Maharashtra Board Class 12 Chemistry Book PDF

https://www.ncertbooks.guru/ https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ The Coordination Committee formed by GR No. Abhyas - 2116/(Pra.Kra.43/16) SD - 4 Dated 25.4.2016 has given approval to prescribe this textbook in its meeting held on 30.01.2020 and it...

https://www.ncertbooks.guru/ https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ The Coordination Committee formed by GR No. Abhyas - 2116/(Pra.Kra.43/16) SD - 4 Dated 25.4.2016 has given approval to prescribe this textbook in its meeting held on 30.01.2020 and it has been decided to implement it from academic year 2020-21. CHEMISTRY STANDARD TWELVE Download DIKSHA App on your smartphone. If you scan the Q.R.Code on this page of your textbook, you will be able to access full text and the audio-visual study material relevant to each lesson provided as teaching and learning aids. 2020 Maharashtra State Bureau of Textbook Production and Curriculum Research, Pune. https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ First Edition : © Maharashtra State Bureau of Textbook Production and 2020 Curriculum Research, Pune - 411 004. The Maharashtra State Bureau of Textbook Production and Curriculum Research reserves all rights relating to the book. No part of this book should be reproduced without the written permission of the Director, Maharashtra State Bureau of Textbook Production and Curriculum Research, ‘Balbharati’, Senapati Bapat Marg, Pune 411004. Committee Illustration Shri. Shubham Gopichand Chavan Dr. Chandrashekhar V. Murumkar, Chairman Cover Dr. Sushama Dilip Joag, (Convener) Shri. Vivekanand S. Patil Dr. Laxman Shamrao Patil, (Co-Convener) Typesetting DTP Section, Textbook Bureau, Pune Dr. Shridhar Pandurang Gejji, Member Dr. Satyawati Sudhir Joshi, Member Co-ordination Dr. Rajashree Vikas Kashalkar, Member Shri. Rajiv Arun Patole Shri. Rajesh Vamanrao Roman, Member Special Officer - Science Section Shri. Rajiv Arun Patole, Member Secretary Chemistry Paper Study group 70 GSM Creamwove Dr. Sujata Sanjay Kale Print Order Dr. Anupa Avinash Kumbhar Dr. Anjali Deepak Ruikar Printer Dr. Dnyaneshwar Dattatraya Kumbhar Shri. Sheetalkumar Sopanrao Bhong Production Shri. Kishor Sadanand Chavan Shri Sachchitanand Aphale Shri. Sachin Ashok Bartakke Chief Production Officer Smt. Archana Sanjeev Kale Shri Liladhar Atram Smt. Pushpalata Babanrao Gangarde Production Officer Smt. Archana Dipak Harimkar Shri. Milind Chandrakant Gaikwad Publisher Shri. Vishnu Rustumrao Deshmukh Shri. Sharad Ajabrao Mankar Shri Vivek Uttam Gosavi Controller Shri. Ritesh Vijay Bijewar Maharashtra State Textbook Shri. Rupesh Dinkar Thakur Bureau, Prabhadevi, Shri. Gajanan Shivajirao Suryawanshi Mumbai - 400 025 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ The Constitution of India Preamble WE, THE PEOPLE OF INDIA, having solemnly resolved to constitute India into a SOVEREIGN SOCIALIST SECULAR DEMOCRATIC REPUBLIC and to secure to all its citizens: JUSTICE, social, economic and political; LIBERTY of thought, expression, belief, faith and worship; EQUALITY of status and    of opportunity; and to promote among them all FRATERNITY assuring the dignity of the individual and the unity and integrity of the Nation; IN OUR CONSTITUENT ASSEMBLY this twenty-sixth day of November, 1949, do HEREBY ADOPT, ENACT AND GIVE TO OURSELVES THIS CONSTITUTION. https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ NATIONAL ANTHEM https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ Preface Dear Students, We welcome you all to std. XII. For the first time, you were introduced to the subject of chemistry discipline in std XI. Chemistry is very vast subject that covers many aspects of our everyday experience. This textbook aims to create awareness and to understand certain essential aspects by the state curriculum framework (NCF) which was formulated in 2005, followed by the national curriculum framework (SCF) in 2010. Based on these two frameworks, reconstruction of the curriculum and prepartion of a revised syllabus has been done and designed now. Three major branches of chemistry, namely organic chemistry, inorganic chemistry and physical chemistry are presented in separate units in this book. Care is taken to have an integrated approach in deliberation of their contents. Chemistry is highly applied subject. In the unit "applied chemistry" along with the application of chemistry in life science and materials in everyday life, the upcoming applied branches, namely nanochemistry and green chemistry are also introduced. You can learn basic principles, understand facts and put them into practice by learning in the classroom and laboratory. The textbook is presented in a simple language with relevant diagrams, graphs, tables, photographs. This will help you to understand various terminology, concepts with more clarity. All the illustrations are in colour form. The new syllabus focuses on the basic principles, concepts, laws based on precise observations, their applications in everyday life and ability to solve different types of problems. The general teaching - learing objectives of the revised syllabus are further determined on the basis of the ‘Principle of constructivism’ i.e. self learning. The curriculum and syllabus is designed to make the students to think independently. The students are encouraged to read, study more through the additional information given in the colored boxes. Activities have been introduced in each chapter. These activities will help to understand the content knowledge on your own efforts. QR code has been introduced for gaining the additional information, abstracts of chapters and practice questions/ activities. The efforts taken to prepare the text book will help the students think about more than just the content of the chemical concepts. Teachers, parents as well as the aspiring condidates preparing for the competitive examinations will be benefited. We look forward to a positive response from the teachers and students. Our best wishes to all ! (Vivek Gosavi) Pune Director Date : 21 February 2020 Maharashtra State Bureau of Textbook Bharatiya Saur : 2 Phalguna 1941 Production and Curriculum Research, Pune 4 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ - For Teachers - Dear Teachers, We are happy to introduce the revised textbook required for each branch of chemistry. Application of chemistry for std. XII. This book is a sincere of this knowledge will help students to understand attempt in continuation with the standard XI book further chapters in each unit. to follow the maxims of teaching as well as develop Each chapter provides solved problems on each a ‘constructivist’ approach to enhance the quality and every concept and various laws. The solved of learning. The demand for more activity based, problems are put into boxes. Teachers should experiential and innovative learning opportunities explain each step of the problems to the student is the need of the time. The present curriculum has and give them practice. been restructured so as to bridge the credibility gap Invite students' participation by making use of the that exists in the experience in the outside world. boxes like ‘Can You Recall’, ‘Do you know?’ Guidelines provided below will help to enrich the Encourage the students to collect related teaching - learning process and achieve the desired information by providing them the websites. learning outcomes. Teaching- learning interactions, processes and To begin with, get familiar with the textbook participation of all students are necessary and so is yourself. your active guidance. The present book has been prepared for Do not use the content of the boxes titled ‘Do you constructivism and activity based learning. know’? for evaluation. Teachers must skilfully plan and organize the Exercises include parameters such as co-relation, activities provided in each chapter to develop critical thinking, analytical reasoning etc. interest as well as to stimulate the thought process Evaluation pattern should be based on the given among the students. parameters. Equal weightage should be assigned Always teach with proper planning. to all the topics. Use different combinations of Use teaching aids as required for the proper questions. understanding of the subject. Illustrative figures are included to help assimilation Do not finish the chapter in short. of new concepts. Teachers should note that students Follow the order of the chapters strictly as listed are not expected to draw all the figures included in in the contents because the units are introduced in this book. As a part of evaluation students can be a graded manner to facilitate knowledge building. asked to draw schematic diagrams/structures and Each unit is structured in a definite manner. It interpret or label other complicated figures. starts from the basic concepts of general chemistry Can you recall ? Remember... Do you know ? Observe and discuss... Use your brain power Can you tell ? Internet my friend Try this... Activity : Can you think ? Cover page : (+) - Limonene and (-) Limonene : major contributors to the characteristic flavours of peelings of oranges and lemons respectively. Image of gold surface : Au (111), obtained by Scanning Tunneling Microscope (STM). Structures of coordination complexes : haemoglobin, chlorophyll and metal -EDTA complex. DISCLAIMER Note : All attempts have been made to contact copy right/s (©) but we have not heard from them. We will be pleased to acknowledge the copy right holder (s) in our next edition if we learn from them. https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ Competency Statements - Standard XII Area/ Unit/ After studying the contents in Textbook students..... Lesson Distinguish crystal structures illustrating unit cell and packing efficiency in cubic systems. Gain information on point defects and band theory in relation to electric and magnetic behavior. Define solubility and rationalise its dependence on various factors. Explain Henry and Raoult's laws. Derive expressions for colligative properties. Learn van’t Hoff factor and its correlation with dissociation constant. Catagorize strong and weak acid bases. Learn Ostwald’s dilution law. Derive Henderson Balch Hassel equation. Explain the role of buffer solutions in controlling of pH. Understand spontaneity of reactions. Know reversible/irreversible processes and PV work. Understand first and second laws of thermodynamics. Work out change in enthalpy, entropy and Gibbs' functions in physical and chemical Physical transformations. Chemistry Apply Hess’s Law in thermochemical equations. State what are strong and weak electrolytes. Define Kohlrausch law and state its importance. Understand functioning of electrolytic and galvanic cells. Write half cell reactions there in. Describe type of electrodes. Derive Nernst equation and understand its importance. Know what are dry cell, lead strong batteries and fuel cells. Describe the electrochemical series and its implications. Define average and instantaneous rate, order and molecularity in kinetics Formulate differential and integral rate laws for zero and first order reactions. Understand basis of collision theory of reaction rates Sketch qualitatively potential energy curve. Understand acceleration of reactions in the presence of catalyst. Solve relevant numerical problems. Write electronic configuration of groups 16, 17, 18 and those of d and f blocks. Correlate atomic properties of elements with electron configuration. Explain the anomalous behaviour of ‘O’ and ‘F’. Understand allotropy in ‘O’ and ‘S’. Draw structures of oxyacids of ‘sulfur’ and ‘halogens’. Write reactions for preparation, chemical properties of O2, O3, SO2, H2SO4, Cl2, HCl, KMnO4, K2Cr2O7. Draw structures of interhalogen and xenon compounds and illustrate their properties. State the methods of preparation with reaction. Know chemistry of the elements belonging to groups 16, 17, 18. Inorganic Understand the principles of metallurgy in extraction of iron. chemistry Compare lanthanoides and actinoides. Enlist properties of the manmade post actinoide elements. Understand Werner theory of coordination compounds. Understand and apply EAN rule for stability of coordination compounds. Understand diverse isomerism in coordination compounds. Use the concept of hybridization for predicting structures and magnetic behaviour of complexes based on the V.B.T. Understand C.F.T. Sketch qualitatively d-orbital splitting diagrams in octahedral and tetrahedral ligand field environments. Distinguish between high spin and low spin complexes. Predict structure, colour and magnetic properties of the complexes based on the C.F.T. https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ State common and IUPAC names of compounds and methods of preparation of halogen derivatives, alcohols, phenols, ethers, aldehydes, ketones, carboxylic acid and amines. Understand structure, chemical properties, laboratory tests and reactions of the above functional groups. Organic Explain acid or base strength of alcohols, phenols, carboxylic acids and amines. Chemistry Explain trends in boiling point and solubility of compounds of above functional groups in terms of intermolecular forces. Understand optical activity, recognize chiral molecules and represent with Fischer projection and wedge formulae. Understand mechanism of nucleophilic substitution reactions and influencing factors. Classify carbohydrates, amino acids, nucleic acids. Represent monosaccharides using the Fischer projection formula. Represent monosaccharides, disaccharides and polysaccharides using the Haworth formula. Correlate properties of carbohydrates to the presence or absence of potential aldehyde group. Learn four level structure of proteins and primary structures nucleic acid. Represent primary structure of dipeptide and tripeptide from data on the terminals. Understand enzyme catalysis and double strand DNA structure. Understand classification of polymers on the basis of source, structure, intermolecular forces, polymerization, number of monomers and biodegradability. Applied Understand addition and condensation polymerization. Chemistry Know properties, structure and preparation of natural rubber, vulcanized rubber, Buna-S, viscose, LDP, HDP, teflon, polyacrylo nitrile, polyamide, polyesters, phenol-formaldehyde resin and PHBV. Understand scope of green chemistry with reference to sustainable development. Recognize twelve principles of green chemistry and their implementation. Correlate the Chemistry knowledge gained so far as pro or counter to the principles of green chemistry. Understand scope and applications of nanochemistry. Gain knowledge of a synthetic method and properties of nanoparticles. Know instrumental techniques for characterization of nanomaterials. CONTENTS Sr. No. Title Page No. 1 Solid State 01-27 2 Solutions 28-46 3 Ionic Equilibria 47-62 4 Chemical Thermodynamics 63-89 5 Electrochemistry 90-119 6 Chemical Kinetics 120-137 7 Elements of Groups 16, 17 and 18 138-164 8 Transition and Inner transition Elements 165-191 9 Coordination Compounds 192-209 10 Halogen Derivatives 210-233 11 Alcohols, Phenols and Ethers 234-253 12 Aldehydes, Ketones and Carboxylic acids 254-281 13 Amines 282-297 14 Biomolecules 298-321 15 Introduction to Polymer Chemistry 322-339 16 Green Chemistry and Nanochemistry 340-352 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ 1. SOLID STATE Can you recall ? Try this... What are the three most common states of matter? How does solid state differ from the other two states ? (Answer with reference to volume, shape, effect of temperature and pressure on these and the motion of constituent particles and interparticle forces.) 1.1 Introduction : As studied earlier, the solid state of matter is characterised by strong Observe the above figure carefully. interparticle forces of attraction. As a result The two types of circles in this figure most solids have definite shape and volume, represent two types of constituent particles which change only slightly with change of a solid. in temperature and pressure. The smallest Will you call the arrangement of particles constituent particles of various solids are in this solid regular or irregular ? atoms, ions or molecules. All such smallest constituent particles of solids will be referred Is the arrangement of constituent to as 'particles' in this chapter. particles same or different in directions → → → 1.2 Types of solids : There are two types AB, CD, and EF ? of solids, namely crystalline solids and amorphous solids. 1.2.1 Crystalline solids : Study of many crystalline solids indicates that they possess Observe and discuss... the following characteristic properties. Collect the following solids : i. There is a regularity and periodicity in grannular sugar, common salt, the arrangement of constituent particles in blue vitriol. crystalline solids. The ordered arrangement Observe a few grannules of these solids of particles extends over a long range. under a magnifying lens or microscope. Discuss your observations with reference ii. Crystalline solids have sharp melting to the following points : (i) Shape of the points, that is, they melt at a definite grannules, (ii) Smoothness of faces of temperature. the grannules and (iii) Angles between iii. All crystalline substances except those various edges of the grannules. having cubic structure are anisotropic. In All the above solids are crystalline solids. other words their properties like refractive Name the properties of crystals that you index, thermal and electrical conductivity, observed in this activity. etc, are different in different directions. Ice, salts such as NaCl, metals such as sodium, gold, copper and materials such as diamond, graphite, ceramics are examples of crystalline solids. 1 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ Glass, plastic, rubber, tar, and metallic glass Do you know ? (metal-metalloid alloy) are a few examples of A single crystal has ordered amorphous solids. (regular and periodic) arrangement of constituent particles throughout its bulk. Use your brain power Majority of crystalline solids, including Identify the arrangements A and B metals, are polycrystalline in nature. as crystalline or amorphous. Single grannule of a polycrystalline solid is made of many single crystals or crystallites packed together with different orientations. Single crystals are difficult to obtain. Diamond is an example of naturally formed single crystal. 1.2.3 Isomorphism and polymorphism Similarity or dissimilarity in crystal structure of different solids is described as isomorphism and polymorphism. i. Isomorphism : Two or more substances having the same crystal structure are said to be 1.2.2 Amorphous solids : The particles of isomorphous. In these substances the chemical a liquid are in constant motion. The stop composition has the same atomic ratio. For action photograph of a liquid describes the example (i) NaF and MgO (ii) NaNO3 and amorphous state. In fact, they are supercooled liquids. Amorphous solids have the following CaCO3 are isomorphous pairs, and have the characteristics. same atomic ratios, 1:1 and 1:1:3, respectively, of the constituent atoms. i. The constituent particles in amorphous solids are randomly arranged. The ii. Polymorphism : A single substance that particles do not have long range ordered exists in two or more forms or crystalline structure, but they do have a short range structures is said to be polymorphous. order. Polymorphs of a substance are formed under ii. Amorphous solids do not have sharp different conditions. For example : Calcite and melting points. They melt gradually aragonite are two forms of calcium carbonate; over a temperature interval. On α-quartz, b-quartz and cristobalite are three heating, amorphous solids gradually and of the several forms of silica. Polymorphism continuously soften and start to flow. occuring in elements is called allotropy. For iii. These solids are isotropic. In other example: three polymorphic (allotropic) forms words, their properties such as refractive of carbon are diamond, graphite and fullerene. index, conductivity are all independent of direction of measurement. They exhibit the same magnitude for any property in every direction. 2 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ 1.3.2 Covalent network crystals Do you know ? Characteristics of covalent network crystals Many crystalline forms of are as follows : silica (SiO2) are found in nature. Three of them are α-quartz, b-quartz and i. The constituent particles in covalent cristobalite network solids are atoms. ii. The atoms in these crystals are linked by a continous system of covalent bonds. The result is a rigid three dimensional network that forms a giant molecule. The entire crystal is a single molecule. iii. As a result of rigid and strongly bonded cristobalite α-quartz structure, covalent network crystals are very hard. In fact they are the hardest and most incompressible of all the materials. These crystals have high melting and boiling points. iv. The electrons are localised in covalent b-quartz bonds and hence are not mobile. As a result, covalent solids are poor conductors 1.3 Classification of crystalline solids : of heat and electricity. Crystalline solids are further classified into For example : diamond, quartz (SiO2), boron four categories : ionic solids, covalent network nitride, carborandum are covalent network solids, molecular solids and metallic solids. solids. 1.3.1 Ionic crystals : Ionic crystals have the following characteristics : Do you know ? i. The constituent particles of ionic crystals Diamond is the hardest known are charged ions. The cations and anions material. may differ in size. Try this... ii. Each ion of a given sign of charge is bonded to ions of opposite charge around Graphite is a covalent solid it by coulomb force. In other words, the yet soft and good conductor of particles of ionic crystals are held by electricity. Explain. electrostatic force of attraction between oppositely charged ions. Can you recall ? iii. Ionic crystals are hard and brittle. They What are structures of have high melting points. diamond and graphite ? iv. These are nonconductors of electricity What are the types of covalent bonds in solid state. However, they are good those link carbon atoms in diamond conductors when melted or dissolved in and graphite ? water. Are all the valence electrons of For example : NaCl, K2SO4, CaF2, KCl are carbon atoms in graphite localized to ionic crystals. specific covalent bonds ? 3 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ c. Intermolecular hydrogen bonds in Remember... solids such as H2O (ice), NH3, HF and Both ionic and covalent so forth. crystals are hard and have high melting and boiling points. We can iii. Because of weak intermolecular attractive use electrical properties to distinguish forces, molecular solids are usually soft between them. Both are insulators at low substances with low melting points. temperature. iv. These solids are poor electrical conductors Ionic solids become good conductors and are good insulators. only at high temperature, above their 1.3.4 Metallic crystals : These are crystalline melting points. solids formed by atoms of the same metallic The conductivity of covalent solids element, held together by a metallic bond. is in general low and increases with Metallic bond : In a solid metal, the valence temperature. However, there is no abrupt electrons are delocalised over the entire crystal rise in conductivity when substance is leaving behind positively charged metal ions. melted. Therefore, metallic crystals are often described as an array of positive ions immersed in a sea 1.3.3 Molecular crystals : Substances such of mobile electrons. The attractive interactions as Cl2, CH4, H2, CO2, O2 on solidfication between cations and mobile electrons constitute give molecular crystals. Crystalline organic the metallic bonds. (For more details refer to compounds are also molecular solids. section 1.9.2) i. The constituent particles of molecular Metallic crystals have the following properties: solids are molecules (or unbonded single atoms) of the same substance. i. Metals are malleable, that is, they can be hammered into thin sheets. Can you recall ? ii. Metals are ductile, that is, they can be What is a hydrogen bond ? drawn into wires. iii. Metals have good electrical and thermal ii. The bonds within the molecules are conductivity. covalent. The molecules are held together Examples : metals such as Na, K, Ca, Li, Fe, by various intermolecular forces of Au, Ag, Co, etc. attraction. (Refer to XI Std. Chemistry The properties of different types of crystalline Textbook, Chapter 10). For example : solids are summarized in Table 1.1. a. Weak dipole-dipole interactions in 1.4 Crystal structure : The ordered three polar molecules such as solid HCl, dimensional arrangement of particles in a H2O, SO2, which possess permanent crystal is described using two terms, namely, dipole moment. lattice and basis. b. Very weak dispersion or London 1.4.1 Crystal, lattice and basis : Lattice is a forces in nonpolar molecules such as geometrical arrangement of points in a three solid CH4, H2. These forces are also dimensional periodic array. A crystal structure involved in monoatomic solids like is obtained by attaching a constituent particle argon, neon. (These substances are to each of the lattice points. Such constituent usually gases at room temperature.) particles that are attached to the lattice points form the basis of the crystal lattice. Crystal 4 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ Table 1.1 : Properties of four types of crystalline solids Type Ionic solids Covalent network Molecular Metallic solids Property solids solids 1. Particles of unit Cations and Covalently bonded Monoatomic Metallic ions in a cell anions atoms or polyatomic sea of electrons molecules 2. Interparticle Electrostatic Covalent bonds London, dipole- Metallic bonds forces dipole forces and/ (attraction between or hydrogen bonds cations and mobile valence electrons) 3. Hardness Hard and brittle Very hard Soft Variable from soft to very hard 4. Melting points High High Low Wide range 6000C to 30000C 12000C to 40000C (-2720C to 4000C) (-390C to 34000C) 5. Thermal and Poor electrical Poor conductors poor conductor of good conductor of electrical conductors in Exceptions : heat and electricity heat and electricity conductivity solid state. Good i. Graphite : good conductors conductor of when melted electricity. or dissolved in ii. Diamond : good water conductor of heat 6. Examples NaCl, CaF2 diamond, silica ice, benzoic acid Na, Mg, Cu, Au lattice is also called space lattice of crystal. The dimensions of unit cell along the three Thus, crystal is the structure that results by axes are denoted by the symbols a, b and c. attaching a basis to each of the lattice points. The angles between these axes are represented It is represented by the following equation. by the symbols ∝, β and γ as shown in Fig 1.1....... + =...... Lattice + Basis = Crystal b a 1.4.2 Unit Cell : The space lattice of a crystal is built up of a three dimensional basic g pattern. This basic pattern is repeated in three dimensions to generate the entire crystal. Fig. 1.1 : Unit cell parameters The smallest repeating structural unit of a crystalline solid is called unit cell. 1.4.3 Types of unit cell : There are four types When the unit cells are stacked together of unit cells. to generate the crystal, each unit cell shares i. Primitive or simple unit cell : In primitive its faces, edges and corners with neighbouring unit cell, the constituent particles are present at unit cells. It is important to understand that the its corners only. geometric shape of a unit cell is same as that ii. Body-centred unit cell : In this type of of the macroscopic crystal. For example, if the unit cell, one constituent particle is present at crystal has cubic shape the unit cell will also the centre of its body in addition to the corner have its constituent particles arranged to form particles. a tiny cube. 5 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ iii. Face-centred unit cell : This unit cell consists of particles at the centre of each of the faces in addition to the corner particles. iv. Base-centred unit cell : This unit cell consists of particles at the centre of any two sc bcc fcc of its opposite faces in addition to the corner particles. 1.4.4 Crystal systems : By mathematical analysis, it has been proved that only fourteen different kinds of space lattices are possible. sc bcc fcc In other words, there are only 14 ways in which similar points can be arranged in a three dimensional order. These 14 lattices, which describe the crystal structure, are called Bravais lattices. sc bcc fcc Fourteen Bravais lattices are divided into seven crystal systems. The possible combinations of lattice point spacings (a, b Fig. 1.2 : Cubic unit cell and c) along three axes and the angles (∝, β and γ) between these axes give rise to seven 1.5.1 Number of particles in cubic unit cells crystal systems. In other words, seven crystal i. Primitive or simple cubic unit cell (sc) : systems are associated with 14 Bravais lattices A simple cubic cell has particles at its eight also called 14 unit cells. corners. When these unit cells are stacked together, particle at each corner of a given unit The seven crystal systems are cell is shared with seven other neighbouring named as cubic, tetragonal, orthorhombic, cubes that come together at that corner. As a rhombohedral, monoclinic, triclinic and result the corner particle contributes its 1/8th hexagonal system. Cubic system will be part to the given unit cell. Thus, a simple cubic discussed in the following section. cell has 1/8 × 8 = 1 particle per unit cell. 1.5 Cubic system : There are three kinds of ii. Body-centred cubic unit cell (bcc) : A unit cells in cubic system : primitive or simple bcc unit cell has eight corner particles and an cubic (sc), body-centred cubic (bcc) and face- additional particle at the centre of the cube. centred cubic (fcc) (Fig. 1.2). One eighth of each particle from eight corners i. Simple cubic unit cell (sc) has a particle at belongs to the given unit cell as mentioned in each of the eight corners of a cube. simple cubic unit cell. ii. Body-centred cubic unit cell (bcc) has The particle at the centre of a given particles at its eight corners and an additional cube is not shared by any other cube. Hence, it particle in the center of the cube. belongs entirely to the given unit cell. Thus bcc unit cell has one particle from eight corners iii. Face-centred cubic unit cell (fcc) has plus one particle in the centre of the cube, particle at the centre of each of six faces in making total of 2 particles per bcc unit cell. addition to the particles at eight corners of the iii. Face-centred cubic unit cell (fcc) : A cube. fcc unit cell has particles at the eight corners plus particles at the centre of its six faces. As described in simple cubic unit cell, one particle 6 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ Do you know ? The names of fourteen Bravais lattices (unit cells) for each of the seven crystal system are shown below. Crystal system Bravais lattices Name unit cell structure 1. Cubic i. simple or primitive ii. body-centred iii. face-centred ii. iii. i. 2. Orthorhombic i. simple or primitive ii. body-centred iii. face-centred iv. base-centred i. ii. iii. iv. 3. Tetragonal i. simple or primitive ii. body-centred i. ii. 4. Monoclinic i. simple or primitive ii. base centred i. ii. 5. Rhombohedral i. simple or primitive 6. Triclinic i. simple or primitive 7. Hexagonal i. simple or primitive 7 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ from eight corners belongs to the given unit cell. Problem 1.1 : When gold crystallizes, it forms face-centred cubic cells. The unit cell Each particle at the centre of the six edge length is 408 pm. Calculate the density faces is shared with one neighbouring cube. of gold. Molar mass of gold is 197 g/mol. Thus,1/2 of each face particle belongs to the given unit cell. From six faces, 1/2 × 6 = 3 Solution : Mn particles belong to the given unit cell. ρ= a3 NA Therefore, fcc unit cell has one corner M = 197 g mol-1, n = 4 atoms for fcc, particle plus 3 face particles, total 4 particles NA = 6.022 ×1023 atoms mol-1, per unit cell. a = 408 pm = 408 ×10-12m = 4.08×10-8 cm Remember... Substitution of these quantities in the equation gives Each corner particle of a cube is shared by 8 cubes, each face ρ= particle is shared by 2 cubes and each edge 197 g mol-1× 4 atom particle is shared by 4 cubes. (4.08×10-8)3 cm3× 6.022×1023 atom mol-1 1.5.2 Relationship between molar mass, = 19.27 g/cm3, 19.27 × 103 kg/m3. density of the substance and unit cell edge length, is deduced in the following steps 1.6 Packing of particles in crystal lattice i. If edge length of cubic unit cell is a, the Constituent particles of a crystalline volume of unit cell is a3. solid are close packed. While describing ii. Suppose that mass of one particle is m and the packing of particles in a crystal, the that there are n particles per unit cell. individual particles are treated as hard spheres. The closeness of particles maximize Mass of unit cell = m × n (1.1) the interparticle attractions. iii. The density of unit cell (ρ), which is same The number of neighbouring spheres as density of the sub-substance is given by that touch any given sphere is its coordination mass of unit cell m×n ρ = volume of unit cell = a3 = density of number. Magnitude of the coordination substance number is a measure of compactness of (1.2) spheres in close-packed structures. The larger the coordination number, the closer are the iv. Molar mass (M) of the substance is given spheres to each other. by 1.6.1 Close packed structures : The three M = mass of one particle × number of particles dimensional close packed structure can be per mole understood conveniently by looking at the = m×NA (NA is Avogadro number) close packing in one and two dimensions. Therefore, m= M/NA (1.3) a. Close packing in one dimension : A close v. Combining Eq. (1.1) and (1.3), gives packed one dimensional structure results by arranging the spheres to touch each other in a nM ρ= (1.4) row (Fig. 1.3 (a)). a3 NA By knowing any four parameters of Eq. (1.4), b. Close packing in two dimensions : A close the fifth can be calculated packed two dimensional (planar) structure results by stacking the rows together such 8 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ that they are in contact with each other. There the free space in this arrangement is less than are two ways to obtain close packing in two in square packing, making it more effecient dimensions. packing than square packing. From Fig.1.3(c) i. Square close packing : One dimensional it is evident that the free spaces (voids) are rows of close packed spheres are stacked over triangular in shape. These triangular voids are each other such that the spheres align vertically of two types. Apex of the triangular voids in and horizontally (Fig. 1.3 (b)). If the first row alternate rows points upwards and downwards. is labelled as 'A' type, being exactly same as the first row, the second row is also labelled as (a) 'A' type. Hence this arrangement is called A, A, A, A..... type two dimensional arrangement. In this arrangement, every sphere touches four neighbouring spheres. Hence, two dimensional coordination number, here, is 4. A square is (b) obtained by joining the centres of these four closest neighbours (Fig. 1.3(b)). Therefore, this two dimensional close packing is called square close packing in two dimension. (c) ii. Hexagonal close packing : Close packed Fig. 1.3 : (a) Close packing in one dimension one dimensional row (Fig. 1.3 (a)) shows that (b) square close packing there are depressions between the neighbouring (c) Hexagonal close packing in two dimension spheres. If the second row is arranged in such c. Close packing in three dimensions : a way that its spheres fit in the depressions of Stacking of two dimensional layers gives rise the first row, a staggered arrangement results. to three dimensional crystal structures. Two If the first row is called 'A' type row, the dimensional square close packed layers are second row, being different, is called 'B' type. found to stack only in one way to give simple Placing the third row in staggered manner in cubic lattice. Two dimensional hexagonal contact with the second row gives rise to an close packed layers are found to stack in arrangement in which the spheres in the third two distinct ways. Accordingly two crystal row are aligned with the spheres in the first structures, namely, hexagonal close packed row. Hence the third row is 'A' type. Similarly (hcp) structure and face centred cubic (fcc) spheres in the fourth row will be alligned with structure are formed. the spheres in the second row and hence the fourth row would be 'B' type. The resulting two i. Stacking of square close packed layers: dimensional arrangement is 'ABAB...' type Stacking of square close packed layers (Fig. 1.3 (c)). In this arrangement each sphere generates a three dimensional simple cubic touches six closest neighbours. Thus, the two structure. Here, the second layer is placed dimensional coordination number in this over the first layer so as to have its spheres packing is 6. A regular hexagon is obtained exactly above those of the first layer (Fig. 1.4). by joining the centres of these six closest Subsequent square close packed layers are spheres (Fig. 1.3 (c)). Hence, this type of two placed one above the other in the same manner. dimensional close packing is called hexagonal In this arrangement, spheres of all the layers close packing in two dimensions. Compared to are perfectly aligned horizontally as well as the square close packing in two dimensions, vertically. Hence, all the layers are alike, and the coordination number in hexagonal close are labelled as 'A' layers. This arrangement packing in two dimensions is higher. Moreover of layers is described as 'AAAA... ' type. The 9 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ structure that results on stacking square close packed layers is simple cubic. Its unit cell is the primitive cubic unit cell (Fig. 1.4). Fig. 1.5 : Two layers of closed packed spheres Fig. 1.4 : Stacking of square closed packed layers It can be seen that in the simple cubic structure, each sphere touches six neighbouring spheres, four in its own layer, one in the layer above and one in the layer below. Hence, coordination number of each sphere is 6. Polonium is the only metal that crystallizes Fig. 1.6 : Tetrahedral void in simple cubic closed packed structure. ii. Stacking of two hexagonal close packed layers : To generate a close packed three dimensional structure, hexagonal close packed layers are arranged in a particular manner. In doing so, spheres of the second layer are placed in the depression of the first layer (Fig. 1.5). If the first layer is labelled as 'A' layer, the second layer is labelled as 'B' layer because the two layers are aligned differently. It is evident from the Fig. 1.5 Fig. 1.7 : Octahedral void that all triangular voids of the first layers are not covered by the spheres of the second Remember... layer. The triangular voids that are covered It is important to note that the by spheres of the second layer generate triangular shapes of depressions in tetrahedral void(Fig. 1.6). A tetrahedral void A and B layer do not overlap. The apices of is surrounded by four spheres. On joining the two triangular depressions in A and B layer centres of these four spheres a tetrahedron is point in opposite directions. formed which encloses the tetrahedral voids (Fig. 1.6). The remaining triangular voids of The depressions in which spheres the first layer have above them the triangular of second layer rest are tetrahedral voids voids of the second layer. The overlapping while the depressions in which no sphere triangular voids from the two layers together rests are octahedral voids. form an octahedral void which is surrounded by six spheres (Fig. 1.7). 10 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ iii. Placing third hexagonal close packed 1 above and 1 below. Hence coordination layer : There are two ways of placing the number of any sphere in sc is 6. third hexagonal close packed layer on the b. In both hcp and ccp/fcc structures that result second. from stacking of hexagonal close packed One way of doing this is to align the layers in two different ways, each sphere is spheres of the third layer with the spheres surrounded by 12 neighbouring spheres, 6 in of the first layer. The resulting pattern of the its own layer, 3 above and 3 below. Hence, the layers will be 'ABAB....'. This arrangement coordination number of any sphere in hcp or results in hexagonal close packed (hcp) ccp/fcc structure is 12. structure (Fig. 1.8(a)). Metals such as Mg, 1.6.3 Number of voids per atom in hcp and Zn, have hcp crystal structure. ccp : The tetrahedral and octahedral voids The second way of placing the third occur in hcp and ccp/fcc structures. There are hexagonal close packed layer on the second two tetrahedral voids associated with each is to cover the octahedral voids by spheres of atom. The number of octahedral voids is half the third layer. In such placing, the spheres of that of tetrahedral voids. Thus, there is one the third layer do not align with the spheres octahedral void per atom. of the second or the spheres of the first layer. The third layer is, therefore, called 'C' layer. Remember... The spheres of the fourth layer get aligned If N denotes number of particles, with the spheres of the first layer. Hence, the then number of tetrahedral voids is fourth layer is called 'A' layer. This pattern 2N and that of octahedral voids is N. of stacking hexagonal close packed layers is called 'ABCABC....'. This arrangement 1.7 Packing efficiency : Like coordination results in cubic close packed (ccp) structure number, the magnitude of packing efficiency (Fig. 1.8(b)). This is same as fcc structure. gives a measure of how tightly particles are Metals such as copper and Ag have ccp (or packed together. fcc) crystal structure. Packing efficiency is the fraction or a percentage of the total space occupied by the spheres (particles). Packing efficiency = Expanded volume occupied by particles in unit cell (a) view (a) ×100 total volume of unit cell (1.5) 1.7.1 Packing efficiency of metal crystal in simple cubic lattice is obtained by the Expanded (b) view (b) following steps. Fig. 1.8 : Formation of hexagonal closed packed Step 1 : Radius of sphere : In simple cubic structures unit cell, particles (spheres) are at the corners 1.6.2 Coordination number in close packed and touch each other along the edge. A face of structure simple cubic unit cell is shown in Fig. 1.9. It is a. In the simple cubic (sc) crystal structure, evident that that results from stacking of square close a = 2r or r = a/2 (1.6) packed layers, each sphere is surrounded by where r is the radius of atom and ‘a’ is the 6 neighbouring spheres, 4 in its own layer, length of unit cell edge. 11 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ a r r Fig. 1.9 : Face of simple cubic unit cell Step 2 : Volume of sphere : Volume of a Fig. 1.10 : bcc unit cell sphere = (4/3π)(r3). Substitution for r from For triangle FED, ∠ FED = 900. Eq. (1.6) gives ∴ FD2 = FE2 + ED2 = a2 +a2 = 2a2 (because Volume of one particle FE = ED = a) (1.8) πa3 = (4/3π) (a/2)3 = (1.7) For triangle AFD, ∠ ADF = 900 6 Step 3 : Total volume of particles : Because ∴ AF2 = AD2 + FD2 (1.9) simple cubic unit cell contains only one particle, Substitution of Eq. (1.8) into Eq. (1.9) yields πa3 AF2 = a2 + 2a2 = 3a2 (because AD = a) volume occupied by particle in unit cell = 6 Step 4 : Packing efficiency or AF = 3 a (1.10) Packing efficiency The Fig. 1.10 shows that AF = 4r. volume occupied by particle in unit cell Substitution for AF from equation (1.10) gives = total volume of unit cell × 100 3 3a = 4r and hence, r = a (1.11) 4 πa3/6 100π 100 × 3.142 Step 2 : Volume of sphere : Volume of sphere = a3 × 100 = 6 = 6 = 52.36% particle = 4/3 π r3. Substitution for r from Eq. (1.11), gives 4 Thus, in simple cubic lattice, 52.36 % of volume of one particle = 3 π ( 3/4a)3 total space is occupied by particles and 47.64% is empty space, that is, void volume. 4 ( 3)3 3 = 3 π× a 1.7.2 Packing efficiency of metal crystal in 64 body-centred cubic lattice 3 π a3 Step 1 : Radius of sphere (particle) : = 16 In bcc unit cell, particles occupy the corners Step 3 : Total volume of particles : Unit and in addition one particle is at the centre of cell bcc contains 2 particles. Hence, volume the cube. Figure 1.10 shows that the particle occupied by particles in bcc unit cell at the centre of the cube touches two corner 2 3 πa3 =2× particles along the diagonal of the cube. 16 To obtain radius of the particle (sphere) 3 πa3 Pythagorus theorem is applied. = (1.12) 8 12 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ Step 4 : Packing efficiency 4 1 3 = 3 π a × ( 2 2) 3 Packing efficiency volume occupied by particles in unit cell π a3 = ×100 = total volume of unit cell 12 2 Step 3 : Total volume of particles : The unit πa3 cell of fcc lattice contains 4 particles. Hence, = 3 3 × 100 = 68 % volume occupied by particles in fcc unit cell 8a π a3 =4× πa = 3 Thus, 68% of the total volume in bcc unit 12 2 32 lattice is occupied by atoms and 32 % is empty space or void volume. Step 4 : Packing efficiency : Packing 1.7.3 Packing efficiency of metal crystal efficiency in face-centred cubic lattice (or ccp or hcp volume occupied by particles in unit cell = total volume of unit cell × 100 lattice) πa3 Step 1 : Radius of particle/sphere : The corner = × 100 = π × 100 = 74% 3 2a 3 32 particles are assumed to touch the particle at the centre of face ABCD as shown in Fig. 1.11. Thus in fcc/ccp/hcp crystal lattice, 74% of the total volume is occupied by particles and 26% The triangle ABC is right angled is void volume or empty space. with ∠ABC = 900. According to Pythagorus theorem, Table 1.3 shows the expressions for AC2 = AB2 + BC2 = a2 +a2 = 2a2 various parameters of particles in terms of unit cell dimension for cubic systems. Use your brain power Which of the three lattices, sc, bcc and fcc has the most efficient packing of particles? Which one has the least efficient packing? Table 1.3 : Edge length and particle parameters in cubic system Fig. 1.11 : fcc unit cell Unit Relation Volume Total (because AB = BC = a) cell between a of one volume and r particle occupied Hence, AC = 2 a (1.13) by Figure 1.11 shows that AC = 4 r. Substitution particles for AC from Eq. (1.13) gives in unit cell a 2a = 4r or r = 2 a = (1.14) 1. sc r = a/2 = πa /6 = 3 πa3/6 = 4 2 2 0.5000a 0.5237 a3 0.5237 a3 Step 2 : Volume of sphere : Volume of one 2. bcc r = 3a/4 = 3 πa /16 3 πa /8 3 3 4 particle = 3 πr3. Substitution for r from 0.4330a = 0.34a3 = 0.68a 3 Eq. (1.14) gives 3. fcc/ r = 2a/4 = πa /12 2 πa /3 2 3 3 4 a 3 Volume of one particle = 3 π ( ) ccp 0.3535a = 0.185a3 = 0.74a3 2 2 13 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ Table 1.4 shows the summary of coordination substitution of M gives xNA number of particles and packing efficiency in Number of particles in 'x' g = ρa3N /n various cubic systems. A Table 1.4 : Coordination number and packing xn = efficiency in systems ρa3 Number of unit cells in 'x' g metal : Lattice Coordination Packing number of atoms efficiency ∴ 'n' particles correspond to 1 unit cell 1. sc 6 : four in the 52.4 % same layer, one xn xn 1 ∴ particles correspond to × directly above ρa3 ρa3 n and one directly unit cells. below x 2. bcc 8 : four in the 68 % ∴Number of unit cells in 'x' g metal = layer below and ρa3 one in the layer Number of unit cells in volume 'V' of above V metal = 3 3. fcc/ccp/ 12 : six in its 74 % a hcp own layer, three above and three Problem 1.2 A compound made of elements below C and D crystallizes in fcc structure. Atoms of C are present at the corners of the cube. 1.7.4 Number of particles and unit cells in Atoms of D are at the centres of faces of the x g of metallic crystal : cube. What is the formula of the compound? The number of particles and the number of Solution: unit cells in given mass of a metal can be i. C atoms are present at the 8 corners. The calculated from the known parameters of unit contribution of each corner atom to the unit cell, namely, number of particles 'n' per unit cell is 1/8 atom. Hence, the number of C cell and volume 'a3' of unit cell. Density (ρ) atom that belongs to the unit cell = 8×(1/8) and molar mass (M) of a metal are related =1 to each other through unit cell parameters as shown below : ii. D atoms are present at the centres of mass six faces of unit cell. Each face-centre ρ= atom is shared between two cubes. Hence, volume number of particles in unit cell M centribution of each face centre atom to the = × unit cell is 1/2 atom. volume of unit cell NA n M The number of D atoms that belong ∴ρ = 3 × a NA to unit cell = 1/2×6 =3 3 a NA ∴M = ρ There are one C atom and three D atoms in n the unit cell. where 'n' is the number of particles in unit cell and 'a3' is the volume of unit cell. ∴ Formula of compound = CD3 Number of particles in 'x' g metal : ∴ Molar mass, M, contains NA particles xNA ∴ x g of metal contains particles. M 14 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ Problem 1.3 : The unit cell of metallic Solution : The atoms of element B form silver is fcc. If radius of Ag atom is 144.4 ccp structure. The number of tetrahedral pm, calculate (a) edge length of unit cell(b) voids generated is twice the number of B volume of Ag atom, (c) the percent of the atoms. volume of a unit cell, that is occupied by Ag Thus, number of tetrahedral voids = 2B atoms, (d) the percent of empty space. The atoms A occupy (1/3) of these Solution: tetrahedral voids. (a) For fcc unit cell, r = 0.3535 a Hence, number of A atoms = 2B×1/3 r = 144.4 pm = 144.4 × 10 m -12 Ratio of A and B atoms = 2/3 B: 1B = 144.4 × 10 cm -10 = 2/3:1 = 2:3 r 144.4 × 10-10cm a = 0.3535 = Formula of compound = A2B3 0.3535 = 4.085 × 10-8 cm 4 Problem 1.5 : Niobium forms bcc structure. (b) Volume of Ag atom = π r3 3 The density of niobium is 8.55 g/cm3 and 4 = × 3.142 × (144.4 × 10-10 cm)3 length of unit cell edge is 330.6 pm. How 3 many atoms and unit cells are present in = 1.261 × 10-23 cm3 0.5 g of niobium? (c) In fcc unit cell, there are 4 Ag atoms Solution: Volume occupied by 4 Ag atoms xn i. Number of atoms in x g niobium = ρa3 = 4×1.26×10-23 cm3 x = 0.5 g, n = 2 (for bcc structure), = 5.044 × 10-23 cm3 ρ = 8.55 g/cm3, Total volume of unit cell = a3 a = 330.6pm = 3.306×10-8cm. = (4.085×10-8 cm)3 Number of atoms in 0.5 g of niobium = 6.817×10 -23 cm 3 0.5 g × 2 = 8.55 g cm-3 × (3.306×10-8 cm)3 Percent of volume occupied by Ag atoms volume occupied by atoms in unit cell = total volume of unit cell × 100 = 3.25×1021 x ii. Number of unit cells in x g = ρa3 5.044×10 cm -23 3 = = 74% 6.817×10-23cm3 Number of unit cells in 0.5 g of niobium (d) Percent empty space = 100 - 74 = 26% =0.5 g × 2 Problem 1.4 : A compound is formed 8.55 g cm × (3.306×10-8cm)3 -3 by two elements A and B. The atoms of = 1.62 ×1021 element B forms ccp structure. The atoms of A occupy 1/3rd of tetrahedral voids. What is the formula of the compound ? 15 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ There are three types of defects: point Problem 1.6 : A compound forms hcp defects, line defects and plain defects. Only structure. What is the number of (a) point defects will be discussed in this chapter. octahedral voids (b) tetrahedral voids (c) total voids formed in 0.4 mol of it. 1.8.1 Point defects : These defects are irregularities produced in the arrangement of Solution : basis at lattice points in crystalline solids. Number of atoms in 0.4 mol = 0.4 × NA There are three major classes of point = 0.4 × 6.022 × 1023 = 2.4098 × 1023 defects: stoichiometric point defects, impurity (a) Number of octahedral voids = number defects and nonstoichiometric point defects. of atoms = 2.4098 × 1023 a. Stoichiometric point defects : Chemical (b) Number of tetrahedral voids formula of a compound shows fixed ratio of number of atoms or number of cations and = 2×number of atoms anions. This fixed ratio is the stoichiometry of = 2×2.4098×1023 the compound. = 4.818×1023 In stoichiometric defect, the stoichiometry (c) Total number of voids remains unchanged. In other words, the ratio of number of atoms or number of cations and = 2.409×1023+ 4.818×1023 anions of compound remains the same as = 7.227 × 1023 represented by its chemical formula. 1.8 Crystal defects or imperfections : The There are four types of stoichiometric real, naturally occurring crystalline substances point defects: vacancy defect, self interstitial do not have perfect crystal structures. They defect, Schottky defect and Frenkel defect. have some disorders or irregularities in the i. Vacancy defect : During crystallization of stacking of atoms. Such irregularities in the a solid, a particle is missing from its regular arrangement of constituent particles of a solid site in the crystal lattice. The missing particle crystal are called defects or imperfections. creates a vacancy in the lattice structure. Thus, Defects are created during the process some of the lattice sites are vacant because of of crystallization. The imperfections are more missing particles as shown in Fig. 1.12. The if the crystallization occurs at a faster rate. It crystal is, then, said to have a vacancy defect. means that the defects can be minimized by The vacancy defect can also be developed carrying out crystallization at a slower rate. when the substance is heated. In fact ideal crystals with no imperfections are possible only at the absolute zero of temperature. Above this temperature no crystalline materials are 100 % pure. They contain defects. Whatever be the nature of a crystal defect, electrical neutrality of the solid is maintained. It is important to note that sometimes defects are to be intentionally created for manipulating the desired properties in Fig. 1.12 : Vacancy defect crystalline solids. 16 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ Due to the absence of particles, the It is interesting to note that in this second mass of the substance decreases. However, case, because of the displacement of a particle a the volume remains unchanged. As a result the vacancy defect is created at its original regular density of the substance decreases. lattice site. At the same time interstitial defect ii. Self interstitial defect in elemental solid results at its new position. We can, therefore, say that in this defect there is a combination of Interstitial sites in a crystal are the vacancy defect and self interstitial defect. spaces or voids in between the particles at lattice points. When some particles of a This defect preserves the density of the crystalline elemental solid occupy interstitial substance because there is neither loss nor sites in the crystal structure, it is called self gain in mass of a substance. interstitial defect. iii. Schottky defect : In an ionic solid, equal This defect occurs in the following two ways : number of cations and anions are missing from their regular positions in the crystal Firstly, an extra particle occupies an empty lattice creating vacancies as shown in Fig. interstitial space in the crystal structure as 1.15. It means that a vacancy created by a loss shown in Fig. 1.13. This extra particle is same of cation is always accompanied by a vacancy as those already present at the lattice points. formed by a loss of anion. Fig. 1.13 : Self interstitial defect Fig. 1.15 : Schottky defect The extra particles increase the total Thus, there exist two holes per ion pair mass of substance without increasing volume. lost, one created by missing cation and the Hence its density increases. other by a missing anion. Such a paired cation- Secondly, in an elemental solid a particle anion vacancy defect is a Schottky defect. gets shifted from its original lattice point and Conditions for the formation of Schottky occupies an interstitial space in the crystal as defect shown in the Fig. 1.14. i. Schottky defect is found in ionic compounds with the following characteristics : High degree of ionic character. High coordination number of anion Small difference between size of cation and anion. The ratio rcation/ranion is not far below unity. Fig. 1.14 : Self interstitial defect 17 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ Consequences of Schottky defect Conditions for the formation of Frenkel As the number of ions decreases, mass defect decreases. However, volume remains Frenkel defect occurs in ionic compounds unchanged. Hence, the density of a with large difference between sizes of substance decreases. cation and anion. The number of missing cations and anions The ions of ionic compounds must be is equal, the electrical neutrality of the having low coordination number. compound is preserved. Consequences of Frenkel defect This defect is found in ionic crystals such as NaCl, AgBr and KCl. As no ions are missing from the crystal lattice as a whole, the density of solid and iv. Frenkel defect : Frenkel defect arises when its chemical properties remain unchanged. an ion of an ionic compound is missing from its regular lattice site and occupies interstitial The crystal as a whole remains electrically position between lattice points as shown in neutral because the equal numbers of Fig. 1.16. cations and anions are present. The cations are usually smaller than This defect is found in ionic crystals like ZnS, anions. It is, therfore, more common to find AgCl, AgBr, AgI, CaF2. the cations occupying interstitial sites. It is b. Impurity defect : Impurity defect arises easier for the smaller cations to accomodate when foreign atoms, that is, atoms different the interstitial spaces. from the host atoms, are present in the crystal lattice. There are two kinds of impurity defects : Substitutional and interstitial impurity defects. i. Substitutional impurity defect : In this defect, the foreign atoms are found at the lattice sites in place of host atoms. The regular atoms are displaced from their lattice sites by impurity atoms. For example : Fig. 1.16 : Frenkel defect Solid solutions of metals (alloys) : Brass is an alloy of Cu and Zn. In brass, host Do you know ? Cu atoms are replaced by impurity of Zn Frenkel defect is not found in atoms. The Zn atoms occupy regular sites pure alkali metal halides because of Cu atoms as shown in Fig. 1.17. cations of allkali metals due to large size cannot occupy interstitial space. It is important to note that the smaller cation is displaced from its normal site to an interstitial space. It, therefore, creates a vacancy defect at its original position and interstitial defect at its new location in the same crystal. Frenkel defect can be regarded as the combination of vacancy defect and interstitial defect. Fig. 1.17 : Brass 18 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ Vacancy through aliovalent impurity : c. Nonstoichiometric defects : Vacancies are created by the addition Nonstoichiometric defect arises when of impurities of aliovalent ions (that is, ions the ratio of number of atoms of one kind to with oxidation state (o.s.) different from that that of other kind or the ratio of number of of host ions) to an ionic solid. cations to anions becomes different from that indicated by its chemical formula. In short, stoichiometry of the compound is changed. It is important to note that the change in stoichiometry does not cause any change in the crystal structure. There are two types of nonstoichiometric defects i. Metal deficiency defect : This defect is possible only in compounds of metals that Fig. 1.18 : Vacancy through aliovalent ion show variable oxidation states. Suppose that a small amount of SrCl2 impurity is added to NaCl during its In some crystals, positive metal ions crystallization. The added Sr2⊕ ions (O.S. + 2) are missing from their original lattice sites. occupy some of the regular sites of Na⊕ host The extra negative charge is balanced by the ions (O.S.+1). presence of cation of the same metal with higher oxidation state than that of missing In order to maintain electrical cation. neutrality, every Sr2⊕ ion removes two Na⊕ ions. One of the vacant lattice sites created by For example, in the compound NiO removal of two Na⊕ ions is occupied by one one Ni ion is missing creating a vacnacy at 2⊕ Sr2⊕ ion. The other site of Na⊕ ion remains its lattice site. The deficiency of two positive vacant as shown in Fig. 1.18. charges is made up by the presence of two Ni3⊕ ions at the other lattice sites of Ni2⊕ ions as ii. Interstitial impurity defect : In this defect, shown in Fig. 1.20. The composition of NiO the impurity atoms occupy interstitial spaces then becomes Ni0.97O1.0 of lattice structure. For example in steel, Fe atoms occupy normal lattice sites. The carbon atoms are present at interstitial spaces, as shown in Fig. 1.19. Fig. 1.20 : Nonstoichiometric Ni0.97O1.0 ii. Metal excess defect : There are two types of metal excess defects. A neutral atom or an extra positive ion occupies interstitial position : ZnO Fig. 1.19 : Stainless steel presents two ways of metal excess defect. In the first case in ZnO lattice one neutral 19 https://www.ncertbooks.guru/  https://www.ncertbooks.guru/ Zn atom is present in the interstitial space Cl ions diffuse to the crystal surface as shown in Fig. 1.21(a) creating vacancies at their regular sites. These Cl ions combine with Na atoms on the surface to form NaCl, by releasing electron from sodium atom. Na + Cl NaCl +e The electrons released diffuse into the crystal and occupy vacant sites of anions as shown in Fig. 1.22. The anion vacant sites occupied by electrons are F-centres or colour- Fig. 1.21 (a) : Neutral Zn atom at interstitial centres. site Fig. 1.21 (b) : Zn2+ ions and electrons at interestitial sites Fig. 1.22 : An F-centre in a crystal In the second case, when ZnO is heated it decomposes as : NaCl shows yellow colour due to the formation of F-centre. The crystal of NaCl Z