I've made a platinum complex but the NMR spectra gave a complicated peaks made of different oligomers. I suspect that I might have got Pt(IV) complex instead of Pt(II). Is there any experiment that I can do to confirm the oxidation state?
I think Hard X-ray Absorption Fine Structure Spectroscopy (light energy range 5 - 20 keV, wavelength range 0.25 - 0.06 nm) would be a good choice
Oxidation states can be measured with all methods including photo electrons: photoelectron absorption or emission spectrocopy, ESCA etc.
Alternatively, if the compound is soluble, you might check the electrochemistry. Pt(II) complexes should give an oxidation in the range of 0 to +1.5 V, Pt(IV) of course not. Pt(IV) complexes in turn shuould be reduced easily.
Furthermore, you can add MeI and see what happens. In case of Pt(II) you will get an oxidative addition, you will see this from the colour change.
Hi,
If you know the geometry you can get it. Pt(IV) is octahedral, whereas Pt(II) is square planar.
You can use 195Pt NMR if you have access to it and someone with experience doing it (it has a large spectral window).
FYI oxidations from Pt(II) to Pt(IV) usually require a decent oxidant. Oxygen only works for very electron-rich Pt(II) complexes (ex: Pt(bpy)Me2, but not Pt(bpy)Cl2). If you put Pt(II) in, it's probably still Pt(II) unless there was a good oxidant (eg. Cl2, R-I, H2O2) in there.
If your compound has a Pt-P bond, you can see the coupling constant in the 13P NMR. Generally if it is less than 2000, it is Pt(IV), but if it is more than 2000 it is Pt(II). It depends on the ligands, though.
-Kyle
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