Correlating the decrease in orbital radius and penetrating power with the increase in l value

Question

I have read that s-orbitals have a stronger penetration effect as compared to p-orbitals and p-orbitals have a stronger penetration effect as compared to d-orbitals, etc.

Therefore, the electrons in an s-orbital are closer to the nucleus of the atom and their ionisation energies are greater than for an electron in a p-orbital. (Example: This is what we use to explain why the energy of ionisation decreases from $\ce{Be}$ to $\ce{B}$).

Everything made perfect sense to me until I read:

The orbital radius slightly decreases with an increase of $l$ value.

Since s-orbitals penetrate more, shouldn't that mean that being closer to the nucleus, they have smaller radius as compared to an orbital with higher $l$ value, say, the p-orbital?

How can we justify this? Or are these two pieces of information not related in this way?

Answer 1

tl;dr

The statement that "the orbital radius slightly decreases with the increase of $l$ value" is true only for orbitals with the same value of $n$ (shell). The penetration of the nucleus by an electron is measured by the relative electron density, which depends on both shell ($n$) and subshell ($l$) of an electron in an atom.

What is Orbital Penetration:

Penetration describes the proximity of electrons in an orbital to the nucleus. Electrons which experience greater penetration experience less shielding and therefore experience a larger Effective Nuclear Charge ($Z_\text{eff}$), but shield other electrons more effectively. Electrons in different orbitals have different wavefunctions and therefore different distributions around the nucleus. However, contrary to what many think, penetration is not the outer electrons penetrating through the shield of the core electrons. It is actually just how well the electrons feel the nucleus. This is similar to the idea of outer electrons penetrating, but not the same. They are not the same because the core electrons have more penetration than the outer electrons since they (the core electrons) feel the strongest pull.

Clarification

As you can see from the definition above, the penetration of a nucleus by an electron is measured by the relative electron density near the nucleus of an atom. It is essentially, how effectively electrons can get close to the nucleus. For instance, below are the electron probability densities for s orbitals and a p orbital.

1. Nodes are regions of zero electron probability (white areas),
2. The orange color corresponds to regions of space where the phase of the wave function is positive, and the blue color corresponds to regions of space where the phase of the wave function is negative.

In a multi-electron system, the penetration of the nucleus by an electron is measured by the relative electron density near the nucleus of an atom for each shell and subshell of an electron. For example, the 2s electron is penetrating the nucleus of an atom more than the 2p electron, because the 2s has more electron density near the nucleus than the 2p electron. The penetration power of an electron, in a multi-electron atom, is dependent on the values of both the shell (n) and subshell (l) of an electron in an atom.

For the same shell value (n) the penetrating power of an electron follows this trend in subshells:

$$\ce{s > p > d > f}$$

And for different values of shell (n) and subshell (l), penetrating power of an electron follows this trend:

$$\ce{1s > 2s > 2p > 3s > 3p > 4s > 3d > 4p > 5s > 4d > 5p > 6s > 4f ...}$$