# Effective Nuclear Charge of Noble Gases

As stated in my textbook, the effective nuclear charge can be calculated by taking the non-valence electrons away from the atomic number, as also stated in the first calculation on Wikipedia: https://en.wikipedia.org/wiki/Effective_nuclear_charge

The textbook also states:

If we assume that the noble gas core is completely shielding, then the 10 inner electrons of the neon core make the effective nuclear charge 10 less than the nuclear charge.

My understanding is it implies noble gases have 0 effective nuclear charge.

However, as many sources state noble gases have 8 valence electrons, shouldn't their effective charges be +8?

I think that this is basically an accounting problem. The full electron configuration of $\ce{Ne}$ is: $1s^22s^22p_x^22p_y^22p_z^2$. So, what does one now define as core electrons? It could be: "all electrons not in the outermost shell", yielding the $1s$ electrons. An alternative is "all electrons in completely filled shells", yielding all 10 electrons as core electrons. (Note that these two definitions yield identical results for almost all non-noble gas elements, with some difficulties surrounding the $f$-block.) Both definitions have some merit, I would prefer the first one, giving an effective charge of $+8$, although I do not know what one does with that number next.
As noted on the wikipedia page linked in the question, to arrive at more meaningful effective nuclear charges (as might be helpful for the systematic creation of basis sets for computational/quantum chemistry), one has to take the effective shielding strength of the different orbital shapes into account. The $d$-block-contraction, as can be seen in the somewhat different chemistry of $\ce{As}$ compared to $\ce{P}$ and $\ce{Sb}$, is due to this. Thus, in the end, a pen-and-paper calculation as alluded to above can be useful to think about some concepts in a qualitative fashion, but more extensive calculation is needed for quantitative reasoning.