Q.Predict the shape and hybridization of the central metal in $[\text{Ni}(\text{CO})_4]$, given that $\text{Ni}$ is in the zero oxidation state and the complex is diamagnetic.

[!TLDR] $\text{Ni}(0)$ is $d^{10}$; four CO ligands give a tetrahedral, $sp^3$-hybridized, diamagnetic complex. [!ANSW …

In $[\text{Ni}(\text{CO})_4]$, $\text{CO}$ is a neutral ligand, so nickel is present in the zero oxidation state, $\text{Ni}(0)$, with the configuration $[\text{Ar}]3d^8 4s^2$ rearranged (on forming the complex) so that all ten valence electrons occupy the $3d$ subshell as $3d^{10}$, leaving the $4s$ and $4p$ orbitals completely empty and available for bonding. With the $3d$ subshell already fully paired ($d^{10}$, no unpaired electrons and no possibility of further pairing needed), nickel uses its four empty valence orbitals — one $4s$ and three $4p$ — hybridizing them into four equivalent $sp^3$ hybrid orbitals, each accepting one lone pair from one of the four …

Determine the oxidation state of the metal from the ligand charges (CO is neutral, so Ni is 0 here); identify the resulting d-electron configuration; count available empt …

Assuming nickel must be $+2$ by default — the oxidation state must always be worked out from the actual charges of the ligands present and the overall charge of the complex, and here, since $\text{CO} …