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This question has to do with the idea of stability and energy. The premise is that systems will tend towards lower energy states, that’s why bonding happens and electrons prefer single orbitals. And that makes sense in all considering the second law of thermodynamics, but it still doesn’t answer what causes the release of energy when bonding. Clearly, the potential energy of each electron is lower when there is a greater charge and a stronger force of attraction between things—it makes sense, as when you have a stronger force of attraction, you have less “pathways” or degrees of freedom, hence less potential energy. It also checks out with the math of Coulomb’s law, as when the radius decreases from bonding, potential energy decreases as well. So by an atom exhibiting a magnetic pull on another and bonding, potential energy of each electron decreases. But the thing is I just don’t get where that loss of potential energy is going—in this scenario, it wouldn’t make sense for it to be randomly emitted. Or is that not even the reason bonds loose energy?

Consider this analogy, and correct me if I’m wrong: Consider I have 2 magnets A and B that are oppositely charged. If I take magnet A and pull it away from magnet B, now I just did work on magnet A and increased the magnetic potential energy of it as well. Now lets say I released magnet A. The potential energy would be converted into kinetic energy and it would tend towards the other magnet. Now, let’s say I catch magnet A before it hits magnet B. The potential energy of Magnet A just decreased in the field because the kinetic energy transferred to my hand, right?

So instead in atoms, the “hand” that stops atoms from ramming into each other completely is other electrons and the repulsion of the nucleus. So it seems all energy is conserved. Each atom is just transferring energy between each other by both pulling on each other. So how does energy get released?

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Consider $$\ce{2 H <=> H2^{*} ->[collision][emission] H2 + energy}.$$

Imagine a hydrogen atom figuratively rolling down to the potential pit of the bond with another hydrogen atom.

Without passing energy elsewhere, it will bounce and appear free again, keeping its initial energy. It more or less performs one molecule vibration at the top energy level which has just enough energy to break the bond.

Like a skateboarder on U-ramp in lossless scenario would appear on the other side of the ramp.

Now imagine the atom collides with another molecular entity during the time it has increased its kinetic energy and decreased the potential one. Or, it emits a photon. The atom then passes a part of its energy away and is then trapped in the potential pit of the bond.

Like if the skateboarder accidentally hits something/somebody during his ride and loses energy. He will not get in such a case to the other side and is "trapped" in the ramp.

In reactions with more than one product, the released potential energy is converted to kinetic energy of products as there is always something handy to push to release the energy, products flying away of each other.

In this way, there is not needed an inert object to pass momentum as in the prior case.

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    $\begingroup$ It's always interesting to see how most people completely take the presence of an inert third body for granted in standard conditions (condensed systems and relatively concentrated gasses), when in fact their absence reshapes chemistry (e.g. high-altitude atmospheric chemistry and astrochemistry) $\endgroup$
    – Nicolau Saker Neto
    Commented Sep 13, 2023 at 8:55
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    $\begingroup$ @NicolauSakerNeto Yes, these chemistry domains are very interesting. Some exotic molecular entities are hard to detect elsewhere. $\endgroup$
    – Poutnik
    Commented Sep 13, 2023 at 8:57
  • $\begingroup$ Our ambient conditions are highly unusual, but they are all we know, so we call everything else exotic. $\endgroup$
    – Karsten ♦
    Commented Sep 14, 2023 at 16:53
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As Poutnik has pointed out the reaction such as $\ce{H + H \rightleftharpoons H_2^\dagger \rightleftharpoons H + H }$ is entirely symmetrical which means that the bond is formed then breaks up again, i.e. energy is conserved as is always the case. The product $\ce{H_2}$ can only be formed if some energy is lost such as by collision with another inert atom or molecule during the small time interval that the nascent $H_2^\dagger$ exists, which is only a few femtoseconds.

If the reaction is atom+diatom, such as $\ce{Cl + H_2 \rightleftharpoons ClH_2^\dagger \rightleftharpoons HCl +H }$ then HCl will form as the H atom will take away some kinetic energy and HCL itself can have different amounts of kinetic, vibrational and rotational energy depending on the initial collision energy. We can picture the reaction as moving along a potential energy surface from reactants to products, see figure.

The picture shows the (LEPS) energy surface for $\ce{Cl + H_2}$, the numbers represent the potential energy and the black line the path the species take, the wiggle represents $\ce{H_2}$ vibrations (bottom) and $\ce{HCl}$ left side after reaction. (Red is high energy, blue low). You can see that $\ce{H2}$ initialy has only a small amount of vibrational energy but $\ce{HCl}$ has a lot more after reaction. The transition state is at the saddle point, the narrow part of the blue shaded contour.

Cl+H2

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