Question

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

$Ni(CO{)}_4$
$Ni\Rightarrow 4s^{2}3d^8$
$Ni(CO{)}_4$

(Refer to Image 1)
As $CO$ is strong ligand, it will paired up all $e^{-}s.$ Thus diamagnetic.

$\left[Ni\right(CN{)}_4{]}^{2-}$
$N{i}^{2+}=4s^{0}3d^8$
$\left[Ni\right(CN{)}_4{]}^{2-}$

(Refer to image 2)
All $e^{-}s$ are paired as $C{N}^{-}$ is a strong field ligand. Thus, it is diamagnetic.

$[NiC{l}_4{]}^{2-}$
$N{i}^{2+}\to 4s^{0}3d^8$

(Refer to Image 3)
Two unpaired $e^{-}s$ present as $C{l}^{-}$ is weak field
ligand. Hence, it is paramagnetic.