A black coloured compound `(A)` on reaction with dilute `H_(2)SO_(4)` gives a gas `(B)` which on passing in a solution of an acid `( C)` gives a white turbidity `(D)`.Gas `(B)` when passed in an acidified solution of a compound `(E)` gives a precipitate `(F)` soluble in dilute `HNO_(3)`.After boiling this solution when an excess of `NH_(4)OH` is added a intense blue coloured compound `(G)` is formed. To this solution on addition of acetic acid and aqueous `K_(4)[Fe(CN)_(6)]` a chocolate brown precipitate `(H)` is obtained.On addition of an aqueous solution of `BaCl_(2)` to an aqueous solution of `(E)` a white precipitate insoluble in dilute `HCl` is obtained. Identify the compounds from `(A)` to `(H)`.

Explanation

To solve the problem step by step, we will identify each compound from (A) to (H) based on the reactions described in the question.

### Step 1: Identify Compound (A)
- The question states that a black colored compound (A) reacts with dilute H₂SO₄ to produce a gas (B).
- Black colored compounds that can react with dilute H₂SO₄ are typically metal sulfides.
- A common black sulfide is Copper(II) sulfide (CuS).

**Conclusion:** (A) = CuS

### Step 2: Identify Gas (B)
- When CuS reacts with dilute H₂SO₄, it produces hydrogen sulfide gas (H₂S) and copper sulfate (CuSO₄).
- Therefore, gas (B) produced is H₂S.

**Conclusion:** (B) = H₂S

### Step 3: Identify Acid (C)
- The gas (B) (H₂S) is passed into a solution of an acid (C) which produces a white turbidity (D).
- H₂S reacts with nitric acid (HNO₃) to produce sulfur and nitrogen oxides, which can cause turbidity due to the formation of sulfur.

**Conclusion:** (C) = HNO₃

### Step 4: Identify White Turbidity (D)
- The white turbidity (D) formed from the reaction of H₂S with HNO₃ is sulfur (S), which appears as a white precipitate.

**Conclusion:** (D) = S (Sulfur)

### Step 5: Identify Compound (E)
- The gas (B) (H₂S) is then passed into an acidified solution of compound (E), which gives a precipitate (F) that is soluble in dilute HNO₃.
- The only acid that fits this description and can form a precipitate with H₂S is Copper(II) sulfate (CuSO₄), which forms copper sulfide (CuS) as a black precipitate.

**Conclusion:** (E) = CuSO₄

### Step 6: Identify Precipitate (F)
- When H₂S reacts with CuSO₄, it forms copper sulfide (CuS), which is a black precipitate.

**Conclusion:** (F) = CuS

### Step 7: Identify Compound (G)
- The black precipitate (CuS) is treated with excess NH₄OH after boiling, which forms an intense blue colored complex.
- This complex is known as tetraamminecopper(II) nitrate, represented as [Cu(NH₃)₄](NO₃)₂.

**Conclusion:** (G) = [Cu(NH₃)₄](NO₃)₂

### Step 8: Identify Compound (H)
- The intense blue colored compound (G) reacts with acetic acid (CH₃COOH) and K₄[Fe(CN)₆] to form a chocolate brown precipitate (H).
- This precipitate is known as copper ferrocyanide, represented as Cu₂[Fe(CN)₆].

**Conclusion:** (H) = Cu₂[Fe(CN)₆]

### Step 9: Identify the White Precipitate from BaCl₂
- The question states that when an aqueous solution of BaCl₂ is added to an aqueous solution of (E), a white precipitate insoluble in dilute HCl is formed.
- This white precipitate is barium sulfate (BaSO₄), confirming that (E) is indeed CuSO₄.

**Final Conclusions:**
- (A) = CuS
- (B) = H₂S
- (C) = HNO₃
- (D) = S (Sulfur)
- (E) = CuSO₄
- (F) = CuS
- (G) = [Cu(NH₃)₄](NO₃)₂
- (H) = Cu₂[Fe(CN)₆]