Degeneracy of second excited state of H-?

This is a question presented in a IIT-JEE 2015 paper I exam. It says,



> 
> . Not considering the electronic spin, the degeneracy of the second excited state (n = 3) of H atom is 9, while the degeneracy of the second excited state of H- is-
> 
> 
> 


I saw a solution which says it's 3.


I think by 9 degenerate energy levels the question meant 9 possible orbitals for M shell. But I am not sure how's that 3 for for H-. 


Please don't delete this question as I haven't found it anywhere.




I think it is important to understand that for hydrogen atom (or any other one-electron system) all orbitals from the same shell have same energy. For instance, $E\_\mathrm{2s} = E\_\mathrm{2p}$, $E\_\mathrm{3s} = E\_\mathrm{3p} = E\_\mathrm{3d}$, etc. Thus,


* The first excited state of hydrogen atom would be one in which either $\mathrm{2s}$ or one of the three $\mathrm{2p}$ orbitals is occupied and it will be 4-fold degenerate: $\mathrm{1s^0 2s^1}$, $\mathrm{1s^0 2p\_x^1}$, $\mathrm{1s^0 2p\_y^1}$, $\mathrm{1s^0 2p\_z^1}$.
* Analogously, the second excited state of hydrogen atom would be one in which either $\mathrm{3s}$ or one of the three $\mathrm{3p}$ or one of the five $\mathrm{3d}$ orbitals is occupied and it will be 9-fold degenerate: $\mathrm{1s^0 2s^0 2p\_x^0 2p\_y^0 2p\_z^0 3s^1}$, $\mathrm{1s^0 2s^0 2p\_x^0 2p\_y^0 2p\_z^0 3s^0 3p\_x^1}$, $\mathrm{1s^0 2s^0 2p\_x^0 2p\_y^0 2p\_z^0 3s^0 3p\_y^1}$, $\mathrm{1s^0 2s^0 2p\_x^0 2p\_y^0 2p\_z^0 3s^0 3p\_z^1}$, plus 5 configurations with only $\mathrm{3d}$-orbitals occupied.


For hydride the situation is different: it is not a one-electron system, so different orbitals from the same shell do *not* have same energy anymore. For instance, $E\_\mathrm{2s} < E\_\mathrm{2p}$, $E\_\mathrm{3s} < E\_\mathrm{3p} < E\_\mathrm{3d}$, etc. Thus,


* The first excited state of the hydride would be one in which one electron populates $\mathrm{1s}$ orbital and another $\mathrm{2s}$ one, i.e. a non-degenerate $\mathrm{1s^1 2s^1}$ state.
* The second excited state of the hydride would be one in which one electron populates $\mathrm{1s}$ orbital and another one of the three $\mathrm{2p}$ ones, i.e. a 3-fold degenerate state: $\mathrm{1s^1 2s^0 2p\_x^1}$, $\mathrm{1s^1 2s^0 2p\_y^1}$, $\mathrm{1s^1 2s^0 2p\_z^1}$.


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In a single electron system like H atom or He+ all orbitals with same principal quantum number (n) have same energy
Therefore the degeneracy refers to the 3 subshells 3s 3p 3d which have same energy 


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