Why does Ni not form low spin octahedral complexes?

Something to do with coordination bonding...

1 Answer
anor277
Feb 1, 2016

Because for a #Ni(II)# complex with octahedral coordination, the #e_g# orbitals are degenerate, and the #t_(2g)# orbitals must be completely filled.

Explanation:

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_(2g)# set (#6e#), 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.

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