Question

Which of the following compounds represents an inverse $2:3$ spinel structure?

• A. $F{e}^{III}\left[F{e}^{II}F{e}^{III}\right]O_4$
• B. $Pb{O}_2$
• C. $A{l}_2{O}_3$
• D. $M{n}_3{O}_4$

Answer 1

The correct option is A $F{e}^{III}\left[F{e}^{II}F{e}^{III}\right]O_4$

Inverse spinel is generally denoted as $A^{+2}{B_2}^{+3}{O}_4$

$A^{+2}$ $\frac{1}{8}$ tetrahedral voids

$A^{+3}$ $\frac{1}{8}$ tetrahedral voids + $\frac{1}{4}$ octahedral voids

$O^{2-}$ at FCC lattice points.

In $F{e}^{+3}\left[F{e}^{+3}F{e}^{+2}\right]O_4$ or ${Fe}_3{O}_4$

$F{e}^{+3}$ present at $\frac{1}{8}$ tetrahedral void and $\frac{1}{4}$ octahedral void ${Z_{Fe}}^{+3}=2$

$F{e}^{+2}$ present at $\frac{1}{8}$ tetrahedral void ${Z_{Fe}}^{+2}=8\times \frac{1}{8}=1$

$O^{2-}$ at FCC lattice points $Z_{eff}=4$

Formula: ${Fe}^{+2}{Fe}_2^{+3}{O}_4\Rightarrow {Fe}_3{O}_4$

Hence, the correct option is $\text{A}$