In dilute alkaline solution,`MnO_(4)^(-)` changes to:

Options

A. `KMnO_(4)^(2-)`
B. `MnO_(2)`
C. `Mn_(2)O_(3)`
D. `MnO`

Explanation

To solve the question regarding the reduction of \( \text{MnO}_4^{-} \) in a dilute alkaline solution, we can follow these steps:

### Step 1: Identify the oxidation states
- In \( \text{MnO}_4^{-} \), manganese (Mn) has an oxidation state of +7.
- In the product \( \text{MnO}_2 \), manganese has an oxidation state of +4.

### Step 2: Write the half-reaction
- The half-reaction for the reduction of \( \text{MnO}_4^{-} \) can be written as follows:
\[ \text{MnO}_4^{-} + \text{e}^- \rightarrow \text{MnO}_2 \]

### Step 3: Balance the equation
- To balance the equation, we need to account for the change in oxidation state and the number of oxygen and hydrogen atoms:
- We have 4 oxygen atoms in \( \text{MnO}_4^{-} \) and 2 in \( \text{MnO}_2 \), which means we need to add 2 water molecules (\( \text{H}_2\text{O} \)) to the right side to provide the additional oxygen.
- This gives us:
\[ \text{MnO}_4^{-} + \text{e}^- + 2 \text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4 \text{OH}^- \]

### Step 4: Add electrons to balance charge
- The left side has a charge of -1 (from \( \text{MnO}_4^{-} \)), while the right side has a charge of -4 (from \( 4 \text{OH}^- \)). To balance the charge, we need to add 3 electrons to the left side:
\[ \text{MnO}_4^{-} + 3 \text{e}^- + 2 \text{H}_2\text{O} \rightarrow \text{MnO}_2 + 4 \text{OH}^- \]

### Step 5: Conclusion
- The product formed when \( \text{MnO}_4^{-} \) is reduced in a dilute alkaline solution is \( \text{MnO}_2 \).

### Final Answer
Thus, in dilute alkaline solution, \( \text{MnO}_4^{-} \) changes to \( \text{MnO}_2 \).

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