# Is [Co(NH₃)₄Cl₂]Cl paramagnetic or diamagnetic?

$$\ce{NH3}$$ is known to be a strong field ligand, while $$\ce{Cl}$$ is known to be a weak field ligand.

Is $$\ce{[Co(NH3)4Cl2]Cl}$$ a high spin complex or a low spin complex?

I assumed this to be a high spin complex. Oxidation state of $$\ce{Co}$$ is $$+3$$. And electronic configuration of $$\ce{Co(III)}$$ is $$\mathrm{[Ar]~3d^6}$$.

Since I assumed this to be a high spin complex — pairing of electrons of $$\mathrm{3d}$$ orbitals will not happen. The $$6$$ pairs of electrons from the ligands must be included in the $$\mathrm{4s}$$, $$\mathrm{4p}$$, $$\mathrm{4d}$$ orbitals — which leaves electrons unpaired in the $$\mathrm{3d}$$ orbital, making this a paramagnetic complex, with hybridisation $$\mathrm{sp^3d^2}$$.

Answer given to this question says otherwise. Hints are welcome!

This question is an excerpt of a question that appeared in the JEE(A) - 2016

• Cl acts as strong field ligand with Co – drake01 Apr 13 '19 at 15:36
• @drake01 Can you link a source? Explanation? – McSuperbX1 Apr 13 '19 at 15:38
• I don't have any source rn, our teacher told this. – drake01 Apr 13 '19 at 15:39
• @drake01 Well then so is the case with this compound: chemistry.stackexchange.com/questions/72685/… – McSuperbX1 Apr 13 '19 at 15:42
• rule of thumb: complexes of Co+3 are low-spin, unless we are talking about hexafluorocobaltat(III) anion. Even the aquacation is low-spin. – permeakra Apr 13 '19 at 20:57

The point is not really whether chloride or ammonia is a strong or weak field ligand, the point is $$\ce{Co^3+}$$ is $$\mathrm{d^6}$$, and virtually all "octahedral" $$\mathrm{d^6}$$ complexes are low spin - essentially some complexes of $$\ce{Fe^2+}$$ and a very small number of fluoro complexes of $$\ce{Co^3+}$$ are the only exceptions to the rule that all $$\mathrm{d^6}$$ octahedral complexes are low spin.

Why is this? Well let's use a bit of Crystal Field Theory (much of below stolen from https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_(Inorganic_Chemistry)/Crystal_Field_Theory/Crystal_Field_Stabilization_Energy). One can consider the crystal field stabilization energy which is how stabilized an ion is compared to the free ion when in a given electronic state electronic and geometry. Here we are interested in high and low spin, and octahedral geometry. Taking a table of these values from the above link:

In the above $$\Delta_\circ$$ is the splitting energy of the orbitals. We can see that low spin $$\mathrm{d^6}$$ has the largest possible stabilization energy of any electronic configuration compared to the high spin case ($$2\Delta_\circ$$-P). Thus unless the splitting is very small octahedral $$\mathrm{d^6}$$ prefers low spin. So how do we get small splitting to see high spin $$\mathrm{d^6}$$? Well small splitting is favoured by

• Low charge first row transition metals - for instance $$\ce{Fe^{2+}}$$
• Very weak field ligands

And what were our exceptions above?

CO>CN->Ethylene Diamine>NH3>NCS-............Cl>....... this is the ligand strength order form the NCERT textbook so we have four strong ligand which will cause splitting of orbitals also called octahedral splitting and pairing takes place aginst the hunds rule making this diamagnetic and due to pairing it is low spin

This is not a full answer, but you should note that halogens occupy the lower end of the spectrochemical series, and they are (almost) never able to act as strong field ligands. The magnitude of crystal field splitting energy of $$Co^{+3}$$ is low, such that any ligands barring the halogens do act as strong field ligands. This makes it a low spin complex. Therefore, in this case, $$\ce{NH3}$$ acts as a strong field ligand.