When a crystalline compound X is heated with `K_(2)Cr_(2)O_(7)` and concentrated `H_(2)SO_(4)`, a reddish brown gas A is evolved. On passing A into caustic soda, a yellow solution of B is formed. A yellow precipitate of C is obtained when a solution B is neutralised with acetic acid and then treated with a lead acetate solution. when X is heated with `NaOH`, a colourless gas is evolved which, when passed into a solution of `K_(2)[HgI_(4)]`, gives a reddish brown precipitate of D. ltBrgt Q. Compound (X) is:

Options

A. `NH_(4)Br`
B. `NH_(4)Cl`
C. `NH_(4)NO_(2)`
D. `NH_(4)NO_(3)`

Explanation

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

### Step 1: Identify the reddish-brown gas A
When the crystalline compound X is heated with potassium dichromate (K₂Cr₂O₇) and concentrated sulfuric acid (H₂SO₄), a reddish-brown gas A is evolved. This gas is likely to be chromyl chloride (CrO₂Cl₂), which is known to produce reddish-brown vapors when chromium compounds react with hydrochloric acid.

**Hint:** Look for reactions involving chromium compounds and acids that produce colored gases.

### Step 2: Determine the composition of compound X
Since gas A is chromyl chloride, we need to consider what could have produced this gas. The presence of ammonium ions (NH₄⁺) is suggested because the gas A will react with caustic soda (NaOH) to form a yellow solution B. This indicates that X is likely to be ammonium chloride (NH₄Cl).

**Hint:** Consider compounds that contain ammonium ions when heated with strong acids and oxidizing agents.

### Step 3: Analyze the formation of yellow solution B
When gas A (chromyl chloride) is passed into caustic soda, a yellow solution B is formed. This yellow solution is likely due to the formation of chromate ions (CrO₄²⁻) from the chromyl chloride reacting with the hydroxide ions from NaOH.

**Hint:** Look for reactions that involve the conversion of chromium compounds in alkaline conditions.

### Step 4: Identify the yellow precipitate C
When solution B is neutralized with acetic acid and treated with lead acetate, a yellow precipitate C is formed. This precipitate is likely lead chromate (PbCrO₄), which is known to form as a yellow precipitate when chromate ions are treated with lead acetate.

**Hint:** Identify common precipitates formed with lead salts and chromate ions.

### Step 5: Identify the colorless gas evolved from X with NaOH
When compound X (NH₄Cl) is heated with sodium hydroxide (NaOH), a colorless gas is evolved. This gas is ammonia (NH₃), which is produced when ammonium salts react with strong bases.

**Hint:** Consider the products of reactions between ammonium salts and strong bases.

### Step 6: Determine the reddish-brown precipitate D
When the colorless gas (ammonia) is passed into a solution of potassium tetraiodomercurate(II) (K₂[HgI₄]), a reddish-brown precipitate D is formed. This indicates the presence of ammonium ions (NH₄⁺) in the solution, as the Nessler's reagent (K₂[HgI₄]) is used to detect ammonium ions.

**Hint:** Look for reagents that specifically react with ammonium ions to form colored precipitates.

### Conclusion
Based on the analysis, the compound X is identified as ammonium chloride (NH₄Cl).

### Final Answer
Compound (X) is: **NH₄Cl (Ammonium Chloride)**