`(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 color at that of (C).
`(v)` A solution of (A) on treatment with a solution of barium chloride gave a white precipitate of compound (F) Which was insoluble in conc. `HNO_(3)` and conc. HCl.
Consider the following statement `:`
`(I) ` Anions of both (B) and (C ) are diamagnetic and have tetradhedral geometray.
`(II)` Anions of both (B) and ( C) are paramagnetic and have tetrahedral geometry.
`(III)` Anions of (B) is paramagnetic and that of (C ) is diamagnetic but both have same tetrahedral geometry.
`(IV)` Green coloured compound (B) in a neutral of acidic medium disproportionates to give (C ) and (D).
Of these select the correct one from the code given `:`

Options

A. `I and III only`
B. `II and III only `
C. `II and IV only `
D. `III and IV only`

Explanation

To solve the question step-by-step, we will analyze each part of the problem and identify the compounds A, B, C, D, E, and F based on the reactions described.

### Step 1: Identify Compound A
- **Given:** A powdered substance (A) on treatment with a fusion mixture gives a green-colored compound (B).
- **Analysis:** The powdered substance is likely manganese(II) sulfate (MnSO4), which is a common powdered form of manganese.
- **Fusion Mixture:** The fusion mixture typically consists of sodium carbonate (Na2CO3) and potassium nitrate (KNO3).
- **Reaction:** When MnSO4 is treated with the fusion mixture, it forms sodium manganate (Na2MnO4), which is green in color.
- **Conclusion:**
- **A = MnSO4**
- **B = Na2MnO4 (green colored compound)**

### Step 2: Identify Compound C
- **Given:** The solution of (B) in boiling water on acidification with dilute H2SO4 gives a pink-colored compound (C).
- **Analysis:** When Na2MnO4 is treated with dilute H2SO4, it forms sodium permanganate (NaMnO4), which is pink in color.
- **Conclusion:**
- **C = NaMnO4 (pink colored compound)**

### Step 3: Identify Compound D
- **Given:** The aqueous solution of (A) on treatment with NaOH and Br2 water gives a compound (D).
- **Analysis:** When MnSO4 is treated with NaOH and Br2, it produces manganese dioxide (MnO2), which is a brown precipitate.
- **Conclusion:**
- **D = MnO2**

### Step 4: Identify Compound E
- **Given:** A solution of (D) in concentrated HNO3 on treatment with lead peroxide at boiling temperature produces a compound (E) which was of the same color as that of (C).
- **Analysis:** MnO2 reacts with concentrated HNO3 and lead peroxide to form permanganic acid (HMnO4), which has a similar color to sodium permanganate (C).
- **Conclusion:**
- **E = HMnO4 (pink colored compound)**

### Step 5: Identify Compound F
- **Given:** A solution of (A) on treatment with a solution of barium chloride gives a white precipitate of compound (F) which was insoluble in concentrated HNO3 and HCl.
- **Analysis:** When MnSO4 reacts with BaCl2, it forms barium sulfate (BaSO4), which is a white precipitate and is insoluble in concentrated HNO3 and HCl.
- **Conclusion:**
- **F = BaSO4 (white precipitate)**

### Summary of Compounds
- **A = MnSO4**
- **B = Na2MnO4 (green)**
- **C = NaMnO4 (pink)**
- **D = MnO2**
- **E = HMnO4 (pink)**
- **F = BaSO4 (white)**

### Evaluation of Statements
1. **Statement I:** Anions of both (B) and (C) are diamagnetic and have tetrahedral geometry.
- **False:** The anion of B (MnO4^2-) is paramagnetic (Mn in +6 state), while C (MnO4^-) is diamagnetic (Mn in +7 state).

2. **Statement II:** Anions of both (B) and (C) are paramagnetic and have tetrahedral geometry.
- **False:** Same reasoning as above.

3. **Statement III:** Anions of (B) is paramagnetic and that of (C) is diamagnetic but both have the same tetrahedral geometry.
- **True:** This statement is correct.

4. **Statement IV:** Green colored compound (B) in a neutral or acidic medium disproportionates to give (C) and (D).
- **True:** Na2MnO4 can disproportionate to form NaMnO4 and MnO2.

### Final Answer
The correct statements are **III and IV**.