# What is the oxidation number of manganese in MnO4-, Mn2O3, and MnO2?

Oct 21, 2015

Here's what you have here.

#### Explanation:

I'll show you how to find manganese's oxidation state in the first two compounds, and leave the last one to you as practice.

In each of those three cases, you can determine the oxidation state of manganese by using the known oxidation state of oxygen and the overall charge of the ion, when that is the case.

Oxygen has a $\text{(-2)}$ oxidation state in these compounds. So, you're dealing with

• the permanganate ion, ${\text{MnO}}_{4}^{-}$

Here, the ion has an overall $\text{(1-)}$ charge, which means that the oxidation numbers of all the atoms that make up the permanganate ion must add up to give $- 1$.

This means that you have - do not forget that you have four oxygen atoms in the permanganate ion!

$O {N}_{\text{Mn" + 4 xx ON_"oxygen}} = - 1$

$O {N}_{\text{Mn}} = - 1 - 4 \times \left(- 2\right)$

$O {N}_{\text{Mn}} = - 1 + 8 = \textcolor{g r e e n}{+ 7}$

In the permanganate anion, manganese has a $\text{(+7)}$ oxidation state.

• manganese(III) oxide, ${\text{Mn"_2"O}}_{3}$

The same strategy applies, only this time you're dealing with a neutral compound, which means that the oxydation numbers of all the atoms must add up to give zero.

This means that you have - keep in mind that you get two manganese atoms and three oxygen atoms in manganese(III) oxide!

$2 \times O {N}_{\text{MN" + 3 xx ON_"oxygen}} = 0$

$2 \times O {N}_{\text{Mn" = 0 - 3 xx ON_"oxygen}}$

$O {N}_{\text{Mn}} = \frac{0 - 3 \times \left(- 2\right)}{2} = \frac{6}{2} = \textcolor{g r e e n}{3}$

Manganese has a $\text{(+3)}$ oxidation state in manganese(III) oxide. Notice that the compound's name illustrates that - the Roman numeral $\text{(III)}$ tells you what the oxidation state of manganese is.

So, use the exact same approach to find the oxidation state of manganese in manganese dioxide, ${\text{MnO}}_{2}$.