why in a covalent bond are "the bonded electrons are in a lower energy state than if the individual atoms held them at the same proximity"?

Also is it correct that " I think when you start pushing two molecules together orbitals between the two start overlapping - forming covalent bonds?"

Essentially why are covalent bonds made? A QM description would be nice but I don't really know QM so a somewhat reduced-to-classical explanation of the QM explanation would also be nice!


4 Answers 4


The covalent bonds form when electrons attached to nearby nuclei can exist in a superposition state where they can partly be on another nucleus. This happens when the electron state they are mixing with is unfilled.

For example, for H2, two hydrogen nuclei are close, there is no electrostatic energy for this in the first approximation because the electron and proton are both spherical electromagnetic sources. But when they come close, and the spin of the two electron is opposite (this is required for binding), each electron will spread out to cover the other atom, overlapping with the other electron (this is allowed because they have opposite spin, and so do not feel Pauli exclusion), and this reduces the kinetic energy of the electron.

The reason is simply that when you allow an electron to wander over a larger space, the kinetic energy always goes down. If you double the size of the space in one direction, the kinetic energy in that direction goes down by a factor of 4. If you consider the two H-atoms as two boxes, doubling the x-size of the box keeping the y and z sizes the same, reduces the kinetic energy from X+X+X to X/4 + X +X or by a factor of 3/4, so the binding energy of two boxes end to end with non-interacting electrons is 1/4 the kinetic energy.

The kinetic energy of an electron in an H-atom is equal to the binding energy (this is the Virial theorem--- the kinetic energy cancels half the potential energy in a 1/r potential to make a binding energy), so you get 1/4 of 12 eV or 3eV of binding energy from this. This is a terrible approximation, because the elecrons repel each other, and the H-atom is not a box, but it shows you that allowing the electron volume to spread gains you a lot of energy on the atomic scale, and it is now plausible that even with repulsion, the electrons will bind, and they do.

  • $\begingroup$ For a hydrogen atom pair, it is seemingly clear that they do not finish paired. We have been fooled because of the hydrogen potential is used as a reference. Instead of desctribing the actual reaction, we have reduced ourselves to describing pairs of reactions. Yes, I said reduced. en.wikipedia.org/wiki/Absolute_electrode_potential $\endgroup$ Commented Aug 6, 2014 at 15:49
  • $\begingroup$ You say, "for H2, two hydrogen nuclei are close" and then say, "But when they come close". Can you elaborate on what you mean? $\endgroup$
    – Kirk Woll
    Commented Nov 5, 2014 at 0:11
  • $\begingroup$ @KirkWoll: If you bring two classical polarizable atoms close, they don't attract. When you bring two quantum atoms close, they do, because the electrons can tunnel to the other atomic volume, reducing their energy. That's it. When they are really close, you get repulsion between the nuclei. $\endgroup$
    – Ron Maimon
    Commented Nov 6, 2014 at 13:01
  • $\begingroup$ This answer is almost right on - We actually find that at long distances, as this answer says, the KE of the electrons decreases as stated, and PE rises (not by much). However at shorter bond lengths, equilibrium bond lengths, the orbitals contract, leading to a reduction in PE, and a rise in KE. At the equilibrium bond length, there is positive KE, but twice as negative PE, leading to -KE binding energy. $\endgroup$ Commented May 14, 2017 at 18:08
  • $\begingroup$ @Ron Maimon I couldn't understand why the kinetic energy decreases when with expansion ? Can you please help me ? $\endgroup$
    – Ankit
    Commented Oct 18, 2020 at 13:08

First, there is no classical explanation for chemical bonding, so any explanation has to include some quantum mechanical ideas.

Second, perhaps the simplest quantum idea here is that electrons exist in energy levels. Continuing the hydrogen example the lowest electron energy level in $H_2$ has less energy than the lowest electron energy level in atomic hydrogen. Since we can get two electrons (with opposite spins) into an electronic energy level, the combination of two $H$ atoms results in the system energy being lowered and energy is thus emitted (usually as heat). To break the $H_2$ molecule apart one must add that energy back into the molecule.

In the helium molecule (which does not exist, but we can calculate it anyway) there are two energy levels, one lower than that for the helium atom and one higher. Their energies average out to the energy of the electronic level in a helium atom. Thus if one attempts to put two helium atoms together to make a helium molecule there is no energy gain or loss. So there is no bonding energy holding $He_2$ together.

That's really all there is to it.


Consider the H$_2$ molecule for definiteness. Binding two H atoms together to form a H$_2$ molecule is impeded in the first line by the coulombian repulsion between the two "naked" H$^+$ nuclei (protons). Establishing a molecular bond has to overcome this primary repulsion.

Protons are heavy, in this case being imagined as localized classical particles. Electrons are light and have to be imagined quantum mechanically as described by a common, more or less localized charge distribution $n(\vec{r})$, with $\int{n(\vec{r})}\ d^3\vec{r}=2$ electrons.

As a first step, try smear out the charge density representing both electrons in the middle of the space between the two protons, making something like a "bridge". This will result in the tendency of the protons to get closer, both of them being attracted to each other by the negative charge density in the middle. However, this purely electrostatic reduction of the H$^+$- H$^+$ repulsion is not enough to guarantee the stability of the H$_2$ molecule.

The decisive contribution to the bond stability comes in the second step and is purely quantum mechanical in origin.

One can guess that, due to electron-electron repulsion, localizing the two electrons within the "bridge" region comes at the cost of an increase in potential kinetic energy. Decisive is the observation that this increase is more than compensated by an increase in the (negative) kinetic potential energy of the electrons, which is maximal when the electrons are localized within the "bridge" region between the protons (no nodes in their wave functions in this case!)

In conlusion: the decisive factor in bond formation it is the more negative potential energy due to electron localization within the "bridge". This is no intuitive result, but a consequence of the celebrated Virial Theorem, and has to be worked out explicitly.

The entire argument holds unaltered even when there is only one electron, namely in the H$_2^+$ molecule ion. A very detailed discussion of these concepts is given by Bader (Atoms in Molecules).

  • 2
    $\begingroup$ Actually the classic reference to learn all this is Ruedenberg's Rev. of Modern Physics 34,326 (1962), but you can get nice pedagogical presentations for free (e.g. google "Rioux why covalent bonds form"). You got the role of kinetic and potential energy reversed. Because of the virial theorem, the decrease of potential energy wins over the increase of kinetic energy: the bonding orbital is contracted except at the bonding coordinate, where is expanded; thus the gain in kinetic energy is not excessive. This guarantees minimization of the total (kinetic+potential) energy. $\endgroup$
    – perplexity
    Commented Sep 2, 2012 at 2:11
  • $\begingroup$ Thank you very much for setting the explanation right. I will try to correct my text accordingly. I knew that it is a lion's pit, but now I will learn this stuff at last. $\endgroup$
    – Lupercus
    Commented Sep 2, 2012 at 7:32
  • $\begingroup$ Rioux is an excellent starter. The paper by Ruedenberg is thorough and contains perhaps more than you would ever want to know, being at the same time a tough and discouraging reading - a historical reference in the line of the Löwdin school. Bader (Section 7.5) is also no easy lecture. He stresses a local form of the virial theorem, which sheds light on the role of the charge-density laplacian in the description of chemical bonding. $\endgroup$
    – Lupercus
    Commented Sep 2, 2012 at 13:53
  • $\begingroup$ This is not a good explanation, even though it appears in the literature. Covalent bonds form because the electron increases it's wandering volume. $\endgroup$
    – Ron Maimon
    Commented Sep 2, 2012 at 21:52
  • $\begingroup$ @RonMaimon: Can you elaborate? And is the above explanation therefore a "legitimate" explanation $\endgroup$ Commented Sep 2, 2012 at 21:56

Here is a more formal, but nevertheless elementary treatment.

The key tool to single out the separate pontential and kinetic energy contributions, for a given averaged total energy $E=<T>+<V>$, is the Virial Theorem (VT). For an atom (one nucleus and its electron cloud) the VT takes the simple form

$$ 2<T>+<V>=0 \ .$$

For a biatomic system (representing two well separated H atoms, but also the covalently bonded H$_2$ molecule) the VT would be stated in "molecular" form as

$$E(R)+R\ \frac{d}{dR}E(R)=-< T > \ (1).$$

It has to be stressed that this VT holds exactly for any arbitrary separation $R$ of the nuclei, for overlapping as well as non-overlapping electron clouds. Here $E(R)$ is the exact and total energy, $<T>$ and $<V>$ are the average values of the kinetic and potential energies of the electrons, averaged with the exact wave function. Naturally, $<T>$ and $<V>$ are also $R$-dependent. Contrary to what one may think at first sight, the $dE/dR$ term is an integral part of the VT, even if the nuclei are kept firmly fixed.

By differentiation of (1) it follows

$$2\frac{dE}{dR}+R\ \frac{d^2E}{dR^2}=-\frac{d}{dR}< T >\ (2) .$$

Pushing the two H atoms towards each other from infinity, if a stable bond is to be formed at a certain separation $R_0$, we ought have for a stable minimum $$d^2E/dR^2>0 \ (3)$$ and a slightly attractive force $$dE/dR\ge 0 \ (4)$$ starting below a certain separation $R\ge R_0$. Conditions $(3,4)$ imply $$\frac{d< T >}{dR}<0 \ (5)$$ while condition $(4)$ alone is satisfied only if $$ 2<T>+<V> \ \le 0 \ \ (6)$$ below some $R\ge R_0$.

Remark 1: If a bond of length $R_0$ is to be formed at all, the kinetic energy of the electrons necessarily has to $increase$ while $R$ is decreased towards $R_0$.

Remark 2: Although the kinetic energy will increase with decreasing $R$, Eq.6 guarantees that the potential energy (negative) will be at least the double of the kinetic energy (only for $R\ge R_0$).

Note: This argumentation is crystal-clearly presented in Cohen-Tannoudji's Quantum Mechanics, Vol.2, p.1191-1199. They also discuss a couple of subtle points related to the physical mechanisms which produce different inequalities at large separations.

  • $\begingroup$ please name your terms - otherwise the equations become useless other than for someone who has already seen them or seen something close enough. What's V? What's T? What's R? $\endgroup$
    – Rhubarb
    Commented Nov 17, 2023 at 10:44

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