The number of unpaired electrons in Ni(CO)_4 | vedclass

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Question

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

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$0$

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(A) The oxidation state of $Ni$ in $Ni(CO)_4$ is $0$. The electronic configuration of $Ni$ $(Z=28)$ is $[Ar] 3d^8 4s^2$. Since $CO$ is a strong field ligand,it causes the pairing of electrons in the $3d$ and $4s$ orbitals. As a result,the $4s$ electrons are promoted to the $3d$ orbital,leading to a fully filled $3d$ subshell $(3d^{10})$ and an empty $4s$ orbital. Therefore,there are $0$ unpaired electrons in $Ni(CO)_4$.

The number of unpaired electrons in $Ni(CO)_4$ is

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