How would we know that #"Ni"("CO")_4# prefers tetrahedral over square planar?

Both have all paired electrons. The only difference is their stability.

#"Ni"("CO")_4# has nickel in its #0# oxidation state, with electron configuration #[Ar] 3d^8 4s^2#. So, we call it a #d^10# complex in the ligand field.

Here is its MO diagram (it is tetrahedral):

Here, the #2e# and #9t_2# orbitals are what we pick out as the #d#-orbital splitting diagram with tetrahedral splitting energy #Delta_t#. The rest comes from ligand field theory.

The square planar splitting diagram (blank) would also be filled completely:

In comparing tetrahedral vs. square planar #d^10#:

In this case, from the basis of the angular overlap method alone, there is no preference between tetrahedral vs. square planar; the destabilization energy relative to the free-ion field is #e_(sigma) = 0# for both.

However, since nickel is a small transition metal, we expect it to favor tetrahedral over square planar, and it does.

Since the #"Ni"# #(n-1)d# and #ns# orbitals are small, they cannot sustain as much electron repulsion as you would see in, say, the corresponding #"Pd"# or #"Pt"# complexes. I discuss this here.

The tetrahedral structure has minimized the anticipated ligand-ligand bonding-pair repulsions, and that is seen in how small #Delta_t# can get.