Solid State Questions — NEET Chemistry

Q1

How many number of tetrahedral voids are formed in

5 \mathrm{~mol}

of a compound having cubic close packed structure? (Choose the correct option)

A

1.550 \times 10^{24}

B

3.011 \times 10^{25}

C

3.011 \times 10^{24}

D

6.022 \times 10^{24}

Correct Answer

Option D

Solution

Number of particles

=5 \mathrm{~N}_{\mathrm{A}}

Number of THV

=2 \times

number of particles, for close packing

\begin{aligned} & =2 \times 5 \mathrm{~N}_{\mathrm{A}} \\ & =10 \mathrm{~N}_{\mathrm{A}} \\ & =10 \times 6.023 \times 10^{23} \\ & =6.023 \times 10^{24} \end{aligned}

Q2

How are edge length '

a

' of the unit cell and radius '

r

' of the sphere related to each other in ccp structure? (Choose correct option for your answer)

A

a=2 r

B

a=r / 2 \sqrt{2}

C

\mathrm{a}=4 \mathrm{r} / \sqrt{3}

D

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

Correct Answer

Option D

Solution

For CCP (FCC)

\begin{aligned} & 4 \mathrm{r}=\sqrt{2} \mathrm{a} \\ & \mathrm{a}=\frac{4 \mathrm{r}}{\sqrt{2}} \\ & \mathrm{a}=2 \sqrt{2} \mathrm{r} \end{aligned}

Q3

A compound is formed by two elements A and B. The elements B forms cubic close packed structure and atoms of A occupy

1 / 3

of tetrahedral voids. If the formula of the compound is

\mathrm{A}_{x} \mathrm{B}_{y}

, then the value of

x+y

is in option

A 4

B 3

C 2

D 5

Correct Answer

Option D

Solution

\mathrm{x+y=5}

Q4

What fraction of one edge centred octahedral void lies in one unit cell of fcc?

A

\dfrac{1}{3}

B

\dfrac{1}{4}

C

\dfrac{1}{12}

D

\dfrac{1}{2}

Correct Answer

Option B

Solution

$\to$ Edge centred octahedral void is shared between four unit cells $\to$ Per unit cell contribution is 1/4

Q5

Choose the correct statement:

A Diamond and graphite have two dimensional network.

B Diamond is covalent and graphite is ionic.

C Diamond is sp 3 hybridized and graphite is sp 2 hybridized

D Both diamond and graphite are used as dry lubricants.

Correct Answer

Option C

Solution

Diamond : $\bullet$ sp 3 hybridised carbon atom $\bullet$ Covalent solid $\bullet$ 3-D structure $\bullet$ Cannot be used as dry lubricant Graphite : $\bullet$ sp 2 hybridised carbon atom $\bullet$ Covalent solid $\bullet$ 3-D structure $\bullet$ Used as dry lubricant

Q6

Copper crystallises in fcc unit cell with cell edge length of 3.608

\times

10

-

8 cm. The density of copper is 8.92 g cm

-

3 . Calculate the atomic mass of copper.

A 63.1 u

B 31.55 u

C 60 u

D 65 u

Correct Answer

Option A

Solution

d = {{ZM} \over {{N_A}{{(a)}^3}}}

Z = 4(FCC), d = 8.92 g cm $-$ 3 , N A = 6.023 $\times$ 10 23 , a = 3.608 $\times$ 10 $-$ 8 cm

M = {{d{N_A}{{(a)}^3}} \over Z}

= {{8.92 \times 6.023 \times {{10}^{23}} \times {{(3.608 \times {{10}^{ - 8}})}^3}} \over 4}

= {{8.92 \times 6.023 \times {{10}^{23}} \times 46.97 \times {{10}^{ - 24}}} \over 4} = {{2523.47 \times {{10}^{ - 1}}} \over 4}

= 630.8 \times {10^{ - 1}} = 63.08 \simeq 63.1\,u

Q7

The correct option for the number of body centred unit cells in all 14 types of Bravais lattice unit cells is :

A 3

B 7

C 5

D 2

Correct Answer

Option A

Solution

$\bullet$ In 14 types of Bravais lattices, body centred unit cell is present in cubic, tetragonal and orthorhombic crystal systems.

$\bullet$ Hence, body centred possible variation is present in three crystal systems.

Q8

Right option for the number of tetrahedral and octahedral voids in hexagonal primitive unit cell are :

A 12, 6

B 8, 4

C 6, 12

D 2, 1

Correct Answer

Option A

Solution

$\bullet$ Number of octahedral and tetrahedral voids formed by N closed packed atoms are N and 2N respectively.

$\bullet$ Each hexagonal unit cell contains 6 atoms therefore, number of tetrahedral and octahedral voids are 12 and 6 respectively.

Q9

An element has a body centered cubic (bcc) structure with a cell edge of 288 pm. The atomic radius is :

A

{{\sqrt 2 } \over 4} \times 288

pm

B

{{4 } \over \sqrt 3} \times 288

pm

C

{{4 } \over \sqrt 2} \times 288

pm

D

{{\sqrt 3 } \over 4} \times 288

pm

Correct Answer

Option D

Solution

For BCC, lattice

\sqrt 3 a = 4r

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

So,

r = {{\sqrt 3 \times 288} \over 4}

Q10

A compound is formed by cation C and anion A. The anions form hexagonal close packed (hcp) lattice and the cations occupy 75% of octahedral voids. The formula of the compound is :

A C 3 A 4

B C 4 A 3

C C 2 A 3

D C 3 A

Correct Answer

Option A

Solution

As anions are in hcp, the number of anions A = 6 Number of cations = 6 $\times$

{{75} \over {100}}

= 6 $\times$

{3 \over 4}

=

{9 \over 2}

So, formula of compound is

{C_{{9 \over 2}}}{A_6}

. $\Rightarrow$ C 9 A 12 $\Rightarrow$ C 3 A 4