I try to understand, from the electromagnetic side only, how the molecule could be more stable than each atom, and concluded that it could not. Where is my mistake?

Here is an illustration :

By using a ruler in order to have an "example" of numerical values, we conclude, by taking the numerical values of distance in the electric field formula (I removed the common factor with $\epsilon_0$, $\pi$, etc.), that the repulsion force is bigger than the attraction force.

So where is the mistake in my reasoning ?

(I try to stay at a very basic approach of fundamental law of electromagnetism, so not using LCAO with quantum mechanics)

enter image description here

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    $\begingroup$ Quantum mechanics was necessary even to explain the existence of an H atom alone, it is not surprising that you cannot describe H-H. Besides that, if you allow for charge distribution one might perhaps find a field distribution that leads to a stable configuration. Note that this goes already - even in loose terms - in the direction of somehow distributed partial negative charges. It would be a rough way towards orbitals. I am very critical about the question in the box as for the stability of the molecule is due to none of them. $\endgroup$
    – Alchimista
    Oct 24, 2020 at 11:44
  • $\begingroup$ @Alchimista : Thank you. What do you mean by "charge distribution" : I put the electron here as "punctual" particle : do you mean replace it by a field=a non punctual particle ? $\endgroup$ Oct 24, 2020 at 11:46
  • $\begingroup$ yes. Kind of "an orbital". I feel you got my point. $\endgroup$
    – Alchimista
    Oct 24, 2020 at 11:47
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    $\begingroup$ You should look into the quantum effect sometimes called the 'exchange' interaction as this adds extra terms missing in any classical description. $\endgroup$
    – porphyrin
    Oct 24, 2020 at 17:00

2 Answers 2


I assume you mean from the classical electrodynamics side only, not from quantum electrodynamics side. By the former, even H atom cannot exist, as the electron would fall along a spiral curve on the nucleus, continually emitting radiation being radially accelerated.

For a hydrogen molecule, both electrons move around both protons, they are not dedicated to the one they came with.

Consider a similar case of charged macro objects, where applying the pre-relativistic classical electrostatics is justified. Imagine 4 respectively charged balls in gravityless vacuum.

The scenario in the bottom picture is one of the less preferred positions, having rather higher energy due short distance of negative charges. But there is plenty of different positions where the whole system would have lower energy than if they were 2 and 2 separately.

Note that even classical system have no general analytic solution for the motion equations for more than 2 objects.

  • $\begingroup$ Thank you. For the system of atom itself, nucleus and its electron, I was not putting the forces on the pictures, sine as you say, this is quantum mechanism that dictates that. So I was only isolating the forces that communicate between the two distinct atoms. $\endgroup$ Oct 24, 2020 at 18:22
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    $\begingroup$ But inter-atom interactions have equally quantum character as nucleus electron interactions. Additionally, aside of classical coulombic interaction, there is Pauli spin interaction $\endgroup$
    – Poutnik
    Oct 24, 2020 at 18:26

Such questions troubled physicists in the early 20th century. Their inability to add more than one electron to the planetary Bohr-model was one of the reasons that quantum mechanics was developed. Nevertheless, continuing to ask this type of questions may help us sharpen our thinking and develop our intuition, as far as it goes (you might also take a look at the simpler Gedanken-experiment about Helium explored here).

In the case at hand, we will not be able to find an analytical solution because this is a four-body problem. However, if you did want to portray the molecule with point charges in a plane then it might be more fruitful to draw an ellipse with —for argument's sake— two electrons in opposite positions ‘chasing each other’. The time-varying force balance will depend on the specific configuration as illustrated with two sketches below.

enter image description here

The intuitively most favourable position would be figure A. In this case, the sum of the attractive forces from the electrons on each proton is higher than the repulsive force between the protons: i.e. there is a net attraction. The least favourable case would seem to be figure B where the electrons pull the protons away from each other in addition to their natural repulsion. One might imagine that the actual force balance is some sort of convoluted average between these extremes — but calculating any such point charge model with multiple electrons has so far proven to be a dead end.

Surprisingly, the parameters of the $\ce{H2}$ molecule can be calculated using classical electrodynamics provided we approach the electron not as a point charge but as a membrane of distributed charge, as proposed by Mills. The shape of the molecule in his theory turns out to be the 3D version of the ellipse pictured above, with the (large) electrons physically encompassing the (small) nuclei. 3D rendering of H2 The internuclear distance is calculated at $\sqrt2$ times the Bohr radius and the binding energy, i.e. the difference between 2 separate $\ce{H}$ atoms and an $\ce{H2}$ molecule, works out to 4.5 eV — in line with the experimental value. It is this more advantageous energy level that makes $\ce{H2}$ a stable molecule.

  • Randell L. Mills, "Classical Quantum Mechanics" Physics Essays, Volume 16: Pages 433-498, 2003 [equation 156 gives the binding energy of $\ce{H2}$]

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