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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?

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2 Answers 2

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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As a simple heuristic under the arguably outdated VSEPR theory, electronegative atoms want to be the furthest away from each other and other electrons as possible due to electric interactions.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 $112^\circ$.

So, when considering hybridization problems, you must remember that hybridization fixes geometry. Thus if any plausible argument for a hybridization must be made, there should be experimental or computed evidence of the geometry. One cannot simply look at a molecule and know what the geometry could be with absolute certainty, but we can make a good guess based on simple heuristics, like those of VSEPR, molecular orbital diagrams, previous experience/similar molecules, etc.

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  • $\begingroup$ Hello, thanks for the explanation, but what about NO , NO2, ClO2 & ClO3? I understand about NO2 and ClO3 that free radical would reside in hybrid orbital due to electronegativity difference, but what about NO & ClO2? BCOZ, they also have lone pair on themselves along with free radical?. $\endgroup$ Nov 1, 2020 at 8:21
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Case - I If oxidation state of central atom is positive then odd electron will participate in hybridization. If oxidation state of central atom is negative then odd electron will not participate in hybridization.

Case - II If lone pair is present with odd electron on central atom then odd electron will not participate in hybridization irrespective of oxidation state of central atom.

Example : (1) NO2 is an example of case-(I) with positive oxidation state of central atom, so odd electron will participate in hybridization. Steric number = 2 σ bonds + 1 odd electron Steric number = 3 Hybridization = sp2 Shape = bent (2) ClO3 is an example of case-(I) with positive oxidation state of central atom, so odd electron will participate in hybridization. Steric number = 3 σ bonds + 1 odd electron Steric number = 4 Hybridization = sp3 Shape = pyramidal (3) ClO2 is an example of case - (II) with one lp and one odd electron on central atom, so odd electron will not participate in hybridization. Steric number = 2 σ bonds + 1lp Steric number = 3 Hybridization = sp2 Shape = bent (4) •CF3 is an example of case-(I) with positive oxidation state of central atom, so odd electron will participate in hybridization. Steric number = 3 σ bonds + 1 odd electron Steric number = 4 Hybridization = sp3 Shape = pyramidal (5) •CH3 is an example of case-(I) with negative oxidation state of central atom, so odd electron will not participate in hybridization. Steric number = 3 σ bonds + o Steric number = 3 Hybridization = sp2 Shape = planar

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