Chemistry Reference
In-Depth Information
Tab l e 5 . 5
The application of the reduction for-
mula to the nine basis vector representation of H
2
O
atomic degrees of freedom defined in Figure 5.2.
C
2v
E
2
σ
v
(
XZ
)
σ
v
(
YZ
)
h
=
4
9
−
1
1
3
C
1
h
C
g
c
χ
(
C
)
χ
i
(
C
)
A
1
9
−
1
1
3
12
3
A
2
9
−
1
−
1
−
3
4
1
B
1
91
−
1
3
8
2
B
2
911
−
3
12
3
There are three translations and three rotations of the molecule as if it were a rigid body.
For any molecule in the point group, the rigid body motions will have the same irreducible
representations. In the standard character tables of Appendix 12 the symbols
x
,
y
,
z
and
R
x
,
R
y
,
R
z
are written in the rightmost columns and can be used to identify the representations
for rigid-body movement and rotation respectively. So, most of the time, it is just a matter
of referring to the character table to find the irreducible representations that should be
removed and so isolate the vibrational mode symbols.
However, to demonstrate how the rigid-body motion conforms to the irreducible rep-
resentations, in this example we will go over the effect of symmetry operations on the
translational and rotational motion of H
2
O.
The three translations are motions along the axes of the coordinate system which follow
the symbols
x
,
y
or
z
in the right-hand column of the character table. The standard
C
2v
table
is reproduced in Table 5.6 and the assignments made for translation are easily checked. For
example, the molecule moving as a whole along the
Y
-axis direction is assigned to the
B
2
representation. To see this we could place a basis vector
y
at the centre of mass of the
molecule parallel to the
Y
reference axis in Figure 5.2, which would represent the motion.
The centre of mass of the molecule lies on the
C
2
axis nearer to the O atom than either of
the two H atoms due to the greater mass of the former.
Tab l e 5 . 6
The standard character table for the C
2v
point
group.
C
2v
E
2
σ
v
(
YZ
)
A
1
11 1 1
z
σ
v
(
XZ
)
x
2
,
y
2
,
z
2
A
2
11
−
1
−
1
R
z
xy
B
1
1
−
1
1
−
1
x
,
R
y
xz
B
2
1
−
1
−
1
1
y
,
R
x
yz
A
y
-vector placed at the centre of mass responds to the symmetry operations in the
same way as the
y
-vector on O in Figure 5.2a: after a
C
2
rotation or reflection in the
σ
v
(
XZ
)
plane it would be reversed, corresponding to a character of
−
1, while the
y
-vector would be
unaffected by the identity operator
E
or a
σ
v
(
YZ
) reflection so that these have a character
of 1. This set of characters is just the
B
2
representation, and so we have confirmed that
motion in the
Y
direction should be assigned to the
B
2
representation.
