Why does Ni not form low spin octahedral complexes?

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Why does Ni not form low spin octahedral complexes?

Something to do with coordination bonding...

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Answer 1 · 2016-02-01T14:02:39

Because for a $N i (I I)$ $Ni(II)$ complex with octahedral coordination, the $e_{g}$ $e_g$ orbitals are degenerate, and the $t_{2 g}$ $t_(2g)$ orbitals must be completely filled.

Explanation:

There are some few octahedral Ni(II) complexes. When the axial and non-axial metal $d$ $d$ orbitals are split by a ligand field, the $e_{g}$ $e_g$ orbitals are degenerate. A $d^{8}$ $d^8$ metal centre fills the $t_{2 g}$ $t_(2g)$ set ( $6 e$ $6e$ ), and the remaining $e_{g}$ $e_g$ pair of orbitals are each singly occupied. A $d^{8}$ $d^8$ 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 $d^{4}$ $d^4$ , $d^{5}$ $d^5$ , $d^{6}$ $d^6$ , and $d^{7}$ $d^7$ configurations give rise to the possibility of high-spin/low-spin electronic configurations.

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Why does Ni not form low spin octahedral complexes?

Something to do with coordination bonding...

1 Answer

Explanation:

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