Among Ni(CO)_4,[Ni(CN)_4]^{2-},and [NiCl_4]^{ | vedclass

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Question

(C) $1$. In $Ni(CO)_4$,$Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing of electrons. It is diamagnetic.$2$

Vedclass · Accepted Answer

$Ni(CO)_4$ and $[Ni(CN)_4]^{2-}$ are diamagnetic and $[NiCl_4]^{2-}$ is paramagnetic.

Vedclass · Answer

(C) $1$. In $Ni(CO)_4$,$Ni$ is in $0$ oxidation state $(3d^8 4s^2)$. $CO$ is a strong field ligand,causing pairing of electrons. It is diamagnetic.$2$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state $(3d^8)$. $CN^-$ is a strong field ligand,causing pairing of electrons. It is diamagnetic.$3$. In $[NiCl_4]^{2-}$,$Ni$ is in $+2$ oxidation state $(3d^8)$. $Cl^-$ is a weak field ligand,so no pairing occurs. It has two unpaired electrons and is paramagnetic.

Among $Ni(CO)_4$,$[Ni(CN)_4]^{2-}$,and $[NiCl_4]^{2-}$,which of the following statements is correct regarding their magnetic properties?

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