Seeking a satisfying explaination for a bond formation problem

[Strafespar]

Hey, I've looked for a little bit just online and in the forums and still have not gotten a definite answer. My question: why does bond formation release energy? There is a reason I am posting this is the Quantum Physics section. I am very certain that the energy released into the environment is from the bound electrons losing energy and emitting photons. The puzzling part is finding how exactly they do that when they form through a ionic, covalent, etc. bond. When I picture the situation I can come up with many different ways the electrons could possibly lose energy, but I'm not sure if there is just one single instance or several. Also if you could point me in a direction to some sources that give extensive explanations on things like this could you provide those because I just can't seem to find much on the internet. You might have already guessed it but I am pretty new to physics and I just encountered this problem in my high school chem class. Thanks in advance, I am really excited to get some responses, as my teachers didn't exactly understand why either.

 

[Kholdstare]

Internet is for information (often incorrect). Go and read a good book on Quantum Physics like Resnick and Eisberg to know the theory.

While forming a bond the two atoms come closer and the two electrons continuously change their wavefunctions to be on the minimum energy (bonding state). They radiate this energy slowly. After the bonding distance is achieved they cannot change anymore, as further change will increase the energy of the bonding state.

 

[Strafespar]

Is the energy radiated continuously?

 

[Kholdstare]

Look at this molecular potential energy curve.
http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch02/ljpot.jpg

the bond formation is just rolling along the curve from large r to the minima point. As there is no discontinuity energy is supposed to be released continuously. However thermal variation will give non-ideal results.

 

[Strafespar]

So if the reaction was incomplete but had started, as in the two particles somehow vanished instantly before the electrons could reach the lowest possible energy, that reaction would have released some energy? In my chem class they say that when an electron falls down to its ground state it only releases one photon equivalent to the energy gained. What does it mean when the energy is released continuously? Thanks.

 

[Kholdstare]

So if the reaction was incomplete but had started, as in the two particles somehow vanished instantly before the electrons could reach the lowest possible energy, that reaction would have released some energy?

Sorry, I don't not get you there.

In my chem class they say that when an electron falls down to its ground state it only releases one photon equivalent to the energy gained. What does it mean when the energy is released continuously?

They did not tell you that it is presumed that the electron belongs to an atom which is not interacting with other atom(s). In the case of bonding the release of energy will depend on the distance between two atoms, which is (at least classically) a continuous variable.

 

[kith]

The bond formation is just rolling along the curve from large r to the minima point. As there is no discontinuity energy is supposed to be released continuously.

This argument is wrong. I would suggest to use the simplest potential, the Coulomb potential of the hydrogen atom (image). There's also no discontinuity in the potential, still the possible energies are discrete.

A naive picture of the situation is this: If the electron is very far away from the proton, it's potential energy is nearly zero. When it gets nearer, it will drop into an allowed state with neagative potential energy (like B_V in the image) by emitting a photon. As it gets nearer and nearer, it climbs down the ladder and emits more photons until it reaches the groundstate. So energy is dissipated by emitting photons which carry away discrete amounts of it.

 

[Kholdstare]

This argument is wrong. I would suggest to use the simplest potential, the Coulomb potential of the hydrogen atom (image). There's also no discontinuity in the potential, still the possible energies are discrete.

A naive picture of the situation is this: If the electron is very far away from the proton, it's potential energy is nearly zero. When it gets nearer, it will drop into an allowed state with neagative potential energy (like B_V in the image) by emitting a photon. As it gets nearer and nearer, it climbs down the ladder and emits more photons until it reaches the groundstate. So energy is dissipated by emitting photons which carry away discrete amounts of it.

He asked about bond formation process. A simple bond formation process is the Hydrogen molecule where two electrons starting from their ground states reach a further lower energy level called bonding level while emitting energy.

 

[kith]

He asked about bond formation process. A simple bond formation process is the Hydrogen molecule where two electrons starting from their ground states reach a further lower energy level called bonding level while emitting energy.

Yes and an even simpler bonding process is the formation of a hydrogen atom by a free electron and a free proton. ;-)

My argumenation applies also to molecules because the vibrational states in the Morse potential are discrete, too.

 

[Kholdstare]

I'm sure the OP was referring to the chemical bond formation process where more than one atoms are involved.

 

[kith]

I'm sure the OP was referring to the chemical bond formation process where more than one atoms are involved.

I can only repeat myself: Also there, the possible (vibrational) states are discrete. The picture of continuous energy dissipation while rolling down the potential is not right, because not every energy value is allowed. It's more like a ladder. Whenever a step is taken, energy is radiated in discrete packages/photons.