(i) A powdered substance (A) on treatment with fusion mixture gives a green coloured compound (B). (ii) The solution of (B) in boiling water on acidification with dilute `H_(2)SO_(4)` gives a pink coloured compound `(C )`.
`(iii)` The aqueous solution of (A) on treatment with NaOH and `Br_(2)-` water gives a compound (D).
(iv) A solution of (D) in conc. `HNO_(3)` on treatment with lead peroxide at boiling temperature produced a compound (E) which was of the same colour at that of (C) . Identify `A , B ,C ,D , E`

Explanation

To solve the problem step by step, we will identify each compound (A, B, C, D, and E) based on the reactions described.

### Step 1: Identify Compound A
The powdered substance (A) reacts with a fusion mixture to produce a green-colored compound (B). The fusion mixture typically consists of sodium carbonate (Na2CO3) and potassium nitrate (KNO3).

When manganese sulfate (MnSO4) reacts with the fusion mixture, it forms sodium manganate (Na2MnO4), which is green in color.

**Thus, A = MnSO4 and B = Na2MnO4.**

### Step 2: Identify Compound C
The green compound (B) is then treated with boiling water and acidified with dilute sulfuric acid (H2SO4).

The reaction of sodium manganate (Na2MnO4) with dilute H2SO4 produces sodium permanganate (NaMnO4), which is pink in color.

**Thus, C = NaMnO4.**

### Step 3: Identify Compound D
Next, the aqueous solution of A (MnSO4) is treated with sodium hydroxide (NaOH) and bromine water (Br2).

This reaction leads to the formation of manganese dioxide (MnO2), along with sodium bromide (NaBr) and sodium sulfate (Na2SO4).

**Thus, D = MnO2.**

### Step 4: Identify Compound E
Finally, the compound D (MnO2) is treated with concentrated nitric acid (HNO3) and lead peroxide (PbO2) at boiling temperature.

This reaction produces permanganic acid (HMnO4), which has the same pink color as compound C.

**Thus, E = HMnO4.**

### Summary of Compounds
- A = MnSO4 (Manganese sulfate)
- B = Na2MnO4 (Sodium manganate)
- C = NaMnO4 (Sodium permanganate)
- D = MnO2 (Manganese dioxide)
- E = HMnO4 (Permanganic acid)