Question

Among $Ni{\left(CO\right)}_4$ and ${\left[Ni{\left(CN\right)}_4\right]}^{2-}$ and $Ni{Cl}_4^{2-}$?

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

Answer 1

Answer: (A)

$Ni{\left(CO\right)}_4$ and ${\left[Ni{\left(CN\right)}_4\right]}^{2-}$ are diamagnetic and $Ni{Cl}_4^{2-}$ is paramagnetic

$Ni(CO{)}_4$

$Ni\Rightarrow 4s^{2}3d^8$

$Ni(CO{)}_4$

(Refer to Image 1)

As $CO$ is strong ligand, all $e^{-}s$ are paired 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 trong ligand. Thus, diamagnetic.

$[NiC{l}_4{]}^{2-}$

$N{i}^{2+}⟶4s^{0}3d^8$

(Refer to Image 3)

Two unpaired $e^{-}s$ present as $C{l}^{-}$ is weak ligand. Hence, it is paramagnetic.