Question

Out of $Ni(CO{)}_4,Ni(CN{)}_4^{2-}$ and $NiC{l}_4^{2-}$ select paramagnetic and diamagnetic.

• A. $Ni(CO{)}_4$ and $Ni(CN{)}_4^{2-}$ are diamagnetic
• B. $NiC{l}_4^{2-}$ is paramagnetic
• C. Both (A) and (B) Correct
• D. None of these

Answer 1

The correct option is C Both (A) and (B) Correct

${Ni\left(CO\right)}_4⟶Ni+4CO$
* The valence shell electronic configuration of ground state $Ni$ atom is $3d^{8}4s^2$.
* All of these $10$ electrons are pushed into $3d$ orbitals and get paired up when strong field $CO$ ligands approach $Ni$ atom. The empty $4s$ and three $4p$ orbitals undergo $s{p}^3$ hybridization and form bonds with $CO$ ligands to give $Ni(CO{)}_4$. Thus ${Ni\left(CO\right)}_4$ is diamagnetic.
${{Ni\left(CN\right)}^{2-}}_4⟶N{i}^{2+}+4C{N}^{-}$
* In ${{Ni\left(CN\right)}^{2-}}_4$, there is $N{i}^{2+}$ ion for which the electronic configuration in the valence shell is $3d^{8}4s^0$.
* In presence of strong field $C{N}^{-}$ ions, all the electrons are paired up. The empty 4d, 3s and two 4p orbitals undergo $ds{p}^2$ hybridization to make bonds with $C{N}^{-}$ ligands in square planar geometry. Thus ${{Ni\left(CN\right)}^{2-}}_4$ is diamagnetic.

${{NiCl}^{2-}}_4⟶N{i}^{2+}+4C{l}^{-}$
* Again in ${{NiCl}^{2-}}_4$, there is $N{i}^{2+}$ ion, However, in presence of weak field $C{l}^{-}$ ligands, $NO$ pairing of d-electrons occurs. Therefore, $N{i}^{2+}$ undergoes $s{p}^3$ hybridization to make bonds with $C{l}^{-}$ ligands in tetrahedral geometry. As there are unpaired electrons in the d-orbitals, ${{NiCl}^{2-}}_4$ is paramagnetic.