Question

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

Answer 1

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

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`"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.