Why is the energy of two ions bonded lower than two ions separated?

[rokku]

How does forming a bond between two ions lower the overall energy of the system? Also how would two hydrogen atoms form if there is proton-proton repulsion and electron-electron repulsion and only proton-electron attraction?

 

[Pythagorean]

In the case of covalent bonding, it's a consequence of the indistinguishability of particles! Griffiths talks about this at the end of his Intro to Quantum book in Chapter 5. Essentially, the superposition that results from treating identical particles (they are treated as distinguishable classically) results in bosons being closer together and fermions (like electrons) being farther apart and once you consider spin, the electrons are able to occupy the "singlet" state where they are more likely to be in between the two nuclei, attracting them towards the center.

 

[Drakkith]

In the case of covalent bonding, it's a consequence of the indistinguishability of particles! Griffiths talks about this at the end of his Intro to Quantum book in Chapter 5. Essentially, the superposition that results from treating identical particles (they are treated as distinguishable classically) results in bosons (nucleus) being closer together and fermions (like electrons) being farther apart and once you consider spin, the electrons are able to occupy the "singlet" state where they are more likely to be in between the two nuclei, attracting them towards the center.

Aren't most nuclei fermions, not bosons?

 

[Pythagorean]

Ah yes, good point. Removed the reference to the nucleus. The electrons concentrating in the middle and attracting the protons is really the key point, I think.

 

[DrDu]

Indistinguishability is not the crucial point, given that you observe covalent bonding already for molecules containing only one electron, namely $\mathrm{H}_2^+$.
Classically you won't expect a big effect, as nuclear-nuclear repulsion and electron-electron repulsion is made up at bonding distances by electron nuclear attraction.
In fact, covalent bonding is an essentially quantum mechanical effect. In the molecule, the electron can move in the potential throughs of two nuclei as compared to only one in the case of a single atom. This increases its positional uncertainty $\Delta x$ along the bond axis and by the uncertainty principle lowers its momentum uncertainty $\Delta p=\hbar/\Delta x$.
Hence also its kinetic energy gets lower although the story doesn't end here.
There is an excellent article by Kutzelnigg, a quantum chemist, on the principle behind bonding:
http://onlinelibrary.wiley.com/doi/10.1002/anie.197305461/abstract

 

[Drakkith]

Indistinguishability is not the crucial point, given that you observe covalent bonding already for molecules containing only one electron, namely $\mathrm{H}_2^+$.
Classically you won't expect a big effect, as nuclear-nuclear repulsion and electron-electron repulsion is made up at bonding distances by electron nuclear attraction.
In fact, covalent bonding is an essentially quantum mechanical effect. In the molecule, the electron can move in the potential throughs of two nuclei as compared to only one in the case of a single atom. This increases its positional uncertainty $\Delta x$ along the bond axis and by the uncertainty principle lowers its momentum uncertainty $\Delta p=\hbar/\Delta x$.
Hence also its kinetic energy gets lower although the story doesn't end here.
There is an excellent article by Kutzelnigg, a quantum chemist, on the principle behind bonding:
http://onlinelibrary.wiley.com/doi/10.1002/anie.197305461/abstract

Huh. I never knew that. Thanks Dru.

 

[Pythagorean]

Indistinguishability is not the crucial point, given that you observe covalent bonding already for molecules containing only one electron, namely $\mathrm{H}_2^+$.
Classically you won't expect a big effect, as nuclear-nuclear repulsion and electron-electron repulsion is made up at bonding distances by electron nuclear attraction.
In fact, covalent bonding is an essentially quantum mechanical effect. In the molecule, the electron can move in the potential throughs of two nuclei as compared to only one in the case of a single atom. This increases its positional uncertainty $\Delta x$ along the bond axis and by the uncertainty principle lowers its momentum uncertainty $\Delta p=\hbar/\Delta x$.
Hence also its kinetic energy gets lower although the story doesn't end here.
There is an excellent article by Kutzelnigg, a quantum chemist, on the principle behind bonding:
http://onlinelibrary.wiley.com/doi/10.1002/anie.197305461/abstract

Interesting read. They explicitly demonstrate how little the "exchange force" plays. I'll note that identical particles is a quantum mechanical effect, not a classical one.

So the next claim Griffiths makes is that in the triplet state, the electron pair are "antibonding" which implies to me that they will prevent covalent bonding in that case. Is that true?

 

[DrDu]

So the next claim Griffiths makes is that in the triplet state, the electron pair are "antibonding" which implies to me that they will prevent covalent bonding in that case. Is that true?

The problem is that e.g. in H_2, starting from the 1s orbitals, you can only form one bounding and one anti-bounding orbital as in H2+. With singlet pairing both electrons can fill the same bounding orbital, while with triplet pairing, one has to be in the anti-bonding orbital, so that no net bond results.

 

[my2cts]

The answer to the original question is much simpler than the above. You can construct a very crude model of 2 positive and 2 negative charges which is bound. Take a square ABCD of side a, with protons at A and C and electrons B and D. The energy is (-4+2sqrt(2))/a, which means binding.
Quantum chemistry will give an accurate number involving e kinetic energy, ep potential energy, e-e repulsion mitigated by e-e correlation. To become more precise you have to use Dirac theory, finite nuclear size, nuclear motion, radiative correction. But that was not the question.