`Cu^(2+)` ions give white precipitate with :

Options

A. potassium iodide solution
B. potassium thiocyanate and saturated solution of `SO_(2)`.
C. excess potassium cyanide solution
D. potassium hydroxide solution

Answer

To solve the question regarding which reagents give a white precipitate with `Cu^(2+)` ions, we can follow these steps:

### Step 1: Identify the Reagents
We need to consider the reagents that can react with `Cu^(2+)` ions to form a precipitate. The common reagents mentioned in the question are:
- Potassium iodide (KI)
- Potassium thiocyanate (KSCN)
- Potassium cyanide (KCN)
- Sodium hydroxide (NaOH)

### Step 2: Analyze the Reaction with Potassium Iodide (KI)
When `Cu^(2+)` ions react with potassium iodide (KI), they undergo a double displacement reaction to form cupric iodide (CuI₂). Cupric iodide is a solid that precipitates out of the solution, and it is white in color.

**Reaction:**
\[ \text{Cu}^{2+} + 2 \text{KI} \rightarrow \text{CuI}_2 (s) \]

### Step 3: Analyze the Reaction with Potassium Thiocyanate (KSCN)
When `Cu^(2+)` ions react with potassium thiocyanate (KSCN), they form a complex, which can also result in the formation of a white precipitate. The complex formed is cupric thiocyanate (Cu(SCN)₂), which is also a white solid.

**Reaction:**
\[ \text{Cu}^{2+} + 2 \text{KSCN} \rightarrow \text{Cu(SCN)}_2 (s) \]

### Step 4: Analyze the Reaction with Potassium Cyanide (KCN)
When `Cu^(2+)` ions react with potassium cyanide (KCN), they initially form cupric cyanide (CuCN), which is a white precipitate. However, in the presence of excess KCN, a soluble complex is formed, and thus no precipitate is observed.

**Initial Reaction:**
\[ \text{Cu}^{2+} + 2 \text{KCN} \rightarrow \text{CuCN} (s) \]

**With Excess KCN:**
\[ \text{CuCN} + \text{KCN} \rightarrow \text{[Cu(CN)_2]}^{2-} \text{ (soluble complex)} \]

### Step 5: Analyze the Reaction with Sodium Hydroxide (NaOH)
When `Cu^(2+)` ions react with sodium hydroxide (NaOH), they form cupric hydroxide (Cu(OH)₂), which is a pale blue precipitate, not white.

**Reaction:**
\[ \text{Cu}^{2+} + 2 \text{NaOH} \rightarrow \text{Cu(OH)}_2 (s) \]

### Conclusion
From the analysis:
- **White precipitate is formed with:**
- Potassium iodide (KI)
- Potassium thiocyanate (KSCN)

Thus, the correct answer is that `Cu^(2+)` ions give a white precipitate with potassium iodide and potassium thiocyanate.

### Final Answer
**The correct options are: Potassium iodide (KI) and Potassium thiocyanate (KSCN).**

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