So as we start Valence Bond Theory in Chemical Bonding from Inorganic Chemistry.
Then we are introduced to reasoning behind bond formation as follows:
As it is a well known fact that $$Potential\;Energy\;\alpha\;\frac{1}{Stability}$$ As in the atoms of molecule (here $H_2$) attractive forces tend to bring molecule closer whereas repulsive forces tend to push them apart.
Magnitude of new attractive forces is more than new repulsive forces and ultimately they reach a critical distance $r_o$ where the attractive forces balance the repulsive forces. The effect of distance is visible in the graph attached. Since maximum lowering is at distance $r_o$ there is a bond formed and this is the bond length (74 pm in case of $H_2$).
enter image description here

Now my problem is that how is the principle of conservation of energy being satisfied? Like how come total energy is constant since it is clear that potential energy of the two atom system varies but its kinetic energy does not compensates for the same.
Even if the energy is radiated then how can it be verified?

Born–Oppenheimer approximation $\;$In which it is supposed that the nuclei, being so much heavier than an electron, move relatively slowly and may be treated as stationary while the electrons move in their field.

It is called a potential energy curve because the kinetic energy of the stationary nuclei is zero.
It is evident that nucleus does move to alter kinetic energy.
Please confirm the energy conservation here.
Thank You

  • $\begingroup$ Who says the kinetic energy does not compensate? Both protons move. If the excited H2 molecule does not pass energy elsewhere while protons bounce together and away again in 1 vibration period, both atoms separate again, having just enough energy for that. $\endgroup$
    – Poutnik
    Commented Oct 18, 2022 at 17:20

3 Answers 3


If the hydrogen atoms travel straight toward each other, they accelerate as they come nearer and potential energy transforms into kinetic. This is not shown in the diagram because you don't plot kinetic energy. It would go up as the atoms get closer, up to the potential energy minimum at 74 pm, then decrease again as the repulsive interactions increase faster than attractive ones.

While it is not explicit, there are implicit assumptions within the drawn potential. For instance, it implies the existence of attractive forces that become substantial at approximately twice the bond length (shown as 74 pm) which would set the nuclei into motion toward each other. Therefore, stating that the nuclei can be stationary isn't quite right unless what is shown is not the total potential (which includes forces that inhibit motion of the nuclei toward each other).

Another piece of information missing from the diagram is the orientation of the individual paths of the nuclei (which presumably are in the gas phase and not being retained by some unknown external potential). The drawn potential is a schematic of what it would look like if you could bring the nuclei toward each other along a line connecting them. In reality the nuclei might take glancing paths past each other in all sorts of random directions rather than carom into each other (sampling the repulsive part of the potential) and instead just deflect each other. In either case energy would be conserved as it is converted into kinetic and then back into potential energy again.

  • 2
    $\begingroup$ Phet has a great simulation of this where you can see the potential energy turning into kinetic. Thus, we can see that energy is conserved. @Buck $\endgroup$
    – Avogadro
    Commented Oct 19, 2022 at 20:27
  • $\begingroup$ @Avogadro very neat and super relevant, thanks for sharing $\endgroup$ Commented Oct 31, 2022 at 10:26

You asked several questions, I will focus on this one:

how is the principle of conservation of energy being satisfied?

Depending on the conditions, the energy is used to heat up the surrounding or dissipates otherwise. You can do experiments to verify that hydrogen molecules have lower potential energy than atoms, or do calculations from first principles.

The potential energy curve makes no assumption about kinetic energy of the nuclei (of course, the electrons always have kinetic energy). The nuclei can be fairly stationary (in a solid at low temperature), or they can move around (translate, rotate, atom pairs vibrate). All these movements are compatible with the potential energy function. As the OP states, the Born-Oppenheimer approximation allows us to calculate the potential energy for a given snapshot with a certain bond distance.


Hydrogen atoms in a vacuum have low attractive forces for each other certainly not enough to generate the potential energy curve. What they do have is intersecting velocity vectors with the associated kinetic energy and momentum. In the collision the kinetic energy is converted into the potential energy of the distorted 1s orbitals. If the KE is high enough repulsion of the protons is paramount. The energy conservation is the original KE of the atoms and the incipient PE in the collision Their sum will be constant. I do not know if this transfer of energy involves excited energy levels or if it is similar to a Raman effect. It seems more of the latter because the removal of energy to form the molecular orbital is by a third body collision not by radiation. It is possible the H2 molecules form in an excited molecular state.

What I am saying is that low temperature hydrogen atoms will not react to form H2 because they will not collide and any collisions will not distort the orbitals enough for rearrangement into molecular orbitals. Imagine a solid block of H atoms at absolute zero as close as possible and unable to react. The epitome of frustration.


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