Question

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 occupied by a monovalent metal $B$. The formula of the oxide is:

• A. $A_2{BO}_2$
• B. $A_2{B}_3{O}_4$
• C. $A{B}_2{O}_2$
• D. $AB{O}_2$

Answer 1

Answer: (C)

$A{B}_2{O}_2$

It is given that the structure of mixed oxide is cubic closed packed.

Therefore, number of $O^{2-}$ ions according to ccp $=4$
So, number of tetrahedral voids $=8$ and octahedral voids $=4$
$A$ ions occupy $\frac{1}{4}$th of tetrahedral voids $=2$
$B$ ions occupy all octahedral voids $=4$
$A:B:O$ $=2:4:4$ $=1:2:2$

Hence, the formula is $A{B}_2{O}_2$.