When `H_(2)S` is passed through `Hg_(2)^(2+)`, we get :

Options

A. `HgS`
B. `HgS + Hg_(2)S`
C. `HgS + Hg`
D. `Hg_(2)S`

Explanation

To solve the question, we need to understand the reaction that occurs when hydrogen sulfide (H₂S) is passed through mercurous ion (Hg₂²⁺). Let's break down the steps involved in this reaction:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactants in this reaction are hydrogen sulfide (H₂S) and mercurous ion (Hg₂²⁺).

2. **Understand the Nature of the Mercurous Ion**:
- The mercurous ion (Hg₂²⁺) consists of two mercury atoms, each in the +1 oxidation state. This is important for determining the products of the reaction.

3. **Write the Reaction**:
- When H₂S is passed through Hg₂²⁺, the reaction can be represented as:
\[ \text{Hg}_2^{2+} + \text{H}_2\text{S} \rightarrow \text{Hg} + \text{HgS} + \text{H}_2 \]

4. **Balance the Reaction**:
- In the products, we have one mercury (Hg) in elemental form and one mercury sulfide (HgS). The hydrogen from H₂S also forms hydrogen gas (H₂).
- The balanced equation is:
\[ \text{Hg}_2^{2+} + \text{H}_2\text{S} \rightarrow \text{HgS} + \text{Hg} + \text{H}_2 \]

5. **Identify the Oxidation and Reduction**:
- In this reaction, one mercury atom is reduced from +1 to 0 oxidation state (Hg), while the other mercury atom remains in the +1 state. The sulfur in H₂S is not oxidized or reduced but forms a compound with mercury.

6. **Conclusion**:
- The products of the reaction when H₂S is passed through Hg₂²⁺ are mercury (Hg), mercury sulfide (HgS), and hydrogen gas (H₂).

### Final Answer:
When H₂S is passed through Hg₂²⁺, we get:
- **HgS (mercury sulfide)**
- **Hg (elemental mercury)**
- **H₂ (hydrogen gas)**