# What is the oxidation number of the Mn in KMnO_4?

Feb 10, 2017

+7

#### Explanation:

As a rule, except for peroxides and superoxides, oxygen is always -2. so if we have 4 oxygens, that is a total of -8.

As a rule alkali metals (group 1 except H) are always in the +1 oxidation state. There is one group 1 metal, K. So that's +1

In order to balance out the total charge to give a net of zero, we have to balance $+ 1 - 8 = - 7$ charges. Therefore the Mn has to be +7.

To solve this sort of question, you need to know the common oxidation states of group 1, group 2, group 17, and period 2 elements. Those don't change, except in rare circumstances. Everything else you can get by subtraction of the known oxidation states.

For example, $S i {O}_{2}$ the silicon is +4.

Feb 10, 2017

We have $M n \left(+ V I I\right)$............

#### Explanation:

Permanganate ion is a potent oxidizing agent, and it is (typically) conceived to accept 5 electrons upon reduction:

$M n {O}_{4}^{-} + 8 {H}^{+} + 5 {e}^{-} \rightarrow M {n}^{2 +} + 4 {H}_{2} O$

This reaction is self-indicating in that permanganate is brightly coloured (deep-red), whereas $M {n}^{2 +}$ is almost colourless. Are charge and mass balanced here?

As to how you calculate the oxidation number, the sum of the oxidation numbers is equal to the charge of the ions, here $- 1$. Now the oxidation number of oxygen is usually $- I I$, and it is here.

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

And thus $M {n}_{\text{oxidation number}} = + V I I$.