Light green (compound 'A')`overset(Delta)(to)"white Residue(B)"overset("high")underset("Temp.")C+D+E`
(i)'D' and 'E' are two acidic gas.
(ii) 'D' is passed through `HgCl_(2)` solution to give yellow ppt.
(iii) 'E' is passed through water first and then `H_(2)S` is passed, white turbidity is obtained.
(iv) A is water soluble and addition of `HgCl_(2)` in it, yellow ppt. is obtained but white ppt does not turn into grey on addition of excess solution of 'A'
Q. 'D' and 'E' are respectively.

Options

A. `SO_(2) and SO_(3)`
B. `SO_(3) and SO_(2)`
C. `SO_(2) and CO_(2)`
D. `CO_(2) and CO`

Answer

To solve the problem step-by-step, we will analyze the information given and derive the answers for 'D' and 'E'.

### Step 1: Identify Compound 'A'
The problem states that compound 'A' is light green in color and is water-soluble. The most common light green compound in this context is hydrated ferrous sulfate, which is represented as FeSO4·7H2O.

**Hint:** Look for common light green compounds that are water-soluble.

### Step 2: Heating Compound 'A'
When hydrated ferrous sulfate (FeSO4·7H2O) is heated, it loses its water molecules and forms anhydrous ferrous sulfate (FeSO4), which is white in color. This white residue is referred to as compound 'B'.

**Hint:** Consider the thermal decomposition of hydrated salts to identify the products.

### Step 3: Further Heating of Compound 'B'
When anhydrous ferrous sulfate (FeSO4) is heated at high temperatures, it decomposes to form iron(III) oxide (Fe2O3), sulfur trioxide (SO3), and sulfur dioxide (SO2). Here, we identify 'C', 'D', and 'E'.

**Hint:** Think about the decomposition products of metal sulfates at high temperatures.

### Step 4: Identify 'D' and 'E'
From the decomposition of FeSO4, we can identify:
- 'D' is sulfur trioxide (SO3)
- 'E' is sulfur dioxide (SO2)

**Hint:** Pay attention to the specific gases mentioned in the problem and their properties.

### Step 5: Verify Conditions for 'D' and 'E'
- **Condition (ii):** 'D' (SO3) is passed through HgCl2 solution to give a yellow precipitate. This is consistent with the formation of mercuric sulfate (HgSO4) from SO3.
- **Condition (iii):** 'E' (SO2) is passed through water and then H2S is added, resulting in white turbidity, which indicates the formation of lead sulfide (PbS) or barium sulfide (BaS).
- **Condition (iv):** The addition of HgCl2 to compound 'A' (FeSO4·7H2O) gives a yellow precipitate, confirming the presence of SO3, but the white precipitate does not turn gray upon the addition of excess solution of 'A', indicating that the precipitate is stable.

**Hint:** Check the chemical reactions and precipitates formed to confirm the identities of 'D' and 'E'.

### Conclusion
Based on the analysis, we conclude that:
- 'D' is sulfur trioxide (SO3)
- 'E' is sulfur dioxide (SO2)

### Final Answer
**D = SO3 (sulfur trioxide), E = SO2 (sulfur dioxide)**