Chemical Forums
Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: mistche20 on January 29, 2012, 04:27:00 AM
-
[Co(H2O)6]2+ is more difficult to oxidize to Co(III) than [Co(NH3)6]2+.
[Co(H2O)6]2+ is a high spin complex, and [Co(NH3)6]2+ a low spin one.
but why is the first one more difficult to oxidize than the second one?
-
The fact that the spins are different probably has a hint in it... The high spin complex has the ligands using up the 4s and 4p orbitals leaving the d orbital free with many unpaired electrons from which one can be removed rather easily to oxidize the metal ion.
On the contrary, since the low spin complex's ligands occupy the d-orbitals, with only one unpaired electron, oxidizing is a bit difficult, because it involves removing an electron from the very same d-subshell.
-
The fact that the spins are different probably has a hint in it... The high spin complex has the ligands using up the 4s and 4p orbitals leaving the d orbital free with many unpaired electrons from which one can be removed rather easily to oxidize the metal ion.
On the contrary, since the low spin complex's ligands occupy the d-orbitals, with only one unpaired electron, oxidizing is a bit difficult, because it involves removing an electron from the very same d-subshell.
shouldn't that be the other way around? for the high spin complex, my guess is that since there are many electrons with the same spin, removing an electron would result in a decrease of exchange energy K. But for the low spin complex, there are less eletrons with the same spin and hence it's easier to remove it??
-
Oh! I wasnt even paying attention to what point I was making :P I'm truly sorry. Seems like I have given the exact wrong answer :P :P