The product obtained on treating `KMnO_(4)` with very strong alkali in absence of any reducing agent is

Options

A. `MnO_(2)`
B. `Mn_(2)O_(7)`
C. `K_(2)MnO_(4)`
D. No reaction

Explanation

To solve the question regarding the product obtained when treating KMnO4 with very strong alkali in the absence of any reducing agent, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
We are given potassium permanganate (KMnO4) and a very strong alkali, which is typically potassium hydroxide (KOH).

2. **Understand the Reaction Conditions**:
The reaction occurs in the absence of any reducing agent. KMnO4 is known to act as a strong oxidizing agent.

3. **Write the Reaction**:
When KMnO4 reacts with KOH, it undergoes oxidation-reduction reactions. KMnO4 oxidizes the hydroxide ions (OH-) from KOH, resulting in the formation of oxygen gas (O2) and a manganese compound.

4. **Determine the Products**:
The manganese in KMnO4 has an oxidation state of +7. In the presence of strong alkali, it can be reduced to +6, forming potassium manganate (K2MnO4).

5. **Balanced Chemical Equation**:
The balanced equation for the reaction can be written as:
\[ 4 \text{KMnO}_4 + 4 \text{KOH} \rightarrow 4 \text{K}_2\text{MnO}_4 + O_2 + 2 \text{H}_2\text{O} \]

6. **Final Product**:
The main product formed from this reaction is potassium manganate (K2MnO4).

### Conclusion:
The product obtained when treating KMnO4 with very strong alkali in the absence of any reducing agent is **K2MnO4**.

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