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406 11MOLECULAR SPECTROSCOPY
�e transitionmoment is given by ∫ ψf∗µ̂ψi dτ, and the symmetry species
of the integrand is found using the direct product; µ̂ transforms as x, y, or
z, which in this case is B3u, B2u, or B1u.�e integral is only non-zero if the
integrand transforms as the totally symmetric irreducible representation,
which is the case here when the component of the dipole is along x
ψf
«
B3u ×
µ̂x
«
B3u ×
ψi
ª
Ag = Ag
�us the π → π∗ transition is symmetry allowed , with the transition
dipole along x .
(b) A carbonyl group (as exempli�ed by that in methanal) is assumed to be-
long to the point group C2v: assume that the H2CO fragment lies in the
xz-plane, with the C–O bond along z. �e π∗ anti-bonding molecular
orbital transforms in the same way as the cartesian function y, that is as
B2.
A non-bonding electron on oxygen is usually considered to be in a 2px
orbital which transforms as B1. When an electron is promoted from n to
π∗, the excited state has symmetry B2 × B1 = A2. �e symmetry of the
integrand for the transition moment is
ψf
ª
A2 ×
µ̂
­
Γx ,y ,z ×
ψi
ª
A1 = A2 × Γx ,y ,z
Because Γx ,y ,z ≠ A2 this product is never A1, so the integral is zero and
the transition is forbidden .
I11.10 �e energy of the HOMO, EHOMO, is reported in the table below, based on
calculations performed with Spartan 10 using the DFT/B3LYP/6-31G*method.
�e experimentally determined energy of the I2–aromatic hydrocarbon charge
transfer bands is also given.
Figure 11.17 shows a plot of EHOMO against hνmax; the best-�t straight line is
also shown.�ere is a modest correlation between the two quantities.
hydrocarbon hνmax/eV EHOMO/eV
benzene 4.184 −6.70
biphenyl 3.654 −5.91
naphthalene 3.452 −5.78
phenanthrene 3.288 −5.73
pyrene 2.989 −5.33
anthracene 2.890 −5.23