The difference in thermal stability of $\ce{CaCO3}$ and $\ce{MgCO3}$ can be explained using Fajans' rules, but why not using the lattice energy method?

Why is the thermal stability of $\ce{CaCO3}$ higher even though the lattice energy decreases down a group (as the size of cations increases)?

  • $\begingroup$ Smaller magnesium ion destabilises the carbonate anion more. Electrostatic and chemical stabilities go against each other here, with the latter having the upper hand. $\endgroup$
    – Poutnik
    Aug 25, 2021 at 8:03

2 Answers 2


The so called "lattice energy" method doesn't work for all compounds, same is the case with Fajan's rule. These are theories which have been developed to explain the characteristics and make it somewhat believable to the general audience but we cannot compare all compounds using one theory or the other because each theory has its own drawbacks.

Lattice energy is directly proportional to the mod of product of charges on individual ions, which is same in this case, further it is inversely proportional to the radius of the resulting compound, and you're right that $\ce{CaCO_3}$ has lower lattice energy implying that it has lower ionic character, which further implies that it should have lesser thermal stability but here's the drawback. Lattice energy isn't the dominating factor when it comes to ionic character.

Diamond has very high lattice energy, you can't say it is ionic, can you?

Fajan's rule along with size compatibility and polarization decides the ionic character of a compound. By Fajan's rule we have: covalent character directly proportional to the extent of polarization, implying that the smaller cation will cause greater distortion of electron cloud as compared to the larger cation. This means that $\ce{CaCO_3}$ would have higher extent of polarization, i.e. higher covalent character. This would cause slight thermal instability, and $\ce{MgCO_3}$ would be relatively more thermally stable.

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    – harry
    Aug 25, 2021 at 12:25

The rules properly predict the relative stabilizes we observe for calcium and magnesium salts, such as the carbonates and hydroxides, but they actually got lucky in that case. If you try to bring in the beryllium compounds, which should be less stable than the magnesium compounds according to these rules, you run into problems. The issue is that kinetics, which the rules do not really account for, in reality enters the picture.

Let's look at the decomposition of beryllium and magnesium hydroxides.

According to the respective Wikipedia articles, beryllium hydroxide decomposes at 400°C whereas magnesium hydroxide does so at only 350°C even though magnesium hydroxide "should,be" more stable.

If you look up the thermochemistry data for $\ce{BeO,Be(OH)2,H2O}$ you find that the standard free energy change for the decomposition

$\ce{Be(OH)2\to BeO + H2O(g)}$

would reach zero at about 125°C, so the oxide and hydroxide would be in equilibrium under one atmosphere water vapor pressure at this temperature. With magnesium in place of beryllium the corresponding equilibrium temperature is higher, about 207°C with the data I used. These temperatures are lower than the observed hydroxide decomposition temperatures but in the "right order" according to Fajan's Rules.

What happens is both decompositions are kinetically hindered, as we would expect with solid phase reactions, and the kinetic hindrance is greater with beryllium than with magnesium. The net result is that beryllium hydroxide is more stable kinetically whereas magnesium hydroxide would have been more stable based on thermodynamics. Because the lattice energy and Fajan rules are generally thermodynamically based, they miss on the kinetically dominated cases given here.


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