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### To Determine Symmetry Orbitals

(i) Determine the group and its symmetry operations.
(ii) Write down the relevant atomic orbitals (2s,2px..) and put them into sets which are not mixed by the symmetry operations.
(iii) Write down the characters of these sets, and then determine the symmetry species
(iv) Determine the symmetry species by projection.
As an example, for H2O, this is what happens to the atomic orbitals under the symmetry operations.

 E C2 2sO 2sO 2sO 2sO 2sO 2pxO 2pxO -2pxO 2pxO -2pxO 2pyO 2pyO -2pyO -2pyO 2pyO 2pzO 2pzO 2pzO 2pzO 2pzO 1sA 1sA 1sB 1sB 1sA 1sB 1sB 1sA 1sA 1sB

We can then write down the characters ( for unchanged/changed and zero otherwise). We must group 1sA,1sB together and add up the  for each of them. We get as follows and can then read off the symmetry species from the character table.

 2sO 1 1 1 1 A1 2pxO 1 -1 1 -1 B1 2pyO 1 -1 -1 1 B2 2pzO 1 1 1 1 A1 1sA,1sB 2 0 0 2 A1+B2

Note that the symmetry of 2px,2py,2pz agrees with the symmetry of x,y,z, given in the group character table.
If we are unable to find the species by inspection (we nearly always can) we may use the following formula.
 (47)
where  denotes the species we are trying to find,  denotes a species for the group, and ai(=0,1,2..) are integers given by
 (48)
g is the number of elements in the group (its order), R are the group elements,  are the characters of the function, and  are the characters from the character table. Applying the formula to 1sA,1sB we get

and thus
=A1+B2
Finally to determine the symmetry orbitals we use a projection formula
 (49)
Thus to find the combination of 1sA and 1sB which has symmetry A1, we put  equal to A1 and apply the projection operator to 1sA. We obtain
Similarly for B2 we obtain
Let us now consider NH3 for which there are three reflection planes and two 120o rotations. The character table is

 C3v E 2C3 3 A1 1 1 1 z A2 1 1 -1 E 2 -1 0 x,y

Here it is apparent that the symmetry sets are 2sN;2pzN;1sA,1sB,1sC;2pxN,2pyN

The difficulty is what happens to the 2pxN,2pyN when they are rotated by 120o. The rule is we have to enter for the character that fraction of the original function that remains after applying the group operation. The diagram helps

and shows that the fraction is . Similarly after reflections the diagram

shows that the entry is . Thus we obtain the following characters for the functions

 E C3 C32 1sA,1sB,1sC 3 0 0 1 1 1 2sN 1 1 1 1 1 1 2pzN 1 1 1 1 1 1 2pxN 1 -1 2pyN 1 1

We can now summarise this table by adding the entries for 2pxN,2pyN together, and observing that the columns for C3,C32 and  are the same ( these are said to form two classes). We get

 E C3 1sA,1sB,1sC 3 0 1 A1+E 2sN 1 1 1 A1 2pzN 1 1 1 A1 2pxN,2pyN 2 -1 0 E

We have identified the symmetry species from the original character table.
To obtain the symmetry orbitals, it is obvious that (1sA+1sB+1sC) will belong to A1. To find the two E components we apply the projection operator to 1sA to obtain
which on being normalised is . If we apply the projection operator to 1sB, we get . An orthogonal combination of these two is . This pair forms an E symmetry pair, as does (2pxN,2pyN). The A1 symmetry orbitals are (1sA+1sB+1sC),2sN,2pzN.

Finally we shall consider Sulphur Hexafluoride which is a more difficult example; it allows a discussion of the effect of d orbitals. We use the 3sS, 3pS and also consider the five 3dS atomic orbitals. We shall only use one of the 2pF orbitals on each F, pointing towards the S atom. Label these as f1,f2...f6. The group is Oh.

The symmetry species may now be evaluated:
3sS....A1g (totally symmetric)
3pS....T1u (read from the group table as x,y,z; it is triply degenerate)
3dS....Eg+T2g (read from the group table as xy,yz,zx;z2,x2-y2)
f1,f2,..f6.....A1g+Eg+T1u (has to be worked using the previously described methods, a little thought convinces one that A1g+T1u must be present)

Next: Orbital Energy Level Diagrams Up: Contents - Previous: Group Theory

Nicholas Handy

1998-09-23