When `KMnO_(4)` is reduced with oxalic acid in acidic solution, the oxidation number of `Mn` changes from

Options

A. `+2` to `+7`
B. `+4` to `+7`
C. `+7` to `+2`
D. `+6` to `+2`

Explanation

To determine the change in the oxidation number of manganese (Mn) when potassium permanganate (KMnO4) is reduced with oxalic acid (H2C2O4) in acidic solution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
The reaction involves potassium permanganate (KMnO4), oxalic acid (H2C2O4), and sulfuric acid (H2SO4). The products formed are potassium sulfate (K2SO4), manganese sulfate (MnSO4), carbon dioxide (CO2), and water (H2O).

2. **Assign Oxidation Numbers**:
- In KMnO4, we need to find the oxidation number of Mn. The oxidation state of K is +1, and the oxidation state of O is -2. The overall charge of the compound is 0.
- The equation can be set up as follows:
\[ +1 + x + 4(-2) = 0 \]
Simplifying this gives:
\[ +1 + x - 8 = 0 \implies x - 7 = 0 \implies x = +7 \]
- Therefore, the oxidation number of Mn in KMnO4 is +7.

3. **Determine the Oxidation Number in the Product**:
- In MnSO4, the oxidation number of Mn can be calculated similarly. The oxidation state of SO4 is -2, and the overall charge is 0.
- The equation is:
\[ x + (-2) = 0 \implies x = +2 \]
- Thus, the oxidation number of Mn in MnSO4 is +2.

4. **Calculate the Change in Oxidation Number**:
- The change in oxidation number of Mn during the reaction is from +7 in KMnO4 to +2 in MnSO4.
- Therefore, the change is:
\[ +7 \rightarrow +2 \]

5. **Conclusion**:
The oxidation number of Mn changes from +7 to +2 when KMnO4 is reduced with oxalic acid in acidic solution.

### Final Answer:
The oxidation number of Mn changes from +7 to +2.

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