0.1 " mol of "`MnO_(4)^(ɵ)` (in acidic medium) can:

Options

A. Oxidise 0.5 " mol of "`Fe^(2+)`
B. Oxidise 0.166 " mol of "`FeC_(2)O_(4)`
C. Oxidise 0.25 " mol of "`C_(2)O_(4)^(2-)`
D. Oxidise 0.6 " mol of "`Cr_(2)O_(7)^(2-)`

Explanation

To solve the problem, we need to determine what 0.1 moles of \( \text{MnO}_4^- \) can do in an acidic medium. The key steps to solve this problem are as follows:

### Step 1: Determine the reaction of \( \text{MnO}_4^- \) in acidic medium
In acidic medium, \( \text{MnO}_4^- \) is reduced to \( \text{Mn}^{2+} \). The oxidation state of manganese changes from +7 in \( \text{MnO}_4^- \) to +2 in \( \text{Mn}^{2+} \).

### Step 2: Calculate the n-factor of \( \text{MnO}_4^- \)
The n-factor is defined as the number of electrons transferred per mole of the substance in a redox reaction. For the reduction of \( \text{MnO}_4^- \) to \( \text{Mn}^{2+} \), 5 electrons are required. Therefore, the n-factor of \( \text{MnO}_4^- \) in this reaction is 5.

### Step 3: Calculate the milliequivalents of \( \text{MnO}_4^- \)
Milliequivalents can be calculated using the formula:
\[ \text{Milliequivalents} = \text{Moles} \times \text{n-factor} \]
For 0.1 moles of \( \text{MnO}_4^- \):
\[ \text{Milliequivalents} = 0.1 \, \text{mol} \times 5 = 0.5 \, \text{milliequivalents} \]

### Step 4: Analyze the options
Now we will analyze the options provided in the question to see how many moles of different substances can be oxidized by 0.5 milliequivalents of \( \text{MnO}_4^- \).

1. **Option A: Oxidize 0.5 moles of \( \text{Fe}^{2+} \)**
- The oxidation of \( \text{Fe}^{2+} \) to \( \text{Fe}^{3+} \) involves 1 electron transfer (n-factor = 1).
- Therefore, 0.5 moles of \( \text{Fe}^{2+} \) would require:
\[ 0.5 \, \text{mol} \times 1 = 0.5 \, \text{milliequivalents} \]
- This option is correct.

2. **Option B: Oxidize 0.11 moles of \( \text{C}_2\text{O}_4^{2-} \)**
- The oxidation of \( \text{C}_2\text{O}_4^{2-} \) to \( \text{CO}_2 \) involves 2 electrons (n-factor = 2).
- Therefore, 0.11 moles of \( \text{C}_2\text{O}_4^{2-} \) would require:
\[ 0.11 \, \text{mol} \times 2 = 0.22 \, \text{milliequivalents} \]
- This option is also correct.

3. **Option C: Oxidize 0.25 moles of \( \text{C}_2\text{O}_4^{2-} \)**
- The oxidation of \( \text{C}_2\text{O}_4^{2-} \) to \( \text{CO}_2 \) involves 2 electrons (n-factor = 2).
- Therefore, 0.25 moles of \( \text{C}_2\text{O}_4^{2-} \) would require:
\[ 0.25 \, \text{mol} \times 2 = 0.5 \, \text{milliequivalents} \]
- This option is also correct.

4. **Option D: Oxidize 0.6 moles of \( \text{Cr}_2\text{O}_7^{2-} \)**
- The \( \text{Cr}_2\text{O}_7^{2-} \) can be reduced, but it cannot be oxidized. Therefore, this option is incorrect.

### Conclusion
The correct options are A, B, and C.