# Molecular orbitals symmetry - states notation

I'm reading some papers and I'm repeatedly seeing the following notation of system states:

Could you, please, explain to me the meaning of it?

E.g. the state $$B^2 \Sigma_u^+$$ - I'm aware, that $$\Sigma$$ is used as a substitute for $$A$$ in both $$C_{\infty h}$$ and $$D_{\infty h}$$. I'm also aware, that $$u$$ stands for "ungerade", i.e. it tells us, that the state is anti-symmetric with respect to the inversion.

But what does the $$B^2$$ before $$\Sigma$$ stand for? And what does $$\Sigma^+$$ mean?

I'm aware of the Mulliken notation, but it doesn't seem to contain these symbols.

The symbol $$B$$ is explained in greater detail here: Letter codes in molecular term symbols. In this context, it indicates that it is the second excited state of the same multiplicity as the ground state. The ground state, labelled with $$X$$, is a doublet; the first doublet excited state is labelled with $$A$$, and the second doublet excited state is labelled with $$B$$.
The superscript $$2$$ denotes the multiplicity, i.e. this is a doublet state, with total spin $$S = 1/2$$ (technically the spin quantum number) such that $$2S + 1 = 2$$. The dinitrogen cation $$\ce{N2+}$$ will have one unpaired electron in its low-lying electronic states, so it's not too surprising that the total spin is $$1/2$$.
Finally, there is a bit more to the symbol $$\Sigma$$ than just being a substitute or a generalisation of $$\mathrm{A}$$. In the linear point groups, this denotes the total orbital angular momentum about the internuclear axis. In this state, there is no orbital angular momentum about this axis, and the relevant quantum number is $$\Lambda = 0$$, which is denoted by the letter $$\Sigma$$. ($$\Lambda = 1$$ is a $$\Pi$$ state, $$\Lambda = 2$$ a $$\Delta$$ state, and so on.)
If you think about the ground state electronic configuration of $$\ce{N2+}$$, it is essentially the same as $$\ce{N2}$$, but with one unpaired electron in the $$\sigma_\mathrm{g}$$ orbital. In order to generate any kind of orbital angular momentum, you need to have an asymmetric population of $$\pi$$-type MOs. Since this isn't the case, the ground state of $$\ce{N2+}$$ has no orbital angular momentum, and is consequently a $$\Sigma$$ state. A similar argument can be applied to the second excited state, which seems to arise from a $$\sigma\to\sigma$$ excitation.
• Just one more question - X, A, B indices provide "indexing" among the states with the same multiplicity, but the same symmetry/angular momentum is not needed? In my example, there are states $\Sigma_g$, $\Pi_u$ and $\Sigma_u$ - are the letter codes really common for all three? – Eenoku Nov 18 '18 at 2:00