# 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. Apr 19, 2016 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, 2016 at 13:14
• This question is very relevant, although not a dupe IMO.
– bon
Apr 19, 2016 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:

In the planar geometry the orbitals are perpendicular and no overlap occurs.

### References

1. Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry, Part A: Structure and Mechanisms, 5th edition.; Springer: New York, 2008. ISBN 978-0-387-68346-1.
• This is a very popular, albeit very organic chemistry, view on that matter. The explanation in terms of Bent's rule might be much easier. Also that hand-wavy argument 'repulses more' cannot be proven, and is likely as wrong as spd hybridisation in hypercoordinate molecules. The second part is actually quite nice though Mar 2, 2020 at 12:42