Question

In a close-packed structure of mixed oxides, the lattice is composed of oxide ions, one-eighth of octahedral voids are occupied by divalent cations while one-half of octahedral voids are occupied by a trivalent cation. The formula of oxide is:

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

Answer 1

The correct option is D $A{B}_2{O}_4$

Let $A$ be the divalent ion and $B$ be the trivalent ion.

In ccp unit cell the effective number of atoms occupying lattice points is-

$8\times \frac{1}{8}+6\times \frac{1}{2}=1+3=4$

Therefore,

No. of $O^{2-}$ ions $=4$

Given that ion $A$ occupies ${1/8}^{th}$ of octahedral void.

Therefore,

No. of $A$ ions $=\frac{1}{8}\times 8=1$

Also given that ion $B$ occupies half of tetrahedral void.

Therefore,

No. of $B$ ions $=\frac{1}{2}\times 4=2$

$A:B:O=1:2:4$

Therefore,

formula of oxide $=A{B}_2{O}_4$

Hence the formula of oxide is $A{B}_2{O}_4$.

Hence, Option "D" is the correct answer.