Question

In normal spinel structure ($A{B}_2{O}_4$), there is a closed packed array of $O^{2-}$ ions. The trivalent cations are present in:

• A. 75% of octahedral voids
• B. 50% of octahedral voids
• C. 12.5% of tetrahedral voids
• D. 25% of octahedral voids

Answer 1

The correct option is B 50% of octahedral voids

For a normal spinel structure, oxide ions forms the ccp arrangements, it occupies the corners and face centres of the cubic unit cell.
$\text{No. of}{O}^{2-}\text{ions in fcc unit cell}=4$
$\therefore$
$\text{Number of tetrahedral voids}=8$
$\text{Number of octahedral voids}=4$

$A^{2+}$ present in $\frac{1}{8}$ of tetrahedral voids and $B^{3+}$ present in $\frac{1}{2}$ of octahedral voids.
Thus, $B^{3+}$ occupies two octahedral voids in an unit cell.
% of $B^{3+}$ present $=\frac{2}{4}\times 100=50\%$
Thus, $50\%$ of octahedral voids are occupied by $B^{3+}$