`Pb(AcO)_(2)+Na_(2)CrO_(4) to PbCrO_(4) darr+2AcONa`

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 between lead(II) acetate and sodium chromate, let's break it down step by step.

### Step-by-Step Solution:

1. **Identify the Reactants**:
The reactants given are lead(II) acetate, \( \text{Pb(AcO)}_2 \), and sodium chromate, \( \text{Na}_2\text{CrO}_4 \).

2. **Write the Reaction**:
The reaction can be written as:
\[ \text{Pb(AcO)}_2 + \text{Na}_2\text{CrO}_4 \rightarrow \text{PbCrO}_4 + 2\text{AcONa} \]

3. **Identify the Products**:
The products of the reaction are lead(II) chromate, \( \text{PbCrO}_4 \), and sodium acetate, \( \text{AcONa} \).

4. **Determine the Nature of the Reaction**:
In this reaction, lead(II) chromate, \( \text{PbCrO}_4 \), is formed as a solid precipitate. This indicates that a new insoluble compound is formed from the reaction of soluble reactants.

5. **Classify the Reaction**:
Since there was no precipitate present on the reactant side and a precipitate (lead(II) chromate) is formed on the product side, this reaction is classified as a **precipitate formation reaction**.

6. **Conclusion**:
Therefore, the correct classification for the reaction \( \text{Pb(AcO)}_2 + \text{Na}_2\text{CrO}_4 \rightarrow \text{PbCrO}_4 + 2\text{AcONa} \) is a **precipitate formation reaction**.