Question

In spinel structure, oxides ions form fcc lattice, whereas $\frac{1}{8}$th of tetrahedral holes are occupied by cations $A^{2+}$ and $\frac{1}{2}$ of octahedral holes are occupied by cations $B^{3+}$ ions. The general formula of the compound having spinel structure is

• A. $A{B}_2{O}_4$
• B. $A_2{B}_2{O}_4$
• C. $A_2{B}_{6}O$
• D. $A_4{B}_{3}O$

Answer 1

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

In spinel structure, oxide ions forms the ccp arrangements, it occupies the corners and face centres of the cubic unit cell.
$\text{No. of}{O}^{2-}\text{ions in unit cell}=N$
$\therefore$
$\text{Number of tetrahedral voids}=2N$
$\text{Number of octahedral voids}=N$

$A^{2+}$ ions occupies $\frac{1}{8}$ of the tetrahedral holes
$\therefore$
$\text{Number of}{A}^{2+}\text{ions in an unit cell}=\frac{1}{8}\times 2N=\frac{N}{4}$

$B^{3+}$ ions occupies $\frac{1}{2}$ of the octahedral holes
$\therefore$
$\text{Number of}{B}^{3+}\text{ions in an unit cell}=\frac{N}{2}$

$\text{The ratio of}{A}^{2+}:B^{3+}:O^{2-}\text{is}=\frac{N}{4}:\frac{N}{2}:N$
Thus, the general formula of spinel structure is $A{B}_2{O}_4$