Chemistry Reference
In-Depth Information
The IR selection rule depends on the fact that the transition dipole moment integral
must be nonzero in order to observe the transition as an IR absorption band. This means
that the integrand for
M
01
must have
A
1
symmetry.
ψ
1
μψ
0
. As explained in
Appendix 6, the ground state will always belong to the
A
1
irreducible representation. The
first excited state vibration will have the same irreducible representation as the underlying
vibrational motion, described by the normal coordinate
q
.
The dipole moment operator has three components with the same symmetry as the sim-
ple
x
,
y
and
z
functions, and these are usually noted in the right-hand columns of character
tables.
To identify those vibrations which cannot have a finite transition dipole moment and
so exclude them from the assignment of spectroscopic bands, we will use the idea of
direct products (introduced in Chapter 4) to look at the symmetry of the integrand in
Equation (6.4) for a few examples.
To use this definition, we require the symmetry of the integrand
H
2
O, C
2
v
We have already found that the symmetric vibration of H
2
O, illustrated in Figure 6.2,
belongs to the
A
1
irreducible representation. The symmetry representations for the com-
ponents of the dipole operator are the same as
x
,
y
and
z
from the
C
2
v
character table
in Appendix 12 the right-hand columns tell us that these belong to the
B
1
,
B
2
and
A
1
irreducible representations. The three components of the dipole operator each give an
integrand; for the symmetric stretch these are
ψ
1
μ
x
ψ
0
⇒
A
1
×
B
1
×
A
1
=
B
1
ψ
1
μ
y
ψ
0
⇒
A
1
×
B
2
×
A
1
=
B
2
(6.5)
ψ
1
μ
z
ψ
0
⇒
A
1
×
A
1
×
A
1
=
A
1
To obtain these products, we multiply, class by class, the characters of the irreducible
representations and then look for a match between the result and a row of the charac-
ter table. In these examples,
A
1
has character 1 for all classes, and so the solutions are
straightforward.
Equation (6.5) says that, to get the
y
-component of
M
01
, we have to integrate a function
with
B
2
symmetry from
, which Figure 6.3 shows must give zero because
the function will have opposite signs either side of the
−∞
to
+∞
σ
v
(
XZ
). Similarly, the
B
1
integrand
generated by
μ
x
has a -1 character under
σ
v
(
YZ
) and so will lead to a zero integral. For the
H
2
O symmetric stretch, only
μ
z
gives an
A
1
integrand.
This is sufficient for us to conclude that the symmetric stretch will lead to an absorption
band and also that this absorption is due to a transition dipole moment aligned with the
Z
-axis in the standard symmetry setting. The physical interpretation of this result can be
understood from a diagram of H
2
O in the standard setting, such as Figure 6.3. As O is more
electronegative than H, both O H bonds will have local dipole moments pointing from O
toward H. In the
A
1
vibration the O H bonds move in phase, and so the
Y
-components of
these dipoles, which are in opposite directions to one another, always cancel out. However,
