Chemistry Reference
In-Depth Information
The multiplication table for H
2
O contains only operations we have already identified:
E
,
C
2
,
σ
v
. This confirms that this set of operations forms a closed group and is a
reassurance that no symmetry operations have been missed.
σ
v
and
2.3.2 Fixed Symmetry Elements
Multiplication Table for NH
3
In the example of H
2
O above we used the idea of a global axis system,
X
,
Y
,
Z
.This
axis system is used to define the positions of the symmetry elements of the molecule
and, once set, the global axis system is not moved by any operations that are carried out.
This means that the symmetry elements should be considered immovable and symmetry
operations only move the atoms in the molecule. This becomes especially important when
molecules with more symmetry elements are considered. For example, ammonia (NH
3
)
has a principal axis of order 3 and three vertical mirror planes, as shown in Figure 2.3.
(b)
(a)
C
3
σ
v
A
σ
v
B
H
2
H
1
H
3
σ
v
C
σ
v
σ
v
σ
v
Figure 2.3
The symmetry elements for ammonia (NH
3
): (a) viewed with the principal axis
in the plane of the page; (b) viewed along the principal axis with the mirror planes labelled
following the text.
In this case the vertical mirror planes are actually equivalent, since they each contain
one N H bond. However, to carry out the multiplication of operations we will add an
additional label to the mirror planes so that they can be distinguished. In the initial config-
uration, we have chosen
σ
v
A
,
σ
v
B
and
σ
v
C
to be the planes containing N
H
1
,N
H
2
and
N H
3
respectively (Figure 2.3b).
This choice sets the location of each mirror plane in space; so, to work out the product
of two operations, it is necessary to hold the planes in place during the whole manoeuvre.
For example, the product
σ
v
A
C
3
1
is illustrated in Figure 2.4. This involves a rotation around
the principal axis which brings H
3
into the
σ
v
A
plane. Reflection by the plane then swaps
H
1
and H
2
. Comparing the final configuration with the start point, it can be seen that H
2
has returned to its original position but H
1
and H
3
have been interchanged, a result that can
be achieved by
σ
v
B
alone, i.e.:
σ
v
A
C
3
1
=
σ
v
B
(2.1)
