Orbital Energy Level Diagrams  Contents -  Group Theory

To Determine Symmetry Orbitals

(i) Determine the group and its symmetry operations.
(ii) Write down the relevant atomic orbitals (2s,2p_x..) 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	$\sigma_v(xz)$	$\sigma_v^{'}(yz)$
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 ($\pm1$ for unchanged/changed and zero otherwise). We must group 1s_A,1s_B together and add up the $\pm1$ 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 2p_x,2p_y,2p_z 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.

$$\Gamma=\sum a_i\Gamma_i$$ (47)

 where $\Gamma$ denotes the species we are trying to find, $\Gamma_i$ denotes a species for the group, and a_i(=0,1,2..) are integers given by

$$a_i=\frac{1}{g}\sum_R\chi^{\Gamma}(R)\chi^{\Gamma_i}(R)$$ (48)

 g is the number of elements in the group (its order), R are the group elements, $\chi^{\Gamma}$ are the characters of the function, and $\chi^{\Gamma_i}$ are the characters from the character table. Applying the formula to 1s_A,1s_B we get
$a_{B_2}=\frac{1}{4}[2\times 1+0\times -1+0\times -1+2\times 1]=1$
$a_{A_1}=\frac{1}{4}[2\times 1+0\times 1+0\times 1+2\times 1]=1$
$a_{B_1}=\frac{1}{4}[2\times 1+0\times -1+0\times 1+2\times -1]=0$
and thus
$\Gamma$=A₁+B₂
Finally to determine the symmetry orbitals we use a projection formula

$$P^{\Gamma_i}=\frac{1}{g}\sum_R\chi^{\Gamma_i}(R)R$$ (49)

 Thus to find the combination of 1s_A and 1s_B which has symmetry A₁, we put $\Gamma_i$ equal to A₁ and apply the projection operator to 1s_A. We obtain $\frac{1}{4}[1\times{\rm 1s}_A+1\times{\rm 1s}_B+1\times{\rm 1s}_B+1\times{\rm1s}_A]=\frac{1}{2}({\rm 1s}_A+{\rm 1s}_B)$ Similarly for B₂ we obtain $\frac{1}{4}[1\times{\rm 1s}_A+-1\times{\rm 1s}_B+-1\times{\rm 1s}_B+1\times{\rm1s}_A]=\frac{1}{2}({\rm 1s}_A-{\rm 1s}_B)$ Let us now consider NH₃ for which there are three reflection planes and two 120^o rotations. The character table is
 

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

 
 
 Here it is apparent that the symmetry sets are 2s_N;2p_zN;1s_A,1s_B,1s_C;2p_xN,2p_yN
 
 The difficulty is what happens to the 2p_xN,2p_yN when they are rotated by 120^o. 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 $-\cos 60^o$. Similarly after reflections the diagram     
shows that the entry is $\pm\cos 60^o$. Thus we obtain the following characters for the functions
 

	E	C3	C32	$\sigma_1$	$\sigma_2$	$\sigma_3$
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	$-\cos 60^o$	$-\cos 60^o$	-1	$\cos 60^o$	$\cos 60^o$
2pyN	1	$-\cos 60^o$	$-\cos 60^o$	1	$-\cos 60^o$	$-\cos 60^o$

 
We can now summarise this table by adding the entries for 2p_xN,2p_yN together, and observing that the columns for C₃,C₃² and $\sigma_1,\sigma_2,\sigma_3$ are the same ( these are said to form two classes). We get
 

	E	C3	$\sigma_v$
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 (1s_A+1s_B+1s_C) will belong to A₁. To find the two E components we apply the projection operator to 1s_A to obtain $\frac{1}{6}[2\times{\rm 1s}_A+-1\times{\rm 1s}_B+-1\times{\rm 1s}_C]$ which on being normalised is $\frac{1}{\sqrt 6}[2\times{\rm 1s}_A-{\rm1s}_B-{\rm 1s}_C]$. If we apply the projection operator to 1s_B, we get $\frac{1}{\sqrt 6}[2\times{\rm 1s}_B-{\rm 1s}_A-{\rm 1s}_C]$. An orthogonal combination of these two is $\frac{1}{\sqrt 6}[2\times{\rm 1s}_A-{\rm1s}_B-{\rm 1s}_C],\frac{1}{\sqrt 2}({\rm 1s}_B-{\rm 1s}_C)$. This pair forms an E symmetry pair, as does (2p_xN,2p_yN). The A₁ symmetry orbitals are (1s_A+1s_B+1s_C),2s_N,2p_zN.

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 3s_S, 3p_S and also consider the five 3d_S atomic orbitals. We shall only use one of the 2p_F orbitals on each F, pointing towards the S atom. Label these as f₁,f₂...f₆. The group is O_h.
  

   
 The symmetry species may now be evaluated:
3s_S....A_1g (totally symmetric)
3p_S....T_1u (read from the group table as x,y,z; it is triply degenerate)
3d_S....E_g+T_2g (read from the group table as xy,yz,zx;z²,x²-y²)
f₁,f₂,..f₆.....A_1g+E_g+T_1u (has to be worked using the previously described methods, a little thought convinces one that A_1g+T_1u must be present)

  
 Orbital Energy Level Diagrams  Contents -  Group Theory

1998-09-23