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.

Very helpful. Thank you!

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Very Helpful, Thanks a lot!

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

Your definition of axes does not follow the right hand coordinate rule

YOU ARE SAVING MY EXAMS, THANK YOU SO MUCH