From the magnetic behaviour of [NiCl_4]^{2-} | vedclass

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(B) For $[NiCl_4]^{2-}$: The oxidation state of $Ni$ is $+2$. The electronic configuration is $[Ar] 3d^8$. Due to the weak field ligand $Cl^-$,it unde

Vedclass · Accepted Answer

$[NiCl_4]^{2-}: Ni^{II}$,tetrahedral; $[Ni(CO)_4]: Ni(0)$,tetrahedral

Vedclass · Answer

(B) For $[NiCl_4]^{2-}$: The oxidation state of $Ni$ is $+2$. The electronic configuration is $[Ar] 3d^8$. Due to the weak field ligand $Cl^-$,it undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry with $2$ unpaired electrons,making it paramagnetic.For $[Ni(CO)_4]$: The oxidation state of $Ni$ is $0$. The electronic configuration is $[Ar] 3d^8 4s^2$. In the presence of the strong field ligand $CO$,the electrons pair up to form $[Ar] 3d^{10} 4s^0$. It undergoes $sp^3$ hybridization,resulting in a tetrahedral geometry with $0$ unpaired electrons,making it diamagnetic.

From the magnetic behaviour of $[NiCl_4]^{2-}$ (paramagnetic) and $[Ni(CO)_4]$ (diamagnetic),choose the correct geometry and oxidation state.

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