[H=1, C=12,O=16, Na=23, P=31,S=32]
`{:("Column-I","Column-II"),((A)4.1 g H_2SO_3,(p)"200 ml of 0.5 N NaOH is used for complete neutralization"),((B)4.9 g H_3PO_4,(q)"200 millimoles of oxygen atoms"),((C )4.5 g "oxalic acid"(H_2C_2O_4),(r)"Central atom has its highest oxidation number"),((D)4.3 g Na_2CO_3,(s)"May react with an oxidising agent"),(,(t)"Shape around central atom is regular"):}`

Explanation

To solve the question, we will analyze each compound in Column I and match it with the appropriate statement in Column II.

### Step-by-Step Solution:

**Step 1: Analyze Compound A - H₂SO₃**
- Given mass: 4.1 g
- Molar mass of H₂SO₃ = (2*1) + (32) + (3*16) = 82 g/mol
- Moles of H₂SO₃ = Mass / Molar mass = 4.1 g / 82 g/mol = 0.050 moles = 50 millimoles
- Equivalent of H₂SO₃ (n-factor = 2) = 50 * 2 = 100 millimoles

**Step 2: Analyze Statement P**
- 200 ml of 0.5 N NaOH = 200 ml * 0.5 eq/L = 100 millimoles
- This matches the equivalent of H₂SO₃.
- **Conclusion:** A matches with P.

**Step 3: Analyze Compound B - H₃PO₄**
- Given mass: 4.9 g
- Molar mass of H₃PO₄ = (3*1) + (31) + (4*16) = 98 g/mol
- Moles of H₃PO₄ = 4.9 g / 98 g/mol = 0.050 moles = 50 millimoles
- Equivalent of H₃PO₄ (n-factor = 3) = 50 * 3 = 150 millimoles

**Step 4: Analyze Statement Q**
- 200 millimoles of oxygen atoms: In 1 mole of H₃PO₄, there are 4 moles of oxygen.
- Therefore, in 50 millimoles of H₃PO₄, there are 50 * 4 = 200 millimoles of oxygen.
- **Conclusion:** B matches with Q.

**Step 5: Analyze Compound C - Oxalic Acid (H₂C₂O₄)**
- Given mass: 4.5 g
- Molar mass of H₂C₂O₄ = (2*1) + (2*12) + (4*16) = 90 g/mol
- Moles of H₂C₂O₄ = 4.5 g / 90 g/mol = 0.050 moles = 50 millimoles
- Equivalent of H₂C₂O₄ (n-factor = 2) = 50 * 2 = 100 millimoles

**Step 6: Analyze Statement P Again**
- 200 ml of 0.5 N NaOH = 100 millimoles, which matches the equivalent of H₂C₂O₄.
- **Conclusion:** C matches with P.

**Step 7: Analyze Statement R**
- Highest oxidation state of carbon in H₂C₂O₄ is +4, so this is not the highest oxidation state.
- **Conclusion:** C does not match with R.

**Step 8: Analyze Compound D - Na₂CO₃**
- Given mass: 4.3 g
- Molar mass of Na₂CO₃ = (2*23) + (12) + (3*16) = 106 g/mol
- Moles of Na₂CO₃ = 4.3 g / 106 g/mol = 0.0406 moles = 40.6 millimoles
- Equivalent of Na₂CO₃ (n-factor = 2) = 40.6 * 2 = 81.2 millimoles

**Step 9: Analyze Statement P Again**
- 200 ml of 0.5 N NaOH = 100 millimoles, does not match.
- **Conclusion:** D does not match with P.

**Step 10: Analyze Statement R Again**
- Highest oxidation state of carbon in Na₂CO₃ is +4, so this is not the highest oxidation state.
- **Conclusion:** D does not match with R.

**Step 11: Analyze Statement T**
- The shape around the central atom (carbon) in Na₂CO₃ is trigonal planar, which is regular.
- **Conclusion:** D matches with T.

### Final Matching:
- A → P
- B → Q
- C → S
- D → T