1

### NEET 2021

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

## Explanation

$\bullet$ 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.
2

### NEET 2021

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

## Explanation

$\bullet$ 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.
3

### NEET 2020 Phase 1

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

## Explanation

For BCC, lattice $\sqrt 3 a = 4r$
$r = {{\sqrt 3 a} \over 4}$
So, $r = {{\sqrt 3 \times 288} \over 4}$
4

### NEET 2019

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
C3A4
B
C4A3
C
C2A3
D
C3A

## Explanation

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

$\Rightarrow$ C9A12

$\Rightarrow$ C3A4