Lets consider a dipole (H-Cl). The positive and negative interactions between dipoles should cancel out, resulting in 0 net intermolecular forces. But " energetically favored lower-energy attractive interactions occur more frequently ". Why are attractive interaction more favorable than their repulsive counterparts?
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Attractive interactions are more favorable up to a certain separation because potential energy is minimized. However, if you were to push the dipoles together too close then repulsion is more favorable. Although this picture of attraction being "more frequent" isn't really the case.
Think of the interactions like a spring. There is a preferred spring length (separation distance) that is equilibrium. If we pull the spring (separation), then the system is "attracted" back to equilibrium. If we push the spring (push together), then the system is "repelled" back to equilibrium.
A system of dipoles will rotate and move around to minimize the energy of the system. But whether things are attracting or repelling just depends on the separation of the dipoles when not in equilibrium.
answered Aug 21, 2018 at 12:28
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$\begingroup$ Then, if a system containing dipoles comes to a state of equilibrium, shouldn't all the repulsive and attractive forces cancel out? If they do, then ideal gas laws should be applicable on them but they are not. I am still confused! $\endgroup$ Commented Aug 21, 2018 at 13:11
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$\begingroup$ Just because the forces cancel out doesn't mean the ideal gas law holds. In fact, the ideal gas law assumes the gas particles don't interact, which is not at all true in this system. $\endgroup$ Commented Aug 21, 2018 at 13:17