`MnO_(4)^(-)` ions can be reduced in strongly alkaline medium to give

Options

A. `MnO_2`
B. `Mn^(2+)`
C. `MnO_4^(2-)`
D. `MnO_3^-`

Answer

To solve the question regarding the reduction of \( \text{MnO}_4^{-} \) ions in a strongly alkaline medium, we can break down the process step by step.

### Step 1: Identify the Reactants and Products
In a strongly alkaline medium, \( \text{MnO}_4^{-} \) ions are reduced. The expected products of this reaction include manganese dioxide (\( \text{MnO}_2 \)), hydroxide ions (\( \text{OH}^- \)), and oxygen gas (\( \text{O}_2 \)).

### Step 2: Write the Unbalanced Reaction
The unbalanced reaction can be written as:
\[ \text{MnO}_4^{-} + \text{OH}^- \rightarrow \text{MnO}_2 + \text{O}_2 + \text{H}_2\text{O} \]

### Step 3: Balance the Reaction
To balance the reaction, we need to ensure that the number of atoms of each element is the same on both sides. The balanced reaction is:
\[ 2 \text{MnO}_4^{-} + 4 \text{OH}^- \rightarrow 2 \text{MnO}_2 + 2 \text{O}_2 + 2 \text{H}_2\text{O} \]

### Step 4: Determine the Oxidation States
Next, we need to determine the oxidation states of manganese in the reactants and products:
- In \( \text{MnO}_4^{-} \):
- Let the oxidation state of Mn be \( x \).
- The equation becomes: \( x + 4(-2) = -1 \)
- Thus, \( x - 8 = -1 \) leading to \( x = +7 \).

- In \( \text{MnO}_2 \):
- Let the oxidation state of Mn be \( y \).
- The equation becomes: \( y + 2(-2) = 0 \)
- Thus, \( y - 4 = 0 \) leading to \( y = +4 \).

### Step 5: Analyze the Change in Oxidation State
The manganese ion is reduced from an oxidation state of +7 in \( \text{MnO}_4^{-} \) to +4 in \( \text{MnO}_2 \). This indicates that the manganese is undergoing reduction.

### Conclusion
The product formed when \( \text{MnO}_4^{-} \) ions are reduced in a strongly alkaline medium is \( \text{MnO}_2 \).

### Final Answer
The correct answer is \( \text{MnO}_2 \).

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