# Why does Ni not form low spin octahedral complexes?

## Something to do with coordination bonding...

Because for a $N i \left(I I\right)$ complex with octahedral coordination, the ${e}_{g}$ orbitals are degenerate, and the ${t}_{2 g}$ orbitals must be completely filled.
There are some few octahedral Ni(II) complexes. When the axial and non-axial metal $d$ orbitals are split by a ligand field, the ${e}_{g}$ orbitals are degenerate. A ${d}^{8}$ metal centre fills the ${t}_{2 g}$ set ($6 e$), and the remaining ${e}_{g}$ pair of orbitals are each singly occupied. A ${d}^{8}$ configuration can only fill the mtal d orbitals the 1 way: high-spin/low-spin configurations ARE NOT an option.
For an octahedral complex, only ${d}^{4}$, ${d}^{5}$, ${d}^{6}$, and ${d}^{7}$ configurations give rise to the possibility of high-spin/low-spin electronic configurations.