In which of the following pairs the precipitates are red and black coloured respectively and both precipitates are soluble in excess `Kl` solution?

Options

A. `Hgl_(2),Hg_(2)l_(2)`
B. `Hgl_(2),Bil_(3)`
C. `Cu_(2)l_(2),Agl`
D. `Cdl_(2),Pbl_(2)`

Explanation

To solve the question, we need to identify pairs of compounds that yield red and black precipitates, respectively, and confirm that both are soluble in excess potassium iodide (KI) solution.

### Step-by-Step Solution:

1. **Identify the Red Precipitate:**
- The red precipitate in question is **Mercury(I) iodide (HgI2)**.
- When HgI2 is treated with excess KI, it forms a soluble complex, **Tetraiodomercurate(II) ion (HgI4^2-)**.

2. **Identify the Black Precipitate:**
- The black precipitate is **Bismuth(III) iodide (BiI3)**.
- When BiI3 reacts with excess KI, it forms a soluble complex, **Bismuth tetraiodide ion (BiI4^-)**.

3. **Confirm Solubility in Excess KI:**
- Both HgI2 and BiI3 are known to form soluble complexes when treated with excess KI.
- HgI2 → HgI4^2- (soluble)
- BiI3 → BiI4^- (soluble)

4. **Conclusion:**
- Therefore, the pair that meets the criteria of having a red precipitate (HgI2) and a black precipitate (BiI3), both of which are soluble in excess KI, is the correct answer.

### Final Answer:
The correct pair is **B (HgI2 and BiI3)**.