Order of Shielding Effect for orbitals

Question

From what I know Shielding effect is the ability of inner electrons to repel outer electrons and reduce the Nuclear charge felt by the outer electrons and this is caused by electron-electron repulsion.

When I read the explanation from a site it said that

Shielding refers to the core electrons repelling the outer rings and thus lowering the 1:1 ratio. Hence, the nucleus has "less grip" on the outer electrons and are shielded from them. Electrons that have greater penetration can get closer to the nucleus and effectively block out the charge from electrons that have less proximity. Because the order of electron penetration from greatest to least is s, p, d, f; the order of the amount of shielding done is also in the order s, p, d, f.

But I don't understand the fact that inner orbitals would have greater shielding effect than the outer orbitals as the closer the electrons are the more the repulsion they will have as it follows the inverse squared law. So, as the outer electrons are closer to other electrons hence they will show more repulsive force than the inner electrons as the outer electrons are closer.

One explanation which I thought was that as the electrons are not stationary and moving at great speeds, the charge could be considered as symmetrically distributed over the volume and as the charge density would be low for outer orbitals they would have less shielding effect.

Is this explanation correct or am I going wrong somewhere?

Source:https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Mechanics/10%3A_Multi-electron_Atoms/Multi-Electron_Atoms/Penetration_and_Shielding

Answer 1

I suggest to look at the data instead of such explanation that you expect, they dont say much on quantity of the effect.

https://en.m.wikipedia.org/wiki/Ionization_energies_of_the_elements_(data_page)

Here you can see that hydrogen and francium only differ about 3.4 times in energy needed to remove one outer electron. Now lets look at the atoms sizes:

https://chem.libretexts.org/Courses/Mount_Royal_University/Chem_1201/Unit_2._Periodic_Properties_of_the_Elements/2.08%3A_Sizes_of_Atoms_and_Ions

About 70% scroll, figure 3.7

So, if we assume that electron is just orbiting charge +1 but at different distance, result is quite similar to what we expect naively, assuming charge of lower electrons and protons in the nucleus just cancels out. As if outer electron orbits around one proton, and the rest of protons and electrons cancel out. Another position to look at this problem from is this trick:

https://en.m.wikipedia.org/wiki/Shell_theorem

Not sure it applies exactly to give solution exactly as if inner electrons and n-1 protons are cancelled, but I assume it is a good starting point as a path to dig deeper into this effect.

I dont think the way effect is described in your link is a good way to describe it, a bit too vague.

Answer 2

the charge could be considered as symmetrically distributed over the volume and as the charge density would be low for outer orbitals they would have less shielding effect.

Not only that, but also, there are correlations between the expected positions of the two electrons measured at the same time. If the charge density is already lower for single inner electron orbital, it is also modified more as we fill the outer orbital.