Q.Which of the following reactions are disproportionation reactions?

Concept: Disproportionation Reaction – a reaction where the same element in one oxidation state simultaneously undergoes oxidation (to a higher state) and reduction (to a lower state).

Step 1: Check oxidation states for each reaction.

1. $Cu^+ \rightarrow Cu^{2+} + Cu$ $Cu^+$: +1 → $Cu^{2+}$: +2 (oxidation) and $Cu$: 0 (reduction). Same element, same starting state → disproportionation.
2. $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$ Mn in $MnO_4^{2-}$: +6 → in $MnO_4^-$: +7 (oxidation) and in $MnO_2$: +4 (reduction). Same element, same starting state → disproportionation.
3. $2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$ Mn in $KMnO_4$: +7 → in $K_2MnO_4$: +6 (reduction) and in $MnO_2$: +4 (reduction). Both products have lower oxidation states — no oxidation of Mn. Oxygen goes from –2 to 0, but that’s a different element. Not a disproportionation (it’s a decomposition). …

What is a disproportionation reaction?

In a disproportionation reaction, a single chemical species — containing an element in one oxidation state — splits into two different products: one where that element is in a higher oxidation state (oxidation) and another where it is in a lower oxidation state (reduction). The key is that the same element from the same starting species is both oxidised and reduced.

Let’s check each reaction by assigning oxidation states and seeing if the condition holds.

1. Reaction (a): $Cu^+ \rightarrow Cu^{2+} + Cu$

• Oxidation state of Cu in $Cu^+$: +1
• In $Cu^{2+}$: +2 (higher → oxidation)
• In $Cu$ (elemental): 0 (lower → reduction)

Both products come from the same starting ion $Cu^+$. So this is a classic disproportionation.

2. Reaction (b): $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$

• In $MnO_4^{2-}$ (manganate ion): let Mn = $x$, O = $-2$ each → $x + 4(-2) = -2$ → $x = +6$
• In $MnO_4^-$ (permanganate): $x + 4(-2) = -1$ → $x = +7$ (higher → oxidation)
• In $MnO_2$: $x + 2(-2) = 0$ → $x = +4$ (lower → reduction)

The same manganate ion ($\text{MnO}_4^{2-}$) is the only source of Mn here. Some of its Mn(+6) goes to +7, some to +4. This is a disproportionation.

3. Reaction (c): $2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$

• In $KMnO_4$: Mn = +7
• In $K_2MnO_4$: Mn = +6 (reduction)
• In $MnO_2$: Mn = +4 (reduction)
• In $O_2$: O = 0 (oxidation from O = $-2$ in $KMnO_4$)

Here, Mn is only reduced (from +7 to +6 and +4) — it does not get oxidised. The oxidation happens to oxygen, not to the same element. So this is not a disproportionation; it is a decomposition (thermal decomposition of $KMnO_4$).

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Method: Oxidation Number Analysis

In a disproportionation reaction, the same element in one oxidation state simultaneously undergoes both oxidation and reduction — its oxidation number both increases and decreases.

Steps:

1. Assign oxidation numbers to each element in the reaction.
2. Identify the element that appears in at least three species (reactant, two products).
3. Check if that element’s oxidation number:
   • Increases in one product (oxidation)
   • Decreases in another product (reduction)
4. If yes → disproportionation. If no → not a disproportionation.

Applying to each reaction:

(a) $Cu^+ \rightarrow Cu^{2+} + Cu$

• Cu in $Cu^+$: +1
• Cu in $Cu^{2+}$: +2 (increase → oxidation)
• Cu in $Cu$: 0 (decrease → reduction)
• ✓ Disproportionation

(b) $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$

• Mn in $MnO_4^{2-}$: +6
• Mn in $MnO_4^-$: +7 (increase → oxidation)
• Mn in $MnO_2$: +4 (decrease → reduction)
• ✓ Disproportionation

(c) $2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$

• Mn in $KMnO_4$: +7
• Mn in $K_2MnO_4$: +6 (decrease)
• Mn in $MnO_2$: +4 (decrease)
• Both products show reduction only — no oxidation of Mn. …

Common Mistakes in Identifying Disproportionation Reactions

Mistake 1: Confusing Disproportionation with Decomposition

The Error: Many students think reaction (c) $2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$ is a disproportionation because one reactant ($KMnO_4$) gives two different products.

Why It's Wrong: In a disproportionation reaction, the same element in one reactant undergoes both oxidation and reduction. Here, oxygen in $KMnO_4$ is getting oxidized (from $-2$ to $0$ in $O_2$), while Mn is getting reduced (from $+7$ to $+6$ and $+4$). But the Mn oxidation and reduction happen from different Mn atoms in the same compound — this is actually a decomposition (thermal decomposition of $KMnO_4$), not a true disproportionation.

How to Avoid: Check if the same element from the same reactant species undergoes both changes. In (c), the Mn changes are from different atoms in the same molecule — that's decomposition, not disproportionation.

Mistake 2: Thinking "One Reactant → Two Products" Always Means Disproportionation

The Error: Students see reaction (a) $Cu^+ \rightarrow Cu^{2+} + Cu$ and think "one reactant, two products = disproportionation" — which is correct here, but they apply this blindly to (c) as well.

Why It's Wrong: In (a), the same Cu⁺ ion undergoes:

• Oxidation: $Cu^+ \rightarrow Cu^{2+} + e^-$ (loss of electron)
• Reduction: $Cu^+ + e^- \rightarrow Cu$ (gain of electron)

This is true disproportionation because the same species ($Cu^+$) is both oxidized and reduced.

How to Avoid: Always assign oxidation numbers to the element in question. If the same element in the same reactant shows both an increase and decrease in oxidation number, it's disproportionation.

Mistake 3: Ignoring the "Same Element in Same Reactant" Rule

The Error: Students mark reaction (b) $3MnO_4^{2-} + 4H^+ \rightarrow 2MnO_4^- + MnO_2 + 2H_2O$ as not a disproportionation because there are multiple reactants.

Why It's Wrong: The key is that Mn in $MnO_4^{2-}$ (oxidation state $+6$) is the only source of Mn. Some Mn atoms go to $+7$ (in $MnO_4^-$) and some to $+4$ (in $MnO_2$). The $H^+$ is just a spectator — it doesn't participate in the redox change.

How to Avoid: Focus only on the element that changes oxidation state. Ignore other reactants that don't undergo redox. Here, Mn in $MnO_4^{2-}$ is the only species undergoing both oxidation and reduction.

Mistake 4: Misidentifying Reaction (d) as Disproportionation

The Error: Students see Mn in different oxidation states and assume it's a disproportionation.

Why It's Wrong: In (d) $2MnO_4^- + 3Mn^{2+} + 2H_2O \rightarrow 5MnO_2 + 4H^+$:

• $MnO_4^-$ (Mn $+7$) is reduced to $MnO_2$ (Mn $+4$)
• $Mn^{2+}$ (Mn $+2$) is oxidized to $MnO_2$ (Mn $+4$)

This is a comproportionation reaction — two different oxidation states of the same element combine to form a single intermediate oxidation state. It's the opposite of disproportionation.

How to Avoid:

• Disproportionation: One oxidation state → two different states (one higher, one lower)
• Comproportionation: Two different oxidation states → one intermediate state

Mistake 5: Forgetting to Check All Species in the Reaction …