repulsion the angle between any two orbitals is 1200 at the central atom and each sp2 orbital gets one electron. Now three fluorine atoms overlap their 2pz orbitals containing unpaired electrons (F9 1s2 2s2 2px 2 2py 2 2pz 1 ) the three sp2 orbitals of ‘B’ that contain unpaired electrons to form three σsp2 -p bonds.

One nitrogen atom and three hydrogen atoms are present in one ammonia molecule. All the three N-H bonds are of same strength and HNH = 1070 48/7

N has electronic configuration 1s2 2s2 2px 1 2py 1 2pz 1 . If three hydrogen atoms overlap their 1s orbitals on the three ‘p’ orbitals of nitrogen atom, they give identical σ p-s bonds but, then the HNH should be equal to 900 where as 1070 48/ . To explain the discrepancy in the bond angle ‘N’ atom is said to undergo sp3 hybridisation. In this process ‘2s’ and 2px , 2py , 2pz orbitals of nitrogen intermix and redistribute into four identical sp3 orbitals. One of the four sp3 orbitals get a pair of electrons and the other three sp3 orbitals get one electron each. Now hydrogen atoms overlap their 1s orbitals containing unpaired electrons the sp3 orbitals of ‘N’ atom containing unpaired electrons to give three σs-sp3 bonds. HNH should be 1090 28/ for sp3 hybridisation. As there is a lone pair in one of the sp3 orbitals, there is a greater lone pair – bond pair repulsion which decreases the bond angle HNH to 1070 48/ .

It is found that HOH is 1040 31/ . The electronic configuration of 8 O is 1s2 2s2 2px 2 2py 1 2pz 1 and 1 H is 1s1 .

Therefore, there should be two σ s-p bonds due to the overlap of ‘s’ orbitals of two hydrogen atoms, the ‘p’ orbitals of oxygen atom which contain unpaired electrons. HOH should be 90° .

But HOH observed is 1040 31/ . To explain this, sp3 hybridisation is suggested for the valence orbitals of ‘O’ atom. One s-orbital (2s) and three ‘p’ orbitals (2px , 2py , 2pz ) intermix and redistribute into four identical sp3 orbitals. As there are six electrons and two sp3 orbitals get pairs and two sp3 orbitals get one electron each. Now, the two sp3 orbitals of ‘O’ atom overlap