The `Mn^(3+)` ion is unstable in solution and undergoes disproportionation reaction to give `Mn^(+2), MnO_(2)`, and `H^(o+)` ion. Write a balanced ionic equation for the reaction.

Options

A. `3Mn^(3+)+4H_(2)O rarr MnO_(2)+Mn^(2+)+8H^(+)`
B. `Mn^(2+)+4H_(2)O rarr MnO_(2) + 4H^(+)`
C. `Mn+2H_(2)O rarr MnO_(2) + 4H^(+)`
D. `2 M n^(3+)+2H_(2)O rarr MnO_(2)+Mn^(2+)+4H^(+)`

Explanation

To write a balanced ionic equation for the disproportionation reaction of the `Mn^(3+)` ion, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction Components**:
The reaction involves `Mn^(3+)`, which will undergo disproportionation to form `Mn^(2+)`, `MnO2`, and `H^+` ions.

2. **Write the Unbalanced Equation**:
The initial unbalanced equation can be written as:
\[ \text{Mn}^{3+} \rightarrow \text{Mn}^{2+} + \text{MnO}_2 + \text{H}^+ \]

3. **Determine Oxidation and Reduction**:
In this reaction, `Mn^(3+)` is both oxidized to `MnO2` and reduced to `Mn^(2+)`.

4. **Write the Oxidation Half-Reaction**:
For the oxidation half-reaction, `Mn^(3+)` is converted to `MnO2`. To balance this:
\[ \text{Mn}^{3+} + 4 \text{H}^+ + 2 \text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4 \text{H}^+ + e^- \]
This shows that one electron is lost in the process.

5. **Write the Reduction Half-Reaction**:
For the reduction half-reaction, `Mn^(3+)` is converted to `Mn^(2+)`. The half-reaction can be written as:
\[ \text{Mn}^{3+} + e^- \rightarrow \text{Mn}^{2+} \]

6. **Balance the Half-Reactions**:
Now we need to combine the two half-reactions. The oxidation half-reaction produces one electron, while the reduction half-reaction consumes one electron. Therefore, we can add them directly:
\[ 2 \text{Mn}^{3+} + 2 \text{H}_2\text{O} \rightarrow \text{Mn}^{2+} + \text{MnO}_2 + 4 \text{H}^+ \]

7. **Final Balanced Equation**:
The final balanced ionic equation for the disproportionation reaction is:
\[ 2 \text{Mn}^{3+} + 2 \text{H}_2\text{O} \rightarrow \text{Mn}^{2+} + \text{MnO}_2 + 4 \text{H}^+ \]