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.
