3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ Related: Why is [PdCl4]2- square planar whereas [NiCl4]2- is tetrahedral? (read the answer) $\endgroup$ – orthocresol Apr 21 '18 at 14:23
2
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

Due to improper shielding by 4d and 5d elements ligand electron get strongly attracted by the nucleus and hence the pairing energy become lower and the complex form low spin complex irrespective of strength of ligand

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.