# Question #f5981

The oxidation number of $M n$ in $M n C {l}_{2}$ is +2 while its oxidation number in $M n {O}_{2}$ is +4.
With a few exceptions, the oxidation number of an element is equal to its charge. In $M n C {l}_{2}$, the charge/oxidation number of the $C l$ atom is -1 since it is in the halogen family, and since the molecule $M n C {l}_{2}$ has 2 $C l$ atoms, it has a -2 oxidation number. Assuming your compound is a neutral compound with no overall charge, $M n$ would have a +2 oxidation number in the compound.
For $M n {O}_{2}$, $O$ has a -2 charge/oxidation number, so 2 $O$ atoms would have a charge of -4. This means $M n$ has an oxidation number of +4 in order to make the neutral compound.