On passing `CrO_2 Cl_2` in water and then adding `(CH_3 COO)_2`Pb, the precipitate formed is

Options

A. `PbCrO_4`
B. `PbCl_2`
C. `CrCl_3`
D. `PbSO_4`

Answer

To solve the question regarding the precipitate formed when passing `CrO2Cl2` in water and then adding `(CH3COO)2Pb`, we can follow these steps:

### Step 1: Understand the Reaction of `CrO2Cl2` in Water
When `CrO2Cl2` is passed in water, it reacts with water to form chromium species. The reaction can be represented as follows:

\[ \text{CrO}_2\text{Cl}_2 + 4 \text{NaOH} \rightarrow \text{Na}_2\text{CrO}_4 + 2 \text{NaCl} + 2 \text{H}_2\text{O} \]

### Step 2: Identify the Product Formed
From the above reaction, we can see that `CrO2Cl2` in the presence of sodium hydroxide produces sodium chromate (`Na2CrO4`).

### Step 3: Reaction with Lead(II) Acetate
Next, we add `(CH3COO)2Pb`, which is lead(II) acetate, to the solution containing sodium chromate. The reaction between sodium chromate and lead(II) acetate can be written as:

\[ \text{Na}_2\text{CrO}_4 + \text{(CH}_3\text{COO)}_2\text{Pb} \rightarrow \text{PbCrO}_4 \downarrow + 2 \text{NaCH}_3\text{COO} \]

### Step 4: Identify the Precipitate
The product of this reaction is lead(II) chromate (`PbCrO4`), which is known to form a yellow precipitate.

### Conclusion
Thus, the precipitate formed when passing `CrO2Cl2` in water and then adding `(CH3COO)2Pb` is lead(II) chromate (`PbCrO4`), which appears as a yellow precipitate.

### Final Answer
The precipitate formed is **Lead(II) Chromate (PbCrO4)**.

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