Question

In a normal spinel type structure, the oxide ions are arranged in ccp whereas $\frac{1}{8}$ tetrahedral holes are occupied by $Z{n}^{2+}$ ions and 50% of octahedral holes are occupied by $F{e}^{3+}$ ions. The formula of the compound is ?

• A. $Z{n}_{2}F{e}_2{O}_4$
• B. $ZnF{e}_2{O}_3$
• C. $Z{n}_{2}F{e}_2{O}_4$
• D. $Z{n}_{2}F{e}_2{O}_2$

Answer 1

The correct option is C

$Z{n}_{2}F{e}_2{O}_4$

Number of O - atoms per unit cell $=\frac{1}{8}\times 8+\frac{1}{2}\times 6=4$
Number of octahedral holes per unit cell $=1\times 4=4$
Number of $F{e}^{3+}$ ions per unit cell $=\frac{50\times 4}{100}=2$
Number of $Z{n}^{2+}$ ions per unit cell $=\frac{1}{8}\times 8=1$
Hence, formula : $ZnF{e}_2{O}_4$