Total spin number 0 together with $B_2$ state - confusion

Question

I'm trying to understand molecular symmetry and I got confused reading the Molpro documentation.

There is an example of a system with $B_2$ symmetry written like this:

$$ wf,10,3,0, $$

which means, that the system has 10 electrons, $B_2$ symmetry and the total spin number is 0.

If I understard the total spin number well, it is 0 only when all electrons are paired in the orbitals (i.e. the system is in singlet state).

But, on the other hand, I thought, that when the electrons are paired, then the orbital is "totally symmetric", i.e. $(\cdot)^2 = A / A_g / A_1$.

So, as far as I understand it, $B_2$ symmetry and total spin number 0 is mutually exclusive.

Is it true or do I understand it wrong? If I do, could you, please, explain my mistake and provide some simple example of the abovementioned configuration?

Answer 1

We can see what a $^1B_2$ will look like by consider the MO diagram for water, a $C_{2v}$ molecule with $10$ electrons.

$\hspace{9ex}$

This is the configuration that we might expect for the ground state, which has $^1A_1$ symmetry. We could get a configuration with $^1B_2$ in a couple ways, namely by exciting form $1b_2$ into $4a_1$ or by exciting from either $2a_1$ or $3a_1$ into $2b_2$ without changing the spin of whatever electron we move. These new configurations would still be singlets because we haven't changed any of the spins and would have $B_2$ symmetry since clearly $B_2\otimes A_1=B_2$.

As to your question about spin multiplicity, if we consider a system with a single electron, we account for both the spin up and spin down configuration by determining the spin multiplicity. The spin multiplicity for $S=\frac{1}{2}$ is $2S+1=2$, a doublet, which gives us exactly the number of spins states we would expect.