11
$\begingroup$

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?

$\endgroup$
  • 5
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ This question is very relevant, although not a dupe IMO. $\endgroup$ – bon Apr 19 '16 at 13:19
8
$\begingroup$

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.
| improve this answer | |
$\endgroup$
  • 3
    $\begingroup$ 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 $\endgroup$ – Martin - マーチン Mar 2 at 12:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.