We determine the geometry and magnetic properties based on Valence Bond Theory:

1. Oxidation State: In $[Ni(CO)_4]$, the ligand Carbonyl (CO) is neutral. Therefore, the oxidation state of Nickel (Ni) is 0.
2. Electronic Configuration: Nickel (Z=28) in the ground state has the configuration $[Ar]3d^8 4s^2$. Since the oxidation state is 0, the configuration remains the same.
3. Ligand Field Effect: CO is a very strong field ligand. Under its influence, the two electrons in the $4s$ orbital are forced to pair up with the electrons in the $3d$ orbitals. This results in a fully filled $3d^{10}$ configuration ($[Ar]3d^{10} 4s^0$).
4. Hybridisation: To accommodate the four CO ligands, the empty $4s$ orbital and the three $4p$ orbitals undergo mixing to form four equivalent $sp^3$ hybrid orbitals.
5. Geometry: $sp^3$ hybridisation corresponds to Tetrahedral geometry.
6. Magnetic Behaviour: Since all the electrons in the $3d$ orbitals are paired ($3d^{10}$), there are no unpaired electrons. Consequently, the complex is Diamagnetic.

Conclusion: The complex is tetrahedral and diamagnetic .