Sulphur on reaction with conc. `H_2SO_4` gives X which on reaction with NaOH gives Y which on further reaction with dil `H_2SO_4` give X again. Then X & Y are respectively:

Options

A. `S, Na_2SO_3`
B. `SO_2, Na_2SO_3`
C. `SO_2, Na_2SO_4`
D. `SO_3, Na_2SO_3`

Answer

To solve the problem, we need to analyze the reactions step by step.

### Step 1: Reaction of Sulphur with Concentrated H2SO4

When sulphur (S) reacts with concentrated sulfuric acid (H2SO4), it produces sulfur dioxide (SO2) gas. This is our first product, which we will denote as X.

**Reaction:**
\[ S + H_2SO_4 \rightarrow SO_2 + H_2O \]
Thus, we identify:
- **X = SO2**

### Step 2: Reaction of X (SO2) with NaOH

Next, we take the product X (SO2) and react it with sodium hydroxide (NaOH). The reaction of sulfur dioxide with sodium hydroxide leads to the formation of sodium sulfite (Na2SO3). This is our second product, which we will denote as Y.

**Reaction:**
\[ SO_2 + 2NaOH \rightarrow Na_2SO_3 + H_2O \]
Thus, we identify:
- **Y = Na2SO3**

### Step 3: Reaction of Y (Na2SO3) with Dilute H2SO4

Finally, we take the product Y (Na2SO3) and react it with dilute sulfuric acid (H2SO4). When sodium sulfite reacts with dilute sulfuric acid, it produces sulfur dioxide (SO2) again.

**Reaction:**
\[ Na_2SO_3 + H_2SO_4 \rightarrow SO_2 + Na_2SO_4 + H_2O \]
This confirms that we regenerate our original product X.

### Conclusion

From the above steps, we can summarize:
- **X = SO2**
- **Y = Na2SO3**

### Final Answer
Thus, the final answer is:
- **X = SO2**
- **Y = Na2SO3**

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