# How to find if the unpaired electron is placed in hybrid orbital or pure orbital in odd electron species?

Odd electron species contain an unpaired electron which can either be placed in hybrid orbital or in pure orbital.

Example: In the case of $\ce{.CF3}$ radical, the unpaired electron is placed in $\mathrm{sp^3}$ hybrid orbital of carbon.

In the case of $\ce{.C(CH3)3}$ radical, the odd electron is placed in pure p orbital of carbon.

How do we determine whether the unpaired electron is placed in hybrid orbiral or pure orbital? What factors govern the same?

When discussing hybrid orbitals, they are justified by the experimentally observed or calculated/simulated molecular geometry. For instance, the $\ce{^.C(CH3)3}$ radical was found to be planar by ab initio calculations (http://www.tandfonline.com/doi/abs/10.1080/00268977600102711). Thus, the molecular shape is best approximated by $\ce{sp^2}$ hybridization and the unpaired electron is most likely to be found in a $\ce{p}$ orbital.
As a simple heuristic under the arguably outdated VSEPR theory, electronegative atoms want to be the furthest away from each other and other electrons as possible due to electric interactions. When considering the $\ce{^.CF3}$ radical, the fluorine atoms are highly electronegative and want to be the furthest they can be from the unpaired electron that holds negative charge. A planar $(\ce{sp^2})$ conformation would make the distance between the unpaired electron and each fluorine atom $90^\circ$. While a pyramidal $(\ce{sp^3})$ conformation would separate the two groups by $109^\circ$, which is much more favorable according to VSEPR theory. The later is what is observed under calculations (http://www.chem.mun.ca/courseinfo/c4420/handouts/handout-03a.pdf) with a $\ce{F-C-F}$ bond angle of $112^\circ$.