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 Table 1. (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.)

Substance Lattice Energy (kJ/mol)
NaI 682
CaI2 1971
MgI2 2293
NaOH 887
Na2O 2481
NaNO3 755
Ca3(PO4)2 10,602
CaCO3 2804
Table: Calculated Lattice Energies. Representative values from CRC Handbook of Chemistry and Physics (2004).

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 1).

According to Coulomb’s law,  E_{\text{p}} = k_e\frac{Q_1Q_2}{r} , 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.

Activity: Comparing Lattice Energies


When looking at Table 1, 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.

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Chemistry 109 Fall 2021 by John Moore, Jia Zhou, and Etienne Garand is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.