Question

How would we know that $Ni{\left( {CO} \right)_4}$ prefers tetrahedral over square planar?

Answer 1

In nickel tetracarbonyl complex, carbonyl ligand acts as a strong ligand and the pairing up of electrons takes place. Thus, there were no d-orbitals leading to the $s{p^3}$ hybridization. $s{p^3}$ hybridization refers to the geometry of tetrahedral and $ds{p^2}$ refers to the geometry of square planar.

Complete answer:
Complex compounds are also known as coordination compounds. These compounds do not dissolve in water or any other organic solvents by retaining their identity. It is the special property of complex compounds. It consists of a central metal atom which can accept electrons and a ligand which can donate electrons.
$Ni{\left( {CO} \right)_4}$ is a complex compound, in which nickel is the central metal atom and the ligand is a carbonyl group. Nickel is an element with atomic number $28$ and has eight electrons in d-orbital and two electrons in s-orbital.
But due to the presence of a strong field ligand, carbonyl the electrons pairing up takes place leading to the formation of the ${d^{10}}$ complex. Thus, d-orbitals were completely filled.
Now the hybridization involved will be $s{p^3}$ due to the completely filled inner d-orbitals. $s{p^3}$ hybridization refers to the geometry of tetrahedral whereas $ds{p^2}$ refers to the geometry of square planar.
Due to the presence of a strong field carbonyl ligand $Ni{\left( {CO} \right)_4}$ prefers tetrahedral over square planar.

Note:
The spectrochemical series was helpful to know the stability of ligands. Ligands like cyanide, carbonyl are considered as strong field ligands. In the presence of strong field ligands, the rules like Pauli’s exclusion principle were deviated and resulted in pairing up of electrons.