I'm beginning to use Molpro software for some computations. But I got completely confused by its description of specification of wavefunction symmetry.

It tells us, that the symmetry of wavefunction is described like $$\text{wf},\text{nelec},\text{irrep},\text{spin},$$

where $\text{nelec}$ is the total number of electrons in the molecule, $\text{irrep}$ is the number of the irreducible representation (list of numbers in Molpro docs) and $\text{spin}$ is the total spin number.

I know, that water belongs to the $C_{2v}$ point group, and I found its character table:

C2v character table
(source: uscupstate.edu)

I know, that 1/-1 values correspond with variance/invariance of orbitals, i.e. $A_1$ representation corresponds to the fact, that $\ce{s}$ and $\ce{p_z}$ orbitals are not inverted by any of the four symmetry operations.

I also know, that the electron configuration of oxygen is $\ce{1s^2 2s^2 2p^4}$ and both hydrogen atoms have $\ce{1s^1}$.

But I'm not sure how am I supposed to choose the irreducible representation of the whole molecule.

In this answer the similar problem is described and $\ce{H2O}$ molecular orbital diagram is given:

MO diagram of water molecule

The answer tells us, that molecular irreducible representation can be obtained like this:

$$(2a_1)^2(1b_2)^2(3a_1)^2(1b_1)^2 = (A_1)^2 \otimes (B_2)^2 \otimes(A_1)^2 \otimes (B_1)^2 = A_1,$$

which is in accordance with the first example in the Molpro docs, where $\ce{H2O}$ wavefunction is specified like WF,10,1;.


  1. So, is this the correct way to find the irreducible representation of the whole molecule? And if it is, could you, please, explain, how to get the molecular orbital representation?
  2. How can I obtain the total spin number?
  • $\begingroup$ Related: chemistry.stackexchange.com/q/58229/41556 $\endgroup$
    – Tyberius
    Feb 25, 2018 at 0:42
  • $\begingroup$ I have used Molpro a decent amount and none of these inputs are actually directly required. Basically the only cases you would want to use them are for open shell systems and charged systems. Charged, you will specify nelec so that it does not simply create a neutral system. For open shell systems or transition metals, you may wish to choose a particular high-spin complex, and so use the total spin input. The symmetry is much easier to specify using the symmetry directive where you just type in the point group. $\endgroup$
    – jheindel
    Feb 27, 2018 at 6:30
  • $\begingroup$ @jheindel I was under impression, that in some cases it's not enough to just specify the point group and that I have to explicitly specify even the irreducible representation of the molecule. $\endgroup$
    – Eenoku
    Feb 27, 2018 at 10:34
  • $\begingroup$ You might be right... I have never encountered these situations. I guess we can wait and see. $\endgroup$
    – jheindel
    Feb 27, 2018 at 16:47

2 Answers 2


Yes, this is the correct way to find the irreducible representation of the molecule. To obtain the energy ordering of the molecular orbitals (which determines the order in which they are filled for the ground state molecule), you need to run a molecular orbital theory computation (e.g. Hartree-Fock or density functional theory).

The spin number in Molpro is not particularly intuitive. For a molecular term symbol such as $^1B_2$, spin information is conveyed as $2S+1$, where $S$ is the total spin quantum number. In the specific case of a $^1B_2$ state, the molecule has a total spin quantum number of $0$. However, Molpro users must specify spin as $2S$.

As noted in the comments, however, specification of the irreducible representation or spin is generally unnecessary for Molpro computations. For Z-matrix inputs, Molpro determines the symmetry automatically, and in most cases this will be the symmetry the user desires. Molpro also automatically sets the spin if the user doesn't specify it. By default, Molpro uses $S=0$ for molecules with an even number of electrons and $S=1$ for those with an odd number of electrons. Since a large number of molecules that are kinetically stable under typical terrestrial conditions have an even number of electrons, the default spin should work well for many users. Molpro users who work on excited electronic states or molecules with unpaired electrons are much more likely to need to specify the spin.


You are using the term "irreducible representation of the whole molecule", but to me this does not make a lot of sense. So I would like to clarify something about this:

Irreducible representations are used to classify different electronic states (or electronic wave functions) of the molecule. The irreducible representation indicates sign changes (and therefore nodal planes) in the wave functions with respect to the symmetry operations of the molecule.

The "whole molecule" does not have sign changes, there are no negative atoms. Therefore it does not make sense to talk about "irreducible representation of the whole molecule". You may argue now whether the "whole molecule" always belongs to the totally symmetric irreducible representation ($A_1$ in case of $C_{2v}$) or does not have one at all. I would just stick with following statement: The molecule has a point group and its electronic states can be classified by irreducible representations.

Answering your questions

1) So, is this the correct way to find the irreducible representation of the whole molecule? And if it is, could you, please, explain, how to get the molecular orbital representation?

Which irreducible representation you need depends on the electronic state you want to calculate. What you showed would be the correct way to get the irreducible representation of a given electron configuration (in this case the ground state). The molecular orbitals can be obtained from a calculation (e.g. HF or DFT), or by symmetry considerations. But the later will only give you qualitative results, no orbital energies.

2) How can I obtain the total spin number?

Just as for the irreducible representation, this depends on the electronic state you are interested in. Usually this is the ground state, and for a closed shell system (like $\ce{H2O}$) we have $S=0$. But in principle it is possible to split up one, or more electron pair(s) and get $S=2,4,\dots$, but this will results in higher total energies (= excited states).


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