# 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. – Tyberius 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? – Eenoku 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. – Tyberius 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? – Eenoku Mar 25 '18 at 20:37
• yeah I'll write something more detailed up when I get the chance. – Tyberius Mar 25 '18 at 20:39

## 1 Answer

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.