Effect of noble gas configuration on polarising power

I don't understand this statement written in my textbook about the polarising power of cations:

If two cations have the same size and charge, then the one with pseudo noble gas configuration ( with 18 electrons in the outermost shell) has greater polarising power than the other with noble gas configuration (with 8 electrons in the outermost shell). This explains why $$\ce{CuCl}$$ is more covalent than $$\ce{NaCl}$$.

Why does this happen?

Recall the fact that the basic difference between an ionic bond and a covalent bond, is basically just charge separation. In ionic bonds, charges are well separated, and the bond arises due to opposite charge attractions. Covalent bonds, on the other hand, have the electrons shared in between them, and it is these electrons that hold the nuclei of the two bonded atoms together.

But sometimes, a covalent bond isn't completely covalent, nor is an ionic bond completely ionic. Even an ionic bond may have some degree of "covalentness" in them. You can compare the covalentness of two ionic bonds with the help of a few rules formulated by Fajans. A few of these rules would be:

Rule 1: Smaller the cation and greater the charge, more the covalent character.

Why is this true? If you consider a cation to be similar to a uniformly charged sphere, the electric field generated at it's surface would be:

$$E=\frac{1}{4\pi \epsilon}\frac{Q}{R^2}$$

Where $Q$ is the charge on the cation, and $R$ is it's radius. It's quite clear now that highly positive cations and smaller cations can generate higher electric fields, and can strongly pull the electrons of the anion into the space between the ions, which gives rise to a partial covalent character.

In your example, it was given that "cation with pseudo-inert gas configuration can polarize better than a one with inert gas configuration". This is very much true, and can be explained using screening effects of orbitals. In the case of $\ce{Na+}$, it's valence electrons are in the $p$ orbitals, while in the case of $\ce{Cu+}$, they are in the $d$ orbitals. It's a known fact that $d$ orbitals shield the electrons from nuclear attractions poorly compared to the $p$ orbitals (see Why do screening effect decreases due to d-orbital? for more details). As a consequence, the nucleus attracts the valence electrons much more strongly in case of $\ce{Cu+}$, leading it to have a smaller size, and more polarizing power.

Rule 2: Greater the size of the anion, more is it's polarizability and greater is the covalent character.

Now the first question that would arise is, what is polarizability and why does it affect covalent character? Polarizability refers to the ability of the anion to get polarized, which is charge separation within the ion. Larger anions allow more room for the electron clouds within them to move around, which causes the cation to polarize it better. See the picture below for better clarity:

The reason why cations with pseudo inert gas configuration have more polarising power is because of the poor shielding effect of the 3d orbitals which increases the ability of cations to distort the electron cloud of anion, thus more polarising power.

Considering the example of NaCl and CuCl: In NaCl, the electrons effectively shield (or screen the outermost electron from the attraction of nucleus) or one can say the shielding is $$100\%$$. Whereas in case of CuCl, there are 10 3d electrons present which ineffectively shields the nucleus, so the value of screening decreases. The increasing order of shielding: s>p>d>f

Thus, CuCl has configuration : $$\ce{[Ar] 3d^10}$$ (pseudo noble gas configuration) NaCl has: $$\ce{1s^2 2s^2 2p^6}$$ or [Ne] (noble gas configuration)

Hence, $$\ce{Cu+}$$ will distort the cloud of $$\ce{Cl-}$$ to a greater extent than the distortion caused by $$\ce{Na+}$$.

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