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

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(C) $1$. $Ni(CO)_4$: The oxidation number of $Ni$ is $0$. Atomic number $= 28$. $Ni = [Ar] 3d^8 4s^2$. Due to the strong field ligand $CO$,electrons p

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) $1$. $Ni(CO)_4$: The oxidation number of $Ni$ is $0$. Atomic number $= 28$. $Ni = [Ar] 3d^8 4s^2$. Due to the strong field ligand $CO$,electrons pair up,resulting in $sp^3$ hybridization (tetrahedral). There are no unpaired electrons,so it is diamagnetic.$2$. $[Ni(CN)_4]^{2-}$: The oxidation number of $Ni$ is $+2$. $Ni^{2+} = [Ar] 3d^8$. $CN^-$ is a strong field ligand,causing pairing of electrons in $3d$ orbitals,resulting in $dsp^2$ hybridization (square planar). There are no unpaired electrons,so it is diamagnetic.$3$. $[NiCl_4]^{2-}$: The oxidation number of $Ni$ is $+2$. $Ni^{2+} = [Ar] 3d^8$. $Cl^-$ is a weak field ligand,so no pairing occurs. It undergoes $sp^3$ hybridization (tetrahedral). There are two unpaired electrons,so it is paramagnetic.

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

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