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

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?

• It might not be the solution to this particular case, but you can have an open shell singlet, where you have two unpaired electrons with different spins. Mar 25 '18 at 14:40
• @Tyberius And otherwise is my understanding correct? And in open-shell systems it is possible to have an orbital with single electron, which has -1/2 spin? Mar 25 '18 at 14:44
• I think otherwise you have the right idea, a fully paired system has to be totally symmetric. My understanding is that the total spin plugged in to determine whether it is a singlet, doublet, etc is taken to be the absolute value of the total spin. Mar 25 '18 at 14:58
• @Tyberius Could you, please, describe your idea in a little more detail and write the answer, so I could accept it? Mar 25 '18 at 20:37
• yeah I'll write something more detailed up when I get the chance. Mar 25 '18 at 20:39

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.