On oxidation with `KMnO_(4)` in acidic medium , `SO_(2)` is oxidised to

Options

A. `SO_(2)`
B. `H_(2)SO_(4)`
C. `SO_(3)^(2-)`
D. `H_(2)S`

Answer

To solve the problem of what `SO2` is oxidized to when treated with `KMnO4` in acidic medium, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**: The reactants are `SO2` (sulfur dioxide) and `KMnO4` (potassium permanganate) in an acidic medium.

2. **Understand the Reaction**: In acidic medium, `KMnO4` acts as a strong oxidizing agent. It can oxidize `SO2` to a higher oxidation state.

3. **Determine the Oxidation States**:
- The oxidation state of sulfur in `SO2` is +4. This can be calculated as follows:
- Let the oxidation state of sulfur be `x`.
- The equation becomes: `x + 2(-2) = 0` (since there are two oxygen atoms, each with a -2 charge).
- Solving gives: `x - 4 = 0` → `x = +4`.
- In `H2SO4` (sulfuric acid), the oxidation state of sulfur is +6:
- The equation becomes: `x + 2(+1) + 4(-2) = 0`.
- Solving gives: `x + 2 - 8 = 0` → `x - 6 = 0` → `x = +6`.

4. **Write the Balanced Reaction**:
- When `KMnO4` is used in acidic medium, it produces `MnO2` and releases nascent oxygen.
- The balanced reaction for the oxidation of `SO2` can be represented as:
\[ 3SO2 + 2KMnO4 + 2H2SO4 \rightarrow 3H2SO4 + 2MnO2 + K2SO4 \]
- Here, `SO2` is oxidized to `H2SO4`.

5. **Conclusion**: Therefore, when `SO2` is oxidized by `KMnO4` in acidic medium, it is converted to `H2SO4`.

### Final Answer:
`SO2` is oxidized to `H2SO4`.

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