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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?

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    $\begingroup$ 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. $\endgroup$
    – Lighthart
    Commented Apr 19, 2016 at 7:17
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    $\begingroup$ 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. $\endgroup$
    – bon
    Commented Apr 19, 2016 at 13:14
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    $\begingroup$ This question is very relevant, although not a dupe IMO. $\endgroup$
    – bon
    Commented Apr 19, 2016 at 13:19

1 Answer 1

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

pyramidal geometry of CF3

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.
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