In the reaction
`SO_(2)+2H_(2)Srarr3S+2H_(2)O`
the substance oxidised is

Options

A. `H_(2)S`
B. `SO_(2)`
C. `S`
D. `H_(2)O`

Explanation

To determine the substance that is oxidized in the reaction \( SO_2 + 2H_2S \rightarrow 3S + 2H_2O \), we need to analyze the oxidation states of the elements involved in the reaction.

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
- Reactants: \( SO_2 \) (sulfur dioxide) and \( H_2S \) (hydrogen sulfide).
- Products: \( S \) (sulfur) and \( H_2O \) (water).

2. **Assign Oxidation States**:
- For \( SO_2 \):
- Sulfur (S) has an oxidation state of +4.
- Oxygen (O) has an oxidation state of -2 (there are two oxygens, contributing -4 total).
- For \( H_2S \):
- Sulfur (S) has an oxidation state of -2.
- Hydrogen (H) has an oxidation state of +1 (there are two hydrogens, contributing +2 total).
- For elemental sulfur \( S \):
- The oxidation state is 0.
- For \( H_2O \):
- Oxygen (O) has an oxidation state of -2.
- Hydrogen (H) has an oxidation state of +1 (two hydrogens contribute +2 total).

3. **Determine Changes in Oxidation States**:
- In \( SO_2 \), sulfur changes from +4 to 0 in elemental sulfur \( S \).
- In \( H_2S \), sulfur changes from -2 to 0 in elemental sulfur \( S \).

4. **Identify the Substance Being Oxidized**:
- Oxidation is defined as the loss of electrons or an increase in oxidation state.
- The sulfur in \( H_2S \) goes from -2 to 0, indicating it is losing electrons (being oxidized).
- Conversely, the sulfur in \( SO_2 \) goes from +4 to 0, indicating it is gaining electrons (being reduced).

5. **Conclusion**:
- The substance that is oxidized in this reaction is \( H_2S \).

### Final Answer:
The substance oxidized is \( H_2S \).

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