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UNIT-1 SOLIDS, QUESTIONS CARRYING ONE MARK:, 1. Which type of solid is anisotropic in nature?, Ans: Crystalline solids are anisotropic in nature, 2. Which type of solids is called as super cooled liquids or pseudo solids?, Ans:Amorphous solids are called super cooled liquids, 3. A solid has a sharp melting point, and then to which type of solids does it, belong?, Ans: Crystalline solids, 4. Which type of solids has long range orderly arrangement of constituent, particles?, Ans: Crystalline solids, 5. Sodium chloride and quartz belong to which type of solid?, Ans: Crystalline solids, 6. A solid shows different values for refractive index when measured in different, directions. - Identify the type of solid, Ans: Crystalline solids, 7. When a solid is cut with a sharp edged tool, they cut into two pieces and the, newly generated surfaces are plain and smooth. – Identify the type of solid., Ans: Crystalline solids, 8. Which type of force of attractions is present between the molecules in polar, molecular solids?, Ans: Dipole –dipole interactions, 9. Which type of force of attractions is present between the molecules in, non-polarmolecular solids?, Ans: London forces or Dispersion forces

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10. Which type of force of attractions is present between the particlesin, ionic solids?, Ans: Electrostatic force of attraction or coulombic force of attraction, 11.Solid SO2 and solid NH3belong to which type of molecular solids?, Ans: Polar molecular solids, 12. What is crystal lattice?, Ans: The regular three dimensional array of lattice points in space is called, crystal lattice, 13. What is a unit cell?, Ans: It is the smallest repeating unit which when arranged in three dimension, gives the crystal lattice., 14. How many types of primitive unit cells are present?, Ans: Three types, 15. What is a primitive cubic unit cell?, Ans: The cubic unit cell in which the particles/atoms are present only at the, eightcorner of the cube is called primitive cubic unit cell., 16. Define the co-ordination number of a particle in solids., Ans:It is the total number of nearest neighboring particles to a given particle., 17. What is the number of octahedral voids generated, if the number of close, packed spheres is N?, Ans: N, 18. What is the number of tetrahedral voids generated, if the number of close, packed spheres is N?, Ans: 2N, 19. What is the co-ordination number of a particle in a tetrahedral void?, Ans:Four, 20. Among Schottky and Frenkel defect, which type of defect decreases, the density of the crystal?, Ans: Schottky defect

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21. What are point defects?, Ans: Deviations from the ideal arrangement around a particular point or an, atomin a crystalline solid, 22. What are F-centers?, Ans: The anionic sites occupied by the unpaired electrons are called F- centre., 23. To which colour potassium chloride crystal turns, when excess potassium, ionis present?, Ans: Violet, 24. Name the type of non-Stoichiometric defect observed when white ZnO, turns yellow on heating., Ans: Metal excess defect, 25. Name the non-Stoichiometric defect responsible for the composition of ferrous, oxide to be Fe0.95O1., Ans:Metal deficiency defect, 26. Which type of point defect is observed when NaCl containing little SrCl2 is, crystallized?, Ans:Impurity defect, 27. Which defect is also called as dislocation defect?, Ans:Frenkel defect, 28. What is doping?, Ans: The process of increasing the conductivity of an intrinsic semiconductors, by adding asuitable impurity is called doping, 29. What type of semiconductors are obtained when silicon doped with, boron impurity?, Ans: p-type semiconductor, 30. Name the unit used to measure magnetic moment., Ans: Am2( 1Bohr magneton= 9.27x10-24 Am2), 31. What are diamagnetic substances?, Ans: These are the substances which are repelled by the magnetic field

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32. What are ferromagnetic substances?, Ans: These are the substances which are strongly attracted by the magnet, 33. How body diagonal and radius of a sphere(r) are related in bcc unit cell?, Ans:4r =√2a, 34. Give an example for Ferromagnetic substance., Ans: Fe Co Ni Gd CrO2, 35. Give an example for Diamagnetic substance., Ans: H2O, NaCl, and C6H6, , QUESTIONS CARRYING TWO MARKS:, 1. How crystalline solids differ from amorphous solids in their melting point?, Ans:Crystalline solids have sharp melting point whereas amorphous solid do not, have a sharp melting point, 2. Write any two differences between crystalline solids and amorphous solids?, Crystalline solid, 3-D long range orderly arrangement, of particles, Sharp Melting point, True solids having definite shape, They have a well-defined cleavage, planes, Anisotropic in nature, , Amorphous solid, No orderly arrangement of constituent, particles, Do not have sharp M P, ( Softening temperature), Pseudo solids having irregular shapes, Do not have cleavage planes, Isotropic in nature, , 3. What is meant by anisotropy? What type of solids show this nature?, Ans: The physical properties like refractive index, coefficient of thermal expansion,, when measured in different directions gives different value for a crystalline, solid hence it is anisotropic in nature., Crystalline solids, 4. What are the nature of particles and the force of attractions between, the particles in non-polar molecular solids?, Ans: In a non-polar molecular solids, the constituent particles are non-polar, moleculeslike H2, Cl2, I2 and even atoms like Ar, Ne, Xe etc., The nature of force of attraction is weak dispersion force or London force.

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5. What are the nature of particles and the force of attractions between, the particles in polar molecular solids?, Ans: In a polar molecular solids, the constituent particles are formed by polar covalent, bond like HCl, SO2., The nature of force of attraction is dipole-dipole attractions, 6. What are the nature of particles and the force of attractions between, the particles in hydrogen bonded molecular solids?, Ans: In a hydrogen bonded molecular solids, the constituent particles are polar, molecules capable of forming hydrogen bond like water., The nature of force of attraction is hydrogen bonding, 7. What are point defects? Mention the types, Ans: Point defects are the irregularities in the arrangement of constituent, particlesaround a point or a lattice site in a crystalline substance., These are of three types., 1. Stoichiometric defects., 2. Non-stoichiometric defect, 3. Impurity defect., 8. What are the differences between Schottky and Frenkel defect?, Schottky defect, a. Shown by ionic solidscontaining, similar-sized cationsand anions, (having high coordination number), , Frenkel defect, a. Shown by ionic solids containing, largedifferences in the sizes of ions,, (having less coordination number), , b. An equal number of cations and, anions are missing to maintain, electrical neutrality, , b. Created when the smaller ion, (usually cation) is dislocated from its, normal site to an interstitial site, , c. Decreases the density of the, substance, , c. No change in density of the crystal., creates a vacancy defect as well as an, interstitial defect .Also known as, dislocation defect, d. Example:, AgCl, AgBr, AgI and ZnS, , d. Example,, NaCl, KCl , CsCl, and AgBr, , 9. What are the nature of particles and the force of attractions between, the particles in ionic solids?, Ans: The nature of the particles is ions (both cation and anion). The nature, of the force of attraction is electrostatic force or coulombic force, 10. What are the nature of particles and type of bonding in network solids?, Ans: The nature of the particles is atoms. The bonding is covalent bond.

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11. Classify the following into polar and non-polar molecular solids:, Ar, HCl, I2 and SO2, Ans: Non-polar molecular solids: Ar,, Polar molecular solids:HCl, SO2, , I2, , 12. Calculate the number of particles present per unit cell in an FCC unit cell., Ans:Contribution of corner particle = 8 x 1/8 = 01, Contribution of a particle at the centreof face = 6 x ½ = 03, Total number particle /unit cell = 04, 13. Calculate the number of particles present per unit cell in a BCC unit cell., Ans:Contribution of corner particle, = 8 x 1/8 = 01, Contribution of a particleat the centre = 1 x 1 = 01, Total number particle /unit cell, = 02, 14. Calculate the number of particles present per unit cell in a simple cubic, unit cell., Ans:Contribution of corner particle = 8 x 1/8 =01, Total number particle /unit cell, = 01, 15. Mention the two characteristics of a unit cell., Ans: Two characteristics of unit cells are, a. Edge length, b. Axial angles, 16. What is the relation between edge length (a) and radius of the sphere (r), infcc unit cell? What is itspacking efficiency?, Ans: The relationship between edge length and radius of the sphere are a=2   2  r, Packing efficiency is 74%, 17. What is the relation between edge length (a) and radius of the sphere (r), in bcc unit cell? What is its packing efficiency?, Ans: The relationship between edge length and radius of the sphere are, Packing efficiency is 68 %, , a=, , !!, , !, , 18. How many tetrahedral and octahedral voids is present, if the number, of sphere is N?, Ans: The number of tetrahedral void is 2N, The number of octahedral void is N, 19. Explain Schottky defect. Give an example., Ans: The defect which arises due to missing of equal number of cations and anions, from the crystal lattice is called Schottky defect. Ex. NaCl, KCl ,CsCl, AgBr

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20. Explain Frenkel defect. Give an example., Ans: The defect in which an ion (generally cation) leaves the original site and, occupies the interstitialsite is called Frenkel defect. E. AgCl, AgBr, AgI, 21. How Schottky defect and Frenkel defect affect the density of the crystal?, Ans: In Schottky defect density of the crystal decreases., In Frenkel defect the density of the crystal remains same., 22. Mention the two types of Non-stoichiometric defects in solids?, Ans: Metal excess defect and metal deficiency defect., 23. What is F- center? What colour is imparted to the NaCl crystal, due to the, presence of excess sodium?, Ans: The anionic sites occupied by the unpaired electrons are called F- Centre, The colour of NaCl crystal is Yellow, 24. Write the formula to calculate the density of the unit cell and explain the, terms., Ans:, z = number of particles present per unit cell, , d=, , 𝒛𝑴, , 𝒂𝟑 𝑵𝑨, ,                M = Molecular mass , d = density, , NA = Avogadro’s number, , a = Edge length., , 25. What are n-type and p-type semiconductors?, Ans:, n-type semiconductor is obtained by doping of the crystal of a group 14 element, such as Si or Ge, with a group 15 element such as P or As(pentavalent)., Conductivity increases due to negatively charged electrons., p-type semiconductor is obtained by doping of the crystal of a group 14 element, such as Si or Ge, with a group 13 element such as B, Al or Ga( trivalent)., Conductivity increases as a result of electron hole, 26. An ionic compound is formed by two elements A and B. The cat ions A are in, ccp arrangement and those of anions B occupy all the tetrahedral voids., What is the simplest formula of the compound?, Ans:, Since cations are in ccp arrangement, the total number cat ions A = 4, The number of tetrahedral voids is double the number of particles = 8, All the tetrahedral voids are occupied by anions B., The number of elements of B = 8, Hence the formula of the ionic compound is A4B8 or AB2

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27. A compound is formed by two elements X and Y. The element X forms ccp and, atoms of Y occupy 1/3 rd of tetrahedral voids. What is the formula of the, compound?, Ans:, Since element X are in ccp arrangement, the number of X per unit cell = 4, The number of tetrahedral void = 8, But only 1/3 rd is occupied by Y, therefore 8 x1/3 = 8/3, Hence the formula of the compound is X4Y8/3 = X12Y8 or X3Y2, 28. Gold(atomic radius=0.144nm)crystallizesin a face centered unit cell., What is the length of the side of the cell?, Ans:, For FCC the edge length and radius of sphere arerelated by the equation,, r = 0.144nm a=2   2  r, a=?, = 2 2    x  0.144  nm, = 2x1.414 x 0.144, = 0.40723nm., 29. Silver forms ccp lattice and X- ray studies of its crystals show that the edge, lengthof its unit cell is 408.6pm. Calculate the density of silver, (atomic mass = 107.9 u), Ans:, , d=, , !", , ! ! !!, , d= 4 x 107.9/(4.08)3 x10-24 x 6.022 x1023, d = 431.6/40.899, d = 10.5528g/cm3, 30. X- ray diffraction studies show that copper crystallizes in an fcc unit cell with, cell edge of 3.6 x10-8cm. In a separate experiment, copperis determined to, have a density of 8.92g/cm3,calculate the atomic mass of copper., Ans:, , d=, , !", ! ! !!, , M = d a3 NA/Z, = 8.92 x(3.6)3x10-24 x 6,022 x1023/4, = 250.61/4, M = 62.6525 u, 31. The edge of fcc unit cell of platinum is 392 pm and density is 21.5 g/cm3,, calculate the Avogadro number., Ans:, !", d = !! !, !, , NA = Z x M/ d a3, = 4 x 195.08/21.5 x (3.92)3x 10—24, = 780.32/1295.08 x10—24, NA= 6.025 x1023

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32. A unit cell of sodium chloride has four formula units. The edge length of the, unit cell is 0.564 nm. What is the density of sodium chloride?, Ans:, , d=, , !", ! ! !!, , d = 4 x 58.5/(5.64)3 x10-24 x 6.022 x1023, d = 234/108.038, d = 2.165 g/cm3, 33. A body centered cubic element having density 10.3 g/cm3, has a edge length, of 314pm. Calculate the atomic mass of the element, (Avogadro’s number= 6.023x1023/mol), Ans:, , d=, , !", ! ! !!, , M = d x a3 xNA/Z, = 10.3 x (3.14)3x 10-24 x 6.022 x1023/2, M = 96.01u, 34. Calcium metal crystallizes in a face centered cubic lattice with edge length, of 0.556nm. Calculate the density of the metal., (Atomic mass of calcium = 40g/mol and Avogadro number= 6.022 x1023mol-1), Ans:, , d=, , !", ! ! !!, , d = 4 x 40/(5.56)3 x10-24 x 6.022 x1023, d = 160/103.50, d = 1.54 g/cm3, 35. Copper crystallizes into afcc lattice with edge length 3.61 x10-8cm., Calculate the density of the of the crystal, (Atomic mass of copper =63.5g/mol and Avogadro number= 6.022 x1023mol-1 ), Ans:, , d=, , !", ! ! !!, , d = 4 x 63.5/(3.61)3 x10-24 x 6.022 x1023, d = 254/28.33, d = 8.9 g/cm3

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36. Silver crystallizes in a face centered cubic structure. If the edge length is, 4.077 x10-8cm and density is 10.5 g/cm3, calculate the atomic mass of silver., , Ans:, , d=, , !", ! ! !!, , M = d a3 NA/Z, = 10.5 x (4.077)3x10-24 x 6,022 x1023/4, = 103.57/4, The atomic mass of silver M = 107.09 u, 37. The density of Li atoms is 0.53g/cm3.The edge length of Li is 3.5 A0. Find out, the number of Li atoms in a unit cell (N0= 6.022 x1023/mol& M= 6.94), Ans:d, , =, , !", , ! ! !!, !, , Z = d x 𝑎 𝑁! /𝑀, = 0.53 x (3.5)3 x10–24x 6.022 x1023/6.94, =2, The number of lithium atoms in unit cell is 2, , Questions carrying THREE marks, 1. Calculate the packing efficiency in simple cubic unit cell, , a = 2r, Volume of the cubic unit cell= a3 = (2r)3= 8r3, !, volume of one particle(sphere) = 𝝅r3, !, The number of particles per unit cell =1, !, Total volume occupied by one sphere = 𝜋𝑟 !, Edge length of the cube =, , !, , Packing efficiency=, , !"#$%  !"#$%&  !""#$%&'  !"  !"#  !"!!"!, ×, !"#$%&  !"  !"#$!  !"#$  !"##, ! !, !! ×!, !, !  !!, , =, ,  , , ×100, = 52.4%, , 100

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2. Calculate the packing efficiency in face centered cubic unit cell, , 2  r, , Edge length of the cube =a=2  , , !, , a3 = 2   2  r, !, volume of one particle(sphere) = 𝝅r3, !, The number of particles per unit cell =4, !, Total volume occupied by four spheres = 4 × 𝜋𝑟 !, Volume of the cubic unit cell=, , edge length of the cube, be ‘a’, In ABC, AC2 = BC2 + AB2, b2 = a2 + a2, b2 = 2a2, b = 2   a, Let the radius of the, atom = r, Length of the diagonal, of ABC, b= 4r, 2  a = 4r, a = 2 2  r, , !, , packing efficiency=, ! !, !! ×!, !, !, , =, , ! !!, , !"#$%  !"#$%&  !""#$%&'  !"  !"#$  !"!!"!#, ×, !"#$%&  !"  !"#$!  !"#$  !"##, , 100, ,  ×    100, =, , !" !, !!, !, !"   !!!, ,  ×    100 =, , 74 %, , 3. Calculate the packing efficiency in body centered cubic unit cell, Edge length of the cube =a=, , !!, !, , Volume of the cubic unit cell=, , a3, , =, , !! !, !, , !, , = 𝝅r3, !, The number of particles per unit cell =2, volume of one particle(sphere), In ABG,, b2 = a2 + a2 ⇒ b2 = 2 a2, In, AGD,, C2 = a2 + b2, C2 = a2 + 2a2, C2 = 3a2⇒ C = 3𝑎, Radius of the atom = r., Length of the body, diagonal, C=4r, 3𝑎 = 4r, !!, a=, !, ,  , , Total volume occupied by two spheres = 2 ×, , packing efficiency=, , =, , ! !, !! ×!, !, !, !, !, !, , !, !, , !"#$%  !"#$%&  !""#$%&'  !"  !"#  !"!!"!#, ×, !"!"#$  !"  !"#$!  !"#$  !"##, ,  ×  100, =, , ! !, !!, !, !" !, !, ! !, , 𝜋𝑟 !, ,  ×  100, , = 68%, , 100

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4.Based on band theory explain conduction in metals, insulators and, semiconductors, Conduction of electricity in metals: In metals, the valence shell is partially filled, so, this valence band overlaps with a higher energy unoccupied conduction band so that, electrons can flow easily under an applied electric field., Conduction of electricity in insulators: In insulators, the valence shell is empty, so, the gap between the valence band and conduction band is very large. so that electrons, cannot flow under an applied electric field., Conduction of Electricity in Semiconductors In semiconductors, the gap between, the valence band and conduction band is so small that some electrons may jump to, the conduction band. Electrical conductivity of semiconductors increases with, increase in temperature. Substances like Si, Ge show this type of behaviour, and are, called intrinsic semiconductors., 5. How are solids classified on the basis of the force of attraction?, Ans:, a. Molecular solids: Particles are held by, a. London forces (in non-polar solids) ex : Benzene, Argon, P4O10, I2, P4, b. Dipole - dipole interaction ( in polar solids), ex: Urea, Ammonia, c. Hydrogen bonding (in hydrogen bonded solids) ex: ice, b. Ionic solids, a. Particles are held by ionic bond, b. Conduct electric current in aqueous solution or molten state, c. Examples: NaCl, MgO, ZnS, d. In solid state, ions are held together by strong electrostatic forces and are, not free to move about within the solid. Hence, ionic solids do not conduct, electricity in solid state. However, in molten state or in solution form, the, ions are free to move and can conduct electricity., c. Covalent or network solids:, a. Particles are held by covalent bonding. Examples SiO2 (quartz), diamond,, d. Metallic solids:, a. Particles are held by metallic bond., b. These are electrical conductors, malleable, and ductile. Examples: Fe, Cu,, 6. What are point defects? Explain the types., Ans: Point defects are the irregularities in the arrangement of constituent particles, around a point or an atom in a crystalline substance. These are of three types., 1. Stoichiometric defects: Do not disturb stoichiometry of the solid., These are also called intrinsic or thermodynamic defects, Ex : Frenkel defect, Schottky defect, 2. Non-stoichiometric defects: This defect alters the stoichiometric ratio of the, constituent elements, i) Metal excess defect, a. Metal excess defect due to anionic vacancies:, b. Metal excess defect due to the presence of extra, cations at interstitial sites:, ii) Metal deficiency defect, a. By cation vacancy, 3. Impurity defect.

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7. What are diamagnetic, paramagnetic and ferromagnetic substances?, 1. Paramagnetic substance: The substance which are attracted by the magnet., The magnetic character is temporary and is present as long as the external, magnetic field is present. Ex; O2, Cu2+, Fe3+, Cr3+ NO., 2. Diamagnetic substance: The substance which are weakly repelled by the magnetic, field TiO2, H2O,NaCl.This property is shown by those substance which contain, fully –filled orbitals (no unpaired electrons), 3. Ferro magnetic substance: The substance which are strongly attracted by the, magnet. They show permanent magnetism even in the absence of magnetic field., Ex : Fe Co Ni Gd& CrO2, 8. An element with molar mass 2.7 x 10-2 kg/mol forms a cubic unit cell with edge, length 405pm. If its density is 2.7 x 103 kg/m3, what is the nature of the cubic, unit cell, Ans:, !", , d=, , ! ! !!, , Z = d x 𝑎 ! 𝑁! /𝑀, , = 2.7 x103 x(405)3 x 10—27 x 6.022 x1023/2.7 x 10-2, =4, Since there are 4 atoms of the element present per unit cell. Hence, the cubic, unit cell must be face centered or cubic close packed structure (ccp), 9. Niobium crystallises in body-centered cubic structure. If density is 8.55g/cm3,, calculate atomic radius of niobium, given that its atomic mass is 93 u., , d=, , Ans:, a3 =, , !", , !", ! ! !!, , !  !!, , = 2 x 93/8.55 x6.022 x1023, = 36.1 x106, a = (36.1)1/3 x102, =330 pm, For BCC r =, r=, , !, !, , !, !, , a, , x 330, , r = 143pm, 10. An element has a body-centered cubic (bcc) structure with cell edge of 288pm., The density of the element is 7.2 g/cm3. How many atoms are present in 208g, of the element?

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Ans:, , d=, , !", ! ! !!, , M = d a3 NA/Z, = 7.2 x (2.88)3x10-24 x 6,022 x1023/2, = 103.57/2, M = 51.78 u, 51.78 g (1mole) contains 6.022 x1023 atoms, Therefore 208g contains 4.01 x 6.022 x1023 = 24.187 x1023 atoms.