Question

In a normal spinel types structure, the oxide ions are arranged in $ccp$ whereas $1 / 8$ 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. $Z n F e_{2}O_{3}$
• C. $Z n F e_{2}O_{4}$
• D. $Z n F e_{2}O_{2}$

Answer 1

Answer: (C)

In one unit cell no. of $O^{2 +}=4$
The no. of $Z n^{2 +}=\frac{1}{8}\times \, 8=1$
The no. of $F e^{3 +}=\frac{1}{2}\times \, 4=2$
$\therefore $ molecular formula of given spinel structure is $Z n F e_{2}O_{4}$