# What are the oxidation states of the atoms in a diatomic gas?

If it is elemental gas, the oxidation number is $0$.
When we do this for ${H}_{2}$, or ${X}_{2}$, or $N \equiv N$, or $O = O$, we get sharing of electrons, and a $0$ oxidation state. Of course, if it is a diatomic gas such as $H - C l$, we break the bond and the electronic charge is distributed to the most electronegative atom, here the halide, i.e. ${H}^{I +}$ and $C {l}^{- I}$. With an interhalogen, say $B r - C l$, we get $B {r}^{I +}$ and $C {l}^{- I}$