Amongst 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 $[Ar] 3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing of $4s$ el

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. The configuration is $[Ar] 3d^8 4s^2$. $CO$ is a strong field ligand,causing pairing of $4s$ electrons into $3d$ orbitals,resulting in $sp^3$ hybridization with no unpaired electrons. Thus,it is diamagnetic.$2$. In $[Ni(CN)_4]^{2-}$,$Ni$ is in $+2$ oxidation state. The configuration is $[Ar] 3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons in $3d$ orbitals,resulting in $dsp^2$ hybridization with no unpaired electrons. Thus,it is diamagnetic.$3$. In $[NiCl_4]^{2-}$,$Ni$ is in $+2$ oxidation state. The configuration is $[Ar] 3d^8$. $Cl^-$ is a weak field ligand,so it does not cause pairing of electrons. This results in $sp^3$ hybridization with two unpaired electrons in $3d$ orbitals. Thus,it is paramagnetic.Therefore,$Ni(CO)_4$ and $[Ni(CN)_4]^{2-}$ are diamagnetic,while $[NiCl_4]^{2-}$ is paramagnetic.

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

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