Chemistry Reference
In-Depth Information
The threefold axis moves all three 1
s
-orbitals around and hence has zeros only on
the diagonal; thus,
χ(C
3
)
=
0(seeEq.(
4.7
)). For the reflection planes, one element
is stabilized, while the other two are interchanged; hence, there is only one nonzero
diagonal element, which will be equal to
+
1 since the orbital is not affected; hence,
χ(σ
1
)
=+
1. The norm of the character string is given by
3
2
1
2
χ
|
χ
=
+
2
×
0
+
3
×
=
12
=
2
|
C
3
v
|
(4.34)
Since the norm is twice the order of the group, we know for certain that the symme-
try of the function space is reducible. In fact, we also predict that it will reduce to
two irreps, each of which occurs once, since the sum of the squares of the multiplici-
ties in Eq. (
4.33
) is equal to 2. These coefficients can now be obtained by calculating
the character brackets, which yields the previous result:
c
A
1
=
1
,c
A
2
=
0
,c
E
=
1.
From Table
4.1
one can also verify that the sum of
χ
A
1
and
χ
E
equals
χ(
1
s)
.This
is sometimes also expressed in a formal way as
Γ
=
A
1
+
E
(4.35)
As a further example, we can take the set of the two in-plane 2
p
-orbitals on nitrogen,
p
x
and
p
y
. Here, the character for the threefold axis is
χ
p
x
,p
y
(C
3
)
=
2 cos
(
2
πi/
3
)
=−
1
(4.36)
Under the
σ
1
plane of symmetry, one of the orbitals is antisymmetric, while the other
is symmetric, and so the trace of the matrix vanishes. The characters of this
p
-set
are thus precisely the ones for the
E
irrep, and we can therefore state that the set
{
p
x
,p
y
}
ˆ
transforms irreducibly as the
E
representation. As a final example, let us
take the alternative set consisting of three counterclockwise displacement vectors of
the hydrogen atoms, along a tangent to the (imagined) circumscribed circle through
the hydrogens, as indicated in Fig.
4.2
. In this example the function space is the set
of three bendings:
φ
A
,φ
B
,φ
C
. We must realize that in this case the function on
which we are operating is actually the distortion itself, and to rotate this distortion in
an active sense means to take the distortion vector and displace it to the next nucleus.
Hence, nuclei are left immobile: only the distortions are moved. One could easily
reconcile this view with the previous orbital rotations by thinking of the distortions
as little tangent
p
-orbitals with positive and negative lobes corresponding to the
head and tail of the vectors, respectively. Hence, as an example, one has
C
3
φ
A
=
φ
B
(4.37)
Since all distortions are cyclically permuted, the character under this generator is
zero. On the other hand, for the reflection plane, one has:
σ
1
φ
A
=−
ˆ
φ
A
σ
1
φ
B
=−
ˆ
φ
C
(4.38)
σ
1
φ
C
=−
ˆ
φ
B
