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

Feb 20, 2017

Well, it has a $V I +$ oxidation state if it is actually $M n {O}_{4}^{2 -}$.

#### Explanation:

$\text{Manganate ion is}$ $M n {O}_{4}^{2 -}$, $M n \left(V I +\right)$.

$\text{Permanganate ion is}$ $M n {O}_{4}^{-}$, $M n \left(V I I +\right)$.

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 $- I I$ in its compounds, and it is here in both these examples.

Thus for $M n {O}_{4}^{2 -}$, $4 \times \left(- 2\right) + M {n}_{\text{oxidation number}} = - 2$.

And thus for $\text{permanganate}$, $M {n}_{\text{oxidation number}} + 4 \times \left(- 2\right) = - 1$

I leave it up to you to solve for $M {n}_{\text{oxidation number}}$ in each instance.

Both are known, and both are accessible oxidation states.