Pauling showed methane has four sp³ hybridized C-H bonds. This is often abbreviated to C in CH₄ is sp³ hybridized. Stretching one bond in CH₄ will change its hybridization [Bent’s rule]. Carbon is no longer at center of a tetrahedron but at center of a triangular pyramid. A tetrahedron [one of the Platonic figures] is a special case of a triangular pyramid. Bonds have no angles, e.g. H₂, O₂. If index of p is now (3+x) the other three C-H bonds are now sp^(3-x/3)hybrids. Interbond angles haven’t changed; they are still tetrahedral. F is most electronegative element thus CH₃F should deviate most from sp³. ¹J_CH for CH₃F is 149 Hz which corresponds to sp^2.35hybridization of its three C-H bonds and sp^4.95for its C-F bond. H-C-F interbond angle is 108^.6^o. C-H and the C-F bond lengths are 110^.3 pm and 141^.3 pm respectively. At other extreme CF₄ is sp³ hybridized. Replacing F with H yields CHF₃. Its ¹J_CH [239^.1 Hz] yields an estimate of sp^1.09for its C-H bond. Thus its three C-F bonds are sp^3.64 hybridized. Observed F-C-H bond angle is 108^.5^o; that calculated using Coulson theorem is 156.4^o [arccos (-1/1.09)]. pKa of CHF₃ is 25-28, similar to acetylene [pKa = 21]. Since pKa correlates with hybridization, sp^1.09for C-H bond of CHF₃ is reasonable. Coulson theorem was derived for a tetrahedron model of C-H bonding; it is a step function and does not fit molecules where a bond is stretched to yield a triangular pyramid. Accordingly, stating interbond angles predict hybridization is FALSE. Rather, change in bond length also determines this. Individual bonds are hybridized, not C atom. Stating C in CH₄ is sp³ hybridized, while convenient, is a logical fallacy. It is a 20^th century Myth analogous to Phlogiston theory which dominated chemistry for most of the 18^th century.

Donald Clarke

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