1. Definition of disproportionation

A disproportionation reaction is one in which the same species undergoes both:

• oxidation and
• reduction

So, one element in a single oxidation state changes into two products containing that element in higher and lower oxidation states.

2. Check each reaction

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

Oxidation state of Cu on left: $+1$

On right:

• in $\mathrm{Cu}^{2+}$, Cu is $+2$
• in $\mathrm{Cu}$, Cu is $0$

So the same species $\mathrm{Cu}^+$ is:

• oxidized: $+1 \to +2$
• reduced: $+1 \to 0$

Hence, (A) is a disproportionation reaction.

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

Find oxidation state of Mn:

• In $\mathrm{MnO}_4^{2-}$: $x+4(-2)=-2 \Rightarrow x=+6$
• In $\mathrm{MnO}_4^{-}$: $x+4(-2)=-1 \Rightarrow x=+7$
• In $\mathrm{MnO}_2$: $x+2(-2)=0 \Rightarrow x=+4$

Thus Mn changes from $+6$ to:

• $+7$ (oxidation)
• $+4$ (reduction)

Same initial species $\mathrm{MnO}_4^{2-}$ undergoes both oxidation and reduction.

Hence, (B) is a disproportionation reaction.

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

Oxidation states:

• In $\mathrm{KMnO}_4$, Mn is $+7$
• In $\mathrm{K}_2\mathrm{MnO}_4$, Mn is $+6$
• In $\mathrm{MnO}_2$, Mn is $+4$

Here Mn only goes from $+7$ to lower oxidation states ($+6$ and $+4$), so Mn is only reduced, not both oxidized and reduced.

Oxygen goes from $-2$ in permanganate to $0$ in $\mathrm{O}_2$, so oxygen is oxidized.

Since the same species of the same element is not simultaneously oxidized and reduced, this is not disproportionation. It is an internal redox reaction.

Hence, (C) is not a disproportionation reaction.

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

Oxidation states of Mn:

• In $\mathrm{MnO}_4^{-}$, Mn = $+7$
• In $\mathrm{Mn}^{2+}$, Mn = $+2$
• In $\mathrm{MnO}_2$, Mn = $+4$

Here:

• Mn in $\mathrm{MnO}_4^{-}$ is reduced: $+7 \to +4$
• Mn in $\mathrm{Mn}^{2+}$ is oxidized: $+2 \to +4$

Two different initial oxidation states of Mn combine to give one intermediate oxidation state. This is comproportionation (or synproportionation), not disproportionation.

Hence, (D) is not a disproportionation reaction.

3. Final selection

Disproportionation reactions are:

• (A)
• (B)

So the correct option is:

$\boxed{\text{A: (A), (B)}}$

4. Comparison with stored answer

Stored correct answer: A

My derived answer: A

They match.