Subject: Summary: Effect of oxidation on electron density

From: "Noel O'Boyle" <noel.oboyle2{at}mail.dcu.ie>
Date: Fri, 22 Apr 2005 10:30:12 +0100

Thanks to everyone who replied to my original question about the
similarity between the change in electron density after oxidation, and
spin density of the resulting oxidised species:
David Close, James Kirkpatrick, Cory Pye, James Eilers and Mikael
Johansson.

The answers are given below.

Regards,
Noel O'Boyle.
--------------------

Original question:
--------------------
Dear CCLers,

I have a question related to visualising the instantaneous change in
electron density accompanying oxidation (of a metal complex).

I am using Gaussian03 and have calculated the electron density of the
molecule in its oxidised state (at the geometry of the native state) and
subtracted this from the electron density of the molecule in its native
state. 

However, I have just noticed that the total spin density of the oxidised
molecule gives an almost identical picture. Is this true in general? (I
don't have a good understanding of what the total spin density shows.)

Regards,
Noel O'Boyle,
University College Dublin.
--------------------

Answers:

David Close:
--------------------
I am not sure I understand the question.  But it may have to do with
electron density versus spin density.  Since you are removing an
electron, you are creating a radical.  True spin densities are defined
as the difference between alpha and beta spin densities.  You can see
this two ways.  The easiest is at the end of the printout where the Sum
of the Mulliken spin densities are printed.  So just looking at the
Isotropic hyperfine couplings tells you which atoms have significant
hyperfine couplings.  To see where these came from you have to do a
population analysis, and at the very end you will get a Gross orbital
population where you can find the difference in alpha and beta spins. 
--------------------

James Kirkpatrick:
--------------------
this is my(probably poor)  understanding of the problem!

The total spin density for the neutral molecule will be 0 everywhere 
because the wave function is "restricted", i.e. wherever there is an 
"up" electron there is also a "down" electron.

For the oxidised molecules, if you use an unrestricted wavefunction the 
spin density will be more or less the spatial distribtuion of 
probability of finding  an unpaired electron (most electrons will be 
more or less paired up in corresponding alpha and beta MOs, so will 
produce no net spin density, but the unpaired alpha HOMO will be 
unpaired), which will be very similar (identical if you use a retricted 
open shell slater determinant?) to the difference in charges betwee the 
neutral and charged molecules.

Does that explain it more or less? Let me know if you get interesting 
answers to this interesting question!
--------------------

James Eilers:
--------------------
If you are doing RHF, yes.
Total spin density is the excess of "alpha electron density" over "beta 
electron density".
So if you remove 1 electron (oxidation) from the HOMO of  a ground  
state singlet  wavefunction and that HOMO  orbital is
fairly similar in the resulting cation (doublet), the density  
difference and the spin density will look very much alike.
--------------------

Cory Pye:
--------------------
Let
alpha = electron density of spin-up electrons
beta  = electron density of spin-down electrons.
alpha+beta = total electron density
alpha-beta = spin density

We will assume a 2N electron system (convention) from which we remove an
alpha
electron.
let a superscript 0,+ denote the native and oxidized species
let a subscript N indicate the number of electrons.

You have calculated
J=(alpha^+_N-1 + beta^+_N) - (alpha^0_N + beta^0_N) .

The spin density is
alpha^+_N-1 - beta^+_N

Note that

J=(alpha^+_N-1 - beta^+_N) + (beta^+_N - alpha^0_N) + (beta^+_N -
beta^0_N)

If the removal of the alpha electron doesn't affect muct the electrons
of
opposite spin, then the final term disappears because
beta^+_N = beta^0_N

The middle term simplifies to beta^0_N - alpha^0_N. If the native
species is
closed-shell, then it has no spin density and the middle term is 0.

Under these conditions, the spin density of the reduced species remains.

J = alpha^+_N-1 - beta^+_N
--------------------

Mikael Johansson:
--------------------
No. Spin density and total electron density difference are two different
and more or less independent creatures, though they _can_ be located to 
similar regions of the molecule (as apparently for you). Spin density is
much more free to change after, for example, oxidation/reduction. The 
electron density is much more restricted as it is bound by classical 
electrostatic forces.

I've done some studies comparing electron and spin density changes
during 
oxidation/reduction of iron porphyrins [1-3]. Perhaps they are of some 
interest.

[1] M.P. Johansson, M.R.A. Blomberg, D. Sundholm, M. WikstrC6m, "Change
in
     electron and spin density upon electron transfer to haem", Biochim.
     Biophys. Acta - Bioenergetics 1553 (2002) 183-187.
[2] M.P. Johansson, D. Sundholm, G. Gerfen, M. WikstrC6m, "The Spin
     Distribution in Low-Spin Iron Porphyrins", J. Am. Chem. Soc. 124
     (2002) 11771-11780
[3] M.P. Johansson, D. Sundholm, "Spin and charge distribution in iron
     porphyrin models: A coupled cluster and density-functional study",
     J. Chem. Phys. 120 (2004) 3229-3236.
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