Manganese (VI) has ability to disproportionate in acidic solution. The difference in oxidation states of two ions it forms in acidic solution is \underline\hspace2cm .

D and F Block Elements PYQ — JEE Main 2022 · 24 Jun · Shift 2 · Q18 with Solution

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Correct answer: 3

We need the disproportionation of manganese in oxidation state $+6$ in acidic medium.
The manganese(VI) ion is manganate, $\mathrm{MnO_4^{2-}}$ .
In acidic solution, manganate disproportionates into:
- permanganate, where manganese is in oxidation state $+7$
- manganese dioxide or ultimately $\mathrm{Mn^{2+}}$ depending on medium, but the standard disproportionation of manganate is: $3\mathrm{MnO_4^{2-}} + 4\mathrm{H^+} \rightarrow 2\mathrm{MnO_4^-} + \mathrm{MnO_2} + 2\mathrm{H_2O}$
Let us check oxidation states of manganese in the products:
- In $\mathrm{MnO_4^-}$ , Mn is $+7$
- In $\mathrm{MnO_2}$ , Mn is $+4$
Therefore, the two ions/species formed have oxidation states $+7$ and $+4$ .
Their difference is: $7 - 4 = 3$

So, the required integer answer is $3$ .