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.
Compound (C ) is also formed by the following reaction

Options

A. `Ba^(2+) + S(2)O_(3)^(2-) rarr E `
B. `Bi^(2+) + S(2)O_(3)^(2-) rarr E `
C. `Bi^(2+) S(2)O_(3) overset(Delta) to E `
D. None of these

Explanation

To solve the given problem, we need to identify the compounds A, B, C, D, E, and F based on the reactions described. Let's break down the information step by step.

### Step 1: Identify Compound A
The question states that a colorless compound (A) gives a heavy white turbidity (B) when placed in water. This suggests that compound A is likely a soluble salt that forms an insoluble precipitate in water.

**Solution:**
- Compound A is Bismuth(III) chloride, \( \text{BiCl}_3 \).
- When \( \text{BiCl}_3 \) is dissolved in water, it forms \( \text{BiOCl} \) (compound B), which is a white precipitate.

### Step 2: Identify Compound B
From the previous step, we identified compound B as the white precipitate formed when A is dissolved in water.

**Solution:**
- Compound B is Bismuth oxychloride, \( \text{BiOCl} \).

### Step 3: Reaction with Concentrated HCl
The question states that compound B gives a clear solution when concentrated HCl is added. This indicates that B dissolves in concentrated HCl.

**Solution:**
- When \( \text{BiOCl} \) is treated with concentrated HCl, it reverts back to \( \text{BiCl}_3 \) (compound A).

### Step 4: Identify Compound C
The question mentions that passing \( \text{H}_2\text{S} \) through a suspension of A or B results in a black precipitate (C).

**Solution:**
- The black precipitate formed is Bismuth(III) sulfide, \( \text{Bi}_2\text{S}_3 \) (compound C).

### Step 5: Identify Compound D
It is stated that concentrated \( \text{H}_2\text{SO}_4 \) added to solid A liberates a gas (D) that is water-soluble and gives a white precipitate with mercuric salts (E).

**Solution:**
- The gas liberated is likely Hydrogen chloride, \( \text{HCl} \) (compound D).
- The white precipitate formed with mercuric salts is Mercurous chloride, \( \text{Hg}_2\text{Cl}_2 \) (compound E).

### Step 6: Identify Compound F
The question indicates that when \( \text{HNO}_3 \) is added to the black precipitate (C), followed by \( \text{H}_2\text{SO}_4 \) and then \( \text{NH}_4\text{OH} \), a white precipitate (F) is formed.

**Solution:**
- The white precipitate formed is likely Bismuth(III) hydroxide, \( \text{Bi(OH)}_3 \) (compound F).

### Summary of Compounds
- Compound A: \( \text{BiCl}_3 \) (Bismuth(III) chloride)
- Compound B: \( \text{BiOCl} \) (Bismuth oxychloride)
- Compound C: \( \text{Bi}_2\text{S}_3 \) (Bismuth(III) sulfide)
- Compound D: \( \text{HCl} \) (Hydrogen chloride)
- Compound E: \( \text{Hg}_2\text{Cl}_2 \) (Mercurous chloride)
- Compound F: \( \text{Bi(OH)}_3 \) (Bismuth(III) hydroxide)