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

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(C) $1$. In $Ni(CO)_4$,$Ni$ is in $0$ oxidation state. The configuration is $3d^8 4s^2$. Due to the strong field ligand $CO$,electrons pair up,resulti

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$Ni(CO)_4$ 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. The configuration is $3d^8 4s^2$. Due to the strong field ligand $CO$,electrons pair up,resulting in $sp^3$ hybridization and diamagnetic behavior.$2$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state $(3d^8)$. $CN^-$ is a strong field ligand,causing pairing of electrons in $3d$ orbitals. It undergoes $dsp^2$ hybridization and 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 undergoes $sp^3$ hybridization with two unpaired electrons,making it 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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