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

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(B) 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

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(B) 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. The $4s$ electrons shift to the $3d$ orbital,resulting in $3d^{10}$ configuration. The hybridization involved is $sp^3$,which corresponds to a tetrahedral geometry.

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

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