Volumetric titrations involving `KMnO_(4)`, are carried out only in presence of dilute `H_(2)SO_(4)` but not in the presence of HCI or `HNO_(3)` This is because oxygen produced from `KMnO_(4)` + dil: `H_(2)SO_(4)`, is used only for oxidizing the reducing agent. Moreover, H2SO4 does not give any oxygen of its own to oxidize the reducing agent. In case HCI is used, the oxygen produced from `KMnO_(4) + HCI` is partly used up to oxidize HCI and in case `HNO_(3)` is used, it itself acts as oxidizing agent and partly oxidizes the reducing agent. `KMnO_(4)`, in various mediums gives following products`Mn^(7+)overset(H^(+))toMn^(2+)`
`Mn^(7+)overset(H_(2)O)toMn^(4+)`
`Mn^(7+)underset(OH^(-))toMn^(6+)`
Q`MnO_(4)^(2-)`(1 mole) in neutral medium disproportionates to,

Options

A. `(2)/(3)`mole of `MnO_(4)^(-)`, and `(1)/(3)`mole of `MnO_(2)`
B. `(1)/(2)`mole of `MnO_(4)^(-)` and` (2)/(3) `mole of `MnO_(2)`
C. `(1)/(3)` mole of `Mn_(2)O_(7)` and mole of `MnO_(2)`
D. `(2)/(3)`mole of `Mn_(2)O_(7)`, and`(1)/(3)` mole of `MnO_(2)`

Answer

To solve the problem of how `MnO4^(2-)` (manganate ion) disproportionates in a neutral medium, we need to follow these steps:

### Step 1: Identify the oxidation states
The first step is to determine the oxidation states of manganese in the manganate ion (`MnO4^(2-)`). In this ion, manganese has an oxidation state of +6.

### Step 2: Understand disproportionation
Disproportionation is a reaction where a single substance is both oxidized and reduced. In this case, `MnO4^(2-)` will be converted into two different manganese species: one with a higher oxidation state and one with a lower oxidation state.

### Step 3: Determine the products
In neutral medium, `MnO4^(2-)` can disproportionate to:
- `MnO4^-` (where manganese has an oxidation state of +7)
- `MnO2` (where manganese has an oxidation state of +4)

### Step 4: Write the balanced equation
To balance the reaction, we can write:
\[ 3 \, \text{MnO}_4^{2-} \rightarrow 2 \, \text{MnO}_4^{-} + \text{MnO}_2 \]

### Step 5: Balance the equation
Now we need to balance the equation in terms of both mass and charge.
- On the left, we have 3 manganese atoms and a total charge of -6.
- On the right, we have 2 manganese atoms in `MnO4^-` and 1 manganese atom in `MnO2`, totaling 3 manganese atoms, and the charge is also -6.

### Step 6: Determine the mole ratio
From the balanced equation, we can see that:
- 3 moles of `MnO4^(2-)` give 2 moles of `MnO4^-` and 1 mole of `MnO2`.
- Therefore, 1 mole of `MnO4^(2-)` will yield:
- \( \frac{2}{3} \) moles of `MnO4^-`
- \( \frac{1}{3} \) moles of `MnO2`

### Step 7: Conclusion
Thus, the final answer to the question is:
- `1 mole of MnO4^(2-)` in neutral medium disproportionates to `2/3 moles of MnO4^-` and `1/3 moles of MnO2`.

### Final Answer
The answer is:
- **2/3 moles of `MnO4^-` and 1/3 moles of `MnO2`**

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