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 colourless inorganic compound (A) is:

Options

A. `Ba(NO_(3))_(2)`
B. `BaCl_(2)`
C. `CuCl_(2)`
D. `CrBr_(3)`

Explanation

To determine the colorless inorganic compound (A) based on the given information, we will analyze each clue step by step.

### Step 1: Identify the Flame Color
The compound (A) imparts a green color to the flame. This is a characteristic of barium salts.

**Hint:** Look for compounds that are known to produce a green flame when burned.

### Step 2: Analyze the Reaction with H₂SO₄
The solution of (A) gives a white precipitate (B) when treated with concentrated sulfuric acid (H₂SO₄). Barium sulfate (BaSO₄) is a well-known white precipitate formed when barium chloride (BaCl₂) reacts with H₂SO₄.

**Hint:** Consider the common precipitates formed with sulfuric acid and identify if they match the characteristics of the compound.

### Step 3: Heating with K₂Cr₂O₇ and H₂SO₄
When (A) is heated with potassium dichromate (K₂Cr₂O₇) and concentrated H₂SO₄, it produces a brown-red vapor (C). This is indicative of the formation of chromyl chloride (CrO₂Cl₂), which is formed when chloride ions react with this mixture.

**Hint:** Recall the chromyl chloride test and the conditions under which it occurs, particularly focusing on the presence of chloride ions.

### Step 4: Reaction of the Gas with NaOH
The brown-red vapor (C) when passed through aqueous sodium hydroxide (NaOH) turns into a yellow solution (D). This yellow solution is likely due to the formation of chromate ions (CrO₄²⁻).

**Hint:** Think about the color changes that occur when chromyl chloride interacts with bases.

### Step 5: Formation of Yellow Precipitate
The yellow solution (D) forms a yellow precipitate (E) with acetic acid and lead acetate. This indicates the presence of chromate ions, which form lead chromate (PbCrO₄), a yellow precipitate.

**Hint:** Identify the common yellow precipitates formed with lead acetate and acetic acid.

### Conclusion
Based on the above analysis, the colorless inorganic compound (A) is **Barium Chloride (BaCl₂)**.

### Final Answer:
The colorless inorganic compound (A) is **Barium Chloride (BaCl₂)**.