Match the compounds listed in column I with characteristic listed in column II.
`{:("Column-I","Column-II"),((A)B_2H_6,(p)"Tetrahedral hybridisation"),((B)Al_2Cl_6,(q)"Trigonal planar hybridisation"),((C )BeCl_2(s),(r)"Empty orbital of central atom participate in hybridisation"),((D)(SiH_3)_3N,(s)"Two types of bonds"),(,(t)"Two type of bond angles"):}`

Explanation

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

### Step 1: Analyze Compound A (B₂H₆)
- **Structure**: B₂H₆ (Diborane) has a structure where two boron atoms are bonded to hydrogen atoms. It has a unique bonding where there are two-center two-electron bonds and three-center two-electron bonds.
- **Hybridization**: The boron atoms undergo sp³ hybridization due to the presence of four sigma bonds (two B-H bonds and two B-B bonds).
- **Matching**:
- (p) "Tetrahedral hybridization" - **Correct** (sp³ hybridization).
- (q) "Trigonal planar hybridization" - **Incorrect**.
- (r) "Empty orbital of central atom participate in hybridization" - **Correct** (boron has empty p orbitals).
- (s) "Two types of bonds" - **Correct** (two-center two-electron and three-center two-electron bonds).
- (t) "Two types of bond angles" - **Correct** (bond angles of 120° and 97°).

### Step 2: Analyze Compound B (Al₂Cl₆)
- **Structure**: Al₂Cl₆ (Aluminum chloride) has a dimeric structure where aluminum atoms are bonded to chlorine atoms.
- **Hybridization**: The aluminum atoms undergo sp³ hybridization as they form four sigma bonds.
- **Matching**:
- (p) "Tetrahedral hybridization" - **Correct** (sp³ hybridization).
- (q) "Trigonal planar hybridization" - **Incorrect**.
- (r) "Empty orbital of central atom participate in hybridization" - **Correct** (aluminum has empty p orbitals).
- (s) "Two types of bonds" - **Correct** (two-center two-electron and three-center two-electron bonds).
- (t) "Two types of bond angles" - **Correct** (different bond angles).

### Step 3: Analyze Compound C (BeCl₂(s))
- **Structure**: BeCl₂ in solid state forms a polymeric structure.
- **Hybridization**: The beryllium atom undergoes sp³ hybridization in the polymeric structure.
- **Matching**:
- (p) "Tetrahedral hybridization" - **Correct** (sp³ hybridization).
- (q) "Trigonal planar hybridization" - **Incorrect**.
- (r) "Empty orbital of central atom participate in hybridization" - **Correct** (beryllium has empty p orbitals).
- (s) "Two types of bonds" - **Incorrect** (only one type of bond).
- (t) "Two types of bond angles" - **Incorrect** (only one bond angle).

### Step 4: Analyze Compound D ((SiH₃)₃N)
- **Structure**: (SiH₃)₃N (Silicon nitride) has a nitrogen atom bonded to three silicon atoms.
- **Hybridization**: The nitrogen atom undergoes sp² hybridization due to back bonding with silicon.
- **Matching**:
- (p) "Tetrahedral hybridization" - **Incorrect**.
- (q) "Trigonal planar hybridization" - **Correct** (sp² hybridization).
- (r) "Empty orbital of central atom participate in hybridization" - **Incorrect** (nitrogen does not have empty orbitals).
- (s) "Two types of bonds" - **Incorrect** (only one type of bond).
- (t) "Two types of bond angles" - **Incorrect** (only one bond angle of 120°).

### Conclusion of Matching:
- **A** (B₂H₆) matches with (p), (r), (s), (t).
- **B** (Al₂Cl₆) matches with (p), (r), (s), (t).
- **C** (BeCl₂) matches with (p), (r).
- **D** ((SiH₃)₃N) matches with (q).

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