Which of the following is/are correctly matched ?

Options

A. `Bil darr to` Black
B. `Cu_(2)l_(2) darr to` White precipitate
C. `Pbl_(2) darr to` Yellow precipitate
D. `Hgl_(2) darr to` Red percipitate

Explanation

To determine which of the given options are correctly matched, we will analyze each compound and its corresponding precipitate color step by step.

### Step 1: Analyze Bismuth Iodide
- **Compound**: Bismuth Iodide (BiI3)
- **Reaction**: When bismuth reacts with iodine, it forms bismuth iodide.
- **Precipitate Color**: Bismuth iodide is known to form a black precipitate.
- **Conclusion**: This option is correctly matched.

### Step 2: Analyze Copper(I) Iodide
- **Compound**: Copper(I) Iodide (Cu2I2)
- **Reaction**: Copper(II) ions (Cu²⁺) can be reduced to Copper(I) ions (Cu⁺) in the presence of iodide ions (I⁻).
- **Precipitate Color**: Copper(I) iodide forms a white precipitate.
- **Conclusion**: This option is correctly matched.

### Step 3: Analyze Lead Iodide
- **Compound**: Lead Iodide (PbI2)
- **Reaction**: Lead(II) nitrate reacts with potassium iodide to form lead iodide.
- **Precipitate Color**: Lead iodide is known to form a yellow precipitate.
- **Conclusion**: This option is correctly matched.

### Step 4: Analyze Mercuric Iodide
- **Compound**: Mercuric Iodide (HgI2)
- **Reaction**: Mercuric chloride reacts with potassium iodide to form mercuric iodide.
- **Precipitate Color**: Mercuric iodide forms a red precipitate.
- **Conclusion**: This option is correctly matched.

### Final Conclusion
All the options provided are correctly matched with their respective precipitate colors.

### Summary of Correct Matches:
1. Bismuth Iodide - Black precipitate (Correct)
2. Copper(I) Iodide - White precipitate (Correct)
3. Lead Iodide - Yellow precipitate (Correct)
4. Mercuric Iodide - Red precipitate (Correct)

Thus, the answer is that all of the options are correctly matched.

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