Two inorganic compounds A and B were heated in a dry test tube. A evolved a colourless gas which turned lead acetate paper black and B evolved a gas which turned lime water milky. The anions in A and B respectively are :

Options

A. `SO_(3)^(2-), CO_(3)^(2-)`
B. `S^(2-), CO_(3)^(2-)`
C. `PO_(4)^(3-), HSO_(3)^(-)`
D. `S^(2-), NO_(3)^(-)`

Answer

To solve the problem, we need to analyze the information given about the two compounds A and B.

### Step-by-Step Solution:

1. **Identify the gas evolved from compound A**:
- The problem states that compound A evolved a colorless gas that turned lead acetate paper black.
- A gas that can turn lead acetate paper black is hydrogen sulfide (H₂S), which indicates the presence of sulfide ions (S²⁻).

2. **Identify the gas evolved from compound B**:
- The problem states that compound B evolved a gas that turned lime water milky.
- A gas that can turn lime water milky is carbon dioxide (CO₂), which indicates the presence of carbonate ions (CO₃²⁻).

3. **Conclude the anions present in compounds A and B**:
- From the analysis, we conclude that the anion in compound A is sulfide (S²⁻) and the anion in compound B is carbonate (CO₃²⁻).

### Final Answer:
- The anions in A and B respectively are:
- A: Sulfide (S²⁻)
- B: Carbonate (CO₃²⁻)

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