In the reaction
`H_(2)S+NO_(2) rarr H_(2)O+NO+S.H_(2)S` is

Options

A. Oxidised
B. Reduced
C. Precipitated
D. None of these

Explanation

To determine whether H₂S is oxidized, reduced, or undergoing another process in the reaction \( H_2S + NO_2 \rightarrow H_2O + NO + S + H_2S \), we need to analyze the changes in oxidation states of the elements involved.

### Step 1: Assign Oxidation Numbers
1. **For H₂S (Hydrogen Sulfide)**:
- Hydrogen (H) has an oxidation state of +1.
- Let the oxidation state of sulfur (S) be \( x \).
- The overall charge of the molecule is 0, so:
\[ 2(+1) + x = 0 \implies x = -2 \]
- Therefore, the oxidation state of H₂S is:
- H: +1
- S: -2

2. **For H₂O (Water)**:
- Hydrogen (H) is +1.
- Oxygen (O) is -2.
- Thus, the oxidation states are:
- H: +1
- O: -2

3. **For NO (Nitric Oxide)**:
- Let the oxidation state of nitrogen (N) be \( y \).
- Oxygen (O) is -2.
- The overall charge of the molecule is 0, so:
\[ y + (-2) = 0 \implies y = +2 \]
- Therefore, the oxidation states are:
- N: +2
- O: -2

4. **For S (Sulfur)**:
- The elemental form of sulfur has an oxidation state of 0.

### Step 2: Analyze Changes in Oxidation Numbers
- **In H₂S**:
- Sulfur changes from -2 (in H₂S) to 0 (elemental S).
- This is an increase in oxidation state, indicating oxidation.

- **In NO₂**:
- Nitrogen in NO₂ has an oxidation state of +4 (since O is -2).
- In NO, nitrogen is +2.
- This indicates a decrease in oxidation state, indicating reduction.

### Step 3: Conclusion
Since sulfur in H₂S is oxidized (its oxidation state increases from -2 to 0), we conclude that:

**H₂S is being oxidized in the reaction.**

### Final Answer
H₂S is being oxidized.

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