10
$\begingroup$

It doesn't seem as though the hybridization model adds anything useful to the discussion of molecular geometry as predicted by the VSEPR model. It's just another way of labeling linear, trigonal planar, and tetrahedral geometry (as $sp$, $sp^2$, and $sp^3$, respectively). Considering that orbital hybridization may not be the most accurate view of molecular bonding, what is the utility of having it around?

$\endgroup$
8
$\begingroup$

I find orbital hybridization of very limited use. Orbital hybridization is a mathematical manipulation of atomic wavefunctions, but that that is the only relationship to a quantum mechanical description of an atom. Presently, we have no evidence that supports the claim that orbital hybridization happens. The reason why I am not in favor of using orbital hybridization to describe chemical phenomena is because it is not a predictive tool. We only know the hybridization of an atom after we know the structure.

VSEPR, however, is a predictive tool, although it has no significant quantum mechanical basis. The VSEPR model is based upon the idea that pairs of electrons (whether bonding or not) will repel one another. Geometrical factors are considered to determine how far away pairs of electrons can be from one another (e.g 180 degrees apart for two pairs, and 120 degrees apart for 3 pairs). The model accurately predicts the H-C-H angle in methane and also predicts that the H-N-H angle in ammonia is closer to 109 degrees than 120 (as is the case in a molecule such as $BF_3$. Because VSEPR assumes ideal conditions, the model starts to fall apart when considering complex molecules. For example, while the VSEPR model correctly predicts that $SF_4$ is seesaw shaped, it can only predict that the structure won't be precisely seesaw (with F-S-F angles of 180, 120 and 90 degrees). It only allows us to suggest that the lone pair on the sulfur will be a "space hog" in a manner of speaking and repel the bonding electrons.

Now, on to the last part of your question why do we still use it? (This is somewhat speculative and opinion-based.) We have to blame the organic chemists here. orbital hybridization terminology ($sp^2, sp^3$, ...) provides a lexicon for describing the bonding environments of carbon atoms that is compact and visually constructive for organic molecules. If I state that there is an $sp^2$ carbon in a molecule, you can begin to visualize what that part of the molecule looks like, that it is a region of planarity, that it cannot be a chiral center, that it likely has a double bond and therefore may undergo certain types of chemical reactions, etc. So while orbital hybridization doesn't provide any insight into molecular geometry, it provides a basis for terminology that chemists can use to transmit meaningful information.

$\endgroup$
  • $\begingroup$ +1. I agree with everything except that hybridization is not predictive. If I've misunderstood what you mean by "predictive," my apologies. Hybridization allows some qualitative predictions, e.g.: we expect H-bearing sp2 atom to be more acidic than an sp3, ceteris paribus, due to additional s-character. Granted, we need to know the structure first. AFAIK, however, that's true for any bonding theory. In MO theory, for example, we still need to know the structure in order to characterize symmetry elements. MO theory is certainly more powerful and more correct, but also more complex. $\endgroup$ – Greg E. Dec 28 '13 at 14:12
  • $\begingroup$ @GregE. Consider an atom of carbon. With MO theory, the same 1 2s and 3 2p orbitals are engaging in bonding regardless of whether the ultimate structure contains a single or double bond; however, in the hybridization model, we must first look at the result and then say, "Oh, because there is a double bond, we must have had $sp^2$ orbitals. Once we can assign a "hybridization", we can make some predictions about the chemistry, I agree. $\endgroup$ – bobthechemist Dec 29 '13 at 13:44
  • $\begingroup$ That's fair. Granted, in MO theory, all atomic orbitals are used in constructing the MOs, not a subset. Still, AFAIK, an assumption (or prior knowledge) of the structure of the molecule must be made. E.g., given three atoms, I can create a plausible molecule which is either linear, bent, or cyclic. The choice will dictate symmetry elements, which in turn determines the valid combinations of AOs. Of course, MO theory will better predict the stability/properties of the resulting structures. In that sense, I would agree. Is there anything I'm missing? $\endgroup$ – Greg E. Dec 29 '13 at 14:47
  • $\begingroup$ It has to do with the order of events. In MO theory, there are tools to determine whether or not orbitals will mix (relative energies and symmetry) before creating a structure. I need to think about a practical example, but that will take a little more coffee.. $\endgroup$ – bobthechemist Dec 29 '13 at 14:55
  • $\begingroup$ I would like to add something to this discussion about whether hybridization is predictive. I recently came across a molecule called Propadiene. Looking at the 2D representation may lead one to assume that the two pairs of hydrogen atoms can rotate with respect to each other. In fact, the 2 Hydrogens on one side are fixed relative to the Hydrogens on the other side, due to the issue of availability of p-orbitals for making pi bonds. $\endgroup$ – Greg Feb 1 '14 at 1:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.