Assuming complete ionization , same moles of which of the following compounds will require the least amount of acidfied `KMnO_(4)` for complete oxidation.

Options

A. `FeSO_(4)`
B. `FeSO_(3)`
C. `FeC_2O_4`
D. `Fe(NO_2)_2`

Explanation

To determine which compound requires the least amount of acidified KMnO4 for complete oxidation, we will analyze the oxidation states and the number of electrons involved in the oxidation of each compound: FeSO4, FeSO3, FeC2O4, and FeNO2 whole twice.

### Step-by-Step Solution:

1. **Identify the Oxidation States**:
- For **FeSO4**:
- Fe is in the +2 oxidation state (Fe²⁺) and can be oxidized to +3 (Fe³⁺).
- Change: Fe²⁺ → Fe³⁺ (1 electron involved).
- For **FeSO3**:
- Fe is in the +2 oxidation state (Fe²⁺) and can be oxidized to +3 (Fe³⁺).
- Additionally, sulfur in SO3 is in the +4 state and can be oxidized to +6 in SO4.
- Change: Fe²⁺ → Fe³⁺ (1 electron) and S +4 → +6 (2 electrons).
- Total: 1 + 2 = 3 electrons involved.
- For **FeC2O4**:
- Fe is in the +2 oxidation state (Fe²⁺) and can be oxidized to +3 (Fe³⁺).
- Each carbon in C2O4 is in the +3 state and can be oxidized to +4 in CO2.
- Change: Fe²⁺ → Fe³⁺ (1 electron) and 2C +3 → 2C +4 (2 electrons).
- Total: 1 + 2 = 3 electrons involved.
- For **FeNO2 whole twice**:
- Fe is in the +2 oxidation state (Fe²⁺) and can be oxidized to +3 (Fe³⁺).
- Nitrogen in NO2 is in the +4 state and can be oxidized to +5 in NO3.
- Change: Fe²⁺ → Fe³⁺ (1 electron) and N +4 → +5 (1 electron).
- Total: 1 + 1 = 2 electrons involved.

2. **Calculate the Total Electrons Required**:
- **FeSO4**: 1 electron
- **FeSO3**: 3 electrons
- **FeC2O4**: 3 electrons
- **FeNO2 whole twice**: 2 electrons

3. **Determine the Compound Requiring the Least Amount of KMnO4**:
- The compound that involves the least number of electrons is **FeSO4**, which requires only 1 electron for oxidation.

### Conclusion:
Thus, the compound that will require the least amount of acidified KMnO4 for complete oxidation is **FeSO4**.

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