Comproportionation Reaction: The Opposite of Disproportionation | CHEMCA

Comproportionation Reaction: The Opposite of Disproportionation

Published by Abhishek Sengar | CHEMCA India

In the Redox Reactions chapter, almost every JEE and NEET student correctly memorizes what a Disproportionation reaction is. But examiners love to test the exact reverse concept to see who is truly paying attention!

If you see a reaction where the same element is both oxidizing and reducing, but it forms only one product, what do you call it? Let's decode this reaction step-by-step.

Video Tutorial: Identifying the Reaction

Watch Abhishek Sengar sir from CHEMCA break down the oxidation states of Bromine to prove why this is NOT a disproportionation reaction.

Step-by-Step Breakdown

The Reaction:

Br^- + BrO₃^- → Br₂

Calculate Oxidation States of Reactants:
- Bromide ion (Br^-): As a monatomic ion, its oxidation state is equal to its charge. OS = -1.
- Bromate ion (BrO₃^-): Let the oxidation state of Br be x. Oxygen is -2.
  x + 3(-2) = -1
  x - 6 = -1 → x = +5.
Calculate Oxidation State of Product:
Bromine gas (Br₂): Because it is in its elemental, uncombined state, its oxidation number is strictly 0.
Analyze the Electron Transfer:
Br^- goes from -1 to 0. This is an increase in oxidation number, meaning it undergoes Oxidation.
BrO₃^- goes from +5 to 0. This is a decrease in oxidation number, meaning it undergoes Reduction.

The Trap:
Because the same element (Bromine) is both oxidizing and reducing, students quickly rush to label this as "Disproportionation". WRONG!

In disproportionation, ONE intermediate oxidation state splits into TWO different (higher and lower) states. In our reaction, TWO extreme oxidation states are converging into ONE intermediate state. This is called a Comproportionation Reaction (also known as Synproportionation).

Fig: Two different oxidation states of the same element reacting to form a single intermediate state.

Summary Definition: Comproportionation is a chemical reaction where two reactants, each containing the same element but with different oxidation numbers, form a product in which the element reaches the exact same intermediate oxidation number.

Practice Questions for JEE & NEET

Make sure you can flawlessly distinguish between these two opposite reaction types with these conceptual checks!

Question 1: Using the exact same element (Bromine), write down a classic Disproportionation reaction to demonstrate the opposite effect.

Answer: 3Br₂ + 6OH^- → 5Br^- + BrO₃^- + 3H₂O

Reasoning:

When Bromine gas (Br₂) reacts with a hot, concentrated alkali (like NaOH), it splits into two different oxidation states:

Reactant: Br₂ (Oxidation State = 0)
Product 1: Br^- (Oxidation State = -1) → This is Reduction.
Product 2: BrO₃^- (Oxidation State = +5) → This is Oxidation.

Because ONE intermediate state (0) split into TWO divergent states (-1 and +5), this is the textbook definition of Disproportionation!

Question 2: Is the reaction Ag²⁺ + Ag(s) → 2Ag⁺ a Comproportionation or Disproportionation reaction?

Answer: Comproportionation.

Reasoning:

Let's map out the oxidation states of Silver (Ag):

Reactant 1: Ag²⁺ (Oxidation State = +2)
Reactant 2: Ag(s) (Oxidation State = 0)
Product: Ag⁺ (Oxidation State = +1)

Since the +2 state reduces to +1, and the 0 state oxidizes to +1, two extreme oxidation states have converged into a single intermediate state (+1). Therefore, it is a Comproportionation reaction!

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