Match the compounds listed in column I with characteristic listed in column II.
`{:("Column-I","Column-II"),((A)"Hypophosphorus acid",(p)"All hydrogen are ioniasable in water"),((B)"Orthoboric acid",(q)"Monobasic acid"),((C )"Hypophosphoric acid",(r)sp^3"hybridised central atom(s)"),((D)"Pyrosulphurous acid",(s)"Lewis acid"),(,(t)"Central atom contains one lone pair of electrons"):}`

Explanation

To solve the problem of matching the compounds listed in Column I with the characteristics in Column II, we will analyze each compound and its properties step by step.

### Step 1: Analyze Hypophosphorus Acid (H₃PO₂)
- **Formula**: H₃PO₂
- **Characteristics**:
- **(p)** All hydrogen are ionizable in water: **False**. Only one hydrogen is acidic.
- **(q)** Monobasic acid: **True**. It has only one ionizable hydrogen.
- **(r)** sp³ hybridized central atom(s): **True**. Phosphorus in H₃PO₂ has four sigma bonds (three with hydrogen and one with oxygen).
- **(s)** Lewis acid: **False**. It does not have a vacant orbital to accept electrons.
- **(t)** Central atom contains one lone pair of electrons: **False**. Phosphorus does not have a lone pair in this structure.

**Matching for A**: (q) and (r)

### Step 2: Analyze Orthoboric Acid (H₃BO₃)
- **Formula**: H₃BO₃
- **Characteristics**:
- **(p)** All hydrogen are ionizable in water: **False**. Not all hydrogens are acidic.
- **(q)** Monobasic acid: **True**. It can donate one proton.
- **(r)** sp³ hybridized central atom(s): **False**. Boron is sp² hybridized due to three sigma bonds.
- **(s)** Lewis acid: **True**. Boron can accept an electron pair.
- **(t)** Central atom contains one lone pair of electrons: **False**. Boron has no lone pairs.

**Matching for B**: (q) and (s)

### Step 3: Analyze Hypophosphoric Acid (H₄P₂O₆)
- **Formula**: H₄P₂O₆
- **Characteristics**:
- **(p)** All hydrogen are ionizable in water: **True**. All four hydrogens can ionize.
- **(q)** Monobasic acid: **False**. It has four ionizable hydrogens.
- **(r)** sp³ hybridized central atom(s): **True**. Both phosphorus atoms are sp³ hybridized.
- **(s)** Lewis acid: **False**. It does not have vacant orbitals.
- **(t)** Central atom contains one lone pair of electrons: **False**. Phosphorus does not have lone pairs.

**Matching for C**: (p) and (r)

### Step 4: Analyze Pyrosulphurous Acid (H₂S₂O₅)
- **Formula**: H₂S₂O₅
- **Characteristics**:
- **(p)** All hydrogen are ionizable in water: **True**. Both hydrogens can ionize.
- **(q)** Monobasic acid: **False**. It has two ionizable hydrogens.
- **(r)** sp³ hybridized central atom(s): **True**. The sulfur atoms are sp³ hybridized.
- **(s)** Lewis acid: **False**. Sulfur does not have vacant orbitals.
- **(t)** Central atom contains one lone pair of electrons: **True**. Sulfur has a lone pair.

**Matching for D**: (p), (r), and (t)

### Final Matching Summary:
- **A**: (q), (r)
- **B**: (q), (s)
- **C**: (p), (r)
- **D**: (p), (r), (t)