The sulfur atom resembles the oxygen atom: they both have an s²p⁴ valence shell. Sulfur atom has 8 more electrons because its n=2 shell is filled. It's valence electrons are in the n=3 shell, 3s²3p⁴. We can use the same representations for S as we used for O:

21.1

All we've changed in the 2D diagram is the color (which is completely arbitrary). The 3D diagram is the same as the one for O.

Since S is similar to O and has two singly occupied 3p orbitals, it also has a nominal valence of 2, answer (b). We expect to be able to form SH and H₂S.

Analogous to O, S has two singly occupied 3p orbitals that are 90° apart, so the nominal bond angle of H₂S is also 90°, answer (d). The bond diagrams and calculated structures for SH and H₂S are shown below. The structures of OH and H₂O are included for comparison.

21.2

We can make a couple basic observations. The SH bond lengths in both compounds are more than 0.3 Å longer than the OH bond lengths, which is a consequence of the 8 extra electrons in the core of S. The bonds are also decidedly weaker, by 20-25%. The bond angle of H₂S is very close to the nominal bond angle of 90°.

In Section 19 we argued that the bond angle of H₂O opens up by more than 10° from the nominal bond angle mostly due to repulsion between the bond pairs. The bond angle of H₂O opens up very little from the nominal bond angle, which implies that there is less repulsion between the bond pairs of H₂S than H₂O. Let's compare the bonding pair orbitals of SH vs OH and then H₂S vs H₂O to see if this logic is supported by the facts. We'll start with the diatomics and look at the orbitals at the minimum separation:

21.3

Note where the H contours cross the bond axis in the two compounds. They almost reach the O nucleus but are much closer to the midpoint of the bond in SH. There is therefore less polarization of the H 1s orbital in SH than in OH. Let's see if this is also true in H₂O and H₂S:

21.4

Here we are looking at just one set of the bond orbital for H₂O and H₂S, on the left side. The polarization is very similar in H₂O and H₂S as in their corresponding diatomic. The more each H 1s orbital is polarized toward O or S, the more it will tend to overlap the other H 1s orbital, which results in greater repulsion. We can see the interference even better if we look at both orbitals superimposed together:

21.5

We can now see quite clearly that the overlap is much worse in H₂O than it is in H₂S. Part of it is because the polarization toward O is more pronounced than the polarization toward S, but we can also see that the longer bond lengths of H₂S mean that the H atoms are farther apart in H₂S than in H₂O and thus the overlap is reduced. The orbitals demonstrate quite well why the nominal bond angle prediction of 90° is so much closer to the real bond angle for H₂S than it is for H₂O.

In the next section, we will look at adding H to the Be atom.