JEE Chemistry 2023 Question with Solution

Given: $K_2Cr_2O_7$ paper is acidified with dilute $H_2SO_4$.

Find: Which gas turns it green.

Acidified $K_2Cr_2O_7$ acts as an oxidizing agent. When it is exposed to $SO_2$, dichromate is reduced to $Cr^{3+}$ ions, which are green in colour.

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 $SO_2$, which corresponds to option D in the given options.

The key observation is the formation of $Cr^{3+}$ ions. Acidified dichromate contains chromium in the $+6$ oxidation state. On reaction with $SO_2$, chromium is reduced to the $+3$ oxidation state.

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