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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?

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