`M^(n+)(aq.)+KI to underset(p pt.)(X darr) underset(KI)overset("Excess")to` ppt. remains insoluble in excess KI solution. Then cation `M^(n+)(aq.)` can be:

Options

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

Answer

To solve the problem, we need to analyze the reaction of the cation \( M^{n+} \) with potassium iodide (KI) and determine which cation forms a precipitate (ppt) that remains insoluble in excess KI solution. Let's go through the options step by step.

### Step 1: Analyze the Reaction with Each Cation

1. **Option 1: Lead (Pb²⁺)**
- Reaction: \( Pb^{2+} (aq) + KI \rightarrow K_{2}PbI_{4} \)
- Result: The product \( K_{2}PbI_{4} \) is soluble in excess KI. Therefore, this option does not satisfy the condition.

2. **Option 2: Copper (Cu²⁺)**
- Reaction: \( Cu^{2+} (aq) + KI \rightarrow CuI_{2} \)
- Result: The product \( CuI_{2} \) forms a white precipitate. When excess KI is added, the precipitate remains insoluble. This option satisfies the condition.

3. **Option 3: Bismuth (Bi³⁺)**
- Reaction: \( Bi^{3+} (aq) + KI \rightarrow K_{2}BiI_{5} \)
- Result: The product \( K_{2}BiI_{5} \) is soluble in excess KI. Therefore, this option does not satisfy the condition.

4. **Option 4: Mercury (Hg²⁺)**
- Reaction: \( Hg^{2+} (aq) + KI \rightarrow K_{2}HgI_{4} \)
- Result: The product \( K_{2}HgI_{4} \) is also soluble in excess KI. Therefore, this option does not satisfy the condition.

### Conclusion
After analyzing all the options, the only cation that forms a precipitate which remains insoluble in excess KI is **Copper (Cu²⁺)**.

### Final Answer
The cation \( M^{n+} \) can be: **Copper (Cu²⁺)**.

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