Question

The cubic unit cell of a mixed oxide is composed of oxide ions in a ccp arrangement. 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. $A_2{B}_3{O}_4$
• B. $A{B}_2{O}_2$
• C. $AB{O}_2$
• D. $A_{2}B{O}_2$

Answer 1

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

The number of $O^{2-}$ ions in cubic close packing $=n$
$\therefore$ Number of octahedral voids $=n$
Number of tetrahedral voids $=2n$
Divalent metal ions '$A$' present $=\frac{1}{4}\times 2n=\frac{n}{2}$
Monovalent metal ions '$B$' present $=n$
$\therefore$ Ratio $A:B:O^{2-}$
$=\frac{n}{2}:n:n$
$=\frac{1}{2}:1:1$
$=1:2:2$
Hence, the formula would be $=A{B}_2{O}_2$