# Why does the binding of DMSO to metal via the S atom (k-S) strengthens the S=O bond?

So, DMSO (dimethyl sulfoxide) can form metal complexes in both $$\mathrm{\kappa O}$$ and $$\mathrm{\kappa S}$$ 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 S and hard metal ions bind to the hard centre O.

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

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

But how is this possible? The sulfur is already electron deficient due to the strongly electronegative O, and should be a $$\ce{\pi}$$-acceptor, and accept electrons from metal into the $$\ce{\pi^*}$$ orbitals. This means that the 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 O is mainly pi donor and the 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 S=O bond in S bonded complexes.]

The pictures of HOMO and LUMO of DMSO: