Question

Calculate the difference in magnetic moment of complexes $\left[Ni\right(CN{)}_4{]}^{2-}$ and $\left[Ni\right(Cl{)}_4{]}^{2-}$.

Answer 1

In both complexes, $Ni$ is as $N{i}^{2+}$ that is $\left[Ar\right]3d^8$.

In $\left[Ni\right(CN{)}_4{]}^{2-}$, the ligand is $C{N}^{-}$ which is a strong field ligand and hence pairing of electrons occur resulting in a square planar complex. Unpaired electrons are 0.

In $[NiC{l}_4{]}^{2-}$, the ligand is $C{l}^{-}$ which is a weak field ligand and hence pairing of electrons does not occur, resulting in tetrahedral complex. Unpaired electrons are 2.


Magnetic moment = $\sqrt{n(n+2)}$

For $[NiC{l}_4{]}^{2-}$, n = 2
$\text{magnetic moment}=\sqrt{2(2+2)}=2.828B.M.$
For $\left[Ni\right(CN{)}_4{]}^{2-}$, n = 0
magnetic moment = 0
Difference in magnetic moment = 2.828 B.M.