When `I^(-)` is oxidised by `MnO_(4)^(-)` in an alkaline medium, `I^(-)` converts into

Options

A. `IO_(3)^(-)`
B. `I_(2)`
C. `IO_(4)^(-)`
D. `IO^(-)`

Answer

To solve the question, we need to analyze the oxidation of iodide ion (I⁻) by permanganate ion (MnO₄⁻) in an alkaline medium.

### Step-by-Step Solution:

1. **Identify the Reactants and Medium**:
- The reactants are iodide ion (I⁻) and permanganate ion (MnO₄⁻).
- The medium is alkaline.

2. **Understand the Role of MnO₄⁻**:
- In an alkaline medium, permanganate ion (MnO₄⁻) acts as an oxidizing agent. This means it will facilitate the oxidation of iodide ions.

3. **Determine the Oxidation Product of I⁻**:
- The iodide ion (I⁻) is oxidized to iodate ion (IO₃⁻) in alkaline conditions. This is a common oxidation reaction where iodide is converted to iodate.

4. **Identify the Reduction Product of MnO₄⁻**:
- While iodide is being oxidized, the permanganate ion (MnO₄⁻) is reduced. In an alkaline medium, MnO₄⁻ is typically reduced to manganese dioxide (MnO₂).

5. **Write the Overall Reaction**:
- The overall reaction can be summarized as:
\[ 3 I^- + 2 MnO_4^- + 4 OH^- \rightarrow 3 IO_3^- + 2 MnO_2 + 2 H_2O \]
- This shows that iodide ions are converted to iodate ions while permanganate ions are reduced to manganese dioxide.

6. **Conclusion**:
- Therefore, when I⁻ is oxidized by MnO₄⁻ in an alkaline medium, I⁻ is converted into IO₃⁻.

### Final Answer:
I⁻ is converted into IO₃⁻.