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Edition : 2019

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ABOUT THE AUTHOR

Dr. Manu Kaushal is a renowned faculty of Chemistry and hasteaching experience of more than 10 years. He has worked as asenior faculty with institutes like Aakash Institute (New Delhi) andHelix Institute (Chandigarh). Through this book, he wants to sharehis vast experience with each and every student in India and helpthem prepare the best for their most important exam.

PREFACEStudents appearing for NEET cannot rely on conventional books and measuresto crack the entrance examination. Keeping in mind the cut-throat competition,it becomes mandatory for all the aspirants to have a thorough exposure of alltypes of problems, especially in Chemistry (Part II).

The objective of this book is to help students get rid of any phobia aboutChemistry. Through this book I have made every possible effort to includevariety of questions in every chapter.

The questions have skillfully been set along with solutions to help studentsdevelop an all round problem solving methodology.

I am thankful to the almighty, my parents, son 'Yuvraj', daughter 'Neyhal' andmy wife for their immense support.

I am also thankful to Mr. Nishant Pathak and Mr. Nitin Sharma for their help inwriting this book.

Finally, I would like to thank GK Publications (P) Ltd. for giving me anopportunity to write this book. For any queries, you can connect me at8968706718 (Whats app only)

Regards

Dr. MANU KAUSHAL

CONTENTS 1. Solid States 1.1 – 1.32

2. Solutions 2.1 – 2.34

3. Electrochemistry 3.1 – 3.24

4. Chemical Kinetics 4.1 – 4.30

5. Surface Chemistry 5.1 – 5.12

6. General Principles and Process of Isolation of Elements 6.1 – 6.8

7. p-Block Elements 7.1 – 7.13

8. d- and f-Block Elements 8.1 – 8.10

9. Co-ordination Compounds 9.1 – 9.30

10. Haloalkanes and Haloarenes 10.1 – 10.33

11. Alcohols, Phenols and Ethers 11.1 – 11.57

12. Aldehydes, Ketones and Carboxylic Acid 12.1 – 12.73

13. Organic Compounds Containing Nitrogen 13.1 – 13.42

14. Biomolecules 14.1 – 14.6

15. Polymers 15.1 – 15.8

16. Chemistry in Everyday Life 16.1 – 16.4

Solid States 1.1

1

Types of Solid, Space Lattice & Its Types1. Which one of the following is anisotropic &

covalent solid?

(a) NaCl (b) Diamond

(c) Copper (d) Tin

2. Identify the mismatch:

(a) Molecular solid Solid CO2

(b) Ionic solid sugar crystals

(c) Covalent solid Solid silicon

(d) Metallic solid tungsten

3. The nature of intermolecular forces in siliconcarbide are:

(a) London forces (b) Electrostatic forces

(c) Metallic forces (d) Covalent forces

4. A match box exhibits :

(a) Cubic geometry

(b) rectangular geometry

(c) Orthorhombic geometry

(d) monoclinic geometry

5. A metal has primitive cubic lattice with unit celledge length of 5 Å. Then, the volume of the unitcell (in cm3) is:

(a) 5 × 10–23 (b) 12.5 × 10–23

(c) 1.25 × 10–24 (d) 2.5 × 10–24

6. Which of the following crystal structures has theparameters as: a = b c & = = = 90º

(a) hexagonal (b) monoclinic

(c) tetragonal (d) orthorhombic

Solid States

CHAPTER

7. The crystal system of a compound with unit celldimensions, a = 0.387 nm, b = 0.387 nm, c = 0.504nm & = = 90º, = 120º is :

(a) Cubic

(b) Hexagonal

(c) Orthorhombic

(d) Rhombohedral

8. Glass is considered a super cooled liquid becauseit:

(a) Can be moulded into various shapes

(b) becomes crystalline on heating

(c) has a tendency to flow, though very slowly

(d) has a sharp melting point.

9. SO2 molecules are held in the crystal lattice by:

(a) Covalent bonds

(b) dipole-dipole interactions

(c) Metallic bonding

(d) dispersion forces

10. Match the crystal system in Column-I with theirpossible variations in column-II.

Column-I Column-II(a) Cubic p. Face centred

(b) Tetragonal q. End centred

(c) Orthorhombic r. Body centred

(a) a – p, r; b – q, r; c – p, q

(b) a – p, r; b – r; c – p, q, r

(c) a – q, r; b – r; c – p, q, r

(d) a – r; b – p, r; c – q, r

1.2 Solid States

11. In a bcc lattice, a unit cell is shared equally by:

(a) 8 unit cells (b) 4 unit cells

(c) 2 unit cell (d) 6 unit cells

12. Metals X, Y & Z crystallise in simple cubic, fcc &bcc unit cells respectively. What would be thecorrect order of metallic radii if volume of eachunit cell is same?

(a) rx > ry > rz (b) rz > rx > ry

(c) rx > rz > ry (d) ry > rx > rz

13. Packing fraction in two dimensional primitivesquare unit cell is:

(a) 4

(b) 2a

(c) 2a

4

(d) 4

14. Which of the following is a characteristic of ionicsolid?

(a) They have low melting & boiling points.

(b) They are always electrical insulators

(c) Ionic solids when dissolved in water or inmolten state, they conduct electricity.

(d) Both (a) & (b).

15. Match the compounds in column–I with theirtypes in column-II

Column-I Column-II(i) H2, Cl2 & I2 p. Hydrogen bonded

molecular solids

(ii) HCl, SO2 q. Non polar molecular

solid

(iii) H2O (ice) r. Polar molecular solid

(a) i – q; ii – r; iii – p

(b) i – r; ii – q; iii – p

(c) i – q; ii – p; iii – r

(d) i – p; ii – q; iii – r

16. In two dimensional square close packing, the co-ordination number is:

(a) 2 (b) 4

(c) 6 (d) 8

17. Which of the following statement is not true aboutamorphous solids?

(a) On heating they may become crystalline at acertain temperature.

(b) They may become crystalline on keeping forlong time.

(c) Amorphous solids can be moulded by heating.

(d) They are anisotropic in nature

18. Which of the following features are not shown byquartz glass?

(a) It is a pseudo solid

(b) Refractive index is same in all directions

(c) It has a definite heat of fusion

(d) It is also called super cooled liquid

19. Match the type of packing given in column-I withthe items in column-II.

Column-I Column-II

(i) square close A. Triangular voids

packing in 2-D

(ii) Hexagonal close B. pattern of spheres is

packing in 2-D. repeated in every

fourth layer

(ABCABC)

(iii) Hexagonal close C. Co-ordination no. = 4

packing in 3-D

(iv) Cubic close D. Pattern of sphere is

packing in 3-D. repeated in alternate

layers (ABABA...)

(a) i – D, ii – A, iii – C, iv – B

(b) i – A, ii – D, iii – C, iv – B

(c) i – B, ii – C, iii – A, iv – D

(d) i – C, ii – A, iii – D, iv – B

20. The axial angles in triclinic crystal system are:

(a) = = = 90º (b) = = ; 90º

(c) 90º (d) = = 90º

21. In tetragonal crystal system, axial distances oredge lengths are:

(a) a = b = c

(b) a = b c

(c) a b c

(d) a b = c

22. Match the unit cell in column-I with their numberof lattice points in column-II

Column-I Column-II(i) End-centred cubic p. 9

(ii) Face centred tetragonal q. 10

(iii) Body-centred cubic r. 14

(a) i – p, ii – q, iii – r

(b) i – r, ii – q, iii – p

(c) i – q, ii – p, iii – r

(d) i – q, ii – r, iii – p

Solid States 1.3

Formula Making

23. A compound is formed by element X & Y. Atomsof X are in simple cubic lattice while atoms Y areat centre of each face. If atom of X from onecorner is missing then the formula is:

(a) X7 Y24 (b) XY

(c) X7 Y3 (d) XY3

24. In a binary compound having cubic unit cells, theanions B are located at the corners of the cube &face centres whereas cations A occupy all thetetrahedral voids. What is the formula of thecompound?

(a) A2 B (b) AB

(c) AB3 (d) A2 B3

25. A compound of A and B crystallizes in a cubiclattice in which ‘A’ atom occupy the lattice pointsat the corners of a cube and two atoms of B occupythe centres of each of the cube faces. Then, theformula of compound is:

(a) AB3 (b) AB4

(c) AB2 (d) AB6

26. If the anions (A) form hexagonal closest packing

& cations (C) occupy only 23

Octahedral voids in

it, then general formula of compound is:

(a) CA (b) CA2

(c) C2 A3 (d) C3 A2

27. The formula of compound made of P, Q, R andS with P occupying all the corners of a cube, twoQ on each face diagnol, two R on each bodydiagnol and S present at the body centre and atall edge centres, is :

(a) PQ11 R4 S8 (b) PQ12 R8 S4

(c) PQ8 R12 S3 (d) PQ4 R4 S12

28. A solid has a structure in which X atoms arelocated at all corners of the cube, Y atoms at alledge centres and Z atom at cube centre, thenformula of the compound is:

(a) ZXY3 (b) XYZ

(c) XYZ3 (d) ZYX3

29. In a solid with ccp arrangement of O2– ions, atomsX are present at body centre and atoms Y atalternate face centres of a unit cell. Then, theformula of the solid is:

(a) XYO4 (b) XYO2

(c) XY2O4 (d) XYO

30. In a ccp lattice of X & Y, the X atoms are presentat corners while Y at all the face centres. If oneof the X atoms from a corner is replaced by Zatom, then the formula of the compound is:

(a) X7 Y24 Z2 (b) X7 Y24 Z

(c) X24 Y7 Z (d) X Y24 Z

31. In a crystalline solid, anions Y– are arranged inccp. Cations X+ are equally distributed b/woctahedral & Tetrahedral voids. If all theoctahedral voids are occupied, then the chemicalformula of the compound is:

(a) XY2 (b) X2Y

(c) XY3 (d) X3Y

32. A compound of iron & oxygen crystallises in ahexagonal close packed array of oxide ions withtwo out of every three octahedral voids occupiedby FeX+ ions. Then the formula of compound is:

(a) FeO (b) Fe3 O4

(c) Fe2 O3 (d) Fe1.94 O

Radius Ratio, Voids & Structures of IonicCompounds33. A compound AB has NaCl type structure. The

face diagnol of the unit cell is 120 2 pm. The

nearest interionic distance between Na+ & Cl–

in pm is :

(a) 30 (b) 120

(c) 60 (d) 45

34. What is true about NaCl structure with edgelength of X pm?

A. Minimum distance between two Na+ is X2

B. Minimum distance between Cl– is X2

C. Co-ordination number of each Cl– is 6.

D. Each Cl– at corner is surrounded by 8Tetrahedral voids.

(a) A, C, D only (b) A & C only

(c) B, C & D only (d) All are correct

35. In NaCl unit cell, if all the ions along one axisjoining opposite face centres are removed, thenthe mass of one unit cell is:

(a) 175.5 amu

(b) 234g

(c) 234 amu

(d) 175.5g

1.4 Solid States

36. The radius ratio of Cs+ to Cl– is 0.90.

If the radius of Cl– is 1.80, the false statementregarding CsCl structure is:

(a) the distance between the centres of two Cl– isgreater than 3.6

(b) the distance between the centres of two Cl–

ions is exactly 3.6

(c) the distance between Cs+ and Cl– is 3.42

(d) both (b) & (c) are correct

37. The number of Na+ ions present at edge centresof unit cell in 58.5g of NaCl is (NA is AvogadrosNumber):

(a) 1.33 NA (b) 0.75 NA

(c) 0.25 NA (d) 6NA

38. A solid AB has rock salt structure. If spheres froma face in a unit cell are removed, then the numberof octahedral voids destroyed (i.e., the voids whichcan no more be regarded as octahedral voids) inan unit cell is/are:

(a) 9 (b) 8

(c) 12 (d) 1

39. Consider the radii:

0.095 nm (Na+), 0.181 nm (Cl–), 0.074 nm (Zn2+)

0.2 nm (S2–), 0.169 nm (Cs+)

Choose the correct statement (s).

(a) Na+ ions are packed in octahedral voidsbetween the planes of close packed Cl– ions.

(b) Zn2+ are packed in tetrahedral voids in ZnS

(c) Cs+ ions are placed in cubic void in CsCl

(d) All are Correct

40. The number of next nearest neighbours of Cs+

ion in Cs+Cl– crystal is:

(a) 12 (b) 8

(c) 6 (d) 4

41. If the volume occupied by a unit cell of NaCl is4.7 × 10–23 ml, the volume of a crystal of NaClweighing 1 gm is:

(a) 2.73 × 1022 ml (b) 8.93 × 10–25 ml

(c) 10.88 ml (d) 0.12 ml

42. In CaF2 (fluorite type structure),

(a) Ca2+ forms fcc and F– occupy all tetrahedralvoids.

(b) Ca2+ forms fcc & F– occupy all octahedral voids.

(c) Ca2+ form bcc & F– occupy all Tetrahedral voids.

(d) None of these

43. Two ions P+ & Q– have radii 88 pm & 200 pmrespectively. In the close - packed crystal ofcompond PQ, the Co-ordination number of P+ is:

(a) 4 (b) 6

(c) 8 (d) None of thses

44. In diamond structure, the packing fraction of unitcell is:

(a) 0.66 (b) 0.69

(c) 0.34 (d) 0.12

45. An octahedral void at body centre (in FCC) isformed by:

(a) 3 face centred atoms + 3 corner atoms

(b) 6 face centred atoms + 8 corners atoms

(c) 6 face centred atoms only

(d) 4 face centred atom + 2 corners atoms

46. An octahedral void at edge centre (in FCC) isformed by:

(a) 6 corner atoms

(b) 2 corner atoms + 4 face centred atoms

(c) 8 corners atoms

(d) 4 corner atoms + 2 face centred atoms

47. A Tetrahedral void (in FCC) is formed by:

(a) 4 face centred atoms

(b) 2 face centred atoms + 2 corner atoms

(c) 3 face centred atoms + 1 corner atom

(d) 6 face centred atoms + 1 corner atom

48. What angle is made by the lines connecting anyZn2+ to any two of its nearest neighbours?

(a) 120º (b) 90º

(c) 109º 28 (d) 60º

49. In fcc crystal lattice, edge length is 400 pm. Then,the diameter of the greatest sphere which can befitted into the interstitial void without distortionof the lattice is:

(a) 152.5 pm (b) 121.3 pm

(c) 148.6 pm (d) 117.1pm

50. A cation ‘X’ fits exactly in trigonal void formed byanions ‘Y’. Then the radius ratio must be:

(a) 0.225 (b) 0.155

(c) 0.732 (d) 1

51. In ionic solids AB & CD, radius ratios are x & yrespectively. If x + y = 0.75 & x–y = 0.25, thencations in AB & CD are respectively, in:

(a) T-voids, O-voids (b) O-voids, T-voids

(c) T-voids, T-voids (d) O-voids, O-voids

Solid States 1.5

52. A metal crystallises as in fcc structure with edgelengh of 300 pm. The radius of smallest atom thatcan fit into the octahedral void of its lattice(without distorting the lattice) is nearly :

(a) 44 pm (b) 88 pm

(c) 50 pm (d) 125 pm

53. In an ionic crystal, the radius of cation A+ & theradius of anion B– is 1.00 & 3.3 pm respectivelythen the approximate edge length of unit cell ofAB is:

(a) 10 pm (b) 3.46 pm

(c) 0.5 pm (d) 4.3 pm54. In a normal spinal type structure, the oxide ions

are arranged in ccp whereas 18

tetrahedral voids

are occupied by Zn2+ ions & 50% of octahedralvoids are occupied by Fe3+ ions. Then, the formulaof the compound is:

(a) Zn2 Fe2 O4 (b) Zn2 Fe2 O3

(c) Zn Fe2 O4 (d) Zn Fe2 O2

55. rNa+ & rCl– represents radius of Na+ & Cl– ions

respectively. If ‘n’ represents the number of NaClunits per unit cell, then the equation to obtainmolar volume is:

(a)

3

Na ClA

8(r r )N

n

(b)

3

Na ClA

3 3(r r )N

2n

(c)

3

Na ClA

2 2(r r )N

n

(d)

3

Na ClA

2 3(r r )N

4n56. Select the correct statement (s) among the

following:

(a) In ZnS structure, Zn2+ occupy alternatetetrahedral voids.

(b) In NaCl structure, the Co. No of Na+ & Cl–

are same.

(c) In NaCl, the next nearest neighbour of Na+

are 12 Na+.

(d) All are correct.

57. The distance between tetrahedral voids in fcc is:

(a) 3a2

(b) a2

(c) a2

(d) All of these

58. In the crystal of CsCl, the nearest neighbours ofeach Cs+ ion are:

(a) 6 Chloride ions (b) 8 Chloride ions

(c) 6 Cesium ions (d) 8 Cesium ions

59. The edge length of unit cell of diamond is 356.7pm. The distance b/w C atoms is

(a) 154.45 pm (b) 318 pm

(c) 297 pm (d) Name of these

Density60. A metal crystallizes in two cubic lattices, fcc &

bcc, whose edge lengths are 3.5 & 3.0 Aºrespectively. The ratio of densities of fcc & bccunit cells is:

(a) 1.295 (b) 1.259

(c) 2.915 (d) None of these

61. Sodium (Na = 23) crystallizes in bcc arrangementwith the interfacial seperation between the atomsat edge 53.6 pm. The density of sodium crystal is:

(a) 3.5 g/cc (b) 2.66 g/cc

(c) 1.19 g/cc (d) None of these

62. An element crystallizes as body centred cubiclattice. Its density is 7.12g/cm3 and the length ofside of unit cell is 2.88 Aº. Then the number ofatoms present in 288g of the element are:

(a) 1.693 × 1023 (b) 1.693 × 1024

(c) 3.386 × 1023 (d) 3.386 × 1024

63. A solid contains A & B atoms in a cubicarrangement in which ‘A’ atoms occupies corners& B the face centres of the unit cell. If the edgelength is 5 Aº, the density of unit cell is (Givenatomic weights of A & B as 60 & 90, respectively)

(a) 7.2 g/cm3 (b) 4.38 g/ cm3

(c) 6.6 g/cm3 (d) 1.8 g/cm3

64. Sodium has a bcc structure with nearestneighbour distance of 365.9 pm. Density of its unitcell is:

(a) 1.5g/ cm3 (b) 3g/cm3

(c) 4g/cm3 (d) 5g/cm3

65. The density of a pure substance ‘A’ whose atomspack in cubic close pack arrangement is 1g/cc. If‘B’ atoms can occupy all tetrahedral voids, Whatis the density of resulting solid in g/cm3?

(At mass of A = 30g/mol & at mass of B = 50 g/mol)

(a) 3.33 (b) 4.33

(c) 6.33 (d) 1.33

1.6 Solid States

66. Copper has fcc crystal structure. Assuming anatomic radius of 130 pm for copper atom, what isthe volume of the unit cell?

(At mass of Cu = 63.54 g/mol)

(a) 4.94 × 10–23 cm2 (b) 6.63 × 10–25 cm3

(c) 7 × 10–30 cm3 (d) None of these

67. An atomic solid crystallizes in a body centred cubiclattice & the inner surface of the atoms atadjacent corner are seperated by 60.3 pm. If theatomic weight of ‘A’ is 48, the density of solid isnearly:

(a) 1.75 g/cc (b) 2.8 g/cc

(c) 3.2g/cc (d) 7.1 g/cc

68. A metal crystallises into two cubic faced fcc & bccwhose edge lengths are 3.5 & 3.0 Aº respectively.The ratio of densities of fcc & bcc unit is cell

(a) 1.26 (b) 3.2

(c) 4.1 (d) 0.23

69. Iron crystallizes in several modifications. At about910k, ‘bcc’ form (called form) undergoestransition to -form with fcc lattice. Assume thatthe distance between the nearest neighbours isthe same in two forms at the transitiontemperature. Then, the ratio of density of -ironto that of -iron at the transition temperature is:

(a) 1.08 (b) 0.74

(c) 0.83 (d) 0.108

70. Calculate mass of a unit cell of an element whichcrystallises in ccp arrangement if its atomic massis 56 amu.

(a) 3.72 × 10–22g (b) 1.86 × 10–22g

(c) 9.3 × 10–23g (d) insufficient data

71. A compound AB has NaCl type structure. Thedensity of AB is 6.25 g/cm3 & its molecular massis 60 g. If the radius of cation A+ is 60 pm, thenradius of anion B– is (NA = 6 × 1023)

(a) 60 pm (b) 140 pm

(c) 120 pm (d) 10 pm

Defects in Crystals72. Which one of the following is not observed when

Na Cl is heated in the atmosphere of Na vapour?

(a) electrons are trapped in anion vacanceis

(b) Non-stoichiometric defect results

(c) Extra Na+ ions occupy interstitial positions

(d) Both (a) and (b)

73. At Curie temperature,

(a) diamagnetic solid changes to paramagnetic

(b) Ferromagnetic solid changes to paramagnetic

(c) Anti ferromagnetic solid changes to paramagnetic

(d) Ferrimagnetic solid changes to paramagnetic

74. The density of crystal always increases in whichof the following defects:

(a) Schottky defect (b) Vacancy defect

(c) Frenkel defect (d) Interstitial defect

75. Which of the following intrinsic defect is shownin the figure below?

A+ A+ A+ B–

B– A+

A+ A2+

A+ A+

B–

B– B–

B– B–

(a) Schottky defect

(b) Frenkel defect

(c) Metal deficiency defect

(d) Metal excess defect

76. Which of the following statement is WRONGabout schottky defect?

(a) Density of the crystal decreases

(b) It is due to occupation of an interstitial site bycation

(c) Stability of crystal decreases

(d) All of these are WRONG.

77. If ‘Y’ is the ratio of the observed to the theoreticaldensities of a solid, then the value of ‘Y’ forschottky defect, interstitial defect & Frenkeldefects present in the solid, respectively, are:

(a) <1, > 1, = 1 (b) > 1, < 1, = 1

(c) < 1, = 1, = 1 (d) < 1, > 1, < 1

78. Carbon (diamond), silicon and copper arecrystalline solids. These are characterised byconduction band and valence band separated byenergy gaps respectively, equal to (Eg)C, (Eg)Si and(Eg)Cu Then

(a) (Eg)C = (Eg)Si = (Eg)Cu

(b) (Eg)C > (Eg)Si > (Eg)Cu

(c) (Eg)C < (Eg)Si < (Eg)Cu

(d) (Eg)C < (Eg)Si > (Eg)Cu

Solid States 1.7

79. Which of the following crystals show Frenkel defect?

(i) AgBr (ii) AgCl

(iii) KBr (iv) ZnS

(a) (i), (ii) & (iv) (b) (i), (ii) & (iii)

(c) (ii), (iii) & (iv) (d) both (iii) & (iv)

80. Which of the following statement regardingelectrical conductivity is/are true?

(a) The electrical conductivity of semi-conductorsincreases with rise in temperature

(b) Conductivity of silicon crystal increases ondoping with gallium

(c) Metals behave like super conductor at verylow temperature

(d) All are correct

81. Which one of the following is a combination ofboth vacancy & interstitial defect?

(a) Frenkel defect (b) Schottky defect

(c) Impurity defect (d) None of these

82. Which of the following are stoichiometric defects?

1. Schottky defect 2. Frenkel defect

3. Interstitial defect 4. Vacancy defect

(a) (2) & (4) only (b) (2) & (3) only

(c) (1), (2) & (4) only (d) All of these

83. Zinc oxide on heating changes to yellow. This isbecause:

(a) Zinc oxide becomes a stoichiometric compound

(b) Zinc oxide is a covalent compound

(c) Zinc oxide shows metal excess defect

(d) It has F-centres

84. The alignment of magnetic dipoles is shown below:

Which of the following property is representedby these dipoles?

(a) Diamagnetism (b) Ferrimagnetism

(c) Ferromagnetism (d) Anti-ferromagnetism

85. Match the types of defects given in column-I withthe statement given in column-II.

Column-I Column-II(i) Impurity defect A. NaCl with anion

vacancies having e–

trapped in them(F-centres)

(ii) Metal excess defect B. FeO with Fe3+

(iii) Metal deficiency C. Na Cl with Sr2+ & somedefect cationic sites are

vacant.

(a) (i – A), (ii – C), (iii – B)

(b) (i – C), (ii – B), (iii – A)

(c) (i – C) (ii – A) (iii – B)

(d) (i – B), (ii – A), (iii – C)

86. Select the incorrect match.

(a) Ferromagnetic substances Fe, Ni, Co

(b) Antiferromagnetic substances FeO, MnO

(c) Ferrimagnetic substance Fe3 O4

(d) None of these is incorrect

Brain Teasers87. How many unit cells are present in a ccp

arrangement of NaCl of mass 1g?

(a) 5.14 × 1021 (b) 1.28 × 1021

(c) 1.71 × 1021 (d) 2.57 × 1021

88. If the radius of the anion in an ionic solid is 200pm, what would be the radius of the cation thatfits exactly into the cubic void?

(a) 146.4 pm (b) 82.8 pm

(c) 45 pm (d) 90 pm

89. If the ratio of co-ordination number of A to thatof B is X:Y, then the ratio of number of atoms ofA to that of number of atoms of B in unit cell is:

(a) X : Y (b) Y : X

(c) X2 : Y (d) Y2 : X

90. Which statement is correct regarding the voids?

(a) Cubic void & octahedral void, both are formedat the centre of cube

(b) Cubic void is located at face centre where asoctahedral void is located at body centre

(c) Octahedral void is located at corner where ascubic void is located at edge centre

(d) Cubic void & octahedral voids are same

91. In XY3, Y– ions have ccp arrangement & X3+ ions

are present in octahedral voids. The fraction oftotal number of voids occupied is:

(a) 13

(b) 16

(c) 19

(d) 1

1292. The total volume occupied in a face centred cubic

unit cell of a metal is (r is the atomic radius):

(a) 312r

3 (b) 316

r3

(c) 320r

3 (d) 324

r3

1.8 Solid States

93. In a sample of FeO when 12% Fe2+ ions getreplaced by Fe3+, the composition of this Wustitesample is:

(a) Fe0.98 O (b) Fe0.93O

(c) Fe0.96O (d) Fe0.94O

94. An alkali metal (at weight = 40) has a bcc structure

with nearest neighbour distance = 3 Aº. Itsdensity in kg m–3 is d × 104. The value of ‘d’ is

(a) 3.32 (b) 1.66

(c) 2.55 (d) 0.83

95. A transition metal M can exist in two oxidationstates + 2 & + 3. It forms an oxide whoseexperimental formula is MXO (where X < 1).

The ratio of number of metal ions in + 3 Oxidationstate to those in + 2 oxidation state is:

(a) 1 x1 x

(b) 1 + 2x

(c) 2 x

2(3x 2) (d)

2(1 x)3x 2

96. If carbon atoms from one of the body diagnol planeof a diamond unit cell are removed, then thereduction in densities is:

(a) 43.75% (b) 50%

(c) 33.3% (d) 62.4%

97. The edge length of diamond unit cell was measuredas 0.4 nm. Then the density of the unit cell is:

(a) 0.250 g/cm3 (b) 2.50 g/cm3

(c) 2.50 kg/cm3 (d) 2.50 g/m3

98. Packing fraction in 2-D hexagonal arrangement is:

(a) 0.6 (b) 0.74

(c) 0.524 (d) 0.906

99. In a non-stoichiometric sample of ferrous oxidewith NaCl structure, the ratio of no. of Fe3+ ionsto Fe2+ ions was found to be 0.15. Calculate thefraction of octahedral sites occupied by vacancies?

(a) 0.9384 (b) 0.0612

(c) 0.75 (d) 0.025

100. In ccp structure of mixed oxides, the oxide ionsoccupy all the corners & all the face centres.Cations ‘A’ occupy 25% Octahedral voids & 12.5%tetrahedral voids are occupied by B. The nearestneighbour distance wrt oxide ion is:

(a) 3a4

(b) a2

(c) a4

(d) 3a8

101. Pervoskite crystallizes with Ca2+ ions at cornersof a cubic unit cell, Ti4+ in its body centre &O2– ions at all the face centres. The co-ordinationnumbr of Ti4+ ion is:

(a) 6 (b) 8

(c) 14 (d) 7

102. Next nearest neighbour of S2– ion in zinc blendestructure is/are:

(a) 6 (b) 8

(c) 12 (d) 10

103. CsCl unit cell has an edge length equal to 4.3 nm.Then, the shortest distance between two Cs+ ions is:

(a) 4.3 nm (b) 8.6 nm

(c) 2.15 nm (d) None of these

104. Three lines are drawn from a single corner of a fccunit cell to meet the other corner such that theyare found to pass through; exactly only octahedralvoid (line one), onvoid of any types (line two) andexactly two tetrahedral void with one octahedralvoid (line three). The lines in same order are:

(a) Edge, face diagonal, body diagonal

(b) Face diagonal, edge, body diagonal

(c) Body diagonal, face diagonal, edge

(d) Edge, body diagonal, face diagonal

105. For a unit cell of edge length 10 Å and havingdensity equal to 1g/cm3, the atomic radius is(given at mass = 150 amu)

(a) 3.53 Å (b) 6.77Å

(c) 12.52 Å (d) 13.10Å

106. The arrangement of spheres in a hexagonal unitcell is shown below. If the labelled atoms aremissing then the number of effective atoms in unitcell left after removing the labelled atoms are:

y

Z

w

XP

(a) 218

(b) 113

(c) 6 (d) 4

Solid States 1.9

107. An ionic compound consists of X+, Y2+ & Z3– ions.Which of the following option (s) is/are Correctwrt their possible structure?

(a) X+ ions occupy alternate corners of a simplecube. Y2+ occupy the body centre & Z3– occupya single tetrahedral void.

(b) Z3– ions form ccp, X+ ions occupy all theoctahedral voids & Y2+ ions occupy half of thetetrahedral voids.

(c) Y2+ ions form hcp, X+ occupy all the octahedralvoids & Z3– ions occupy half of the tetrahedralvoids.

(d) Both b & c are correct

108. A monoatomic solid element exists in a cubiccrystal. If its atomic radius is 1 Å & the ratio ofpacking fraction & density is 0.1 cm3/gm, thenthe atomic mass of the element is:

(a) 8

(b) 7.5

(c) 64

(d) 32

109. Ratio of effective No. of atoms (Z) for a 2Dhexagonal close packed unit cell (I) to 3Dhexagonal centred unit cell (II) is :

1 2

6 3

45(I)

(II)

(a) 23

(b) 1

(c) 12

(d) 13

110. XY2 has ideal fluorite structure. The incorrectstatement is/are:

1. each X2+ at corner is sorrounded by 18Tetrahedral voids

2. each X2+ is sorrounded by 6 octahedral voids

3. shortest distance between two Y– ions is halfof edge length

4. shortest distance between X2+ & Y– is half ofedge length

(a) only 3 (b) 1, 3 & 4 only

(c) 1 & 4 only (d) 1, 2 & 4 only

111. In a face centred cubic lattice, atoms A are at thecorners & atoms B at the face centres. If atom Bis missing from one of the face centred points,the formula of the ionic compound is:

(a) AB2 (b) A5B2

(c) A2 B3 (d) A2 B5

112. In a monoclinic unit cell, the relation of sides andangles, are respectively:

(a) a = b c and = = = 90º

(b) a b c and = = = 90º

(c) a b c and = = 90º

(d) a b c and 90º

113. Ice crystallises in a hexagonal lattice havingvolume of the unit cell as 132 × 10–24 cm3. Ifdensity is 0.92g cm–3 at a given temperature, thennumber of H2O molecules per unit cell are:

(a) 2 (b) 3

(c) 4 (d) 6

114. An oxide has a ccp structure in which oxide ionsare present at all lattice points (face centres &corners) whereas Na ions are located at all theedge centres and W ion at the body centre. Then,the oxidation state of ‘W’ is:

(a) + 2 (b) + 3

(c) + 4 (d) + 5

115. The packing fraction in zinc blende (sphalerite)structure unit cell is (assuming ideal crystal)

(a) 0.749 (b) 0.385

(c) 0.575 (d) 0.912

116. A crystalline solid AB has NaCl type structurewith radius of B– ion as 250 pm. Which of thefollowing cation can be made to slip into thetetrahedral site of crystal of A+ B– ?

(a) C+ (ionic radius = 180 pm)

(b) D+ (ionic radius = 90 pm)

(c) E+ (ionic radius = 200 pm)

(d) G+ (ionic radius = 150 pm)

1.10 Solid States

117. In FCC unit cell, if on one of the face, unoccupiedspace between the atoms is 20 pm, what will bethe edge length of this unit cell?

20 pm

(a) 68.25 pm (b) 100 pm

(c) 143.88 pm (d) 190 pm

118. If NaCl is doped with 10–3% AlCl3, what is theconcentration of cation vacancies?

(a) 6.02 × 1018 (b) 12.04 × 1018

(c) 2 (d) 12.04 × 1020

119. Three layers of bcc are shown as:

5

2

34

1

Z

Y

X

7

89

6

Which one of the following is correct?

(a) Distance between atoms labelled 6 & 7 = 2R

(b) Distance between atoms labelled 5 & 3 = 4R(c) Ratio of distance between 7 & 9 and that

between 7 & 5 = 83

(d) Ratio of distance between 4 and 5 & that

between 8 and 5 = 38

120. If NaCl is doped with AlCl3 and has 5 mole%vacancies. Then the formula of the compound is:

(a) Na0.925 Al0.025Cl (b) Na0.075 Al0.025Cl

(c) Na0.975 Al0.05Cl (d) Na0.045 Al0.05Cl

121. The packing efficiency of the two dimensionalsquare unit cell shown below is:

a(a = 2r)

(a) 78.5% (b) 66.2%

(c) 20% (d) 52.4%

122. If a solid A+ B– having ZnS structure is heated sothat the ions along two of the axis passing throughthe face centre, particles are lost & bivalent ion(Z) enters here to maintain electrical neutralityso that the new formula becomes AX BY ZC. Thenvalue of X + Y + C is:

(a) 7 (b) 9

(c) 5 (d) 6

123. In a fcc unit cell, atoms are numbered as shownbelow.

The atoms not touching each other are (Atomnumbered 3 is face centre of front face)

3 4

2

1

(a) 3 & 4 (b) 1 & 3

(c) 1 & 2 (d) 2 & 4

124. Which of the following expression is correct forpacking fraction of NaCl if all the ions along theface are diagonally removed?

(a)

3 3

3

13 16r r

3 38(r r )

(b)

3 3

3

13 4r r

3 38(r r )

(c)

3 3

3

16 13r r

3 38(r r )

(d)

3 3

3

4 13r r

3 38(r r )

125. The composition of a sample of wustite is Fe0.93O1.00. Which of the following statement(s) is/arecorrect wrt this compound?

(a) The percentage of Fe3+ by mass is 11.5%

(b) The ratio of number of Fe3+ ions to the numberof Fe2+ ions is 0.17.

(c) The weight of Fe3+ ion is 7.84g.

(d) Both (b) & (c) are correct.

126. In the CaF2 structure, if the radius ratio is

31 ,

2

then how many ions does each cation

touch?

(a) 4 anions (b) 12 cations

(c) 8 anions (d) both (b) and (c)

127. The packing fraction for spinel structure (ideal)is :

(a) 0.768 (b) 0.56

(c) 0.356 (d) 0.16

Solid States 1.11

128. Certain metal crystallizes in bcc lattice. Thedistance between the nearest neighbours is 433pm. Then, the distance between the next nearestneighbours is:

(a) 400 pm

(b) 500 pm

(c) 600 pm

(d) 650 pm

129. The carbon-carbon bond length in diamond is173.2 pm. If diamond’s structure were consideredto be ZnS type & having radii equal to half thebond length, then what would be its expecteddensity?

(a) 2.49 g/cm3

(b) 1.76 g/cm3

(c) 7.65 g/cm3

(d) 8.8 g/cm3

130. If NaCl is doped with 10–3 mol% of AlCl3, theconcentration of cation vacancies is:

(a) 12.04 × 1018

(b) 6.02 × 1016

(c) 6.02 × 1020

(d) 3.01 × 1018

131. The figure given below shows the location ofatoms in a crystallographic plane in fcc lattice,which of the following plane will show thisarrangement?

(a) Face plane (b) Body diagonal plane

(c) Rectangular plane (d) Any imaginary plane

132. Which of the following statement is NOT trueabout hexagonal close packing in 3D?

(a) The co-ordination No. is 12.

(b) It has 74% packing efficiency

(c) Tetrahedral voids of the 2nd layer are coveredby the spheres of the 3rd layer.

(d) In this arrangement, spheres of the fourthlayer are exactly aligned with those of the firstlayer.

133. The tetrahedral voids formed by ccp arrangementof Cl– ions in rock salt structure are –(a) Occupied by Na+ ion

(b) Occupied by Cl– ion(c) Occupied by either Na+ or Cl–ion(d) Vacant.

1. (b) 2.(b) 3. (d) 4. (c) 5. (b) 6. (c) 7. (b) 8. (c) 9. (b) 10. (b)

11. (d) 12. (c) 13. (a) 14. (c) 15. (a) 16. (b) 17. (d) 18. (c) 19. (d) 20. (c)

21. (b) 22. (d) 23. (a) 24. (a) 25. (d) 26. (c) 27. (b) 28. (a) 29. (a) 30. (b)

31. (b) 32. (c) 33. (b) 34. (d) 35. (a) 36. (b) 37. (b) 38. (a) 39. (d) 40. (c)

41. (d) 42. (a) 43. (b) 44. (c) 45. (c) 46. (b) 47. (c) 48. (c) 49. (d) 50. (b)

51. (b) 52. (a) 53. (a) 54. (c) 55. (a) 56. (d) 57. (d) 58. (b) 59. (a) 60. (b)

61. (c) 62. (d) 63. (b) 64. (a) 65. (b) 66. (a) 67. (a) 68. (a) 69. (a) 70. (a)

71. (b) 72. (c) 73. (b) 74. (d) 75. (c) 76. (b) 77. (a) 78. (b) 79. (a) 80. (d)

81. (a) 82. (d) 83. (c) 84. (b) 85. (a) 86. (d) 87. (c) 88. (a) 89. (b) 90. (a)

91. (c) 92. (b) 93. (c) 94. (b) 95. (d) 96. (a) 97. (b) 98. (d) 99. (b) 100. (a)

101. (a) 102. (c) 103. (a) 104. (a) 102. (a) 106. (b) 107. (d) 108. (a) 109. (c) 110. (c)

111. (d) 112. (c) 113. (c) 114. (d) 115. (a) 116. (b) 117. (a) 118. (b) 119. (c) 120. (a)

121. (a) 122. (a) 123. (c) 124. (a) 125. (d) 126. (d) 127. (a) 128. (b) 129. (a) 130. (a)

131. (a) 132. (d) 133. (d)

1.12 Solid States

3. Silicon carbide has a structure similar to diamondwhere every alternate ‘C’ is replaced by silicon.

5. Volume of unit cell = a3 (5 × 10–8 cm)3

125 × 10–24 cm3

9. SO2 has a net dipole moment & is hence polar.

So, SO2 molecules are held in lattice by dipole-dipole interactions.

11. In any lattice, a unit cell is shared equally by 6unit cells.

12. Volume of unit cell = a3

Same volume of each unit cell means that edgelength of each unit cell is same.

For simple cubic, For fcc,

rx

rx

ry

ry

2rx = a 2ry = 2a2

x

ar

2 ry = 2a

2 2

y

ar

2 2

yr 0.353a

For bcc,

rz

rz

2rz = 3a2

z

3ar

4

zr 0.433a

13. a

a = 2r

P.F = Area covered by atoms

Total area

= 2 2 2

2 2 2

1 r r ra (2r) 4r 4

16.

22.

End - Centred Cubic

Face Centred tetragonal

Body Centred Cubic

23.

Locations occupied by X atoms = 7 corners

(1 corner atom is missing)

Effective No. of X atoms = 17

8

Y atoms are at locations = 6 (6 face centres)

Effective No. of Y atoms = 16 3

2

378

X Y

7 24X Y

Solid States 1.13

24. B-located atCorners

Face Centers

8 × 18

= 1

6 × 12

= 3

So, effective 4 B– ions are present per unit cell.

Tetrahedral voids formed = 8

So, effective A+ cations = 8

Formula is A8 B4 = A2B

25. A Corners 1

8 18

B 2 atoms a each face centre 1

12 62

Formula is AB6.

26. Effective No. of atoms in hcp (3D) = 6

So, A– = 6

Effective No. of Octahedral voids = 6

No. of octahedral voids occupied by C+ = 2

63

So, formula is 4 6 2 3C A C A

27. P at all corners 1

8 18

Q 2 on each face diagnol 1

24 122

R 2 on each body diagnol (2 × 4 = 8)

S one at body centre + at all edge centres

Effective no. of S atoms = 1 + 12 × 14

= 4

So, formula is P1 Q12 R8 S4

28. X atoms at all corners 1

8 18

Y atoms at all edge centres 112 3

4

Z atom 1 (at cube centre)

So, formula is X1 Y3 Z1.

29. O2– in ccp Effective atoms = 4

(fcc) (Corners = 18

8 + face centres

= 1

6 42

)

X (at body centre) = 1

Y (at alternate face centres) = 12 1

2

So, formula is X1 Y1 O4

30. Effective no. of X atoms = 17

8

(one corner is removed)

Effective no. of Z atom = 11

8

Effective no. of Y atoms = 1

6 32

7 3 1 7 24 1/8 /8X Y Z X Y Z

31. Effective Y– in ccp (or fcc) = 4

1 1

8 (from corners) 6 (form face centres)8 2

No. of Octahedral Voids formed = 4

No. of Octahedral Voids occupied by X+ = 4 (given)

No. of T. Voids (Tetrahedral) formed = 8

No. of T. Voids occupied by X+ = 4

( Cations equally distributed)

So, formula is X8 Y4 = X2 Y

32. No. of effective O2– ions = 6

No. of effective O. Voids = 6

No. of O. voids occupied = 4

by Fex+

Total Anionic charge = – 12

Total Cationic charge = + 12

Wrt Fex+ = 4

4x = 12

x 3

Fe3+ O2– = Fe O2 3

33.

CaA

Ba

Apply pythagoras theorem

For ABC

BC2 = AB2 + AC2

BC2 = a2 + a2

BC 2 a

1.14 Solid States

So, face diagnol (BC) = 2a

2a 120 2

a 120 pm

For NaCl, Na+ are at O. voids & Cl– occupy fcclocations.

a2

rNa+

rCl–

rNa+ + rCl– =

a2

rNa+ + rCl– =

1202

= 60

So, nearest interionic distance between Na+ &Cl– = 60 pm.

34. Na+ occupy Octahedral Voids & Cl– occupy fcclattice points.

A

Na+ Z

YNa+

Apply Pythagoras Theorem for ZAY –

ZY2 = AZ2 + AY2

ZY2 = 2 2X X

2 2

ZY2 = 2 2X X

4 4

ZY2 = 22X

4

X

ZY2

So, min. distance between two Na+ is X2 .

For ideal NaCl structure (i.e., Na+ exactly fittingin Octahedral void), Anion-anion contact ismaintained.

At corner

Cl (at face center)–

Cl –

2r– = 2 X X2 2

So, minimum distance between Cl– = X2

Wrt ‘C’ statement

Cl–

Cl–, sorrounded by 6 Na+ ions

Co-ordination No. wrt Cl– = 6

Wrt ‘d’ statement

Each Cl– at corner is sorrounded by 8 T.V. at 8Body diagnols which emerge from corner.

35.

In NaCl structure

Na+ occupy all O.V.

Cl– occupy fcc locations

Cl–Corners

Face Centers

8 × 18

= 1

6 × 12

= 3

Effective Cl– ions per unit cell = 4

So, Octahedral voids present (Effective) = 4

Effective Na+ ions per unit cell = 4

Hence, formula units of NaCl

Per unit cell = 4

After removal of ions

Na+

Cl–

Cl–

2 face centred Cl– ions

and one Na+ at body centre (O. V.) is lost.

Solid States 1.15

Now, effective Cl– ions

Face center

Corners

8 × 18

= 1

4 × 12

So, effective total Cl– ions = 3

Effective Na+ ions = 1

12 34

So, formula = Na3 Cl3

Mass of unit cell = 3 × 23 + 3 × 35.5

= 69 + 106.5

= 175.5 amu

36. For CsCl,

Cs

Cl

r0.90

r

Csr

0.901.80

Cs

r 0.90 1.80 1.62

wrt Cs+ Cl–, Cl– is at corners & Cs+ at body centre(in cubic void)

O Cl –

Cs+

Since, radius ratio is > 0.732. So, Cs+ is not exactlyfitting & hence ANION-ANION Contact is notmaintained.

Wrt the present structure,

body centred Cs+Cl at corner–

r– + r+ = 3a2

1.80 + 1.62 = 3

a2

3.42 2a

1.732

3.94 a

Also, Cl–Cl–

So, distance between 2 Cl– ions = 3.94

Also, distance between Cl– & Cs+ = 1.80 + 1.62

= 3.42

37. Wrt NaCl structure, Na+ occupy all O. Voids.

Effective Na+ at edge centres = 112 3Na

4

3 Effective Na+ are present at edge centres perunit cell & 4 formula units of NaCl are presentper unit cell.

4 formula units of NaCl Correspondto 3 Effective

Na+ at edge centres

A

58.54 gm

N Correspond

to 3 Effective Na+ at edge

centres

58.5 gm Correspondto

A

358.5

458.5

N

= A

3N 58.5

58.5 4

= A A

3N 0.75 N

4

38. Na+ occupy all O.Voids.

O.Voids are present at all edge centres & bodycentre in fcc.

17

8

9 B

5632

D 4 C

A

E

If Cl– at A, B, C, D & E locations are destroyed,then O.Voids at edge centres 1, 2, 3, 4, 5, 6, 7 & 8are destroyed. Also, O.void labelled 9 (at bodycentre) is also destroyed.

39. Radius ratio

(for NaCl) =

Na

Cl

r 0.0950.524

r 0.181

For, Octahedral void, rr

= 0.414 – 0.732

& 0.524 lies between the range.

1.16 Solid States

So, Na+ are in O. voids.

For ZnS,

2

2

Zn

s

r 0.0740.37

r 0.2

Now, for tetrahedral void, rr

= 0.225 – 0.414 &

0.37 lies in the range. So, Zn2+ ions occupytetrahedral voids.

For Cs+Cl–,

Cs

Cl

r 0.1690.933

r 0.181

For cubic void, rr

= 0.732 – 1.00 & 0.933 lies

within this range. So, Cs+ is present in cubicvoid.

Hence, all the three statements are true.

40. Nearest neighbour of Cs+ are Cl– ions. Nextnearest neighbour of Cs+ will be 6 Cs+ ions at adistance ‘a’.

a/2 a/2

Cs+

41. One unit cell of NaCl has 4 formula units of NaCland

NA formula units of NaCl weigh = 58.5g

4 formula units of NaCl weigh = A

58.54g

N

So, A

58.54g

N NaCl has vol. = 4.7 × 10–23 ml

1 gm NaCl has vol. = 23

A

4.7 104

58.5N

= 23

234.7 106.02 10

58.5 4

= 28.294

0.12 ml234

43. Calculating the radius ratio, we get:

r 88

0.44r 200It lies in the range, 0.414 – 0.732.

So, Co-ordination No. is 6.

44. Diamond has ZnS type structure where all S2–

ions & Zn2+ ions are replaced by C atoms.

Location of C atomsat alternate tetrahedralvoids (4)

at face centers at corners

8 × 18

= 1 6 × 12

= 3

So, there are effective (1 + 3 + 4) 8 C atoms perunit cell.

Hence, volume occupied by 8C atoms

= 348 r

3 ...(1)

Total volume of the cube = a3

Also,

C atom at cornerC atom in Tetrahedral void

r + r = 3a4

3a

2r4

3a

r8

Packing fraction = Volume occupied by atomsTotal Volume of cube

P.F =

3

3

3 3

4 3a4 88 r 3 83 0.34a a

45.O. Void

face centred atoms

Solid States 1.17

48. Zn2+ is in Tetrahedral void.

49. FCC has 2 voids, tetrahedral & octahedral.Greatest sphere can be accomodated only ingreatest void (octahedral). To fit in O.Void withoutdistortion,

void

(atom packed in fcc)

rr = 0.414 ...(1)

For fcc 2r = 2a2

2r = 2

4002

r = 2

4004

r(radius of atom in fcc) = 100 2Now, placing the value in equation 1, we get

voidr0.414

100 2

rvoid = 0.414 × 100 2rvoid = 0.414 × 100 × 1.414 = 58.53 pm

diameter of greatest sphere that can fit

= 2 × 58.53 = 117 pm50.

A C

Br–

r–r–

r–

r– r–

In ABC,

cos 30º = r

r r

3 r2 r r

Reciprocating both sides

2 r rr3

2 r rr r3

2 r1

r3

2 3 rr3

r0.155

r

51. Given

x + y = 0.75 ...(1)

x – y = 0.25 ...(2)

Add (1) and (2)

2x = 1

x = 1

0.52

Place volume of x in (1)

0.5 + y = 0.75

y 0.25

So, x 0.5

A

B

r0.5

r

...(3)

C

D

r0.25

r

...(4)

From equation 3, radius ratio = 0.5 which liesbetween 0.414 – 0.732.

So, cation is in octahedral void.

From eqn. 4, radius ratio = 0.25 which lies b/w0.225 – 0.414, So, cation is in Tetrahedral void.

52. For exactly fitting in octahedral void,

void

metal atom

r0.414

r ...(1)

Also, in fcc

2r = 2a2

(where r is radius of metal atoms packed in fcc)

2r = a2

r = 3002 2

(a = 300 pm (given))

r = 150

106.08 PM2

Placing the value of radius of metal atom inequation 1.

voidr0.414

106.08

rvoid = 0.414 × 106.08 = 43.91 44pm