Question Ch9.pdf

Chapter

9

The Solid State Solutions SECTION - A Objective Type Questions 1.

Which of the following crystal is represented by a  b  c and  90°? (1) Orthorhombic

(2) Monoclinic

(3) Triclinic

(4) Tetragonal

Sol. Answer (3) a  b  c,   90°  Is the parameter for crystal triclinic. 2.

Copper belongs to a crystal system represented by the crystal dimensions as (1)  =  =  = 90º, a = b = c (2) 

   , a = b = c

(3)  =  = 90º, 

 90º, a = b = c

(4)  =  = = 90º, a

bc

Sol. Answer (1) Cu  belongs to cubic crystal which has dimensions a = b = c and  =  =  = 90° 3.

What is the relation between diamond and graphite? (1) Polymorphous

(2) Isomer

(3) Isotope

(4) Isomorphous

Sol. Answer (1) Diamond and graphite are polymorphous because both have similar chemical composition but different arrangement of constituent particles i.e., carbon. 4.

Maximum possible numbers of two dimensional and three dimensional lattices are respectively (1) 5 and 14

(2) 7 and 14

(3) 14 and 4

(4) 5 and 13

Sol. Answer (1) Two dimensional lattices are = 5 [Square, Rectangle, Rhombus, Parallelogram, Hexagonal] Three dimensional lattices are = 14 [Bravais lattices] Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

2 5.

The Solid State

Solution of Assignment

A compound formed by element A and B crystallizes in the cubic structure, where A atoms are at the corners of a cube and B atoms are at the centre of the body. The formula of the compounds is (1) AB

(3) A2B3

(2) AB2

(4) AB3

Sol. Answer (1) A  corners of cube  Total 8 atoms at corner, which have contribution at corner So  8 

1 8

1  1  number of atoms of A per unit cell. 8

B  centre of body  at centre of body one atom contributed completely. So,  1 × 1 = 1 i.e., A1B1 = AB 6.

A solid with formula ABC3 would probably have (1) A at body centre, B at face centres and C at corners of the cube (2) A at corners of cube, B at body centre, C at face centre (3) A at corners of hexagon, B at centres of the hexagon and C inside the hexagonal unit cell (4) A at corner, B at face centre, C at body centre

Sol. Answer (2) ABC3 At corners number of atom = 8 

1  1 (contribution at corner) 8

A B

At body centre = 1 × 1 = 1

At face centre =

6

1 2



(Total face)

3

C

(Contribution at face)

ABC3 7.

A solid ABC has A, B and C arranged as below. The formula of solid is

C



B

A

 (1) ABC

(2) AB2C2

(3) A2BC

(4) AB8C2

Sol. Answer (1) A  present at body centre in given figure A

So, its number = 1 × 1 = 1 (contributed completely) B present at corners =

8



(Total atom)

1 8

1

B

(Contribution)

C  Present at two opposite face 

2



(Total 2 atoms at face)

1 2

(Contribution at each face)

1

 C

So, formula ABC Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

8.

The Solid State

3

An alloy of copper, silver and gold is found to have copper constituting the ccp lattice. If silver atoms occupy the edge centres and gold is present at body centre, the alloy has a formula (1) Cu4Ag2Au

(2) Cu4Ag4Au

(3) Cu4Ag3Au

(4) CuAgAu

Sol. Answer (3) 1 ⎧ ⎫ ⎪⎪Present at corners = 8 × 8 = 1 ⎪⎪ Cu  ccp ⎨ ⎬ Total 4 Cu ⎪Present at centres of each face = 6 × 1  3 ⎪ ⎪⎭ 2 ⎩⎪

Ag  at edge centre 

1 4



12

(Total atoms at edge)

⇒3

(contribution)

Au  at body centre  contributed completely = 1 × 1  1 Cu4Ag3Au1 9.

In a face centred cubic arrangement of A and B atoms, atoms of A are at the corner of the unit cell and atoms of B are at the face centres. One of the A atom is missing from one corner in unit cell. The simplest formula of compound is (1) A7B3

(2) AB3

(3) A7B24

(4) A7/8B3

Sol. Answer (3) A

8



(Total atoms)

1 8

1

(contribution)

B  face centre 

1 2



6

(Total atoms)

3

(contribution)

Now, one atom is missing from one corner  So, 7



1 8

(contribution at corner )



7 8

8

(total atoms)

–1=7

A

A 7 B3  A 7B24 8

10. In any ionic crystal A has formed cubical close packing and B atoms are present at every tetrahedral voids. If any sample of crystal contain ‘N’ number of B atoms then number of A atoms in that sample is (1) N

(2)

N 2

(3) 2 N

(4)

2N

Sol. Answer (2)

1 1 ⎡ ⎤ A  ccp  total atoms = 4 ⎢8   6   4 ⎥ 2 ⎣ 8 ⎦ B  at tetrahedral voids  double of number of atoms  2 × 4 = 8 given B  8  N. So, number of A atoms is half the number of (B) atoms, i.e.,

8 ⎛N⎞ ⇒ 4 or ⎜ ⎟  4 2 ⎝2⎠

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4

The Solid State

Solution of Assignment

11. A binary solid A+B– has a structure with B– ions constituting the lattice and A+ ions occupying 25% tetrahedral holes. Formula of the solid is (1) A2B

(2) AB

(3) AB2

(4) AB4

Sol. Answer (3) A+B–  Rock salt type  f.c.c.  number of atoms = 4 B–

 constituting the lattice  i.e.,  4

⎡ Tetrahedral void ⇒ 2  no. of atoms ⎤ ⎢ ⎥  2 4  8 ⎣ ⎦

A+  25% of tetrahedral void 

25 8 ⇒ 2 100

A2+B4– or AB2 12. In a crystalline solid anions B are arranged in cubic close packing. Cation A are equally distributed between octahedral and tetrahedral voids. If all the octahedral voids are occupied, the formula for the solid is (1) AB

(2) AB2

(3) A2B

(4) A2B3

Sol. Answer (3)

⎡ ⎤ 1 1 ⎥ B  ccp  4 ⎢ 8   6  (N) ⎢ 8 2 ⎥ ⎣⎢(corners) (face centred) ⎦⎥ ⎧Octahedral ⇒ no. of atoms  N  4 ⎫ A⎨ ⎬ given all octahedral voids are occupied ⎩Tetrahedral ⇒ double of no. of atoms(2N) ⇒ 8 ⎭

and A - equally distributed between both

⎧4(octahedral) ⎫ voids so, A ⎨ ⎬ equal no. ⇒ total = 8 ⎩4(tetrahedral)⎭ (according to octahedral) A8B4  A2B 13. In a cubic close packed structure of mixed oxides, the lattice is made up of oxide ions, one-eighth of tetrahedral voids are occupied by divalent (X2+) ions, while one-half of the octahedral voids are occupied by trivalent ions (Y3+), then the formula of the oxide is (1) XY2O4

(2) X2YO4

(3) X4Y5O10

(4) X5Y4O10

Sol. Answer (1)

1 1⎤ ⎡ O2  ccp ⇒ 4 ⎢8   6  ⎥ 2⎦ ⎣ 8 X+2 

1 8

× 8(tetrahedral void)  1

(given)

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Solution of Assignment

Y+3 

1 2

The Solid State

5

× 4 (octahedral void)  2

(given)

So, formula X1Y2O4 14. Titanium crystallizes in a face centred cubic lattice. It reacts with carbon or hydrogen interstitially by allowing atoms of these elements to occupy holes in the host lattice. Hydrogen occupies tetrahedral holes but carbon occupies octahedral holes the formula of titanium carbide and hydride are (1) TiC2, TiH4

(2) TiC, TiH2

(3) Ti3C, TiH2

(4) TiC2, TiH

Sol. Answer (2)

1 1⎤ ⎡ Ti  ccp  fcc  4 ⎢8   6  ⎥ 8 2 ⎣ ⎦ given  carbon  at octahedral holes = equal to no. of atoms  4 So, Ti4C4  TiC given  hydrogen  at tetrahedral holes

 double of number of atoms 2×4=8

Ti4H8  TiH2 15. The site labelled as ‘X’ in fcc arrangement is

X

(1) Face with

1 contribution 4

(3) Corner with

1 contribution 4

(2) Edge with

1 contribution 4

(4) Tetrahedral void with

1 contribution 8

Sol. Answer (2)

⎛ 1⎞ In given figure (X) is present at centre of edge which contributed ⎜ ⎟ ⎝4⎠ 16. A unit cell is obtained by closed packing layers of atoms in ABAB....... pattern. The total number of tetrahedral and octahedral voids in the unit cell are respectively (1) 6, 12

(2) 8, 4

(3) 4, 8

(4) 12, 6

Sol. Answer (4) In ABAB....... i.e., hexagonal close packing, number of atoms = 6. So, tetrahedral voids = 6 × 2 = 12 octahedral voids = 6 1 of the tetrahedral voids and cations 6 B occupy one-third of the octahedral voids. The probable formula of the compound is

17. In certain solid, the oxide ions are arranged in ccp. Cations A occupy

(1) ABO3

(2) AB2O3

(3) A2BO3

(4) A2B2O3

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6

The Solid State

Solution of Assignment

Sol. Answer (1) O–2  ccp  4 (no. of atoms) 1 8  8  6 (tetrahedral void) 6

A

given

1 4  4  3 (octahedral) 3

B

given

A 8 B 4 O 4 or ABO3 6

3

18. A solid has a structure in which A atoms are located at the cube corners of the unit cell, B atoms are located at the cube edges of unit cell and the C atoms at the body centre. Formula of the compound (2) C2AB3

(1) CAB3

(3) CA3B

(4) C2A3B

Sol. Answer (1) A  corners  8  B  edges = 12 

1 1 8

1 3 4

 A1  B3

C  body centre  1 × 1 = 1 . C1 AB3C 19. If ‘a’ is the length of unit cell, then which one is correct relationship? (1) For simple cubic lattice, Radius of metal atom =

a 2

3a 4

(2) For bcc lattice, Radius of metal atom = a

(3) For fcc lattice, Radius of metal atom =

2 2

(4) All of these Sol. Answer (4)

for S.C.C.,

a r 2

r

r 

2r = a r = a/2

a

3a = Body diagonal

r for B.C.C., r 

3 a 4

2r



4r = 3a r

2a = face diagonal

r for F.C.C., r 

a 2 2

2r



r = 3a 4

4r = 2a r

a 2 r= 4 = 2 2

So, all three options are correct. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

The Solid State

7

20. For face centered cubic structure edge length ‘a’ can be related with radius ‘r’ as (1) a  r  2

(3) a  2 2 r

(2) a = r

(4) a 

4 3

r

Sol. Answer (3)

2a  4r

For f.c.c. a

4r 2



2

⇒

2

4 2r  2 2r 2

21. A crystalline solid AB adopts sodium chloride type structure with edge length of the unit cell as 745 pm and formula mass of 74.5 a.m.u. The density of the crystalline compound is (1) 2.16 g cm–3

(2) 0.99 g cm–3

(3) 1.88 g cm–3

(4) 1.197 g cm–3

Sol. Answer (4) (density) ⇒

⇒

ZM

for NaCl  f.c.c. (Z = 4)

NA  a 3

4  74.5 6.022  10

23

 (745  10 –10 )3

M = 74.5 amu

 1.202 g/cm3 a = 745 pm = 745 × 10–12 m = 745 × 10–10 cm 22. If radius of an octahedral void is r and atomic radius of atoms assuming cubical close packing is R. Then the relation between r and R is (1) r = 2R

(2) r = 1.414 R

(3) r = 0.414 R

(4) r 

R 2

Sol. Answer (3) For octahedral void r = 0.414 R 23. Polonium adopts cubic structure with edge length of cube being 0.336 nm. The distance between the polonium atoms which lie at the corners along the body diagonal is (1) 0.336 nm

(2) 0.291 nm

(3) 0.582 nm

(4) 0.481 nm

Sol. Answer (3)

3a

3a  3  0.336 nm = 0.581 nm

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8

The Solid State

Solution of Assignment

24. CsCl crystallises in a cubic cell that has a Cl– at each corner and Cs+ at the centre of the unit cell. If radius of Cs+ is 1.69 Å and rCl– =1.81 Å, what is the edge length of unit cell? (1) 3.50 Å

(2) 4.04 Å

(3) 2.02 Å

(4) 1.01 Å

Sol. Answer (2)

Cl

–

Distance between Cl–(corner) and Cs+ (centre) is half of body diagonal

–

r

+

+

r

3a  r  r– 2

3a

Cs

3 a  1.69  1.81 2 a  4.04 Å

25. Ice crystallises in a hexagonal lattice having the volume of unit cell as 132 × 10–24 cm3. If density is 0.92 g cm–3 at a given temperature, then number of H2O molecules per unit cell is (1) 1

(2) 2

(3) 3

(4) 4

Sol. Answer (4) ZM



MH2O  18

NA  a 3

0.92 

Z

Z  18

a3 = volume of unit cell  132 × 10–24 cm3

6.022  1023  132  10 –24

0.92  6.022  1023  132  10 –24 18

 4.0 26. For tetrahedral co-ordination, the radius ratio (r+/r–) should be (1) 0.414 – 0.732

(2) 0.732 – 1.0

(3) 0.156 – 0.225

(4) 0.225 – 0.414

Sol. Answer (4)

For tetrahedral void the radius ratio

r r–

 0.225  0.414

27. The radius of the Na+ is 95 pm and that of Cl– ion is 181 pm. The co-ordination number of Na+ will be (1) 4

(2) 6

(3) 8

(4) Unpredictable

Sol. Answer (2)

r r

–



95  0.52 (this value is present in between normal range so, C.N. remain 6) 181

generaly the C.N. of NaCl is 6 when

r r–

 0.414  0.732

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Solution of Assignment

The Solid State

9

28. Lithium metal has a body centred cubic structure. Its density is 0.53 g cm–3 and its molar mass is 6.94 g mol–1. Calculate the edge length of a unit cell of Lithium metal (1) 153.6 pm

(2) 351.6 pm

(3) 527.4 pm

(4) 263.7 pm

Sol. Answer (2) 

ZM

a3 

NA  a 3

a3 

 z = 2 (for B.C.C.)

ZM NA  

2  6.94 6.022  1023  0.53

= 4.34 × 10–23 a = (4.34 × 10–23)1/3 ~ 351.6 pm

29. What is the volume of a face centred cubic unit cell, when its density is 2.0 g cm–3 and the molar mass of the substance is 60.23 g mol–1? (1) 4 × 10–22 cm3

(2) 2 × 10–22 cm3

(3) 44 × 10–22 cm3

(4) 22 × 10–22 cm3

Sol. Answer (2) 

ZM

a3 = volume of unit cell = V

NA  a 3



ZM NA  V

V

ZM 4  60.23  NA   6.022  1023  2

Z = 4  for f.c.c.

 20 × 10–23  2 × 10–22 g/cm3 30. The number of octahedral sites in a cubical close pack array of N spheres is (1) N/2

(2) 2 N

(3) 4 N

(4) N

Sol. Answer (4) Number of octahedral sites  number of spheres = N 31. For a solid with the following structure, the coordination number of the point B is

(A)

(1) 3

(2) 4

(B)

(3) 5

(4) 6

Sol. Answer (4) (B) is located at edge centre which shown C.N. = 6 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

10

The Solid State

Solution of Assignment

32. The empty space between the shaded balls and hollow balls as shown in the diagram is called

(1) Hexagonal void

(2) Octahedral void

(3) Tetrahedral void

(4) Double triangular void

Sol. Answer (2) The empty space in figure is octahedral void because it is surrounded by six spheres. 33. A mineral having formula AB2 crystallises in the cubic close packed lattice, with the A atoms occupying the lattice points. Hence coordination number of A and B atoms are (1) 4, 8

(2) 4, 4

(3) 8, 8

(4) 8, 4

Sol. Answer (4) AB2  CaF2 type of crystal which has C.N. = 8 : 4. 34. KF has NaCl type of structure. The edge length of its unit cell has been found to be 537.6 pm. The distance between K+ F– in KF is (1) 26.88 pm

(2) 268.8 pm

(3) 2688 pm

(4) Unpredictable

Sol. Answer (2)  – For NaCl (fcc) and KF (r  r ) 



a  interionic distance. 2

537.6 2

 268.8 pm 35. Which of the following features is false regarding the structure of CsCl? (1) It has bcc arrangements

(2) For each ion coordination number is 8

(3) For each ion coordination number is 6

(4) The radius ratio (r+/r–) is 0.93

Sol. Answer (3) It has B.C.C. arrangement For CsCl,

C.N. = 8 r r–

 0.93

So, third option is incorrect because C.N. of CsCl is 8 not 6. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

The Solid State

11

36. Which type of solids will have the highest melting point? (1) Ionic crystals

(2) Network covalent solid

(3) Molecular solids

(4) Metallic crystals

Sol. Answer (2) Network covalent solids have highest melting point due to network structure in which bonds are tightly held like SiO2.

Si

O

O Si

Si

O

O O

Si O

Si

O

Si

Layered structure so, it require very high temperature to melt it. 37. The mass of unit cell of Na2O is (1) Twice the formula mass of Na2O

(2) Four times the formula mass of Na2O

(3) Six times the formula mass of Na2O

(4) Thrice the formula mass of Na2O

Sol. Answer (2) Na2O  Has 4 atoms per unit cell So, the mass of unit cell is  4 × formula mass of Na2O 38. In normal spinel structure there is a closed packed array of O2– ions. The trivalent cations are present in (1) 75% of octahedral voids

(2) 50% of octahedral voids

(3) 12.5% of tetrahedral voids

(4) 25% of octahedral voids

Sol. Answer (2) In normal spinel structure

T.V +2 1 1A  occupy of T.V. 8 (12.5%)

+2

+3 2

–2 4

A B O

1 ⎫  1⎪ ⎪ 8 ⎬4 1  6  3⎪ ⎪⎭ 2

8

1 +3 2B i.e., O.V. will be occupied 2 (50%) T.V. = tetrahedral void O.V. = octahedral void 39. The C–C and Si–C interatomic distances are 154 pm and 188 pm. The atomic radius of Si is (1) 77 pm

(2) 94 pm

(3) 114 pm

(4) 111 pm

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12

The Solid State

Solution of Assignment

Sol. Answer (4) Diameter = 154 pm (internuclear distance) 154 ⇒ 77 pm so, radius of one carbon atom  2

Diameter (internuclear distance)  188 so, radius of Si + radius of carbon = 188

C

C

154 pm Si

C

Radius of Si + 77 = 188 188 pm

rSi = 188 – 77  111 pm

40. What is the coordination number of Rb+ in RbBr unit cell if ionic radii of Rb+ and Br– ions being 148 and 195 respectively? (1) 6

(2) 4

(3) 8

(4) 12

Sol. Answer (3) r

Rb

r

Br –



148  0.75 195

This ratio (0.75) is more than 0.732 so, it is cubic structure which has C.N. = 8 41. A crystalline solid AB has NaCl type structure with radius of B– ion is 250 pm. Which of the following cation can be made to slip into tetrahedral site of crystals of A+B– ? (1) P+ (radius = 180 pm)

(2) Q+ = (radius = 56 pm)

(3) R+ = (radius = 200 pm)

(4) S+ = (radius = 150 pm)

Sol. Answer (2) For NaCl type

r r

 0.414  0.732

r  0.414 250 r+ = 0.414 × 250 r+ = 103.5 This is radius of void so, the cation which has equal to this radius or smaller to this radius can easily fit into it that is option (2) Q+ = 56 pm which is smaller than radius of void thats why it can fit into it. 42. Number of formula units in unit cell of MgO (rock salt), ZnS (zinc blende) and Pt (fcc) respectively (1) 4, 3, 2

(2) 4, 3, 4

(3) 4, 4, 4

(4) 4, 3, 1

Sol. Answer (3) MgO = 4 ZnS = 4 Pt (f.c.c.) = 4 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

The Solid State

13

43. An element crystallises in a ‘bcc’ lattice. Nearest neighbours and next nearest neighbours of the elements are respectively (1) 8, 8

(2) 8, 6

(3) 6, 8

(4) 6, 6

Sol. Answer (2)

[4 atoms above and 4 atoms below]

In B.C.C.

(3)

Nearest neighbours = 8

(2) (1)

(4) (5)

(6)

Next nearest neighbours = 6 (6 edge atoms at corner)

44. The total number of elements of symmetry in a cubic crystal is (1) 9

(2) 23

(3) 10

(4) 14

Sol. Answer (2) Total number of elements of symmetry = 23 Rectanglular plane of symmetry = 3 + Diagonal plane of symmetry = 6 + centre of symmetry = 1 + axis of symmetry = 13 [3 fold = 4, 2 - fold = 6, 4 fold = 3]  axis of symmetry. 45. A crystal may have one or more planes of symmetry as well as one or more than one axis of symmetry but it has (1) Two centres of symmetry

(2) Only one centre of symmetry

(3) No centre of symmetry

(4) Three centres of symmetry

Sol. Answer (2) A crystal may have more than plane of symmetry and axis of symmetry but centre of symmetry is only one. 46. Which of the following statement is correct? (1) On increasing temperature the coordination number of solid remains unchanged (2) On increasing pressure the coordination number of solid increases (3) On increasing temperature the coordination number of solid increases (4) On increasing pressure the coordination number of solid decreases Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

14

The Solid State

Solution of Assignment

Sol. Answer (2) On increasing pressure the C.N. of solid increases because on increasing pressure atoms/particles come closer to each other i.e., C.N. increases [closeness increases] and [K.E. decreases] and on increasing temperature K.E. of particles increases by which particles go far away i.e., C.N. decreases [closeness decreases] [C.N.  closeness of atoms i.e., coordination number] 47. Pyroelectric crystals produce feeble electric current (1) On deformation

(2) On dissolving in a solvent

(3) On heating

(4) On sublimation

Sol. Answer (3) Pyroelectric crystal produce feeble electric current on heating. 48. Zinc oxide on heating changes to yellow. This is because (1) Zinc oxide is a stoichiometric compound

(2) Zinc oxide is a covalent compound

(3) Zinc oxide shows metal excessive defect

(4) It shows metal deficiency defect

Sol. Answer (3)  ZnO   Zn2  2e – 

1 O2 2

It causes excess of Zn in the lattice causing metal excess (defect)  it changes its colour 49. F-centres in an ionic crystals are (1) Lattice sites containing electrons

(2) Interstitial sites containing electrons

(3) Lattice sites that are vacant

(4) Interstitial sites containing cations

Sol. Answer (1) In some defect, negative ions may be missing from their lattice sites leaving holes in which electrons remain trapped to maintain electrical neutrality. The holes with entrapped electrons are called F-centres. They impart colour to the crystal lattice e.g., heating crystals of NaCl in vapours of Na 50. When an element of group 14 is doped with an element of group 15 (1) p-type of semi-conductors are formed

(2) n-type of semi-conductors are formed

(3) Zeolites are formed

(4) Electrolytes are formed

Sol. Answer (2) Group-14 element e.g.  Si  4-valence electrons.

Si

Si

Si

Doped with group-15 element like (P) (phosphorus)

Si

Si

P

–

4e s of (P) can form bond – – with 4e s of Si but one e remain free, so, it n-type of semi conductor due to negative – nature of e .

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Solution of Assignment

The Solid State

15

51. Antiferromagnetic property is given as (1)

(2)

(3)

(4)

Sol. Answer (3) If magnetic moments are aligned in compensatory way so that all cancel out each other and net magnetic .

moment becomes zero, these are antiferromagnetic substances

52. Substances which are magnetic but having less magnetic moment than theoretically calculated value are called (1) Ferromagnetic

(2) Ferrimagnetic

(3) Antiferromagnetic

(4) Diamagnetic

Sol. Answer (2) When magnetic moments of domains are aligned in parallel and antiparallel directions in unequal number, it  less magnetic moment than

results in some net magnetic moment. Ferrimagnetic theoritically calculated value. 53. In antiferromagnetism (1) Alignments of magnetic moments is additive

(2) Alignments of magnetic moments in one direction is compensated by alignments in the opposite directions (3) Alignments of magnetic moments does not take place (4) Alignments of magnetic moments varies with the nature of the material Sol. Answer (2)  equal and opposite. So net magnetic moment = 0.

In antiferromagnetism 

54. Which is true about Piezoelectric crystals? (1) They produce an electric current on heating (2) They produce an electric current when a mechanical stress is applied (3) They are insulators (4) They are magnetic in nature Sol. Answer (2) Piezoelectric crystals produce an electric current when a mechanical stress is applied. 55. When a crystal having rock salt type geometry is heated in the presence of it’s metal vapour then defect in it will be (1) Stoichiometric defect

(2) Metal excess defect

(3) Anion excess defect (4) Frenkel defect

Sol. Answer (2) Metal excess defect  heated in the presence of its metal vapour e.g., heating of NaCl in Na vapour. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

16

The Solid State

Solution of Assignment

56. The mass percentage of Fe3+ ion present in Fe0.93 O1.00 is (1) 15%

(2) 5.5%

(3) 10.0%

(4) 11.5%

Sol. Answer (4)

Fe0.93O1.00 according to charge balance

Let

+3

Fe (0.93 – x)

2x + 3[0.93 – x] = 2 x = 0.79

+2

Fe x

Total mass of Fe0.93O = 56 × 0.93 + 16 = 68.08 amu

+2

Fe = x = 0.79 +3 Fe = 0.14

mass of Fe+3 = 0.14 × 56 = 7.84 amu mass percentage of Fe 3  7.84  100  11.5% 68.08

57. If 1 mole of NaCl is doped with 10–3 mole of SrCl2. What is the number of cationic vacancies per moe of NaCl? (1) 10–3 mole–1

(2) 6.02 × 1018 mole–1

(3) 1050 mole–1

(4) 6.02 × 1020 mole–1

Sol. Answer (4) 1 mole of SrCl2  creates one mole cation vacancy. given 10–3 mole of SrCl2  creates 10–3 × NA number of vacancy. = 6.02 × 1020 mol–1 58. Which of the following is incorrect statement about the Bragg’s equation n = 2d sin ? (1) n, represents order of reflection

(2) , represents wavelength of X-rays used

(3) , represents angle of glance

(4) d, represents distance between two parallel planes

Sol. Answer (3) n = 2dsin  = incident angle not angle of glance.

SECTION - B Objective Type Questions 1.

In crystalline solids, which of the following element of symmetry is not present? (1) Axis of symmetry

(2) Angle of symmetry

(3) Centre of symmetry (4) Plane of symmetry

Sol. Answer (2) In crystalline solid angle of symmetry is not present. 2.

Amorphous solids have (1) Orderly arrangement of atoms

(2) Repeating unit of unit cell

(3) Long range of melting point

(4) Anisotropy

Sol. Answer (3) Amorphous solids have

not ordered arrangement. not repeating units long range of melting point Isotropy

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Solution of Assignment

3.

The Solid State

17

The type of crystal system shown is

b a

(1) Cubic

(2) Orthorhombic

a

(3) Monoclinic

(4) Tetragonal

Sol. Answer (4) For tetragonal a = b  c Two sides are equal and one side is unequal. 4.

In a unit cell, atoms A, B, C and D are present at corners, face-centres, body-centre and edge-centre respectively in a cubic unit cell. The total number of atoms present per unit cell is (1) 4

(2) 8

(3) 15

(4) 27

Sol. Answer (2) A  at corners = 8 B  at face centres = 6 C  body centre = 1 D  edge centre = 12 Total atoms in a cube = 8 + 6 + 1 + 12 = 27 A = 8

1 1 8

B = 6

1 3 2

C=1×1=1

1 3 4 Total = 8 atoms per unit cell D = 12 

5.

In a unit cell, atoms A, B, C and D are present at half of total corners, all face-centres, body-centre and one third of all edge-centres respectively. Then formula of unit cell is (1) AB3CD3

(2) ABCD

(3) AB6C2D4

(4) AB6C2D2

Sol. Answer (4) Corner A 

1 1 4  8 2

Face centre B 

1 6  3 2

Total C = 1 Total D =

1 4 1 4

formula A 1 B3 C1D1 2

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18 6.

The Solid State

Solution of Assignment

In a unit cell, atoms A, B, C and D are present at corners, face-centres, body-centre and edge-centres respectively. If atoms touching one of the plane passing through two diagonally opposite edges are removed, then formula of compound is (2) ABD2

(1) ABCD2

(3) AB2D2

(4) AB4D5

Sol. Answer (4) Given atoms of one diagonal planes are to be removed.  4 atoms from corner, 2 atom from face centre and 1 atom from body centre will be removed. 1 1  [4 atoms removed from corner] 8 2

Total A  4  Total B 

1  4  2 [2 atoms are removed from face centre] 2

Total C = 0 [1 atom removed from body centre] Total D 

1 5  10  [2 atom removed from edge] 4 2

A 1 B2C0D 5  A1B 4D5 2

7.

2

In a CsCl structure, if edge length is x, then distance between one Cs atom and one Cl atom is

a 3 2

(1)

(2)

a 3 4

(3) a 2

(4)

a 2

Sol. Answer (1)

Cl

–

3a 2

+

Cs

3a

 [a = x]

3a 3x  2 2 8.

The correct statement about, CCP structure is (1) Packing fraction = 26%

(2) Coordination number = 6

(3) Unit cell is face centred cubic

(4) AB–AB type of packing

Sol. Answer (3)

packing fraction = 74% ccp C.N. = 12 Unit cell is fcc ABCABC......... Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

9.

The Solid State

19

In a NaCl structure, if positions of Na atoms and Cl atoms are interchanged, then in the new unit cell (1) Na atom is present at body centre

(2) Cl atom is present at face centre

(3) Na atom is present in tetrahedral voids

(4) Cl atom is present in octahedral voids

Sol. Answer (4) In NaCl structure Na+ present at edge centres [octahedral voids] and as well as body centre. Cl– present at corners and centre of each face. Given if atom are interchanged then Na+ comes at corners and centre of each face. Cl– comes at edge centres (octahedral voids) and at body centre. 10. If radius of a metal atom (A) is 5 pm and radius of an electronegative atom (B) is 20 pm, then in the unit cell (1) A in octahedral voids, B in FCC unit

(2) A in FCC unit, B in tetrahedral void

(3) A in BCC unit, B in cubic void

(4) A in tetrahedral void, B in FCC unit

Sol. Answer (4) r

5 1  ⇒  0.25 4 r – 20 This ratio comes in between 0.225 – 0.414 i.e., tetrahedral void. So, A in tetrahedral void due to cation (small size).

So, B in f.c.c. 11. A metal can be crystallized in both BCC and FCC unit cells whose edge lengths are 2 pm and 4 pm respectively. Then ratio of densities of FCC and BCC unit cells is (1)

1 4

(2) 4

(3)

1 16

(4) 16

Sol. Answer (1) 

ZM NA  a 3

3 F.C.C. ZF.C.C. aB.C.C.   3 B.C.C. ZB.C.C aF.C.C



(M, NA = constant)

[given, aB.C.C = 2 pm; aF.C.C. = 4 pm]

4 (2)3 48 1  3  ⇒ 2 (4) 2  64 4

12. In a unit cell containing X2+, Y3+ and Z2– where X2+ occupies 1/8th of tetrahedral voids, Y3+ occupies 1/2 of octahedral voids and Z2– forms ccp structure. Then formula of compound is (1) X2Y4Z

(2) XY2Z4

(3) XY3Z4

(4) X4YZ2

Sol. Answer (2) ⎡ ⎤ 1 1 ⎥ ccp  number atoms per unit cell = 4  ⎢ 8   6  ⎢ 8 2 ⎥ (Z2 ) ⎣⎢(corners) (face centre) ⎦⎥

Total tetrahedral void = 8 Total octahedral void = 4 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

20

The Solid State

Given, X 2  Y 3 

Solution of Assignment

1 1 of T.V.   8  1 8 8

1 1 of O.V. =  4  2 2 2

So, formula Z42 X12 Y23  XY2 Z4 13. On rising temperature and decreasing pressure in CsCl solid (1) C.N. of metal ion increases from 6 to 8 (2) Number of formula unit per unit cell (Z) changes from one to four (3) Density of unit cell is increased (4)

r (radius ratio) is increased r

Sol. Answer (2) T , P   NaCl

CsCl

C.N.  8 :8

C.N.  6 : 6

(Z  1)no.of atom

(Z  4) due to F.C.C.

due to simple cubic

On increasing temperature and decreasing pressure closeness of atoms in CsCl crystal decreases because ⎧ r ⎫ ⎪ (radius ratio)  decreases ⎪ r K.E. increases. So, that ⎨  ⎬ due to closeness  ⎪density of unit cell decreases ⎪ ⎩ ⎭

14. In a ccp type structure, if half of the face-centred atoms are removed, then percentage void in unit cell is approximately (1) 54%

(2) 46.25%

(3) 63%

(4) 37%

Sol. Answer (1) In ccp,  74% space is occupied by 4 atoms (when all corner and face centre atoms are located) Given Half of face centred atoms are removed. So, total face centred atoms = 6 half removed it. becomes = 3 Now contribution for corners remain same i.e., 8  for face centred = (3)  Total 

1 3  2 2

1 =1 8

3 5 1 2 2

For, 4 - atoms  74% (space occupied) 1 - atom  for

74% 4

5 74% 5   46.25% atom  4 2 2

i.e., % void = 100 – occupied space = 100 – 46.25 = 54% Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

The Solid State

21

15. In a BCC unit cell, if half of the atoms per unit cell are removed, then percentage void is (1) 68%

(2) 32%

(3) 34%

(4) 66%

Sol. Answer (4) In B.C.C.  68% space is occupied by 2 atoms. If half of atoms per unit cell is removed then it becomes = 2 – 1 = 1. So, 2 atom  68% 1 atom 

68  34% (occupied space) 2

Now, % void = 100 – occupied space = 100 – 34 = 66% 16. Number of atoms per unit cell, if atoms are present at the corner of unit cell and 2 atoms at each body diagonal (1) 9

(2) 10

(3) 6

(4) 4

Sol. Answer (1) Given = atoms are present at corners and 2 atoms at each body diagonal. In normal total 4 – body diagonal are present in a simple cube. So, Total no. of atoms on body diagonal = 4 × 2 = 8 for corners = 8 

1 1 8

Body diagonal = 4 × 2 = 8 Total = 9 17. Number of unit cells in 10 g NaCl

(1)

1.5  1024 58.5

(2)

2.5  1023 58.5

(3)

5.6  1020 58.5

(4)

5.6  1021 58.5

Sol. Answer (1) In NaCl, 4 atoms per unit cell i.e., 4 atoms in one unit cell or 1 atom in

or



mole 

1 unit cell 4

or NA atoms = (1 mole) 

NaCl  23  35.5  58.5 10 58.5

1  NA unit cell 4

10 1 10 mole   NA  unit cell 58.5 4 58.5 1.5  1024 unit cell 58.5

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22

The Solid State

Solution of Assignment

18. Some of the molecular solids upon heating produces small amount of electricity, hence solid is (1) Piezoelectric

(2) Pyroelectric

(3) Ferrielectric

(4) Ferroelectric

Sol. Answer (2) Pyroelectric are the molecular solids which produces small amount of electricity upon heating. 19. NaCl becomes paramagnetic at high temperature due to (1) Formation of F-centre

(2) Molten state

(3) Change in oxidation state

(4) Conversion of Na+ to Na

Sol. Answer (1) NaCl becomes paramagnetic at high temperature in the presence of Na - vapour due to migration of Cl– ion to the surface from lattice by which vacancy is produced known as F-centre in which e– is trapped from outer Na vapour i.e., some free electrons comes in lattice that’s why it becomes paramagnetic. 20. Second nearest neighbour in CsCl solid (1) 8

(2) 6

(3) 16

(4) 10

Sol. Answer (2) In CsCl  first nearest neighbours is (8) due to four atoms above at corners and four atoms below at corners. Second nearest neighbours is 6 due to six edge corner atoms of second unit cell. (3)

(2) (1)

(4) (6) (5)

+

Cs

21. The ratio of number of rectangular plane and diagonal plane in a cubic unit cell (1) 1 : 2

(2) 3 : 1

(3) 2 : 3

(4) 3 : 4

Sol. Answer (1) Total number of rectangular plane = 3. Number of diagonal plane = 6. ratio = 3 : 6 = 1 : 2 22. In a calcium fluoride structure, the co-ordination number of cation and anion is respectively. (1) 6, 6

(2) 8, 4

(3) 4, 4

(4) 4, 8

Sol. Answer (2) CaF2  AB2 type  C.N. = 8 : 4 because Ca+2 ions form ccp structure and F– ions in all tetrahedral holes. for ccp atoms = 4 Tetrahedral holes = 8 Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

The Solid State

23

23. In which of the following defect density increases? (1) Schottky defect

(2) Frenkel defect

(3) F-centre

(4) Impurity defect

Sol. Answer (4) In impurity defect density increases because some impurities will come from outside which increases mass. So, density increases.

impurity added Normal crystal

Impurity mass increases = density increases

24. Glass is a (1) Micro-crystalline solid (2) Super cooled liquid

(3) Gel

(4) Polymeric mixture

Sol. Answer (2) Glass is a super cooled liquid or amorphous solid because on heating its arrangement of atoms become improper i.e., become slightly liquid nature.

SECTION - C Previous Year Questions 1.

The vacant space in bcc lattice unit cell is (1) 23%

(2) 32%

[Re-AIPMT-2015] (3) 26%

(4) 48%

Sol. Answer (2) 2.

The correct statement regarding defects in crystalline solids is

[Re-AIPMT-2015]

(1) Frenkel defect is a dislocation defect (2) Frenkel defect is found in halides of alkaline metals (3) Schottky defects have no effect on the density of crystalline solids (4) Frenkel defects decrease the density of crystalline solids Sol. Answer (1) 3.

A given metal crystallizes out with a cubic structure having edge length of 361 pm. If there are four metal atoms in one unit cell, what is the radius of one atom? [AIPMT-2015] (1) 108 pm

(2) 40 pm

(3) 127 pm

(4) 80 pm

Sol. Answer (3) 4.

If a is the length of the side of a cube, the distance between the body centered atom and one corner atom in the cube will be [AIPMT-2014] (1)

2 3

a

(2)

4 3

a

(3)

3 a 4

(4)

3 a 2

Sol. Answer (4)

3a  4r (BCC) 2r 

3a 2

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24 5.

The Solid State

Solution of Assignment

A metal has a fcc lattice. The edge length of the unit cell is 404 pm. The density of the metal is 2.72 g cm–3.The molar mass of the metal is (NA Avogadro's constant=6.02×1023 mol–1) [NEET-2013] (1) 30 g mol–1

(2) 27 g mol–1

(3) 20 g mol–1

(4) 40 g mol–1

Sol. Answer (2) 

ZM NA  a

  NA  a3 Z

M

3



a = 404 × 10–12 × 100 cm = 404 × 10–10 cm

2.72  6  1023  (404  10–10 )3 4

= 26.9 ~ 27 g/mol

Z = 4  for f.c.c. 6.

The number of carbon atoms per unit cell of diamond unit cell is (1) 8

(2) 6

(3) 1

[NEET-2013] (4) 4

Sol. Answer (1) Number of carbon atoms per unit cell of diamond unit cell is 8. It has ZnS like structure in which Zn+2 located at half of tetrahedral voids = 4. S–2 located at ccp  8 

1  8

(corners)

6

1 2

4

(face centre)

In diamond all carbon replaces Zn+2 and S–2 so, some (C) at half of T.V. = 4 and some (C) at ccp = 4, total = 8. 7.

A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 408 pm. The diameter of the metal atom is [AIPMT (Prelims)-2012] (1) 144 pm

(2) 204 pm

(3) 288 pm

(4) 408 pm

Sol. Answer (3) For F.C.C. r 

a 2 2

diameter = 2r =

8.



408 2 2

2  408 2 2

pm

 288.5 pm

The number of octahedral void(s) per atom present in a cubic close-packed structure is [AIPMT (Prelims)-2012] (1) 2

(2) 4

(3) 1

(4) 3

Sol. Answer (3) Number of O.V. equal to number of atoms 4 atoms in unit cell  4 O.V. So, for 1 atom  1 O.V. 9.

Structure of a mixed oxide is cubic close packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are occupied by a monovalent metal B. The formula of the oxide is [AIPMT (Mains)-2012] (1) ABO2

(2) A2BO2

(3) A2B3O4

(4) AB2O2

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Solution of Assignment

The Solid State

25

Sol. Answer (4)

⎛ ⎞ 1 1 ⎟  6 Given oxide ions = ccp = 4  ⎜ 8  ⎜⎜ 8 2 ⎟⎟ ⎝ corner face centres ⎠ 1 1 th of T.V. by A+2   8 (Total T.V. = 8) 4 4 2 O.V. by B+  4 × 1  4 A2+2

B4+

O4–2

(Total O.V. = 4)



2:4:41:2:2 AB2O2 10. A solid compound XY has NaCl structure. If the radius of the cation is 100 pm, the radius of the anion (Y–) will be [AIPMT (Mains)-2011] (1) 241.5 pm

(2) 165.7 pm

(3) 275.1 pm

(4) 322.5 pm

Sol. Answer (1) XY has NaCl structure 100 r– r– 

r r–

⇒ 0.414  0.732

 0.414 100  241.5 pm 0.414

11. AB crystallizes in a body centred cubic lattice with edge length 'a' equal to 387 pm. The distance between two oppositively charged ions in the lattice is [AIPMT (Prelims)-2010] (1) 335 pm

(2) 250 pm

(3) 200 pm

(4) 300 pm

Sol. Answer (1) B.C.C., d = 2r (r+ + r–) 

3a 3  387   335 pm 2 2

12. Among the following which one has the highest cation to anion size ratio? (1) Cs

(2) CsF

(3) LiF

[AIPMT (Mains)-2010] (4) NaF

Sol. Answer (2)

r



Cs

r F Size of Cs+ is maximum and size F– is minimum so ratio is highest. 13. Lithium metal crystallises in a body centred cubic crystal. If the length of the side of the unit cell of lithium is 351 pm, the atomic radius of the lithium will nearly be [AIPMT (Prelims)-2009] (1) 152 pm

(2) 75 pm

(3) 300 pm

(4) 240 pm

Sol. Answer (1) a = 351 pm B.C.C. , r 

3a 3  351   151.9 pm 4 4

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26

The Solid State

Solution of Assignment

14. Copper crystallises in a face-centred cubic lattice with a unit cell length of 361 pm. What is the radius of copper atom in pm ? [AIPMT (Prelims)-2009] (1) 157

(2) 181

(3) 108

(4) 128

Sol. Answer (4) F.C.C. r  r

a 2 2

361  127.65  128 2  1.414

15. If 'a' stands for the edge length of the cubic systems : simple cubic, body centred cubic and face centred cubic, then the ratio of radii of the spheres in these systems will be respectively [AIPMT (Prelims)-2008] (1) 1a : 3a : 2a

(2)

1 3 1 a: a: a 2 4 2 2

(3)

1 1 a a : 3a : 2 2

(4)

1 3 2 a: a: a 2 2 2

Sol. Answer (2) S.C.C  r 

a 2

3a 4

B.C.C.  r  F.C.C.  r 

a 2 2

a 3a a : : 2 4 2 2

16. Percentage of free space in a body centred cubic unit cell is (1) 28%

(2) 30%

(3) 32%

[AIPMT (Prelims)-2008] (4) 34%

Sol. Answer (3) Occupied space in B.C.C. = 68% free space = 100 – 68  32% 17. With which one of the following elements silicon should be doped so as to give p-type of semiconductor ? [AIPMT (Prelims)-2008] (1) Boron

(2) Germanium

(3) Arsenic

(4) Selenium

Sol. Answer (1) Si = 4 valence electrons (group-14 element)

Si

Si

Si

doped with three valence electrons (B)  group-13 element

Si

Si

B

vacancy/hole is created which has positive charge so known as p-type

Three electrons of (B) can form bond. So, that one site of (B) is vacant Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

The Solid State

18. Which of the following statements is not correct ?

27

[AIPMT (Prelims)-2008]

(1) The number of Bravais lattices in which a crystal can be categorized is 14 (2) The fraction of the total volume occupied by the atoms in a primitive cell is 0.52 (3) Molecular solids are generally volatile (4) The number of carbon atoms in a unit cell of diamond is 4 Sol. Answer (4) Option (4) is incorrect because number of atoms in an unit cell of diamond is 8 not 4. 19. The fraction of total volume occupied by the atoms present in a simple cube is (1)

 4

(2)

 6

(3)



[AIPMT (Prelims)-2007] (4)

3 2

 4 2

Sol. Answer (2) Packing fraction 

volume of total lattice points total volume of unit cell

 a = 2r (for S.C.C.) 

1

4 3 4 3 r r  3 3   3 3 6 a (2r)

20. If NaCl is doped with 10–4 mol % SrCl2, the concentration of cation vacancies will be (NA = 6.02×1023 mol–1) [AIPMT (Prelims)-2007] (1) 6.02×1014 mol–1

(2) 6.02×1015 mol–1

(3) 6.02×1016 mol–1

(4) 6.02×1017 mol–1

Sol. Answer (4) One mole SrCl2 causes one mole cation vacancy. So, 10–4% mole SrCl2

10 –4 mole cation vacancy. 100

Number of cation vacancy 

10 –4  NA  6.02  1023 6 = 6.02 × 1017 mol–1 100

21. CsBr crystallises in a body centred cubic lattice. The unit cell length is 436.6 pm. Given that the atomic mass of Cs = 133 and that of Br = 80 amu and Avagadro number being 6.02 × 1023 mol–1, the density of CsBr is [AIPMT (Prelims)-2006] 3

(1) 42.5 g/cm

3

3

(2) 0.425 g/cm

(3) 8.25 g/cm

(4) 4.25 g/cm3

Sol. Answer (4) a = 436.6 pm



Cs = 133, Br = 80 amu



ZM NA  a 3

213  1 6  1023  (436.6  10 –10 )3

Total mass = 133 + 80 = 213 Z = 1 for CsBr (S.C.C.)



213 ⇒ 4.26 g/cm3 8.32  6

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28

The Solid State

Solution of Assignment

22. The appearance of colour in solid alkali metal halides is generally due to (1) F-centres

(2) Schottky defect

(3) Frenkel defect

[AIPMT (Prelims)-2006] (4) Interstitial positions

Sol. Answer (1) Due to F-centres 23. In a face-centered cubic lattice, a unit cell is shared equally by how many unit cells? (1) 8

(2) 4

(3) 2

[AIPMT (Prelims)-2005]

(4) 6

Sol. Answer (4) 24. Ionic solids, with Schottky defects, contain in their structure (1) Cation vacancies only

(2) Cation vacancies and interstitial cations

(3) Equal number of cation and anion vacancies

(4) Anion vacancies and interstitial anions

Sol. Answer (3) In Schottky defect equal number of cation and anion vacancies are present. +

Na Cl

–

–

Cl

+

Na

+

Na Cl

–

–

Cl Na

+

one cation is removed

Na

+

Cl removed

Cl

–

–

+

Na

+

Na

Cl

Cl

–

–

Cation vacancy

removed To maintain neutrality one anion should also be removed +

Na

Cl

– +

Na

+

Na

–

Cl

Cl

–

cation vacancy

anion vacancy 25. The number of atoms in 100 g of an FCC crystal with density d = 10 g/cm3 and cell edge equal to 100 pm, is equal to (1) 2 × 1025

(2) 1 × 1025

(3) 4 × 1025

(4) 3 × 1025

Sol. Answer (3) 

ZM NA  a 3

Z = 4 for f.c.c,. a = 100 × 10–12 × 102 = 100 × 10–10 cm, w = 100 g.

Molar mass 

  NA  a3 10  6  1023  (100  10–10 )3  = 1.5 g Z 4

Number of atoms  moles × NA 

w 100  NA ⇒  6  1023  4  1025 M 1.5

26. An element (atomic mass = 100 gm/mol) having BCC structure has unit cell edge 400 pm. The density of element is (1) 7.289 gm/cm3

(2) 2.144 gm/cm3

(3) 10.376 gm/cm3

(4) 5.188 gm/cm3

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Solution of Assignment

The Solid State

29

Sol. Answer (4) 

ZM

⇒

⇒

⇒

z = 2  (B.C.C.)

NA  a 3

2  100 6  10

23

a = 400 × 10–12 m

 (4  108 )3

2  100 6  10

23

= 400 × 10–12 × 100 cm

 64  1024

1  102  10 3  64

= 4 × 10–8 cm

 0.0052 × 103  5.2 g/cm3 27. If we mix a pentavalent impurity in a crystal lattice of germanium, what type of semiconductor formation will occur? (1) n-type semiconductor (2) p-type semiconductor

(3) Both (1) & (2)

(4) None of these

Sol. Answer (1) Ge  14th group element  4 valence electrons.

pentavalent impurity Ge

Ge

Ge

e.g., phosphorus – five valence (e s)

Ge

Ge

P

free electron negative charged, so, n-type

28. The intermetallic compound LiAg crystallizes in cubic lattice in which both lithium and silver have coordination number of eight. The crystal class is (1) Face-centred cube

(2) Simple cube

(3) Body-centred cube

(4) None of these

Sol. Answer (2) In simple cubic C.N. = 8 Li+  located in void of body centre.

Li+

Ag  at corners due to large size. 29. Schottky defect in crystals is observed when (1) Density of the crystal is increased (2) Unequal number of cations and anions are missing from the lattice (3) An ion leaves its normal site and occupies an interstitial site (4) Equal number of cations and anions are missing from the lattice Sol. Answer (4) Equal number of cations and equal number of anions missing from lattice. 30. The edge length of rock salt type unit cell is 508 pm. If the radius of the cation is 110 pm, the radius of the anion assuming NaCl type structure is (1) 144 pm

(2) 398 pm

(3) 288 pm

(4) 618 pm

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30

The Solid State

Solution of Assignment

Sol. Answer (1) (r   r – ) 

a = interionic distance for f.c.c. 2

508 2 – r = 254 – 110  144 pm

110 + r– =

31. The second order Bragg diffraction of X-rays with  = 1.00 Å from a set of parallel planes in a metal occurs at an angle 60°. The distance between the scattering planes in the crystal is (1) 2.00 Å

(2) 1.00 Å

(3) 0.575 Å

(4) 1.15 Å

Sol. Answer (4) n = 2dsin

n  2  second order

2 × 1 = 2 × dsin60°

  60º,   1 Å (given)

dsin60° = 1

3 2



sin60 

d

3 2  1, d   1.15 Å 2 3

32. In crystals of which one of the following ionic compounds would you expect maximum distance between centres of cations and anions? (1) CsI

(2) CsF

(3) LiF

(4) LiI

Sol. Answer (1) Cs+  largest cation and I– largest anion among their group. So, distance between them is also very high. 33. In cube of any crystal A-atom placed at every corners and B-atom placed at every centre of face. The formula of compound (1) AB

(2) AB3

(3) A2B2

(4) A2B3

(3) 12

(4) 4

Sol. Answer (2)

1 ⎫  1⎪ 8 ⎪ (Contribution) AB ⎬ 3 ⎪ 1 B  6  (face) ⇒ 3 ⎪ ⎭ 2 A  corners = 8 ×

34. Coordination number in ABAB... type arrangement is (1) 6

(2) 8

Sol. Answer (3)

C.N. = 12

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Solution of Assignment

The Solid State

31

35. The pyknometric density of sodium chloride crystal is 2.165 × 10 3 kg m –3 while its X-ray density is 2.178 × 103 kgm–3. The fraction of unoccupied sites in sodium chloride crystal is (2) 5.96 × 10–2

(1) 5.96

(3) 5.96 × 10–1

(4) 5.96 × 10–3

Sol. Answer (4) Unoccupied sites  X-ray density – pyknometric density = 2.178 × 103 – 2.165 × 103 = 0.013 × 103 % of unoccupied sites 

0.013  103 0.013  103  X-ray density 2.178  103

 5.96 × 10–3 36. A compound formed by elements X and Y crystallizes in a cubic structure in which the X atoms are at the corners of a cube and the Y atoms are at the face-centres. The formula of the compound is (2) X3Y

(1) XY3

(3) XY

(4) XY2

Sol. Answer (1)

1 1 8

⎫ ⎪ ⎪ (Contribution) ⎬ XY3 ⎪ 1 Y  face centre  6  (Contribution) ⇒ 3 ⎪ 2 ⎭ X  corners = 8 ×

37. In a face-centered cubic lattice, a unit cell is shared equally by how many unit cells? (1) 2

(2) 4

(3) 6

(4) 8

Sol. Answer (3)

SECTION - D Assertion-Reason Type Questions 1.

A : In NaCl structure, the interionic distance is a/2. (a = Unit cell edge length) R : NaCl forms face centered cubic unit cell.

Sol. Answer (2) In NaCl

r  r 

a  True 2

F.C.C.  true But (R) is not explanation of Assertion. 2.

A : The co-ordination number of Ca F2 is 8 : 4. R : Ca2+ ions occupy ccp lattice while F– ions occupy 50% octahedral voids and 50% tetrahedral voids.

Sol. Answer (3) (i) C.N. of CaF2 is 8 : 4  true (ii) Ca+2 occupy ccp lattice while F– ions occupy all tetrahedral holes not 50% O.V. and 50% T.V. i.e., reason is wrong. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

32 3.

The Solid State

Solution of Assignment

A : The number of spheres are equal to the number of octahedral void as well as tetrahedral void. R : Octahedral void and tetrahedral void has equal size.

Sol. Answer (4) (i) Number of spheres = number of O.V. Number of T.V. = 2 × number of sphere (ii) Octahedral and tetrahedral voids does not have equal size. So, (A) & (R) are wrong 4.

A : In Schottky defect, density of crystal decreases. R : Equal number of cations and anions are missing in Schottky defect.

Sol. Answer (1) In Schottky defect density decreases because equal number of cations and equal number of anions are missing i.e., A & R  true and correct explanation. 5.

A : If a tetrad axis is passed through the unit cell of NaCl and all ions are removed which are touching to tetrad axis then the formula of NaCl becomes Na3Cl4. R : Only one Na+ is removed not the Cl–.

Sol. Answer (4) If tetrad axis are passed through the unit cell of NaCl two Cl– ions and one Na+ will be removed. Formula unit will be Cl–  8 

Na 

1 1  4  3 8 2

–

Cl ion

1  12  3 4

+

Na ion

i.e., Na3Cl3 NaCl

–

Cl ion

Both statements are wrong.

6.

A : A particle at the corner of CCP unit cell has

1 th of its contribution to the unit cell. 8

R : In any space lattice, the corner of the unit cell is always shared by the eight unit cell. Sol. Answer (3) Assertion is true but reason is wrong because unit cell is always shared by eight unit cell only for cubic crystal not for all space lattice. 7.

A : Glass belongs to the category of covalent network solid. R : Unit cell of glass is hexagonal.

Sol. Answer (4) Glass belongs to amorphous solid not have an unit cell, so both A & R are wrong. 8.

A : NaCl shows Schottky defect at room temperature. R : NaCl shows ‘F centre’ at high temperature.

Sol. Answer (2) Both are true but not correct explanation. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

Solution of Assignment

9.

The Solid State

33

A : Fe3O4 is ferrimagnetic at room temperature but becomes paramagnetic at 850 K. R : The magnetic moment in Fe3O4 are aligned equally in parallel and antiparallel directions which on heating randomise.

Sol. Answer (3) A is true but R is wrong because in ferrimagnetic magnetic moment are aligned inequally. 10. A : In molecular solids the lattice points are occupied by the atoms or molecules. R : Molecular solids are generally sublime. Sol. Answer (2) Both are true but not correct explanation. 11. A : Silicon is insulator at 0 K but semiconductor at room temperature. R : Conductivity of silicon at room temperature is due to electronic defect. Sol. Answer (1) Silicon is semiconductor  true. at room temperature due to electronic defect (doping). 12. A : Amorphous solids are isotropic. R : Amorphous solids are not rigid. Sol. Answer (2) Not explanation. 13. A : In NaCl coordination number of Cl– ion is 6 but in CsCl coordination number of Cl– ion is 8. R : Ionic radii changes with type of lattice. Sol. Answer (3) In NaCl C.N. = 6, CsCl = 8 and ionic radii is not change with type of lattice. 14. A : All crystals of same substance possess the same elements of symmetry. R : The size of crystal of same substance may vary depending upon the conditions of crystallisation. Sol. Answer (2) Both are true but not correct explanation. 15. A : AgBr shows both Schottky and Frenkel defect. R : AgBr is a crystalline solid. Sol. Answer (2) AgBr shows both Schottky and Frenkel defects it is a crystalline solid. 16. A : Number of carbon atoms per unit cell in diamond is 8. R : The structure of diamond is similar to ZnS. Sol. Answer (1) Diamond shows C.N. = 8 it has ZnS type structure. Aakash Educational Services Pvt. Ltd. - Regd. Office : Aakash Tower, Plot No.-4, Sector-11, Dwarka, New Delhi-75 Ph.011-47623456

34

The Solid State

Solution of Assignment

17. A : The coordination number of ionic compound depends upon radius ratio. R : Higher the coordination number higher will be stability. Sol. Answer (2) C.N. of ionic compounds depends upon radius ratio, more the radius ratio, more will be C.N. and higher will be stability. 18. A : Number of rectangular plane in a cubic crystal is 3. R : Rectangular planes passes through corner to corner of unit cell. Sol. Answer (3) Total rectangular planes are 3 and rectangular plane passes through opposite face. 19. A : ccp is more efficient than hcp. R : Packing fraction is different in both cases. Sol. Answer (4) CCP and HCP both have same packing efficiency. 20. A : Coordination number of both Na+ and Cl– in NaCl is 6. R : Second coordination number of Cl– in the NaCl unit is 12. Sol. Answer (2) A & R both are correct but not correct explanation.







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