No, the reaction proceeds to form more products., , 7.22, , 3 × 10–4 molL–1, , 7.23, , 0.149, , 7.24, , a) – 35.0kJ, b) 1.365 × 106, , 7.27, , [PH ]eq = [PBr ]eq = 2.5 × 10–2bar, [PHBr] = 10.0 bar, , 7.30, , b) 120.48, , 7.31, , [H2]eq = 0.96 bar, , 7.33, , 2.86 × 10–28 M, , 7.34, , 5.85x10–2, , 7.35, , NO2–, HCN, ClO4, HF, H2O, HCO3–, HS–, , 7.36, , BF3, H+, NH4+, , 7.37, , F–, HSO4–, CO32–, , 7.38, , NH3, NH4+, HCOOH, , 7.41, , 2.42, , 7.42, , 1.7 x 10–4M, , 7.43, , F = 1.5 x 10–11, HCOO–= 5.6 × 10–11, CN–= 2.08 x 10–6, , 7.44, , [phenolate ion]= 2.2 × 10–6, α = 4.47 × 10–5 , α in sodium phenolate = 10–8, , 7.45, , [HS ]= 9.54 x 10–5, in 0.1M HCl [HS–] = 9.1 × 10–8M, [S2–] = 1.2 × 10–13M, in 0.1M, HCl [S2–]= 1.09 × 10–19M, , 7.46, , [Ac–]= 0.00093, pH= 3.03, , 7.47, , [A–] = 7.08 x10–5M, Ka= 5.08 × 10–7, pKa= 6.29, , 7.48, , a) 2.52 b) 11.70 c) 2.70 d) 11.30, , 7.49, , a) 11.65 b) 12.21 c) 12.57 c) 1.87, , 7.50, , pH = 1.88, pKa = 2.70, , 7.51, , Kb = 1.6 × 10–6, pKb = 5.8, , 7.52, , α = 6.53 × 10–4, Ka = 2.35 × 10–5, , 7.53, , a) 0.0018 b) 0.00018, , 7.54, , α = 0.0054, , 7.55, , a) 1.48 × 10–7M,, , b) 0.063, , c) 4.17 × 10–8M, , d) 3.98 × 10–7, , 7.56, , a) 1.5 × 10–7M,, , b) 10–5M,, , c) 6.31 × 10–5M, , d) 6.31 × 10–3M, , 7.57, , [K+] = [OH–] = 0.05M, [H+] = 2.0 × 10–13M, , 2, , 2, , –, , –, , Download all NCERT books PDFs from www.ncert.online, , 2019-20

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258, , CHEMISTRY, , 7.58, , [Sr2+] = 0.1581M, [OH–] = 0.3162M , pH = 13.50, , 7.59, , α = 1.63 × 10–2, pH = 3.09. In presence of 0.01M HCl, α = 1.32 × 10–3, , 7.60, , Ka = 2.09 × 10–4 and degree of ionization = 0.0457, , 7.61, , pH = 7.97. Degree of hydrolysis = 2.36 × 10–5, , 7.62, , Kb = 1.5 × 10–9, , 7.63, , NaCl, KBr solutions are neutral, NaCN, NaNO2 and KF solutions are basic and, NH4NO3 solution is acidic., , 7.64, , (a) pH of acid solution= 1.9, , 7.65, , pH = 6.78, , 7.66, , a) 12.6, , 7.67, , Silver chromate S= 0.65 × 10–4M; Molarity of Ag+ = 1.30 x 10–4M, , b) 7.00, , (b) pH of its salt solution= 7.9, , c) 1.3, , Molarity of CrO42– = 0.65 × 10–4M; Barium Chromate S = 1.1 × 10–5M; Molarity of, Ba2+ and CrO42– each is 1.1 × 10–5M; Ferric Hydroxide S = 1.39 × 10–10M;, Molarity of Fe3+ = 1.39 × 10–10M; Molarity of [OH–] = 4.17 × 10–10M, Lead Chloride S = 1.59 × 10–2M; Molarity of Pb2+ = 1.59 × 10–2M, Molarity of Cl– = 3.18 × 10–2M; Mercurous Iodide S = 2.24 × 10–10M;, –, , Molarity of Hg22+ = 2.24 × 10–10M and molarity of I = 4.48 × 10–10M, 7.68, , Silver chromate is more soluble and the ratio of their molarities = 91.9, , 7.69, , No precipitate, , 7.70, , Silver benzoate is 3.317 times more soluble at lower pH, , 7.71, , The highest molarity for the solution is 2.5 × 10–9M, , 7.72, , 2.43 litre of water, , 7.73, , Precipitation will take place in cadmium chloride solution, , Download all NCERT books PDFs from www.ncert.online, , 2019-20