# What is the oxidation number of iodine in I_2?

The oxidation number of $I$ in the iodine molecule is $0$.
By definition, oxidation number is the charge left on the central atom when all its bonds are (conceptually) cleaved with the charge going to the most electronegative atom. So let's cleave the $I - I$ bond:
$I - I \rightarrow 2 I \cdot$ . We get two iodine atoms, two neutral iodine radicals, $2 \times I \cdot$. Since there is no charge, the oxidation number is a big fat $0$. On the other hand, when we do the same operation with the interhalogen $I - C l$, the charge goes to the more electronegative atom, $C l$:
$I - C l \rightarrow {I}^{+} + C {l}^{-}$. To give formal oxidation states of ${I}^{+}$ of iodine in the molecule, and ${I}^{-}$ for chlorine (note these are Roman numerals). The same oxidation states would pertain for $B r - C l$, as chlorine is more electronegative than bromine. What are the oxidation numbers of fluorine and oxygen in $O {F}_{2}$? Please report back the answers here.