Question

In a compound $A{B}_6{O}_4,$ oxide ions are arranged in ccp and cations A are present in octahedral voids. Cations B are equally distributed among octahedral and tetrahedral voids. The fraction of the octahedral voids occupied is :

• A. $0.25$
• B. $0.5$
• C. $0.75$
• D. $1$

Answer 1

The correct option is D $1$

In $A{B}_6{O}_4$, 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, $A{B}_6{O}_4$,In one unit cell,
Number of cation A is 1
Number of cations B are 6
Cations A are present in octahedral void and cations B are equally distributed among octahedral and tetrahedral. So, one A and 3 out of 6 B cations occupies octahedral sites.
$\text{Fraction of octahedral sites occupied}=\frac{4}{4}=1$