`Pb(NO_(3))_(2)+2KI to PbI_(2)darr+2KNO_(3)`

Options

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

Answer

To solve the question regarding the reaction of lead(II) nitrate with potassium iodide, we will analyze the reaction step by step.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants in the given reaction are lead(II) nitrate, \( \text{Pb(NO}_3\text{)}_2 \), and potassium iodide, \( \text{KI} \).

2. **Write the Chemical Equation**:
The reaction can be written as:
\[ \text{Pb(NO}_3\text{)}_2 + 2\text{KI} \rightarrow \text{PbI}_2 + 2\text{KNO}_3 \]

3. **Determine the Products**:
When lead(II) nitrate reacts with potassium iodide, lead iodide \( \text{PbI}_2 \) is formed, which is a yellow precipitate, along with potassium nitrate \( \text{KNO}_3 \).

4. **Classify the Reaction**:
Since a solid product (precipitate) is formed from the reaction of two aqueous solutions, this reaction is classified as a **precipitate formation reaction**.

5. **Conclusion**:
The correct option that matches this reaction is option 1, which states that it is a precipitate formation reaction.

### Final Answer:
The reaction \( \text{Pb(NO}_3\text{)}_2 + 2\text{KI} \rightarrow \text{PbI}_2 + 2\text{KNO}_3 \) is classified as a **precipitate formation reaction**.