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

Answer 1

Answer: (D)

According to $ccp$, structure,
Number of $O^{2-}$ ions = 4
$\therefore$ tetrahedral void = 8
and octahedral void = 4
$\because$ A ions occupy l/4th of tetrahedral voids
$\therefore$ Number of A ions = $\frac{1}{4}\times 8$ = 2
Again B ions occupy all the tetrahedral voids
$\therefore$ Number of B ion = 4
A : B : O = 2 : 4 : 4 = 1 : 2 : 2
$\therefore$ the formula of oxide = $A{B}_2{O}_2$