[Tubulin]

I just recently had a computer lab where we calculated the reaction energies for various reactions. One was anionic hydroxyl to radical hydroxyl plus an electron, OH- -> OH + e- . Comparing to experimental values this resulted in a -310 kJ/mol difference.

One question I have to answer is why electron correlation is important for this reaction. The way I see it, OH- does not have any unpaired electrons (a closed-shell?) while radical OH does have an unpaired electron (open-shell).

I guess that Hartree-Fock does not account for opposite spins (which would be paired electrons?) but does account for parallel spin. Although I do have some issues accurately explaining the above results. Any help would be much appreciated.

EDIT: After thinking some more about it, I guess Hartree-Fock (HF) makes a bigger mistake with molecules with paired electron. The radical hydroxyl has an unpaired electron, and have a smaller mistake. Because of this, I think that the HF mistakes does not cancel out when I calculate the energy for the reaction (reactant - product). Does anyone know if I am on the right track? Thanks in advance.

[Corribus]

In the Hartree-Fock approximation, electron orbitals are assumed to be independent of each other.  That is not to say it treats one electron as if the others don't exist, but it assumes that any given electron is interacting with the others through an average potential "background".  We say the HF method neglects "electron correlation" because the method assumes that the energies/motions of one electron are not correlated to the energies/motions of other electrons in the system.  Formally, the Hamiltonian for a multielectron wavefunction is approximated as the product of multiple single electron wavefunctions.  Effective Hamiltonians devised by the HF approximation are solved by an interative method called 'self-consistent field'. 

Electron correlation will be important in any system that has more than one electron operating in the same spatial volume element.  Both the OH radical and OH anion qualify here.  I'm not sure what kind of H-F calculation you are doing, but a simple restricted H-F calculation will not work on an open shell (unpaired electron) system.  Therefore I'm assuming you are using an unrestricted H-F (UHF) calculation.  Either way, ignoring electron correlation will introduce considerable error for both the OH anion and radical.  In short, I expect it will do a pretty craptastic job with both molecules.  When you say "Comparing to experimental results", what do you mean exactly - are you compared to the overall DH of the reaction?  I'm not sure how you do this because your reaction as written includes a "naked electron", which doesn't really exist in nature.  If you compare the energies of each molecule (OH anion and OH radical) with respective literature values, which molecule is closer?

Multielectron systems can be solved directly by the variational method, but this is generally cumbersome and only useful for simple systems.  More importantly, to do so abandons the atomic/molecular orbital concept, which would have other significant drawbacks to chemists.  Therefore the HF approximation is still used ubiquitously in computation chemistry - it's just that H-F orbitals are subjected to perturbations (such as Moller-Plesset or configuration interaction) to help account for electron-correlation effects.

[Tubulin]

Thanks for the answer.

I was propably very unclear in my initial post.

What we did in the computer lab was to calculate the QM energies for various molecules using H-F and B3LYP, amoung them OH- and OH radical. We where then presented with various reactions, amoung them OH^-  OH + e^-. We then calculated the energy of the reaction by subtracting the QM energies for the reactant from the product.

From my understanding when calculating such energies of reaction using HF, the errors made often cancel each other out. But in the instance of the hydroxyl anion and the hydroxyl radical, the hydroxyl radical contains one less electron pair than anionic hydroxyl. If I am correct, this would mean that the error made by H-F would not cancel each other out in this case when calculating the energy of the reaction, as it seem to do for most reactions where the amount of electron pairs are equal on both sides.

We got some experimental energies for the reaction given in the lab protocol, supposedly taken from http://webbook.nist.gov/. Comparing to this value we got a large error for the OH^-  OH + e^- reaction for the H-F calculated energies. When comparing to B3LYP we did not see such errors, since I believe that it takes into account electron correlation. My main issue is whether my train of thought for the energy of the reaction is correct in the case of the H-F calculations, that the errors for OH radical is less than the error for anionic OH, therefore the errors does not cancel out when calculating the energy for the reaction.

The purpose of this exercise was just to underline the importance for electron correlation.

I hope this was clearer

[Corribus]

I see, that makes a little more sense.  Two things: first, as I wrote above, a regular HF calculation won't work well at all with a free radical. So you should check and see if your calculation was unrestricted (or restricted open shell).  If it wasn't, you probably have your answer right there.  Second, bear in mind that the HF energy always deviates from the true value in the same direction: the HF energy is always greater than the true energy.  This is often called the HF Limit.  It is a natural consequence of the variational method for approximating wavefunctions.  The point being that the sign of the deviation from the true value for the energy change of the reaction should, I imagine, depend on which participating molecule (your reactant or product) is deviating more.  What I mean is the following:

Suppose you have a simple reaction A B, with energy change ΔE determined by E_B - E_A.  Now if you calculate the ΔE by the HF method you describe, you can define a error, call it Q, related to ΔE(exp) - ΔE(calc), with ΔE(exp) = E_B(exp) - E_A(exp) and ΔE(calc) = E_B(calc) - E_A(calc). Here, "calc" and "exp" refers to calculated and experimental, respectively.  However you know that E_A(calc) - E_A(exp) > 0, and likewise for E_B, so the sign of Q should tell you which of A or B performs more poorly for the HF calculation, because Q = (E_B(exp) - E_B(calc)) - (E_A(exp) - E_A(calc)).  If Q > 0, A performed worse because the absolute magnitude of (E_A(exp) - E_A(calc)) > absolute magnitude of (E_B(exp) - E_B(calc)); if Q < 0, B was worse for the opposite reason. 

I can't quite tell from your post what the sign of Q is in this case (you said deviation of -310 kJ/mol, but is that the experimental minus the calculated or the calculated minus the experimental?).

(I hope I got all those signs right...)

[Tubulin]

Thank you very much for your answer.

Sadly, I cannot seem to access WebMO where we did all our calculation to check which H-F method we used, although I suspect it was not unrestricted.

The -310 kJ/mol deviation is ΔE(cal) - ΔE (exp) so in my case i guess it would means that A performed worse than B. This seems to support that the error made by H-F was less for the hydroxyl radical than for the anionic hydroxyl.