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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&imgrefurl=https://chemistryonline.guru/carbonium-ion/&tbnid=1bNjv-81SP8QxM&vet=1&docid=MPke176Wg2AS2M&w=1280&h=720&source=sh/x/im

Thomas Prévost
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Desai
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  • https://images.app.goo.gl/q8mvwqGhvbyYRyVX6 – Desai Dec 16 '20 at 09:53
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    Yes, I agree that the attached groups due to the sp2 hybridisation, are aligned into a planar geometry, but the vacant p-orbital must remain perpendicular to the plane of the sp2 orbitals. So, that means the p-orbital being dumbbell shaped, should be above and below the plane of the sp2 bonds. – Desai Dec 16 '20 at 09:56
  • Yes, that is just a representation I could find off the internet. The point is, when it is a FREE RADICAL (having a single electron in P-orbital) it is going to repel any bonded lone-pair containing atoms like for example halogens. – Desai Dec 16 '20 at 10:18
  • I even took it as we were discussing of carbocations. I delete all the above :) Do the same otherwise is a mess. You'll get the answer.... – Alchimista Dec 16 '20 at 10:38
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    .... from here https://chemistry.stackexchange.com/questions/49754/what-is-the-hybridisation-of-trifluoromethyl-free-radical – Alchimista Dec 16 '20 at 10:42
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    And as suggested in a comment therein, a look at https://en.m.wikipedia.org/wiki/Bent%27s_rule – Alchimista Dec 16 '20 at 10:59
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    Please don't link through Google tracking algorithms to images. If possible and allowed, please include them directly in your post. Otherwise try a permanent URL. – Martin - マーチン Dec 16 '20 at 18:52
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    I return here specifically for you. While the explanation given by your teacher might be not accurate (as for the material posted below) however it is not inaccurate because the reasoning leading to your doubt. It doesn't matter that the e is repelling equally from above and belove. It matters that with a more pyramidal or sp3 like geometry all repulsions are minimised. – Alchimista Dec 17 '20 at 10:23
  • As already pointed out by Martin to not use Google links, here is the relevant meta post to give you the context: https://chemistry.meta.stackexchange.com/questions/4830/can-outdated-versions-of-answers-be-removed/4832#4832 – Nilay Ghosh Dec 19 '20 at 12:03

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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).

Jan
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