Question

In a closed packed structure of mixed oxides, the lattice is composed of mixed oxides ions. One-eighth of tetrahedral voids are occupied by divalent cations $\left(A^{2+}\right)$ while one half of octahedral voids are occupied by trivalent cations $\left(B^{3+}\right)$. The formula of mixed oxide is :

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

Answer 1

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

$O^{2-}$ ions form close packing. The number of octahedral voids will be equal to the number of $O^{2-}$ ions and the number of tetrahedral voids will be equal to twice the number of $O^{2-}$ ions.
One-eight of tetrahedral voids are occupied by $A^{2+}$ ions. Hence, the number of $A^{2+}$ ions will be $\frac{2}{8}=\frac{1}{4}$ of $O^{2-}$ ions.
One half of the octahedral voids are occupied by $B^{3+}$ ions.
Hence, the number of $B^{3+}$ ions will be $\frac{1}{2}$ of $O^{2-}$ ions.
Hence, $A:B:O=\frac{1}{4}:\frac{1}{2}:1$ or $1:2:4$
Hence, the formula of mixed oxide is $A{B}_2{O}_4$