The geometry possessed by [Ni(CO)_4] is | vedclass

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

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tetrahedral

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(A) In $[Ni(CO)_4]$,the oxidation state of $Ni$ is $0$. The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$. Due to the presence of a strong field ligand $(CO)$,the electrons pair up,resulting in $3d^{10}$ configuration. The $4s$ and $4p$ orbitals undergo $sp^3$ hybridization. Therefore,the geometry of $[Ni(CO)_4]$ is tetrahedral.

The geometry possessed by $[Ni(CO)_4]$ is

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