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

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(C) $1$. In $Ni(CO)_4$,the oxidation state of $Ni$ is $0$. The configuration is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairin

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) $1$. In $Ni(CO)_4$,the oxidation state of $Ni$ is $0$. The configuration is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairing of electrons,resulting in $3d^{10}$ configuration. Thus,it is diamagnetic.$2$. In $[Ni(CN)_4]^{2-}$,the oxidation state of $Ni$ is $+2$. The configuration is $[Ar] 3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons in $3d$ orbitals,resulting in $d^8$ configuration with no unpaired electrons. Thus,it is diamagnetic.$3$. In $[NiCl_4]^{2-}$,the oxidation state of $Ni$ is $+2$. The configuration is $[Ar] 3d^8$. $Cl^-$ is a weak field ligand,so it does not cause pairing. There are $2$ unpaired electrons in the $3d$ orbitals. Thus,it is paramagnetic.

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

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