Number of electrons transferred in each case when `KMnO_(4)` acts as an oxidising agent to give `MnO_(2),Mn^(+2), Mn(OH)_(3)` and `MnO_(4)^(2-)` are respectively:

Options

A. `3,5,4 and 1`
B. `4,3,1 and 5`
C. `1,3,4 and 5`
D. `5,4,3 and 1`

Explanation

To determine the number of electrons transferred when KMnO₄ acts as an oxidizing agent to produce different manganese compounds, we need to analyze the changes in oxidation states of manganese in each case.

### Step-by-Step Solution:

1. **Identify the oxidation state of Mn in KMnO₄:**
- In KMnO₄, manganese (Mn) has an oxidation state of +7.

2. **Reduction of KMnO₄ to MnO₂:**
- In MnO₂, Mn has an oxidation state of +4.
- Change in oxidation state: +7 to +4.
- Number of electrons transferred:
\[ 7 - 4 = 3 \text{ electrons} \]

3. **Reduction of KMnO₄ to Mn²⁺:**
- In Mn²⁺, Mn has an oxidation state of +2.
- Change in oxidation state: +7 to +2.
- Number of electrons transferred:
\[ 7 - 2 = 5 \text{ electrons} \]

4. **Reduction of KMnO₄ to Mn(OH)₃:**
- In Mn(OH)₃, Mn has an oxidation state of +3.
- Change in oxidation state: +7 to +3.
- Number of electrons transferred:
\[ 7 - 3 = 4 \text{ electrons} \]

5. **Reduction of KMnO₄ to MnO₄²⁻:**
- In MnO₄²⁻, Mn has an oxidation state of +6.
- Change in oxidation state: +7 to +6.
- Number of electrons transferred:
\[ 7 - 6 = 1 \text{ electron} \]

### Summary of Electrons Transferred:
- KMnO₄ to MnO₂: 3 electrons
- KMnO₄ to Mn²⁺: 5 electrons
- KMnO₄ to Mn(OH)₃: 4 electrons
- KMnO₄ to MnO₄²⁻: 1 electron

Thus, the number of electrons transferred in each case is:
- **3, 5, 4, and 1** respectively.

### Final Answer:
The correct sequence of electrons transferred is **3, 5, 4, 1**.

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