`Light green (Compound 'A') overset(Delta)(rarr)White Residue '(B)' underset("Temp")overset("High")(rarr)C+D+E`
`i)` 'D' and 'E' are two acidic gases.
ii) 'D' is passed through `HgCl_(2)` solution to give yellow pt.
`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, white ppt is obtained but white ppt does not turn into grey on addition of excess solution of 'A'.
'C' is soluble in

Options

A. dil. HCl
B. dil.`H_(2)SO_(4)`
C. Conc.`CH_(3)COOH`
D. Boiled conc. HCl

Answer

To solve the question step by step, we will analyze the information provided and deduce the compounds involved.

### Step 1: Identify Compound 'A'
- **Given Information**: Compound 'A' is light green and is water-soluble.
- **Conclusion**: The light green color and water solubility suggest that Compound 'A' is likely to be **Ferrous Sulfate Heptahydrate (FeSO4·7H2O)**.

### Step 2: Heating Compound 'A'
- **Process**: Upon heating, Compound 'A' loses its water of crystallization.
- **Result**: This results in the formation of **Anhydrous Ferrous Sulfate (FeSO4)**, which is a white residue (Compound 'B').

### Step 3: Further Heating of Compound 'B'
- **Process**: When anhydrous FeSO4 is heated further at high temperatures, it decomposes.
- **Result**: The decomposition produces **Iron(III) Oxide (Fe2O3)**, **Sulfur Dioxide (SO2)**, and **Sulfur Trioxide (SO3)**. Here, **C = Fe2O3**, **D = SO2**, and **E = SO3**.

### Step 4: Analyze the Properties of 'D' and 'E'
- **D (SO2)**: It is an acidic gas and can react with **HgCl2** to form a yellow precipitate (mercury(II) sulfite).
- **E (SO3)**: It is also an acidic gas and reacts with water to form sulfuric acid (H2SO4). When H2S is passed through this solution, it forms a white precipitate of barium sulfide (BaS), indicating the presence of sulfate ions.

### Step 5: Analyze the Reaction of 'A' with HgCl2
- **Given Information**: When HgCl2 is added to a solution of 'A', a white precipitate is formed, which does not turn grey upon the addition of excess 'A'.
- **Conclusion**: The white precipitate is likely **Mercury(II) Sulfate (HgSO4)**, which does not turn grey because it is not reduced by excess ferrous ions.

### Step 6: Identify Compound 'C'
- **Given Information**: Compound 'C' is soluble in dilute HCl, dilute H2SO4, concentrated acetic acid, and boiled concentrated HCl.
- **Conclusion**: Since 'C' is **Fe2O3**, it is known to be soluble in hydrochloric acid (HCl) and reacts to form ferric chloride (FeCl3).

### Final Conclusion
- **Compound A**: FeSO4·7H2O (light green)
- **Compound B**: FeSO4 (white residue)
- **Compound C**: Fe2O3 (soluble in HCl)
- **Compound D**: SO2 (acidic gas)
- **Compound E**: SO3 (acidic gas)

### Summary of the Solution
- **A**: FeSO4·7H2O
- **B**: FeSO4
- **C**: Fe2O3
- **D**: SO2
- **E**: SO3