# What is the second excited state of $H^{-}$?

Not considering the electronic spin, the degeneracy of the second excited state of H-atom is 9, while the degeneracy of the second excited state of $$H^{-}$$ is :

The only thing that I need to solve this question is to know the second excited state of $$H^{-}$$. According to the answer, the second excited state of $$H^{-}$$ is $$1s^1 2s^0 \color{blue}{2p}^1$$. This seems to follow the Aufbau's principle( makes sense as $$H^{-}$$ is multielectronic species). But I have a couple of doubts regarding this configuration.

1. I learnt that Aufbau's principle is used to write Electronic Configuration of atoms in their ground state. Is it right to use it for writing down configuration in the excited states?
2. I learnt something called "Selection Rules" according to which only those transitions are allowed for which $$\Delta l = ±1$$. So from $$1s$$ orbital, election must go to $$2p$$ (1st excited state ) and then to $$3s$$ or $$3d$$ orbital (2nd excited state). But this logic fails from this question. Where is the fallacy?
• Selection rules are about whether ana atom can get from one state to another: the question is about the nature of a state - whether it's accessible or not from the ground state doesn't come into it and doesn't need to be worried about Jul 30 at 16:10
• "According to the answer, the second excited state of $H^−$ is $1s^1 2s^0 2p^1$" How is that the second excited state? Don't your excitation energies only depend on n?
– hft
Aug 2 at 4:52

You are thinking too much into selection rules. Selection rules do not change the energy, existence, or numbering of states. If you can write down a state, it exists, and its energy doesn't depend on the sequence of transitions you need to get there from the ground state. Ideally, you would use a properly quantum calculation of the energy, but this case is simple enough that the general rule $$1s\to2s\to2p$$ works fine. This rule is for energy calculation and thus does not interact with selection rules.
For more detailed reasoning, you could do this: Start with the $$H^-$$ ground state $$1s^2.$$ Now you want to find the first excited state. You cannot move the electrons within the $$n=1$$ shell, so move one electron up to $$n=2$$ (and assume the energy associated with changes of $$n$$ are larger than any other energy considered). This is not a physical transition that would be subject to selection rules; you are simply enumerating possible states. Now you have a choice between $$1s^1 2s^1$$ and $$1s^1 2p^1.$$ In $$H,$$ $$2s^1$$ and $$2p^1$$ are degenerate, so roughly both of them are the first excited state, but in $$H^-$$ the $$1s^1$$ breaks the degeneracy. Specifically $$s$$ orbitals are more penetrating than $$p$$ orbitals (get closer to the nucleus) and are lower in energy when there are electrons in lower shells shielding the nucleus. Thus the first excited state is $$1s^1 2s^1$$ and the second is $$1s^1 2p^1.$$ The degeneracy of this state is easy to tell.
Mostly as an aside: some selection rules are not exact ("rigorous") and especially the one you mentioned can be circumvented. The conditions under which selection rules are derived are not always present and violations of those conditions generically lead to violations of the selection rules. In this case, you have a forbidden $$1s\to2s$$ transition due to the rule $$\Delta l=\pm1.$$ This rule applies to transitions due to radiation from changes in the electric dipole moment. When such a mechanism is possible, it dominates, but if one isn't possible then a slower mechanism may allow the transition anyway. I can't parse out if higher order radiation would allow this transition (see Wikipedia's table), but there are yet other mechanisms (particularly collisions) that should be able to make it.