Question

A first row low-spin transition metal octahedral complex (d5) in the +3 state is reduced to d6. Will a distortion occur in this process?

Answer 1

My argument was that a distortio would occur due to asymmetry of electrons in the t2g. However, i am starting to get really confused since it's saying +3 state. I would still argue though, based on Jahn-teller, that since I have 4 electrons on the t2g, and 2 in the eg set, distortion would occur to reduce this degeneracy?

Answer 2

well, since its a d5 low spin [since it's +3, it's Fe3+], it would have all 5 electrons in the t2g orbitals it would have some distortion, but once it got reduced the electron would be added to the t2g orbitals, eliminating the distortion since the electron distribution would be equivalent amongst degenerate orbitals (that is if it remained low spin). And jahn teller distortion would not be observed.

Answer 3

Oh yeah! I don't think I saw low-spin. I kept re-reading why you put "low-spin". I got one question that i'm wondering why i got only partial credit. U mind clarifying one more, dude?

Answer 4

hah yeah the crucial thing was the low-spin bit.

yeah no problem, i might not know though haha

Answer 5

@aaronq

Nickel (II) vs Platinum (II).
Why is Pt (II) found ONLY in square planar form, but Ni (II) is found in several? For instance hexammine nickel (II) but platnium forms tetrammine platinum (II). Both are d8.

ADDITIONAL INFO:
Ni (II) = 132 kJ/mol
Pt (II) = 359 kJ/mol

my argument was that energy wise, it was less costly for Nickel, hence it could account for the formation of different geometries for Ni but not for Pt complex? I only got partial credit for this answer :/
I can't think of another way or better way to explain this...

Answer 6

hey man i'm gonna look into this question, i'm been sick this whole week (i missed a test and pellet) so my brain is all foggy, ill get back to you on this

Answer 7

Oh man. That sounds like me last week!! LOL. I know the exact feel :/ feel better, mate.