# How to explain the excited states in the dinitrogen cation?

The $$\ce{N2^+}$$ molecule has the molecular orbitals $$\ce{\sigma(1s)^2\; \sigma^*(1s)^2\; \sigma(2s)^2\; \sigma^*(2s)^2\; \pi(2p_x)^2 \; \pi(2p_y)^2 \; \sigma(2p_z)^1,}$$ which can be seen on the following picture: So, the ground state has $$\mathrm{A_g}$$ symmetry in $$D_\mathrm{2h}$$ point group, i.e. $$\Sigma_\mathrm{g}^+$$ in $$D_\mathrm{\infty h}$$.

## But what will the excited states look like?

I'd expect, that the first excited electrons will be the ones, which "need" the smallest amount of energy to get into a higher state.

So, the first electron will be excited from $$\ce{\pi(2p_x)}$$ or $$\ce{\pi(2p_y)}$$ orbitals to $$\ce{\sigma(2p_z)}$$ like this: The first excited state will have $$\mathrm{B_{3u}}$$ or $$\mathrm{B_{2u}}$$ symmetry in $$D_\mathrm{2h}$$.

The second excited state would be I.e. it would have the $$\mathrm{B_{3g}}$$ or $$\mathrm{B_{2g}}$$ symmetry.

Is my expectation correct or will the electrons be excited in a different order? And what will the 3rd and 4th excited states look like?

• Not a full answer, but strictly speaking its only an approximation that you could obtain an excited state from exciting from an occupied to a virtual orbital. If we consider a model where we do promote an electron, that promotion should be followed by a step where all the orbitals relax to better conform to the new configuration of electrons. So you really can't excite into a different orbital because as soon as you move an electron, you change the shape/energy of all the orbitals of the molecule. – Tyberius Oct 29 '18 at 19:48
• Just a note on convention. Capital letters are reserved for the spectroscopic state/wavefunction symmetry, while we label orbitals/electronic configuration with lowercase letters. (e.g. a_g rather than A_g ) – PJ R Oct 30 '18 at 2:01

Excited states make things a bit more complicated. The electron configuration only represent the HF ground state, and changing the occupied orbitals will have an effect on all the orbitals (they are correlated). So this picture becomes more and more inaccurate the higher lying the electronic state. What you do to resolve this is expand the electronic state in many configurations (which all most belong to the same IRREP).

For diatomic molecules there is a lot of spectroscopic data available. So let's check it: $$\ce{N2+}$$ in the NIST Chemistry WebBook

The states in that table are ordered by energy, with the ground state at the bottom. So scrolling down shows us the 3 lowest lying states: $$\rm X ^2\Sigma_g^+$$, $$\rm A^2\Pi_u$$, $$\rm B^2\Sigma_u^+$$.

Basically, this means you are right about the first state. To explain why we need a bit of symmetry. The first excited state has $$\Pi_u$$ symmetry, in the $$D_{\infty h}$$. Since we are working with $$D_{2h}$$ here, we will need to reduce it. Sparing the details, the result is $$\Pi_u$$ becomes $$B_{2u}\oplus B_{3u}$$, which means it will show up once in each IRREP, which is fine because $$\Pi$$ states are doubly degenerate anyway (just as $$\pi$$ orbitals are doubly degenerate).

The other state ($$\Sigma_u^+$$) becomes $$B_{1u}$$. This one you can for example get by excitation from the anti-bonding $$2\sigma_u$$ into $$3\sigma_g$$, with respect to the ground state configuration (see this reference). So there is an electron "jumping over" the $$1\pi_u$$ orbitals.

Note, that the ordering of the states can change if you start to move away from the equilibrium distance towards dissociation.

• By "the other state ($\Sigma_u^+$)" you mean that one electron from $\sigma^*(2s)$ will get excited to $\sigma(2p_z)$, while both electrons in $\pi(2p_x)$ stay in place? I.e. the electron will "jump over" $\pi(2p_x)$ and $\pi(2p_y)$ orbitals? – Eenoku Oct 29 '18 at 20:10
• Yes, this is indeed what happens here. Not really straight-forward to spot. Excited states are more complex ... – Feodoran Oct 29 '18 at 20:13