`Pb(CH_3COO)_2` gives …….colour with `H_2S`

Options

A. Orange
B. Red
C. Black
D. White

Answer

To solve the question regarding the reaction of lead(II) acetate, \( \text{Pb(CH}_3\text{COO)}_2 \), with hydrogen sulfide, \( \text{H}_2\text{S} \), we will follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are lead(II) acetate \( \text{Pb(CH}_3\text{COO)}_2 \) and hydrogen sulfide \( \text{H}_2\text{S} \).

### Step 2: Write the Reaction Equation
When lead(II) acetate reacts with hydrogen sulfide, it forms lead(II) sulfide \( \text{PbS} \) and acetic acid \( \text{CH}_3\text{COOH} \). The balanced chemical equation for the reaction is:

\[ \text{Pb(CH}_3\text{COO)}_2 + \text{H}_2\text{S} \rightarrow \text{PbS} + 2\text{CH}_3\text{COOH} \]

### Step 3: Identify the Product
The product formed from this reaction is lead(II) sulfide \( \text{PbS} \).

### Step 4: Determine the Color of the Product
Lead(II) sulfide \( \text{PbS} \) is known to form a black precipitate when it is produced in a reaction.

### Conclusion
Therefore, when lead(II) acetate \( \text{Pb(CH}_3\text{COO)}_2 \) reacts with hydrogen sulfide \( \text{H}_2\text{S} \), it gives a black color due to the formation of lead(II) sulfide \( \text{PbS} \).

### Final Answer
**Black color**

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