`K_(2)[Cd(CN)_(4)]+H_(2)S to CdS darr+2KCN + 2HCN`

Options

A. For precipitate formation formation reaction
B. For precipitate dissolution reaction
C. For precipitate exchange reaction
D. For no reaction

Answer

To analyze the reaction \( K_2[Cd(CN)_4] + H_2S \rightarrow CdS \downarrow + 2KCN + 2HCN \), we will break down the steps to determine the type of reaction it represents.

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
- **Reactants**: \( K_2[Cd(CN)_4] \) and \( H_2S \)
- **Products**: \( CdS \), \( 2KCN \), and \( 2HCN \)

2. **Examine the Formation of Precipitate**:
- In the reaction, \( CdS \) is formed as a product. \( CdS \) is known to be insoluble in water, which means it will precipitate out of the solution.

3. **Determine the Nature of the Reaction**:
- Since there is no precipitate on the reactant side (both \( K_2[Cd(CN)_4] \) and \( H_2S \) are soluble), but a precipitate (\( CdS \)) is formed on the product side, this indicates a precipitate formation reaction.

4. **Conclusion**:
- The reaction is classified as a **precipitate formation reaction** because it starts with soluble reactants and ends with the formation of an insoluble product (precipitate).

### Final Answer:
The type of reaction is a **precipitate formation reaction**.