Question

The complex showing a spin only magnetic moment of 2.82 BM is:

• A. $\left[Ni\right(CO{)}_4]$
• B. $[NiC{l}_4{]}^{2-}$
• C. $\left[Ni\right(PP{h}_3{)}_4]$
• D. $\left[Ni\right(CN{)}_4{]}^{2-}$

Answer 1

The correct option is B $[NiC{l}_4{]}^{2-}$

The $3d$ electrons in $Ni$ atom remain unpaired in the presence of weak field ligand.
Here $C{l}^{-}$ is weak field ligand whereas others are strong field ligand which can pair up the electrons in $3d$ orbital.
In $[NiC{l}_4{]}^{2-}$, $Ni$ is in +2 oxidation state with configuration, $\left[Ar\right]3d^8$.
$[NiC{l}_4{]}^{2-}$ is paramagnetic having two unpaired electrons.

${\mu }_{\text{spin}}=\sqrt{n(n+2)}=\sqrt{2(2+2)}$
$=2\sqrt{2}=2\times 1.414=2.82BM$