Identify the geometry and magnetic behavior of the complex...

The oxidation state of $\text{Ni}$ in $[\text{Ni}(\text{CO})_4]$ is $0$.

The electronic configuration of $\text{Ni}(0)$ is $[\text{Ar}] 3d^8 4s^2$.

Since $\text{CO}$ is a strong field ligand, it causes the pairing of electrons. The two $4s$ electrons are forced into the $3d$ orbitals, resulting in a completely filled $3d^{10}$ configuration.

The empty $4s$ and three $4p$ orbitals undergo $sp^3$ hybridization to accommodate the four electron pairs from the four $\text{CO}$ ligands.

Due to $sp^3$ hybridization, the geometry of the complex is tetrahedral.

Since all electrons are paired, the complex is diamagnetic.

Answer: tetrahedral geometry and diamagnetic