# Oxidation Number Of Manganese In Mno4

+7

## 3 Answers 3 Sorted by:

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.

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