When copper sulphate solution is treatd with potasium iodide and excess of hypo solution is added in resulting solution, a white precipitate is formed. The white ppt. is due to formation off:

Options

A. `Na_(2)S_(4)O_(6)`
B. `CuI_(2)`
C. `CuI`
D. `NaI`

Answer

To solve the question regarding the formation of a white precipitate when copper sulfate solution is treated with potassium iodide and excess hypo solution is added, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Reactants**:
- The reactants involved are copper sulfate (CuSO₄) and potassium iodide (KI).

2. **Write the Reaction**:
- When copper sulfate reacts with potassium iodide, the following reaction occurs:
\[ \text{CuSO}_4 + 2 \text{KI} \rightarrow \text{CuI}_2 + \text{K}_2\text{SO}_4 + \text{I}_2 \]
- Here, copper(I) iodide (CuI) is formed along with potassium sulfate (K₂SO₄) and iodine (I₂).

3. **Add Excess Hypo Solution**:
- The hypo solution is sodium thiosulfate (Na₂S₂O₃). When iodine is present in the solution, it reacts with sodium thiosulfate:
\[ \text{I}_2 + 2 \text{Na}_2\text{S}_2\text{O}_3 \rightarrow 2 \text{NaI} + \text{Na}_2\text{S}_4\text{O}_6 \]
- This reaction reduces iodine to iodide ions (NaI).

4. **Identify the White Precipitate**:
- The white precipitate formed in the initial reaction is due to the formation of copper(I) iodide (CuI).
- The compound CuI is known to be a white precipitate.

5. **Conclusion**:
- Therefore, the white precipitate formed in the reaction is due to the formation of copper(I) iodide (CuI).

### Final Answer:
The white precipitate is due to the formation of **CuI (copper(I) iodide)**.