ChemPhile - 7-1-2005 at 08:39
For example, [Cr(HzO)6]2+ ion
mick - 7-1-2005 at 11:27
The S orbitol is circular. The P ( where the electron might be) orbitol is not. The D/E/F orbitol can be doughnut shapped.
mick
[Cr(H2O)6]2+ at high-spin state
Mephisto - 7-1-2005 at 12:24
You've got 4 electrons in 5 3d orbitals in normal Cr2+. In high-spin configuration 3 of them (occupied by single electrons) are t2g (low) and 2
orbitals are on the higher eg-niveau (one of them has no electron, the other has one single electron).
JohnWW - 7-1-2005 at 17:03
My understanding is that low spin states in transition metal cations, where the number of d electrons present is substantially less than 10, occur
only where there are particularly strong ligands, which can cause spin-pairing in preference to distribution of unpaired electrons among the d
orbitals. This enables electron pairs from the strong ligands to occupy more d orbitals. In the case of water-solvated [Cr(H2O)6]++, and other
aqua-cations, it would almost certainly be high-spin. Complexes with the likes of amines, phosphines, and organic sulfides are much more likely to be
low-spin.