`KMnO_(4)` acts as on oxidising agent in alkaline medium. When alkaline `KMnO_(4)` is treated with `KI`, iodide ion is oxidised to `…………….`

Options

A. `I_2`
B. `IO^(-)`
C. `IO_3^(-)`
D. `IO_4^(-)`

Explanation

To solve the question regarding the oxidation of iodide ions by alkaline KMnO4, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants in this reaction are potassium permanganate (KMnO4) and potassium iodide (KI) in an alkaline medium.

2. **Understand the Role of KMnO4**:
KMnO4 is known to act as a strong oxidizing agent, especially in alkaline conditions. In this case, it will oxidize the iodide ions (I⁻) present in KI.

3. **Determine the Oxidation Process**:
The iodide ion (I⁻) is oxidized to iodate ion (IO3⁻). The oxidation state of iodine in I⁻ is -1, and in IO3⁻, it is +5.

4. **Write the Balanced Reaction**:
The balanced chemical reaction in alkaline medium can be expressed as:
\[ 2 KMnO_4 + 3 KI + 4 H_2O \rightarrow 2 MnO_2 + 3 KIO_3 + 4 KOH \]
Here, KMnO4 is reduced to MnO2, and I⁻ is oxidized to IO3⁻.

5. **Conclusion**:
Therefore, when alkaline KMnO4 reacts with KI, the iodide ion is oxidized to iodate ion (IO3⁻).

### Final Answer:
The iodide ion is oxidized to **IO3⁻** (iodate ion).

---