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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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