`KMnO_(4)` oxidises `I` - in acidic and neutralffaintly alkaline solution,respectively,to :

Options

A. `IO_(3^(-)) and IO_(3^(-))`,
B. `I_(2)` and `IO_(3)^(-),`
C. `I_(2) and I_(2),`
D. `IO_(3^(-)) and I_2,`

Answer

To solve the question regarding the oxidation of iodide ions (I-) by potassium permanganate (KMnO4) in different media, we need to analyze the reactions that occur in acidic and neutral or faintly alkaline solutions.

### Step-by-Step Solution:

1. **Identify the Medium**: The question specifies two different media: acidic and neutral (or faintly alkaline).

2. **Reaction in Acidic Medium**:
- In an acidic medium, potassium permanganate (KMnO4) acts as a strong oxidizing agent.
- The half-reaction for the oxidation of iodide ions (I-) in acidic conditions can be written as:
\[ 2MnO_4^- + 10I^- + 16H^+ \rightarrow 2Mn^{2+} + 5I_2 + 8H_2O \]
- Here, I- is oxidized to iodine (I2).

3. **Reaction in Neutral or Faintly Alkaline Medium**:
- In a neutral or faintly alkaline medium, the behavior of KMnO4 changes.
- The half-reaction in this medium can be represented as:
\[ 2MnO_4^- + 4I^- + 4H_2O \rightarrow 2MnO_2 + 2IO_3^- + 8OH^- \]
- In this case, iodide ions (I-) are oxidized to iodate ions (IO3^-).

4. **Conclusion**:
- In acidic medium, I- is oxidized to I2.
- In neutral or faintly alkaline medium, I- is oxidized to IO3^-.

### Final Answer:
- The oxidation products of I- by KMnO4 in acidic and neutral (or faintly alkaline) solutions are:
- **Acidic Medium**: I2
- **Neutral/Faintly Alkaline Medium**: IO3^-