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. The octahedral voids are occupied by a monovalent metal B. The formula of the oxide is

• A. $ABO_2$
• B. $A_2B_3O_4$
• C. $A_2BO_2$
• D. $AB_2O_2$

Answer 1

Answer: (D)

$AB_2O_2$

Answer 2

Suppose number of ${O^{2-}}$ ions in close packing = $n$
$\therefore$ No. of octahedral voids = $n$
No. of tetrahedral voids = $2n$
Divalent metal ions A present = $\frac{2n}{4} =\frac{n}{2}$
Monovalent metal ions $B$ present = $n$
$\therefore$ Ration $A : B : O^{-2} =\frac{n}{2} : n : n =\frac{1}{2} : 1 : 1 = 1 : 2 : 2$
Hence, formula = $AB_2O_2$