On treating a compound with warm dil. `H_2 SO_4`, gas X is evolved which turns `K_2 Cr_2 O_7` paper acidified with dil. `H_2 SO_4` to a green compound Y. X and Y respectively are -

Options

A. ` X = SO_2 , Y = Cr_2 O_3`
B. `X = SO_3, Y = Cr_2 O_3`
C. `X= SO_2 , Y = Cr_2 (SO_4)_3`
D. `X = SO_3 , Y = Cr_2 (SO_4)_3`

Answer

To solve the problem step by step, let's analyze the information given and derive the answer.

### Step 1: Identify the Reaction
We are treating a compound with warm dilute sulfuric acid (H₂SO₄). The compound likely contains sulfur, as we are looking for a gas that evolves and can react with K₂Cr₂O₇.

**Hint:** Look for compounds that release gases when treated with acids, particularly those containing sulfur.

### Step 2: Determine the Evolved Gas
When a compound containing sulfur is treated with dilute sulfuric acid, the most common gas that can evolve is sulfur dioxide (SO₂). This gas is known to reduce potassium dichromate (K₂Cr₂O₇) in acidic conditions, leading to a color change.

**Hint:** Recall that SO₂ is a reducing agent and can change the color of potassium dichromate from orange to green.

### Step 3: Reaction with K₂Cr₂O₇
The gas X (which we identified as SO₂) will react with the acidified K₂Cr₂O₇. The orange color of K₂Cr₂O₇ indicates the presence of Cr(VI), which is reduced to Cr(III) in the presence of SO₂, resulting in a green compound, which we can identify as chromium(III) sulfate (Cr₂(SO₄)₃).

**Hint:** Understand the oxidation states of chromium and how they change during the reaction.

### Step 4: Identify the Green Compound
The green compound Y formed from the reaction of SO₂ with K₂Cr₂O₇ in acidic medium is chromium(III) sulfate, Cr₂(SO₄)₃. This compound is responsible for the green color observed on the filter paper.

**Hint:** Remember that the color change indicates the formation of a new compound, which is often a product of a reduction reaction.

### Step 5: Conclusion
From the analysis:
- The gas X evolved is sulfur dioxide (SO₂).
- The green compound Y formed is chromium(III) sulfate (Cr₂(SO₄)₃).

Thus, the final answer is:
- X = SO₂
- Y = Cr₂(SO₄)₃

### Final Answer
X and Y respectively are:
- X = SO₂
- Y = Cr₂(SO₄)₃