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

Options

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

Answer

To solve the given question, we will analyze the reaction step by step and identify the type of reaction it represents.

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
The given reaction is:
\[ \text{Pb(NO}_3\text{)}_2 + 2\text{KI} \rightarrow \text{PbI}_2 \downarrow + 2\text{KNO}_3 \]
Here, the reactants are lead(II) nitrate \((\text{Pb(NO}_3\text{)}_2)\) and potassium iodide \((\text{KI})\). The products are lead(II) iodide \((\text{PbI}_2)\) and potassium nitrate \((\text{KNO}_3)\).

2. **Observe the Formation of Precipitate**:
In this reaction, lead(II) iodide \((\text{PbI}_2)\) is indicated with a downward arrow, which signifies that it is a precipitate. This means that it forms as a solid during the reaction.

3. **Determine the Type of Reaction**:
- **Precipitate Formation Reaction**: This type of reaction occurs when two solutions react to form an insoluble solid (precipitate).
- In our case, there was no precipitate present in the reactants (as both reactants are soluble in water), but a precipitate (PbI2) is formed in the products.

4. **Conclusion**:
Since a precipitate is formed from the reactants, this reaction is classified as a **precipitate formation reaction**.

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

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