If Be forms two covalent bonds with two chlorine atoms, one bond should be σ2s-3p due to the overlap of ‘2s’ orbital of Be, the ‘3pz ’ orbital of one chlorine atom and the other bond should be σ2p-3p due to the overlap of ‘2px ’ orbital of Be atom the 3p orbital of the other chlorine atom. As the orbitals overlapping are different, the bond strengths of two Be-Cl must be different. But, both bonds are of same strength and ClBeCl is 1800 . To explain the discrepancies like this a phenomenon called ‘hybridisation of atomic orbitals’ was proposed by Linus Pauling (1931).

Hybridisation is a phenomenon of intermixing of atomic orbitals of almost equal energy which are present in the outer shells of the atom and their reshuffling or redistribution into the same number of orbitals but with equal properties like energy and shape.

Be atom in its excited state allows its 2s orbital and 2px orbital which contain unpaired electrons to intermix and redistribute to two identical orbitals. As per Hund’s rule each orbital gets one electron. The new orbitals based on the types of orbitals that have undergone hybridisation are called sp orbitals. The two sp orbitals of Be get separated by 1800 . Now, each chlorine atom comes with its 3pz 1 orbital and overlaps it the sp orbitals of Be forming two identical Be-Cl bonds (σsp-p bonds) . ClBeCl = 1800 . Both the bonds are of same strength. Formation of BF3 molecule

5 B has electronic configuration 1s2 2s2 2px 1 . As it has one unpaired electron (2px 1 ) it should form only one covalent bond to give B-F molecule. But we get practically BF3 molecule. To explain this, it is suggested that, i. Boron (B) first undergoes excitation to get electronic configuration 1s2 2s1 2px 1 2py 1 . ii. As it forms three identical B-F bonds in BF3 , it is suggested that excited ‘B’ atom undergoes hybridisation. There is an intermixing of 2s, 2px , 2py orbitals and their redistribution into three identical orbitals called sp2 hybrid orbitals. For three sp2 orbitals to get separated to have minimum