Question

In a close-packed structure of mixed oxides, the lattice is composed of oxide ions, one-eighth of tetrahedral voids are occupied by divalent cations $A$ while one-half of octahedral voids are occupied by trivalent cations $B$. The formula of the 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 $A{B}_2{O}_4$

The lattice is compound of oxide ions. So, number of oxide ions in the lattice $=\frac{8}{8}=1$

Number of ions of $A=\frac{1}{8}\times numberoftetrahedralvoids$

$=\frac{1}{8}\times 2\times numberofoxideions=\frac{1}{8}\times 2\times 1=\frac{1}{4}$

Number of ions of $B=\frac{1}{2}\times numberofoctahedralvoids$

$=\frac{1}{2}\times numberofoxideions=\frac{1}{2}\times 1=\frac{1}{2}$

So, $A:B:O\Rightarrow A:B:O$

$\frac{1}{4}:\frac{1}{2}:1$ $1:2:4$

So, formula of compound is $A{B}_2{O}_4$