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(D) For $[Ni(CO)_4]$: The oxidation state of $Ni$ is $0$. The electronic configuration is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it cau

Vedclass · Accepted Answer

$[Ni(CO)_4]$ diamagnetic,$[NiCl_4]^{2-}$ paramagnetic

Vedclass · Answer

(D) For $[Ni(CO)_4]$: The oxidation state of $Ni$ is $0$. The electronic configuration is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairing of electrons,resulting in $sp^3$ hybridization with $0$ unpaired electrons,making it diamagnetic.For $[NiCl_4]^{2-}$: The oxidation state of $Ni$ is $+2$. The electronic configuration is $[Ar] 3d^8$. Since $Cl^-$ is a weak field ligand,it does not cause pairing of electrons,resulting in $sp^3$ hybridization with $2$ unpaired electrons,making it paramagnetic.

Select the option with the correct property -

Similar Questions

Nickel combines with a uninegative monodentate ligand $(X^{-})$ to form a paramagnetic complex $[NiX_4]^{2-}$. The hybridisation involved and the number of unpaired electrons present in the complex are respectively:

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