What is the oxidation state of Mn in product formed by the oxidising action of `KMnO_(4)` inneutral medium ?

Answer

To determine the oxidation state of manganese (Mn) in the product formed by the oxidizing action of potassium permanganate (KMnO4) in a neutral medium, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**: In a neutral medium, potassium permanganate (KMnO4) acts as an oxidizing agent. The product formed from this reaction is manganese dioxide (MnO2).

2. **Write the Balanced Reaction**: The balanced chemical equation for the reaction of KMnO4 in neutral medium can be written as:
\[ 2 KMnO_4 + 2 H_2O \rightarrow 2 MnO_2 + 2 KOH + O \]
Here, nascent oxygen (O) is also produced.

3. **Determine the Oxidation State of Mn in MnO2**:
- In MnO2, we need to find the oxidation state of manganese (Mn).
- Let the oxidation state of Mn be \( x \).
- The oxidation state of oxygen (O) is -2.
- Since there are two oxygen atoms, the total contribution from oxygen is \( 2 \times (-2) = -4 \).
- The compound is neutral, so the sum of oxidation states must equal zero:
\[ x + (-4) = 0 \]
- Solving for \( x \):
\[ x - 4 = 0 \implies x = +4 \]

4. **Conclusion**: The oxidation state of manganese (Mn) in the product MnO2 is +4.

### Final Answer:
The oxidation state of Mn in the product formed by the oxidizing action of KMnO4 in neutral medium is +4.