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Given that -F=dU/dx, you've just redefined the question. WHY is there a lower potential energy there?
I understand that this is too advanced to explain fully, but could you give an idea why overlapping orbitals= low potential energy? Is it because the opposite charges of the system are closer together overall?
Sounds like a plausible theory, though. I'll keep this open.
Energy tends to be released, classically, when charged particles 'jerk', and there would be a slight jerk as they almost collide, move a little back, oscillating dampedly.
Consider a hydrogen molecular ion -- \[H_2^+ \] Consisting of two protons and an electron. The question is whether the energy of the ion is greater than or less than the energy of a hydrogon atom plus a free proton. Very, very qualitatively speaking, the potential energy resulting from the interaction of the two protons is positive but the potential energy resulting from the interaction of the free proton and the electron is negative, and so there exists a point in at which the positive contribution from the proton-proton interaction is outweighed by the negative contribution from the proton-electron interaction, and the potential energy of the whole system decreases slightly. This is what is responsible for the minimum in potential energy.
So the negative contribution to the potential energy is basically proton-electron attraction?
And far off, they attract each other negligibly, I assume. Anyway, thank you again, for the umpteenth time.
The potential energy function looks some thing like this:|dw:1346957926183:dw| And sure, no problem.