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!


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  • $\begingroup$ 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 Jun 16 '14 at 14:39

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?


In the molecular orbital theory, the valence electrons are considered to be associated with all the nuclei in the molecule.Thus the atomic orbitals from different atoms be centered to combined to produce molecular orbitals.

Electrons may be considered either as particles or waves. An electron in an atom may therefore be described as occupying an atomic orbital, or by a wave function $\psi$, which is a solution to the Schrodinger wave equation.Electrons in a molecule are said to occupy molecular orbitals. The wave function describing a molecular orbital may be obtained by one of the two procedures:

  • Linear Combination of Atomic Orbitals(LCAO)
  • United Atom Method


Consider two atoms A and B which have atomic orbitals described by the wave functions $\psi_A$ and $\psi_B$. If the electron clouds of these two atoms overlap when the atoms approach, then the wave function for the molecule( molecular orbital $\psi_{AB}$) can be obtained by a linear combination of the atomic orbitals $\psi_A$ and $\psi_B$: $$\psi_{AB}=N(c_1\psi_A+c_2\psi_b)$$ where N is a normalizing constant choosen to ensure that the probability of finding an electron in the whole of the space is unity, and $c_1$ and $c_2$ are constants choosen to give a minimum energy for $\psi_{AB}$.If atoms A and B are similiar, then $c_1$ and $c_2$ will have similiar values.If atoms A and B are the same, then $c_1$ and $c_2$ are equal.

The probability of finding an electron in a volume of space $dv$ is $\psi^2dv$. So the probability density for the combination of the two atoms as above is related to the wave function squared: $$\psi_{AB}^2=(c_1^2\psi_A^2+2c_1c_2\psi_A\psi_B + c_2^2\psi_B^2)$$ If we examine the three terms on the right of the equation, the first term $c_1^2\psi_A^2$ is related to the probability of finding an electron on atom A if A is an isolated atom. The third term $c_2^2\psi_B^2$ is related to the probability of finding an electron on atom B if B is an isolated atom. The middle term becomes increasingly important as the overlap between the two atomic orbitals increases, and this term is called the overlap integral. This term represents the main difference between the electron clouds in the individual atoms and in the molecule. The larger this term the stronger the bond.

example, s-s combination of orbitals

Suppose the atoms A and B are hydrogen atom; then the wave functions $\psi_A$ and $\psi_B$ describe the 1s atomic orbitals on the two atoms. Two combinations of the wave function $\psi_A$ and $\psi_B$ are possible: - Where the signs of the two wave functions are the same. - Where the signs of the two wave functions are different.

f one of the wave function $\psi_A$ is arbitarily assigned a +ve sign, the other may be +ve or -ve.) Wave functions which have the same sign may be regarded as waves that are in phase, which when combined add up to give a larger resultant wave.Similiarly wave functions of different signs correspond to waves that are completely out of phase and which cancel each other by destructive interference.

$$\large\text{The signs + and - refer to the signs of the wave functions, which}\\ \large\text{ determine their symmetry, and have nothing to do with the}\\ \large\text{electrical charges.}$$

The two combinations are: $$\psi_g=N(\psi_A+\psi_B)$$, and $$\psi_u=N(\psi_A+(-\psi_B))\equiv N(\psi_A-\psi_B)$$ $$\large\text{The latter equation should not be regarded as the }\\ \large\text{summation of the wave functions and not the}\\ \large\text{ mathematical difference between them}$$

When a pair of atomic orbitals $\psi_A$ and $\psi_b$ combine, they give rise to a pair of molecular orbitals $\psi_g$ and $\psi_u$. The number of molecular orbitals produced must be always equal to the number of atomic orbitals invovled.The function $\psi_g$ leads to increases electron density in between the nuclei, and is therefore a bonding orbital. It is lower in energy than the original orbitals. Conversly $\psi_u$ results in two lobes of opposite sign cancelling and hence giving zero electron density in between the nuclei. This is an antibonding molecular orbital which is higher in energy .

Atomic orbitals may hold up to two electrons(provided that they have opposite spins) and the same applies to molecular orbital .In case of two hydrogen atoms combining, there are two electrons to be considered, one from the 1s orbital of atom A and one from the 1s orbital of atom B.When combined, these two electrons both occupy the bonding molecular orbital $\psi_g$. It is only because the system is stabilized in this way that a bond is formed.

Consider the hypothetical case of two He atoms combining. The 1s orbitals on each He conatin two electrons, making a total of four electrons to put into molecular orbitals.Two of the electrons occupy the bonding MO, and two occupy the antibonding MO.Thus overall there is no saving of energy so $He_2$ is not formed.

You are misconceptually trying to mix orbitals and electrons. Two electrons will always form only one of the bonding, anti-bonding or non-bonding orbitals, whereas the whole orbital may form not just one.

If you want to ask about these graphs ask another question.

  • $\begingroup$ sorry, misunderstood, rm comment $\endgroup$ – Martin Jun 21 '14 at 8:30
  • $\begingroup$ I still see the sum of one wave function squared and some other wave function squared, that is either positive (bonding) or zero (anti-bonding). How would to atomic orbital wave function combine with different result? I'm sorry for all the questions, but I have absolutely no background in Schrodinger equation or quantum mechanics, I just stumbled upon this rather unclear concept in my organic chemistry book. ($\sin x$ was just what the book seemed to show. I think they only meant the middle part or something) $\endgroup$ – Jori Jun 21 '14 at 15:20
  • $\begingroup$ Perhaps I even have a better idea, I'll include those pages where they explain it. Perhaps that will make my difficulties more clear. $\endgroup$ – Jori Jun 21 '14 at 15:23
  • $\begingroup$ Here is an excerpt from the book I'm working with: s000.tinyupload.com/?file_id=02891213342135625346 (pdf format). $\endgroup$ – Jori Jun 21 '14 at 15:35
  • $\begingroup$ @Jori Are you satisfied with detail or I should cut short it, it was too much so please ask about graphs in another question. $\endgroup$ – RE60K Jun 22 '14 at 14:24

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 Jun 9 '14 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 Jun 9 '14 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 Jun 10 '14 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 Jun 10 '14 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 Jun 10 '14 at 16:46

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!


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