Reduction of the metal centre in aqueous permanganate ion involves

Options

A. 3 electrons in neutral medium
B. 5 electrons in neutral medium
C. 3 electrons in alkaline medium
D. 5 electrons in alkaline medium

Explanation

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

### Step-by-Step Solution:

1. **Identify the Species**: The species in question is the permanganate ion, \( \text{MnO}_4^- \).

2. **Determine the Oxidation State of Manganese in Permanganate**:
- In \( \text{MnO}_4^- \), let the oxidation state of manganese be \( x \).
- The equation is:
\[ x + 4(-2) = -1 \quad \text{(since the overall charge is -1)} \]
- Solving for \( x \):
\[ x - 8 = -1 \implies x = +7 \]

3. **Reduction in Acidic Medium**:
- In acidic medium, \( \text{MnO}_4^- \) is reduced to \( \text{Mn}^{2+} \).
- The oxidation state of manganese in \( \text{Mn}^{2+} \) is +2.
- The change in oxidation state is:
\[ 7 \to 2 \quad \text{(change of 5)} \]
- Therefore, 5 electrons are involved in this reduction.

4. **Reduction in Basic Medium**:
- In basic medium, \( \text{MnO}_4^- \) is reduced to \( \text{MnO}_4^{2-} \).
- The oxidation state of manganese in \( \text{MnO}_4^{2-} \) is calculated as follows:
\[ x + 4(-2) = -2 \quad \text{(since the overall charge is -2)} \]
- Solving for \( x \):
\[ x - 8 = -2 \implies x = +6 \]
- The change in oxidation state is:
\[ 7 \to 6 \quad \text{(change of 1)} \]
- Therefore, 1 electron is involved in this reduction.

5. **Reduction in Neutral Medium**:
- In neutral medium, \( \text{MnO}_4^- \) is reduced to \( \text{MnO}_2 \).
- The oxidation state of manganese in \( \text{MnO}_2 \) is:
\[ x + 2(-2) = 0 \quad \text{(since the overall charge is 0)} \]
- Solving for \( x \):
\[ x - 4 = 0 \implies x = +4 \]
- The change in oxidation state is:
\[ 7 \to 4 \quad \text{(change of 3)} \]
- Therefore, 3 electrons are involved in this reduction.

### Conclusion:
- In acidic medium, 5 electrons are involved.
- In basic medium, 1 electron is involved.
- In neutral medium, 3 electrons are involved.

### Final Answer:
The reduction of the metal center in aqueous permanganate ion involves:
- 5 electrons in acidic medium,
- 1 electron in basic medium,
- 3 electrons in neutral medium.