Question

Which of the following has a square planar structure?
A. $[Ni{(CO)_4}]$
B. ${[Ni{(Cl)_4}]^ - }$
C. ${[Ni{(CN)_4}]^{2 - }}$
D. All of the above

Answer 1

To solve this question, first we need to understand what a square planar structure is? We can define square planar as a molecular shape that results in the formation of four different bonds and two lone pairs on the central atom of a molecule.

Complete step by step answer:
As we know that in a square planar molecular structure, the central atom is surrounded by its constituent atoms. These atoms are further used to form the corners of a square on the same plane.
Here, in the case of ${[Ni{(CN)_4}]^{2 - }}$, $Ni$ is in the $+ 2$ oxidation state which means that $Ni$ has ${d^8}$ configuration. Also, we can observe that there are $4C{N^ - }$ ions in ${[Ni{(CN)_4}]^{2 - }}$. So with the help of that, it is clear that it can either have a tetrahedral geometry or square planar geometric structure.
As we are aware that the $C{N^ - }$ ion is a strong field ligand in comparison to others due to the pairing of unpaired $3d$ electrons. Therefore, it is evident that ${[Ni{(CN)_4}]^{2 - }}$ undergoes $ds{p^2}$ hybridization (option C). As all the electrons are paired in ${[Ni{(CN)_4}]^{2 - }}$. Thus, it is diamagnetic in nature.

So, the correct answer is Option C .

Note:
For answering the above question that $ds{p^2}$ orbital hybridization is a square planar we need to know that there are inner orbital complexes in which the electrons get paired up due to the presence of a strong field ligand. Thus, the electron pairs of those ligands will occupy one d-orbital, one s-orbital and then 2-p orbitals. Therefore, it gives us a square planar structure.