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Having 3 sigma bonds in a similar manner to CH3 free radical it should also have hybridisation sp2. However, if we look at its shape, it is pyramidal and not planar unlike CH3 free radical(which is sp2 hybridised) which signifies that it should have sp3 hybridisation. But how is this possible because the 3 sigma bonds will bond with 3 hybrid orbitals, from where did the third p orbital came?

If its hybridisation is sp3, then why is it?

Thank you for answering the question

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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 Apr 19 '16 at 7:17
  • $\begingroup$ chemistry007.blogspot.de/2015/01/… (upon googling CH3 vs CF3 radical) $\endgroup$ – ssavec Apr 19 '16 at 7:19
  • $\begingroup$ 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 Apr 19 '16 at 13:14
  • $\begingroup$ This question is very relevant, although not a dupe IMO. $\endgroup$ – bon Apr 19 '16 at 13:19
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There are two important effects that contribute to the pyramidalization of $CF_3$ 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 $sp^2$ hybridization) than in the pyramidal geometry ($sp^3$): enter image description here

There is also a hyperconjugation-type stabilization in the pyramidal geometry, between the orbital that contains the lone pair and the C-F antibonding orbital

enter image description here

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

Carey, Sundberg: Advanced Organic Chemistry, Part A, 5th Edition.
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