`[X]+H_(2)SO_(4) to [Y]` a colourless gas with irritating smell, `[Y]+K_(2)Cr_(2)O_(7)+H_(2)SO_(4) to` green solution.
`[X]` and `[Y]` are respectively :

Options

A. `SO_(3)^(2-),SO_(2)`
B. `Cl^(-),HCl`
C. `S^(2-),H_(2)S`
D. `CO_(3)^(2-),CO_(2)`

Explanation

To solve the problem, we need to identify the substances [X] and [Y] based on the given reactions and their characteristics.

### Step-by-Step Solution:

1. **Identify the Colorless Gas with Irritating Smell:**
- The problem states that [Y] is a colorless gas with an irritating smell. The most common gas that fits this description is sulfur dioxide (SO₂).

2. **Determine the Precursor [X]:**
- The first reaction is given as [X] + H₂SO₄ → [Y]. Since we have established that [Y] is SO₂, we need to find [X].
- A common precursor that can produce SO₂ when reacted with sulfuric acid (H₂SO₄) is sulfite ions (SO₃²⁻). Therefore, [X] is likely to be sulfite (SO₃²⁻).

3. **Confirm the Reaction:**
- The reaction can be written as:
\[ \text{SO}_3^{2-} + \text{H}_2\text{SO}_4 \rightarrow \text{SO}_2 + \text{H}_2\text{O} \]
- This confirms that when sulfite reacts with sulfuric acid, it produces sulfur dioxide, which is a colorless gas with an irritating smell.

4. **Second Reaction with Potassium Dichromate:**
- The next part of the question states that [Y] reacts with K₂Cr₂O₇ and H₂SO₄ to produce a green solution.
- The reaction is:
\[ \text{SO}_2 + \text{K}_2\text{Cr}_2\text{O}_7 + \text{H}_2\text{SO}_4 \rightarrow \text{Cr}_2(SO_4)_3 + \text{K}_2\text{SO}_4 + \text{H}_2\text{O} \]
- In this reaction, sulfur dioxide (SO₂) reduces the dichromate ion (Cr₂O₇²⁻) to chromium(III) ions (Cr³⁺), which gives a green solution.

5. **Final Identification:**
- From the above deductions, we conclude that:
- [X] = SO₃²⁻ (sulfite ion)
- [Y] = SO₂ (sulfur dioxide)

### Final Answer:
- [X] is SO₃²⁻ (sulfite ion) and [Y] is SO₂ (sulfur dioxide).