Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Can anyone give me a quantum mechanical explanation of the theory of valence? (i.e. why atoms bond just enough to have a complete orbital)

EDIT: To clarify, I already have an idea of why atoms bond, and the molecular orbit makes sense to me. The problem is that the valence bond theory provides a simple and fairly accurate description for when a molecule is stable. To make this more concrete: why is CH_4 more stable than CH_3 or CH_5. Looking up the former on wikipedia, I see a better way of phrasing the question might be: why are free radicals reactive? (Should I change the title?)

share|improve this question
    
It has to do with the Lie Algebra of the angular momentum operator for the electrons circling the nuclei. My advice is to pick up any decent book on QM and read the section on angular momentum. –  Matt Calhoun Mar 2 '11 at 7:03

2 Answers 2

All these rules are approximate, and all of them follow from Schrödinger's equation that offers a much more accurate, quantitative result.

Bonds in molecules may arise not just from "valence bonds", which remain localized in atoms, but also from "molecular orbitals" which are localized across the whole molecule. The latter kind of bond is more general and the methods to study it are newer.

Back to the valence bonds. Energetically, the bonds occur because they allow the pair of atoms to reduce its energy. Just like for individual atoms, pairs of atoms are more stable when a whole orbital is filled because the other states that could be filled are separated by a gap and they have a higher energy.

So whenever it's possible to divide the electrons among the pair of atoms so that both atoms have full shells, then all the electrons with low enough energy have been added, to minimize the ratio of energy per electron, and the addition of an extra electron would increase the energy by a much higher amount, which suggests that the previous full-shell solution is at least a local minimum of the energy per electron.

Because the interactions between the electrons themselves are complicated, this superficial wording can't really replace a calculation - quantum mechanics for many electrons, which deals with a $3N$-dimensional wave function. In fact, in this picture, even in the Hartree-Fock approximation, many other emergent phenomena occur such as "orbital hybridization" - the emergence of totally new kinds of orbitals that are appropriate to describe the molecule.

share|improve this answer

I will try to give a very qualitative explanation of chemical bonds like the so called covalent bonds. Let's take the example of hydrogen molecule where two hydrogen atoms share two electrons. Now, if the two nuclei of the two hydrogen atoms are separated by a distance, there is a potential barrier between them. Classically the electrons can not cross that barrier, but quantum mechanics allows them to cross it due to tunneling effect. The electrons therefore wander back and forth continuously through the barrier and each of them spends almost equal times in close proximity of the two nuclei. Now quantum mechanical calculation will show that the potential energy of this system is much lower than that of the system if they were far apart. There is an optimum distance between the nuclei when the potential energy of the system is minimum. Naturally the system at that distance will be most stable. This is how covalent bond forms. In other molecules it may be the case that the two atoms are different. In that case electrons spend a little more time to the proximity of the relatively positive nucleus at the most stable (i.e. least potential energy) configuration.

There are other kinds of bonds as well. The mathematical treatments for them differ but each follows from the rules of quantum mechanics and in all types of bonds the common thing is, the potential energy of the system has to be minimum for the stable configuration.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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