Question

In the spinel structure, oxides ions are cubical-closest packed whereas $1/8^{\text{th}}$ of tetrahedral voids are occupied by ${\text{A}}^{2+}$ cation and $\frac{1}{2}$ of octahedral voids are occupied by ${\text{B}}^{3+}$ cations. The general formula of the compound having spinel structure is:

• A. ${\text{A}}_4{\text{B}}_2{\text{O}}_2$
• B. ${\text{A}}_2{\text{B}}_2{\text{O}}_4$
• C. ${\text{A}}_2{\text{B}}_4{\text{O}}_2$
• D. ${\text{AB}}_2{\text{O}}_4$

Answer 1

The correct option is D ${\text{AB}}_2{\text{O}}_4$

${\text{Effective number of O}}^{2-}\text{in the unit cell}$
$=8\times \frac{1}{8}+6\times \frac{1}{2}=4$
${\text{Effective number of A}}^{2+}\text{in the unit cell}$
$=8\times \frac{1}{8}=1$
${\text{Effective no. of B}}^{3+}\text{in the unit cell}$
$=4\times \frac{1}{2}=2$
$\therefore {\text{the general formula is AB}}_2{\text{O}}_4$.