A small amount of a light green coloured compound `X` is heated in a test tube. In the beginning, it loses some water and then gas(es) `Z` with suffocating smell come(s) out. The vapours of gas(es) are collected and dissolved in water. The solution turns blue litmus red. The residue `Y` left in the test tube turns reddish brown. `X, Y` and `Z` could be respectively

Options

A. `PbSO_4, Pb_2O_3 and SO_3`
B. `FeSO_(4).7H_2O, Fe_2O_3 and SO_2, SO_3`
C. `Na_2SO_4. 10H_2O , Na_2SO_4 and SO_2`
D. `Pb(NO_3)_2, PbO_2 and NO_2, N_2O_4`

Explanation

To solve the problem, we need to identify the compounds X, Y, and Z based on the given information about their properties and reactions.

### Step-by-Step Solution:

1. **Identify Compound X**:
- The problem states that compound X is a light green colored compound that loses some water upon heating.
- A common light green compound that fits this description is **Iron(II) sulfate heptahydrate**, which has the chemical formula **FeSO4·7H2O**.
- When heated, it loses water of crystallization.

**Hint**: Look for hydrated compounds that are light green in color and lose water when heated.

2. **Heating of Compound X**:
- Upon heating FeSO4·7H2O, it first loses water and then decomposes to produce gas Z.
- The gas released has a suffocating smell, which indicates the presence of sulfur oxides.

**Hint**: Consider the gases produced from the thermal decomposition of iron(II) sulfate.

3. **Identify Gas Z**:
- The gases Z released during the decomposition of FeSO4 are likely to be **sulfur dioxide (SO2)** and **sulfur trioxide (SO3)**.
- Both of these gases have a suffocating smell and are known to be acidic in nature.

**Hint**: Think about the products of sulfur-containing compounds when heated.

4. **Collecting Gas Z**:
- The vapors of gas Z are collected and dissolved in water.
- The solution turns blue litmus paper red, indicating that the solution is acidic.

**Hint**: Remember that non-metal oxides, such as SO2 and SO3, form acidic solutions when dissolved in water.

5. **Identify Residue Y**:
- After the heating process, the residue left in the test tube is described as reddish-brown.
- The residue formed from the further heating of FeSO4 is **Iron(III) oxide**, which has the formula **Fe2O3**. This compound is known for its reddish-brown color.

**Hint**: Consider the oxidation states of iron and the products formed from the decomposition of iron(II) sulfate.

### Conclusion:
Based on the analysis, we can conclude:
- **Compound X** is **FeSO4·7H2O** (Iron(II) sulfate heptahydrate).
- **Gas Z** is a mixture of **SO2** (sulfur dioxide) and **SO3** (sulfur trioxide).
- **Residue Y** is **Fe2O3** (Iron(III) oxide).

Thus, the final answer is:
- **X = FeSO4·7H2O**
- **Y = Fe2O3**
- **Z = SO2 and SO3**