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

Jan 15, 2017

The oxidation number of $\text{Mn}$ 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 $\text{O}$ 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 $\text{O}$ is -2.

We write the oxidation number of the element above its symbol and the total for 3 $\text{O}$ 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 $\stackrel{\textcolor{b l u e}{\text{+1")("K")"Mn"stackrelcolor(blue)("-2")("O}}}{_} 4$
$\textcolor{w h i t e}{m m l l} \stackrel{\textcolor{b l u e}{\text{+1")color(white)(mmmmm)stackrelcolor(blue)("-8}}}{\textcolor{w h i t e}{m l}}$

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

The number under $\text{Mn}$ must be +7.

This gives $\stackrel{\textcolor{b l u e}{\text{+1")("K")"Mn"stackrelcolor(blue)("-2")("O}}}{_} 4$
$\textcolor{w h i t e}{m m m m l} \stackrel{\textcolor{b l u e}{\text{+1")color(white)(mm)stackrelcolor(blue)("+7")(color(white)(ll))stackrelcolor(blue)("-8}}}{\textcolor{w h i t e}{m l}}$

There is only one $\text{Mn}$ atom, so its oxidation number is +7.

This gives $\stackrel{\textcolor{b l u e}{\text{+1")("K")stackrelcolor(blue)("+7")"Mn"stackrelcolor(blue)("-2")("O}}}{_} 4$
$\textcolor{w h i t e}{m m m m l} \stackrel{\textcolor{b l u e}{\text{+1")color(white)(mm)stackrelcolor(blue)("+7")(color(white)(ll))stackrelcolor(blue)("-8}}}{\textcolor{w h i t e}{m l}}$

Jan 19, 2017

The metal has a $+ V I I$ oxidation state in $M n {O}_{4}^{-}$, $\text{permanganate ion}$.

#### 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.