When `H_2 S` is passed through acidified `KMnO_4`, we get

Options

A. `K_(2)SO_(3)`
B. `MnO_(2)`
C. `KHSO_(3)`
D. sulphur

Explanation

To solve the question, we need to analyze the reaction of hydrogen sulfide (H₂S) with acidified potassium permanganate (KMnO₄).

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants in this reaction are hydrogen sulfide (H₂S) and acidified potassium permanganate (KMnO₄). The acidic medium is typically sulfuric acid (H₂SO₄).

2. **Determine Oxidation States**:
- In H₂S, sulfur (S) has an oxidation state of -2.
- In KMnO₄, manganese (Mn) has an oxidation state of +7 (since K is +1 and each O is -2).

3. **Redox Reaction**:
- In this reaction, H₂S is oxidized, and KMnO₄ is reduced.
- The oxidation state of sulfur in H₂S changes from -2 to 0 (elemental sulfur).
- The oxidation state of manganese changes from +7 in KMnO₄ to +2 in MnSO₄.

4. **Products Formation**:
- The products formed from this reaction are manganese(II) sulfate (MnSO₄), potassium hydrogen sulfate (KHSO₄), and water (H₂O).
- The overall balanced reaction can be represented as:
\[ 5 \text{H}_2\text{S} + 2 \text{KMnO}_4 + 3 \text{H}_2\text{SO}_4 \rightarrow 5 \text{S} + 2 \text{MnSO}_4 + KHSO_4 + 6 \text{H}_2\text{O} \]

5. **Conclusion**:
When H₂S is passed through acidified KMnO₄, the products formed are manganese(II) sulfate (MnSO₄), potassium hydrogen sulfate (KHSO₄), and water (H₂O), along with elemental sulfur (S).

### Final Answer:
When H₂S is passed through acidified KMnO₄, we get MnSO₄, KHSO₄, H₂O, and elemental sulfur (S).