Question

Not considering the electron 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:

Answer 1

Energy order of orbitals of $H$ is decided by only principle quantum number $\left(n\right)$
while energy order of $H^{-}$ is decided by $(n+l)$ rule:
Electronic configuration of $H^{-}$ is $1s^2$.
$H^{-}=1s^{2}2s^{0}2p^0$
Its second excited state is $2p$
And degeneracy of $2p$ is $3$.