Match the species in Column I with the type of hybrid orbitals in Column II.
`{:(,"Column I",,"Column II"),((i),BF_(3),(a),sp^(3)d),((ii),H_(2)O,(b),sp^(2)),((iii),PCl_(5),(c),sp^(3)d^(2)),((iv),SF_(6),(d),sp^(3)):}`

Answer

To solve the matching of species in Column I with the type of hybrid orbitals in Column II, we will analyze each species and determine its hybridization.

### Step-by-Step Solution:

1. **Identify the hybridization of BF₃ (Boron Trifluoride)**:
- Boron has an atomic number of 5, with the electron configuration of 1s² 2s² 2p¹.
- In BF₃, boron forms three bonds with fluorine atoms, which requires it to use three sp² hybrid orbitals (one 2s and two 2p orbitals).
- **Match**: (i) BF₃ → (b) sp²

2. **Identify the hybridization of H₂O (Water)**:
- Oxygen has an atomic number of 8, with the electron configuration of 1s² 2s² 2p⁴.
- In H₂O, oxygen forms two bonds with hydrogen and has two lone pairs, leading to sp³ hybridization (one 2s and three 2p orbitals).
- **Match**: (ii) H₂O → (d) sp³

3. **Identify the hybridization of PCl₅ (Phosphorus Pentachloride)**:
- Phosphorus has an atomic number of 15, with the electron configuration of 1s² 2s² 2p⁶ 3s² 3p³.
- In PCl₅, phosphorus forms five bonds with chlorine atoms, which requires it to use sp³d hybrid orbitals (one 3s, three 3p, and one 3d orbital).
- **Match**: (iii) PCl₅ → (a) sp³d

4. **Identify the hybridization of SF₆ (Sulfur Hexafluoride)**:
- Sulfur has an atomic number of 16, with the electron configuration of 1s² 2s² 2p⁶ 3s² 3p⁴.
- In SF₆, sulfur forms six bonds with fluorine atoms, which requires it to use sp³d² hybrid orbitals (one 3s, three 3p, and two 3d orbitals).
- **Match**: (iv) SF₆ → (c) sp³d²

### Final Matches:
- (i) BF₃ → (b) sp²
- (ii) H₂O → (d) sp³
- (iii) PCl₅ → (a) sp³d
- (iv) SF₆ → (c) sp³d²