Reddish - brown (chocolate ) ppt. is formed with :

Options

A. `Cu^(2+) and Fe(CN)_(4)^(4-)`
B. `Ba^(2+)` and `SO_(4)^(2+)`
C. `Pb^(2+)` and `I^(Θ)`
D. None of these

Explanation

To solve the question of which compound forms a reddish-brown (chocolate) precipitate, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Key Components**: The question mentions a reddish-brown (chocolate) precipitate. We need to identify which of the given options can lead to the formation of such a precipitate.

2. **Analyze the Given Options**:
- **Cu²⁺ and Fe(CN)₆⁴⁻**: This combination is known to form a reddish-brown precipitate, specifically Cu₂[Fe(CN)₆], which is often referred to as a chocolate color compound.
- **Ba²⁺ and SO₄²⁻**: This combination forms barium sulfate (BaSO₄), which is a white precipitate, not reddish-brown.
- **Pb²⁺ and I⁻**: This combination forms lead(II) iodide (PbI₂), which is yellow, not reddish-brown.
- **None of these**: This option implies that none of the previous combinations form a reddish-brown precipitate.

3. **Confirm the Correct Reaction**:
- The correct reaction to form the reddish-brown precipitate would be:
\[ \text{CuSO}_4 + \text{K}_4[\text{Fe(CN)}_6] \rightarrow \text{Cu}_2[\text{Fe(CN)}_6] + \text{K}_2\text{SO}_4 \]
- This reaction shows that copper(II) sulfate reacts with potassium ferrocyanide to form the chocolate-colored precipitate Cu₂[Fe(CN)₆] and potassium sulfate.

4. **Evaluate the Options**:
- The first option (Cu²⁺ and Fe(CN)₆⁴⁻) is the only one that leads to the formation of the reddish-brown precipitate.
- The other options do not yield a reddish-brown precipitate.

5. **Conclusion**: The reddish-brown (chocolate) precipitate is formed with Cu²⁺ and Fe(CN)₆⁴⁻.

### Final Answer:
The reddish-brown (chocolate) precipitate is formed with **Cu²⁺ and Fe(CN)₆⁴⁻**.