Manganese (VI) has ability to disproportionate in acidic solution. The difference in oxidation states of two ions it forms in acidic solution is______

${\mathrm{MnO}}_4^{2-}$ disproportionate in a neutral or acidic solution to give ${\mathrm{MnO}}_4^{-}$ and ${\mathrm{Mn}}^{+4}$

$$\begin{gathered} 3{\mathrm{MnO}}_4^{2-}+4{\mathrm{H}}^{+}⟶2{\mathrm{MnO}}_4^{-}+{\mathrm{MnO}}_2+2{\mathrm{H}}_2\mathrm{O} \\ \text{O.S. of }\mathrm{Mn} \text{in }{\mathrm{MnO}}_4^{-}=+7 \\ \text{ O.S. of }\mathrm{Mn}\text{ in }{\mathrm{MnO}}_2=+4 \\ \text{ difference }=3 \end{gathered}$$