This may sound like a repetition of this: When do oppositely charged ions (say Na$^+$ and Cl$^-$) stop attracting each other because of their electron shells?

(I asked that too...), but that was about another detail that, though relevant, is a different subject and doubt.

I think the question says it. I am toying with coding a primitive atom simulator, and I have no clue on how to start dealing with the forces involved when dealing with the attraction-repulsion of ions. Say, Na$^+$ and Cl$^-$. What is relevant for me is just the charge of the nucleus (nucleus dimensionless, for my purposes), the radius and charge of the electron shell of 2 atoms.

What do I do? I tried to think of two ions as two positive balls separated by 2 negative walls, and before I wrote any equations to verify stuff, my mind stopped at 'the walls will repel more than the overall ions will attract, because they will always be closer' (We use Coulomb, right?). Then it goes back to 'hold on... but at a distance D, they DO react as differently charged bodies and DO attract'. Help please? Thank you all in advance.


If the distance between the ions was very large compared with the electrons' orbital radius in the cation, the small charge difference between the cation's nucleus and electron cloud would give a net attractive force on the anion. There is some distance of separation where the force is balanced, and once the anion gets that close it is no longer accelerated towards the cation. Instead it overshoots this equilibrium point until it is decelerated to instantaneous rest, at which point the ions move apart until an overshoot past the equilibrium eventually leads to another reversion.

This oscillation is similar to the $\ddot{x}=-\omega^2 x$ behaviour you see for a spring, where $x$ is its displacement relative to equilibrium. The high mass of each nucleus ensures the length scale for this oscillation will be quite small, so for all intents and purposes the ions are locked at an equilibrium separation. And that distance is still small enough for us to think of the ion pair as a bound state, since it will take an ample amount of energy to separate them, just as it takes a lot of energy to snap a spring.

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  • $\begingroup$ Many thanks :) So, to get the attraction force, I might consider my ions as point charges and Coulomb them, but how do I get the repulsion force? This is the part killing me, cuz if I use the same treatment, the forces cancel each other and nothing moves :P $\endgroup$ – Jorge Al Najjar Mar 9 '17 at 19:10
  • $\begingroup$ @JorgeAlNajjar If two atoms were very close their outermost electrons would repel each other over a much shorter distance than the distance from either cloud to the other nucleus. $\endgroup$ – J.G. Mar 10 '17 at 0:00
  • $\begingroup$ Sure! but how? Coulomb will give me that the ions attract according to: Fattraction=(9*10^9) * (1.6*10^-19)*(-1.6*10^-19)/d^2 (d being the distance between nuclei, cuz Im considering the ions as points here). If I apply the same thing to the electron shells, Frepulsion=(9*10^9)*(1.6*10*10^-19)*(1.6*18*10-19)/(d-R1-R2)^2, charges increase(10 electrons in Na, 18 in Cl), distance diminishes, and repulsion force is always bigger, meaning, what am I doing wrong? :) $\endgroup$ – Jorge Al Najjar Mar 10 '17 at 17:06
  • $\begingroup$ One of your formulae used electrons' charges, the other unit charges instead of nuclear charges. That may be your problem. $\endgroup$ – J.G. Mar 10 '17 at 19:15

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