(i)A powdered substance (A) on treatment with fusion mixture `(Na_2CO_3+KNO_3)` gives a green coloured compound (B).
(ii)The solution of (B) in boiling water on acidification with dilute `H_2SO_4` gives a pink coloured compound ( C).
(iii)The aqueous solution of (A) on treatment with excess of NaOH and Bromine water gives a compound (D)
(iv)A solution of (D) in conc. `HNO_3` on treatment with `PbO_2` at boiling temperature produced a compound (E) which was of the same colour at that of (C ).
Sum of oxidation number of central atom of A,B,C,D & E is:

Explanation

To solve the problem step by step, we will identify the compounds mentioned, their reactions, and calculate the oxidation states of the central atom (manganese) in each compound.

### Step 1: Identify Compound A and Reaction to Form B
- **Given:** A powdered substance (A) reacts with a fusion mixture of `Na2CO3 + KNO3` to give a green compound (B).
- **Identified Compound A:** Manganese(II) sulfate, `MnSO4`.
- **Reaction:**
\[ 2 \text{MnSO}_4 + 2 \text{Na}_2\text{CO}_3 + 2 \text{KNO}_3 \rightarrow 2 \text{Na}_2\text{MnO}_4 + 2 \text{KNO}_2 + 2 \text{Na}_2\text{SO}_4 + 2 \text{CO}_2 \]
- **Compound B:** Sodium manganate, `Na2MnO4` (green color).

### Step 2: Reaction of B with Dilute H2SO4 to Form C
- **Given:** The solution of B in boiling water on acidification with dilute `H2SO4` gives a pink compound (C).
- **Reaction:**
\[ 3 \text{Na}_2\text{MnO}_4 + 2 \text{H}_2\text{SO}_4 \rightarrow 2 \text{NaMnO}_4 + 2 \text{MnO}_2 + 2 \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O} \]
- **Compound C:** Sodium permanganate, `NaMnO4` (pink color).

### Step 3: Reaction of A with NaOH and Bromine Water to Form D
- **Given:** Aqueous solution of A treated with excess NaOH and bromine water gives compound D.
- **Reaction:**
\[ \text{MnSO}_4 + 4 \text{NaOH} + \text{Br}_2 \rightarrow \text{MnO}_2 + 2 \text{Na}_2\text{SO}_4 + 2 \text{NaBr} + 2 \text{H}_2\text{O} \]
- **Compound D:** Manganese dioxide, `MnO2`.

### Step 4: Reaction of D with Concentrated HNO3 and PbO2 to Form E
- **Given:** A solution of D in concentrated `HNO3` on treatment with `PbO2` at boiling temperature produces compound E, which is the same color as C.
- **Reaction:**
\[ 2 \text{MnO}_2 + 4 \text{HNO}_3 + 2 \text{PbO}_2 \rightarrow 2 \text{HMnO}_4 + 2 \text{Pb(NO}_3)_2 + 2 \text{H}_2\text{O} \]
- **Compound E:** Manganic acid, `HMnO4` (pink color).

### Step 5: Calculate the Oxidation States of Manganese in Each Compound
1. **Compound A (MnSO4):**
- Oxidation state of Mn: \( x + 4(-2) = 0 \) → \( x = +2 \)

2. **Compound B (Na2MnO4):**
- Oxidation state of Mn: \( x + 4(-2) = -1 \) → \( x = +6 \)

3. **Compound C (NaMnO4):**
- Oxidation state of Mn: \( x + 4(-2) = -1 \) → \( x = +7 \)

4. **Compound D (MnO2):**
- Oxidation state of Mn: \( x + 2(-2) = 0 \) → \( x = +4 \)

5. **Compound E (HMnO4):**
- Oxidation state of Mn: \( x + 4(-2) = -1 \) → \( x = +7 \)

### Step 6: Sum of the Oxidation States
- Sum = Oxidation state of A + B + C + D + E
- Sum = \( +2 + 6 + 7 + 4 + 7 = 26 \)

### Final Answer:
The sum of the oxidation numbers of the central atom of A, B, C, D, and E is **26**.