When a KI solution is added to a metal nitrate, a black precipitate is produced which dissolves in an excess of KI to give an organge solution. The metal ion is:

Options

A. `Hg^(2+)`
B. `Bi^(3+)`
C. `Cu^(2+)`
D. `Pb^(2+)`

Explanation

To solve the problem, we need to identify the metal ion that produces a black precipitate when potassium iodide (KI) is added to a metal nitrate solution, and that this precipitate can dissolve in excess KI to form an orange solution.

### Step-by-Step Solution:

1. **Identify the Metal Ions**: The question provides four potential metal ions:
- Hg²⁺ (Mercury)
- Bi³⁺ (Bismuth)
- Cu²⁺ (Copper)
- Pb²⁺ (Lead)

2. **Reaction with Potassium Iodide**:
- When KI is added to a solution of the metal nitrate, we need to determine which metal ion forms a black precipitate.
- Bismuth nitrate (Bi(NO₃)₃) reacts with KI to form bismuth iodide (BiI₃), which is known to produce a black precipitate.

3. **Formation of the Black Precipitate**:
- The reaction can be written as:
\[ \text{Bi(NO}_3\text{)}_3 + 3 \text{KI} \rightarrow \text{BiI}_3 (s) + 3 \text{KNO}_3 \]
- Here, BiI₃ is the black precipitate formed.

4. **Dissolution in Excess KI**:
- The black precipitate (BiI₃) can dissolve in excess KI to form a complex ion:
\[ \text{BiI}_3 + 2 \text{KI} \rightarrow \text{K}_2\text{BiI}_5 \]
- The resulting solution is orange due to the formation of the K₂BiI₅ complex.

5. **Conclusion**:
- Since we have identified that the metal ion that produces a black precipitate upon the addition of KI and forms an orange solution in excess KI is bismuth (Bi³⁺), we conclude that the metal ion is:
\[ \text{Bi}^{3+} \]

### Final Answer:
The metal ion is **Bi³⁺ (Bismuth)**.

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