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. 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 will analyze the information provided about the compounds and their reactions.

### Step 1: Identify Compound A
- **Observation**: Compound A is described as light green and water-soluble.
- **Conclusion**: The light green compound A is likely to be **ferrous sulfate (FeSO4·7H2O)**, which is known to be light green in color and soluble in water.

**Hint**: Look for common light green compounds that are soluble in water; ferrous sulfate is a good candidate.

### Step 2: Heating Compound A
- **Observation**: Upon heating, compound A gives a white residue B.
- **Reaction**: When ferrous sulfate is heated, it loses water of crystallization and forms anhydrous ferrous sulfate (FeSO4), which is a white solid.
- **Conclusion**: The white residue B is **anhydrous ferrous sulfate (FeSO4)**.

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

### Step 3: Further Heating of Compound B
- **Observation**: Compound B (FeSO4) on further heating produces compounds C, D, and E.
- **Reaction**: The decomposition of anhydrous ferrous sulfate at high temperatures yields:
- C: **Iron(III) oxide (Fe2O3)**
- D: **Sulfur trioxide (SO3)**
- E: **Sulfur dioxide (SO2)**

**Hint**: Look for common products formed from the thermal decomposition of ferrous sulfate.

### Step 4: Identify Acidic Gases D and E
- **Observation**: D and E are acidic gases.
- **Conclusion**: D is **SO3** and E is **SO2**, both of which are acidic gases.

**Hint**: Recall that sulfur oxides are common acidic gases.

### Step 5: Reaction of D and E with HgCl2
- **Observation**: D (SO3) and E (SO2) are passed through a solution of HgCl2, resulting in a yellow precipitate.
- **Reaction**: The reaction of SO3 with HgCl2 produces **basic mercury sulfate (HgSO4)**, which is a yellow precipitate.

**Hint**: Consider the reactions of sulfur oxides with metal halides to identify precipitates.

### Step 6: Further Reaction of E with Water and H2S
- **Observation**: When E (SO2) is passed through water and then H2S, white turbidity is obtained.
- **Conclusion**: This indicates the formation of **mercury sulfide (HgS)**, which is a white precipitate.

**Hint**: Think about the reactions of sulfur dioxide in water and with hydrogen sulfide.

### Step 7: Analyze the Final Observation
- **Observation**: Compound A is water-soluble, and when HgCl2 is added, a yellow precipitate forms. However, the white precipitate does not turn grey upon the addition of excess solution of A.
- **Conclusion**: The yellow precipitate formed is **basic mercury sulfate (HgSO4)**.

**Hint**: The behavior of precipitates in the presence of excess reactants can help confirm their identity.

### Final Answer
The yellow precipitate in the above observation is **Basic Mercury Sulfate (HgSO4)**.