When `KMnO_4` reacts with `I^-` in a slightly alkaline and acidic medium, the respective products obtained are

Options

A. `Mn^(2+) + IO_3^(-),MnO_2 + I_2`
B. `MnO_2 + I_2 , Mn^(2+) + IO_3^(-)`
C. `MnO_2 + IO_3^(-), Mn^(2+) + I_2`
D. `Mn^(2+) + I_2 , MnO_2 + IO_3^(-)`

Answer

To determine the products obtained when potassium permanganate (KMnO4) reacts with iodide ions (I^-) in slightly alkaline and acidic mediums, we can analyze each reaction step by step.

### Step 1: Reaction in Slightly Alkaline Medium
1. **Identify the reactants**: KMnO4 and I^- in a basic medium.
2. **Write the balanced chemical equation**:
- The reaction can be represented as:
\[ 2 \text{MnO}_4^- + \text{H}_2\text{O} + 2 \text{I}^- \rightarrow 2 \text{MnO}_2 + 2 \text{OH}^- + \text{IO}_3^- \]
3. **Identify the products**:
- In this reaction, the iodide ion (I^-) is oxidized to iodate ion (IO3^-), and manganese is reduced to manganese dioxide (MnO2).

### Step 2: Reaction in Acidic Medium
1. **Identify the reactants**: KMnO4 and I^- in an acidic medium.
2. **Write the balanced chemical equation**:
- The reaction can be represented as:
\[ 10 \text{I}^- + 2 \text{MnO}_4^- + 16 \text{H}^+ \rightarrow 2 \text{Mn}^{2+} + 8 \text{H}_2\text{O} + 5 \text{I}_2 \]
3. **Identify the products**:
- In this reaction, the iodide ions (I^-) are oxidized to iodine (I2), and manganese is reduced to manganese ions (Mn^2+).

### Conclusion
- In slightly alkaline medium, the products are:
- MnO2 and IO3^-
- In acidic medium, the products are:
- Mn^2+ and I2

Thus, the final answer is:
- **Products in alkaline medium**: MnO2 and IO3^-
- **Products in acidic medium**: Mn^2+ and I2

### Final Answer:
The respective products obtained are:
- **MnO2 + IO3^- (in alkaline medium)**, **Mn^2+ + I2 (in acidic medium)**.

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