On treating a compound with warm dil. H₂SO₄, gas X is evolved which turns K₂Cr₂O₇ paper acidified with dil. H₂SO₄ to a green compound Y. X and Y respectively are :

Solution

Chemistry ma’am · 4 steps

Dichromate turning green means something reduced it, and in acid the chromium stays dissolved as a sulphate.

1. Identify gas X. A sulphite (SO₃²⁻) reacts with warm dilute H₂SO₄ and gives off SO₂ gas: SO₃²⁻ + 2H⁺ → SO₂ + H₂O. Sulphur here is in the +4 state, so SO₂ can still be oxidised. That is why it works as a reducing agent.
2. What the paper shows. Orange Cr₂O₇²⁻ (Cr in +6) is reduced to green Cr³⁺ (Cr in +3), while SO₂ is oxidised to sulphate (S in +6). The paper is acidified with H₂SO₄, so the Cr³⁺ ends up as chromium(III) sulphate.

   K₂Cr₂O₇ + H₂SO₄ + 3SO₂ → K₂SO₄ + Cr₂(SO₄)₃ + H₂O

3. Check electrons. Three SO₂ lose 3 × 2 = 6 electrons. Two Cr atoms gain 2 × 3 = 6 electrons. The equation balances, and green Cr₂(SO₄)₃ is the product Y.
4. So. X is SO₂ and Y is Cr₂(SO₄)₃, which is option (d).

The idea to keep. A gas that turns acidified dichromate green is a reducing gas such as SO₂. In acid the green product is the Cr³⁺ salt of that acid, here Cr₂(SO₄)₃, not the oxide.

Answer from the NTA final answer key. Solution by PrepWiser.