Question

Oxidation number of $Fe$ in $F{e_3}{O_4}$ is:
A: $\dfrac{1}{2}$
B: $\dfrac{2}{6}$
C: $\dfrac{8}{3}$
D: $\dfrac{3}{2}$

Answer 1

Hint: Oxidation number is defined as the charge which is present on an atom if the compound was ionic. A molecule is composed of two or more atoms. Among those atoms, the most electronegative element will possess a negative oxidative oxidation state.

Complete step by step answer:
We know that oxidation number is defined as the charge which is present on an atom if the compound was ionic. In other words we can say that the oxidation number of an element is defined as the number of electrons gained or lost by an atom in a compound. If an atom gains electrons then it is assigned negative oxidation number and when an atom loses electron it is assigned positive oxidation number. Electronegativity of an element is defined as the ability of that atom to attract a shared pair of electrons. In $F{e_3}{O_4}$ molecule there are three iron atoms and four oxygen atoms. Among $Fe$ and $O$ we know that $O$ is more electronegative. So, it will pull a shared pair of electrons towards itself. Since it is gaining electron oxidation, the number of oxygen will be negative. Oxidation state of one oxygen atom is $ - 2$. There are a total four oxygen atoms. so, the oxidation state of all the oxygen atoms in this molecule is $ - 8$. Let the oxidation number of one atom of $Fe$ be $x$. As a whole, the compound is neutral. So, the sum of oxidation numbers will be zero. Therefore,
$3x - 8 = 0$
$3x = 8$
$x = \dfrac{8}{3}$
So, the oxidation number of $Fe$ in $F{e_3}{O_4}$ is $\dfrac{8}{3}$ and the answer to this question is option C.

Note:
Oxygen is among one of the most electronegative elements. Fluorine is the only element which is more electronegative than oxygen. All the compounds of oxygen in which fluorine is not present, the oxidation number of a single atom of oxygen will be $ - 2$ but if oxygen and fluorine both are present in a compound then oxygen will possess a positive oxidation state.