A mixture consists `(A)` (red solid) and `(B)` (colourless solid) which gives lilac colour in flame.
(a)Mixture gives black precipitate `( C)` on passing `H_(2)S (g)`.
(b)`( C)` is soluble in aquaregia and on evaporation of aquaregia and adding `SnCl_(2)` gives greyish black precipitate `(D)`.
The salt solution with `NH_(4)OH` gives a brown precipitate.
The sodium extract gives yellow precipitate with `AgNO_(3)` solution which is insoluble in dilute ammonia solution.
Identify `(A)` and `(B)`, and the precipitates `(C )` and `(D)`.

Explanation

To solve the problem, we need to identify the compounds (A) and (B) based on the given clues, and also identify the precipitates (C) and (D). Let's break down the information step-by-step:

### Step 1: Identify the Components of the Mixture
The mixture consists of a red solid (A) and a colorless solid (B) that gives a lilac color in the flame. The lilac color in the flame indicates the presence of potassium ions (K⁺).

**Hint:** Lilac flame is characteristic of potassium.

### Step 2: Determine the Identity of (A) and (B)
Since (A) is a red solid and (B) is a colorless solid, we can deduce:
- (A) is likely potassium iodide (KI) because it gives a lilac flame due to K⁺.
- (B) is likely mercuric iodide (HgI₂), which is a colorless solid.

**Hint:** Look for compounds that contain potassium and mercury.

### Step 3: Analyze the Reaction with H₂S
The mixture gives a black precipitate (C) when H₂S gas is passed through it. The black precipitate formed is mercurous sulfide (HgS).

**Hint:** H₂S typically forms black precipitates with mercury compounds.

### Step 4: Solubility in Aqua Regia
The black precipitate (C) is soluble in aqua regia. When treated with aqua regia, HgS is converted to mercuric chloride (HgCl₂).

**Hint:** Aqua regia can dissolve noble metals and their sulfides.

### Step 5: Reaction with SnCl₂
Upon evaporation of aqua regia and adding SnCl₂, a grayish-black precipitate (D) is formed. This grayish-black precipitate is elemental mercury (Hg).

**Hint:** SnCl₂ reduces HgCl₂ to elemental mercury.

### Step 6: Reaction with NH₄OH
The salt solution with NH₄OH gives a brown precipitate. This indicates the presence of mercurous hydroxide (Hg₂(OH)₂), which is brown.

**Hint:** NH₄OH can precipitate hydroxides of metals.

### Step 7: Sodium Extract with AgNO₃
The sodium extract gives a yellow precipitate with AgNO₃, which is insoluble in dilute ammonia. This yellow precipitate is silver iodide (AgI), indicating the presence of iodide ions (I⁻).

**Hint:** Yellow precipitates with AgNO₃ often indicate the presence of iodide ions.

### Conclusion
Based on the deductions:
- (A) is Potassium Iodide (KI)
- (B) is Mercuric Iodide (HgI₂)
- (C) is Mercurous Sulfide (HgS)
- (D) is Elemental Mercury (Hg)

### Final Answers:
- **(A)**: KI
- **(B)**: HgI₂
- **(C)**: HgS (black precipitate)
- **(D)**: Hg (grayish-black precipitate)