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Rajini
Jun8-10, 07:17 AM
hello all,
we know that

1) Fe-O-O is end-on binding of O to Fe and
2) [see attachment] is side-on binding of O to Fe (please assume vertical lines/pipe as a single bond).

Now my doubt is:
Is it completely and always true to name 1 as superoxo and 2 as peroxo ?
I am just asking whether the other name for 1 is superoxo and for 2 is peroxo.
thanks

alxm
Jun8-10, 09:02 AM
Is it completely and always true to name 1 as superoxo and 2 as peroxo ?


Nope. It depends on the metal, its redox state, and whatever other ligands you might have.

See e.g. this (http://www.pnas.org/content/100/7/3635.abstract) paper, and many others from Tolman and Cramer.
(Also, since you seem interested in metalloenzymes, I should mention Siegbahn and Blomberg's work too, since they've done more theoretical studies of metalloenzymes, especially oxygen-activating ones, than anyone else)

Being a theorist, I would hesitate to characterize them as one or the other until I'd done a calculation and seen the spin-density distribution.

Rajini
Jun8-10, 09:33 AM
Ahh.
I only want to know for Fe !

Rajini
Jun8-10, 09:39 AM
For eg.,
in this paper by Que:
doi:10.1038/nature07371
can i say the structure in Fig. 2, bottom-left, as side-on rather than Fe(III)-peroxo ?
they mention it as Fe(III)-peroxo .

alxm
Jun8-10, 11:02 AM
It's a peroxo binding side-on to Fe(III), yes.

But let's just be clear, there's a sliding scale here.
You always have a resonance between for instance,
Fe(I)-O2 <-> Fe(II)-superoxo <-> Fe(III)-peroxo
After all, you're just moving electrons, which can easily "be in several places at once".

Quantum-mechanically you never see an exact integer number of electrons. So when you say it's "Fe(III)-peroxo" you're just saying it's mostly like that.
But there's so much going on with spin states, ferromagnetic coupling, etc, that it's difficult to say offhand. Which is of course why they do theoretical studies of this stuff in the first place.

In my experience, the binding (side-on vs end-on) is more due to the redox state of the metal and its other ligands, than due to the redox state of the oxygen.
(And the binding, in turn does influence what redox state the oxygen ends up in. )

Rajini
Jun10-10, 07:50 AM
Hi i got the essence from your reply.
thanks