`MnO_(4)^(2-)` in acidic solution undergoes disportionation to give `"…....................."` and `"…..................."`.

Answer

To solve the problem of the disproportionation of \( \text{MnO}_4^{2-} \) in acidic solution, we will follow these steps:

### Step 1: Understand Disproportionation
Disproportionation is a reaction where a single substance is both oxidized and reduced, resulting in two different products. In this case, \( \text{MnO}_4^{2-} \) will undergo disproportionation.

### Step 2: Identify the Oxidation States
In \( \text{MnO}_4^{2-} \), manganese (Mn) has an oxidation state of +6. During the disproportionation, one manganese atom will be reduced to a lower oxidation state, while another will be oxidized to a higher oxidation state.

### Step 3: Determine the Products
In acidic solution:
- One manganese will be reduced to \( \text{Mn}^{2+} \) (oxidation state +2).
- The other manganese will be oxidized to \( \text{MnO}_4^{-} \) (oxidation state +7).

### Step 4: Write the Balanced Reaction
The balanced reaction for the disproportionation of \( \text{MnO}_4^{2-} \) in acidic solution can be written as:
\[ 2 \text{MnO}_4^{2-} \rightarrow \text{Mn}^{2+} + \text{MnO}_4^{-} + 2 \text{H}_2\text{O} \]

### Final Answer
The products of the disproportionation of \( \text{MnO}_4^{2-} \) in acidic solution are:
- \( \text{Mn}^{2+} \)
- \( \text{MnO}_4^{-} \)

---