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?
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


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.