Question

$[NiC{l}_4{]}^{2-}$ is paramagnetic while $\left[Ni\right(CO{)}_4]$ is diamagnetic though both are tetrahedral. Why?

Answer 1

Magnetic nature

If the coordination compound have unpaired electrons in their d-orbitals then that will be paramagnetic in nature and if compound have zero unpaired electrons then that will be diamagnetic in nature.

Effect of ligand on electronic configuration

The difference in the magnetic behaviour is due to the nature of ligands present. In case $[NiC{l}_4{]}^{2-}$ the oxidation state of nickel is $+2$ and also $C{l}^{-}$ is a weak ligand.

Thus, its configuration becomes -


Here, $[NiC{l}_4{]}^{2-}$ have $s{p}^3$ hybridization and have a tetrahedral geometry and also it have $2$ unpaired electrons so the nature is paramagnetic.

In the case of $\left[Ni\right(CO{)}_4],$ the oxidation state of nickel is $0.$ So, its configuration is $3d^{8}4s^2.$ We also know that $CO$ is a strong ligand, thus the configuration of nickel becomes -


Here $\left[Ni\right(CO{)}_4]$ also has $s{p}^3$ hybridization and tetrahedral geometry but it has zero unpaired electrons. So its nature is diamagnetic.