• First statement:

    In the molecular orbital theory, the valency electrons are considered to be associated with all the nuclei in the molecule. - Concise Inorganic Chemistry by J.D.Lee.

  • Second statement:

    In molecular orbital theory, electrons are treated as spreading throughout the entire molecule: every electron contributes to the strength of every bond. - Elements of Physical Chemistry by Peter Atkins.

Two different sources are actually saying the same thing in regard to MO theory: electron density is spread over the entire molecule, okay that is quite right & agreeable.

But the way the fact is presented as if it were the sole property of MO theory. Doesn't the electron density span over the entire molecule in VB theory also?

Check this pic taken from Peter Atkin's above mentioned book:

enter image description here

This is the electron density in $\ce H_2$ as predicted by VB theory; notice how the resultant wavefunction after bonding is spread over the entire molecule & this implies the electron density is spread over the entire molecule. So, even in VB theory, the electron density is spread over the nuclei of the molecule. So, why those special writings above?

Why is it that the spreading of electron density over the nuclei of the molecules specially in MO theory? Is not this exhibited even in VB theory also? Or am I missing something?

Can anyone please help me whether the spreading of the electron density is the sole property of Mo theory as is put up by the two authors? Is not it exhibited in VB theory also?

  • 2
    $\begingroup$ Aren't your other questions asking the same thing? chemistry.stackexchange.com/q/33879/4945 chemistry.stackexchange.com/q/33951/4945 $\endgroup$ Dec 27, 2015 at 10:16
  • $\begingroup$ @Martin-マーチン: Same thing? How can they be?? Of course, they are related to each other, but how can it be the same thing? $\endgroup$
    – user5764
    Dec 27, 2015 at 10:18
  • $\begingroup$ @Martin-マーチン: At one, the difference between the two theories is highlighted while the later deals why resonance isn't required for MO theory. Here, I'm asking whether the property of spreading the electron density over the entire molecule is solely exhibited in MO theory or as I am saying, also exhibited in VB theory also. Of course, I would appreciate if you elaborate how the previous queries have been reflected again here. $\endgroup$
    – user5764
    Dec 27, 2015 at 10:22
  • $\begingroup$ To be clear: the electron density does spread over the whole molecule in both models. As ifilot's answer describes, the difference is in whether all AO's potentially contribute to a given MO, or whether just AO's on neighboring atoms are used to construct that MO. $\endgroup$
    – hBy2Py
    Dec 27, 2015 at 12:10

1 Answer 1


In MO theory, you construct a molecular orbital by a linear combination of atomic orbitals. Because those atomic orbitals are part of different atoms, in this way, the valence electrons are associated with all the atoms in the molecule. So even if you consider a molecular orbital that is bonding between, let's say, atom 1 and atom 2 in the molecule, that molecular orbital has also contributions from all the other atomic orbitals (besides those that arise from atom 1 or atom 2).

In contrast, in VB theory, a bond between atom 1 and atom 2 can only have atomic orbitals of atom 1 and atom 2. The bond between two atoms is then considered to arise from two weakly coupled atomic orbitals. I have to point out though that modern valence bond theory now complements molecular orbital theory and no longer adheres to the valence bond idea that electron pairs are localized between two specific atoms in a molecule but that they are distributed in sets of molecular orbitals which can extend over the entire molecule.

Given the discussion below, I would like to add to my answer a simple example. Let us consider the $\ce{H2O}$ molecule. Within the framework of Hartree-Fock and using a STO-3G basis set, the $2\sigma$ orbital of water can be expanded in terms of the atomic orbitals as follows: $$ \psi_{2 \sigma} = (-0.4330)\; \phi_{\textrm{O},1s} + (-0.4686)\; \phi_{\textrm{O},2s} + (0.2369)\; \phi_{\textrm{O},2p_y} + (0.2685)\; \phi_{\textrm{H1},1s} + (0.2685)\; \phi_{\textrm{H2},1s}. $$

From this expansion you can see that multiple atomic orbitals contribute to the same molecular orbital. This is a molecular orbital that forms, in a way, a multi-atom bond between O and the two H atoms. You could also say that this molecular orbital contributes to both of the O-H bonds (in fact, more molecular orbitals than just the $2\sigma$ do).

  • $\begingroup$ which do not arise from atom 1 or atom 2- what? Could you please clarify that? $\endgroup$
    – user5764
    Dec 27, 2015 at 11:26
  • $\begingroup$ I was not very clear here. I have edited my answer. I meant to say: "besides those that arise from atom 1 and atom 2" $\endgroup$
    – Ivo Filot
    Dec 27, 2015 at 11:28
  • $\begingroup$ You meant to say there are contributions of atomic orbitals which are not of 1 & 2? $\endgroup$
    – user5764
    Dec 27, 2015 at 11:37
  • 1
    $\begingroup$ Any molecular orbital within MO theory is a linear combination of atomic orbitals. In mathematical terms: $\psi_{i}=\sum_{j} c_{j} \phi_{j}$ where $\phi_{j}$ are the atomic orbitals, $c_{j}$ the corresponding coefficients and $\psi_{i}$ the molecular orbital. $j$ runs over all the atomic orbitals and not only the orbitals of the atoms to which the bond adheres. So for the description of a molecular orbital, you use the complete basis set that consists of all atomic orbitals in that molecule. Hence, all these atomic orbitals contribute. $\endgroup$
    – Ivo Filot
    Dec 27, 2015 at 11:56
  • 1
    $\begingroup$ @user36790 Because these examples are diatomic molecules--the only atoms the AO's can come from are necessarily participating in a bond. For something larger, say, ethane, a given $\sigma$ bond MO may have contributions from AOs centered on atoms all over the molecule. $\endgroup$
    – hBy2Py
    Dec 27, 2015 at 12:03

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