`Hg_(2)^(2+)` when reacts with `H_(2)S`, black ppt. (A) formed which when reacts with `Na_(2)S` followed by filtration leaving behind black ppt. (B). The filitrate with `H^(+)` gives black ppt. (C). A, B and C are:

Options

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

Answer

To solve the problem, we need to analyze the reactions involving the compound \( \text{Hg}_2^{2+} \) when it interacts with \( \text{H}_2\text{S} \) and \( \text{Na}_2\text{S} \).

### Step-by-Step Solution:

1. **Reaction of \( \text{Hg}_2^{2+} \) with \( \text{H}_2\text{S} \)**:
- When \( \text{Hg}_2^{2+} \) reacts with \( \text{H}_2\text{S} \), it produces a black precipitate. The reaction can be represented as:
\[ \text{Hg}_2^{2+} + \text{H}_2\text{S} \rightarrow \text{Hg} + \text{HgS} \downarrow \]
- Here, \( \text{HgS} \) (mercury sulfide) is the black precipitate formed, which we will denote as precipitate (A).

2. **Reaction of the precipitate with \( \text{Na}_2\text{S} \)**:
- The black precipitate \( \text{HgS} \) formed can further react with \( \text{Na}_2\text{S} \). The reaction is:
\[ \text{HgS} + \text{Na}_2\text{S} \rightarrow \text{Na}_2\text{S}_2 + \text{Hg} \downarrow \]
- This results in another black precipitate (B), which is simply \( \text{Hg} \) (elemental mercury).

3. **Filtration and Reaction of the Filtrate with \( \text{H}^+ \)**:
- After filtration, the remaining solution (filtrate) can be treated with \( \text{H}^+ \). The reaction is:
\[ \text{Hg}^{2+} + \text{H}^+ + \text{S}^{2-} \rightarrow \text{HgS} \downarrow \]
- This produces another black precipitate (C), which is again \( \text{HgS} \).

### Summary of Precipitates:
- **Precipitate A**: \( \text{HgS} \) (black precipitate formed initially)
- **Precipitate B**: \( \text{Hg} \) (black precipitate formed after reaction with \( \text{Na}_2\text{S} \))
- **Precipitate C**: \( \text{HgS} \) (black precipitate formed after adding \( \text{H}^+ \) to the filtrate)

### Final Answer:
- A is \( \text{HgS} \)
- B is \( \text{Hg} \)
- C is \( \text{HgS} \)

The correct answer for the question is:
- A = \( \text{HgS} \)
- B = \( \text{Hg} \)
- C = \( \text{HgS} \)