The number of moles of oxalate ions oxidised by one mole of `MnO_(4)^(-)` ion in acidic medium is :

Options

A. `(5)/(2)`
B. `(2)/(5)`
C. `(3)/(5)`
D. `(5)/(3)`

Answer

To determine the number of moles of oxalate ions oxidized by one mole of `MnO4^(-)` ion in acidic medium, we can follow these steps:

### Step 1: Understand the Reaction
In acidic medium, the permanganate ion `MnO4^(-)` acts as a strong oxidizing agent. It gets reduced while oxidizing other species, in this case, oxalate ions `(C2O4^2-)`.

### Step 2: Determine the N-factor of `MnO4^(-)`
In acidic medium, the N-factor (number of electrons gained or lost per mole) for `MnO4^(-)` is 5. This means that one mole of `MnO4^(-)` can accept 5 moles of electrons.

### Step 3: Determine the N-factor of Oxalate Ion `(C2O4^2-)`
The oxalate ion `(C2O4^2-)` is oxidized to carbon dioxide `(CO2)`. The N-factor for oxalate ion is 2 because each oxalate ion loses 2 electrons during oxidation.

### Step 4: Set Up the Relationship
Let `X` be the number of moles of oxalate ions oxidized by 1 mole of `MnO4^(-)`. The relationship can be established based on the N-factors:
- For `MnO4^(-)`: 1 mole gives 5 equivalents.
- For `C2O4^2-`: `X` moles give `2X` equivalents.

### Step 5: Equate the Equivalents
Since the equivalents of oxidizing and reducing agents must be equal:
\[ 5 \text{ (from MnO4)} = 2X \text{ (from oxalate)} \]

### Step 6: Solve for `X`
To find `X`, rearrange the equation:
\[ X = \frac{5}{2} \]

### Conclusion
Thus, the number of moles of oxalate ions oxidized by one mole of `MnO4^(-)` ion in acidic medium is:
\[ \frac{5}{2} \text{ moles} \]