Question

Among $Ni(CO{)}_4,[Ni(CN{)}_4{]}^{2-}$ and $NiC{l}_4^{2-}$:

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

Answer 1

The correct option is A $Ni(CO{)}_4$ and $\left[Ni\right(CN{)}_4{]}^{2-}$ are diamagnetic and $NiC{l}_4^{2-}$ is paramagnetic.

Electronic configuration:
$Ni-\left[Ar\right]3d^{8}4s^2$
$N{i}^{2+}-\left[Ar\right]3d^8$
When the ligand is strong field ligand it will pair all the unpaired electrons and hence diamagnetic.
When the ligand is weak field ligand it won't pair the unpaired electrons and hence paramagnetic.
$Ni(CO{)}_4$ and $\left[Ni\right(CN{)}_4{]}^{2-}$ is diamagnetic and $NiC{l}_4^{2-}$ is paramagnetic.