1

### NEET 2013

The number of carbon atoms per unit cell of diamond unit cell is
A
6
B
1
C
4
D
8

## Explanation

Total number of carbon atoms per unit cell

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

### NEET 2013

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
(NA Avogadro's constant = 6.02 $\times$ 1023 mol$-$1)
A
27 g mol$-$1
B
20 g mol$-$1
C
40 g mol$-$1
D
30 g mol$-$1

## Explanation

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
3

### AIPMT 2012 Prelims

The number of octahedral void(s) per atom present in a cubic close-packed structure is
A
1
B
3
C
2
D
4

## Explanation

Number of octahedral voids is same as number of atoms.
4

### AIPMT 2012 Mains

Structure of a mixed oxide is cubic close packed (ccp). The cubic unit cell of mixed oxide is composed of oxide ions. One fourth of the tetrahedral voids are occupied by divalent metal A and the octahedral voids are cooupied by a monovalent metal B. The formula of the oxide is
A
ABO2
B
A2BO2
C
A2B3O4
D
AB2O2

## Explanation

Number of atoms in cubic close packing = 4 = O2-

Number of tetrahedral voids = 2 × N = 2 × 4

Number of A2+ ions = 8 $\times$ ${1 \over 4}$ = 2

Number of octahedral voids = Number of B+ ions = N = 4

Ratio of ions will be,

O2– : A2+ : B+ = 4 : 2 : 4 = 2 : 1 : 2

Formula of oxide = AB2O2