In molecular orbital theory, the fact that a bonding and antibonding molecular orbital pair have different energies is accompanied by the fact that the energy by which the bonding is lowered is less than the energy by which antibonding is raised, i.e. the stabilizing energy of each bonding interaction is less than the destabilising energy of antibonding. How is that possible if their sum has to equal the energies of the combining atomic orbitals and conservation of energy has to hold true?

"Antibonding is more antibonding than bonding is bonding."

For example, the fact that $\ce{He2}$ molecule is not formed can be explained from its MO diagram, which shows that the number of electrons in antibonding and bonding molecular orbitals is the same, and since the destabilizing energy of the antibonding MO is greater than the stabilising energy of bonding MO, the molecule is not formed. This is the common line of reasoning you find at most places.

  • $\begingroup$ Usually, when atoms form bounds they release energy (heating, radiation etc). So, the energy conservation law holds, but this kind of sum rule does not work. Vice versa, one has to spend energy to get dissociation of the molecule. $\endgroup$
    – freude
    May 30, 2013 at 12:31
  • $\begingroup$ stabilizing energy of each bonding is less than the destabilising energy of antibonding. Now how is that possible if their sum has to equal the energies of the combining atomic orbitals has it? $\endgroup$
    – Alex
    Jul 29, 2013 at 21:31
  • 1
    $\begingroup$ The reason is because there is a term for AO overlap that appears in the normalization constant: $\frac{1}{\sqrt{2-2S}}$ vs. $\frac{1}{\sqrt{2+2S}}$, or something. It will take some time to write it up, as it is a mathematical derivation. $\endgroup$
    – Eric Brown
    Sep 7, 2013 at 3:53
  • $\begingroup$ @EricBrown Non mathematical but an intuitive reason would do. $\endgroup$ Sep 10, 2013 at 3:08
  • 3
    $\begingroup$ related chemistry.stackexchange.com/questions/39431 $\endgroup$ Sep 9, 2016 at 7:20

5 Answers 5


Mathematical Explanation

When examining the linear combination of atomic orbitals (LCAO) for the $\ce{H2+}$ molecular ion, we get two different energy levels, $E_+$ and $E_-$ depending on the coefficients of the atomic orbitals. The energies of the two different MO's are: $$\begin{align} E_+ &= E_\text{1s} + \frac{j_0}{R} - \frac{j' + k'}{1+S} \\ E_- &= E_\text{1s} + \frac{j_0}{R} - \frac{j' - k'}{1-S} \end{align} $$

Note that $j_0 = \frac{e^2}{4\pi\varepsilon_0}$, $R$ is the internuclear distance, $S=\int \chi_\text{A}^* \chi_\text{B}\,\text{d}V$ the overlap integral, $j'$ is a coulombic contribution to the energy and $k'$ is a contribution to the resonance integral, and it does not have a classical analogue. $j'$ and $k'$ are both positive and $j' \gt k'$. You'll note that $j'-k' > 0$.

This is why the energy levels of $E_+$ and $E_-$ are not symmetrical with respect to the energy level of $E_\text{1s}$.

Intuitive Explanation

The intuitive explanation goes along the following line: Imagine two hydrogen nuclei that slowly get closer to each other, and at some point start mixing their orbitals. Now, one very important interaction is the coulomb force between those two nuclei, which gets larger the closer the nuclei come together. As a consequence of this, the energies of the molecular orbitals get shifted upwards, which is what creates the asymmetric image that we have for these energy levels.

Basically, you have two positively charged nuclei getting closer to each other. Now you have two options:

  1. Stick some electrons between them.
  2. Don't stick some electrons between them.

If you follow through with option 1, you'll diminish the coulomb forces between the two nuclei somewhat in favor of electron-nucleus attraction. If you go with method 2 (remember that the $\sigma^*_\text{1s}$ MO has a node between the two nuclei), the nuclei feel each other's repulsive forces more strongly.

Further Information

I highly recommend the following book, from which most of the information above stems:

  • Peter Atkins and Ronald Friedman, In Molecular Quantum Mechanics; $5^\text{th}$ ed., Oxford University Press: Oxford, United Kingdom, 2011 (ISBN-13: 978-0199541423).
  • 5
    $\begingroup$ I highly recommend using a different textbook (maybe any other - If you would like a list of good reads, I am happy to compile it). The energy of the orbitals has nothing to do with nuclear repulsion. Hence your given equations describe the ground and the exciting state. The generality of this example should also be disputed very much. It might hold for simple diatomic molecules, but in general MO are formed from more than only two AO. $\endgroup$ Apr 8, 2014 at 11:38
  • 5
    $\begingroup$ @Martin I tried to taylor this answer specifically for diatomic molecules. And maybe I should have left it more fuzzy in terms of coulomb interactions (because those between the electrons will have a much larger impact). However, an intuitive explanation for the asymmetry was asked for, so I tried to provide one. The way this site works, you should try to provide a better answer, which then gets upvoted. In the best case, mine gets un- and yours accepted. $\endgroup$
    – tschoppi
    Apr 10, 2014 at 7:14
  • $\begingroup$ @Martin-マーチン I would be interested by two or three (or more if you want) items from your list of good reads. $\endgroup$
    – The Quark
    Feb 4, 2023 at 10:19

The sum of the energies over all orbitals is constant. The lower energy in a bonded state arises from the fact that not all MOs are occupied by electrons. Because the energy of the electrons is lowered, the total system energy is lowered by bonding. See http://www.dlt.ncssm.edu/tiger/diagrams/bonding/BondingEnergy.gif for an illustration.

Edit: This phenomenon is due to the nuclear-nuclear repulsion of the two (or more) atoms that are binding. See Ball, D.W. Physical Chemistry p. 408


There is no such thing as conservation of orbital energies at different fixed values of $R$, the internuclear distance. Obviously, at $R \rightarrow \infty$ the energies are the same regardless of the linear combinations you use (there is no interaction and the "molecular" energy is exactly the same as the sum of the "atomic" energies); obviously too, the energies raise a lot if you make $R$ very small: you have the fusion barrier that doesn't allow you to convert your initial atomic orbitals of atoms A and B into the atomic orbitals of atom AB. Energy is obviously conserved in a dynamical process where two atoms approach and you have kinetic energy in the collisional coordinate and so on and so forth (you can immediately see that two atoms cannot form a diatomic molecule is something else does not take away energy from the system).


Another possible (and slightly more simplistic) explanation given by Fleming (2009) is based on inter-electronic repulsion, when describing the molecular orbitals of the $\ce {H2}$ molecule:

In addition, putting two electrons into a bonding orbital does not achieve exactly twice the energy-lowering of putting one electron into it. We are allowed to put two electrons into the one orbital if they have opposite spin, but they still repel each other, because they have the same sign and have to share the same space; consequently, in forcing a second electron into the $\sigma$ orbital, we lose some of the bonding we might otherwise have gained.

However, note that the author cited this as an add-on. The primary reason, as discussed in greater detail by the author, is still what was mentioned by orthocresol here. Also, this does not explain the case for the interaction of two unfilled orbitals since the absence of any electrons would mean that there is no inter-electronic repulsion.


Fleming, I. (2009). Molecular Orbitals and Organic Chemical Reactions. United Kingdom: John Wiley & Sons, Ltd.


Here is an explanation that, I believe, solves most of the problems raised by others on this thread.


When we bring the atoms closer from infinity, electrostatic repulsions come into play between the electrons of the two atoms (as Martin - マーチン pointed out under tschoppi's answer, the energy of the orbitals has nothing to do with nuclear repulsion). These inter-electronic repulsions increase the potential energy of the system.

Since the energy increases, the average energy of the bonding and anti-bonding molecular orbitals also increases. Thus, the bonding molecular is not lowered in energy (on the molecular orbital diagram) to the extent expected, and the anti-bonding molecular orbital is raised in energy to a higher extent than expected. That is why "antibonding is more antibonding than bonding is bonding."

(Somewhat) Mathematical Explanation

For simplicity, let us assume a symmetrical, diatomic molecule $X_2$ (we do not know how many electrons are present in it). Let us also consider only the $1s$ atomic and molecular orbitals.

When we talk of the energies of atomic orbitals, we talk of atoms ($X$'s) at infinite separation. Let the energy of the $1s$ atomic orbital of $X$ be $E_X$. Then the total energy of the $1s$ atomic orbitals in two $X$ atoms is

$$E_i = 2E_X$$

However, when we talk of the energies of molecular orbitals, we must remember that the atoms are now a finite distance away. Let us consider the bonding and anti-bonding molecular orbitals separately.

Let the ideal/expected energy (without taking into account the inter-electronic repulsion) of the $\sigma 1s$ molecular orbital be $E_1$. However, there will be inter-electronic repulsions present in the bonding molecular orbital; at least, more so than at infinite separation of atoms. Let the extra energy due to repulsions be $E_{e_1}$. Then the energy of the $\sigma 1s$ molecular orbital will be

$$E_\sigma = E_1 + E_{e_1}$$

Again, let the ideal/expected energy of the $σ^* 1s$ molecular orbital be $E_2$. However, there will still be inter-electronic repulsions present in the bonding molecular orbital; at least, more so than at infinite separation of atoms (we do not need to know whether the repulsions are greater in the bonding or anti-bonding molecular orbital; just that they exist). Let the extra energy due to repulsions be $E_{e_2}$. Then the energy of the $σ^* 1s$ molecular orbital will be

$$E_{\sigma ^*} = E_2 + E_{e_2}$$

Therefore, the total energy of the $1s$ molecular orbitals is

$$E_f = E_\sigma + E_{\sigma ^*} = (E_1 + E_{e_1}) + (E_2 + E_{e_2}) = (E_1 + E_2) + (E_{e_1} + E_{e_2})$$

The term $(E_{e_1} + E_{e_2})$ arises due to the change in distance between atoms, i.e. from infinite to finite separation. Therefore it is the change in potential energy of the system, or the work done in bringing the two atoms to the final finite distance.

$$W = E_{e_1} + E_{e_2}$$

Now, we can apply the law of conservation of energy:

$$E_i + W = E_f$$ $$2E_X + (E_{e_1} + E_{e_2}) = E_\sigma + E_{\sigma ^*}$$ $$E_\sigma + E_{\sigma ^*} - 2E_X = E_{e_1} + E_{e_2} > 0$$ $$E_\sigma + E_{\sigma ^*} > 2E_X$$ $$\frac{E_\sigma + E_{\sigma ^*}}2 > E_X$$

Therefore, the average energy of the bonding and anti-bonding molecular orbitals is greater than that of the atomic orbitals.

To put it another way,

$$E_\sigma + E_{\sigma ^*} > 2E_X$$ $$E_{\sigma ^*} - E_X = E_X - E_\sigma$$

Therefore, "antibonding is more antibonding than bonding is bonding."

Note that had we not considered inter-electronic repulsions, i.e. the work done to bring the atoms closer, we would find the anti-bonding molecular orbital to be "just as anti-bonding" as the bonding molecular orbital is "bonding". This is the mistake made in applying the law of conservation of energy directly.

As for the case of unfilled orbitals (pointed out by Tan Yong Boon in their answer), unfilled orbitals don't really "exist" until they are filled by electrons. Orbitals have energy only due to the electrons in them, and thus we cannot say anything about the energy of unfilled orbitals (the inter-electronic repulsion idea also cannot be used here).


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