`MnO_(4)^(2-)` (`1` mole) in neutral aqueous medium is disproportionate to

Options

A. `(2)/(3) " mol of "MnO_(4)^(ɵ) and (1)/(3) " mol of "MnO_(2)`
B. `(1)/(3) " mol of "MnO_(4)^(ɵ) and (2)/(3) " mol of "MnO_(2)`
C. `(1)/(3) " mol of "Mn_(2)O_(7) and (2)/(3) " mol of "MnO_(2)`
D. `(2)/(3) " mol of "Mn_(2)O_(7) and (1)/(3) " mol of "MnO_(2)`

Explanation

To solve the problem of the disproportionation of \( \text{MnO}_4^{2-} \) in a neutral aqueous medium, we can follow these steps:

### Step 1: Write the balanced equation for the disproportionation reaction.
In a neutral aqueous medium, \( \text{MnO}_4^{2-} \) can disproportionate into \( \text{KMnO}_4 \) and \( \text{MnO}_2 \). The balanced reaction can be written as:
\[ 3 \text{K}_2\text{MnO}_4 + 2 \text{H}_2\text{O} \rightarrow 2 \text{KMnO}_4 + 1 \text{MnO}_2 + 4 \text{KOH} \]

### Step 2: Determine the moles of products formed from 1 mole of \( \text{MnO}_4^{2-} \).
From the balanced equation, we see that:
- 3 moles of \( \text{K}_2\text{MnO}_4 \) produce 2 moles of \( \text{KMnO}_4 \) and 1 mole of \( \text{MnO}_2 \).

To find the amount produced from 1 mole of \( \text{MnO}_4^{2-} \):
- For \( \text{KMnO}_4 \):
\[ \text{Moles of } \text{KMnO}_4 = \frac{2}{3} \text{ moles from 1 mole of } \text{K}_2\text{MnO}_4 \]
- For \( \text{MnO}_2 \):
\[ \text{Moles of } \text{MnO}_2 = \frac{1}{3} \text{ moles from 1 mole of } \text{K}_2\text{MnO}_4 \]

### Step 3: Summarize the products formed.
Thus, from 1 mole of \( \text{MnO}_4^{2-} \), we get:
- \( \frac{2}{3} \) moles of \( \text{KMnO}_4 \)
- \( \frac{1}{3} \) moles of \( \text{MnO}_2 \)

### Conclusion:
The final products of the disproportionation of 1 mole of \( \text{MnO}_4^{2-} \) in neutral aqueous medium are:
- \( \frac{2}{3} \) moles of \( \text{KMnO}_4 \)
- \( \frac{1}{3} \) moles of \( \text{MnO}_2 \)