##### Document Text Contents

Page 1

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

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]

Solutions

Chapter 9

The Solid State

Page 2

2 The Solid State Solution of Assignment

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

5. 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 (2) AB

2

(3) A

2

B

3

(4) AB

3

Sol. Answer (1)

A corners of cube Total 8 atoms at corner, which have contribution at corner

1

8

So

1

8 1

8

number of atoms of A per unit cell.

B centre of body at centre of body one atom contributed completely.

So, 1 × 1 = 1

i.e., A

1

B

1

= AB

6. A solid with formula ABC

3

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)

ABC

3

At corners number of atom =

1

8 1

8

(contribution at corner) A

At body centre = 1 × 1 = 1 B

At face centre = (Total face)

(Contributionat face)

1

6 3

2

C

ABC

3

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

A

C

B

(1) ABC (2) AB

2

C

2

(3) A

2

BC (4) AB

8

C

2

Sol. Answer (1)

A present at body centre in given figure

So, its number = 1 × 1 = 1 (contributed completely) A

B present at corners =

(Total atom)

(Contribution)

1

8 1

8

B

C Present at two opposite face (Total 2atomsat face)

(Contribution at each face)

1

2 1

2

C

So, formula ABC

Page 17

17Solution of Assignment The Solid State

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

3. The type of crystal system shown is

b

a

a

(1) Cubic (2) Orthorhombic (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 =

1

8 1

8

B =

1

6 3

2

C = 1 × 1 = 1

D =

1

12 3

4

Total = 8 atoms per unit cell

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) AB

3

CD

3

(2) ABCD (3) AB

6

C

2

D

4

(4) AB

6

C

2

D

2

Sol. Answer (4)

Corner

1 1

A 4

8 2

Face centre

1

B 6 3

2

Total C = 1

Total D =

1

4 1

4

formula

1 3 1 1

2

A B C D

AB

6

C

2

D

2

Page 18

18 The Solid State Solution of Assignment

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

6. 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

(1) ABCD

2

(2) ABD

2

(3) AB

2

D

2

(4) AB

4

D

5

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.

Total

1 1

A 4

8 2

[4 atoms removed from corner]

Total

1

B 4 2

2

[2 atoms are removed from face centre]

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

Total

1 5

D 10

4 2

[2 atom removed from edge]

1 2 0 5 1 4 5

2 2

A B C D A B D

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

(1)

a 3

2

(2)

a 3

4

(3) a 2 (4)

a

2

Sol. Answer (1)

Cs

+

Cl

–

3a

3a

2

[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)

ccp

packing fraction = 74%

C.N. = 12

Unit cell is fcc

ABCABC.........

Page 33

33Solution of Assignment The Solid State

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

9. A : Fe

3

O

4

is ferrimagnetic at room temperature but becomes paramagnetic at 850 K.

R : The magnetic moment in Fe

3

O

4

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.

Page 34

34 The Solid State Solution of Assignment

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

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.

� � �

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

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]

Solutions

Chapter 9

The Solid State

Page 2

2 The Solid State Solution of Assignment

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

5. 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 (2) AB

2

(3) A

2

B

3

(4) AB

3

Sol. Answer (1)

A corners of cube Total 8 atoms at corner, which have contribution at corner

1

8

So

1

8 1

8

number of atoms of A per unit cell.

B centre of body at centre of body one atom contributed completely.

So, 1 × 1 = 1

i.e., A

1

B

1

= AB

6. A solid with formula ABC

3

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)

ABC

3

At corners number of atom =

1

8 1

8

(contribution at corner) A

At body centre = 1 × 1 = 1 B

At face centre = (Total face)

(Contributionat face)

1

6 3

2

C

ABC

3

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

A

C

B

(1) ABC (2) AB

2

C

2

(3) A

2

BC (4) AB

8

C

2

Sol. Answer (1)

A present at body centre in given figure

So, its number = 1 × 1 = 1 (contributed completely) A

B present at corners =

(Total atom)

(Contribution)

1

8 1

8

B

C Present at two opposite face (Total 2atomsat face)

(Contribution at each face)

1

2 1

2

C

So, formula ABC

Page 17

17Solution of Assignment The Solid State

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

3. The type of crystal system shown is

b

a

a

(1) Cubic (2) Orthorhombic (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 =

1

8 1

8

B =

1

6 3

2

C = 1 × 1 = 1

D =

1

12 3

4

Total = 8 atoms per unit cell

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) AB

3

CD

3

(2) ABCD (3) AB

6

C

2

D

4

(4) AB

6

C

2

D

2

Sol. Answer (4)

Corner

1 1

A 4

8 2

Face centre

1

B 6 3

2

Total C = 1

Total D =

1

4 1

4

formula

1 3 1 1

2

A B C D

AB

6

C

2

D

2

Page 18

18 The Solid State Solution of Assignment

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

6. 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

(1) ABCD

2

(2) ABD

2

(3) AB

2

D

2

(4) AB

4

D

5

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.

Total

1 1

A 4

8 2

[4 atoms removed from corner]

Total

1

B 4 2

2

[2 atoms are removed from face centre]

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

Total

1 5

D 10

4 2

[2 atom removed from edge]

1 2 0 5 1 4 5

2 2

A B C D A B D

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

(1)

a 3

2

(2)

a 3

4

(3) a 2 (4)

a

2

Sol. Answer (1)

Cs

+

Cl

–

3a

3a

2

[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)

ccp

packing fraction = 74%

C.N. = 12

Unit cell is fcc

ABCABC.........

Page 33

33Solution of Assignment The Solid State

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

9. A : Fe

3

O

4

is ferrimagnetic at room temperature but becomes paramagnetic at 850 K.

R : The magnetic moment in Fe

3

O

4

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.

Page 34

34 The Solid State Solution of Assignment

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

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.

� � �