`1` mole of ferric oxalate is oxidised by `x` mole of `MnO_(4)^(-)` in acidic medium, Hence value of `x` is:

Options

A. 1.2
B. 1.6
C. 1.8
D. 1.5

Explanation

To solve the problem of how many moles of \( \text{MnO}_4^{-} \) are required to oxidize 1 mole of ferric oxalate in acidic medium, we can follow these steps:

### Step 1: Understand the oxidation states
In acidic medium, the oxidation state of manganese in \( \text{MnO}_4^{-} \) is +7, and it gets reduced to +2. This means that each \( \text{MnO}_4^{-} \) ion accepts 5 electrons during the reduction process.

**Hint:** Remember that the change in oxidation state helps determine the number of electrons transferred.

### Step 2: Determine the composition of ferric oxalate
Ferric oxalate has the formula \( \text{Fe}_2(\text{C}_2\text{O}_4)_3 \). Each oxalate ion (\( \text{C}_2\text{O}_4^{2-} \)) contains two carbon atoms, and in this case, there are three oxalate ions in ferric oxalate.

**Hint:** Identify the components of the compound to understand how many moles of oxalate are present.

### Step 3: Calculate the number of electrons released by oxalate
Each oxalate ion (\( \text{C}_2\text{O}_4^{2-} \)) can be oxidized to carbon in a higher oxidation state (e.g., \( \text{C}^{+4} \)). For each oxalate ion, 2 electrons are released (1 electron per carbon atom). Since there are 3 oxalate ions in ferric oxalate, the total number of electrons released is:
\[ 3 \text{ oxalate ions} \times 2 \text{ electrons/oxalate} = 6 \text{ electrons} \]

**Hint:** Calculate the total number of electrons released based on the number of oxalate ions.

### Step 4: Relate the moles of \( \text{MnO}_4^{-} \) to the moles of ferric oxalate
The total number of moles of \( \text{MnO}_4^{-} \) required can be calculated using the equivalence concept. The number of equivalents of \( \text{MnO}_4^{-} \) must equal the number of equivalents of ferric oxalate.

Using the formula:
\[ \text{Number of equivalents} = \text{moles} \times \text{valence factor} \]
For \( \text{MnO}_4^{-} \):
- Moles = \( x \)
- Valence factor = 5 (since it accepts 5 electrons)

For ferric oxalate:
- Moles = 1
- Valence factor = 6 (since it releases 6 electrons)

Setting the equivalents equal:
\[ x \times 5 = 1 \times 6 \]

**Hint:** Use the relationship between moles and valence factors to set up an equation.

### Step 5: Solve for \( x \)
Now, we can solve for \( x \):
\[ x \times 5 = 6 \implies x = \frac{6}{5} = 1.2 \]

**Hint:** Isolate \( x \) to find the number of moles of \( \text{MnO}_4^{-} \) needed.

### Conclusion
The value of \( x \) is 1.2, meaning 1.2 moles of \( \text{MnO}_4^{-} \) are required to oxidize 1 mole of ferric oxalate in acidic medium.

**Final Answer:** \( x = 1.2 \) moles of \( \text{MnO}_4^{-} \)