Like with diatomic molecules, a consideration of the symmetry of the molecule and the symmetry of involved atomic orbitals yields valuable information on the molecular orbitals. As an approximation to molecular orbitals, we will deal with linear combinations of atomic orbitals. Finding the molecular orbitals is facilitated by introducing the aspect of symmetry. Therefore, we establish "symmetry adapted linear combinations" of atomic orbitals, the socalled symmetry orbitals.
Taking the water molecule as an example, we will illustrate this method. We assume a bent structure with identical distances HO (questions on geometry are answered by the rules of Walsh). The molecule falls into point group C_{2v}. The molecular orbitals of H_{2}O need to show this symmetry and the symmetry operations for C_{2v} affect these orbitals in the same way the respective irreducible representations are affected. For the mentioned point group, all irreducible representations or symmetry species are onedimensional; thus the values +1 and −1 in the character table for C_{2v} indicate whether an MO behaves symmetric or antisymmetric with respect to
an operation.

An MO of symmetry a_{2} (we use minuscules letters for single electron states, i.e. orbitals, and majuscules for the total electronic states of molecules) is therefore symmetric with respect to identity E and a rotation on a twofold axis C_{2} (If there is a distinct axis, it is taken as axis z). In contrast, it is antisymmetric with respect to a reflection on the planes xz and yz. None of the four symmetry operations affects the framework of the molecule itself.
We would be free to choose an orientation with the two hydrogen atoms in the plane xz or in plane yz normally, but, by convention, the second possibility is preferred and both hydrogen atoms lie within the plane yz and axis z bisects the angle HOH.
Fig.1: Atomic orbitals of the H_{2}O molecule. The orbital p_{x} is perpendicular to the plane yz. 
With respect to point group C_{2v}, this set of orbitals is already symmetry adapted as we will demonstrate for the atomic orbital p_{x}. Applying any of the symmetry operations of the group to the orbital's wave function, we ask whether the sign of the wave function remains constant or changes.
Not surprisingly, identity E leaves the wave function unaffected, therefore 1 is the value noted down for p_{x}.
E  C_{2}  σ_{v}  σ'_{v} 
1 
Next we imagine a rotation of 180° about the principle axis and check whether there is a change of sign for the respective wave function. As this is the case, the entry below the symmetry operation C_{2} p_{x} is 1.
E  C_{2}  s_{v}  σ'_{v} 
1  1 
If the atomic orbital is submitted to reflection on the plane xz, we find the wave function unaltered; therefore the entry below σ_{v}(xz) is 1.
E  C_{2}  σ_{v}(xz)  σ'_{v} 
1  1  1 
If the atomic orbital is submitted to reflection on the plane yz, we obtain the wave function with reversed sign, therefore the entry below σ_{v}(yz) is −1.
E  C_{2}  s_{v}  σ_{v}'(yz) 
1  1  1  1 
If we compare the values recieved with the entries in the character table of point group C_{2v}, we recognize it to be identical with one set of characters in a row of the table. This correspondence is fundamental for the classification of orbitals to a certain symmetry species. In our example, b_{1} is obtained for p_{x}. In this way, atomic orbitals and the irreducible representations of C_{2v} are linked.
atomic orbital  s  p_{x}  p_{y}  p_{z} 
irreducible representation  a_{1}  b_{1}  b_{2}  a_{1} 
In contrast to the orbitals of the oxygen atom, the AOs h_{1} and h_{2} are not symmetry adapted as C_{2} or σ(xz) convert h_{1} in h_{2} and vice versa. But it is easy to recognize that linear combinations like
h_{s} = ^{1}/_{√2} (h_{1} + h_{2}) 
h_{a} = ^{1}/_{√2} (h_{1} − h_{2}) 
are symmetry adapted. To find the correspondence between the obtained orbital and the symmetry species, we apply the operations. The result is shown in the table below.
atomic orbital  h_{s}  h_{a} 
irreducible representation  a_{1}  b_{2} 
Using the six symmetryAOs s, p_{x} p_{y}, p_{z},h_{s} and h_{a}, an equal number of MOs can be formed. Note that identical symmetry is a precondition for such combinations.
As there is only one atomic orbital of symmetry species b_{1}, it is not introduced in any linear combination and therefore we regarded p_{x} as a molecular orbital of H_{2}O as well. Three AOs are of symmetry species a_{1}. Therefore s, p_{z} and h_{s} are combined to yield three molecular orbitals. Finally, p_{y} and h_{a} belong to b_{2} and are thus combined to yield two molecular orbitals
1a_{1} = c_{11}s + c_{12}p_{z} + c_{13}h_{s} 
2a_{1} = c_{21}s + c_{22}p_{z} + c_{23}h_{s} 
3a_{1} = c_{31}s + c_{32}p_{z} + c_{33}h_{s} 
1b_{1} = p_{x} 
1b_{2} = c_{44}p_{y} + c_{45}h_{a} 
2b_{2} = c_{54}p_{y} + c_{55}h_{a} 
The coefficients c_{ik} remain to be determined. It is obvious that, for the two b_{2} orbitals, these coefficients will have the same sign (we denote this combination with 1b_{2}) and in the other, they will have opposite signs as orthogonality is imposed on the resulting molecular orbitals. Fig. 2 shows these MOs schematically with hatched areas for positive values and blank areas for negative values of the function.
Fig. 2:
Representation of the orbitals 1b_{2} and the 2b_{2} in H_{2}O. 
In a situation where several orbitals of the involved atoms fall in one symmetry species (like a_{1} in our example), each pair of atomic orbitals A or B_{n} usually yields one bonding and one antibonding linear combination. The remaining atom orbitals become nonbonding MOs.
bonding:  1a_{1}  1b_{2} 
non bonding:  2a_{1}  1b_{1} 
antibonding:  3a_{1}  2b_{2} 
We have derived the electronic configuration of the ground state of the water molecule without introducing the aspect of energy. The sequence of MOs in orbital energy is dependent on
Fig. 3: Qualitative representation of the energy of the valence molecular orbitals in the molecule H_{2}O. 
If, instead of eight electrons, we had twelve to be distributed over the six orbitals, the number of fully occupied bonding and antibonding orbitals would be equal and no stable molecule could exist. For the stability of a molecule it is normally required that the number of doubly occupied bonding orbitals is equal or higher to the number of doubly occupied antibonding MOs. The difference between these numbers is the number of bonds in the molecule.
An additional important aspect in our context is to consider whether there is at all the possibility to form covalent bonds. If the hydrogen orbitals h_{1} and h_{2} were of substantially higher energy, we would have linear combinations of approximately this kind
1a_{1} ≈ c_{11}s + c_{12}p_{z} 
1b_{2} ≈ p_{y} 
I.e. the electrons were located in the sphere of the oxygen atom and would not contribute to bonding. If we now fill the molecular orbitals 1a_{1}, 1b_{2}, 2a_{1} and 1b_{1}, we would recieve the oxygen anion O^{2−} and the hydrogen atoms left as protons. But, due to electronic repulsion, there is a limit for the transfer of electrons. This aspect is not revealed by the scheme of MOs. For compounds AB_{n} with strongly asymetric distribution of charges, this scheme could lead to wrong conclusions.
In principle, to understand the electronic configuration of H_{2}O and other molecules of point group
Kimball table for molecules AB_{2} of symmetry C_{2v}
for atoms B in the plane yz 

Atom  C_{2v}  a_{1}  a_{2}  b_{1}  b_{2} 
A  s
p d 
1
1 2 
0
0 1 
0
1 1 
0
1 1 
B_{2}  σ
π 
1
1 
0
1 
0
1 
1
1 
The table predicts the way atomic orbitals split and transform to irreducible representations of C_{2v}. For atom A, the sorbital becomes a_{1}, the three porbitals become a_{1}, b_{1} and b_{2}, the five dorbitals transform to two a_{1}orbitals, one a_{2}, one b_{1} and one b_{2}orbital. The entries for B_{2} display the kind of irreducible representations emerging from σ and πAOs respectively. The names σ and π are used with reference to the connecting lines between atoms A and B. Orbitals with rotational symmetry about this axis are classified as σ and orbitals with a nodal plane containing this axis as π. For example,
Applying the Kimball table to the water molecule, we take the s, p and σ states and state that
Often, antibonding molecular orbitals are labelled by an asterisk.
Auf diesem Webangebot gilt die Datenschutzerklärung der TU Braunschweig mit Ausnahme der Abschnitte VI, VII und VIII.