Molecular symmetry in assigning IR vibrational modes for polyatomic molecules

Molecular symmetry in assigning IR vibrational modes for polyatomic molecules

If we can categorise a molecule into a point
group, we can use this symmetry to simplify the process of describing the vibrational
modes of a molecule. Let’s look at water as an example. Water is described by the C2V
point group, and the C2V character table is shown.
As a brief reminder, the character table shows the name of the point group, all of the symmetry
operations that belong to this point group – in this case E is the identity operator
– which does nothing – C2 is rotation operator which rotates the molecule about an axis by
180 degrees; and two mirror operators which reflect the molecule in the plane, in this
case the xz plane and the yz plane. The value h here is the order of the group, which we
can calculate this easily by summing up the numbers before the symmetry operations: 1+
1 + 1 + 1. In the left hand column, we have the symmetry species – these are irreducible
representations, which are described by A, B, etc. Finally the table itself has characters;
the number 1 meaning symmetric and -1 meaning anti-symmetric. The right hand columns list
the functions – these are important as the first column of functions contain x, y, z
and relate to infrared activity. The second column list quadratic functions, and relate
to Raman activity as we will see later. The process involves the determination of
a reducible representation that describes the vectors of movement of each of the atoms
in the molecule, where movement includes translational, rotational and vibrational. Once we have done
this, we will remove the translational and rotational movements, and thus use the remaining
information to arrive at the vibrational modes of the molecule.
As mentioned, water has C2v symmetry. The first step is to define the x, y, and z axis
as shown. The second step is to describe the reducible
representation for all possible movements of the atoms in water. When we complete an
operation contained in the character table, if the vector remains unchanged, we assign
it a value of 1. If the vector moves, we assign it a value of zero, and if the vector is reversed
we assign a value of -1. Let’s take each one in turn: the identity
operator leaves the molecule unchanged. Therefore each atom remains unchanged, and the three
vectors on each atom remain unchanged. Three atoms each with three vectors, means there
are 9 unchanged vectors, so a value of +9 is entered in our reducible representation.
For a C2 rotation, the hydrogen atoms switch positions, while the oxygen rotates in the
same position. What happens to each vector? For hydrogen, the vectors have completely
moved, so each of these 6 vectors returns a value of zero. For oxygen, let’s take each
axis in turn. Rotating the molecule does not change the z-axis; so it returns a value of
+1. For the x and y axis, the vector rotates 180 degrees to the opposite direction, so
these return a value is -1. Thus the overall total is +1 -1 -1 giving a total of -1.
For the mirror reflection in the xz plane, the molecule is bisected. Again the hydrogens
exchange, so their vectors return a value of zero. The x and z axis for oxygen are in
the mirror plane, so they remain unchanged, returning a value of +1 each. And the y-axis
of oxygen is reflected in the plane, so its direction changes 180 degrees, and is assigned
a value of -1, giving an overall total of +1.
Finally the mirror plane in the yz plane, which is the plane of the computer screen,
the atoms stay in the same position, and only the x-axis vectors of each of the three atoms
change direction by 180 degrees, and hence we have +6 – 3=+3.
Therefore the reducible representation represents all of the modes of movement for the water
molecule. We now need to see what combination of irreducible representations give rise to
this reducible representation. This can occasionally be done by eye, but it is easiest to use a
simple formulaic approach, which effectively factors out the number of each of the irreducible
representations contained in the reducible representation.
We can say that the number N of each irreducible representations can be obtained by multiplying
the character of the irreducible representation by the character in the reducible representation
by the number of in front of the class, all divided by the order of the group, h. This
sounds more complex than it really is. Let’s take each one example.
For A1, the number is 9 x 1 x 1 (for E) + -1 x 1 x 1 (for C2) + 1 x 1 x 1 (for the first
mirror operator) + 3 x 1 x 1 (for the second. All of sums up to twelve divided by 4 which
gives us 3. So there are 3 A1 irreducible representations contained in this reducible
representation. We can repeat this as shown. Pause here if
you would like to try these yourself. We see now that we have one A2, two B1, and
three B2 irreducible representations joining the three A1 in the reducible representation.
Therefore the total number of modes of movement is given by these. We are only interested
in vibrational modes so we can remove translational and rotational movements. So we remove a mode
for B1 for the x, a B2 for the y, and an A1 for the z. For rotational, we subtract out
B2, B1, and A2 for the three rotational movements Rx Ry and Rz. Note that if there were two
translational or rotational movements for a particular irreducible representation, we
would only subtract it once (for example if x and y were in the same line).
Once these are subtracted, we are left with the vibrational motions. We can see we have
three modes, as expected, two A1 and a B2. Finally, we can check whether these are infrared
active. If they have an x, y, or z next to it in the function column, it is IR active,
as it involves a displacement and hence a change in dipole moment. In the case of A1,
it has a z value, and therefore it is infrared active. In the case of B2, it has a y value,
and therefore it is infrared active. Indeed, as they also have quadratic functions associated
with them, they are also Raman active. What are these three modes? The B2 is anti-symmetric
with respect to the principal rotation axis, and from the spectrum we will see that it
occurs at 3756 wavenumbers. The first A1 is a symmetric stretch, at 3657, and the second
is a bend, which is symmetric. This appears at 1595 in the spectrum.
One final thing to note when we are applying it to other molecules. In this case, we looked
at changing vectors, where the vector had changed by 180 degrees. There are some scenarios
where the vector changes, but by a smaller amount (e.g. 120 degrees due to a C3 rotation).
The actual value we assign is more correctly cosine of the angle alpha. Therefore for a
c3 rotation, where the vector rotates by 120 degrees, cosine (120) is – 0.5, so that is
the value is assigned. You can see for the two values we have already considered, where
the vector remained unchanged, and the vector changed direction, you can see where we get
these values – for a value of of 180 degrees: cos(180) gives is -1, as used before. A value
of 360 degrees (ie vector doesn’t change direction) results in cos(360)=1, as used before.

6 Replies to “Molecular symmetry in assigning IR vibrational modes for polyatomic molecules”

  1. That's great ….. please where i can found tables to show where all Mullikan symbols appear (for example A1 appear at 1700 and 3700 cm-1) or how i can calculate it. thanks a lot

Add a Comment

Your email address will not be published. Required fields are marked *