What is the geometry of [Ni(CO)_4]? | vedclass

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Question

(A) In $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$. The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong fiel

Vedclass · Accepted Answer

Tetrahedral

Vedclass · Answer

(A) In $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$. The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes pairing of electrons in the $3d$ orbital. The $4s$ electrons shift to the $3d$ orbital,resulting in a $3d^{10}$ configuration. The hybridization is $sp^3$,which corresponds to a tetrahedral geometry.

What is the geometry of $[Ni(CO)_4]$?

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