Question

Structure of a mixed oxide is cubic close packed (c. c. p.). 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. $A_2{B}_3{O}_4$
• B. $A{B}_2{O}_2$
• C. $AB{O}_2$
• D. $A_{2}B{O}_2$

Answer 1

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

$O^{-2}$ ions form CCP, therefore $4O^{-2}$ ions are present per unit cell.
$\therefore$ No. of tetradral voids $=8$

Tetrahedral voids occupied by $A^{+2}=\frac{1}{4}\times 8=2$
Also, no. of octahedral voids present $=4$
Octahedral voids occupied by $B^{+}=4$
$\therefore$ Formula of oxide could be $=A_2{B}_4{O}_4\text{or}A{B}_2{O}_2$