Computational Chemistry

electrons) arenotequivalent in energy (notdegenerate). This is an experimental
fact that can be shown by photoelectron spectroscopy [ 26 ]. Instead of four orbitals of
the same energy we have three degenerate orbitals and one lower in energy (and of
course the almost undisturbed 1score orbital of carbon). This surprising arrange-
ment is a consequence of the fact that symmetry requires one combination (i.e. one
MO) of carbon and hydrogen orbitals (essentially a weighted sum of the C2sand the
four H1sorbitals) to be unique and the other three AO combinations (the other three
MOs) to be degenerate (they involve the C2pand the H1sorbitals) [ 26 , 27 ]. It must
be emphasized that although the methane valence orbitals areenergeticallydiffer-
ent, the electron and nuclear distributionistetrahedrally symmetrical – the molecule
indeed has Td(Section 2.6) symmetry. The four MOs formed directly from AOs
are thecanonicalMOs. They aredelocalized(spread out over the molecule), and do
not correspond to the familiar four bonding Csp^3 /H1sMOs, each of which is
localized between the carbon nucleus and a hydrogen nucleus. However, the
canonical MOs can be mathematically manipulated to give the familiar localized
MOs (Section 5.2.3.1).

x

z

y

2 px

2 py

2 pz

2 s

C

H

H

H

H

+ 4 H atoms

(no hybridization)

C

H

H H

H

One C-H bond somehow skewed,

and different in length and strength

from the others, which are mutually

perpendicular, like the p orbitals

WRONG RIGHT

The four C-H bonds

tetrahedrally oriented

and of equal length and

strength

Fig. 4.6 Hybridization is not needed to explain bonding, e.g. the tetrahedral geometry of methane

4.3 The Application of the Schr€odinger Equation to Chemistry by H€uckel 105