# Oxidation state of Mn in MnO₄⁻

I have tried to understand what the oxidation state for Mn in $$\ce{MnO4^-}$$ is. So far i have understood that O have the oxidation state of -2 so because there is 4 O the number turns to -8. What happens to the minus at O? and what is the oxidation state for Mn?

## 3 Answers

The sum of the oxidation states of all of the atoms needs to equal the total charge on the ion.

For example, for $$\ce{NH4+}$$, if each hydrogen atom has an oxidation state of +1, and the overall charge is +1, then we can solve for the oxidation state of nitrogen:

\begin{aligned} 4(+1) + OS_\ce{N} &= +1\\ OS_\ce{N} &= +1 -4(+1) \\ OS_\ce{N} &= -3 \end{aligned}

Similarly, we can do $$\ce{SO4^2-}$$. If each oxygen atom has an oxidation state of -2, and the overall charge is -2, we can solve for the oxidation state of sulfur:

\begin{aligned} 4(-2) + OS_\ce{S} &= -2\\ OS_\ce{S} &= -2 -4(-2) \\ OS_\ce{S} &= +6 \end{aligned}

Seeing as Oxygen has an oxidation state of -2 and you have 4 Oxygen atoms, the total charge contributed by Oxygen = -8. Because the total charge of this compound = -1, then in this case Mn = +7 +7 - 8 = -1

As you said Oxygen has the oxidation state of -2 so if we add up all the oxygens they have a -8 oxidation state.

As the oxidation state of the ion is equal to its charge, the Mn has to "balance" the -8 to get it up to -1. Therefore the oxidation state of Mn in Mno4- is +7.