I came up with a question to arrange thermal stability order of $\ce{NaF}$, $\ce{MgF2}$ and $\ce{AlF3}$ and I think the answer is $\ce{NaF>MgF2>AlF3}$ because $\ce{Na+}$ has largest ionic radius among the cations(anion is same) and also NaF has the greatest ionic character. And as we know greater the ionic character, the greater is the thermal stability.

But the answer is exactly opposite that is $\ce{NaF<MgF2<AlF3}$. Where am I wrong?

Do lattice enthalpy also have a role to play in determining the thermal stability of metal chlorides and fluoride? If yes, then how?

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    $\begingroup$ I what you understand by thermal stability, I have a feeling it's not what you think it is. $\endgroup$
    – Mithoron
    Apr 11, 2019 at 18:02
  • $\begingroup$ @Mithoron According to the most widely prescribed Chemistry textbooks (published by the NCERT) in India, thermal stability is measured by the melting/boiling point of an element/compound relative to other members in a group/period/family of compounds (and whether something decomposes on heating or doesn't exist stably at room temperature). The terminology isn't very accurate and some books are known to contain errors. $\endgroup$
    – Truffle
    Apr 12, 2019 at 7:45
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    $\begingroup$ @Kartik you can't trust NCERT just because it is the most widely prescribed chemistry textbook. They are known to contain notorious errors. $\endgroup$ Apr 12, 2019 at 8:52
  • $\begingroup$ @Nilay Ghosh. Who's trusting it? Just clarifying how the context in which the term thermal stability is used in the question came about. Your comment helps that explain even more! $\endgroup$
    – Truffle
    Apr 12, 2019 at 8:56

1 Answer 1


Metal fluoride thermal stability trends are largely dictated by lattice enthalpies; what matters is the relative size of the cation in consideration (the charge/radius ratio - a measure of polarizability - is also a key factor). Smaller cations can better hold small anions like fluoride. The larger the lattice enthalpy factor, the tougher thermal decomposition becomes. As atomic size decreases across a period, the order of stability is AlF3 > MgF2> NaF.

  • $\begingroup$ But as Al3+ has the smallest size wouldn't three F- electrons result in very large magnitude of inter electronic repulsion, thus decreasing its stability? $\endgroup$ Apr 11, 2019 at 12:23
  • $\begingroup$ @suhridisen Several ionic compounds such as AlF3 have high thermal stability, as exemplified by AlF3's melting point (1290° C). This is due to the electrostatic attraction between the cation and the anion. Even though there are several fluorine electrons around the small Al, the strong attraction between F electrons and the nucleus overcome the repulsive forces. Also, fluorine being highly electronegative instead of having its electrons wandering around the Al ion would keep much of them to itself while Al achieves a stable electronic configuration. $\endgroup$
    – Truffle
    Apr 11, 2019 at 17:03

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