Solid State — Page 2 | NEET Chemistry

Q11

Iron exhibits bcc structure at room temperature. Above 900°C, it transforms to fcc structure. The ratio of density of iron at room temperature to that at 900°C (assuming molar mass and atomic radii of iron remains constant with temperature) is

A

{{\sqrt 3 } \over {\sqrt 2 }}

B

{{4\sqrt 3 } \over {3\sqrt 2 }}

C

{{3\sqrt 3 } \over {4\sqrt 2 }}

D

{1 \over 2}

Correct Answer

Option C

Solution

The density of a substance is given by the formula:

\rho = \dfrac{Z \times M_w}{N_A \times V}

Here, $Z$ is the number of atoms in one unit cell, $M_w$ is the molar mass, $N_A$ is Avogadro’s number, and $V$ is the volume of the unit cell.

1.

For body-centered cubic (bcc) structure: In bcc, $Z = 2$ and the relation between atomic radius $r$ and lattice parameter $a$ is:

r = \dfrac{\sqrt{3}a}{4}

Volume of one unit cell,

V = a^3

2.

For face-centered cubic (fcc) structure: In fcc, $Z = 4$ and the relation between atomic radius $r$ and lattice parameter $a$ is:

r = \dfrac{a}{2\sqrt{2}}

Volume of one unit cell,

V = a^3

3.

Ratio of densities: Since molar mass and atomic radius (hence atomic mass per atom) remain constant, the ratio of densities between bcc and fcc forms can be written as:

\dfrac{\rho_1}{\rho_2} = \dfrac{Z_1 \times V_2}{Z_2 \times V_1}

Substituting the values:

\dfrac{\rho_1}{\rho_2} = \dfrac{2 \times \dfrac{\sqrt{3}\pi a^3}{16}}{4 \times \dfrac{\pi a^3}{12\sqrt{2}}}

Simplifying,

\dfrac{\rho_1}{\rho_2} = \dfrac{2 \times \sqrt{3} \times \sqrt{2} \times 12}{4 \times 16}

Therefore,

\dfrac{\rho_1}{\rho_2} = \dfrac{3\sqrt{3}}{4\sqrt{2}}

Hence, the ratio of densities of iron in bcc form (at room temperature) to fcc form (above 900°C) is

\dfrac{3\sqrt{3}}{4\sqrt{2}}.

Q12

In calcium fluoride, having the fluorite structure, the coordination numbers for calcium ion (Ca 2+ ) and fluoride ion (F

-

) are

A 4 and 2

B 6 and 6

C 8 and 4

D 4 and 8

Correct Answer

Option C

Solution

In fluorite (CaF 2 ) structure, C.N. of Ca +2 = 8 C.N. of F – = 4.

Q13

The ionic radii of A + and B

-

ions are 0.98

\times

10

-

10 m and 1.81

\times

10

-

10 m. The coordination number of each ion in AB is

A 8

B 2

C 6

D 4

Correct Answer

Option C

Solution

From radius ratio,

{{{r_ + }} \over {{r_ - }}} = {{0.98 \times {{10}^{ - 10}}} \over {1.81 \times {{10}^{ - 10}}}}

= 0.541 It lies in the range of 0.414 to 0.732 hence coordination number of each ion will be 6 as the compound will have NaCl type structure i.e., octahedral arrangement.

Q14

Lithium has a bcc structure. Its density is 530 kg m

-

3 and its atomic mass is 6.94 g mol

-

. Calculate the edge length of a unit cell of lithium metal. (N A = 6.02

\times

10 23 mol

-

1 )

A 527 pm

B 264 pm

C 154 pm

D 352 pm

Correct Answer

Option D

Solution

For bcc structure we have, Z = 2 Given, $\rho$ = 530 kg m –3 atomic mass of Li = 6.94 mol –1 N A = 6.02 × 10 23 mol –1 $\rho$ = 530 kg m –3 =

{{530 \times 1000} \over {{{\left( {100} \right)}^3}}} =

0.53 g cm –3 Also, $\rho$ =

{{z \times At.mass} \over {{N_A} \times {a^3}}}

$\Rightarrow$

{a^3} = {{Z \times At.mass} \over {{N_A} \times \rho }}

=

{{2 \times 6.94} \over {6.02 \times {{10}^{23}} \times 0.53}}

= 43.5 $\times$ 10 -10 cm 3 $\Rightarrow$

a

= 352 × 10 –10 cm = 352 pm

Q15

The correct statement regarding defects in crystalline solids is

A Frenkel defects decrease the density of crystalline solids

B Frenkel defect is a dislocation defect

C Frenkel defect is found in halides of alkaline metals

D Schottky defects have no effect on the density of crystalline solids.

Correct Answer

Option B

Solution

Frenkel defect is a dislocation defect as smaller ions (usually cations) are dislocated from normal sites to interstitial sites.

Frenkel defect is shown by compounds having large difference in the size of cations and anions hence, alkali metal halides do not show Frenkel defect.

Also, Schottky defect decreases the density of crystal while Frenkel defect has no effect on the density of crystal.

Q16

The vacant space in bcc lattice unit cell is

A 48%

B 23%

C 32%

D 26%

Correct Answer

Option C

Solution

Packing efficiency of bcc lattice = 68% Hence, empty space = 32%.

Q17

A given metal crystallises 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?

A 80 pm

B 108 pm

C 40 pm

D 127 pm

Correct Answer

Option D

Solution

As Z = 4, so, structure is fcc. Hence, r =

{a \over {2\sqrt 2 }} = {{361} \over {2\sqrt 2 }}

= 127.65 pm = 127 pm

Q18

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

A

{2 \over {\sqrt 3 }}a

B

{4 \over {\sqrt 3 }}a

C

{{\sqrt 3 } \over 4}a

D

{{\sqrt 3 } \over 2}a

Correct Answer

Option D

Solution

The distance between the body centered atom and one corner atom is

{{\sqrt 3 } \over 2}a

.

Q19

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 ( N A Avogadro's constant = 6.02

\times

10 23 mol

-

1 )

A 27 g mol

-

1

B 20 g mol

-

1

C 40 g mol

-

1

D 30 g mol

-

1

Correct Answer

Option A

Solution

for fcc Z = 4. d =

{{ZM} \over {{N_A}{a^3}}}

$\Rightarrow$ M =

{{d{N_A}{a^3}} \over Z}

=

{{2.72 \times 6.023 \times {{10}^{23}} \times {{\left( {404 \times {{10}^{ - 10}}} \right)}^3}} \over 4}

= 27 g mol –1

Q20

The number of carbon atoms per unit cell of diamond unit cell is

A 6

B 1

C 4

D 8

Correct Answer

Option D

Solution

Total number of carbon atoms per unit cell =

8 \times {1 \over 8} + 6 \times {1 \over 2} + 4

= 8