I know how to predict magnetic nature when the compound contains only strong field or only weak field ligands. But in $\ce{[Co(NH3)4Cl2]Cl},$ $\ce{NH3}$ is a strong ligand whereas $\ce{Cl-}$ is a weak ligand.

So, how do we know if the ligands are causing the electrons to pair up or not? I understand VBT and CFT to predict the hybridisation and magnetic nature of compounds.

  • 1
    $\begingroup$ It is low spin. Nearly all d6 is low spin. See chemistry.stackexchange.com/questions/112675/… which is essentially a duplicate of this $\endgroup$
    – Ian Bush
    Sep 15, 2019 at 11:44
  • $\begingroup$ @IanBush Quite the opposite. Iron(II) is $\mathrm d^6$ but most of iron(II) is high spin. $\endgroup$
    – Jan
    Oct 17, 2019 at 8:16
  • $\begingroup$ Agreed Fe2+ is by far the most likely of d6 species to be high spin, but everything else is low spin with very, very few exceptions. And the question is about Cobalt. $\endgroup$
    – Ian Bush
    Oct 17, 2019 at 9:10

1 Answer 1


The ammine ligand is not really a strong-field ligand. It is not a strong π acceptor which is characteristic of strong-field ligands such as the nitrosyl cation or carbonyl. Instead, it is a weak π donor. Quite unlike other strong-field ligands such as cyanide, there are a great number of high-spin complexes with ammine ligands. Ammine (and aqua) ligands are better termed intermediate field. They can stabilise low-spin complexes but they typically also support high-spin environments.

Understanding when ligands cause the electrons to pair is very non-trivial but essentially guesswork. This is exemplified by some complexes being isolated in both the high-spin and low-spin states as is the case in $\ce{[Fe(ptz)6]^2+}$ ($\ce{ptz}$ being 1-propyl-1H-tetrazol-κN4). Furthermore, the metal has even more influence on the question with higher oxidation states typically stabilising low-spin configurations while lower oxidation states often favour high-spin complexes. The only clear cases are 4d and 5d transition metals which can be safely assumed low-spin unless stated otherwise.

With that said, cobalt(III) complexes feature a combination of a high (-ish) oxidation state of the central metal combined with a $\mathrm d^6$ environment which generally favours low-spin a bit more as it allows for maximum electronic stabilisation. Therefore, cobalt(III) complexes can be guessed as low-spin if they are not entirely made up of weak-field ligands. Ultimately, however, the experiment or calculation will have to answer.


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