Group Tables

What are group tables?

Once the point group of a molecule has been determined, we can look up the group table of that point group. This is a convenient way of writing down all the symmetry operations present in a molecule of that point group, and of showing which orbitals present in the molecule will have which symmetries.
Using the group table will allow us to deduce what the vibrational spectrum of the molecule will look like if we were to use IR or Raman Spectroscopy to analyse it.

A simple table to start with is the C_2v table, such as we would use for H₂S.

The top line of the table tells us that the molecule has four symmetry operations: Identity E, C₂ rotation (which defines the z axis), and two s_v mirror planes.
A₁, A₂, B₁ and B₂ on the left-hand side of the table are labels for the different symmetry elements present in the molecule. The most important thing about these labels is that atomic orbitals can only hybridise when they have the same symmetry label. The numbers 1 and -1 in the body of the table describe what happens to the orbitals in the molecule when different symmetry operations are performed - a 1 means the orbital is unchanged, a -1 means that it is in the same position but with reversed sign, and a zero (which can be found on other tables) means that it is in a completely different place. There will be more on this later.
The symbols to the right of the table describe how the axes behave under different symmetry types, and which symmetry labels apply to which orbitals.

Symmetry of orbitals

The table above tells us the symmetry of all the orbitals on the sulphur atom. The orbitals will behave like the mathematical functions used to express them:

Note that the d_z² orbital behaves as the mathematical function 2z² - x² - y². This can sometimes be confusing.

Let us look at another example of a group table, this time for a tetrahedral molecule belonging to point group T_d, such as CH₄.

There are three major differences between the T_d table and the table for C_2v:

There are now some numbers in the top row of the table. This is used to refer to multiple operations of the same kind. Thus there 8 C₃ rotations in a tetrahedral molecule, two about each of the four C-H bonds. You may find it helpful to check that you can find them all on a model of the molecule.
Similarly, there are three C₂ rotations (one about each axis), six S₄ axes, and six s_d mirror planes, for a total of 24 symmetry operations.

There are some new symbols for the symmetry labels in the left-hand column, E and T. E is used to refer to a doubly degenerate system. This has the entry 2 in the identity, E, column. T is used to refer to a triply degenerate system, with 3 in the E column.

Several functions are bracketed together in the right-hand columns. This means that not only do they have the same symmetries, they are also degenerate. For example, in any molecule belonging to the point group T_d, the p_x, p_y and p_z orbitals are degenerate. This is commonly seen in molecules which possess a great deal of symmetry.

Molecular Orbital Diagrams

Once we know how to use group tables to identify which atomic orbitals have which symmetry elements, we can use this information to deduce which orbitals are able to hybridise with each other to form molecular orbitals and chemical bonds. This allows us to construct the molecular orbital diagrams for the molecules we analyse.

We will start with the water molecule, H₂O, which is a simple example since it has a small group table and only contains sigma-bonds. We determine the molecular orbital diagram by following a series of rules:
Place the z-axis along the highest C axis of rotation, in this case C₂, and place the x-axis in the plane of the molecule. Look at the group table.

It is immediately clear from the table that the oxygen s- and p_z orbitals have A₁ symmetry, the p_x is B₁, and the p_y is B₂.
We must now deduce how the hydrogen orbitals will interact with the oxygen orbitals, and to do this we must determine their symmetry elements, relative to the molecule as a whole.

The hydrogen s-orbitals are s-type (i.e. they will only form s-bonds, not p-bonds), so can be called H(s).
We must look at what happens to these orbitals under the various symmetry operations present in the water molecule:

Under E, they both remain unaffected, so they contribute 2. (Two orbitals remaining in the same places.)

Under C₂, they both move to different positions. H¹ exchanges places with H². Neither remain unaffected and so they contribute 0.

Under s_v(xz), which is reflection in the plane of the molecule, they both remain unaffected and so the contribution is 2.

Finally, under s_v(yz), they both move to different positions. Neither remains where it was, so the contribution is 0.

Compare this last table with the group table for C_2v. It does not match any of the entries in the table.
The entries in the C_2v table are called irreducible representations. The 2 0 2 0 result we obtained for the hydrogen orbitals is called a reducible representation because it consists of a sum of A₁ and B₁.

This means that the two 1s atomic orbitals of the hydrogen atoms generate two molecular orbitals of symmetry A₁ and B₁. The A₁ orbital can form a bond with an oxygen orbital of symmetry A₁, i.e. the 2s or 2p_z orbitals. The B₁ orbital can bond with an oxygen orbital of symmetry B₁, i.e. the 2p_x orbital. We will look at these in more detail later, when we determine the explicit forms of these orbitals.

General method of splitting reducible representations

For more complicated molecules, we need a general way of splitting reducible representations. This can be done by following a series of rules.

1. Determine the order of the group. In this case it is 4.
This can be determined in either of two ways:

a). Count up the total number of symmetry operations.
E, C₂, s_v(xz), and s_v(xz) = 4 in this case.
Be aware that some group tables have more than one of some symmetry operations, e.g. the 8C₃ in the point group T_d.

b). Sum the squares of the entries in the E column, i.e.
1² + 1² + 1² + 1² = 4


2. Work out the contribution made by each symmetry element to the reducible representation.

First A₁.

We simply multiply the numbers in each column together, then sum the products of the columns.
(1 x 1 x 2) + (1 x 1 x 0) + (1 x 1 x 2) + (1 x 1 x 0) =4

Remember that the first 1 refers to the number of operations in each column. In this case it is always 1, but other point groups may have different numbers. This is very easy to overlook.

Next we divide our answer by the order of the group. In this case, 4/4 = 1. This means that there is one A₁ contribution.
Note that this must always give an integer. If it had not done so, we would know that a mistake had been made in deriving the 2 0 2 0.

We now repeat this step for each symmetry element in the group table:

A₂.

(1 x 1 x 2) + (1 x 1 x 0) + (1 x -1 x 2) + (1 x -1 x 0) = 0
There is no A₂ contribution.

B₁.

(1 x 1 x 2) + (1 x -1 x 0) + (1 x 1 x 2) + (1 x -1 x 0) = 4
There is one B₁ contribution.

B₂.

(1 x 1 x 2) + (1 x -1 x 0) + (1 x -1 x 2) + (1 x 1 x 0) = 0
There is no B₂ contribution.

Hence the combination is A₁ and B₁.

How to construct a molecular orbital diagram