When people say "The energy of an orbital depends on its ℓ value due to screening effect" , what is screening what?

Question

My book says:

Both the attractive and repulsive interactions depend upon the shell and shape of the orbital in which the electron is present. For example electrons present in spherical shaped, s orbital shields the outer electrons from the nucleus more effectively as compared to electrons present in p orbital. Similarly electrons present in p orbitals shield the outer electrons from the nucleus more than the electrons present in d orbitals, even though all these orbitals are present in the same shell.

Here, they've mentioned outer electrons. What exactly do they mean by those?

Because when I tried to reason out how shielding affects the value of ℓ, I thought that 4s had more orbitals between it and the nucleus (everything between the nucleus and 3d, and then 3d too), and thus 4s should be more shielded than 3d.

Answer 1

Note that orbitals are theoretical constructs and not physical objects. The term orbital is primarily the wave function describing the particular electron quantum state. But it can be also understood as this quantum state itself or the 3D geometrical representation of the spatial presence probability of such an electron.

An outer electron is mostly more distant from the nucleus than an inner electron. Therefore the mean radial repulsion of the outer one by the inner one aims outwards.

There is said mostly above, as there is significant probability that e.g. 2s or 2p electrons are closer to the nucleus than 1s electrons, even if vice versa is (much) more probable. That is because of orbital geometrical overlapping.

The resulting effect is like if the outer electron were attracted by a nucleus of less than actual charge. This effect is called nucleus shielding/screening.

The effect intensity for 2 considered orbitals depends on their mutual probabilistic overlapping (more overlapping = less screening) and inner orbital shape (closer to spherical symmetry = better screening)

Electrons in s orbitals have significant occurance probability near nucleus and in inner regions. Therefore, they are less shielded than electrons in other orbitals with the same $n$ and have the lowest energy of them.

OTOH, due their spherical symmetry, they are the best in shielding, unless overlap level overrules that.

Answer 2

If I understand your question correctly, you are asking how the value the orbital angular momentum $\ell$ affects the shielding -- or in other words the energy. To answer this question, let's first look at the hydrogen atom. In the hydrogen atom, a single electron is bound to a single proton. Because of the simplicity of the system, the potential energy is purely Coulombic, that means that $V\propto r^{-1}$. As a consequence, the different $\ell$ levels in hydrogen for the same principal quantum number $n$ are degenerate (ignoring relativistic effects). This probably makes sense, because if you have only one electron there is no shielding.

Now, let's look at a multielectron atom. When we look at atoms with more than one electron, we still often use the solutions of the Schrödinger equation of the hydrogen atom, that is, we use one-electron wavefunctions and ignore the interaction between the different electrons. Although this might sound like a very rough assumption, it is not too bad in practice. If you consider one of the electrons of a multielectron atom, the other $Z-1$ electrons together with the nucleus of charge $+Z$ give a nett positive charge close to $1e$.

The situation is thus very similar to the case of the hydrogen atom. A single electron bounded to a core with a positive charge. The description will get closer to that of hydrogen if the electron under consideration is in a highly excited state corresponding to a large clasical radius so that the electron effectively sees a single core. For lower electronic states, the electron can penetrate the "cloud" of other electrons, resulting in a slightly different potential energy.

But why does this energy then depends on $\ell$? Well, if you look at the radial part of the Schrödinger equation with partial wave expansion (known as the Coulomb equation), there is a potential energy term in the equation of the form $$ \frac{\ell(\ell+1)}{r^2}. $$ This centrifugal barrier makes it more difficult for higher $\ell$ states to get close to the nucleus and experience the effect of the other electrons. As a consequence, orbitals with $\ell\ge 3$ behave almomst like those of hydrogen (meaning that they are degenerate and have a so-called quantum defect close to zero).