Obviously, there are forces of attraction between the Cl- anions and the Na+ cations. But aren't there also forces of repulsion between the Na+ atoms (look diagonally between the grey atoms)? Likewise for Cl-? Why is this not considered?
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1$\begingroup$ Well, you just considered them, but I don't see the results. $\endgroup$– MithoronCommented Sep 28 at 12:06
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$\begingroup$ Fun fact: had we forgot to consider them, the lattice energy of any lattice would turn out infinite. $\endgroup$– Ivan NeretinCommented Sep 28 at 13:31
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2 Answers
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Before asking why, ask if. They are obviously considered.
But for $\ce{NaCl}$ case, the nearest same ions are $\sqrt{2}$ times farther than the nearest opposite ions, therefore the respective repulsion force is two times weaker than the attractive one. Similarly for farther ions.
In average, the opposite ions are closer than the same ions (similarly as in salt solutions), what leads to positive lattice energy (negative lattice formation enthalpy).
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We very much do consider those repulsions, and for good measure all the interactions between more distant ion pairs. These are all rolled up into a factor in the lattice energy of ionic compounds, called the Madelung constant. The reader is referred to the Wikipedia link for all the boring mathematical details.
Less boring is the fact that the ionic lattices themselves are built to include the repulsions between like ions: they are not close-packed precisely because contact between like-charged ions is to be avoided; in terms of the math having such contacts would give a smaller Madelung constant and thus less lattice stabilization.
