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Explain sp2 hybridization by taking BCl3 as an example.

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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 2px1 2py1

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 σsp-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 σsp-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°.

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