Question

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

Answer 1

The hybridization $ds{p^2}$ give square planar structure and hybridization $s{p^3}$ gives tetrahedral structure. Shape of the molecule and hybridization is dependent upon the nature of ligands attached to the metal atom.

Complete Solution:
- The ligand field theory introduces us to two types of ligands. Ligands when come in contact with the free metal atom, the repulsion of electrons between the ligands and metal atom, splits the orbitals in two levels. The two types of ligands are viz, Strong field ligand and Weak field ligand. The spectrochemical series ranks the ligands according to the energy difference ${\Delta _o}$between the ${t_{2g}}$ and ${e_g}$ orbitals in their octahedral complexes. When a ligand produces large splitting energy it is called strong field ligand and ligands that produce small splitting energy are called weak field ligand.
The spectrochemical series is:
${I^ - } < B{r^ - } < C{l^ - } < NO_3^ - < {F^ - } < O{H^ - } < {H_2}O < Pyridine < N{H_3} < NO_2^ - < C{N^ - } < CO$
The electronic configuration of $N{i^{2 + }}$is $[Ar]4{s^0}3{d^8}$.

- When strong field ligand is present with $N{i^{2 + }}$, then the electrons in 3d get paired, and this leaves one subshell in d-orbital empty so one pair electrons of ligands get filled in the left d-orbital, making the hybridization $ds{p^2}$. Therefore, the complex is said to have a square planar structure.
- When weak field ligand is present with $N{i^{2 + }}$, then electrons get filled in the higher orbitals, and then get paired, now no subshell in d-orbital is free to accept electrons from ligands, so they go into the next available orbitals that is s- and p-orbital, making the hybridization $s{p^3}$. The structure of the following compound will be tetrahedral.
- Since in the given options, Metal is same, and option C and A both have strong field ligand, but metal atom in option A do not have +2 oxidation state, so, all the subshells of d-orbital will be filled, therefore, hybridization will be $s{p^3}$. So only option C has a strong field and has hybridization $ds{p^2}$, so the structure will be Square planar.

So, the correct answer is “Option C”.

Note: With strong field ligands, the energy gap is larger than the pairing energy, so electrons get paired in the lower orbital instead of getting into the orbital with higher energy. With weak field ligands, the energy gap is less than that of pairing energy, so electrons first get filled in both the orbitals then, they get paired.