# Distinguishing between different geometries of bromochlorobis(triphenylphosphine)platinum(II) using ¹⁹⁵Pt NMR

### Question

The question off which I am basing my question assumes that $$\ce{[PtClBr(PPh3)2]}$$ can exist in either the square-planar or tetrahedral geometry. If one would use $$\ce{^{195}Pt}$$ NMR spectroscopy, how would the spectra of the possible molecules differ from each other? The effect of the $$\ce{Cl}$$ and $$\ce{Br}$$ can be disregarded.

### My reasoning

For the square planar geometry, either the cis or trans isomer is possible. I am assuming that you won't be able to distinguish between the square planar trans molecule and the tetrahedral molecule since the $$\ce{PPh3}$$ ligands in each molecule are equivalent. The $$\ce{^{195}Pt}$$ NMR spectra for these two orientations will thus be the same, a single triplet.

Will the square planar cis molecule also produce the same spectrum or will there be second-order coupling present since the $$\ce{PPh3}$$ ligands are non-equivalent? How would the $$\ce{^{195}Pt}$$ NMR spectra then look for the cis isomer?