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.

  • $\begingroup$ Please add the source $\endgroup$ – Avnish Kabaj Apr 21 '18 at 13:26
  • $\begingroup$ I'm not able to recall the exact source, but I do remember reading the exact statement. $\endgroup$ – Ritwik Das Apr 21 '18 at 13:29
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    $\begingroup$ I've been taught the same fact, and am interested in an answer to this question. Sadly, I wasn't able to verify the fact from both JD Lee and VK Jaiswaal inorganic :( $\endgroup$ – Gaurang Tandon Apr 21 '18 at 13:41
  • $\begingroup$ @GaurangTandon Even I searched all my books and was unable to find anything regarding this concept. $\endgroup$ – Ritwik Das Apr 21 '18 at 13:45
  • $\begingroup$ Related: Why is [PdCl4]2- square planar whereas [NiCl4]2- is tetrahedral? (read the answer) $\endgroup$ – orthocresol Apr 21 '18 at 14:23

The 4d and 5d orbitals are much more diffused as compared to 3d orbitals due to which their shielding effect is much poorer than that of 3d electrons. As a result, there is an increased attraction between the ligands and the nucleus of the central metal. Due to this, any ligand no matter weak or strong is able to pair up the electrons ( if required ) i.e. the pairing energy becomes less than crystal field splitting energy ... which is the condition of a low spin complex.


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