(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 l

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(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.The $4s$ electrons shift to the $3d$ orbital,resulting in a fully filled $3d$ subshell $(3d^{10})$.Therefore,there are $0$ unpaired electrons.