Componds `(A)` and `B` are treated with dilute `HCl` separately. The gases liberated are `Y` and `Z` respectively. `Y` turns acidified `K_2 Cr_2 O_7` paper green while `Z` turns lead acetate paper black. The compounds `A` and `B` are respectively :

Options

A. `Na_2 S` and `Na_2 SO_3`
B. `Na_2 SO_3` and `Na_2 S`
C. `NaCl` and `Na_2 CO_3`
D. `Na_2 SO_3` and `Na_2 SO_4`

Explanation

To solve the problem, we need to identify the compounds A and B based on the gases Y and Z produced when they react with dilute HCl.

### Step 1: Analyze the information given
- Compound A produces gas Y when treated with dilute HCl, which turns acidified potassium dichromate green.
- Compound B produces gas Z when treated with dilute HCl, which turns lead acetate paper black.

### Step 2: Identify gas Y
- The gas Y that turns acidified potassium dichromate green is likely sulfur dioxide (SO₂). This is because sulfur dioxide can reduce potassium dichromate (K₂Cr₂O₇) in acidic medium, leading to the formation of chromium(III) ions, which are green in color.

### Step 3: Identify compound A
- Since gas Y is sulfur dioxide, we need to find a compound that produces sulfur dioxide when reacted with dilute HCl. Sodium sulfide (Na₂S) reacts with dilute HCl to produce sulfur dioxide:
\[ \text{Na}_2\text{S} + 2\text{HCl} \rightarrow \text{SO}_2 + 2\text{NaCl} + \text{H}_2\text{O} \]
- Therefore, compound A is sodium sulfide (Na₂S).

### Step 4: Identify gas Z
- The gas Z that turns lead acetate paper black is hydrogen sulfide (H₂S). Hydrogen sulfide reacts with lead acetate to form lead sulfide (PbS), which is black:
\[ \text{H}_2\text{S} + \text{Pb(C}_2\text{H}_3\text{O}_2)_2 \rightarrow \text{PbS} \downarrow + 2\text{HC}_2\text{H}_3\text{O}_2 \]

### Step 5: Identify compound B
- Since gas Z is hydrogen sulfide, we need to find a compound that produces hydrogen sulfide when reacted with dilute HCl. Sodium sulfide (Na₂S) can also produce hydrogen sulfide under certain conditions:
\[ \text{Na}_2\text{S} + 2\text{HCl} \rightarrow \text{H}_2\text{S} + 2\text{NaCl} \]
- Therefore, compound B is also sodium sulfide (Na₂S).

### Conclusion
The compounds A and B are:
- Compound A: Sodium sulfide (Na₂S)
- Compound B: Sodium sulfide (Na₂S)

### Summary
The answer to the question is:
- Compound A: Sodium sulfide (Na₂S)
- Compound B: Sodium sulfide (Na₂S)