How does the anti-bonding eg orbital (eg*) affect the bonding of ligands?
As you would expect from increasing occupancy of antibonding orbitals, the result is weaker metal-ligand bonding. This is one reason why the octahedral geometry is not as strongly favored for metals with greater numbers of d electrons. For a d8 metal such as the Ni(II) in the example, square planar and octahedral complexes are both common, and planar structures tend to be favored if the ligand is strong enough to result in a low-spin planar complex (as is often the case with Ni(II)). Only a weak field ligand like H2O will tend to form octahedral complexes with Ni(II).
A semi-quantitative approach to this issue is found in &quot;structure energy&quot; approximations and determination of &quot;structure preference energy&quot;. Although that approach is not rigorously quantitative, it will show you trends.
It is important to note, though, that even with full occupancy of the eg antibonding orbitals (for example in a d10 metal like Zn(II)), octahedral complexes are possible. One can think of the sigma bonding in that case as being essentially entirely via the s and p metal orbitals, similar to a p-block central atom such as S in SF6.