A colourless (A) when place into water a heavy white turbidly of (B) solid (A) gives a close solution in conesolution in cone `HCI` when `HCI` solution is added to clear solution water ,(B) forms again (B) dissolves in dilute `HCI`. When `H_(2)S` is passed through a sespension of (A) or (B), a black precipitate (C ) forms , (C ) is insolves in yellow ammonium sulphide `(NH_(4))_(2)S` , cone `H_(2)SO_(4)` added to solid (A) liberates gas (D) gas (D) is water soluble and gives white precipitate with mercuric salts (E ) and not mercuric salt .The black precipitate (C ) dissolves in `HNO_(3), (1,1)` to give a solution to which `H_(2)SO_(4)` is added followed by addition of `NH_(4)OH` when a white precipitate (F) is formed (E ) gives a black ppt , (G) with solution of sodium stannite.
When compound (E ) reacts with `NH_(4)OH` , then product is a

Options

A. White ppt
B. Black ppt
C. yellow ppt
D. Green ppt

Explanation

To solve the problem step by step, we will analyze the information provided and deduce the compounds involved in the reactions.

### Step 1: Identify Compound A
- **Given**: A colorless compound (A) when placed in water forms a heavy white turbidity (B).
- **Analysis**: The description indicates that A is likely a salt that forms an insoluble precipitate in water. A common compound that fits this description is **Bismuth chloride (BiCl₃)**, which forms **Bismuth oxychloride (BiOCl)** when it reacts with water.

### Step 2: Identify Compound B
- **Given**: Solid A gives a clear solution in concentrated HCl, and when HCl is added to the clear solution, B forms again.
- **Analysis**: When BiOCl is treated with concentrated HCl, it can dissolve back to form BiCl₃. Thus, **B = BiOCl**.

### Step 3: Identify Compound C
- **Given**: When H₂S is passed through a suspension of A or B, a black precipitate (C) forms.
- **Analysis**: The black precipitate formed when H₂S is passed through BiCl₃ or BiOCl is **Bismuth sulfide (Bi₂S₃)**. Thus, **C = Bi₂S₃**.

### Step 4: Analyze Gas D
- **Given**: Concentrated H₂SO₄ added to solid A liberates gas D, which is water-soluble and gives a white precipitate with mercuric salts (E).
- **Analysis**: When concentrated H₂SO₄ is added to BiCl₃, it can produce **Chlorine gas (Cl₂)**, which is water-soluble. The white precipitate with mercuric salts indicates the formation of **Mercurous chloride (Hg₂Cl₂)**. Thus, **D = Cl₂** and **E = Hg₂Cl₂**.

### Step 5: Identify Compound F
- **Given**: The black precipitate (C) dissolves in HNO₃ to give a solution that forms a white precipitate (F) when NH₄OH is added.
- **Analysis**: When Bi₂S₃ is treated with HNO₃, it forms bismuth nitrate, which upon treatment with NH₄OH gives **Bismuth hydroxide (Bi(OH)₃)** as a white precipitate. Thus, **F = Bi(OH)₃**.

### Step 6: Analyze Reaction of E with NH₄OH
- **Given**: When compound E reacts with NH₄OH, it produces a product.
- **Analysis**: When mercurous chloride (Hg₂Cl₂) reacts with NH₄OH, it forms **Mercuric amidochloride (HgNH₂Cl)**, which is a black precipitate.

### Conclusion
- **Final Compounds**:
- A = BiCl₃
- B = BiOCl
- C = Bi₂S₃
- D = Cl₂
- E = Hg₂Cl₂
- F = Bi(OH)₃
- G = HgNH₂Cl

### Summary of the Final Answer
The product formed when compound E reacts with NH₄OH is **HgNH₂Cl** (Mercuric amidochloride), which is a black precipitate.

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