(i)A blue coloured compound `(A)` on heating gives two product `(B)` & `( C)`.
(ii)A metal `(D)` is deposited on passing hydrogen through heated `(B)`.
(iii)The solution of `(B)` in `HCl` on treatment with the `[Fe(CN)_(6)]^(4-)` gives a chocolate brown coloured precipitate of compound `(E)`.
(iv)`(C )` turns lime water milky which disappears on continuous passage of `(C )` forming a compound `(F)`.
Identigfy `(A)` to `(F)` and give chemical equations for the reactions at step (i) to (iv).

Explanation

To solve the problem step by step, we will identify the compounds (A) through (F) and provide the corresponding chemical equations for each reaction described in the question.

### Step 1: Identify Compound (A)
- **Given:** A blue colored compound (A) on heating gives two products (B) and (C).
- **Solution:** The blue colored compound is Copper(II) carbonate, \( \text{CuCO}_3 \).
- **Chemical Equation:**
\[ \text{CuCO}_3 (s) \xrightarrow{\text{heat}} \text{CuO} (s) + \text{CO}_2 (g) \]
Here, (B) is Copper(II) oxide (\( \text{CuO} \)) and (C) is Carbon dioxide (\( \text{CO}_2 \)).

### Step 2: Identify Compound (D)
- **Given:** A metal (D) is deposited on passing hydrogen through heated (B).
- **Solution:** When hydrogen is passed over heated Copper(II) oxide (\( \text{CuO} \)), it reduces to Copper metal.
- **Chemical Equation:**
\[ \text{CuO} (s) + \text{H}_2 (g) \xrightarrow{\text{heat}} \text{Cu} (s) + \text{H}_2\text{O} (g) \]
Here, (D) is Copper metal (\( \text{Cu} \)).

### Step 3: Identify Compound (E)
- **Given:** The solution of (B) in HCl on treatment with \( [\text{Fe(CN)}_6]^{4-} \) gives a chocolate brown colored precipitate of compound (E).
- **Solution:** Copper(II) ions from \( \text{CuO} \) react with \( [\text{Fe(CN)}_6]^{4-} \) to form Copper(I) ferrocyanide, which is chocolate brown.
- **Chemical Equation:**
\[ \text{Cu}^{2+} (aq) + [\text{Fe(CN)}_6]^{4-} (aq) \rightarrow \text{Cu}_2[\text{Fe(CN)}_6] (s) \quad \text{(chocolate brown precipitate)} \]
Here, (E) is Copper(I) ferrocyanide.

### Step 4: Identify Compound (F)
- **Given:** (C) turns lime water milky which disappears on continuous passage of (C) forming compound (F).
- **Solution:** Carbon dioxide (\( \text{CO}_2 \)) reacts with lime water (calcium hydroxide, \( \text{Ca(OH)}_2 \)) to form calcium carbonate (\( \text{CaCO}_3 \)), which is milky. On prolonged passage of \( \text{CO}_2 \), it forms calcium bicarbonate (\( \text{Ca(HCO}_3)_2 \)), which is colorless.
- **Chemical Equations:**
1. Initial reaction:
\[ \text{CO}_2 (g) + \text{Ca(OH)}_2 (aq) \rightarrow \text{CaCO}_3 (s) \quad \text{(milky)} \]
2. Prolonged reaction:
\[ \text{CaCO}_3 (s) + \text{CO}_2 (g) + \text{H}_2\text{O} (l) \rightarrow \text{Ca(HCO}_3)_2 (aq) \quad \text{(colorless)} \]
Here, (F) is Calcium bicarbonate (\( \text{Ca(HCO}_3)_2 \)).

### Summary of Compounds
- (A) = Copper(II) carbonate (\( \text{CuCO}_3 \))
- (B) = Copper(II) oxide (\( \text{CuO} \))
- (C) = Carbon dioxide (\( \text{CO}_2 \))
- (D) = Copper metal (\( \text{Cu} \))
- (E) = Copper(I) ferrocyanide
- (F) = Calcium bicarbonate (\( \text{Ca(HCO}_3)_2 \))