# 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
• chemistry007.blogspot.de/2015/01/… (upon googling CH3 vs CF3 radical) – ssavec Apr 19 '16 at 7:19
• 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

There are two important effects that contribute to the pyramidalization of $$\ce{CF3^·}$$ radical:

1. electron-electron repulsion;

2. hyperconjugation.

The unpaired electron on carbon repulses more with the lone pairs on the fluorine atoms in the planar geometry (predicted by $$\mathrm{sp^2}$$ hybridization) than in the pyramidal geometry $$(\mathrm{sp^3}):$$

There is also a hyperconjugation-type stabilization in the pyramidal geometry, between the orbital that contains the lone pair and the $$\ce{C-F}$$ antibonding orbital:

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 – Martin - マーチン Mar 2 at 12:42