I've been using 'Chemical Structure and Reactivity: An integrated approach' by James Keeler and Peter Wothers to study periodicity, among other sources. However, there seem to be some commonly accepted contradictions used, and I don't understand how to reconcile these ideas:

1) D-block electrons shield poorly (s>p>d>f)

2) Using Slater's scale, d block electrons would, in fact, shield as well as most other (n-1) orbitals. Even though this is a rough scale, I take it to mean that d block electrons do have a notable shielding effect.

3) D-block contraction is the effect of increased Zeff due to the poor shielding of d-block electrons, and the addition of an equal number of protons.

Therefore, for example, down Group 13: the filling of the d-block orbital corresponds to an extra increase in Zeff with respect to the 4s/4p compared to the 3s/3p orbital. The 4s/4p orbitals will therefore be lower in energy than expected: shielding does not completely offset the extra protons.

4) Across the first row d-block elements, the energy of the 4s orbital does not fall very much. This is because the increase in nuclear charge is offset almost entirely by the addition of a shielding d-block electron, implying almost perfect shielding.

These four points seem, to me, to be essentially using d-block electrons at both ends of the spectrum when convenient. Is there a way to reconcile these ideas, or where have I gone wrong in my logic?


Let us read Keeler and Wothers's Chemical Structure and Reactivity carefully. (It's really a great book. I wish I had used it in the past.) We excerpt the relevant paragraphs that detail the points you mention, and highlight particularly important statements regarding screening. All page numbers are taken from the second edition.

Point 2, on Slater's rules.

[Energy ordering:] (1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) ...

  1. If the electron being considered is in an ns or np orbital, then electrons in the next lowest shell (i.e. that with (n-1)) each contribute 0.85 to $S$. Those electrons in lower shells (i.e. (n-2) and lower) contribute 1.00 to $S$.

  2. If the electron being considered is in an nd or nf orbital, all electrons below it in energy contribute 1.00 to $S$ [the shielding constant]. (p. 262)

Point 3, on Z$_\text{eff}$ and d-block contraction.

...there is a downwards kink in the [s-block] orbital energy on moving from the 3s to the 4s [orbitals]. [...] This difference is due to the filling of the 3d orbitals between calcium and gallium. The effects of all the extra protons in the nucleus is to cause an additional lowering in the energy of the valence orbitals for the p-block elements in Period 4. However, the effect is not really drastic: the 4s and 4p orbitals in gallium do not seem to have experienced the full effect of the ten extra protons. We interpret this by saying that the effect of the extra protons on the orbital energies of the 4s and 4p has, to a large extent, been cancelled out by the electrons that have been added in the 3d orbitals.

We could rephrase all of this by looking at the problem from the perspective of effective nuclear charges rather than orbital energies. The increase in Z$_\text{eff}$ for gallium is slightly greater than might have been expected by comparison with boron and aluminium because of the filling of the d-block. (p. 266)

Point 4, on orbital energies.

...between scandium (Sc) and copper (Cu) electrons are being added to 3d orbitals, and such electrons form an effective screen for the 4s electrons. To put it another way, as far as the 4s electrons are concerned, on moving from one element to the next between scandium and copper, the effect of the extra proton is largely cancelled out by the addition of the electron to the lower 3d orbital. Consequently the energy of the 4s falls rather slowly. (p. 261)

The outer electron in potassium occupies the 4s AO, rather than the 3d. This is because the 4s orbital penetrates to the nucleus more effectively than does the 3d, resulting in the energy of the 4s being lower than that of the 3d. [...] ...both the 3s and the 3p have subsidiary maxima close in to the nucleus, but the first maximum for the 3s is very much closer in. The 3d has no such subsidiary maxima, and therefore is much less penetrating than 3s or 3p. ...the electrons in the 4s do not screen the 3d particularly well since much of the electron density from the 4s is further out from the nucleus than that from the 3d.

From scandium onwards the energies of both the 4s and 3d AOs drop steadily: this is simply the result of the increase in nuclear charge not being quite offset by the increase in electron-electron repulsion. Put another way, the electrons in the 3d sub-shell do not screen one another particularly well. Although both AOs fall in energy, the 4s falls less steeply, which can be explained by noting that this AO is quite well screened by the 3d electrons, not least as these are in a lower shell. (p. 627)

We see that Keeler and Wothers are at least self-consistent regarding points 2, 3, and 4: 3d electrons are effective at shielding 4s and 4p electrons and not effective at shielding 3d electrons. I believe, therefore, that the contradiction lies in point 1, which should be corrected as follows:

d-block electrons shield poorly s or p electrons in the same shell

I suspect the statement that d-block electrons shield poorly is taught in reference to the concept of penetration. If we look at plots of radial probability density for the 3s, 3p, and 3d orbitals, we might naively conclude, on the basis of mean distance from nucleus, that 3d electrons are closest to the nucleus, which means that they should shield 3s and 3p electrons well. In reality, because of penetration, this is not the case. Thus we conclude that 3d electrons shield (s or p electrons in the same shell) poorly, and the misconception arises if we try to extend this poor shielding to 4s electrons and the like.


If u see the radial probability distribution function of nd orbital and that of ns orbital u will notice small humps are more in case of s orbital than that of d orbital. Because of this reason s orbital are more 'penenetrating' towards nucleus than that of a d orbital. So the screening constant value for a d orbital is much less compared to that of s orbital. Again if we consider the angular probability distribution function then we can see that d orbital are very much distributed in space compared to that of a s orbital which is spherically symmetric. Due to this reason d orbital registers a lower value of screening constant.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.