[Corribus]

When we see the slope of the graph we find two points (74 pm and infinity) where slope is zero meaning that Net force is zero which make sense as Force of attraction = Force of Repulsion for 74 pm and no force at infinity. Similarly at infinity the Potential energy is taken to be zero (relatively) but at another point somewhere close to 25 pm we can see that the Potential energy is again zero. What does being zero at that point mean?

It means nothing. The value of the potential energy by itself tells you nothing about how the system will evolve. It is the gradient (slope, in 2D) that forecasts how the system will change if you start from a certain point. If you're standing on the side of a mountain looking across a valley to a plateau with a nice campsite on it, does it mean anything significant that you happen to be at the same elevation as the campsite? Not really, other than that you've got to go down first into the valley and then climb all the back up the other side before you can enjoy the fire.

As you said the bond will form when the attraction energy is greater than repulsion energy. But as we know at 74 pm the net force is 0 so how is the bond forming?

At the risk of confusing you further: it is too simple to reduce bond formation to considering when attraction energy is greater than repulsion energy. Bonding between two atoms involves more factors than electrostatic (Coulombic) interactions. To understand bonding one must also consider the effects of electron exchange, which is a purely quantum mechanical phenomenon based on symmetry requirements of the wavefunction. I suspect this may be beyond your level, but Coulomb integrals, exchange integrals, and overlap integrals all contribute to the final form of the potential surface and therefore the final bond length and strength. It is not possible to see this by just looking at the final potential energy function.

You might check out this discussion of the potential surface of the simpler molecules, H₂⁺. You will see that in this simple molecule, electron-proton attraction cannot override the proton-proton repulsion term by itself. Exchange interactions are an essential component of forming a quantum mechanical bond in this molecule, and most other molecules as well. (Exchange also plays a critical role in forming antibonding orbitals, which determine many chemical and physical properties of molecular systems.)

In short, you may say that bond formation is favorable at certain H---H nuclear separations (given the right conditions) because the potential energy is lower at those values of R than the potential energy for two infinitely separated H atoms, and systems tend to move towards states of lower potential energy. The role that attraction and repulsion play in determining those potential energies is more complicated, so while it may be helpful to think about it in those terms, it's not really what's going on at a quantum level except at very small nuclear separations, where the repulsion term is overwhelmingly large compared to everything else.

[iamvts]

When we see the slope of the graph we find two points (74 pm and infinity) where slope is zero meaning that Net force is zero which make sense as Force of attraction = Force of Repulsion for 74 pm and no force at infinity. Similarly at infinity the Potential energy is taken to be zero (relatively) but at another point somewhere close to 25 pm we can see that the Potential energy is again zero. What does being zero at that point mean?

It means nothing. The value of the potential energy by itself tells you nothing about how the system will evolve. It is the gradient (slope, in 2D) that forecasts how the system will change if you start from a certain point. If you're standing on the side of a mountain looking across a valley to a plateau with a nice campsite on it, does it mean anything significant that you happen to be at the same elevation as the campsite? Not really, other than that you've got to go down first into the valley and then climb all the back up the other side before you can enjoy the fire.

As you said the bond will form when the attraction energy is greater than repulsion energy. But as we know at 74 pm the net force is 0 so how is the bond forming?

At the risk of confusing you further: it is too simple to reduce bond formation to considering when attraction energy is greater than repulsion energy. Bonding between two atoms involves more factors than electrostatic (Coulombic) interactions. To understand bonding one must also consider the effects of electron exchange, which is a purely quantum mechanical phenomenon based on symmetry requirements of the wavefunction. I suspect this may be beyond your level, but Coulomb integrals, exchange integrals, and overlap integrals all contribute to the final form of the potential surface and therefore the final bond length and strength. It is not possible to see this by just looking at the final potential energy function.

You might check out this discussion of the potential surface of the simpler molecules, H₂⁺. You will see that in this simple molecule, electron-proton attraction cannot override the proton-proton repulsion term by itself. Exchange interactions are an essential component of forming a quantum mechanical bond in this molecule, and most other molecules as well. (Exchange also plays a critical role in forming antibonding orbitals, which determine many chemical and physical properties of molecular systems.)

In short, you may say that bond formation is favorable at certain H---H nuclear separations (given the right conditions) because the potential energy is lower at those values of R than the potential energy for two infinitely separated H atoms, and systems tend to move towards states of lower potential energy. The role that attraction and repulsion play in determining those potential energies is more complicated, so while it may be helpful to think about it in those terms, it's not really what's going on at a quantum level except at very small nuclear separations, where the repulsion term is overwhelmingly large compared to everything else.

Thanks for providing clarity on this complicated topic. As I high schooler, yes it does go above my level but I get the general idea behind it and that more factors are involved in it. Thanks for your time. Also I have a request, if you have time can you please provide a over simplification for it as it might be much more of a help. Thanks for your effort.

[Corribus]

Also I have a request, if you have time can you please provide a over simplification for it as it might be much more of a help.

Not sure what you mean.

[iamvts]

Also I have a request, if you have time can you please provide a over simplification for it as it might be much more of a help.

Not sure what you mean.

Well, I meant can you explain me what's happening at different critical points in the graph using some kind of over simplification wherever needed. It has created a mess in my head. I just wanted to sort it out. Thanks for your effort and time.

[Corribus]

I think it's convenient to fall back on analogies using gravity because gravity is a potential field that people intuitively understand.

So, suppose your potential energy surface is a physical surface made of wood and your molecule is a marble. You place the marble at the far end of your surface (large values of x). Assume no friction. What happens? Because the surface is sloped everywhere, the marble starts to move to the left because systems naturally prefer to move toward states with lower potential energy. The gravitational force the marble experiences depends on the slope under it at any point. (The gravity force is actually the same everywhere, but if you know any physics, a large part of the force experienced is canceled out by the surface pushing back, and the amount canceled depends on the tangent vector at the point of contact.) Initially that force is weak where the slope is small, but as the marble moves, the force gets larger and larger because the slope gets steeper. Also, as the marble loses potential energy (goes down in elevation), that potential energy is converted into kinetic energy. The marble moves faster. That is, until it reaches the bottom of the surface, and which point it continues moving (because it has a large kinetic energy at this point), and starts to go up the hill on the other side. Now it is gaining potential energy and losing kinetic energy. Eventually all that kinetic energy will be converted back into potential energy (really far to the left) and it will come to a brief stop, before turning back around and going to the right again.

Consider two scenarios:

(1) In the first scenario, when you initially place the marble at the far end of the track, you give it a hard flick with your finger in the direction of the low point of the track, imparting a large amount of kinetic energy (KE). This doesn't change anything about the way the marble behaves except that after the marble passes the bottom of the track, it will go quite far against the steep slope at small x values because it has a lot of extra KE. Remember that because there's no friction, the total energy (kinetic plus potential) is constant. So when the marble hits its high point, it will turn back around, go through the minimum again, up the other side and back to where it started. And it will keep going because it has that excess kinetic energy that you imparted by flicking it. The point is: the marble doesn't get trapped in the well because it has enough kinetic energy overcome gravity and get back out.

(2) In the second scenario, rather than flicking the marble you just set it down, so it basically starts with no kinetic energy. Everything here is the same but the marble only comes right back exactly where it started.

Now in reality there is friction. Friction basically siphons a little bit of the energy away from the marble as it moves. Can you imagine what happens in a world with friction? In scenario 1, the initial kinetic energy is so large that, although the marble loses a little bit of energy to friction, it still has enough kinetic energy to get back out of the well and go on its merry way. In scenario 2, it doesn't. So the marble almost but not quite completes a whole circuit, comes almost but not quite back to where it started. And then it falls back toward the bottom of the well. This cycle repeats many times, each cycle the marble losing a bit more energy to friction, each time the marble making it a little less far up each side of the well, until finally it comes to a rest at the bottom of the well.
 
By a crude approximation, a molecular potential well behaves the same way. Except the potential in this case is formed by electrostatic and quantum interactions between two atoms. Suppose you have two hydrogen atoms cruising toward each other with some initial speed. They begin to feel an attraction toward each other that becomes stronger as they get closer. The attraction is due both to charge considerations (between electrons around one hydrogen and the nucleus of the other, say) and exchange interactions (no real classical analog, so you can just think of it as quantum effects that make electrons happy to be close to each other). But there's also a pushing effect because the two proton cores are the same charge. At some point, that repulsion starts to overwhelm the attraction between electrons and protons and the cozy quantum effects. That point is the bottom of the potential well. Of course, the two atoms have a lot of kinetic energy so they keep rushing toward each other, but now a lot of that kinetic energy is being rapidly converted into potential energy to fight that repulsion effect. When the atoms run out of kinetic energy, they are as close as they are going to get, but the repulsion is still there, so they start moving away from each other, building up speed. But as they do so, the repulsion gets weaker and then as they pass the minimum of the potential surface the attractiveness starts to take hold again, which tugs them back together.

Here again we apply the two scenarios.

In scenario one, the two atoms started off with so much kinetic energy (maybe there was a lot of heat) that they collide (more or less), bounce of each other, and go flying away faster than the attraction force can pull them back together. No molecule formed.

In scenario two, they started with very little kinetic energy and so they barely have enough energy to match the attractive forces that try to pull them back together after they come back out of the well. If the system loses any energy at all during the process (due to "friction"), then the attraction wins and they start going towards each other again... and then they cycle back and forth around the low point of the potential, coming closer together and then going further apart as potential energy is repeatedly converted back and forth with kinetic energy like a (an)harmonic oscillator. At this point the two particles are trapped/bound because they don't have enough energy to get out of the well. That's what we call a molecule.

A few last points.

First, lone atoms usually have enough kinetic energy to avoid being trapped. But, lone atoms are never lone. Typically it is collisions with other (spectator) particles that cause the "friction" needed to trap them into a molecular potential well. So say two hydrogen atoms are rushing towards each other and suddenly when they are at about 74 pm apart a nearby molecule happens to b-ump into them. Some of the kinetic energy of the hydrogen atoms is transferred to the molecule. Now the two hydrogen atoms don't have enough kinetic energy to escape their attraction and a new hydrogen molecule is formed.

Second, even when a pair of hydrogen atoms becomes trapped, they still retain a lot of kinetic energy - just not enough to overcome their attraction and escape the well. So they are perpetually in a state where they try to fly apart, then come back together until they are pushed strongly by repulsion, then fly apart again, then come back together, always cycling back and forth around the average value determined by the bottom of the well (like a marble rolling back and forth in friction free world). This is why molecules vibrate, and the equilibrium bond length is the value at the bottom of the well around which the atoms cycle. Collisions still happen and sometimes the system can absorb or release radiation which can cause the atoms to pick up or release kinetic energy - essentially causing the molecules to vibrate faster or slower. Of course in quantum world only certain vibration frequencies/energies are allowed. That is the basis for vibrational spectroscopy.

Finally, it's good to know that unlike in the marble/track system, molecular systems can never be at rest. No matter how much kinetic energy the atoms lose due to "friction", they can never stop at the bottom of the well. The minimum energy they can have is called the zero point energy and is required by the laws of quantum physics (i.e., we can't know exactly where a quantum particle is, and two quantum particles cannot occupy exactly the same point in space with the same properties). However, while the system can never come to rest, the system CAN absorb enough energy to cause the atoms to go flying apart... the minimum energy required to do that is (you guessed it) the bond dissociation energy, the difference between the zero point energy and the top of the potential well. Note that this is not the same as the depth of potential well due to the zero point energy (the molecule can never be at the very bottom of the well). Any input of energy less than the bond dissociation energy will not provide enough kinetic energy to the bound atoms for them to overcome their electrostatic and exchange attractions.

[iamvts]

"Of course, the two atoms have a lot of kinetic energy so they keep rushing toward each other, but now a lot of that kinetic energy is being rapidly converted into potential energy to fight that repulsion effect."


Can you give me a rough insite of how is potential energy fighting repulsion effect?


"the bond dissociation energy, the difference between the zero point energy and the top of the potential well. Note that this is not the same as the depth of potential well due to the zero point energy (the molecule can never be at the very bottom of the well). Any input of energy less than the bond dissociation energy will not provide enough kinetic energy to the bound atoms for them to overcome their electrostatic and exchange attractions."


Please re-explain the zero point energy and the above paragraph, It seems i cant get hold of it even after reading it 7 times.


Thank you for dedicating so much of your valuable time to some random folk on the Internet. Thank you very much.

[Corribus]

Can you give me a rough insite of how is potential energy fighting repulsion effect?

I think maybe it would help you to better understand what potential energy is. When you lift an object against gravity, you are doing work against the gravitational force. That work energy is stored in the object you lifted. We call this potential energy because the energy is still there, waiting to be used if the opportunity arises. When you drop the object (i.e., remove the force that's holding the object up), the potential energy is then converted back to kinetic energy by the restoring force of gravity until another force is encountered (the ground, say).

In the case of two hydrogen atoms: they have a certain amount of kinetic energy, and when they approach each other, work is done against the electromagnetic force - specifically the repulsive force between two positive charges. The particles can only do as much work against the force as they have kinetic energy. Once the kinetic energy is gone, all that energy has been stored as potential energy. Then the electromagnetic force uses that potential energy to push the particles in the opposite direction along the direction of the force... until they encounter an opposite force in the form of attraction and quantum mechanical exchange.

Please re-explain the zero point energy and the above paragraph, It seems i cant get hold of it even after reading it 7 times.

Imagine a pendulum, which is a nearly harmonic oscillator that you might be familiar with in the macro-world. If you draw the weight back and let it go, the weight will swing back and forth repeatedly, converting kinetic energy to potential energy and potential energy to kinetic energy. Friction will gradually siphon off this energy and the weight will go a little less high with every swing. A long time later, it will finally come to rest in a state in which both the potential energy and kinetic energy are zero (with respect to the frame of reference).

A quantum mechanical pendulum doesn't work like this. There are two key differences. First, energy states  aren't continuous. So rather than the weight almost imperceptibly lowering with each swing, what you might observe is the pendulum swinging back up to exactly the same height for a random number of times, and then suddenly the next time it would swing a lot less high, and so on. Second, eventually the pendulum would reach a point where it would never get lower, it would just keep swinging forever just above the minimum potential energy forever (or until someone disturbed it). Which is to say, the final state would not have potential energy and kinetic energy both equal to zero - there would be a residual energy. Even at absolute zero temperature, quantum systems never stop. This is the zero point energy. This follows from the uncertainty principle. Which I guess introduces a third difference: you'd never really be able to pinpoint an exact location or momentum for the pendulum, because there's a fundamental uncertainty of position with respect to momentum for quantum particles. It is worth emphasizing then that some of the examples I've given over the last page are grounded in classical physics approximations. (Do atoms really crash into each other and bounce in the other direction? Not as we usually understand it. We can never know exactly when or where such events happen.) 

[iamvts]

Thank you so much for your help. If it wasn't for you, I wouldn't have understood this. Finally I can sleep with peace. I don't know how to thank you for your time and effort. Thank you very much sir.

[Corribus]

My pleasure.