The stability of the +II oxidation state in lead has been widely attributed to the so-called “inert pair effect”: the 6s subshell is stabilized through relativistic contraction due to the near-c speed of its electrons. Now, Pekka Pyykkö states in his 1988 article (p. 585) that

In addition to the 6s2 inert pair discussed above, analogous compounds (such as GeO and SnO) are known for rows 4 and 5. While the relativistic 5s stabilization is not entirely negligible, the 4s stabilization due to the “d-block contraction” probably overweighs relativity on row 4.

How does the d-block contraction cause a change in valency? I know that due to poor shielding by d electrons, atomic radii across the fourth period don’t increase particularly. But how is it responsible for the increasing stability of the +II oxidation state in group 14 (before getting to lead, where relativistic effects play a significant role)?

Would I be wrong if I said that the cause for the increasing stability of the s electrons in germanium and tin was the existence of a full d subshell (absent in, say, silicon), which, however slightly, shields the nuclear charge for the p electrons better than for the s electrons? Or what is it that happens, exactly?

Finally, according to the orbital potential energies listed on page 16 of Appendix B of Miessler and Tarr's Inorganic Chemistry (5th ed.), the 5s electrons of tin aren’t more stable than those of silicon – their energy (–14.56 eV) is higher than the energy of the 3s electrons in the Si atom (–14.89 eV), even if only by a small amount. Compare this to germanium and lead, whose valence s orbitals are both more stable than the silicon’s 3s (–16.05 eV and –15.12 eV, respectively). How is it, then, that tin forms divalent compounds while silicon doesn’t? Is Pyykkö correct in associating the Sn valency with 5s stabilization?

  • 3
    $\begingroup$ An excellent question, you poked my interest. Indeed, tin's outer shell electrons are less bound than that of silicon. However, energy of formation of tin-element bonds is generally lower than for silicon. My guess is that this is because of d-subshell interference: its interference lengthens covalent bonds, preventing effective orbital overlap, thus tin compounds are shifted towards ionic spectrum, while silicon compounds are mostly covalent. $\endgroup$
    – permeakra
    Jan 1, 2016 at 17:02
  • $\begingroup$ Indeed, it seems that the bond length is the reason, but I don’t think it is related to the d-subshell (see my answer below). $\endgroup$
    – Corundum
    Jan 2, 2016 at 11:16
  • $\begingroup$ Drago wrote a paper that discussed this exact issue: J. Phys. Chem. 1958, 62 (3), 353–357 The conclusion is of course the same as what you have written. It arises from a combination of the stabilisation of the s subshell, and the difficulty of forming covalent bonds. $\endgroup$ Mar 27, 2017 at 22:36

1 Answer 1


I feel very ashamed to answer my own question, but having found a possible answer, I don’t see why I shouldn’t share it with the community.

I would like to start by bringing in this table of orbital energies (from the appendix I linked to in the question) as a reference point for further explanations:

$$\begin{array}{ccc} \hline \text{Element} & \text{Energy of outer s orbital / eV} & \text{Energy of outer p orbital / eV} \\ \hline \text{Carbon} & -19.43 & -10.66 \\ \text{Silicon} & -14.89 & -7.78 \\ \text{Germanium} & \mathbf{-16.05} & -7.54 \\ \text{Tin} & -14.56 & -7.01 \\ \text{Lead} & \mathbf{-15.12} & -6.81 \\ \hline \end{array}$$

As you can see, the energies of the p orbitals get higher and higher as you go down the group, while the energies of the s orbitals don't.

The reason why, in general, orbitals rise in energy when going down a group is that even though the nuclear charge is bigger (which has a stronger effect than the rise in the quantum number $n$), the increased screening effect of added electrons prevails over this rise.

Germanium has such a low 4s electron energy because of the so-called d-block contraction. Since it has 10 added protons due to the existence of the d-block, and the d electrons don’t shield this increased nuclear charge well, the s electrons are stabilised. The p electrons don’t feel this effect as much because they are shielded more than the s electrons (less penetration), thus experiencing the rise in the nuclear charge less, while being shielded additionally by the stabilised s electrons (and presumably by the d electrons as well, to a lesser extent).

The lowered energy of the 6s orbital of lead is the result of the equivalent f-block contraction, amplified by the relativistic contraction of the s shell. The p electrons don’t have a noticeable change for the same reason they didn’t in germanium.

Now, tin has d electrons too, so it will experience the d-block contraction – but to a far more limited extent, since the change in nuclear charge is not as big in percentage. To illustrate this concept, it is useful to bring in this formula:

$$E \approx -13.6 \left(\frac{Z}{n}\right)^2 \mathrm{~eV}$$

where $E$ is the energy of the electron (in the hydrogen-like atom!), $Z$ the nuclear charge and $n$ the principal quantum number. Since it refers to hydrogen-like atoms, this formula obviously doesn’t bring in the screening effect. To see why the d-block contraction doesn’t have such a strong influence on the tin atom, we can use this formula to calculate how the non-screened, non-relativistic energy grows from Si to Sn:

$$\begin{array}{cc} \hline \text{Element} & \text{Non-screened energy of outer orbital / eV} \\ \hline \ce{Si} & -296.178 \\ \ce{Ge} & -870.4 \\ \ce{Sn} & -1360 \\ \hline \end{array}$$

As you can see, the percentual change from Si to Ge is much higher than the change from Ge to Sn – thus, the increase in nuclear charge from Ge to Sn doesn’t have an enormous effect on the electron energy, while the screening effect between Ge and Sn grows steadily: the orbital energy is lowered in the many-electron atom. Hence, the Sn s shell is stabilised, but to a lesser extent.

Now, why does Sn form divalent compounds so easily, when Ge and Si have lower s energies and don’t? There’s a passage in Greenwood’s Chemistry of the Elements (2nd ed), pp 226-7 that explains it:

The term “inert-pair effect is somewhat misleading since it implies that the energy required to involve $n\mathrm{s^2}$ electrons in bonding increases in the sequence $\ce{Al} < \ce{Ga} < \ce{In} < \ce{Tl}$. […] This is not so […]. The explanation lies rather in the decrease in bond energy with increase in size from Al to Tl so that the energy required to involve the s electrons in bonding is not compensated by the energy released in forming the 2 additional bonds.

Indeed, that is what happens: the $\ce{Sn-X}$ bonds are longer than the $\ce{Si-X}$ bonds, and thus have smaller energy, as permeakra pointed out in his/her comment. However, I believe this is not due to the interference of the d-subshell, but simply to the larger Sn nuclear charge. This increased nuclear charge amplifies the repulsion to the nuclei of the other species (say, oxygen), thus lengthening the bond. Since the s electrons have a very low energy compared to the p ones, they do not participate in this bond – they cannot overlap.

In short, tin forms divalent compounds because:

  1. Its s subshell is more stable than the p subshell, because of screening differences;
  2. Its bonds are longer, due to the increased nuclear charge: thus, the s electrons don’t bond.

The combined effect is smaller in germanium compounds (Ge is more tetravalent), because even though the s shell is more stabilised, the bonds are shorter; it is stronger in lead compounds (Pb is more divalent) because the 6s shell is more stabilised than the Sn 5s shell and the bonds are even longer.


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