# Magnetic behaviour of complex compounds

Which one of the following cyanide complexes would exhibit the lowest value of paramagnetic behaviour?

1. $\ce{[Cr(CN)6]^{3-}}$
2. $\ce{[Mn(CN)6]^{3-}}$
3. $\ce{[Fe(CN)6]^{3-}}$
4. $\ce{[Co(CN)6]^{3-}}$

Here I determined central atom configuration and it is $\mathrm{3d^3}, \mathrm{3d^4}, \mathrm{3d^5}, \mathrm{3d^6}$ for each option in order. $\ce{CN-}$ is a strong-field ligand so electron will get paired hence option 2 and 4 will have least magnetic behaviour

But only option 4 is correct. Why?

Magnetic behaviour of complex compounds are determined on the basis of magnetic moment, rather the extent of paramagnetism is measured in terms of the magnetic moment μ.

μ = √n(n+2) BM , where n notates the number of unpaired electrons.

First lets give a look towards the number of unpaired electron each of the central metal atom has:

Here the question is asking to determine the complex compound which shows the lowest paramagnetic behaviour. As Co^3+ has no unpaired electron, its μ = 0, hence it is diamagnetic, and hence shows shows lowest paramagnetic behaviour among the above complexes. And best of luck for jee'17. Me too appearing the exam.

Hope it helps

• why does pairing occur only once, in $\ce{Mn^3+}$? @Yb609 – Supernova Mar 22 '17 at 16:15
• Because 2 vacant d orbitals are required by the CN- . Basically hybridisation is just a method to explain the molecular geometry. Don't think that molecular geometries are based on hybridisation. Its to true. Hybridisation is just a theory to explain molecular geometries. It has been found experimentally that [Mn(CN)6]^3- shows octahedral geometry and hence to explain that we use the concept of hybridisation and hence we take it as a dirty rule that for coordination number 6, in presence of strong ligand electrons are paired up so as to create 2 vacant d orbitals. – Yb609 Mar 22 '17 at 16:47
• And hence the above rule is a conclusion of what we have experimentally observed. – Yb609 Mar 22 '17 at 16:47
• Oh, thats why this happens, then what about in complexes like $\ce{[V(H2O)6]}$? in case you know, there was a question in the new section, pls do post it there, Im eager to know that too, thanks @Yb609 – Supernova Mar 22 '17 at 16:50
• @QuarkyLittleThing ya i have also seen that question. This same reason works out there as well. That guy who asked that question has a misconception that the molecular structures are given by hybridisation. Hybridisation is not a phenomenon. It is just a concept to explain. You too know that in chemistry experiments are done first and then we give theories (like hybridisation) to explain that.Similarly in [V(H2O)6]^3- case, experimentally it was found to be octahedral low spin complex and hence to explain that we say the hybridisation is d2sp3. – Yb609 Mar 22 '17 at 17:15

This is because in Co+3 there is no unpaired electrons .i.e all paired up (due to the strong ligand's effect) due to which it will be diamagnetic in nature and hence it will have the least paramagnetic behaviour.