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I've recently come across a source where it stated that all 4d and 5d metals form low spin complexes irrespective of the strength of ligand.

I'm not sure of the authenticity of the source and couldn't figure out the reason for this phenomenon, thereby enquiring about it.

I feel that the splitting energy is higher than pairing energy in this case, but I am not able to justify it properly.

Ritwik Das
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2 Answers2

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While the vast majority of the compounds and complexes of the heavier transition metals are low spin (where that concept has any meaning) $\ce{PdF2}$ is one exception, which is unusual not only this way but also having $\ce{Pd^2+}$ in an octahedral environment (see e.g. Wikipedia). Thus your source is not quite correct.

As for the reason why 2nd and 3rd row transition metals are more likely to form low spin complexes than the lighter elements, the reason is given in the answer linked above in the comments. In fact, while the question may be different, the answer is almost a duplicate.

andselisk
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Ian Bush
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You can simply remember that CFSE of 4d and 5d series is far more than that of 3d series. Therefore, However strong the ligand, The pairing energy will always be lesser than the CFSE. So mostly, all the complexes of these two series are inner orbital complexes. And yes the reason for high CFSE is diffused state of 4d and 5d orbitals. Order of CFSE: 5d>4d>3d.

Suchetan
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