n-Factor of `KMnO_(4)` in neutral medium is

Options

A. 6
B. 5
C. 4
D. 3

Answer

To find the n-factor of KMnO₄ in a neutral medium, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definition of n-Factor**:
The n-factor in a redox reaction is defined as the number of moles of electrons lost or gained per mole of the substance. In this case, we are looking for the n-factor of KMnO₄.

2. **Identify the Reaction in Neutral Medium**:
In a neutral medium, KMnO₄ reacts to form MnO₂. The balanced reaction can be represented as:
\[ KMnO_4 + OH^- \rightarrow MnO_2 + K^+ + H_2O \]

3. **Determine the Oxidation States**:
- For KMnO₄, the oxidation state of manganese (Mn) can be calculated as follows:
\[ x - 8 = -1 \quad \text{(where x is the oxidation state of Mn)} \]
Solving gives:
\[ x = +7 \]
- For MnO₂, the oxidation state of Mn is:
\[ x - 4 = 0 \]
Solving gives:
\[ x = +4 \]

4. **Calculate the Change in Oxidation State**:
The change in oxidation state for Mn from KMnO₄ to MnO₂ is:
\[ +7 \text{ (in KMnO₄)} \rightarrow +4 \text{ (in MnO₂)} \]
This indicates that Mn is reduced, gaining electrons.

5. **Determine the Number of Electrons Gained**:
The change in oxidation state from +7 to +4 represents a gain of 3 electrons:
\[ \text{Change} = 7 - 4 = 3 \text{ electrons} \]

6. **Calculate the n-Factor**:
Since there is one manganese atom in KMnO₄, the n-factor is equal to the number of electrons gained:
\[ \text{n-factor} = 3 \]

### Final Answer:
The n-factor of KMnO₄ in a neutral medium is **3**.

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