What would be the physical meaning of the difference between the MO wave function and VB wave function of $\ce{H2}$?
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2$\begingroup$ What do you mean by physical meaning? Molecular orbital theory and valence bond theory are two approaches to the electronic structure of a molecule. Carried out to infinity with the same prerequisites they are identical. $\endgroup$– Martin - マーチン ♦Commented Oct 23, 2015 at 5:46
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$\begingroup$ I mean if you took the actual wave functions of MO and of VB for H2 and subtracted them mathematically, you'll end up with a mathematical expression which represents some sort of physical meaning. Assuming not to infinity, what would this difference represent? $\endgroup$– Plex ASMCommented Oct 23, 2015 at 6:18
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2$\begingroup$ Both theories are mathematical representations of the same thing. If there are any differences they are due to approximations or incompleteness and they do not have any meaning whatsoever. $\endgroup$– Martin - マーチン ♦Commented Oct 23, 2015 at 6:59
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3$\begingroup$ chemistry.stackexchange.com/a/38144/186 $\endgroup$– WildcatCommented Oct 23, 2015 at 9:42
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I'll give a somewhat simplistic answer. VB assumes that all bonding is between only two atoms. MO bonding considers bonding as a molecular phenomenon.
In $\ce{H2}$ there isn't really much difference. However consider benzene, $\ce{C6}\ce{H6}$. VB theory would have three double carbon bonds and three single carbon bonds which isn't really true. So VB bonding then hybridizes over the six carbon atom ring to predict the aromatic character of the molecule. However in MO theory the aromatic character is fundamental to the model.
From wikipedia
"In particular, valence bond theory correctly predicts the dissociation of homonuclear diatomic molecules into separate atoms, while simple molecular orbital theory predicts dissociation into a mixture of atoms and ions. For example the molecular orbital function for dihydrogen is an equal mixture of the covalent and ionic valence bond structures and so predicts incorrectly that the molecule would dissociate into an equal mixture of hydrogen atoms and hydrogen positive and negative ions."
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$\begingroup$ VB and MO will predict the same structure as the wave functions are essentially the same, if the description of complete. Differences do not carry any physical meaning and this is what the question is about. $\endgroup$– Martin - マーチン ♦Commented Oct 24, 2015 at 19:12
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1$\begingroup$ Also hybridization is part of VB not MO theory and VB theory does not predict alternating single and double bonds for benzene, it makes use of the concept of resonance to suggest a structure that is essentially the same as that suggested by MO. $\endgroup$– bonCommented Oct 24, 2015 at 19:31
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$\begingroup$ @Bon - You're right about hybridization being part of Vb theory. I fixed that part. $\endgroup$– MaxWCommented Oct 24, 2015 at 19:46
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$\begingroup$ @Martin - "VB wave function" seems odd to me. (Granted my chemistry degree was earned when the four elements were air, water, earth and fire.) To me VB theory is more about the linear combination of atomic orbitals. $\endgroup$– MaxWCommented Oct 24, 2015 at 19:50
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$\begingroup$ Well, the aromatic character of benzene is not exactly "fundamental" to MO theory - it is actually derived from the model. You can draw MO diagrams for the $\pi$ system of benzene without actually knowing that is it aromatic, and figure out from the diagrams that it is aromatic - which is the origin of Huckel's famous $(4n+2)$ rule... I mean, it's probably really just how you look at it. If you can get your hands on it, see Ian Fleming's Molecular Orbitals and Organic Chemical Reactions for a description. $\endgroup$ Commented Oct 24, 2015 at 19:54