# What is the hybridisation of trifluoromethyl free radical?

Having three σ bonds in a similar manner to $$\ce{CH3^·}$$ free radical, $$\ce{CF3^·}$$ should also have $$\mathrm{sp^2}$$ hybridisation. However, if we look at its shape, it is pyramidal and not planar like $$\ce{CH3^·}$$ free radical (which is $$\mathrm{sp^2}$$-hybridised), which signifies that $$\ce{CF3^·}$$ should have $$\mathrm{sp^3}$$ hybridisation.

But how is this possible because the three σ-bonds will bond with three hybrid orbitals? Where does the third p-orbital come?

If its hybridisation is $$\mathrm{sp^3},$$ then why is it?

• 1. Geometry drives hybridization, not the other way around. 2. An odd of consequence of #1 is that non-integer exponents in the sp/sp2/sp3 notation have physical relevance. – Lighthart Apr 19 '16 at 7:17
• As Lighthart suggested, it is neither sp2 or sp3. It will be somewhere inbetween. Exactly where is an interesting question which I don't have the answer to right at the minute. – bon Apr 19 '16 at 13:14
• This question is very relevant, although not a dupe IMO. – bon Apr 19 '16 at 13:19

Hyperconjugation between the orbital that contains the lone pair and the $$\ce{C-F}$$ antibonding orbital contributes to the pyramidalization of $$\ce{CF3^·}$$ radical: