When `SO_2` is passed through a solution of `H_2 S` in water :

Options

A. Sulphuric acid is formed
B. A clear solution is formed
C. Sulphur acid is precipitated
D. No change is observed

Explanation

To solve the question regarding the reaction that occurs when \( SO_2 \) is passed through a solution of \( H_2S \) in water, we can follow these steps:

### Step 1: Identify the Reactants
The reactants in this scenario are sulfur dioxide (\( SO_2 \)) and hydrogen sulfide (\( H_2S \)).

### Step 2: Write the Balanced Chemical Equation
When \( SO_2 \) reacts with \( H_2S \), the reaction can be represented as:
\[ SO_2 + 2H_2S \rightarrow 3S + 2H_2O \]
In this reaction, sulfur (\( S \)) is formed as a precipitate, and water (\( H_2O \)) is produced.

### Step 3: Analyze the Reaction Type
This reaction is classified as a disproportionation reaction. In a disproportionation reaction, a single substance is both oxidized and reduced.

### Step 4: Determine the Oxidation States
- In \( SO_2 \), sulfur has an oxidation state of +4.
- In \( H_2S \), sulfur has an oxidation state of -2.
- In the products, elemental sulfur (\( S \)) has an oxidation state of 0.

### Step 5: Identify Oxidation and Reduction
- The sulfur in \( SO_2 \) is reduced from +4 to 0.
- The sulfur in \( H_2S \) is oxidized from -2 to 0.

### Step 6: Conclusion
Since sulfur is formed as a precipitate, the correct option regarding the outcome of this reaction is that sulfur is precipitated.

### Final Answer
The correct option is that sulfur is precipitated.

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