Precipitation of `Cd^(2+)` and `Cu^(2+)` takes place in presence of (a)____by (b)____.

Answer

To solve the question regarding the precipitation of `Cd^(2+)` and `Cu^(2+)`, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Ions**: The ions we are dealing with are `Cd^(2+)` (Cadmium ion) and `Cu^(2+)` (Copper ion). Both of these ions are part of the same group in the periodic table, specifically group 2 of the cations.

2. **Determine the Precipitating Agent**: To precipitate these ions, we need to identify a suitable precipitating agent. In qualitative inorganic analysis, `HCl` (Hydrochloric acid) is commonly used to create an acidic environment that allows for the precipitation of certain metal ions.

3. **Use of H2S**: The actual precipitation of `Cd^(2+)` and `Cu^(2+)` occurs when we introduce `H2S` (Hydrogen sulfide) gas. `H2S` reacts with the metal ions to form insoluble sulfides.

4. **Write the Precipitation Reactions**:
- For Cadmium:
\[ \text{Cd}^{2+} + \text{S}^{2-} \rightarrow \text{CdS (s)} \quad \text{(Cadmium sulfide precipitate)} \]
- For Copper:
\[ \text{Cu}^{2+} + \text{S}^{2-} \rightarrow \text{CuS (s)} \quad \text{(Copper sulfide precipitate)} \]

5. **Final Answer**: Therefore, the precipitation of `Cd^(2+)` and `Cu^(2+)` takes place in presence of **dilute HCl** by **H2S gas**.

### Summary:
- The final answer to fill in the blanks is:
- (a) dilute HCl
- (b) H2S gas