Question

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Answer 1

CO is a strong pi-acceptor, hence it's crystal field splitting will be strong field, weak spin. While water (I believe a sigma-donor), on the otherhand, would be a weak field, high spin, ligand, so your delta would be smaller.
I'm just a little confused as to what the geometry of iron porphyrin is. Is it square planar??

Answer 2

And by displacement, I assume you're referring to jahn-teller distortion?

Answer 3

thanks for replying , dude. Well, the geometry would be octahedral, axial ligands will be N's, the only space vacant (or occupied by the H2O molecule is the z^2) this is where CO would bind. The displacement refers to the Fe(II) being just underneath the plane of the porphyrin (unsure if this would also be called JT distortion or entatic state), but closer to being on the plane when bound to the CO (low spin) than the H2O (high spin). So i guess my question is more towards the effect of the distorted geometry on the reduction potential and whether being low spin or high spin affect the reduction potential, if so how? lol

Answer 4

Well, if it's high spin (weak field), it would probably be easier to reduce, due to lower energy field splitting. Whereas a low spin (strong field) CO would not be so easy to reduce due to it's pi-antibonding properties. Stronger bond and removal of electron density.
As far as it's geometry, I would say to do a crystal field splitting diagram for Fe (II), If the electrons. I think you can see whether it's geometry would be distorted based on it's electron filling on the T2g and Eg, since asymmetry usually causes a geometric distortion.

Also, idk if that's the charge, but i think 2+ is the preferred state of Iron and since you're reducing?

Answer 5

Depending on it's crystal field splitting, you can kind of see "how" it will want to reduce it's geometry. If there's asymmetry on it's Dz^2 or Dx^2-y^2, then it might either go an axial decompression or elongation. Those are the usual distortions that occur for octahedral species.

Answer 6

I don't think im explaining myself correctly. soo many things coming into my head all at once. Lol.

Answer 7

haha no don't worry, this stuff gets confusing without visuals, i find. But i get what you mean. There were 2 hypotheses as for the change in geometry i was presented in class, one was that Fe(III) goes to Fe(II), or that Fe(II) goes from HS to LS, i'm focusing on the second option because they're talking about CFT... so basically because it's high spin, it's able to lose the ligand easier (because the bond is weaker)?

Answer 8

so if an electron is added to a lower energy orbital the Ered will increase?

Answer 9

Eg is antibonding remember. let me switch browsers n draw CFSP

Answer 10

Eg is always anti bonding?

Answer 11

the one on the left would be for a low spin, and on the right, would be for high spin.

Answer 12

you would fill it based on hunds rule. pairing of the lower energy first. HOWEVER, notice that for low spin, the energy of the eg set is too high up. the pairing would have to continue ont he t2g. for instance a d6 (such as iron)