D5.4 Lattice Energy

Refer back to Activity: Analyzing Formation of an Ion Pair. The third step involved bringing one Li⁺ ion close to one F⁻ ion, which lowered the energy. When a large number of oppositely charged ions are brought close together to form a crystal lattice, the energy is lowered even more, because more ions are close together in a structure where attractive forces predominate over repulsive forces.

The lattice energy of an ionic compound is defined as the decrease in energy when the ionic crystal lattice forms from the separated ions. Lattice energy can be calculated using a modified form of Coulomb’s law; some calculated lattice energies are in the table below. (Note that because lattice energy is defined as a decrease in energy, a large positive lattice energy corresponds to a large negative ΔE, that is, a very stable (low energy) ionic solid.)

Lattice energies have a wide range of values depending on which ions make up a compound. Consider, for example, why the lattice energy increases as you go from NaI, to CaI₂, to MgI₂ (first three rows in Table: Calculated Lattice Energies).

According to Coulomb’s law, , the lattice energy is directly proportional to the sizes of the charges on the ions and inversely proportional to the distance between them (the sum of their ionic radii). Thus, we can compare NaI, CaI₂, and MgI₂ with respect to magnitudes of charges and sizes of ions.

Magnitudes of Charges: NaI consists of 1+ ions and 1− ions; CaI₂ and MgI₂ consist of 2+ ions and 1− ions.

Because the magnitude of Q₁Q₂ is larger, the lattice energies of CaI₂ and MgI₂ should be larger than for NaI.

Sizes of ions: All three compounds contain I⁻ anions, so we only need to compare cation radii.

The radii of Ca²⁺ and Na⁺ are similar. This can be estimated from the periodic table: Na⁺ and Ca²⁺ are both smaller than K⁺, Na⁺ because it is above K⁺ in the periodic table, and Ca²⁺ because it is isoelectronic with K+ but has more protons in the nucleus. It is reasonable to assume that the radii are not very different, and the ionic-radius table confirms this: Na⁺ 116 pm; Ca²⁺ 114 pm. Thus, r is not a major factor affecting lattice energy for NaI and CaI₂.

The radius of Mg²⁺ is significantly smaller than for Ca²⁺ (and Na⁺) because Mg is directly above Ca in the periodic table and the ions have the same charge. Thus, MgI₂ should have a larger lattice energy than CaI₂, which is consistent with the values in the table.

When looking at Table: Calculated Lattice Energies, you may have noticed that lattice energies for compounds containing 2+ and 2− ions are nearly quadruple those for similar compounds containing 1+ and 1− ions of similar size (compare NaNO₃ and CaCO₃). This observation is consistent with the general rule that lattice energies are highest for substances with small, highly charged ions.