A colourless inorganic compound (A) imparts a green colour to the flame. Its solution gives a white ppt. (B) with `H_(2)SO_(4)`. When heated with `K_(2)Cr_(2)O_(7)` and conc. `H_(2)SO_(4)`, a brown red vapour/gas (C) is formed. The gas/ vapour when passed through aqueous NaOH solution, it turns into a yellow solution (D) which forms yellow precipitate (E) with `CH_(3)COOH` and `(CH_(3)COO)_(2)Pb`. with reference to above information, answe the following questions.
Q. The yellow ppt. formed when (D) reacts with `CH_(3)COOH and (CH_(2)COO)_(2)Pb` is:

Options

A. `PbI_(2)`
B. `PbCrO_(4)`
C. `BaCrO_(4)`
D. `AgBr`

Explanation

To solve the problem step by step, we will analyze the information provided and deduce the required compound.

### Step 1: Identify Compound A
- **Given Information**: A colorless inorganic compound (A) imparts a green color to the flame.
- **Analysis**: The green flame indicates the presence of barium ions (Ba²⁺). Thus, compound A is likely to be barium chloride (BaCl₂).

### Step 2: Reaction with H₂SO₄
- **Given Information**: The solution gives a white precipitate (B) with H₂SO₄.
- **Analysis**: When barium chloride reacts with sulfuric acid (H₂SO₄), it forms barium sulfate (BaSO₄), which is a white precipitate. Therefore, B is BaSO₄.

### Step 3: Heating with K₂Cr₂O₇ and H₂SO₄
- **Given Information**: When heated with K₂Cr₂O₇ and concentrated H₂SO₄, a brown-red vapor/gas (C) is formed.
- **Analysis**: The reaction of barium chloride with potassium dichromate (K₂Cr₂O₇) in the presence of concentrated sulfuric acid produces chromyl chloride (CrO₂Cl₂), which is a brown-red vapor. Thus, C is CrO₂Cl₂.

### Step 4: Reaction of Gas C with NaOH
- **Given Information**: The gas/vapor (C) when passed through aqueous NaOH solution turns into a yellow solution (D).
- **Analysis**: The chromyl chloride (C) reacts with NaOH to form sodium chromate (Na₂CrO₄), which is a yellow solution. Therefore, D is Na₂CrO₄.

### Step 5: Reaction of Solution D with Acetic Acid and Lead Acetate
- **Given Information**: The yellow solution (D) forms a yellow precipitate (E) with CH₃COOH and (CH₃COO)₂Pb.
- **Analysis**: When sodium chromate (D) reacts with acetic acid (CH₃COOH) and lead acetate [(CH₃COO)₂Pb], it forms lead chromate (PbCrO₄), which is a yellow precipitate. Thus, E is PbCrO₄.

### Conclusion
The yellow precipitate formed when D reacts with CH₃COOH and (CH₃COO)₂Pb is **lead chromate (PbCrO₄)**.

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