Submitted by Nancy Williams / Scripps College, Pitzer College, Claremont McKenna College on Thu, 09/25/2008 - 19:12
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OK, so we all love the homonuclear diatomics, and how the 2p sigma is *below* the 2p pi for O2 and F2, almost equal at N2, and above 2p pi for C2, B2, etc.

 CO looks a lot like N2, but it's asymmetric...the 2p sigma is the HOMO. Disturbingly, the gap seems to be *larger* than in N2...why is that?!? Being isoelectronic, it seems like it should be about the same, but there should be *more* contribution from the oxygen orbitals in the *bonding* orbitals since O is more electronegative than C.

It gets weirder for NO...while I can't find a photoelectron spectrum, all the written diagrams I can find place the sigma well above the pi. Weirdly, a low-level Spartan (TM) calc shows the sigma *between* the pi orbitals (yes, I understand they unpaired electron in the pi* breaks the symmetry). It seems like if anything, NO should be closer to O2 than N2, and N2 has nearly equal energies....

What gives here? What do I not understand about sp mixing that is causing me to get this so wrong?

Adam Johnson / Harvey Mudd College
the problem _may_ be your "low" level of theory?  Can you crank up the basis sets?
Thu, 09/25/2008 - 19:43 Permalink
Jason D'Acchioli / University of Wisconsin-Stevens Point

In reply to by Adam Johnson / Harvey Mudd College

Hi all,

This is my first post, so please bear with me. Do you have any qualitative --or quantitative-- MO diagrams you can post? I would agree that "low" levels of theory would influence your results. What level of theory / basis sets did you use? In terms of sp mixing, I take it you're looking at this in terms of VB theory as opposed to MO theory?

 Thanks!

Tue, 10/07/2008 - 22:37 Permalink
Nancy Williams / Scripps College, Pitzer College, Claremont McKenna College

In reply to by Jason D'Acchioli / University of Wisconsin-Stevens Point

Alas, I'm embarrassed to say that we were just using the very low-brow levels of HF calculation available in Student Spartan (3-21G and 6-31G*). We were kinda hoping that for neutral diatomic molecules in the second row of the periodic table, that the HOMO would be about right...we very well may have been too naive on this point. 

We've been meaning to corral our computational chemist neighbor across the street (Bob Cave at Harvey Mudd College) to help take these up a notch (or ten), but we've been swamped by the semester.

Sun, 10/12/2008 - 17:49 Permalink
Mookie Baik / Indiana Univeristy Bloomington

In reply to by Nancy Williams / Scripps College, Pitzer College, Claremont McKenna College

Guys,

Now let's not beat up HF/6-31G*, which is a decent level of theory for this. I am pretty sure that the following is what is going on. Just to be calibrated, I state some of the obvious things - please don't yell at me for stating the obvious. I promise the last part will not be obvious:

(1) If we totally neglect second-order mixing, i.e. we don't consider hybridization of the s and p orbitals, we'd get 1(sig) and 1(sig)* out of the 2s orbs and then 2(sig) and 2(sig)* out of the 2p orbitals. Without invoking second-order mixing, you'd always have 2(sig) to be lower in energy than the 1(pi) and 2(pi) MOs, which are the degenerate in-phase pi-MOs, of course.

(2) Invoking 2nd order mixing, 1(sig), which used to be simply 2s-2s in-phase, becomes sp-sp in-phase with the big-lobes pointing at each other. 1(sig)* becomes also lower in energy, because that one used to be 2s-2s antibonding and is now sp-sp out-of-phase with the tiny lobes pointing at each other that are of course out-of-phase.

(3) There is no free-lunch, and thus 2(sig) that used to be really low in energy (below pi), because it was 2pz-2pz in-phase before second-order mixing, has to pay the price of 1(sig) and 1(sig)* going down. It is still bonding in character, but the lobes are the tiny ones of the sp-hybrid. The big loser is 2(sig)* that is out-of-phase and is pointing the big lobes at each other.

(4) In case of N2, 2(sig) is either slightly above 1(pi)/2(pi) or slightly below (doesn't really matter for the overall picture). When you compare CO to N2, what you are doing more than anything else is you are increasing both the s and p orbital energies on one end (N => C) and decreasing those energies on the other side (N => O).The result is this:

The 2(sig) is now energetically closer to C2s (because the 2s orb is now higher in C than in N) and also energetically closer to O2p (because the 2p orb in O is lower than in N). Consequently, 2(sig) has more C2s character and also more O2pz character. Remember now that the second-order mixing contribution to 2(sig) is from 1(sig)*, i.e. it is out-of-phase, so you are seeing more C2s -- O2p antibonding interaction, which pushes 2(sig) higher in energy than what one may expect just from a first-order MO diagram where there is no hybridization.

(5) NO is a different matter altogether, because the closed shell MO-diagram is not really a good representation. One would also expect to see more multi-configurational nature showing up here.

OK - I think what I write above is a little difficult to see when you don't have figures and actual numbers to compare to. I am having my postdoc and a grad student of mine prepare a short writeup on this - we can also give you a full set of computational results on this using HF, DFT, CCSD(T) and CASSCF. I promise to post this within TWO weeks (announcing this will put pressure on me and my students to actually do it!).

Mon, 10/20/2008 - 22:06 Permalink
Adam Johnson / Harvey Mudd College

In reply to by Mookie Baik / Indiana Univeristy Bloomington

The Leadership Council just had a weekend meeting where we learned (or relearned) quantum mechanics and practiced using webMO/Gaussian.  I am looking forward to your answer WITHIN TWO WEEKS!  Hopefully you will post it as a learning object somewhere on the site.

Thanks, and welcome to VIPEr,

Adam

Tue, 10/21/2008 - 12:10 Permalink
Nancy Williams / Scripps College, Pitzer College, Claremont McKenna College

In reply to by Mookie Baik / Indiana Univeristy Bloomington

Thanks-that actually does make a lot of sense, though I'm going to have to sit down with some well-drawn out MO diagrams before I try to reproduce that argument for a student! 
Tue, 10/21/2008 - 15:27 Permalink
Mookie Baik / Indiana Univeristy Bloomington

In reply to by Adam Johnson / Harvey Mudd College

Oh, yes - I will post the writeup as a case-study on here and will drop a note right in this forum to let you know. A student of mine (back then in high school) has also done a very interesting study on the electronic structure of CO - you'd think that we know everything that there is to be known about CO at this point (!).

I am thinking about writing up that work together with this story as a paper for J. Chem. Ed. - I was thinking that the CO story is too specialized to be of real interest to anyone, but am now encouraged by the discussion here. My students are working on the writeup now, while I am having endless fun with proposal writing..... grmpf....

Mookie 

Wed, 10/22/2008 - 23:54 Permalink
Mookie Baik / Indiana Univeristy Bloomington

Dear All,

You won't believe this: Turns out most of the modern quatum chemical methods do NOT get F2 right! HF, MP2, CCSD(T) all compute that the 5(sig) is actually higher in energy than 1(pi) - note that I have changed labels now. 1(sig) and 2(sig) are the in-phase and out-of-phase combos of 1s, 3(sig) and 4(sig) are those of 2s and 5(sig) and 6(sig) are made of the pz-orbitals in the first-order-MO diagram.

We need CASSCF to get the expected ordering - thinking after the fact, this is maybe not that surprising, since we do expect strong correlation effects, but I am still intrigued. We are plowing through this - my postdoc found a paper by Bickelhaupt, Nagle and Klemm by the way that deals with some of what we are discussing (J. Phys. Chem. A 2008, 112, 2437) - they do a good job on quantitatively explaning what I hadnwaved above. They do "choose" to neglect the pi's though. We are still working on this.

Sat, 11/01/2008 - 19:29 Permalink