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In books it is commonly written that whenever a chemical bond is formed the energy of the molecules/atoms gets lowered and hence energy is released. This is generally explained by diagrams like this

enter image description here

But I'm really unable to understand this process, I accept that before bonding the potential energy of the molecules were more and after bonding the potential is less and hence this difference of energy must be released but how does this lowering of potential energy happens? Let's just take an example of electrostatic situation (and as far as I know its the electric potential energy only that get's lowered and hence some energy gets released), say I have an electron at the origin and a proton at $x=x_1$ and then somehow it is moved to $x=x_2$ like this

enter image description here We know from equation of electrostatics that potential energy of proton at $x=x_1$ is $$ U_1 = \frac{1}{4\pi \epsilon_0}~\frac{-e^2}{x_1} \\ \\ \textrm{and at} ~x=x_2 \\ U_2 = \frac{1}{4\pi \epsilon_0}~\frac{-e^2}{x_2}$$ Since, $x_1$ is greater than $x_2$ (from figure) therefore $U_1 \gt U_2$. So, we energy at $x_2$ is less therefore there must be some release of energy, but how? If we let the field to do the work then no energy will be released because our proton gonna get the kinetic energy, if we do the work to move it from $x_1$ to $x_2$ then according the to the laws of physics I will do the negative work and hence energy will transfer from the proton to me. So, how does energy ever get released?

How in chemistry the energy gets released by the formation of bonds? I know that we can say without intricacies that potential energy gets lowered and hence the difference of energy is the energy that we get, but how does this happens? If we have some cation and then we put some anion in the reaction mixture then, of course, cations will get attracted towards the anion and hence will gain kinetic energy and according to me that's the only possible way to lower the potential energy. Is that kinetic energy that we measure as heat and say that the reaction is exothermic?

Any help or clarification will be much appreciated.

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    $\begingroup$ Yes, in some cases it is just the kinetic energy that we perceive as heat. In other cases, photons are an important part of the picture. $\endgroup$ – Ivan Neretin Jan 15 at 9:39
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    $\begingroup$ That first picture above is a horrible supersimplification. Can you guess why? And please say wherefrom you have it. Whoever made it deserves a good public shaming. $\endgroup$ – Karl Jan 15 at 20:05
  • $\begingroup$ @Karl I’m sorry but I’m unable to see What you are pointing out. $\endgroup$ – Knight Jan 16 at 3:38
  • $\begingroup$ The curve looks like this should be a transition state, but the atoms above do certainly not depict one. $\endgroup$ – Karl Jan 16 at 7:31
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    $\begingroup$ @Alchimista No, definitely NOT, sorry. A transition state is a bound state. If you put in enough energy to break all bonds at once, the whole thing flies apart. You show such pictures if you want your students to learn facts, but never understand. Im sorry, but that is a despicable attitude for a teacher or textbook author. It comes from thinking students are idiots, which is a self fulfilling prophecy. $\endgroup$ – Karl Jan 16 at 9:00
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Let us assume the simplest case to form a bond from two atoms (assumed to be in their ground state) that approach each other from an infinite distance. For convenience, we put the potential energy of the system at zero here. If the atoms come closer, they start to interact via long-range forces (such as van der Waals forces) and the potential energy is slightly lowered. If the atoms come even closer, their wave functions start to overlap and the potential is reduced rather steeply as compared to the long-range forces. Moving the atoms even closer increases the potential as the electrons will start to repel each other. The potential energy as a function of the interatomic distance can be approximately described by a Morse potential (the blue curve in the figure below (taken from Wikipedia))

Morse potential in blue (from Wikipedia)

In the figure you also see horizontal lines in the Morse potential which represent the energies of the vibrational states that are supported by the potential. Now that we know what the potential looks like we can take our two atoms at infinite distance and move them towards each other, that is, the total energy $E_\text{tot}(r)=E_\text{Morse}(r)+E_\text{kin}(r)>0$. As the atoms get closer, the potential energy gets lower and the velocity and kinetic energy increase, the total energy is - of course - conserved. When the atoms come so close that they start to repel each other (when they hit the repulsive "wall" on the left), they fly away in opposite direction and end up at infinity again! This is the same result that you would find with an ideal marble track (no friction) that is similarly shaped to the Morse potential: the marble will roll down, first accelerate, then climbs up the wall on the left and rolls back to where it started. In order to from a bond (or to stop the marble in the middle) we need to remove part of the energy of the system. As Ivan Neretin pointed out, this can be done by emitting a photon. When the atoms are separated at $r_e$, the release of a photon with energy $h\nu=D_0+E_\text{kin}$ will result in a loss of energy and the atoms will be trapped in the potential. (In our marble track, friction can play a similar role)

This process is called radiative association and it is believed that it played an important role in the chemistry of the early Universe. However, not all molecules can emit a photon. There are some rules - selection rules - which have to be fulfilled. The molecule needs to have a dipole moment for instance. Therefore $\text{H}_2$ molecules are unlikely to be formed via radiative association.

Another, more likely, method to remove the energy is to have a third atom (or molecule) colliding with the other two atoms. This atom can then take away the excess energy in the from of kinetic energy in a so-called three-body collision.

When you have more complex molecules that collide to form bonds, the excess energy can be distributed over the different degrees of freedom that the molecule possesses. For instance, the molecule's vibrational and/or rotational energy is increased which results in an increase in temperature.

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You may simply say that energy is needed to break a bond. Breaking a bond is endothermic. So the inverse is exothermic. Energy is released when a bond is formed.

Apparently you want to just discuss what is happening when an electron approaches a proton from far away (x1 in your drawing) to a shorter distance (x2). The electron is supposed to have no kinetic energy at the beginning, in x1. When arriving at x2, the difference in potential energy has been transformed into kinetic energy. When the electron arrives at the x2 position, it has a great velocity. It will not stay here. It will continue its way until it will nearly touch the proton at the abscissa x3 < x2. At this point, all its energy is potential again (no kinetic energy) : it will be repelled back to higher x values. It will go back exactly to the initial point x1, where the kinetics energy was zero, then start again the same oscillation to x3 and back, again and again. The only way of stopping this oscillation is a collision with a third body, to which the electron gives a part of its kineticenergy. This will stabilize its movement. After a couple of collisions, the electron will get to the stablest position at x = 0.5 Å, as it is in a stable H atom.

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  • $\begingroup$ See, the title asks "why", but the question asks "how". $\endgroup$ – Ivan Neretin Jan 15 at 10:19
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    $\begingroup$ I request you to please read my question. $\endgroup$ – Knight Jan 15 at 10:32
  • $\begingroup$ @adeshmishra "The only way of stopping this oscillation is a collision with a third body" does answer your question, I think. The answer by Paul includes a different possibility in special cases, namely emitting a photon. $\endgroup$ – Karsten Theis Feb 14 at 16:09
  • $\begingroup$ @KarstenTheis I’m reading Paul’s answer again and again for understanding it properly therefore I’m taking some time. $\endgroup$ – Knight Feb 14 at 16:18

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