(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 )`.
The oxidation state of central metal ions of (A), (B) and (C ) compounds are respectively `:`

Options

A. `+II, +VI and +VII`
B. `+II, +VI and +VI`
C. `+II, +VII and +VII`
D. `+VI, +VII and +VII`

Explanation

To solve the problem, we need to identify the powdered substance (A), the green-colored compound (B), and the pink-colored compound (C), and then determine the oxidation states of the central metal ions in each of these compounds.

### Step 1: Identify Compound A
The powdered substance (A) is mentioned to react 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 this fusion mixture, it produces sodium manganate (Na2MnO4), which is a green-colored compound.

**Thus, Compound A is MnSO4.**

### Step 2: Identify Compound B
From the reaction of MnSO4 with the fusion mixture, we identified that the green-colored compound (B) is sodium manganate.

**Thus, Compound B is Na2MnO4.**

### Step 3: Identify Compound C
Next, we need to determine what happens when compound B (Na2MnO4) is treated with dilute sulfuric acid (H2SO4). The reaction of sodium manganate with dilute H2SO4 produces sodium permanganate (NaMnO4), which is a pink-colored compound.

**Thus, Compound C is NaMnO4.**

### Step 4: Determine Oxidation States
Now we will find the oxidation states of manganese (Mn) in each of the compounds A, B, and C.

1. **For Compound A (MnSO4)**:
- The sulfate ion (SO4^2-) has a charge of -2.
- Let the oxidation state of Mn be x.
- The equation is: x + (-2) = 0
- Therefore, x = +2.
- **Oxidation state of Mn in A = +2.**

2. **For Compound B (Na2MnO4)**:
- Sodium (Na) has an oxidation state of +1.
- Let the oxidation state of Mn be y.
- The equation is: 2(+1) + y + 4(-2) = 0
- Therefore, 2 + y - 8 = 0 → y = +6.
- **Oxidation state of Mn in B = +6.**

3. **For Compound C (NaMnO4)**:
- Sodium (Na) has an oxidation state of +1.
- Let the oxidation state of Mn be z.
- The equation is: +1 + z + 4(-2) = 0
- Therefore, 1 + z - 8 = 0 → z = +7.
- **Oxidation state of Mn in C = +7.**

### Final Answer
The oxidation states of the central metal ions of compounds A, B, and C are:
- **A (MnSO4): +2**
- **B (Na2MnO4): +6**
- **C (NaMnO4): +7**

### Summary
- Compound A: MnSO4 → Oxidation state of Mn = +2
- Compound B: Na2MnO4 → Oxidation state of Mn = +6
- Compound C: NaMnO4 → Oxidation state of Mn = +7