Match List I with List II and select the correct answer using the codes given below the lists `:`
`{:("List "I,,"List "II),(a.(SiH_(3))_(3)N,,i. "3 centre -2-electron bond"),(b.BF_(3),,ii. sp^(3)-"hybridization"),(c. SiO_(2),,iii.p pi-p pi " bond"),(d.B_(2)H_(6),,iv.p pi-d pi " bond"):}`
`:`

Explanation

To solve the problem of matching List I with List II, we will analyze each compound in List I and determine its corresponding bond type or hybridization from List II.

### Step-by-Step Solution:

1. **Identify Compound (a): SiH₃₃N**
- **Analysis**: Silicon (Si) has vacant d-orbitals. The lone pair of electrons on the nitrogen atom can undergo back bonding with Si, leading to the formation of a p pi-d pi bond.
- **Match**: This corresponds to option (iv) "p pi-d pi bond".
- **Result**: a → iv

2. **Identify Compound (b): BF₃**
- **Analysis**: Boron (B) in BF₃ is sp² hybridized. The bonding involves the p orbitals of boron and fluorine, leading to p pi-p pi bonds.
- **Match**: This corresponds to option (iii) "p pi-p pi bond".
- **Result**: b → iii

3. **Identify Compound (c): SiO₂**
- **Analysis**: Silicon dioxide (SiO₂) has silicon in an sp³ hybridization state. The structure involves tetrahedral geometry with four bonds formed.
- **Match**: This corresponds to option (ii) "sp³ hybridization".
- **Result**: c → ii

4. **Identify Compound (d): B₂H₆**
- **Analysis**: In diborane (B₂H₆), there are three-center two-electron bonds (banana bonds) formed between the boron and hydrogen atoms.
- **Match**: This corresponds to option (i) "3 center-2-electron bond".
- **Result**: d → i

### Final Summary of Matches:
- a → iv (p pi-d pi bond)
- b → iii (p pi-p pi bond)
- c → ii (sp³ hybridization)
- d → i (3 center-2-electron bond)

### Final Answer:
The correct matching is:
- a - iv
- b - iii
- c - ii
- d - i

Thus, the final answer in code form is: 4 3 2 1.

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