# What is the transition-metal oxidation state in potassium permanganate, K^(+)[MnO_4]^(-)?

May 27, 2017

For $M n {O}_{4}^{-}$, we has $M n \left(V I I +\right)$..........

#### Explanation:

Manganese is an element with a very large oxidation manifold. And as always, $\text{oxidation number}$ is the charge left on the central atom, when all the bonding atoms are removed, with the charge assigned to the most electronegative atom. Now oxygen generally assumes an oxidation state of $- I I$ in its compounds, and it certainly does so here. And thus:

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

Clearly manganese has a formal oxidation number of $+ V I I$.

And typically, $M n \left(V I I +\right)$ is reduced to $M n \left(I I +\right)$. And this is a highly useful redox couple, because while $M n {O}_{4}^{-}$ is deep red/purple in colour, $M {n}^{2 +}$ is almost colourless (because its ${d}^{5}$ spin state result in spin forbidden electronic transitions). And to represent the reduction, we could write the formal reduction reaction:

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 {e}^{-} \rightarrow M {n}^{2 +} + 4 {H}_{2} O \left(l\right)$