According to the definition, steric number = number of atoms it is attached + lone pair.
In that way, I get 4 for methyl free radical which means sp3 hybridization . But that's wrong ! Why is this happening ?
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$\begingroup$ The point is that in methyl radical there is not a lone pair and a direct application of the formula for steric number does not apply, not smoothly. The unpaired electron can reside in a non hybrid p orbital without much repulsion with the bonding paired electrons. As such the C atom is sp2 and the geometry is quasi planar trigonal. However the barrier to a pyramidal molecular geometry is low. If H atoms are substituted by more electronegative atom or radicals (e.g. in CF3 radical) geometry is then pyramidal. Qualitatively, you can see this as if an unpaired electron count about 1/2 $\endgroup$ – Alchimista Oct 18 '17 at 13:38
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$\begingroup$ Then how do I arrive at sp2 hybridization ? What else technique to use ? $\endgroup$ – Tilak Maddy Oct 18 '17 at 17:25
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$\begingroup$ Vsepr gives (with exceptions) reliable predictions but for electron pairs! Is it in its name. The trick of assigning about 0.5 to an unpaired electron can be useful while comparing species which formally differs only in the number of electrons like NO2+ NO2, or to rationalise the case CH3-, CH3 radical, CH4. The case of CH3 radical is indeed a special one, and its geometry cannot be really predicted this way. $\endgroup$ – Alchimista Oct 18 '17 at 18:27
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$\begingroup$ Meantime I saw as I expected that basically the same question has been answered chemistry.stackexchange.com/questions/19370/… it should be clear in every book that vsepr is a first approximation... $\endgroup$ – Alchimista Oct 18 '17 at 18:36
