(i)A black coloured compound `(B)` is formed on passing `H_(2)S` through the solution of a compound `(A)` in `NH_(4)OH`
(ii)`(B)` on treatment with `HCl` and potassium chlorate or aquaregia gives `(A)`
(iii)`(A)` on treatment with `KCN` gives a buff/reddish-brown coloured precipitate which dissolves in excess of this reagent forming a compound `( C)`.
(iv)The compound `( C)` is changed into a compound `(D)` when its aqueous solution is boiled in air.
(v)The solution of `(A)` was treated with excess of `NaHCO_(3)` & then with bromine water. On cooling & shaking for some time, a green colour of compound `(E)` is formed.No change is observed on heating.
Identify `(A)` to `(E)` and give chemical equations.

Explanation

To solve the problem step by step, we will identify the compounds A to E based on the provided information and chemical reactions.

### Step 1: Identify Compound (A)
From the first statement, we know that a black colored compound (B) is formed when H₂S is passed through a solution of compound (A) in NH₄OH. Since H₂S typically reacts with metal salts to form metal sulfides, compound (A) must be a metal salt. Given the context, we can deduce that compound (A) is cobalt(II) chloride, which has the formula:

**(A) = CoCl₂ (Cobalt(II) chloride)**

### Step 2: Identify Compound (B)
When H₂S reacts with cobalt(II) chloride in the presence of NH₄OH, it forms cobalt(II) sulfide, which is a black precipitate. Therefore, we can identify compound (B) as:

**(B) = CoS (Cobalt(II) sulfide)**

### Step 3: Identify Compound (C)
The third statement indicates that compound (A) reacts with KCN to produce a buff/reddish-brown precipitate that dissolves in excess KCN. The reaction of cobalt(II) chloride with KCN forms a complex ion. The compound formed is:

**(C) = K₄[Co(CN)₆] (Potassium hexacyanocobaltate(II))**

### Step 4: Identify Compound (D)
The fourth statement mentions that compound (C) is converted into compound (D) when its aqueous solution is boiled in air. The oxidation of the hexacyanocobaltate(II) complex in air leads to the formation of hexacyanocobaltate(III):

**(D) = K₃[Co(CN)₆] (Potassium hexacyanocobaltate(III))**

### Step 5: Identify Compound (E)
The fifth statement describes the reaction of compound (A) with NaHCO₃ followed by bromine water, resulting in a green-colored compound (E). The reaction proceeds as follows:

1. Cobalt(II) chloride reacts with NaHCO₃ to form a cobalt carbonate complex.
2. The green color indicates the formation of a cobalt(II) carbonate complex.

Thus, we can identify compound (E) as:

**(E) = Na₃[Co(CO₃)₃] (Sodium hexacarbonatocobaltate(II))**

### Summary of Compounds
- **(A) = CoCl₂ (Cobalt(II) chloride)**
- **(B) = CoS (Cobalt(II) sulfide)**
- **(C) = K₄[Co(CN)₆] (Potassium hexacyanocobaltate(II))**
- **(D) = K₃[Co(CN)₆] (Potassium hexacyanocobaltate(III))**
- **(E) = Na₃[Co(CO₃)₃] (Sodium hexacarbonatocobaltate(II))**

### Chemical Equations
1. **Formation of (B)**:
\[ \text{CoCl}_2 + \text{H}_2\text{S} \rightarrow \text{CoS} \downarrow + 2\text{HCl} \]

2. **Formation of (A) from (B)**:
\[ \text{CoS} + 4\text{HCl} + \text{KClO}_3 \rightarrow \text{CoCl}_2 + \text{NOCl} + 2\text{H}_2\text{O} \]

3. **Formation of (C)**:
\[ \text{CoCl}_2 + 6\text{KCN} \rightarrow \text{K}_4[\text{Co(CN)}_6] + 2\text{KCl} \]

4. **Formation of (D)**:
\[ \text{K}_4[\text{Co(CN)}_6] + \text{O}_2 + \text{H}_2\text{O} \rightarrow \text{K}_3[\text{Co(CN)}_6] + 4\text{OH}^- \]

5. **Formation of (E)**:
\[ \text{CoCl}_2 + 3\text{NaHCO}_3 \rightarrow \text{Na}_3[\text{Co(CO}_3)_3] + 3\text{HCl} + 3\text{CO}_2 + 3\text{H}_2\text{O} \]