Oxidation state of Mn in MnO₄⁻
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:

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

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:

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