[pioyi]

Hello everyone! It seems nice in here.

I have been searching for a couple of days now and I keep finding unsatisfying information.
My book (not a universally known one) attempts to explain the fact that fluorine forms a weaker halogen elemental bond than chlorine by giving 2 reasons:
1) Closely packed electrons (I guess you have already heard about that)
2) Due to the lack of d orbitals in fluorine. All other halogens have these, which in turn participate in "the hybridization" (?) thus strengthening the halogen-halogen bond.

° So, does it really make sense for Cl₂ to hybridize? Most sources say that hybridization doesn't take place (makes sense right? We are only forming one bond here. Why would we waste so much energy to make all the orbitals degenerate? What are we gaining energetically?).

° How exactly do d orbitals strengthen the bond? Is something like Pi backbonding (as referred to transition metals) taking place here? It makes me very uncomfortable to think about the d orbitals here because everything is so closely packed together (But I guess this logic doesn't apply here. Nitrogen can form a triple bond with ease and a very strong one indeed). Chlorine is not even that big of an atom, do the p-orbitals really share their inert electron pairs with the neighboring halogen's p orbitals?

Thanks for your time, I hope I didn't overlook anything.

[Hunter2]

d orbitals no need here. But flourine and also chlorine are sp^3 hybridisiest. The atomes have 3 free pairs of electrons and one is paired to another atome.

Compare the main atomes of  methane, ammonia, water, and hydrogenflouride
The same with Silane,  phosphane, hydrogensulfide and hydrochloric.

[pioyi]

So are they indeed sp3 hybridized?

I understood all of the examples that you wrote down. But here we have something different. It's a linear molecule and it only forms one bond. Also you gave HCl as an example and it resembles diatomic chlorine. But 100% of the sources say that HCl is unhybridized.

Because user azmanam said in https://www.chemicalforums.com/index.php?topic=37990.0 that it's really not.
He went on to give some bibliography: http://books.google.com/books?id=x6ct1xjyJpoC&pg=PA371&lpg=PA371&dq=vsepr+theory+chlorine&source=bl&ots=uB-xVEvcO3&sig=fvrKp9FyofKCWXYTEMaESxYUYiE&hl=en&ei=p3sWS6HoMtCWlAfHvvzMBQ&sa=X&oi=book_result&ct=result&resnum=3&ved=0CA4Q6AEwAg#v=onepage&q=vsepr%20theory%20chlorine&f=false

If I understood correctly d orbitals don't participate in the bond. So the book that I have at hand must have made a serious mistake

From stack exchange of the chemists: https://chemistry.stackexchange.com/questions/60238/what-is-the-hybridization-of-diatomic-nitrogen-diatomic-anything-like-oyxgen-b a users goes on to say that there is no need for hybridization to occur. I find his explanation very logical.

I know hybridization is just a concept to make sense of nature, but is it really applicable in here? Hybridization is useful whenever multiple bonds are formed, right? Hybridization of diatomic molecules seems quite silly to me. Does it really make sense for hybridization to be present in a single-bonded system?

Thank you for your time!

[Corribus]

My view is that the first explanation is sufficient without the need to invoke hybridization. Pauling formulated hybridization to help explain molecular geometry in simple organic molecules. It has been extended with success to bigger molecules and in some particular cases for inorganic systems, but I don't find the hybridization to have much use beyond this. It is worth keeping in mind that, historically, that was never the purpose of the model. Therefore it should be no surprise that more sophisticated computational chemistry models have since cast a lot of doubt on the usefulness of hybridization in many systems where it was previously applied, particularly when d-orbitals are brought into the picture. Insofar as halogens usually only bond to one other atom, and so geometry isn't really at issue, I see very little use for hybridization here. I think that applying the model here is beyond it's originally intended purpose, even if it seems to give an prediction that is consistent with observation.

That said - since, in quantum mechanics, no system with more than one electron is exactly solvable, we are invariably required to make approximations to quantitatively or qualitatively model molecular systems. In that sense, no bonding model is "correct". We will always be able to point to some deficiency of a model in explaining some aspect of what we observe. I personally feel that hybridization requires making a lot of assumptions (not the least of which is the energetic requirements for orbital mixing) without any good basis to do so, but it nevertheless is used often because it seems to give answers that (qualitatively) match what we observe. That's not a very good way to go about understanding chemistry at a fundamental physical level IMO. But if you don't care about that, and it helps you make good predictions most of the time, then does it matter? (Well, other than that "most of the time" is not "all of the time".)

You aren't going to be able to "prove" that hybridization does or does not contribute to halogen bonding. This is primarily because hybridization isn't something real that can or cannot be "proven". If you do enough digging in the primary literature, maybe you'll find that complex computation chemistry calculations have something to say about whether d-hybridization is consistent with the fundamental physics of quantum mechanics as applied to halogen atoms.* Or not. But we're talking a qualitative understanding here. So, if you like the explanation, use it. Nobody is going to be able to prove it wrong.

I will point out two things, though.

1) I'm not sure what you mean by "fluorine does not have d-orbitals".
2) Your hybridization explanation may help you understand differences between fluorine and the other halogens, but does it help you understand quantitative differences in bonding characteristics (strength, say, however you want to measure it) throughout the series (i.e., the trend)?

Particularly, point 2 may be something to consider when evaluating the relative merits of the two choices you offered in your first post.

*I suspect you won't, because it's not a particularly important question. People who want to quantitatively predict molecular structures and bonding properties don't use hybridization because they know it's useless in that regard.

[pioyi]

Hello Corribus! Thanks a lot for your remarks.

You are right I understand what you are saying. Models are here to be used in the most appropriate scenarios and hybridization isn't really helpful here.

1) Yes this is not a logical thing to say. Although I still have a lot to learn, I know that they are functions and they aren't property. I rephrase the passage of the book to: Fluorine can't have electrons in the d orbitals because of the energetic difference of the 3d and 2p orbitals.

So in your opinion, how do d orbitals strengthen the bond of Cl₂ if they do at all?


Later addition: I think after all these questions I have found my answer! Computational chemistry findings were really helpful as you advised. Some researchers did experiments and finally concluded: This shows the effect mainly comes from the d functions but f and g functions also contribute. These tests confirmed our theory that the peak of BO and dissociation energy at Cl₂ is due to exchange polarization.

They found:
Bond order of F2: 0.982
Bond order of Cl2: 1.357 (!)
Bond order of Br2: 1.264
[...]

Now the only thing that is missing, is for me to understand exchange polarization on a very very basic level! I think I will succeed in that. Thank you for your help,  Corribus and Hunter2

[Corribus]

I think something should be clarified. Orbital mixing depends on symmetry compatibility and energy difference. Mathematically, you can mix in d-orbital character any time you want, even in the case of carbon, but the 3d orbitals are considerably higher in energy than the 3p and 3s orbitals, much less the 2d and 2s orbitals, so their contribution will be small. This is the same reason we don't include 1s orbitals, even though in principle we could do so. In organic molecules, geometries of so-called sp3 hybridized carbons always have some deviation from the "perfect" bond angles predicted by the model. One reason is that the (qualitative) model assumes only a limited number of participating AOs; for perfect accuracy, every orbital should be included.

It would be fair to say that as you go down the period from fluorine to iodine, d-orbital contributions will likely become more important in determining the bonding characteristics, if only because the d-orbitals become closer in energy to the valence electron s- and p-orbitals. it is a simple matter of math: the closer in energy two orbitals are, the more they interact, and the relation is not nonlinear. For that matter, as you go down the group, relativistic effects become more important as well! So really it is not a matter of "do they or don't they" but rather "how much". Still, I think other effects play a significant role in what happens to halogen bond strength as you make your way down the periodic table. Bond strength depends not only on the input AO energies but also overlap, and it is clearly the case the overlap gets poorer as the nucleus gets heavier. 

Consider this: the bond strengths of F2, Cl2, Br2, and I2, are ~ 157, ~242.6, ~193.8, ~152.5 kJ/mol.
Bond lengths go like: 144, 199, 228, 267 pm

There is more than one effect at play here. We could surely give a bunch of ad hoc rationalizations to the trends, all of which would probably be, to varying degrees, correct. d-orbitals would probably not be among the first things I would leap to. Nevertheless, if we were to calculate these values with computational packages, we could probably even quantify how d-orbital character changes as you go down the periodic table. It might be a fun exercise, but I'm not sure what we would learn from it. Maybe there's a point where there is "enough" d character that we could feel confident calling the atoms "d-hybridized", even though all these molecules have similar geometries to a first approximation. But I guess that would be subjective.