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MCQEasyJEE 2023Potassium Dichromate & Permanganate

JEE Chemistry 2023 Question with Solution

K2Cr2O7K_2Cr_2O_7 paper acidified with dilute H2SO4H_2SO_4 turns green when exposed to:

  • A

    Carbon dioxide

  • B

    Sulphur trioxide

  • C

    Hydrogen sulphide

  • D

    Sulphur dioxide

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: K2Cr2O7K_2Cr_2O_7 paper is acidified with dilute H2SO4H_2SO_4.

Find: Which gas turns it green.

Acidified K2Cr2O7K_2Cr_2O_7 acts as an oxidizing agent. When it is exposed to SO2SO_2, dichromate is reduced to Cr3+Cr^{3+} ions, which are green in colour.

Cr2O72−+3SO2+2H+→2Cr3++3SO42−+H2OCr_2O_7^{2-} + 3SO_2 + 2H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + H_2O

Therefore, the gas that turns the paper green is Sulphur dioxide, so the correct option is D.

The solution labels the correct option as A, but its own explanation and reaction clearly identify SO2SO_2, which corresponds to option D in the given options.

Why the colour changes

The key observation is the formation of Cr3+Cr^{3+} ions. Acidified dichromate contains chromium in the +6+6 oxidation state. On reaction with SO2SO_2, chromium is reduced to the +3+3 oxidation state.

The Cr3+Cr^{3+} ion has a green colour, so the paper turns green. Hence the test is positive for Sulphur dioxide.

Common mistakes

  • Confusing the oxidizing agent and reducing agent. Here acidified K2Cr2O7K_2Cr_2O_7 is reduced, while SO2SO_2 is oxidized. Track both species together to avoid reversing the reaction role.

  • Choosing Hydrogen sulphide because it is also a reducing gas. The provided solution specifically shows reduction by SO2SO_2 and the given options must be matched to that working, so select Sulphur dioxide.

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