Symmetry of Normal Modes
The normal modes of a molecule must belong to one of the symmetry species of the molecular point group, and this proves to be a highly convenient method of classifying them, allowing rapid prediction of the form of the vibrational spectra of the molecule.
The determination of the symmetry species of the normal modes is a totally standard procedure which may be applied quite generally. It is most simply explained by an example, so we shall here work out the symmetry species of the normal modes of vibration of the water molecule, which belongs to C2v:
|C2v||E||C2||σv||σv‘||h = 4|
|A1||1||1||1||1||z, z2, x2, y2,|
|B1||1||-1||1||-1||x, xz, Ry|
|B2||1||-1||-1||1||y, yz, Rx|
a) Work out the characters of the atoms composing the molecule under each of the symmetry operations of the molecule. This is very simply the number of atoms left unchanged by each operation, as each such atom has a character of one. All atoms which are moved by a symmetry operation have a character of zero. (Note that since atoms have no sign, the complication of having to consider whether or not the symmetry operation changes the sign of an object, in which case it has a character of -1, does not arise.)
Thus, for water, we obtain:
i.e. None of the atoms are moved by the identity operation or the σv‘ reflection, which is the reflection in the plane of the molecule. In the C2 rotation of water, only one atom (the oxygen atom) is unmoved by the operation, as it lies on the rotational axis. Similarly, only the oxygen atom is unmoved by the σv operation, as it lies in the mirror plane. Note the symbol ΓN merely indicates that the characters in the table are derived using the N atoms of the molecule as a basis.
b) Carry out the same procedure for the three Cartesian axes, x, y, and z. Note that in this case it is possible for an axis to become minus itself, so the possibility of an axis having a negative character under an operation exists. Since nothing changes under the identity, the character under E is always three for this stage. For the symmetry operations in C2v, we obtain:
Γxyz indicates that the characters are those for the x y and z axes. All axes remain the same under E, each contributes 1 so the overall character is three. Under C2, the axis coincident with the C2 axis remains the same so has a character of 1, but both the other axes are turned into minus themselves so have characters of -1.
The overall character under this operation is thus 1 + (-1) + (-1) = -1
In general, the character of the three Cartesian axes under a Cn operation is 1 + 2cos(360/n).
For both mirror planes, two axes are in the plane and remain unchanged so have character 1, while the other is reflected in the plane to become minus itself and has character -1. The overall character of the three axes is thus 1 + 1 + (-1) = 1 for any mirror plane.
For a Sn operation, the character of the three axes is given by 2cos(360/n) – 1.
c) For each class of symmetry operation, multiply together the characters of ΓN and Γxyz , to give the characters under each operation of Γ3N .( This represents the characters of all 3N possible vibrations, i.e. the characters of the displacements of the N atoms in any of the three Cartesian directions.) For this example, we obtain:
d) It is now necessary to break these characters down into the different symmetry species (A1, A2 etc) that they represent. The number of times a symmetry species X appears in Γ3N (which we shall represent by n(X)) can always be calculated by the following method:
For each class of symmetry operation, multiply together the number of members of the class, the character of Γ3N , and the character of X. Sum the values for the different symmetry operations, and divide the result by the order of the group, h. Repeat for all the different symmetry species. Thus for water, we obtain the results:
n(A1) = ( (1 x 9 x 1) + (1 x -1 x 1) + (1 x 1 x1) + (1 x 3 x 1) ) / 4 = 3
n(A2) = ( (1 x 9 x 1) + (1 x -1 x 1) + (1 x 1 x -1) + (1 x 3 x -1) ) / 4 = 1
n(B1) = ( (1 x 9 x 1) + (1 x -1 x -1) + (1 x 1 x 1) + (1 x 3 x -1) ) / 4 = 2
n(B2) = ( (1 x 9 x 1) + (1 x -1 x -1) + (1 x 1 x -1) + (1 x 3 x 1) ) / 4 = 3
which allows us to write Γ3N = 3A1 + A2 + 2B1 + 3B2 i.e. the character of Γ3N under the symmetry operations of the group shows that the possible vibrations of the molecule have the symmetry species indicated.
Note that the above method, if performed correctly, cannot give negative or fractional coefficients for the symmetry species.
e) We now take into account the fact that three of the possible vibrations are translations, and three are rotations (two if the molecule is linear). We do this by subtracting the symmetry species that correspond to the translations (which are the same as the symmetry species of x, y, and z), and the symmetry species which correspond to rotations (which are the same as those of Rx, Ry and Rz.) Conveniently, the symmetry species required are indicated in the character table.
For water, we must subtract A1, A2, 2B1 and 2B2. This gives the symmetry species of the 3N – 6 normal modes of water as: Γ3N-6 = 2A1 + B2.