Reduction of metal centre in aqueous permanganate ion involves
(1) 5 electrons in neutral medium
(2) 5 electrons in acidic medium
(3) 3 electrons in neutral medium
(4) 3 electrons in alkaline medium

Options

A. 1, 2
B. 2, 3
C. 1, 3
D. 2, 3, 4

Answer

To solve the question regarding the reduction of the metal center in the aqueous permanganate ion, we will analyze the different mediums: acidic, neutral, and alkaline.

### Step-by-Step Solution:

1. **Understanding the Permanganate Ion**:
- The permanganate ion is represented as \( \text{MnO}_4^- \).
- In this ion, manganese (Mn) is in the +7 oxidation state.

2. **Reduction in Acidic Medium**:
- In acidic medium, the reduction of permanganate ion can be represented as:
\[ \text{MnO}_4^- + 8 \text{H}^+ + 5 \text{e}^- \rightarrow \text{Mn}^{2+} + 4 \text{H}_2\text{O} \]
- Here, 5 electrons are required to reduce Mn from +7 to +2.

3. **Reduction in Neutral Medium**:
- In neutral medium, the reduction can be represented as:
\[ \text{MnO}_4^- + 3 \text{e}^- + 2 \text{H}_2\text{O} \rightarrow \text{MnO}_2 + 2 \text{OH}^- \]
- In this case, 3 electrons are needed to reduce Mn from +7 to +4.

4. **Reduction in Alkaline Medium**:
- In alkaline medium, the reduction can be represented as:
\[ \text{MnO}_4^- + \text{e}^- \rightarrow \text{MnO}_4^{2-} \]
- Here, only 1 electron is required to reduce Mn from +7 to +6.

5. **Conclusion**:
- From the analysis:
- In acidic medium: 5 electrons
- In neutral medium: 3 electrons
- In alkaline medium: 1 electron
- Therefore, the correct answer to the question is **(2) 5 electrons in acidic medium**.