What is the oxidation number of the ##"Mo"## in ##"MoO"_4^(2-)##?

The oxidation number of polyatomic ions is equal to the charge on the ion. The oxidation number of ##"Mo"## in ##"MoO"_4^(2-)## will be ##+6##

The of all the atoms in a compound must add up to the charge of that compound.

##"O"= -2##

##"Mo"= x##

The charge of the molybdate anion, ##"MoO"_4^(2-)##, is ##2-##, therefore

## x + 4* (-2) = -2##

##x-8 = -2 implies x = -2 + 8 = +6##

Therefore, ##"Mo"## has an oxidation number of ##+6## in the molybdate anion.