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In a normal spinel type structure, the oxide ions are arranged in ccp, whereas $1 / 8$ tetrahedral holes are occupied by $\mathrm{Zn}^{2+}$ ions and $50 \%$ of octahedral holes are occupied by $\mathrm{Fe}^{3+}$ ions. The formula of compound is
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$\mathrm{ZnFe}_2 \mathrm{O}_4$
In one unit cell, number of $\mathrm{O}^{2-}=4$
The number of $\mathrm{Zn}^{2+}=\frac{1}{8} \times 8=1$
The number of $\mathrm{Fe}^{3+}=\frac{1}{2} \times 4=2$
$\therefore$ Molecular formula of given spinel structure is $\mathrm{ZnFe}_2 \mathrm{O}_4$
The number of $\mathrm{Zn}^{2+}=\frac{1}{8} \times 8=1$
The number of $\mathrm{Fe}^{3+}=\frac{1}{2} \times 4=2$
$\therefore$ Molecular formula of given spinel structure is $\mathrm{ZnFe}_2 \mathrm{O}_4$
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