Question

How do you find the oxidation numbers of the elements in `MnO4^(2-)`?

Answer 1

Well, it has a `VI+` oxidation state if it is actually `MnO_4^(2-)`.

Explanation:

`"Manganate ion is"` `MnO_4^(2-)`, `Mn(VI+)`.

`"Permanganate ion is"` `MnO_4^(-)`, `Mn(VII+)`.

In both instances, the sum of the oxidation numbers of the individual elements must equal the charge on the ion. The oxidation number of oxygen is usually `-II` in its compounds, and it is here in both these examples.

Thus for `MnO_4^(2-)`, `4xx(-2)+Mn_"oxidation number"=-2`.

And thus for `"permanganate"`, `Mn_"oxidation number"+4xx(-2)=-1`

I leave it up to you to solve for `Mn_"oxidation number"` in each instance.

Both are known, and both are accessible oxidation states.