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Is the energy released when the electron is attached to the Cl atom enough to ionize the Na atom? To find out, let’s compare the ionization energy of Na to the electron affinity of Cl:

Na → Na + + e - Ionization energy = 496 kJ/mol

Cl + e - → Cl - Electron affinity = -349 kJ/mol

Our answer is no. Taking an electron from a Na atom and giving it to a Cl atom costs a good amount of energy in total. This seems to suggest that the electron should not leave the Na atom and join the Cl atom, so NaCl shouldn’t form ions and therefore shouldn’t form a stable compound.

We haven’t considered one key factor, however. The energy comparison above leads to the formation of independent positive and negative ions which don’t interact with each other after the reaction is complete. But in reality, the Na+ and Cl- ions are very close to one another and attracted to one another. Coulomb’s law tells us that this significantly lowers the energy. And there is even more to consider. A crystal of NaCl does not consist of a single Na + and a single Cl - . Instead, it is an entire array of many positive and negative ions. Each positive ion is surrounded by several negative ions. And each negative ion is surrounded by the same number of positive ions. (It turns out that number is 6.) Coulomb’s law tells us that we get a huge lowering of energy from having all these opposite charges adjacent to one another. This energy is called the “lattice energy” and it is very large, -787 kJ/mol. This is much more than the energy deficit for ionizing both atoms, and accounts easily for the bonding in NaCl.

The bonding in NaCl is thus different than the covalent bonding in, say, HF or the metallic bonding in, say, Cu metal. For obvious reasons, we refer to this type of bonding as “ionic bonding.”

Before concluding that ionic bonding is responsible for the stability of NaCl, we need to ask about the other primary property of NaCl mentioned above. Specifically, NaCl is brittle and not malleable. This is quite different from the property of a metal. In a metal, we could rearrange the atoms, for example by bending or by deforming with a hammer, and the atoms remain strongly bonded. But we cannot bend NaCl crystals, and if we hit them with a hammer, the bonding is destroyed as the crystal shatters. We simply cannot rearrange the atoms. It is clear that the bonding in NaCl depends very much on the arrangement of the atoms.

If we think about our ionic bonding model, this makes perfect sense. For the ionic bonding to work, the negative ions must remain surrounded by the positive ions and vice versa. Any attempt to rearrange these ions will result in positive ions adjacent to positive ions and negative ions next to negative ions. This will create strong repulsions, and the solid will fall apart. Ionic bonding thus accounts for the brittleness of NaCl.

So far, we’ve only looked at ionic bonding in NaCl as an example, but since we’ve seen that different covalent bonds have different energies, perhaps different ionic bonds have different energies. We compare different salts to see if there are different lattice energies in the ionic bonds. [link] shows a set of lattice energies for salts formed from alkali metals (Li, Na, K, Rb) and halogens (F, Cl, Br, I). There are some clear trends in these data. The largest lattice energy corresponds to the combination of the two smallest ions, Li + and F - . The lattice energy decreases when either or both of the ions are larger, with the smallest being for RbI, consisting of the two largest ions.

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Source:  OpenStax, Concept development studies in chemistry 2013. OpenStax CNX. Oct 07, 2013 Download for free at http://legacy.cnx.org/content/col11579/1.1
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