What is the geometry of a tri-substituted methyl free radical? My teacher explained that free radicals could have a geometry that is planar or pyramidal(tending to tetrahedral) based on whether or not it is substituted by lone-pair containing atoms.
For example,
$\ce{(CH3)_3C}$ would exhibit sp2 hybrid geometry of planar.
But when tri-substituted with Flourine/Chlorine, it will exhibit a geometry that is tending towards tetrahedral(pyramidal) because the free radical electron in the P-orbital repels the lone-pairs on the Halogen.
It does not seem convincing because, the P-orbital is above and below the plane, then why does the electron in P-orbital repel the groups into a geometry tending towards tetrahedral?
That's like as if the electron was present only on the top and was repelling the groups down.
Here is a picture that might help visualise the p-orbital above and below the plane: https://www.google.com/imgres?imgurl=https://chemistryonline.guru/wp-content/uploads/2016/09/carbonium-1280x720.jpg&amp;imgrefurl=https://chemistryonline.guru/carbonium-ion/&amp;tbnid=1bNjv-81SP8QxM&amp;vet=1&amp;docid=MPke176Wg2AS2M&amp;w=1280&amp;h=720&amp;source=sh/x/im
A priori, I would expect substituted methyl radicals to adopt a planar structure due to Bent’s rule.
However, I consider it possible for methyl radicals with electron-withdrawing substituents to adopt a trigonal-pyramidal structure too. In this case, Bent’s rule would imply that the electron-withdrawing substituents are ‘more electron-withdrawing’ than the radical is electron-poor. In such a case, it might be beneficial to distort the planar structure, causing rehybridisation and thus resulting in $\mathrm{sp}^n$ orbitals across the board so that all orbitals together add up to $\mathrm{sp^3}$. This structure may not be perfectly pseudotetrahedral but close enough to have a significant s-contribution to the singly-occupied orbital.
Note that $\mathrm{sp^3}$-type radicals are not inherently unstable (unlike $\mathrm{sp^3}$-type cations) as can be seen in bicyclo[2.2.2]octa-1-yl radicals (or bridgehead radicals in general). This points to $\mathrm{sp^3}$ radicals being accessible and thus the finer electronic environments mattering.

Addendum: Ron has shared links to two other questions he answered (1, 2) which argue the case that methyl radicals would be (slightly) pyramidal rather than planar in their ground state. So my a priori expectation might be wrong. I believe it might be very hard to prove the structure conclusively, as it is expected for a pyramidal radical to rapidly inverse as nitrogen does in ammonia and its derivatives so that any non-short timescale measurements would result in an averaged structure (which would be planar).