# Magnetic nature of tetraamminedichlorocobalt(III) chloride

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.

• It is low spin. Nearly all d6 is low spin. See chemistry.stackexchange.com/questions/112675/… which is essentially a duplicate of this – Ian Bush Sep 15 '19 at 11:44
• @IanBush Quite the opposite. Iron(II) is $\mathrm d^6$ but most of iron(II) is high spin. – Jan Oct 17 '19 at 8:16
• 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. – Ian Bush Oct 17 '19 at 9:10

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.