7
$\begingroup$

So, DMSO (dimethyl sulfoxide) can form metal complexes in both $\ce{\kappa-O}$ and $\ce{\kappa-S}$ mode i.e. binding with the oxygen or the sulfur respectively. The general explanation given is that softer metal ions bind to the soft centre $\ce{S}$ and hard metal ions bind to the hard centre $\ce{O}$.

From the IR data it can be inferred that the binding of DMSO to a metal ion via $\ce{O}$ causes the $\ce{S=O}$ bond stretching frequency to go down. For example the $\ce{S=O}$ stretch in free DMSO (liquid) is $\pu{1055 cm-1}$, whereas the stretch in $\ce{[Co(dmso)6]+}$ appears at $\pu{950 cm-1}$. This makes sense to me because the oxygen can only act as $\pi$ donor (apart from the usual $\sigma$ donation), and donate electrons to the metal $\mathrm{t_{2g}}$ from bonding $\pi$ orbitals, which will weaken the bond between $\ce{S}$ and $\ce{O}$.

However, when DMSO binds to a metal ion via $\ce{S}$, the $\ce{S=O}$ bond stretching frequency goes up. For example, in $\ce{PdCl2(dmso)2}$ shows the $\ce{S=O}$ stretch at $\pu{1116 cm-1}$.

But how is this possible? The sulfur is already electron deficient due to the strongly electronegative $\ce{O}$, and should be a $\pi$-acceptor, and accept electrons from metal into the $\pi^*$ orbitals. This means that the $\ce{S=O}$ bond should also be weakened in this case. But this does not happen. Why? Does it have something to do with the molecular orbitals of DMSO?

[Note: According to this post, the $\ce{O}$ is mainly $\pi$-donor and the $\ce{S}$ $\pi$-acceptor for DMSO, and I agree with the argument in that answer. But it does not say anything about the strengthening of the $\ce{S=O}$ bond in $\ce{S}$ bonded complexes.]

The pictures of HOMO and LUMO of DMSO respectively:

DMSO HOMOHOMO and LUMO of DMSO

$\endgroup$
5
  • 2
    $\begingroup$ Just guessing here, but I think you're confusing yourself by thinking of the S-O bond as having a lot of pi character. It's essentially just a sigma bond strengthened by the ionic interaction of S+ and O-. That's why the geometry of DMSO is more like trigonal pyramid than trigonal planar. So in kS, as long as S donates more density to the metal via a sigma interaction than it gets back in a pi interaction, it becomes more charged and the ionic component of the S-O bond strengthens. With kO binding, the charge on O decreases, weakening the ionic component of the S-O bond. $\endgroup$
    – Andrew
    May 16, 2021 at 20:22
  • $\begingroup$ @Andrew I am not sure that there can be full +1 and -1 charges on S and O though. The electronegativity of S is 2.58 which is nearly around C (2.55). If C=O bonds have a lot of pi-character then S=O bond can too. Is it possible to estimate how much of S=O bond is pi and how much is ionic? $\endgroup$
    – S R Maiti
    May 16, 2021 at 21:20
  • $\begingroup$ -1 For seemingly so well researched question, to think of S-O bond as double, not dative? Seriously? $\endgroup$
    – Mithoron
    May 16, 2021 at 21:41
  • 2
    $\begingroup$ The C=O bond is not a dative bond like the S->O, where the S still has a lone pair on it after forming the S-O sigma bond. That's why acetone is planar, but DMSO is pyramidal. Also look at the HOMO, which clearly has a nonbonding lone pair on S. I'm sure the charges aren't fully -1 and +1, but certainly the O is negative and the S is positive because of the dative nature of the bond. $\endgroup$
    – Andrew
    May 16, 2021 at 21:42
  • $\begingroup$ @Andrew Ok, that makes sense I think. Would you be willing to make it into an answer? If you do I will accept it. $\endgroup$
    – S R Maiti
    May 17, 2021 at 7:38

1 Answer 1

2
+100
$\begingroup$

The confusion seems to arise from the incorrect statement that the S-O bond has pi character and that there exist $\pi$ and $\pi^*$ molecular orbitals in DMSO. Instead, DMSO (despite the way it is often drawn) is most accurately depicted as having three single bonds and one lone pair on the the S atom (clearly visible as a dominant element in the HOMO shown in the question). The trigonal pyramidal geometry (also shown in the question) is consistent with this description.

The S-O bond is a dative bond, meaning that the S atom provides both electrons rather than having one electron come from each atom. That results in a formal positive charge on S and negative charge on the O. The O therefore is best represented as having one bond (to S) and three lone pairs, similar to the oxygen atom in an alkoxide ion. Because of the charge separation, the S-O bond also has a strong ionic component.

[Aside: so-called hypervalent sulfur compounds are typically drawn with double bonds that are actually combined sigma + ionic bonds rather than sigma + pi bonds. DMSO is not the only one. Keep this in mind in the future when trying to understand properties of sulfur-containing molecules.]

We now consider what happens when DMSO interacts with a transition metal. Similar to the O of an alkoxide ion, the O atom in DMSO can interact with metals as both a $\sigma$ and $\pi$ donor because of the multiple lone pairs. Both the $\sigma$ and $\pi$ donations result in decreased electron density on O, reducing its charge and weakening the S+/O- ionic interaction. The weaker ionic interaction means that the bond overall weakens, consistent with the decrease in the observed stretching frequency. There is no need for an S-O $\pi$ interaction to exist in order for this to happen.

If instead the S atom interacts with the metal, it does so primarily as a $\sigma$ donor via its one lone pair. There may be capacity for a $\pi$ acceptor interaction with $\sigma^*$ orbitals, as has been proposed for ligands such as triphenylphosphine, but if this occurs, it is with the $\sigma^*$ orbital associated with the S-C bonds, not the S-O bond, consistent with the LUMO shown in the question, which has a large contribution from the py orbital on S (ie the p orbital in the plane of the C atoms). Thus there is no weakening of the S-O $\sigma$ interaction due to $\pi$ acceptance.

As long as the $\sigma$ donation is stronger than the $\pi$ acceptance (which is what we expect), there is a net loss of electron density from S, which further increases the positive charge on the already electron deficient S atom. Increased charge means an increased strength in the ionic S+/O- interaction, so the S-O bond overall is stronger, consistent with the observed increase in the S-O stretching frequency.

$\endgroup$

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.