Number of moles of `MnO_(4)^(-)` required to oxidise one mole of ferrous oxalate completely in acidic medium will be

Options

A. 0.6 mole
B. 0.4 mole
C. 7.5 mole
D. 0.2 mole

Explanation

To determine the number of moles of \( \text{MnO}_4^{-} \) required to oxidize one mole of ferrous oxalate completely in acidic medium, we can follow these steps:

### Step 1: Identify the oxidation states
- **For \( \text{MnO}_4^{-} \)**:
- The oxidation state of manganese (Mn) in \( \text{MnO}_4^{-} \) can be calculated as follows:
\[ x + 4(-2) = -1 \implies x - 8 = -1 \implies x = +7 \]
- Therefore, the oxidation state of Mn in \( \text{MnO}_4^{-} \) is +7.

- **For ferrous oxalate \( \text{FeC}_2\text{O}_4 \)**:
- The oxidation state of iron (Fe) in ferrous oxalate is +2.
- The oxidation state of carbon (C) in oxalate \( \text{C}_2\text{O}_4^{2-} \) can be calculated as follows:
\[ 2x + 4(-2) = -2 \implies 2x - 8 = -2 \implies 2x = 6 \implies x = +3 \]
- Therefore, the oxidation state of C in \( \text{C}_2\text{O}_4^{2-} \) is +3.

### Step 2: Determine the changes in oxidation states
- **Oxidation of Fe**:
- From +2 (in \( \text{Fe}^{2+} \)) to +3 (in \( \text{Fe}^{3+} \)) involves a change of 1 electron.

- **Oxidation of C**:
- From +3 (in oxalate) to +4 (in \( \text{CO}_2 \)) involves a change of 1 electron for each carbon atom.
- Since there are 2 carbon atoms in oxalate, the total change for carbon is 2 electrons.

### Step 3: Calculate total electrons transferred
- The total number of electrons transferred during the oxidation of 1 mole of ferrous oxalate is:
\[ \text{Total electrons} = 1 \text{ (from Fe)} + 2 \text{ (from C)} = 3 \text{ electrons} \]

### Step 4: Determine the number of moles of \( \text{MnO}_4^{-} \) required
- Each mole of \( \text{MnO}_4^{-} \) can accept 5 electrons (as Mn changes from +7 to +2).
- To find out how many moles of \( \text{MnO}_4^{-} \) are needed to provide 3 electrons:
\[ \text{Moles of } \text{MnO}_4^{-} = \frac{\text{Total electrons required}}{\text{Electrons accepted by one mole of } \text{MnO}_4^{-}} = \frac{3}{5} = 0.6 \]

### Conclusion
The number of moles of \( \text{MnO}_4^{-} \) required to oxidize one mole of ferrous oxalate completely in acidic medium is **0.6 moles**.

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