Related is this question: https://chemistry.stackexchange.com/questions/2586/molecular-and-atomic-orbitals-and-antibonding.

In my textbook on organic chemistry they represent an orbital as a 2d wave function to explain the formation of bonding and anti-bonding molecular orbitals (I displayed it here as a simple $\sin x$ function, just for illustration. I understand real wave function may be quite different).

Example wave function, sin x

They say the bonding atomic orbitals can either combine in phase, resulting in a bonding molecular orbital (the electron density becomes larges between the molecules, creating an electrostatic astringent force exerted by the electron in the center on the nuclei of the combining atoms), here represented as $2\sin x$

Example bonding molecular orbital wave function, 2sin x

or combine out of phase resulting in a nodal plane between the nuclei: an anti-bonding molecular orbital. Here represented as $\sin x + \sin (x+\pi)$

Example anti-bonding molecular orbital wave function, sin x + sin (x + pi)

Now the question: how can a single orbital, that is either in phase or out of phase with the orbital of the bonding atom (i.e. the other atom) form both a bonding molecular orbital and an anti-bonding molecular orbital, while it is clear that an orbital is either in one phase or an other and not both?

Related is, what this 2d diagram actually represents (related question, but not quite the same: https://chemistry.stackexchange.com/questions/12397/meaning-of-depiction-of-atomic-orbitals).

Please ask if anything is unclear. Your help is appreciated!

  • $\begingroup$ I think the crux of the question is that the orbital must be either in-phase or out-of-phase. I don't think this is the case at all. Unfortunately, I haven't thought about this question in a while, so I don't have a good answer for you at the moment as to why that is. $\endgroup$
    – chipbuster
    Commented Jun 16, 2014 at 14:39

2 Answers 2


Perhaps a good way to understand this is through the example of the $H_2$ molecule. Before delving into the molecule, however, let us examine the Hydrogen atom more closely. The $1s$ state of the Hydrogen atom is filled by one electron, but it can take two at most because of spin. Now, the other Hydrogen atom also has one atom in the $1s$ state and can again accommodate at most two.

If we were to bring the two atoms closer together, and then ultimately form a molecule, do you think that the number of states changes? Of course not! The molecule again must be able to accommodate a total of four electrons (two from each atom). (We are ignoring higher energy states for now).

When we think about the Hydrogen molecule, there is a very strong constraint on what the wavefunction must look like. This is because the Hydrogen molecule possesses parity or reflection symmetry. This means that when I switch the atoms, I must either get a $+$ sign or a $-$ sign on the wavefunction. (See the figure below.)

h2 molecule mirror plane
(source: tu-bs.de)

To be more explicit, let's represent the $1s$ states on the Hydrogen atoms 1 and 2 when they are far apart by $|1s_{1}\rangle$ and $|1s_{2}\rangle$. When we bring the atoms close together, the reflection symmetry constraint imposes a severe constraint. Let's see how we can combine the wavefunctions of the individual atoms to satisfy this constraint. It turns out that the following combinations work quite well:

\begin{equation} |\psi_B\rangle = |1s_1\rangle + |1s_2\rangle \end{equation} \begin{equation} |\psi_A\rangle = |1s_1\rangle - |1s_2\rangle \end{equation}

where $B$ and $A$ denote bonding and anti-bonding respectively. Now look what happens when I switch the places of the two molecules by applying the reflection operation, $R$:

\begin{equation} R|\psi_B\rangle = |1s_2\rangle + |1s_1\rangle = +|\psi_B\rangle \end{equation} \begin{equation} R|\psi_A\rangle = |1s_2\rangle - |1s_1\rangle = -|\psi_A\rangle \end{equation}

This set of wavefunctions therefore satisfies the constraint that we set by imposing reflection symmetry. Now it turns out that because the reflection symmetry, the wavefunctions must look something like this:


Again, each orbital can accommodate two electrons for a total of four, exactly the number of states we started out with. As a rule of thumb, the energy of the orbital with the greater curvature is usually the more energetic one, which is the anti-bonding one in this case. Therefore, the bonding orbital will be filled by the two electrons.

I think that the way you are trying to visualize the problem is flawed. You should not think of the standing wave as a sine wave extending through space infinitely. It is really cut off by the potential of the hydrogen atom and looks like the red and green curves above for the individual atoms. When you bring them closer together and they mix you get the purple and blue curves. Hope this helps!

  • $\begingroup$ beautiful answer! $\endgroup$ Commented Dec 16, 2023 at 4:57

I think you may be asking yourself the wrong question. It may be better to ask the more general questions of (1) why the electron orbital shapes are what they are, and (2) how do we mathematically described them?

The electron orbital (or wavefunction) is the standing wave setup by boundary conditions. So, try to get a good picture of standing wave patterns. Check out these cool java apps:

one dimensional standing waves: http://www.grandinetti.org/Teaching/Chem121/Lectures/QMMatter/StringWaveDemo

two dimensional standing waves: http://www.grandinetti.org/Teaching/Chem121/Lectures/QMMatter/CircularMembraneWaveDemo

three dimensional standing waves: http://www.grandinetti.org/Teaching/Chem121/Lectures/QMMatter/HydrogenAtomOrbitalDemo

Next, realize that molecular orbitals are the standing wave patterns for the electron wavefunction when it is attracted to multiple positively charged nuclei. How do you mathematically described these standing wave patterns? Chemist do this with linear combinations of atomic orbitals.

Check out these notes for more details:


  • $\begingroup$ I really appreciate your helpful links and explanation, but it is kind of beyond be (at least currently). I'm a hobbyist organic chemist following Clayden's work in my free time. Now Clayden gives a brief and simple introduction to VSEPRT (VSEPR theory) and kind of assigns these things to be true. I don't think his intention was actually that the reader would wonder too deep about what was actually explained. Being myself, however, I couldn't resist wondering these seemingly inconsistent statements. (continued) $\endgroup$
    – Jori
    Commented Jun 9, 2014 at 16:09
  • $\begingroup$ As far as he explains it, orbitals are solutions to the Schrödinger equation \Psi for different quantum numbers. Such a solutions describes the electron density in 3d space. Then later he shows these diagrams, without explaining them further. My difficulty with these diagrams is 1) that I cannot imagine how these diagrams represent the 3d solution to the \Psi and what it even means for an orbital to be in phase or out of phase, 2) how an orbital can be both in phase and out of phase at the same time to form bonding and anti-bonding molecular orbitals. I hope I'm a bit clearer now :-) . $\endgroup$
    – Jori
    Commented Jun 9, 2014 at 16:15
  • $\begingroup$ The atomic orbitals are the standing wave pattern for the electron's wavefunction when it is bound to a single positively charged nucleus. When the electron is bound to two positively charged nuclei the atomic orbitals (wavefunctions) are no longer mathematical solutions to the Schrodinger equation. The new solution is the molecular orbital. Chemists "pretend" that the molecular orbitals can be written as linear combinations of atomic orbitals. It gives them a good approximate solution. So, I would suggest you approach the problem differently. $\endgroup$
    – pjg
    Commented Jun 10, 2014 at 16:35
  • $\begingroup$ Ask yourself, how should the molecular orbitals look? Remember that the lowest energy orbital will have the longest wavelength and the fewest nodes. The next highest energy orbital will have a shorter wavelength and an additional node. So, imagine bring two atomic s orbitals together to make an ellipsoidal orbital with the two nuclei at the ellipse's focal points. That would be a good start for the lowest energy molecular orbital. $\endgroup$
    – pjg
    Commented Jun 10, 2014 at 16:43
  • $\begingroup$ Now consider the next higher energy molecular orbital. It has to have a shorter wavelength. You'll want to have a nodal surface in the wavefunction. That means one side will have a positive phase and the other side has a negative phase. How are you going to make that orbital? Just bring together two s orbitals, but this time make one 180 degrees out of phase. When they come together you'll get the desired MO with the nodal surface inbetween the nuclei. $\endgroup$
    – pjg
    Commented Jun 10, 2014 at 16:46

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