# Why is the methyl radical planar?

Why is the methyl radical planar? The VSEPR theory would predict an angle between 120° and 109.5°, while it is actually 120°.

• 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. ;-)
– Karl
Aug 12 '15 at 17:55
• – user7951
Aug 12 '15 at 19:37

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.