# What is the highest possible effective nuclear charge as a percentage of nuclear charge an electron in any neutral atom can experience?

I was wondering about the following statement "The higher the principal quantum number of the orbital the electron is in, the lower the nuclear charge it experiences." I then argued that this statement is false because when an atom has more electrons, it also has more protons and as the orbitals become more diffuse, the attraction of the protons would significantly outweigh inter-electronic repulsion at higher atomic numbers.

This led me to think about the following question: What is the highest possible (nuclear charge an electron can experience divided by the nuclear charge of the atom)? My guess is that this electron would certainly belong to one of the more diffuse subshells. (i.e. d or f)

Please note that the quantity I am looking at is not effective nuclear charge but it is the effective nuclear charge divided by the nuclear charge.

A high principal quantum number does not necessarily mean more electrons and/or protons. For example Hydrogen can be in an electronic excited state with $n=4$ and higher, as can be observed for example in the Balmer series. So for a specific element your cited statement remains true.
According to Slater's rules the second $1s$ electron screens with $0.3$ (all other screening constants are larger). Therefore the ratio you are looking for is $\frac{Z_{eff}}{Z}=\frac{Z-0.3}{Z}$ and converges to $1$ for large atomic numbers $Z$.
The only way to get even less screening would be to have no further electrons. As you restricted the question to neutral atoms we are left with Hydrogen. The charge divided by the nuclear charge of the atom, an electron can experience here is $\frac{Z_{eff}}{Z}=1$ for any excited state (principle quantum number) of the Hydrogen atom.