If there is only crystal field splitting and no Jahn-Teller distortion (In a octahedral environment).
Dear Sir,
In general, for perfect Oh-symmetry, the d-state (formerly 10-fold degenerated in the
atom of spherical symmetry) is split into a 6-fold degenerated t2g state and a 4-fold
degenerated eg state. Here, degeneration includes also spin quantum numbers.
Since you are asking for the 7th electron I assume that the ligands are negatively
charged. Then, the lower lying t2g state is fully occupied by electron #1 ... #6. Since all
levels in the eg have the same energy (perfect Oh symmetry !!!) neither the dz2-r2
state nor the dx2-y2 state must be preferred when forming the wavefunction
of the 7th electron. With proper normalization, the wavefunction must read as
|#7> = 1/sqrt(2) |dz2-r2> + 1/sqrt(2) |dx2-y2>
(or any other combination of signs), resulting in
= 1/2 +0 +0 + 1/2 = 1
Note that this is only valid if there are no other symmetry breaking effects, such
as Van Vleck mechanism, spin-orbit coupling etc. !!!!
Regards, FM
- Badges
- Science topic
- Similar topics
- Chemistry
- Nanotechnology
- Nanostructures
- Thin Films