`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'.
Yellow ppt in the above observation is

Options

A. Mercuric oxide
B. Basic mercury(II) sulphite
C. Basic mercury (II) sulphate
D. Mercuric iodine

Answer

To solve the question step-by-step, we need to analyze the information given and deduce the identities of the compounds involved:

### Step 1: Identify Compound A
- **Observation**: Compound A is light green in color.
- **Conclusion**: The light green compound is likely hydrated ferrous sulfate, which is represented as \( \text{FeSO}_4 \cdot 7 \text{H}_2\text{O} \).

### Step 2: Heating Compound A
- **Observation**: When heated, Compound A (hydrated ferrous sulfate) loses water and forms a white residue B.
- **Chemical Reaction**:
\[ \text{FeSO}_4 \cdot 7 \text{H}_2\text{O} \xrightarrow{\text{heat}} \text{FeSO}_4 + 7 \text{H}_2\text{O} \]
- **Conclusion**: The white residue B is anhydrous ferrous sulfate (\( \text{FeSO}_4 \)).

### Step 3: Further Heating of Residue B
- **Observation**: Upon further heating of residue B, it decomposes to form products C, D, and E.
- **Chemical Reaction**:
\[ 2 \text{FeSO}_4 \xrightarrow{\text{heat}} \text{Fe}_2\text{O}_3 + \text{SO}_2 + \text{SO}_3 \]
- **Conclusion**:
- C is \( \text{Fe}_2\text{O}_3 \) (iron(III) oxide).
- D is \( \text{SO}_3 \) (sulfur trioxide).
- E is \( \text{SO}_2 \) (sulfur dioxide).

### Step 4: Analyzing Gas D
- **Observation**: D (sulfur trioxide) is passed through \( \text{HgCl}_2 \) solution, resulting in a yellow precipitate.
- **Chemical Reaction**:
\[ \text{SO}_3 + \text{HgCl}_2 \rightarrow \text{Hg}_2\text{SO}_4 + \text{HCl} \]
- **Conclusion**: The yellow precipitate formed is basic mercury(II) sulfate, represented as \( \text{Hg}_2\text{SO}_4 \).

### Step 5: Analyzing Gas E
- **Observation**: E (sulfur dioxide) is passed through water followed by \( \text{H}_2\text{S} \), resulting in white turbidity.
- **Chemical Reaction**:
\[ \text{SO}_2 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_3 \]
\[ \text{H}_2\text{SO}_3 + \text{H}_2\text{S} \rightarrow \text{white precipitate (lead sulfide)} \]
- **Conclusion**: The white turbidity indicates the formation of a sulfide, likely lead sulfide.

### Step 6: Water Solubility of A
- **Observation**: Compound A is water-soluble and forms a white precipitate with \( \text{HgCl}_2 \), which does not turn grey upon excess addition.
- **Conclusion**: This confirms that the white precipitate is indeed not turning grey, indicating it is not metallic mercury but rather the basic mercury sulfate.

### Final Conclusion
- The yellow precipitate formed when D (sulfur trioxide) is passed through \( \text{HgCl}_2 \) solution is basic mercury(II) sulfate, \( \text{Hg}_2\text{SO}_4 \).

### Answer:
The yellow precipitate in the above observation is **basic mercury(II) sulfate**.

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