Degeneracy of second excited state of H-?
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, $E_\mathrm{2s} &lt; E_\mathrm{2p}$, $E_\mathrm{3s} &lt; E_\mathrm{3p} &lt; E_\mathrm{3d}$, 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: $\mathrm{1s^1 2s^0 2p_x^1}$, $\mathrm{1s^1 2s^0 2p_y^1}$, $\mathrm{1s^1 2s^0 2p_z^1}$.