Explanation

The given compound is an ionic compound

It is made up of:

• one potassium
• one manganese,
• four oxygen atoms.

Oxidation numbers of potassium (K) and oxygen (O) are +1 and -2 respectively.

Let us assume that the oxidation number of Mn is $x$.

Since ionic compounds are neutral, the net charge on them is zero.

(Oxidation number of K) + (Oxidation number of Mn) + (4 $\times$ The oxidation number of O) = $0$

(1)$\times$(K) + (1)$\times$(Mn) + 4$\times$(O) = $0$

$(1)$$\times$(+1) + (1)$\times$($x$) + 4$\times$(-2) = $0$

$x$ = +7

Hence, the oxidation number of $\text{Mn}$ in ${\text{KMnO}}_4$ is +7.