Question

Amongst $Ni(CO{)}_4,[Ni(CN{)}_4{]}^{2-}andNiC{l}_4^{2-}:$

• A. $Ni(CO{)}_{4}andNiC{l}_4^{2-}$ are diamagnetic; and $\left[Ni\right(CN{)}_4{]}^{2-}$ is paramagnetic.
• B. $\left[Ni\right(CN{)}_4{]}^{2-}andNiC{l}_4^{2-}$ are diamagnetic; and $Ni(CO{)}_4$ is paramagnetic.
• C. $Ni\left(CO{)}_{4}and\right[Ni(CN{)}_4{]}^{2-}$ are diamagnetic; and $NiC{l}_4^{2-}$ is paramagnetic.
• D. $Ni(CO{)}_{4}isdiamagnetic;[Ni(CN{)}_4{]}^{2-}$ and $NiC{l}_4^{2-}$ is paramagnetic.

Answer 1

The correct option is C

$Ni\left(CO{)}_{4}and\right[Ni(CN{)}_4{]}^{2-}$ are diamagnetic; and $NiC{l}_4^{2-}$ is paramagnetic.

$Ni(CO{)}_4=Ni+4CO$
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, $Ni(CO{)}_4$ is diamagnetic.



$\left[Ni\right(CN{)}_4{]}^{2-}=N{i}^{2+}+4C{N}^{-}$
In $\left[Ni\right(CN{)}_4{]}^{2-},$ 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 $\left[Ni\right(CN{)}_4{]}^{2^{-}}$ is diamagnetic.



$NiC{l}_4^{2-}=N{i}^{2+}+4C{l}^{-}$
Again in $NiC{l}_4^{2-},$ 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, $NiC{l}_4^{2-}$ is paramagnetic.