`Cu^(2+) `gives white ppt. of (a)____ with (b) ____ and deep blue colour of _____(c ) ______with (d) ____.

Answer

To solve the question, we need to identify the compounds formed when Cu²⁺ ions react with specific reagents, as well as the colors of the resulting compounds.

1. **Identifying the White Precipitate**:
- When Cu²⁺ ions react with potassium iodide (KI), they form a compound called copper(I) iodide, which is represented as Cu₂I₂. This compound precipitates out of the solution, forming a white precipitate.
- Therefore, for part (a), the answer is **Cu₂I₂**.
- For part (b), the answer is **KI**.

2. **Identifying the Deep Blue Color**:
- When Cu²⁺ ions react with excess ammonia (NH₃), they form a complex ion known as tetraamminecopper(II), represented as Cu(NH₃)₄²⁺. This complex is characterized by its deep blue color.
- Thus, for part (c), the answer is **Cu(NH₃)₄²⁺**.
- For part (d), the answer is **excess of ammonia**.

Putting it all together, we have:
- (a) Cu₂I₂
- (b) KI
- (c) Cu(NH₃)₄²⁺
- (d) excess of ammonia

### Final Answer:
Cu²⁺ gives white ppt. of (a) **Cu₂I₂** with (b) **KI** and deep blue colour of (c) **Cu(NH₃)₄²⁺** with (d) **excess of ammonia**.