Solid State — Page 3 | JEE Chemistry

Q21

A metal crystallises in a face centred cubic structure. If the edge length of its unit cell is ‘a’, the closest approach between two atoms in metallic crystal will be :

A

2\sqrt 2 a

B

\sqrt 2 a

C

{a \over {\sqrt 2 }}

D 2a

Correct Answer

Option C

Solution

For a fcc unit cell

r = {{\sqrt 2 a} \over 4}

$\therefore$ Closest distance (2r) =

{{\sqrt 2 a} \over 4} = {a \over {\sqrt 2 }}

Q22

A crystal is made up of metal ions 'M1' and 'M2' and oxide ions. Oxide ions form a ccp lattice structure. The cation 'M1' occupies 50% of octahedral voids and the cation 'M2' occupies 12.5% of tetrahedral voids of oxide lattice. The oxidation numbers of 'M1' and 'M2' are, respectively :

A + 3, + 1

B + 4, + 2

C + 1, + 3

D + 2, + 4

Correct Answer

Option D

Solution

O–2 ions form ccp $\Rightarrow$ O4. M1 = 50% octahedral void =

{{50} \over {100}} \times 4

= 2 M2 = 12.5% tetrahedral void =

{{12.5} \over {100}} \times 8

= 1 So formula is : (M1)2(M2)1O4 Let charge on M1 and M2 are +x and +y respectively.

And O4 has -8 charge.

As crystal is neutral.

So metals must have +8 charge in total. $\therefore$ +2x + y = 8 ....(

1) By checking options we found eq (1) satisfy when x = +2 y = +4

Q23

An element crystallises in a face-centred cubic (fcc) unit cell with cell edge

a

. The distance between the centres of two nearest octahedral voids in the crystal lattice is :

A

a

B

{a \over 2}

C

{a \over {\sqrt 2 }}

D

{\sqrt 2 a}

Correct Answer

Option C

Solution

Distance between nearest octahedral voids (O.V.) =

\sqrt {{{\left( {{a \over 2}} \right)}^2} + {{\left( {{a \over 2}} \right)}^2}}

=

{a \over {\sqrt 2 }}

Q24

A solid having density of 9

\times

103 kg m–3 forms face centred cubic crystals of edge length

200\sqrt 2

pm. What is the molar mass of the solid? [Avogadro constant

\cong

6

\times

1023 mol–1 ,

\pi

\cong

3]

A 0.0305 kg mol–1

B 0.4320 kg mol–1

C 0.0216 kg mol–1

D 0.0432 kg mol–1

Correct Answer

Option A

Solution

$\rho$ =

{{Z \times M} \over {{a^3} \times {N_A}}}

$\Rightarrow$ 9 $\times$ 103 =

{{4 \times M} \over {\left( {200\sqrt 2 \times {{10}^{ - 12}}} \right) \times 6 \times {{10}^{23}}}}

$\Rightarrow$ M = 0.0305 kg mol–1

Q25

An element has a face-centred cubic (fcc) structure with a cell edge of

a

. The distance between the centres of two nearest tetrahedral voids in the lattice is :

A

a

B

{3 \over 2}a

C

{a \over 2}

D

\sqrt 2 a

Correct Answer

Option C

Solution

In FCC, tetrahedral voids are located on the body diagonal at a distance of

{{\sqrt 3 a} \over 4}

from the corner. Together they form a smaller cube of edge length

{a \over 2}

.

Q26

Select the correct statements (A) Crystalline solids have long range order. (B) Crystalline solids are isotropic (C) Amorphous solid are sometimes called pseudo solids. (D) Amorphous solids soften over a range of temperatures. (E) Amorphous solids have a definite heat of fusion. Choose the most appropriate answer from the options given below.

A (A), (B), (E) only

B (B), (D) only

C (C), (D) only

D (A), (C), (D) only

Correct Answer

Option D

Solution

(A) Crystalline solids have definite arrangement of constituent particles and have long range order.

(C), (D) Different constituent particles of an amorphous solid have different bond strengths and soften over a range of temperatures.

Q27

Lithium forms body centred cubic structure. The length of the side of its unit cell is 351 pm. Atomic radius of the lithium will be :

A 75 pm

B 300 pm

C 240 pm

D 152 pm

Correct Answer

Option D

Solution

For

BCC

structure

\sqrt 3 \,a = 4r

r = {{\sqrt 3 } \over 4}a = {{\sqrt 3 } \over 4} \times 351 = 152

Q28

Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 4.29 Å . The radius of sodium atom is approximately :

A 5.72 Å

B 0.93 Å

C 1.86 Å

D 3.22 Å

Correct Answer

Option C

Solution

In

bcc

the atoms touch along body diagonal $\therefore$

2r + 2r = \sqrt 3 a

$\therefore$

r = {{\sqrt 3 a} \over 4} = {{\sqrt 3 \times 4.29} \over 4} = 1.857\mathop A\limits^ \circ

Q29

A compound is formed by two elements

\mathrm{X}

and

\mathrm{Y}

. The element

\mathrm{Y}

forms cubic close packed arrangement and those of element

\mathrm{X}

occupy one third of the tetrahedral voids. What is the formula of the compound?

A

\mathrm{XY}_{3}

B

\mathrm{X_3Y}_{2}

C

\mathrm{X_3Y}

D

\mathrm{X_2Y}_{3}

Correct Answer

Option D

Solution

A compound is formed by two elements

\mathrm{X}

and

\mathrm{Y}

. The element

\mathrm{Y}

forms a cubic close-packed (CCP) arrangement, and element

\mathrm{X}

occupies one third of the tetrahedral voids. In a CCP structure, there are 4 atoms of

\mathrm{Y}

in a unit cell. This means there are 8 tetrahedral voids in the CCP structure. Since

\mathrm{X}

occupies one third of the tetrahedral voids, there are

\frac{1}{3} \times 8 = \frac{8}{3}

atoms of

\mathrm{X}

in the formula unit. The ratio of

\mathrm{X}

to

\mathrm{Y}

in the formula unit is

\frac{8}{3} : 4 = \frac{2}{3} : 1

. Multiplying both parts of the ratio by 3, we get

2 : 3

. So, the formula of the compound is

\mathrm{X_2Y_{3}}

Q30

Copper crystallizes in fcc with a unit cell length of 361 pm. What is the radius of copper atom?

A 108 pm

B 127 pm

C 157 pm

D 181 pm

Correct Answer

Option B

Solution

For

fcc

unit cell,

4r = \sqrt 2 \,\,a;\,r = {{\sqrt 2 \times 361} \over 4} = 127\,\,pm