Answers on the internet suggest that tetracyanonickelate ($\ce{([Ni(CN)4]^{2-}}$) with square planar structure is diamagnetic because $\ce{CN-}$ is a strong field ligand and causes pairing, but my book says that VBT does not distinguish between strong and weak field ligands. So can anyone explain the logic using VBT?


According to VBT, for $\ce{[Ni(CN)_4]^{2-}}$, because it is a square planar geometry, it will experience $\ce{dsp^2}$ hybridization. The following process and result will be the following:

enter image description here

The diagram clearly shows that using VBT, we get that $\ce{[Ni(CN)_4]^{2-}}$ is diamagnetic. However, in general, VBT is not used often to describe these bonds, and more and more complex theories of bonding have been developed, although all involve approximations/assumptions. One of them is crystal field theory which is what the internet answers were suggesting.

In terms of crystal field theory. $\ce{Ni^{2+}}$ has a $\ce{d^8}$ configuration, and the crystal field splitting diagram is square planar. Since $\ce{CN-}$ creates large splitting, only four orbitals will be filled in the following diagram:

enter image description here

We will find that only $\ce{d_{yz}}$, $\ce{d_{xz}}$, $\ce{d_{z^2}}$, and $\ce{d_{xy}}$ will be filled as $\ce{d_{x^2-y^2}}$ is too high in energy. Since there are 8 electrons, all of the 4 orbitals will be fully filled and this makes it diagmagnetic according to crystal field theory.

  • $\begingroup$ please see chemistry.stackexchange.com/questions/76726/… $\endgroup$ – Mithoron May 5 at 15:55
  • $\begingroup$ It is dsp2 because that is the hybridization of square planar geometries according to VBT. If it is tetrahedral in the case of $\ce{[Nicl_4]^{2-}}$, it will undergo sp3 hybridization. $\endgroup$ – M.L May 5 at 15:58
  • $\begingroup$ @Mithoron I am not sure if you saw my comment but see this chemistry.stackexchange.com/questions/58012/… $\endgroup$ – M.L May 5 at 22:57
  • $\begingroup$ And you saw Martin's comment there? VBT isn't what you think - it's part of class of computational methods - about as good approach as MOT and the calculations say there's almost always only negligible mixing of p and d orbitals - no actual hybridisation of d and p orbital together. $\endgroup$ – Mithoron May 5 at 23:11
  • $\begingroup$ @Mithoron I know, that is what my second paragraph explains that VBT is not generally used to describe these bonds and so on. I know there are other theories like MOT, CFT, and Ligand Bond Theory (Which I posed a question about that hasn't been answered). It's just the question asked about how would VBT be used to model the bonding in $\ce{Ni(CN)_4}^{2-}$ $\endgroup$ – M.L May 5 at 23:18

@M.L has already provided the solution as per VBT, but that answer seems to have some flaws when it comes to CFT (looks like he has corrected the mistakes now). So, I would like to take some time to answer this question using CFT.

According to crystal field theory, the degeneracy of the orbitals of a same orbital is lost in presence of a ligand and the pattern in which they are distorted depends upon the nature of the ligand. The spectrochemical series helps us understand the nature of a ligand.

In general there are two types of ligands, strong field and weak field. Strong field ligands have a greater power to distort the degeneracy of the sub-orbitals whereas weak field have a weaker power of distortion.

As per the spectrochemical series, $\ce{CN-}$ is a strong field ligand and thus creates a greater distortion in the degeneracy of $\mathrm d$ orbitals.

The central metal ion here is the $\ce{Ni^2+}$ with its $\mathrm d^8$ electronic configuration.

A $\mathrm d^8$ system prefers a square planar approach of ligands if the ligands are strong field. In such a condition the the degeneracy is disrupted and the energy diagram of the $\mathrm d$ orbitals is as below:

Square planar approach

Now when trying to fill this diagram with electrons we have to take into consideration that an electron in the $\mathrm d_{x^2-y^2}$ orbital can pair up with the electron in $\mathrm d_{xy}$ in case of a strong field ligand. This is because the extent of splitting done by a strong field ligand is so high that it can overcome the energy gained due to pairing of electrons.

Thus the energy diagram filled with electrons looks as following:


Now the ligands donate their lone pairs in the $3\mathrm d_{x^2-y^2}$, $4\mathrm s$, $4\mathrm p_x$ and $4\mathrm p_y$ sub-orbitals.

In reality, no hybridization takes place but if forced to say the hybridization of this complex, we can say $\mathrm{dsp}^2$ because of the orbitals involved in the $\sigma$ bonds between the metal and ligands.

As you can see the energy diagram does not contain any unpaired electron and thus the complex is diamagnetic.


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