Question

Structure of a mixed oxide is cubic close packed $\left(ccp\right)$. 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:
(AIPMT 2012)

• A. $AB{O}_2$
• B. $A_{2}B{O}_2$
• C. $A{B}_2{O}_2$
• D. $A_2{B}_3{O}_4$

Answer 1

The correct option is C $A{B}_2{O}_2$

$O^{2-}\text{makes ccp}$

$\text{So, number of}{O}^{2-}\text{ions}=4$
$\text{Tetrahedral voids}=2\times 4=8$
$\text{Octahedral voids}=4$

$\frac{1}{4}\text{of tetrahedral void}=\frac{1}{4}\times 8=2\text{occupied by}{A}^{2+}$

And octahedral voids = 4 occupied by $B^{+}$
$A:B:O=2:4:4$
$\text{So, the formula of oxide is}A{B}_2{O}_2$.