Question

In a compound $P{Q}_2{O}_4,$ oxide ions are arranged in ccp and cations P are present in octahedral voids. Cations Q are equally distributed among octahedral and tetrahedral voids. The fraction of the octahedral voids occupied is

• A. 1/4
• B. 1/2
• C. 1/3
• D. 2/3

Answer 1

The correct option is B 1/2

In $P{Q}_2{O}_4$, oxide ions are arranged in ccp lattice. We know, ccp lattice has a face centred unit cell.
Therefore, the number of particles present in unit cell are 4.
Thus,
$\text{Number of oxide ions in unit cell}=4$
$\text{Number of octahedral voids}=4$
$\text{Number of tetrahedral voids}=2\times 4=8$
From the formula of compound, $P{Q}_2{O}_4$, the number of cation P in one unit cell is 1 and number of Q cation in one unit cell is 2.
Cations P are present in octahedral void and cations Q are equally distributed among octahedral and tetrahedral. So, one P and one of 2Q (totally 2) atoms occupies octahedral sites.
$\text{Fraction of octahedral sites occupied}=\frac{2}{4}=\frac{1}{2}$