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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?

andselisk
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James Hunt
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  • 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 07:17
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    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
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    This question is very relevant, although not a dupe IMO. – bon Apr 19 '16 at 13:19