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 \( KMnO_4 \) in a neutral medium, we can follow these steps:

### Step 1: Identify the oxidation state of manganese in \( KMnO_4 \)
- The formula for \( KMnO_4 \) can be broken down as follows:
- Potassium (K) has a +1 oxidation state.
- Each oxygen (O) has a -2 oxidation state.

Using the formula:
\[ \text{Oxidation state of Mn} + 4 \times (-2) + 1 = 0 \]
Let the oxidation state of Mn be \( x \):
\[ x - 8 + 1 = 0 \implies x - 7 = 0 \implies x = +7 \]

### Step 2: Determine the product formed in neutral medium
- In a neutral medium, \( KMnO_4 \) is reduced to \( MnO_2 \).

### Step 3: Identify the oxidation state of manganese in \( MnO_2 \)
- In \( MnO_2 \):
\[ \text{Oxidation state of Mn} + 2 \times (-2) = 0 \]
Let the oxidation state of Mn be \( y \):
\[ y - 4 = 0 \implies y = +4 \]

### Step 4: Calculate the change in oxidation state
- The change in oxidation state of Mn from \( +7 \) to \( +4 \) is:
\[ \Delta \text{Oxidation State} = +7 - (+4) = +3 \]

### Step 5: Determine the number of electrons transferred
- Since the manganese is reduced, it accepts 3 electrons. Therefore, the number of electrons transferred is 3.

### Step 6: Calculate the n-factor
- The n-factor is defined as the total number of electrons transferred per formula unit of the compound. Since only one manganese atom is involved in this change, the n-factor of \( KMnO_4 \) in neutral medium is:
\[ \text{n-factor} = 3 \text{ (change in oxidation state)} \times 1 \text{ (number of manganese atoms)} = 3 \]

### Final Answer
The n-factor of \( KMnO_4 \) in neutral medium is **3**.

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