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:
(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:
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;
.
Questions
- 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?
- How can I obtain the total spin number?