Question

Explain sp² hybridization by taking BCl₃ as an example.

Answer 1

The electronic configuration of 'B' in ground state is 1s² 2s² 2p¹ with only one unpaired electron. Since the formation of three bonds with chlorine atoms require three unpaired electrons, there is promotion of one of 2s electron into the 2p sublevel by absorbing energy. 

Thus Boron atom gets electronic configuration: 1s² 2s² 2p_x¹ 2p_y¹ . 

However to account for the trigonal planar shape of this BCl₃ molecule, sp² hybridization before bond formation was put forwarded.

In the excited state, Boron undergoes sp² hybridization by using a 2s and two 2p orbitals to give three half filled sp² hybrid orbitals which are oriented in trigonal planar symmetry. 

Boron forms three σ_sp-p bonds with three chlorine atoms by using its half filled sp² hybrid orbitals. Each chlorine atom uses it's half filled p-orbital for the σ-bond formation. Thus the shape of BCl₃ is trigonal planar with bond angles equal to 120°.

Boron forms three σ_sp-p bonds with three chlorine atoms by using its half filled sp² hybrid orbitals. Each chlorine atom uses it's half filled p-orbital for the σ-bond formation.

Thus the shape of BCl₃ is trigonal planar with bond angles equal to 120°.