Question

Oxidation number of $Fe$ in $F{e}_3{O}_4$ are:

• A. +1 and +2
• B. +1 and +3
• C. +2 and +3
• D. +2 and +4

Answer 1

Answer: (C)

$F{e}_3{O}_4$ contains $Fe$ atoms of both $+2$ and $+3$ oxidation number.
It is a stoichiometric mixture of Ferrous $(FeO)$ and Ferric $\left(F{e}_2{O}_3\right)$
oxides combined as $FeO$. $F{e}_2{O}_3$.
Oxidation state of oxygen in $FeO$ and $F{e}_2{O}_3$ is $-2$.
Let the oxidation state of $Fe$ in $FeO$ and $F{e}_2{O}_3$ be $x$ and $y$ respectively.
As we know that the sum of the oxidation states of all the atoms or ions in a neutral compound is zero.
Therefore,
Oxidation state of $F$ e in $FeO-$
$x+(-2)=0$
$\Rightarrow x=2$
Oxidation state of $F$ in $F{e}_2{O}_3-$
$2(y)+3(-2)=0$
$\Rightarrow 2y=6$
$\Rightarrow y=3$
Hence the oxidation number of $Fe$ in $F{e}_3{O}_4$ are $+2$ and $+3$ respectively.