Question

$F{e}_{0.95}O$ can be due to presence of iron in $+2$ and $+3$ oxidation numbers. Then iron present in $+3$ oxidation state will be

Answer 1

Let $x$ atoms of $F{e}^{+3}$ be present. Then atoms of $F{e}^{+2}=(0.95-x)$

Since the net charge on the molecule is zero,

$\Rightarrow x\times (+3)+(0.95-x)\times 2+(-2)=0$

$\Rightarrow x=0.1$

Hence the compound may be represented as $Fe\left(III{)}_{0.1}Fe\right(II{)}_{0.85}O$