If d-electrons are such poor shielders, why do trends increase more gradually across the d-block than the s or p-block?

Question

If I understand correctly, the shielding effect of d- (and f-) electrons seems to be much poorer than those of s- and p-electrons, due to the fact that they are less penetrating, have less electron density near the nucleus and so on (as can be seen from radial distribution functions).

However, trends for ionization energy, electron affinity, and electronegativity are MUCH more gradual than for s and p-blocks. This means that the effective nuclear charge ALSO increases much more gradually in the d-block, as confirmed by Slater and Clementi values. If d-electrons were poor shielders, surely this would mean that Z-eff would be much higher?

Doesn't this mean that, say for period 4, 3d-electrons are much more effective at shielding the 4s-electrons, than the 4s-electrons are at shielding 4p-electrons? Why does this happen?

Is there another factor at play? Is it because we are adding electrons to an "inner shell" that the shielding becomes more effective? Is Z-eff = Z - S an incorrect guideline to use?

Answer 1

Your question is similar to what I had in mind like a week back.

I know some textbooks do state a contradiction — that d-electrons have a poor screening effect, but then quickly use the explanation that due to increasing screening effect of nuclear charge from left to right in d-block elements (due to more electrons coming in the d-orbitals), trends are gradual.

Let me clear your mind of this false statement. Remember: it is not the screening effect that leads to a gradual change in trends across d-block elements, but rather the repulsion between d-electrons and outermost s-electrons.

The follwing example from the 1st series of d-block elements should make this clear to you. As nuclear charge increases, electrons come in the 3d-orbitals, which is not the outermost. The outermost is the 4s orbital, already filled up with 2 electrons.

Now, from left to right in this period, the increased nuclear charge will tend to hold on to the outermost 4s electrons tighter, but the d-electrons have an opposite effect. They repel these 4s electrons, and hence net force felt by the 4s electrons decreases. So, atomic size is increases more than expected, and hence it is easier to remove these electrons too (less ionisation energy)

Hence, the increase in ionization energy, reduction is size, etc. is much more gradual while traversing from left to right in a period.