Why is the methyl radical planar? The VSEPR theory would predict an angle between 120° and 109.5°, while it is actually 120°.
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1$\begingroup$ VSEPR is about "electron PAIR repulsion". I would say it does not predict anything for radicals. Or at least be very careful to not make definite statements. ;-) $\endgroup$– KarlAug 12, 2015 at 17:55
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$\begingroup$ Related: Shapes of methyl and silyl radicals $\endgroup$– user7951Aug 12, 2015 at 19:37
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1 Answer
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This question unfortunately illustrates the significant shortcomings of the VSEPR model. This model is based upon classical electrostatic arguments. The molecular geometry will seek to maximize the distances between the repulsive electrons in the valence (outer) shell, that are completely localized in their chemical bonds and their lone pairs/singly occupied molecular orbitals.
This model works well in some cases. However, when quantum mechanics comes into play, it falls pretty flat, as it totally neglects (the really important) quantum effects. A better way of finding the molecular geometry is to find the geometry that minimizes the total energy of the molecular orbitals. In this case, quantum mechanical calculations find that a set of three $\mathrm{sp^2}$-hybridized molecular orbitals arising from a $D_\mathrm{3h}$ symmetry has a lower total energy than breaking the symmetry to form the pyramidal $C_\mathrm{3v}$ geometry.
However, this does not mean that all $\ce{\overset{.}{C}X3}$ radicals follow this pattern. Put an extremely electronegative substituent on in place of $\ce{H},$ e.g. in $\ce{CF3},$ and the situation changes. The lowest energy geometry is now the pyramidal one.
Sadly, quantum chemistry doesn't always care to follow general rules.
