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:


  • 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 '21 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 '21 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 '21 at 21:41
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    $\begingroup$ 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 '21 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 '21 at 7:38

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.


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