Đang tải... (xem toàn văn)
https://1drv.ms/b/s!AmkCsf2WlV7n1WoyCT5nnvO18-29?e=EB8hdN
Krishna's CHEM STRY for By Alok Bariyar Ph.D (CSIR fellow), M.Sc.(IIT Delhi), G.A.T.E., N.E.T Ex Nuclear Scientist Bhabha Atomic Research Centre, Mumbai, India Prakashan Media Pvt Ltd KRISHNA HOUSE, 11, Shivaji Road, Meerut-250 001 (U.P.), India CHEMISTRY for Syllabus PHYSICAL CHEMISTRY Basic Mathematical Concepts: Functions; maxima and minima; integrals; ordinary differential equations; vectors and matrices; determinants; elementary statistics and probability theory Atomic and Molecular Structure: Fundamental particles; Bohr's theory of hydrogen-like atom; wave-particle duality; uncertainty principle; Schrodinger's wave equation; quantum numbers; shapes of orbitals; Hund's rule and Pauli's exclusion principle; electronic configuration of simple homonuclear diatomic molecules Theory of Gases: Equation of state for ideal and non-ideal (van der Waals) gases; Kinetic theory of gases; MaxwellBoltzmann distribution law; equipartition of energy Solid State: Crystals and crystal systems; X-rays; NaCl and KCl structures; close packing; atomic and ionic radii; radius ratio rules; lattice energy; Born-Haber cycle; isomorphism; heat capacity of solids Chemical Thermodynamics: Reversible and irreversible processes; first law and its application to ideal and nonideal gases; thermochemistry; second law; entropy and free energy; criteria for spontaneity Chemical and Phase Equilibria: Law of mass action; Kp, Kc, Kx and Kn; effect of temperature on K; ionic equilibria in solutions; pH and buffer solutions; hydrolysis; solubility product; phase equilibria-phase rule and its application to onecomponent and two-component systems; colligative properties Electrochemistry: Conductance and its applications; transport number; galvanic cells; EMF and free energy; concentration cells with and without transport; polarography; concentration cells with and without transport; Debey-Huckel-Onsagar theory of strong electrolytes Chemical Kinetics: Reactions of various order; Arrhenius equation; collision theory; transition state theory; chain reactions normal and branched; enzyme kinetics; photochemical processes; catalysis Adsorption: Gibbs adsorption equation; adsorption isotherm; types of adsorption; surface area of adsorbents; surface films on liquids 10 Spectroscopy: Beer-Lambert law; fundamental concepts of rotational, vibrational, electronic and magnetic resonance spectroscopy ORGANIC CHEMISTRY 11 Basic Concepts in Organic Chemistry and Stereochemistry: Electronic effects (resonance, inductive, hyperconjugation) and steric effects and its applications (acid/base property); optical isomerism in compounds with and without any stereocenters (allenes, biphenyls); conformation of acyclic systems (substituted ethane/n-propane/n-butane) and cyclic systems (mono- and di-substituted cyclohexanes) 12 Organic Reaction Mechanism and Synthetic Applications: Chemistry of reactive intermediates (carbocations, carbanions, free radicals, carbenes, nitrenes, benzynes etc ); Hofmann-Curtius-Lossen rearrangement, Wolff rearrangement, Simmons-Smith reaction, Reimer-Tiemann reaction, Michael reaction, Darzens reaction, Wittig reaction and McMurry reaction; Pinacol-pinacolone, Favorskii, benzilic acid rearrangement, dienone-phenol rearrangement, Baeyer-Villeger reaction; oxidation and reduction reactions in organic chemistry; organometallic reagents in organic synthesis (Grignard, organolithium and organocopper); Diels-Alder, electrocyclic and sigmatropic reactions; functional group inter-conversions and structural problems using chemical reactions 13 Qualitative Organic Analysis: Identification of functional groups by chemical tests; elementary UV, IR and 1H NMR spectroscopic techniques as tools for structural elucidation 14 Natural Products Chemistry: Chemistry of alkaloids, steroids, terpenes, carbohydrates, amino acids, peptides and nucleic acids (iii) 15 Aromatic and Heterocyclic Chemistry: Monocyclic, bicyclic and tricyclic aromatic hydrocarbons, and monocyclic compounds with one hetero atom: synthesis, reactivity and properties INORGANIC CHEMISTRY 16 Periodic Table: Periodic classification of elements and periodicity in properties; general methods of isolation and purification of elements 17 Chemical Bonding and Shapes of Compounds: Types of bonding; VSEPR theory and shapes of molecules;hybridization; dipole moment; ionic solids; structure of NaCl, CsCl, diamond and graphite; lattice energy 18 Main Group Elements (s and p blocks): General concepts on group relationships and gradation in properties; structure of electron deficient compounds involving main group elements 19 Transition Metals (d block): Characteristics of 3d elements; oxide, hydroxide and salts of first row metals; coordination complexes: structure, isomerism, reaction mechanism and electronic spectra; VB, MO and Crystal Field theoretical approaches for structure, color and magnetic properties of metal complexes; organometallic compounds having ligands with back bonding capabilities such as metal carbonyls, carbenes, nitrosyls and metallocenes; homogenous catalysis 20 Bioinorganic Chemistry: Essentials and trace elements of life; basic reactions in the biological systems and the role of metal ions, especially Fe2+, Fe3+, Cu2+ and Zn2+; structure and function of hemoglobin and myoglobin and carbonic anhydrase 21 Instrumental Methods of Analysis: Basic principles; instrumentations and simple applications of conductometry, potentiometry and UV-vis spectrophotometry; analysis of water, air and soil samples 22 Analytical Chemistry: Principles of qualitative and quantitative analysis; acid-base, oxidation-reduction and complexometric titrations using EDTA; precipitation reactions; use of indicators; use of organic reagents in inorganic analysis; radioactivity; nuclear reactions; applications of isotopes Weightage of Inorganic, Organic & Physical Chemistry Questions Asked in Last Three Years JAM 2016 JAM 2018 JAM 2017 INORGANIC INORGANIC INORGANIC 17 Q PHYSICAL 10 Q 13 Q 27 Q 23 Q PHYSICAL 20 Q 20 Q 28 Q ORGANIC PHYSICAL 22 Q ORGANIC ORGANIC Physical Chemistry : 23 Questions Physical Chemistry : 27 Questions Physical Chemistry : 28 Questions Organic Chemistry : 20 Questions Organic Chemistry : 20 Questions Organic Chemistry : 22 Questions Inorganic Chemistry : 17 Questions Inorganic Chemistry : 13 Questions Inorganic Chemistry : 10 Questions (iv) Preface DEDICATED TO LORD KRISHNA – Publishers & Author I am happy to present this book entitled "Chemistry for IIT JAM" It has been written according to the Latest syllabus to fulfil the requirement of students The book is written with the following special features: It is written in a simple language so that all the students may understand it easily It has an extensive and intensive coverage of all topics The complete syllabus has been divided into sections having 22 Chapters Sufficient Numerical Problems, Subjective Questions and Objective type questions with Hints & Solutions given at the end of each chapter will enable students to understand the concept Sufficient Mock Test Papers have also been added for the practice of students First of all I want to express my sincere gratitude to Dr S.B.P Sinha, Prof J.C Ahluwalia, Prof N.K Jha for their invaluable guidance, immense interest and constant encouragement for the successful completion of the work I am also thankful to Bandana Bariyar, Abhishek Bariyar, Purnima Sinha & family for their kind help at many occasions I am extremely grateful to my respected and beloved parents whose incessant inspiration guided us to accomplish this work I also express gratitude to my respected family members for their moral support I am immensely thankful to Mr S.K Rastogi, Managing Director, Mr Sugam Rastogi, Executive Director, Mrs Kanupriya Rastogi, Director and entire team of Krishna Prakashan Media (P) Ltd., for taking keen interest in getting the book published The originality of the ideas is not claimed and criticism and suggestions are invited from the students, teaching community and other readers Author (v) Exam Pattern Paper Specific Instructions The examination is of hours duration There are a total of 60 questions carrying 100 marks The entire paper is divided into three sections, A, B and C All sections are compulsory Questions in each section are of different types Section - A contains a total of 30 Multiple Choice Questions (MCQ) Each MCQ type question has four choices out of which only one choice is the correct answer Questions Q.1-Q.30 belong to this section and carry a total of 50 marks Q.1-Q.10 carry mark each and Questions Q.11 - Q.30 carry marks each Section - B contains a total of 10 Multiple select Questions (MSQ) Each MSQ type question is similar to MCQ but with a difference that there may be one or more than one choice(s) that are correct out of the four given choices The candidate gets full credit if he/she selects all the correct answers only and no wrong answers Questions Q.31-Q.40 belong to this section and carry marks each with a total of 20 marks Section - C contains a total of 20 Numerical Answer Type (NAT) questions For these NAT type questions, the answer is a real number which needs to be entered using the virtual keyboard on the monitor No choices will be shown for these type of questions Questions Q.41-Q.60 belong to this section and carry a total of 30 marks Q.41-Q.50 carry mark each and Questions Q.51-Q.60 carry marks each In all sections, questions not attempted will result in zero mark In Section-A (MCQ), wrong answer will result in NEGATIVE marks For all mark questions, 1/3 marks will be deducted for each wrong answer For all marks questions, 2/3 marks will be deducted for each wrong answer In Section - B (MSQ), there is NO NEGATIVE and NO PARTIAL marking provisions There is NO NEGATIVE marking in Sections-C (NAT) as well Source : jam.iitb.ac.in (vi) Common Mistakes Done By Students Avoiding the Basic Concepts : Students generally start their study with reference books and other advanced study materials They forget that to fight a battle in the field it's important to learn basic concepts of the fight, which is applicable to IIT JAM as well Not working on conceptual terms tend to make students puzzled when the question is slightly twisted Tomorrow Never Comes : It's a natural tendency of humans to say, “I'll it from tomorrow onwards” Making a schedule alone is not enough to make student win this battle, you must work on it with full dedication To create an effective timeschedule, save the important dates for the exam and plan up your activities accordingly Self Confidence : Self confidence is the key to success, you can lose even a small fight if you have doubts about yourself On the other hand, self-confidence is the only key that can lead you to win over the battles Giving a little space to doubts leads to hesitation and lack of confidence Have trust in yourself and make a positive mind set about whatever you have prepared is more than enough for exams This will increase your confidence and you will perform well in the exam Mock Tests : Mock Tests are practice tests used as a mirror to your skill sets They help you in finding out the true analysis of your performance & knowledge level Helping you in increasing your speed, they also improve your answer giving skills with greater accuracy One should never skip these tests as they are the most important part of preparation Skipping Revision : Taking too much stress or no stress, both can be harmful to a student Many students panic a lot which worsens their performance in the exams and many of them are so relaxed that they not bother even about the revision Students must understand that the preparation they have done might be enough but revision is of equal importance Revision Tips Start Revising Early: Always make a time table for revision well before the exams A time table can include which topics to revise to score good marks Divide time accordingly for all the subjects Make Summary Notes: By making summary notes students can focus on key points of important topics Take Short Breaks: During revision take short breaks after every hour not every 10 minutes so that you feel energetic Have a Proper Sleep: Having less sleep could impact your next day revision and could lead to less concentration for revision Research has shown that lack of sleep leads to clogging of brain and makes us forget sooner Revision Room and Sitting Conditions: Revise in a room that is quiet and you don't get distracted from outside noises Use proper chair and table for revision Don't try to revise in bed Wash your face and eyes with water to avoid sleeping state Lastly, be confident before exams with all the revision you have done Remember revision is a skill that leads to improvement of memory and enahnces learning ability of students Practice makes one perfect but effective Revision Techniques put excellence in ones perfectness (vii) Contents SECTION - A Chapter : Basic Mathematical Concepts (A-01-A-40) Chapter : Atomic and Molecular Structure (B-01-B-24) Chapter : Theory of Gases (C-01-C-16) Chapter : Solid State (D-01-D-16) Chapter : Chemical Thermodynamics .(E-01-E-20) Chapter : Chemical and Phase Equilibria (F-01-F-28) Chapter : Electrochemistry .(G-01-G-20) Chapter : Chemical Kinetics .(H-01-H-28) Chapter : Adsorption (I-01-I-08) Chapter 10 : Spectroscopy (J-01-J-16) SECTION - B Chapter : Basic Concepts in Organic Chemistry and Stereochemistry (A-01-A-32) Chapter : Organic Reaction Mechanism and Synthetic Applications (B-01-B-44) Chapter : Qualitative Organic Analysis .(C-01-C-24) Chapter : Natural Products Chemistry .(D-01-D-36) Chapter : Aromatic and Heterocyclic Chemistry (E-01-E-40) SECTION - C Chapter : Periodic Table (A-01-A-24) Chapter : Chemical Bonding and Shapes of Compounds (B-01-B-20) Chapter : Main Group Elements (s and p blocks) (C-01-C-60) Chapter : Transition Metals (d Block) (D-01-D-32) Chapter : Bioinorganic Chemistry (E-01-E-08) Chapter : Instrumental Methods of Analysis .(F-01-F-12) Chapter : Analytical Chemistry (G-01-G-40) Mock Test Papers Latest Examination Paper (viii) Section A Chapter 1: Basic Mathematical Concepts Chapter 2: Atomic and Molecular Structure Chapter 3: Theory of Gases Chapter 4: Solid State Chapter 5: Chemical Thermodynamics Chapter 6: Chemical and Phase Equilibria Chapter 7: Electrochemistry Chapter 8: Chemical Kinetics Chapter 9: Adsorption Chapter 10: Spectroscopy A-3 C HAPTER Section-A Basic Mathematical Concepts Syllabus • Functions • Maxima and minima • Integrals • Ordinary differential equations • Vectors and matrices • Determinants • Elementary statistics and probability theory INTRODUCTION TO FUNCTION Let A & B be two sets & let there exist a rule or manner or correspondence ’f ’ which associates to each element of A, a unique element in B Then f is called a function or mapping from A to B.It is denoted by the symbol f : A → B or A f B which reads “f ” is a function from A to B or f maps A′ toB So it can be concluded that f:A→B Input Output y = f(x) One many ⇒ Not a function Many one ⇒ Function x y A B INTERVALS The set of numbers between any two real numbers is called interval Types of intervals : A relation R from a set A to a set B is called a function if (i) Representation : [a, b] i e., a ≤ x ≤ b Each element of A is associated with some element of B (ii) Each element of A has unique image in B Some important points : To every x there is one & only one y i.e a b c Closed intervals : Open intervals : Representation : ] a, b [ or (a, b) i e., a < x < b Semiclosed or semi open : [a, b] → a ≤ x < b Not a function [a, b] → a < x ≤ b) Example 1: Solve a 2 3 b Not a function c a b c a b c d Is a function − 2x x −4 7− x + ≥ − x and ≥ −3 −3 Solution : 6x − + 2x − 7−x ≥1 ⇒ ≥ 1− x & −6 −3 7−x ≥ −3 ⇒ 8x − 17 ≤ 6x − & or 2x ≤ 11 11 x≤ 11 x ε −∞, 2 11/2 & − x ≤ − 10 & x ≥ 10 & x ε [10, ∞) 10 Is a function Hence, no common value so x ε φ A-4 EXERCISE A + Solve −6 ≤ (1 − x ) ≤ Solve −3 ≤ − 2x ≤ & −4 ≤ − 5x < + –∞ –1 WAVY CURVE METHOD If a, b = ⇒ either a = or b = But if a ⋅ b > ⇒ if we conclude a > or b > it is wrong Similarly, if (x − 1) (x − 2) > 0, then this doesn’t mean (x − 1) > or (x − 2) > These wrong interpretation can be corrected by wavy curve method x ε [1, 4] Example 3: Solve ( x + 1)2 ( x − 2)3 ( x − 4)4 ( x − 3)2 ( x − 7)5 –1 f (x ) = n2 (x − a k ) m1 m2 (x − bk )mp (x − a1) (x − a 2) (x − b1) (x − b2) (≥ 0, ≤ 0) Step I : First arrange all values of x at which either numerator or denominator becomes zero Step II : Values of x at which numerator becomes zero should be marked with dark circle So wave will not change the direction at these points + So Step IV : From right to left, draw a wavy curve (Beginning above the number line in case of value of f (x ) is positive in step III otherwise form below the number line) Passing throughly all the marked points so that when passes through a points (exponent or power whose factor is odd) intersect the number line & when passing thoroughly a point (exponent whose corresponds factor is even) the curve doesn’t intersect the real line & remain on the same side of real line Step V : The appropriate intervals are chosen in accordance with the sign of inequality (the function f (x ) is positive wherever the curve is above the number line It is negative if the curve is found below the number line) Their union represents the solution of inequality x =1 x=3 – +∞ Example 2: Solve 2x + x − 1< x + 4x + or x − 3x − < x − 3x − < x = & x = −1 – + – ∞ –∞ + – – + ∞ x ε [0, 1] ∪ [3, ∞] Example 5: Solve ( x − 1)3 ( x − 2)5 ( x − 7) ( x − 3)2 ( x − 1)5 x ≤0 Solution : Don’t cancel common factor, instead, add their power Here for x − 1, + = i.e., even wave will not change the direction + –∞ + + + – – ∞ EXERCISE B Solve (x − 1) (x − 2) ≥0 (x − 3) (x − 4) p2 ≥ 16 f (x ) = (x − 4) (x − 3x + 2) (x − 1) (x 2) (x + 3x + 2) for which f (x ) is positive x ε [1, 3] (x − 4) (x + 1) < or – Solution : + + Example 4: Solve ( x − 1)3 ( x − 2)2 ( x − 3)2015 x ( x − 4)2020 ≥ Example 1: Solve ( x − 1) (3 − x ) ≥ Solution : + x ε [−∞, 1] ∪ [−1, 2] ∪ (7, ∞) Step III : Values of x at which denominator becomes zero should be marked with empty circle Check the value of f (x ) for any real number greater than the right most marked number or the number line – ∵ Exponents of factor of −1, 3, & are even, Method –∞ Check f (x) for x > 7, f(8) > –∞ –1 − x = or nk W h e r e n1, n n k ; m1, m2, mp a r e r e a l n u m b e r s & a1, a 2, a k , b1, b2 bk are any real numbers such that a1 ≠ b j where i = 1, 2, & j = 1, 2, p Solution : Roots are x − = or ≥0 Solution : Mark points with dark circle for which numerator is zero & point of discontinuity (at which denominator becomes zero) with empty circle In order to solve the inequalities of the form n1 +∞ – y < y (1 − y) 2x + x − ≥ Answers: x ε [−∞ 1] ∪ [2, 3] ∪ [4, ∞] p ε [−∞, 4] ∪ [4, ∞] x ε R { −2, − 1, 2, 0} yεφ C-14 HINTS AND SOLUTIONS MULTIPLE CHOICE QUESTIONS 46 52 (c) Gases show minimum deviation from ideal gas behaviour at low P and high T 53 (b) van der Waal’s equation for real gas is an2 P + (V − nb) = nRT V (a) Volume of cylinder before dent V1 = nRT 100 × 0.0821 × 300 = = 30.8 L P 40 × Volume of cylinder after dent = V2 At high pressure for one mole of gas it is reduced to V1 n W = = V2 n2 W2 V2 = V1 × P= W2 30.8 × 90 = W1 100 RT V−b ∴ Slope of P vs T graph is = 27.7 L 47 R V−b (c) ∵100 ml blood has 0.02 g O2 and 0.08 g CO2 MULTIPLE SELECT QUESTIONS ∴ 10,000 ml blood has g O2 and g CO2 Refer properties of real & ideal gas Molecules of He will collide more frequently as they are light & have large average molecular speed Dec in pressure As T is const, So KE is also const U msp α Effective volume = nb, pressure correction = Using ∴ For × 0.0821 × 310 PV = nRT, for O2 , × V = 32 VO = 1.59 litre CO2 , × V = × 0.0821 × 310 44 VCO = 4.62 litre ∴ & fraction possessing very low velocity will dec T = 27.7 L 48 PREVIOUS YEAR QUESTIONS (c) PV = RT P∆V = R∆T (for one mole) ∆V R RV V = = = ∆T P RT T ∆V δ = = V∆T T 49 50 51 3RTH MH (c) 3RTN2 MN 50 28 × = = 800 t (a) Vi = V0 1 + ; where V0 is volume at zero degree 273 V V centigrade Use = to get this T1 T2 (c) ∴ RTC R × 8a × 27 b2 = PC VC 27 Rb × a × 3b RTC = PC VC MULTIPLE CHOICE QUESTIONS 3RT M RMS speed = The ratio of different types of molecular speeds is, RMS : Average speed : MPS 1.224 : 1.128 : High temperature Low temperature ¦¦¦ C HAPTER D-1 Section-A Solid State Syllabus • Crystals, crystal systems • X-rays • NaCl and KCl structures • Close packing • Atomic and ionic radii • Radius ratio rules • Lattice energy • Born-Haber cycle • Isomorphism • Heat capacity of solids INTRODUCTION Solid-state Chemistry, is the study of the synthesis, structure, and properties of solid phase materials, particularly of non-molecular solids The main area of study involves different types of lattice structures, determination of interplanar spacings and the type of defects introduced – intrinsic or extrinsic It therefore has a strong overlap with crystallography, ceramics, metallurgy, thermodynamics, solid state physics, materials science and even electronics with a focus on the synthesis of novel materials and their simulation DIFFERENT BASIC TERMS UNIT CELL Solid is defined as, “that form of matter, which possesses rigidity and hence possesses a definite shape and a definite volume.” Solids broadly are of two types : A unit cell is the smallest repeating unit of a solid representing all the properties of the solid A unit cell is the building block of a solid just like amino acids are building blocks of proteins CRYSTALLINE SOLID CRYSTAL SYSTEM of constituent particles Regular arrangement of the constituent particles of a crystalline solid in three dimensional space is called space lattice.According to Bravais (1848), geometrical and symmetrical considerations give rise to only seven possible crystal system or lattice structures for the unit cells These are : AMORPHOUS SOLID Cubic Solid is said to be amorphous if distribution of the constituent particles is not altogether regular Even in the amorphous state there is probably some regularity but the regularity is restricted to few angstroms only Hexagonal Tetragonal Solid is said to be crystalline if the various constituent particles like atoms, ions or molecules are arranged in a definite geometric pattern within the solid giving rise to a highly ordered arrangement Orthorhombic Rhombohedral Monoclinic Triclinic BRAVAIS LATTICES Crystalline solids can be further classified into four types : ionic , Bravais also showed that there would be only four possible ways of arranging the atoms or ions in a unit cell These are: covalent, metallic and molecular solids Primitive or simple Body centered Face centered End centered Table 1: Different crystal system and bravais lattices S.No Crystal system Bravais lattices Parameters of unit cell Intercepts Crystal angle Cubic Primitive, Face centered, Body centered = a=b=c α = β = γ = 90° Orthorhombic Primitive, Face centered, Body centered, End centered = a≠b≠c α = β = γ = 90° Tetragonal Primitive, Body centered =2 a=b≠c α = β = γ = 90° Monoclinic Primitive, End centered = a≠b≠c α = γ = 90° , β ≠ 90° α ≠ β ≠ γ ≠ 90° Triclinic Primitive = a≠b≠c Hexagonal Primitive = a=b≠c α = β = 90°, γ = 120° Rhombohedral Primitive = a=b=c α = γ = 90° , β ≠ 90° Total = 14 D-2 So total number of unit cells possible are × = 28 Bravais predicted that out of the 28 possible unit cells, only 14 unit cells exist in practice These 14 unit cells which exists in nature are referred as Bravais lattices DIFFERENT TYPES OF UNIT CELL AND CLOSE PACKING The important cubic unit cells are : Primitive cubic Body centered cubic Face centered cubic There is one more important type of unit cell which is not cubic, rather it is hexagonal-known as hcp PRIMITIVE CUBIC (SIMPLE CUBIC) A cubic unit cell is said to be primitive if all the eight corners of the cube are occupied by the same atoms and these atoms are not found anywhere else in the cube.Adjoining figure shows the unit cell view of a primitive unit cell Fig 1: Simple cubic arrangement Packing Fraction (PF) : It is defined as “the ratio of the volume occupied by the spheres in a unit cell to the volume of the unit cell.” Volume occupied by the sphere in a unit cell PF = Volume of unit cell = 4 πr πr For a primitive cubic unit cell, PF = = = 0.52 a3 (2r )3 For a PC, no of atoms per unit cell = × 2× ∴ PF = 4 πr × πr 3 = = 0.68 a3 4r 3 FACE-CENTERED CUBIC UNIT CELL A unit cell is said to be ideal facecentered cubic unit cell if the same atoms or ions are present at all the corners of the cube and are also present at the centre of each square face If the atoms present at the corners and at the face centered positions are of different types, then the unit cell is Fig 3: FCC arrangement called pseudo FCC The FCC unit cell view is shown in adjoining sketch It must be understood that the corner atoms are not in contact with each other and that the face-centered atoms are in contact with the four corner spheres The rank or number of atoms per unit cell for FCC would be as each corner sphere contributes 1/8 th to the unit cell while each face-centered sphere contributes 1/2 to the unit cell Moreover, since the face-centered atoms are in contact with four corner atoms, the face diagonal of the cube, which is, 2a is equal to 4r For a FCC, no of atoms per unit cell = × 4× ∴ PF = 1 + × = 4 πr × πr 3 = = 074 a3 4r 2 HEXAGONAL CLOSE PACKED UNIT CELL In this type of lattice structure, the spheres are arranged in the following manner, such that number of formula units per unit cell is BODY-CENTERED CUBIC UNIT CELL A unit cell is said to be ideal body centered cubic unit cell if the same atoms or ions are present at all the eight corners of a cube and is also present at the center of the cube c a Fig 4: 3D HCP arrangement Fig 2: BCC arrangement For a BCC, no of atoms per unit cell = 8× +1= Some more structural features of hcp arrangement are shown diagrammatically D-3 ABAB stacking sequence 3D Projection c 2D Projection A A sites Top layer B sites Middle layer A sites Bottom layer D E B a Fig 7: Cross section through planar triangle site Fig 5: Structural features of HCP The coordination number of hcp and fcc is 12 and its packing fraction is the same as fcc, 0.74 The value of inter layer distance (c) is 1.633a CO-ORDINATION NUMBER Co-ordination number gives rise to a tetrahedral geometry Angle ABC is the tetrahedral angle of 109° 28' and hence the angle ABD is half of this, that is 54°44′ In the triangle ABD, B B A B A C CC A D A 54°44' C Fig 8: Cross section through tetrahedral site Fig 6: The hcp lattice (left) and the fcc lattice (right) DENSITY OF CRYSTAL LATTICE In order to calculate the density of crystal lattice, one calculates the density of the unit cell mass of unit cell volume of unit cell number of effective atoms in the unit cell × mass of each atom = volume of unit cell number of effective atoms in theunit cell × atomic mass = volume of unit cell × Avogadro’ s number Z×M = N AV × a Density of unit cell (ρ) = ATOMIC AND IONIC RADIUS AND RADIUS RATIO The ratio of the radii of the cation to that of the anion is termed as the radius ratio The radius ratio depends upon the co-ordination number of a cation or a void Different co-ordination numbers and corresponding radius ratio are discussed below, sin ABD = 0.8164 r− = AD / AB = r+ + r− Taking reciprocals, r+ + r− r We get, − r+ r− = = 1.225 rearranging 0.8164 = 0.225 CO-ORDINATION NUMBER Co-ordination number yields an octahedral geometry A cross section through an octahedral site is shown in the adjacent figure and the smaller positive ion (of radius r + ) touches six larger negative ions (of radius r − ) Importantly, only four negative ions are shown in fig and one is above and one below the plane of the paper It is obvious that AB = r + + r − and BD = r − The angle ABD is 45° in the triangle ABD cos ABD = 07071 = (BD / AB) = CO-ORDINATION NUMBER The adjacent figure shows the smaller positive ion of radius r + is in contact with three larger negative ions of radii r − It can be seen that AB = BC = AC = 2r − , BD = r − + r + Further, the angle ABC is 60°, and the angle DBE is 30° By trigonometry, cos 30° = (BE / BD) BD = (BE / cos 30° ), r + + r − = r − / cos 30° − B 45° D C Fig 9: Cross section through octahedral site − = (r / 0.866) = r × 1155 r + = (1155 r − ) − r − = 0155 r − , Hence A r+ r− = 0155 r− r+ + r− D-4 Rearranging, r+ r − i.e., NaCl The other substances having this kind of a structure are halides of all alkali metals except cesium halides and oxides of all alkaline earth metals except beryllium oxide If a is the edge length of this unit cell, then = 0.414 The packing efficiency of NaCl is, 4× = 4 πr+ + × πr−3 3 a3 rNa + + rCl − = a (Using 2a = 4r and r+ / r− = 0.52) = 079 KCl shows similar traits Limiting radius r+ ratio, =x r− Fig.10: Rock salt structure CESIUM HALIDE STRUCTURE Co-ordination number Shape Example x < 0155 Linear BeF2 0155 ≤ x < 0225 Planar triangle AlCl 0225 ≤ x < 0414 Tetrahedron ZnS, diamond 0414 ≤ x < 0732 Square planar PtCl 24 − 0414 ≤ x < 0732 Octahedron NaCl 0732 ≤ x < 0999 Body centered cubic CsCl Chloride ions are arranged in a primitive cubic manner while the cesium ion occupies the centre of the unit cell There is one chloride ion and one cesium ion per unit cell Therefore, the formula is CsCl The co-ordination number of cesium is eight and Fig 11: Cesium that of chloride ions is also eight Other halide structure substances which exist in this kind of a structure are various halides of cesium If a is the edge length of the unit cell, then it can be said that, rCs+ + rX− = ISOMORPHISM Existence of different substances in one crystalline form is known as isomorphism and the substances showing isomorphism are known as ‘isomorphous’ In order to form isomorphous crystals two substances must have the same chemical formulation, they must contain atoms which have corresponding chemical properties and the sizes of corresponding atoms should be similar These requirements ensure that the forces within and between molecules and ions are approximately similar and result in crystals that have the same internal structure So, Na SO4 and Ag SO4 are isomorphous as the two compounds exist in hexagonal crystalline form Other examples are, KBF4 and BaSO4 existing in orthorhombic crystal structure, ZnSO4 and NiSO4 both exist in orthorhombic structure, CaCO3 and NaNO3 both exist in trigonal lattice structure etc Some of the characteristics of isomorphous substances are : Isomorphic substances have same atomic ratio Empirical formula of isomorphic substances is same They have different chemical and physical properties When their solutions are mixed, they form mixed type of crystals 3a (Here, X is a halogen) FLUORITE STRUCTURE Calcium ions are present in face centred cubic manner and fluoride ions are present in all the tetrahedral voids Therefore, there are four calcium ions and eight fluoride ions per unit cell So,the formula is Ca F8 , or CaF2 The co-ordination number of fluoride ions is four as these are present in tetrahedral voids and thus the co-ordination number of calcium ions is eight Other substances which exist in this kind of structure are UO2 , ThO2 etc.If a is the edge length of the unit cell, then, it can be represented in terms of radii of constituent ions as, 3a rCa 2+ + rF − = Fig 12: Fluorite structure SOME IONIC SOLIDS ANTI-FLUORITE STRUCTURE In the following structures, a black circle would denote a cation and a white circle would denote an anion In any solid of the type Ax By , the ratio of the co-ordination number of A to that of B would be y : x Oxide ions are present in face centered cubic manner and lithium ions are present in all the tetrahedral voids There are four oxide ions and eight lithium ions per unit cell This unit cell is just the reverse of Fluorite structure, in the sense that, the positions of cations and anions are interchanged Other substances which exist in this kind of a structure are Na O, K O, Rb O etc If a is the edge length of this unit cell, then ROCK SALT STRUCTURE (NaCl AND KCl STRUCTURE) Cl − is constituting a face centred cubic unit cell and Na + ions are present in the octahedral voids The co-ordination number of each Na + is and therefore that of Cl − would also be Moreover, there are 4Na + ions and 4Cl − ions per unit cell The formula is Na Cl rLi + + rO2 − = 3a D-5 Lattice enthalpy + – + M X M +X Ionization potential Enthalpy of formation M (s) Fig 13: Zinc blende structure Bond disssociation Sulphide ions are face centered and zinc is present in alternate Here,formation of NaCl is considered to explain Born Haber cycle, tetrahedral voids Formula is Zn S i.e, ZnS Coordination number of Zn is and that of sulphide is also Other substance that exists in this kind of a structure is BeO and diamond If a is the edge length of Na(s) + Cl(g) number of C-atom per unit cell in diamond is + = LATTICE ENERGY Lattice energy of an ionic crystal is the amount of energy released when cations and anions in their gaseous state are brought together from infinite separation to form a crystal It can also be defined as the amount of energy consumed when one mole of an ionic solid is dissociated into its ionic constituents U = lattice energy The theoretical treatment of ionic lattice energy was given by M Born and A Lande According to this treatment, the potential U E.A I.E Diamond has the same lattice structure as that of zinc blende The NaCl(s) ∆Hdiss Na(g) 3a ∆H°f 1/2Cl2(g) ∆Hsub this unit cell, then it can be expressed as, −u X (g) Fig 14: Born Haber cycle ZINC BLENDE STRUCTURE M + (g ) + X − (g ) → MX (s) Electron affinity Enthalpy of Vaporization 1/2 X2(g) + M (s) rZn 2+ + rS2 − = – Na+(g) – + Cl (g) Fig 15: Born Haber cycle for NaCl Applying concept of Born Haber cycle, it can be said that the overall energy change in a process depends only on the energy of the initial and final states and not on the route taken Hence, ∆H ° f = ∆H sub + I E + ∆H diss + E A + U Hence from above energy diagram one can calculate any unknown value such as lattice energy, ionization energy, electron gain enthalpy etc by putting other various known values X-RAYS AND BRAGG’S EQUATION X-ray which reflects from the surface of a substance has travelled less distance than an X-ray which reflects from a plane of atoms inside the crystal as shown below in fig 16 energy of an ion pair, M + , X − in a crystal separated by a distance r is considered The coulombic electrostatic energy of attraction is given by, U att (r ) = Z +Z −e 4π ∈0 r Since Z − is negative, the electrostatic energy is negative with respect to energy at infinite separation and becomes increasingly so as the θ θ z A θ θ C d interionic distance decreases BORN HABER CYCLE Born and Haber devised this cycle It relates the lattice energy of a crystal to other thermochemical data The elements in their standard state are first converted to gaseous atoms, then to ions and finally packed into the crystal lattice.The general Born Haber cycle is shown below for a MX type salt, B Fig 16: Reflection of X-rays from two parallel planes The penetrating X-ray travels down to the internal layer, reflects, and travels back over the same distance before being back at the surface The distance travelled depends on the separation of the layers and the angle at which the X-ray entered the material For this wave to be in phase with the wave which reflected from the surface it needs to have travelled a whole number of wavelengths while inside the material Bragg expressed this in an equation now known as Bragg's law: D-6 Bragg’s Law : Fractional intercepts : a / a, ∞ / a, ∞ / a i.e., 1, ∞, ∞ n λ = 2d sin (θ) Step : Take the reciprocals of the fractional intercepts : where, λ is the wavelength of the rays θ is the angle between the incident rays and the surface of the crystal d is the spacing between layers of atoms and constructive interference occurs when n is an integer (whole number) As stated, the reflected waves from different layers are perfectly in phase with each other and produce a bright point on a piece of photographic film Otherwise the waves are not in phase, and will either be missing or faint MILLER INDICES AND INTERPLANAR SPACING z Miller indices define characterize crystal planes of a lattice structure Miller indices are a set of integers (h, k , l ) which are used to define crystallographic a planes and correspond to interception of the unit cell vectors along various y a axes in a crystal The Miller indices of a a face of a crystal are inversely proportio- x nal to the intercepts of that face on the Fig.17: Cubic various axes The following treatment of arrangement of planes the procedure used to assign the Miller Indices is a simplified one and only a cubic crystal system (one having a cubic unit cell with dimensions a × a × a) will be considered The procedure is most easily illustrated using an example so we will first consider the following surface/plane: z y (a, 0, 0) x Fig 18: (a, 0, 0) plane Step : Identify the intercepts on the x–, y – and z– axes: In this case the intercept on the x-axis is at x = a [at the point (a, 0, 0)], but the surface is parallel to the y − and z-axis strictly therefore there is no intercept on these two axes but we shall consider the intercept to be at infinity (∞) for the special case where the plane is parallel to an axis The intercepts on the x −, y − and z-axis are thus, Intercepts : a, ∞, ∞ Step : Specify the intercepts in fractional co-ordinates: Co-ordinates are converted to fractional co-ordinates by dividing by the respective cell-dimension For example, a point (x , y, z ) in a unit cell of dimensions a × b × c has fractional co-ordinates of ( x/a , y/b , z/c ) In the case of a cubic unit cell each co-ordinate will simply be divided by a This gives, This final manipulation generates the Miller indices which (by convention) should then be specified without being separated by any commas or other symbols The Miller indices are also enclosed within standard brackets (….) when one is specifying a unique surface such as that being considered here The reciprocals of and ∞ are and respectively, thus yielding Miller indices : (100) So the surface or plane illustrated is the (100) plane of the cubic crystal Example (i) : The (110) surface z Assignment Intercepts : a, a, ∞ Fractional intercepts : 1, 1, ∞ (0, a, 0) y Miller indices : (110) (a, 0, 0) x Fig 19: 110 plane Example (ii): The (111) surface z (0, 0, a) Assignment Intercepts : a, a, a Fractional intercepts : , , (0, a, 0) y Miller indices : (111) The (100), (110) and (111) surfaces (a, 0, 0) considered above are the so-called x Fig 20: 210 plane low index surfaces of a cubic crystal system (the “low” refers to the Miller indices being small numbers or in this case) These surfaces have a particular importance but there an infinite number of other planes that may be defined using Miller index notation We shall just look at one more … z Example (iii) : The (210) surface Assignment Intercepts : / a, a, ∞ Fractional interscepts: / 2, 1, ∞ Miller indices : (210) (1/2 a, 0, 0) (0, a, 0) y x Fig 21: 111 plane INTERPLANAR SPACING IN A CRYSTAL SYSTEM It can be easily shown that in a crystal, the interplanar distance d hkl is given by …(1) / (d hkl )2 = (h / a)2 + (k / b)2 + (l / c)2 here h, k , l are the Miller indices of the planes and a, b, c are the corresponding dimensions of the cell Generally, in case of a cubic system, a = b = c so that from eq (1), …(2) d hkl = a / [h + k + l ]1/ D-7 HEAT CAPACITY OF SOLIDS For a tetragonal system, which is shown here a=c Dulong-Petit law, which was discovered empirically, states that the molar heat capacity assumes the value R for matter in a crystalline solid phase In practice, for solid metallic chemical elements at room temperature, molar heat capacities range from about 2.8 R to 3.4 R Large exceptions at the lower end involve solids composed of relatively low-mass, tightly bonded atoms, such as beryllium at 2.0 R and diamond at only 0.735 R The Dulong-Petit limit results from the equipartition theorem, and as such is only valid in the classical limit of a microstate continuum, which is a high temperature limit For light and non-metallic elements like Be, B, as well as most of the common molecular solids based on carbon compounds at standard ambient temperature, quantum effects may also play an important role, as they in multi-atomic gases These effects usually combine to give heat capacities lower than R per mole of atoms in the solid c a a Fig 22: Tetragonal system so that / (d hkl )2 = (h + k ) / a + l / c …(3) For an orthorhombic system, as shown below, a=b =c c a b Fig 23: Orthorhombic system So, / (d hkl )2 = h / a + k / b2 + l / c …(4) Limitations : Despite its simplicity, Dulong–Petit law offers fairly good prediction for the specific heat capacity of many solids with relatively simple crystal structure at high temperatures This is because in the classical theory of heat capacity the heat capacity of solids approaches a maximum of 3R per mole of atoms, due to the fact that full vibrational-mode degrees of freedom amount to degrees of freedom per atom each corresponding to a quadratic kinetic energy term and a quadratic potential energy term By the equipartition theorem, the average of each quadratic term is 1/2kT, or 1/2RT per mole Multiplied by degrees of freedom and the two terms per degree of freedom, this amounts to 3R per mole heat capacity.The Dulong–Petit law fails at room temperatures for light atoms bonded strongly to each other, such as in metallic beryllium, and in carbon as diamond Here, it predicts higher heat capacities than are actually found, with the difference due to higher-energy vibrational modes not being populated at room temperatures in these substances SOLVED EXAMPLES Example 1: Gold crystallizes in a fcc unit cell If edge length is ‘a’, calculate the closest distance between the gold atom and an impurity atom if it occupies : = 1.225 × (ii) A tetrahedral void AB = (ii) An octahedral void without any change in volume of the unit cell AC = r Solution : (i) If A, B, are the gold atoms and C is impurity atom, r and r' are the radius of gold atom and impurity atom respectively, we have by sin rule A 35.265 sin 109.47 sin 35.265 = 2r r + r′ ⇒ Now, 109.47° 8r B C AC = r + r ′′ B 2a = 4r ⇒ r = (r + r ′ ) = 2r r′ = r − r = 0.414 r 2a a = Example 2: Use the Born-Haber cycle and the following data to calculate the electron affinity of chlorine 2a 2a + 0.225 × 2r A (r′ ′ = radius of impurity atom) r + r ′ ′ = 1.414r = 1.414 × r ′ = 0.225 r 3a For octahedral void (i) C 2a = ∆Hf (RbCl) = –10.2.9 kcal/mol 2a Ionization energy of Rb = 95 kcal/mol ∆H sub (Rb) = +20.5 kcal/mol D-8 B.E (Cl2 ) = 54 kcal/mol Lattice energy of RbCl = −166 kcal/mol Solution : Number of B − ions in unit cell = × + = 8 ∴ Number of A+ ions in unit cell = 4, Solution : Rb(s) + –102.9 Cl2(g) 27 20.5 Cl(g) ∆H = + E.A 95 Rb(g) RbCl(s) –166 Cl–(g) Rb+(g) ∆H = + EA = − 102.9 − (20.5 + 27 + 95 − 166) = − 79 kcal/mol Thus, formula is A2B ∴ Ans (c) Example 8: A substance has density of kg dm −3 and it crystallizes in fcc lat tice with edge length equal to 700 pm, then the molar mass of the substance is : (a) 74.50 g mol −1 (c) −1 56.02 g mol Solution : ρ= (b) 103.30 g mol −1 (d) 65.36 g mol −1 Z ×M N AV × a Example 3: Metallic gold crystallizes in face centred lattice with edge length 4.070 Å Closest distance between gold atoms is : (a) 2.035 Å (b) 8.140 Å (c) 2.878 Å (d) 1.357 Å (since, effective number of atoms in unit cell = ) Solution : Closest distance in fcc lattice ∴ = a 4.070 = =2.878 Å 2 ∴ Ans (c) 2= 4× M 6.023 × 1023 × (7 × 10−8)3 M = 103.30 g mol −1 ∴ Ans (b) Example 9: For a solid with the following structure, the co-ordination number of the point B is : Example 4: Radius of Na + is 95 pm and Cl − is 181 pm Predict co-ordination number of Na + : (a) (b) (c) (d) unpredictable rNa + 95 Solution : = = 0.524 rCl − 181 i.e., in between 0.414 to 0.732 and thus co-ordination no is (A) ∴ Ans (b) Example 5: If the unit cell length of sodium chloride crystal is 600 pm, then its density will be : (a) 2.165 g/cm3 (b) 3.247 g/cm3 1.79 g/cm3 (d) 1.082 g/cm3 Z ×M × 58.5 Solution : ρ = = N AV × a 6.023 × 1023 × (6 × 10−8)3 (c) = 1.79 g/cc ∴Ans (c) Example 6: Edge length of M + X − (fcc structure) is 7.2 Å Assuming M + − X − contact along the cell edge, radius of X − ion is (rM+ = 1.6 Å) : (a) 2.0 Å (c) 2.8 Å Solution : (b) 5.6 Å (d) 3.8 Å 2(rM + + rX − ) = 7.2 rX − = 2.0 Å ∴ ∴ Ans (a) Example 7: A solid A+B − has the B − ions arranged as below If the A+ ions occupy half of the tetrahedral sites in the structure The formula of solid is : (a) AB (b) AB (c) A2B (d) A3B (a) (b) (B) (c) (d) Solution : It is evident from figure that B occupies octahedral voids and thus co-ordination number is six ∴ Ans (d) Example 10: An element A crystallizes in FCC form with atomic radius of 1.5 Another element B also crystallizes in FCC form with atomic radius of 1.15 An alloy of both the elements is formed which has a BCC structure Find (i) the density of the alloy and (ii) the percentage change in volume (Given, atomic mass of A = 40 and atomic mass of B = 60) Solution : For alloy, in body centred cavity, Radius of atom at body centre 115 = = 076 Radius of atom at corner 1.5 This means that the atom at body centre is in contact with corner atoms but corner atoms are not in contact with each other ∴ a = 2(1.5 + 115 ) = 5.3 5.53 a= = 3.06Å 1732 Volume of unit cell of alloy = (3.06 × 10−8)3 = 28.65 × 10−24 cm Since each FCC unit cell has effective atoms, they would give us unit cells of the alloy (each unit cell having effective atoms, one from each FCC unit cell) on mixing ∴ Total volume of unit cells of alloy = × 28.65 × 10−24 = 11.46 × 10−23 cm D-9 Density of alloy = (1 × 40) + (1 × 60) 6.023 × 1023 × (3.06 × 10−8)3 = 579 g / cm ∴ ∴ Volume of unit cell of element For element A, B = (3.25 × 10−8)3 = 3.43 × 10−23 cm 2a = 4rA × 1.5 a= = 4.24Å ∴ × 115 a' = = 3.25Å Total volume of A + B = (7.62 × 10−23) + (3.43 × 10−23) = 11.05 × 10−23 cm Volume of unit cell of element ∴ Increase in volume A = (4.24 × 10−8)3 = 7.62 × 10−23 cm = 11.46 × 10−23 − 11.05 × 10−23 = 0.41 × 10−23 cm For element B, ∴ Percentage increase in volume = 2a'= 4rB 0.41 × 10−23 11.05 × 10−23 × 100 = 371 % EXERCISE SUBJECTIVE QUESTIONS The edge length of a cubic unit cell of an element (atomic mass = 95.54) is 313 pm and its density is 10 gcm −3 Calculate the (i) How many K + ions and how many Br − ions are in each unit cell ? (ii) Assuming the additivity of ion radii, what is a? (iii) Calculate the density of a perfect KBr crystal (iv) What minimum value of r+ / r− is needed to prevent anion-anion contact in this structure? atomic radius A metal crystallizes into two cubic phases, face centered cubic (FCC) and body centred cubic (BCC), whose unit cell lengths are 3.5 and 3.0 Å, respectively Calculate the ratio of densities of FCC and BCC A metallic element crystallizes into a lattice containing a sequence of layers of ABABAB… Any packing of spheres leaves out voids in the lattice What percentage by volume of this lattice is empty space? Copper has a face centred cubic structure with a unit cell edge length of 3.61 Å The size of the larger atoms which are present in the interstices of the copper atoms is 0.55 Å Find whether copper atoms are in contact with each other or not? Potassium crystallizes in a body-centered lattice with a unit cell, a = 5.20 Å 10 (i) What is the distance between nearest neighbour? (ii) What is the distance between next-nearest neighbours? (iii) How many nearest neighbours does K atom have? (iv) How many next nearest neighbours does each K atom have? (v) What is the calculated density of crystalline K? Metallic gold crystallizes in the face centered lattice, the length of the cubic unit cell, is a = 4.07 Å (i) What is the closest distance between gold atoms ? (ii) How many "nearest neighbours" does each gold atom have at the distance calculated in (a) (iii) What is the density of gold? (iv) Prove that the packing fraction of gold is 0.74 BaTiO3 crystallizes in the pervoskite structure This structure may be described as a cubic lattice, with barium ions occupying the corners of the unit cell, oxide ions occupying the face centres and titanium ions occupying the centres of the unit cells (i) If titanium is described as occupying holes in the BaO lattice, what type of the hole does it occupy? (ii) What fraction of the holes of this type does it occupy ? (iii) Can you suggest a reason why it has certain holes of this type but not the other holes of the same type 11 Cesium may be considered to form interpenetrating simple primitive cubic crystals The edge length of unit cell is 412 pm Determine: (i) the density of CsCl, and (ii) the ionic radius of Cs+ if the ionic radius of Cl − is 181 pm The NaCl lattice has FCC unit cell KBr crystallizes in this lattice (rK + = 33 Å, rBr − = 1.95 Å) The ZnS structure is cubic The unit cell may be described as a face centered sulphide ion sublattice with zinc ions in the centers of alternating minicubes made by partitioning the main cube into eight equal parts (i) How many nearest neighbours does each Zn2+ have ? (ii) How many nearest neighbours does each S2− have ? (iii) What angle is made by the lines connecting any Zn2+ to any two of its nearest neighbour? (iv) What minimum r+ / r− ratio is needed to avoid anion-anion contact ? Closest cation-anion pairs are assumed to touch Li forms a body centered cubic lattice If the lattice constant (edge length) is 3.5 × 10−10 m and the observed density is 5.3 × 102kg m −3, calculate the percentage occupancy of lattice points by Li metal 12 In a cubic crystal of CsCl (density = 3.97 g / cm 3), all the corners are occupied by Cl − ions with Cs+ at the centre and vice-versa Calculate the distance between the neighbouring Cs+ and Cl − ions What is the radius ratio of two ions? MULTIPLE CHOICE QUESTIONS The unit cell present in ABCABC… closest packing of atoms is : (a) Hexagonal (b) Tetragonal (c) Face-centred cube (d) Primitive cube The number of nearest neighbours around each particle in a face-centred cubic lattice is : (a) (b) (c) (d) 12 In the closest packing of atoms, there are : (a) One tetrahedral void and two octahedral voids per atom (b) Two tetrahedral voids and one octahedral voids per atom (c) Two of each tetrahedral and octahedral voids per atom (d) One of each tetrahedral and octahedral voids per atom Which of the following statement(s) is/are correct in the rock-salt structure of an ionic compound ? D-10 (a) Co-ordination number of cation is four whereas that of anion is six (b) Co-ordination number of cation is six whereas that of anion is four (c) Co-ordination number of each cation and anion is four (d) Co-ordination number of each cation and anion is six 15 A solid has a structure in which W atoms are located at the corners of a cubic lattice, O atom at the centre of edges and Na atom at centre of the cube The formula for the compound is : (a) NaWO2 (b) NaWO3 (c) Na2WO3 (d) NaWO4 16 A certain sample of cuprous sulphide is found to have composition Cu1 8S, because of incorporation of Cu 2+ ions in the lattice What is the mole % of Cu 2+ in total copper content in this crystal? (a) 99.8% (b) 11.11% (c) 88.88% (d) None of these If the anions (A) form hexagonal closest packing and cations (C) occupy only 2/3 octahedral voids in it, then the general formula of the compound is : (a) CA (b) CA (c) C2A (d) C3A 17 In the closest packing of atom A (radius, ), the radius of atom B that can be fitted into tetrahedral void is : (a) 0.155 (b) 0.225 (c) 0.414 (d) 0.732 The composition of a sample of Wustite is Fe 0.93O What is the percentage of iron present as Fe 3+ in total iron? (a) 15.05% (b) 25% (c) 35% (d) 45% 18 Which of the following expressions is correct in the case of a sodium chloride with cell edge length, a ? (a) rc + = a (b) rc + = a / (c) rc + = 2a (d) rc + = 2a In diamond, carbon atom occupy fcc lattice points as well as alternate tetrahedral voids If edge length of the unit cell is 356 pm, then diameter of carbon atom is : (a) 77.07 pm (b) 154.14 pm (c) 251.7 pm (d) 89 pm 19 In an atomic bcc unit cell what fraction of edge is not covered by atoms? (a) 0.32 (b) 0.16 (c) 0.134 (d) 0.268 20 The packing efficiency of a simple cubic crystal with an interstitial atom exactly fitting at the body center is : (a) 0.48 (b) 0.52 (c) 0.73 (d) 0.91 21 An atomic solid crystallizes in a body centre cubic lattice and the inner surface of the atoms at the adjacent corner are separated by 60.3 pm If the atomic weight of A is 48, then density of the solid, is nearly : (a) 2.7 g/cc (b) 5.07 g/cc (c) 3.5 g/cc (d) 1.75 g/cc 22 The density of solid argon (M Ar = 40g / mol) is 1.68g / ml at 40 K If the argon atom is assumed to be a sphere of radius The edge length of unit cell of sodium chloride is 564 pm If the size of Cl − ion is 181 pm, the size of Na+ ion would be : (a) 101 pm (b) 167 pm (c) 202 pm (d) 383 pm Which one of the following schemes of ordering close packed sheets of equal sized spheres not generate close packed lattice? (a) ABCABC (b) ABACABAC (c) ABBAABBA (d) ABCBCABCBC 10 The co-ordination no of cation and anion in Fluorite, CaF2 and Rutile, TiO2 are respectively : (a) 8, and 6, (b) 6, and 4, (c) 6, and 8, (d) 4, and 2, 11 A solid is formed and it has three types of atoms X, Y and Z X forms a FCC lattice with Y atoms occupying all the tetrahedral voids and Z atoms occupying half the octahedral voids The formula of the solid is : (a) X2Y4 Z (b) XY2Z (c) X4 Y2Z (d) X4 YZ 12 1.50 × 10–8 cm, what % of solid Ar is apparently empty space? ( N A = × 1023) (a) 35.64 24 In the face centered cubic unit cell, the closest packed layers are perpendicular to : (a) The face of the unit cell (b) The face diagonal of the unit cell (c) Edges of the unit cell (d) The body diagonal of the unit cell 25 The density of a pure substance ‘A’ whose atoms are packed in cubic close pack arrangement, is 1g / cc If B atoms can occupy tetrahedral void and if all the tetrahedral voids are occupied by ‘B’ atom then the density of resulting solid in g/cc is [atomic mass ( A) = 30 g / mol and atomic mass (B) = 50 g / mol] : (a) 3.33 (b) 4.33 (c) 2.33 (d) 5.33 26 The interstitial hole is called tetrahedral because (a) It is formed by four spheres (b) Partly same and partly different (b) The size of tetrahedral void is smaller than that of octahedral void (d) The size of tetrahedral void may be greater or smaller or equal to that of octahedral void depending upon the size of atoms How many unit cells are present in 5.0 gm of crystal AB (formula mass of AB = 40) having rock salt type structure? (N A = Avogadro’s no.) (a) N A (b) 01 (c) 4N A (d) None of these NA 14 The intermetallic compound LiAg crystallizes in cubic lattice in which both lithium and silver have co-ordination number of eight The crystal class is : (a) Simple cubic (b) Body centred cubic (c) Face centred cubic (d) None of these (d) None of these A body centered cubic lattice is made up of hollow spheres of B Spheres of A are present in hollow spheres of B Radius of sphere A is half of radius of sphere B What is the ratio of total volume of spheres of B unoccupied by A in a unit cell and volume of unit cell? 3π 7π (a) (b) (c) (d) None of these 64 128 24 (a) The size of tetrahedral void is greater than that of octahedral void 13 (c) 74 23 In the closest packing of atoms : (c) The size of tetrahedral void is equal to that of octahedral void (b) 64.36 D-11 (c) It is formed by four spheres the centres of which form a regular tetrahedron (d) None of the above three (d) In a face centered cubic unit cell, each face centered atom is in contact with the face centered atoms of all those faces which meet at this face 27 At the limiting value of radius ratio r+ / r−, the : (a) Forces of attraction are larger than forces of repulsion (b) Forces of attraction are smaller than forces of repulsion (c) Forces of attraction and repulsion are just equal (d) None of these 38 In an ionic solid, anions occupy the corners and face centres of a cubic unit cell, whereas, the cations occupy all the octahedral voids as well as half of the tetrahedral voids The effective number of all types of cations in the unit cell are : (a) (b) (c) (d) 10 28 For an ionic crystal of the general formula AX and co-ordination number 6, the value of radius ratio will be : (a) Greater than 0.73 (b) In between 0.73 and 0.41 (c) In between 0.41 and 0.22 (d) Less than 0.22 39 If the unit cell length of sodium chloride crystal is 600 pm, then its density will be : 29 The tetrahedral voids formed by ccp arrangement of Cl − ions in rock salt structure are : (a) Occupied by Na+ ions 30 The number of Na+ ions filling all octahedral voids in NaCl structure are : (a) (b) 12 (c) 13 (d) 14 31 Potassium has a bcc structure with nearest neighbour distance 4.52 Å Its atomic weight is 39 Its density will be : 33 34 35 (c) g / cm 40 (a) 454 kg / m (b) 804 kg / m (c) 852 kg / m (d) 910 kg / m The co-ordination number of hexagonal closest packed (hcp) structure is : (a) 12 (b) 10 (c) (d) (b) 3.2 g / cm 3 (d) g / cm A compound XY crystallizes in bcc lattice with unit cell edge-length of 480 pm If the radius of Y − is 225 pm, then the (b) Occupied by Cl − ions (c) Occupied by either Na+ or Cl − ions (d) Vacant 32 (a) 2.2 g / cm radius of X + is : (a) 190.70 pm (b) 225 pm 41 (c) 127.5 pm (d) None of these At very low temperature, SO2 crystallizes in a hexagonal lattice If the cross sectional area of unit cell is 17.77 × 10−16 cm 2, the lattice constant ‘b’ is 7.41Å and density is 0.805 g/cc then the number of SO2 molecules per unit cell is : (a) (b) (c) (d) 42 How many moles of atoms are present on the surface of cm long iron rod (assume cubic shape rod) having FCC unit cell with edge length 0.556 nm (a) 2.57 × 10−8 (b) 6.23 × 10−7 43 (c) × 106 (d) 2.24 × 10−2 A metallic element exists as a cubic structure The edge length of the unit cell is 2.9 Å The density of the metal is 7.2 g / cc The The mass of a unit cell of CsCl corresponds to : number of unit cells in 100 g of the metal would be : (a) 1Cs+ and 1Cl − (b) 1Cs+ and 6Cl − (c) 4Cs+ and 4Cl − (d) 8Cs+ and 1Cl − (a) 57 × 1021 A mineral having the formula AB2 crystallizes in the cubic close packed lattice, with the A atoms occupying the lattice points The co-ordination number of the A atoms, that of B atoms and the fraction of the tetrahedral sites occupied by B atoms are : (a) 8, 4, 100% (b) 2, 6, 75% (c) 3, 1, 25% (d) 6, 6, 50% 44 36 A compound (AB) consisting of cation (A) (radius = 1.0 Å) and anion (B) (radius = 2.5 Å) will probably crystallize in the : (a) NaCl type structure (b) CsCl type structure (c) ZnS type structure (d) CaF2 type structure 37 Which of the following statement is not true? (a) In a face centered cubic unit cell, a face centered atom is in contact with four atoms at the corner of the same face of the cube (b) In a body centered unit cell, the corner atoms are in contact with each other, irrespective of the size of the body centered atom (c) In a primitive cubic unit cell, the packing fraction is 0.52, if all the lattice points are occupied and no other atom is inside the unit cell NaH crystallizes in the same structure as that of NaCl The edge length of the cubic unit cell of NaH is 4.88 Å and the ionic radius of Na+ is 0.95 Å, then the ionic radius of H − ion is : (a) 3.27 Å (b) 1.49 Å (c) 1.72 Å (d) 2.98 Å 45 What will be the distance between a tetrahedral hole and octahedral hole present at the body diagonal of FCC unit cell? 3a 3a 3a (a) 3a (b) (c) (d) 46 In an ionic compound M 2+ X 2− in the crystalline state, the If the positions of Na+ and Cl − are interchanged in NaCl, the crystal lattice with respect to Na+ and Cl − is: (a) Both f.c.c (b) Both b.c.c (c) f.c.c and b.c.c (d) b.c.c and f.c.c (b) 16 × 1031 (c) 57 × 1023 (d) 16 × 1033 co-ordination number of M 2+ will be when the radius ratio rX − / rM + is approximately in the range : (a) 0.414 to 0.732 (b) 1.36 to 2.4 (c) 2.4 to 4.6 (d) 4.6 to 47 A substance AX BY crystallizes in a face centered cubic (FCC) lattice in which atoms ‘A’ occupy each corner of the cube and atoms ‘B’ occupy the centers of each face of the cube Identify the correct composition of the substance AX BY : (a) AB3 (b) A4 B3 (c) A3B (d) Composition cannot be specified 48 In a solid ‘AB’ having the NaCl structure, ‘A’ atoms occupy the corners of the cubic unit cell If all the face-centered atoms along one of the axes are removed, then the resultant stoichiometry of the solid is : (a) AB2 (b) A2B (c) A4 B3 (d) A3B4 D-12 49 50 In which of the following crystals alternate tetrahedral voids are occupied? (a) NaCl (b) ZnS (c) CaF2 (d) Na 2O (a) Height of hexagonal unit is × (b) Empty space is 34% (c) Form by CsCl (d) Packing fraction is 0.74 The limiting radius ratio of the complex [Ni(CN)4 ]−2 (with dsp hybridization of Ni) is : (a) 0.225 – 0.414 (c) 0.155 – 0.225 10 Select the correct statement(s) : (a) The C.N of cation occupying a tetrahedral hole is (b) C.N of cation occupying a octahedral hole is (c) Bragg’s equation deals with interplanar spacings (d) hcp is ABCABC … arrangement Which of the following crystals shows : co-ordination? (a) CaF2 (b) BeF2 (c) SiO2 (d) PbO2 Select the correct statement(s): (a) CsCl is a bcc lattice (b) Diamond has the same lattice structure as ZnS (c) hcp and ccp structures have the same co-ordination number 12 (d) On increasing pressure, co-ordination number of CsCl decreases to that of NaCl Which of the following crystals have : co-ordination? (a) CsCl (b) AlFe (c) MnO (d) NH Br Which of the following compounds represent a normal : spinel structure? (a) MgAl 2O4 (b) CO3O4 (c) Zn(TiZn)O4 (d) Ni(CO)4 The spacing of lattice planes designated as dhkl is correctly related to the grazing angle of reflection (θ) of the first order in the way: 2a (a) sin θ = (h + k + l 2)1/ λ λ (b) sin θ = (h + k + l 2)1/ 2a 2a (c) θ = sin −1 [h + k + l 2]1/ λ 2a (d) θ = sin −1 [h + k + l 2] λ 1/ Select the correct statement(s): (a) When two forms of solid exist together in equilibrium with each other at a particular temperature, which depends only on pressure, the property is called enantiotropy (b) The temperature at which two forms of solid exist in equilibrium with each other, which depends only on pressure, is called transition temperature (c) In case of monotropy, the transition point lies below melting point (d) In case of enantiotropy, the transition temperature lies below boiling point & above melting point Which of the following is hardest substance other than diamond? (a) Neutron (b) Lonsdaleite (c) Wurtzite (d) Admantium vibrenate In hexagonal close packing in 3D, which of the following is/are correct? The correct logical option among the following is (are) : (a) PtCl 2− : square planer ; Zns : linear (b) 0.414 – 0.732 (d) None of these (b) Elements : allotropy ; compounds : polymorphism (c) AA (2D) : 76% ; AB (2D) : 94% (d) Rochellesalt : ferroelectricity ; Leadzirconate : piezoelectricity MULTIPLE SELECT QUESTIONS a PREVIOUS YEAR QUESTIONS SUBJECTIVE QUESTIONS 2013 (i) Write the correct order of lattice energy for LiX, X = F , Cl, Br and I (ii) A first order reflection from (111) plane is observed for LiX with 2θ = 24.6° (X-ray of wavelength 1.54 Å) Assuming LiX to be a cubic crystal system, calculate the length of the side of the unit cell in Å 2011 The edge length of unit cell of a metal having molecular weight 75g / mol is 5Å which crystallizes in cubic lattice If the density is 2g / cc then find the radius of metal atom ( N A = × 1023) Give the answer in pm 2010 Draw a properly labeled unit cell diagram of CsCl Show through calculations that there is only one CsCl per unit cell 2006 Draw the unit cell of CsCl lattice Draw the (100) and (110) planes separately and indicate the position of cesium and chloride ions MULTIPLE CHOICE QUESTIONS 2017 The number of normal modes of vibration in napthalene is : (a) 55 (b) 54 (c) 48 (d) 49 2016 The coordination numbers of Cs + and Cl − ions in the CsCl structure, respectively, are: (a) 4,4 (b) 4,8 (c) 6,6 (d) 8,8 2015 At what angle(s) of incidence, X-rays of wavelength 5.0 Å will produce diffracted beam from the (110) planes in a simple cubic lattice with a = 10 Å ? (a) 6.8° (b) 10.2° (c) 20.7° (d) 45.0° 2008 In CsCl structure the number of Cs + ions that occupy second nearest neighbour location of a Cs ions is : D-13 (a) (c) 10 2016 (b) (d) 12 In a diffraction experiment with X-rays of wavelength 1.54 Å, a diffreaction line corresponding to 2θ = 20.8° is observed The inter-planar separation in Å is ……… The potential energy of interaction between two ions in an ionic Z Z compound is given by U = 1389.4 kJ mol −1 Assuming r/ Å 2007 The ionic radii of Ca 2+ and F – are 100 pm and 133 pm respectively The co-ordination number of Ca 2+ in the ionic solid will be : (a) (b) (c) (d) that CaCl is linear molecule of length 5.6 Å, the potential energy 2006 If the values of Madelung constants of the following compounds are equal, then their lattice energy values decrease in the order: (a) KCl > NaF > CaO > Al 2O3 (b) Al 2O3 > CaO > NaF > KCl (c) NaF > KCl > CaO > Al 2O3 (d) Al 2O3 > CaO > KCl > NaF for CaCl molecule in kJ mol −1 is ……… 2015 The ionic radii of Cs + and Cl − ions are 181 and 167 pm, respectively The Born exponents for the He, Ne, Ar, Kr and Xe configurations are 5, 7, 9, 10 and 12, respectively If the value of NUMERICAL ANSWER TYPE (NAT) QUESTIONS ANe is 2.45 × 10−4 J m, the lattice energy (in KJ mol −1) of CsCl 4πε0 2017 according to Born-Lande equation is ……… Silver crystallizes in a face-centered cubic lattice The lattice parameter of silver (in picometer) is ……… [Given : Avogadro’s number = 6.023 × 1023 mol −1, molar mass of 2013 CaO, VO and MnO have octahedral co-ordination of the metal ions in a rock-salt structure The correct increasing order of their lattice enthalpies is ………… A hypothetical element (atomic weight = 300) crystallizes in a simple cubic lattice For this crystal, the first order X-ray diffraction with wavelength of Å appears at an angle of 30° The density of silver = 107.87 g mol −1 and density of crystal = 10.5 g cm −3] The separation of 123 planes (in nm) in an orthorhombic cell with a = 0.25 nm, b = 0.5 nm and c = 075 nm is ……… (final answer should be rounded off to two decimal places) the crystal is …… g cm −3 [Avogadro number, N A = 6.02 × 1023] ANSWERS SUBJECTIVE QUESTIONS 135.5pm 1.26 26% No (i) 4.5 Å (ii) 5.20 Å (iii) (iv) (v) 0925 g / cm3 (i) 2.878 Å (ii) 12 (iii) 194 g / cm2 (i) Octahedral holes (ii) (i) g / cm3 (ii) rCs + = 17579 PM (iii) In case of other octahedral holes, the proximity of two cations Ba2 + and Ti4 + would be electrostatically unfavourable 10 (i) (ii) (iii) 109°28′ (iv) 0.225 11 97.79% 12 (i) Four of each (ii) 6.56 Å (iii) 2.8 g / cm3 (iv) 0.414 357 Å, 0.728 MULTIPLE CHOICE QUESTIONS (c) (d) (b) (d) (c) (b) (b) (a) (c) 10 (a) 11 (a) 12 (b) 13 (d) 14 (b) 15 (b) 16 (b) 17 (a) 18 (b) 19 (c) 20 (c) 21 (d) 22 (b) 23 (d) 24 (b) 25 (b) 26 (c) 27 (c) 28 (b) 29 (d) 30 (c) 31 (d) 32 (a) 33 (a) 34 (a) 35 (a) 36 (c) 37 (b) 38 (c) 39 (c) 40 (a) 41 (d) 42 (a) 43 (c) 44 (b) 45 (b) 46 (b) 47 (a) 48 (d) 49 (b) 50 (b) D-14 MULTIPLE SELECT QUESTIONS a,b,c b,c a,b,c a,b,d a,b b,d (c) (a) (a) (b) a,b b,c a,d 10 b,d PREVIOUS YEAR QUESTIONS MULTIPLE CHOICE QUESTIONS (c) (d) NUMERICAL ANSWER TYPE (NAT) QUESTIONS 408-409 0.14-0.15 VO > MnO > CaO 100 4.2-4.3 1738-1734 −638.0 to −636.0 HINTS AND SOLUTIONS MULTIPLE SELECT QUESTIONS Use Brag’s eq & Miller indices Refer definations of each Although neutron is 20 times harder than diamond, but it is th subatomic particle option is fictional Refer derivation of hexagonal close packing in 3D 10 b & d Using nλ = 2d sin θ; we get d = × 10−8 cm And density = × 300 N A × (5 × 10−8)3 (As d = × 10−8 cm = edge length) Lattice energy α charge α interionic distance PREVIOUS YEAR QUESTIONS NUMERICAL ANSWER TYPE (NAT) QUESTIONS MULTIPLE CHOICE QUESTIONS 1 The stability is decided on the basis of CFSE values Use d = × 107.87 N A × a3 ¦¦¦ ... ORGANIC Physical Chemistry : 23 Questions Physical Chemistry : 27 Questions Physical Chemistry : 28 Questions Organic Chemistry : 20 Questions Organic Chemistry : 20 Questions Organic Chemistry :... electron system Since the electrostatic force balances the centrifugal force, for the stable electron in the orbit i.e., centrifugal force = electrostatic force mv kZe …(2) ∴ = r r2 According to... electromagnetic force The protons and neutrons in the nucleus are attracted to each other by a different force, the nuclear force, which is usually stronger than the electromagnetic force repelling