Q.How many electrons in an atom with atomic number 105 can have $(n+l)=8$?

In the hydrogen atom (a single electron), orbital energy depends only on the principal quantum number $n$, so all orbitals sharing one $n$ - $2s$ and $2p$; $3s$, $3p$ and $3d$; and so on - are exactly degenerate (equal in energy): $1s<2s=2p<3s=3p=3d<4s=4p=4d=4f<\dots$

In multi-electron atoms, electron-electron repulsion makes the Schrödinger equation too complex to solve exactly, so orbital energies are instead ranked approximately using the empirical (n + l) rule: the lower the value of $(n+l)$, the lower the orbital's energy; and if two orbitals share the same $(n+l)$, the one with smaller $n$ has the lower energy. Applying this systematically reproduces the real multi-electron filling order,

notably placing $4s$ (n+l = 4) below $3d$ (n+l = 5) even though $3d$ has the smaller $n$. …