A disproportionation reaction is a chemical reaction in which a single substance is simultaneously reduced and oxidized, forming two different products. To identify disproportionation reactions, one should look for a specified element in the reactant that gains electrons (reduction) and for the same element that lose electrons (oxidation).

Here's an analysis of each reaction :

(A) $\mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}$

Here, copper (Cu) is in the +1 oxidation state and forms two products, where it is oxidized to the +2 state in $\mathrm{Cu}^{2+}$ and reduced to the 0 state in Cu. This is a classic example of disproportionation. So, (A) is true for disproportionation.

(B) $3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{MnO}_4^{-}+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$

In this reaction, $\mathrm{MnO}_4^{2-}$ (where Mn is in +6 oxidation state) is converted into $\mathrm{MnO}_4^{-}$ (where Mn is in +7 oxidation state, oxidation has occurred) and $\mathrm{MnO}_2$ (where Mn is in +4 oxidation state, reduction has occurred). Here, again, we have the same element, manganese (Mn), undergoing both reduction and oxidation. Hence, (B) is also a disproportionation reaction.

(C) $2 \mathrm{KMnO}_4 \longrightarrow \mathrm{K}_2 \mathrm{MnO}_4+\mathrm{MnO}_2+\mathrm{O}_2$

For the reaction with potassium permanganate, $\mathrm{KMnO}_4$, manganese starts in the +7 oxidation state. It forms $\mathrm{K}_2 \mathrm{MnO}_4$ where Mn is in +6 state (reduction) and $\mathrm{MnO}_2$ where Mn is in +4 state (further reduction), but no increase in oxidation state of manganese is observed. Instead, oxygen is released, so the manganese is not being oxidized in any of its products; it is only reduced twice. This is not a disproportionation reaction because the same element should be undergoing both oxidation and reduction, which is not the case here. Hence, (C) is not a disproportionation reaction.

(D) $2 \mathrm{MnO}_4^{-}+3 \mathrm{Mn}^{2+}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 5 \mathrm{MnO}_2+4 \mathrm{H}^{+}$

In this reaction, the $\mathrm{MnO}_4^{-}$ ion (where Mn is at +7 oxidation state) and $\mathrm{Mn}^{2+}$ ion (where Mn is at +2 oxidation state) react to form $\mathrm{MnO}_2$ (where Mn is at +4 oxidation state). This reaction is not a case of disproportionation; it is a comproportionation or synproportionation reaction, where two species with the same element in different oxidation states produce a product with the element at an intermediate oxidation state.

Therefore, the disproportionation reactions among the given choices are:

• (A) $\mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}$


• (B) $3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{MnO}_4^{-}+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}$

Option A would be the correct choice: (A), (B).