Question

all free radicals are sp2 hybridisation expect one have sp3 hybridisation??????

Answer 1

Suppose a carbon free radical , i.e. CH3

Then on calculating total electron pair ,we get

e pair = 1/2( 4+3)

e pair =3.5

Or, also we can say that it have 3 e pair in bonded condition and 1 electron is in unbonded condition(because ‘.5’ represents half of a pair of electron). And as we know that for 3 e pair sp2 hybridization exist ,which consist of 1 unhybridized p orbital.



As obvious from the above figure, that all the electrons got their suitable places, so we can say that carbon free radical is sp2 hybridized.

Answer 2

The unpaired electron on carbon repulses more with lone pairs on the flurorine atom in the planer geometry (predicted by sp2 hybridization) than in the pyramidal geomatery sp3. There is also a hyperconjugation-type stabilization in the pyramidal geomatery, between the orbital that contains the lone pair and the C-F antibonding orbital.