`KMnO_(4)` acts as an oxidising agent in acidic medium. The number of moles of `KMnO_(4)` that will be needed to react with one mole of sulphide ions in acidic solution is

Options

A. `2/5`
B. `3/5`
C. `4/5`
D. `1/5`

Explanation

To solve the problem, we need to determine how many moles of KMnO₄ are required to react with one mole of sulfide ions (S²⁻) in an acidic medium.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between KMnO₄ and sulfide ions in acidic medium can be represented as follows:
\[ \text{KMnO}_4 + \text{S}^{2-} \rightarrow \text{S} + \text{Mn}^{2+} \]
Here, KMnO₄ is reduced, and sulfide ions are oxidized to sulfur.

2. **Determine Oxidation States**:
- In KMnO₄, manganese (Mn) has an oxidation state of +7.
- In the sulfide ion (S²⁻), sulfur has an oxidation state of -2.
- In the product, elemental sulfur (S), the oxidation state is 0.
- The oxidation state changes for Mn is from +7 to +2 (a change of 5), and for S from -2 to 0 (a change of 2).

3. **Calculate n-Factors**:
- The n-factor for KMnO₄ (Mn) is the change in oxidation state, which is:
\[ n_{\text{KMnO}_4} = 7 - 2 = 5 \]
- The n-factor for sulfide ions (S²⁻) is:
\[ n_{\text{S}^{2-}} = 0 - (-2) = 2 \]

4. **Set Up the Equivalence Equation**:
The equivalents of KMnO₄ must equal the equivalents of sulfide ions:
\[ \text{Equivalents of KMnO}_4 = \text{Equivalents of S}^{2-} \]
Using the formula for equivalents:
\[ \text{Equivalents} = \text{Moles} \times n \]
We can express this as:
\[ \text{Moles of KMnO}_4 \times n_{\text{KMnO}_4} = \text{Moles of S}^{2-} \times n_{\text{S}^{2-}} \]

5. **Substituting Values**:
Given that we have 1 mole of sulfide ions:
\[ \text{Moles of KMnO}_4 \times 5 = 1 \times 2 \]
Simplifying this gives:
\[ \text{Moles of KMnO}_4 \times 5 = 2 \]

6. **Solve for Moles of KMnO₄**:
\[ \text{Moles of KMnO}_4 = \frac{2}{5} \]

### Final Answer:
The number of moles of KMnO₄ needed to react with one mole of sulfide ions in acidic solution is \(\frac{2}{5}\).

---