Chemistry Q70 – JEE-MAIN 2026

Home Chemistry Redox ReactionsQuestion #555

A paper dipped in a dil.

H_{2} {SO}_{4}

solution of 'X' upon treatment with

{SO}_{2}

gas turns into green. The compound 'X' is:

Clue 1 of 3: Redox Reaction Identification

The problem describes a color change from an unknown compound 'X' to green upon reaction with ${SO}_{2}$ in an acidic medium, indicating a redox reaction where 'X' is an oxidizing agent.

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Step 1: Identify the nature of the reaction✦ Active

The problem states that a solution of 'X' in dilute $H_{2} {SO}_{4}$ turns green upon treatment with ${SO}_{2}$ gas. Sulfur dioxide ( ${SO}_{2}$ ) typically acts as a reducing agent. Therefore, 'X' must be an oxidizing agent that gets reduced to a green product.

Step 2: Evaluate the options based on redox properties and color changes○ Expand

Let's examine the given options:

1. $KI-starch$ : Used to detect oxidizing agents (turns blue-black due to $I_{2}$ formation), not green. ${SO}_{2}$ would not oxidize $I^{-}$ .

2. ${KMnO}_{4}$ (Potassium Permanganate): A strong oxidizing agent (purple). In acidic medium, it is reduced to colorless ${Mn}^{2 +}$ ions, not green.

3. ${Pb(CH}_{3} {COO)}_{2}$ (Lead Acetate): Not a strong oxidizing agent and does not produce a green color with ${SO}_{2}$ .

4. $K_{2} {Cr}_{2} O_{7}$ (Potassium Dichromate): A strong oxidizing agent (orange). In acidic medium, it is reduced to green ${Cr}^{3 +}$ ions.

Step 3: Confirm the reaction for the correct option○ Expand

The reaction of dichromate with sulfur dioxide in acidic medium is:

{Cr}_{2} O_{7}^{2 -} (orange) + 3 {SO}_{2} + 2 H^{+} \to 2 {Cr}^{3 +} (green) + 3 {SO}_{4}^{2 -} + H_{2} O

This reaction perfectly matches the description in the question, where the orange dichromate solution turns green due to the formation of ${Cr}^{3 +}$ ions. Therefore, compound 'X' is $K_{2} {Cr}_{2} O_{7}$ .

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Chemistry Question 70 – JEE-MAIN 2026

Clue 1 of 3: Redox Reaction Identification

🧠 Full Solution Path