Question

What is the oxidation number of `"Mn"` in `"KMnO"_4`?

Answer 1

The oxidation number of `"Mn"` is +7.

Explanation:

We assign the oxidation number per a set of rules.

For this question, the important rules are:

1. The oxidation number of `"O"` in compounds is usually -2, but it is -1 in peroxides.

2. The oxidation number of a Group 1 element in a compound is +1.

3. The sum of the oxidation numbers of all of the atoms in a neutral compound is 0.

Rule 1 states that the oxidation number of `"O"` is -2.

We write the oxidation number of the element above its symbol and the total for 3 `"O"` atoms below the symbol.

This gives `"KMn"stackrelcolor(blue)("-2")("O")_4`
`color(white)(mmmmmmmll)stackrelcolor(blue)("-8"color(white)(mm))`

Rule 2 states that the oxidation number of K is +1.

This gives `stackrelcolor(blue)("+1")("K")"Mn"stackrelcolor(blue)("-2")("O")_4`
`color(white)(mmll)stackrelcolor(blue)("+1")color(white)(mmmmm)stackrelcolor(blue)("-8")color(white)(ml)`

Rule 3 states the numbers along the bottom must add up to zero.

The number under `"Mn"` must be +7.

This gives `stackrelcolor(blue)("+1")("K")"Mn"stackrelcolor(blue)("-2")("O")_4`
`color(white)(mmmml)stackrelcolor(blue)("+1")color(white)(mm)stackrelcolor(blue)("+7")(color(white)(ll))stackrelcolor(blue)("-8")color(white)(ml)`

There is only one `"Mn"` atom, so its oxidation number is +7.

This gives `stackrelcolor(blue)("+1")("K")stackrelcolor(blue)("+7")"Mn"stackrelcolor(blue)("-2")("O")_4`
`color(white)(mmmml)stackrelcolor(blue)("+1")color(white)(mm)stackrelcolor(blue)("+7")(color(white)(ll))stackrelcolor(blue)("-8")color(white)(ml)`

Answer 2

The metal has a `+VII` oxidation state in `MnO_4^-`, `"permanganate ion"`.

Explanation:

See this link for more of the same. Remember that oxidation number is a factitious entity, however, it does help us to balance redox reactions.

[And also here.](https://socratic.org/questions/how-do-you-find-the-oxidation-number-of-the-central-atom)