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

We need to find the difference in oxidation states of the two ions formed when Manganese(VI) disproportionates in acidic solution. Manganese in the +6 oxidation state exists as the manganate ion: $$MnO_4^{2-}$$ (where Mn is +6).

In disproportionation, the same species is simultaneously oxidized and reduced. In acidic solution, $$MnO_4^{2-}$$ disproportionates as:

$$3MnO_4^{2-} + 4H^+ \to 2MnO_4^- + MnO_2 + 2H_2O$$

In $$MnO_4^-$$ (permanganate ion): Let the oxidation state of Mn be $$y$$. Then $$y + 4(-2) = -1$$, giving $$y = +7$$. In $$MnO_2$$ (manganese dioxide): Let the oxidation state of Mn be $$z$$. Then $$z + 2(-2) = 0$$, giving $$z = +4$$.

Difference in oxidation states = $$+7 - (+4) = 3$$. The answer is 3.