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QJEEM24Ch44e2: JEE Main 2024 Chemistry - Class 11 JEEM24Ch44e2 Solution: Which of the following reactions are disproportionation…

Which of the following reactions are disproportionation reactions?

(A) Cu+→Cu2++Cu\mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}

(B) 3MnO42−+4H+⟶2<br><br>MnO4−+MnO2+2H2O3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow 2 <br><br>\mathrm{MnO}_4^{-}+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}

(C) 2KMnO4⟶K2MnO4+MnO2+O22 \mathrm{KMnO}_4 \longrightarrow \mathrm{K}_2 \mathrm{MnO}_4+\mathrm{MnO}_2+\mathrm{O}_2

(D) 2MnO4−+3Mn2++2H2O⟶5MnO2+4H+2 \mathrm{MnO}_4^{-}+3 \mathrm{Mn}^{2+}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow 5 \mathrm{MnO}_2+4 \mathrm{H}^{+}

Choose the correct answer from the options given below :

Multiple Choice Options

  • A (A), (B)
  • B (A), (D)
  • C (B), (C), (D)
  • D (A), (B), (C)

Step-by-Step Verified Solution

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) Cu+→Cu2++Cu\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 Cu2+\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) 3MnO42−+4H+⟶2MnO4−+MnO2+2H2O3 \mathrm{MnO}_4^{2-}+4 \mathrm{H}^{+} \longrightarrow 2 \mathrm{MnO}_4^{-}+\mathrm{MnO}_2+2 \mathrm{H}_2 \mathrm{O}

In this reaction, MnO42−\mathrm{MnO}_4^{2-} (where Mn is in +6 oxidation state) is converted into MnO4−\mathrm{MnO}_4^{-} (where Mn is in +7 oxidation state, oxidation has occurred) and MnO2\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) 2KMnO4⟶K2MnO4+MnO2+O22 \mathrm{KMnO}_4 \longrightarrow \mathrm{K}_2 \mathrm{MnO}_4+\mathrm{MnO}_2+\mathrm{O}_2

For the reaction with potassium permanganate, KMnO4\mathrm{KMnO}_4 , manganese starts in the +7 oxidation state. It forms K2MnO4\mathrm{K}_2 \mathrm{MnO}_4 where Mn is in +6 state (reduction) and MnO2\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) 2MnO4−+3Mn2++2H2O⟶5MnO2+4H+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 MnO4−\mathrm{MnO}_4^{-} ion (where Mn is at +7 oxidation state) and Mn2+\mathrm{Mn}^{2+} ion (where Mn is at +2 oxidation state) react to form MnO2\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) Cu+→Cu2++Cu\mathrm{Cu}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Cu}

  • (B) 3MnO42−+4H+⟶2MnO4−+MnO2+2H2O3 \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).

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