`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 color of the product formed when lead(II) acetate, \( \text{Pb(CH}_3\text{COO)}_2 \), reacts with hydrogen sulfide (\( \text{H}_2\text{S} \)), we can follow these steps:

### Step 1: Write the Reaction
The first step is to write the balanced chemical reaction for the interaction between lead(II) acetate and hydrogen sulfide.

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

### Step 2: Identify the Products
From the reaction, we can see that lead(II) acetate reacts with hydrogen sulfide to produce lead(II) sulfide (\( \text{PbS} \)) and acetic acid (\( \text{CH}_3\text{COOH} \)).

### Step 3: Determine the Color of the Product
Next, we need to identify the color of lead(II) sulfide (\( \text{PbS} \)). It is well-known that lead(II) sulfide is a black solid.

### Step 4: Conclusion
Thus, when lead(II) acetate reacts with hydrogen sulfide, the color observed is black due to the formation of lead(II) sulfide.

### Final Answer
The answer to the question is that \( \text{Pb(CH}_3\text{COO)}_2 \) gives a **black** color with \( \text{H}_2\text{S} \).