For an ionic compound, is it always true that the greater the thermal stability, the greater the lattice energy. E.g. for 2 ionic compounds MX and MY, if MX has a higher thermal stability than MY, does MX necessarily have a higher lattice energy?


2 Answers 2


1) Lattice energy is the measure of the stability of an ionic solid.

It is the energy required to separate the ions in one mole of a solid ionic compound into gaseous ions.

The weaker the bond between the ions, the easier it will be to break that bond and separate the ions into gaseous ions and the lesser will be the lattice energy.

Therefore, the stronger the bond between the ions, the more is the lattice energy of the ionic compound.

2 )Thermal stability depends upon the amount of heat required to break the ionic compound. Clearly, the stronger the bond between the ions is, the more heat will be required to break the bond and as a result, the thermal stability of the compound increases.

Therefore, stronger the bond, more is the thermal stability of the ionic solid.

In general, we can say that as thermal stability increases, the lattice energy also increases, but they also also depend upon the crystal structure of the solid and other factors.

So, you cannot say with certainty that if the thermal stability is high, then lattice energy will be high too.

Hence, the answer is no. If MX has a higher thermal stability than MY, then it may or may not have higher lattice energy than MY.

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    $\begingroup$ That correlation is not true. There's a very simple counter-example: $\ce{MgCO3}$ vs $\ce{BaCO3}$. $\ce{MgCO3}$ has a larger lattice energy ($\ce{Mg^2+}$ smaller than $\ce{Ba^2+}$), but lower thermal stability (decomposes at a lower temperature). $\endgroup$ Feb 16, 2017 at 13:59
  • $\begingroup$ You mean the part where I said one implies the other and vice versa? Yeah, I didn't take into account the crystal structures. $\endgroup$
    – Arishta
    Feb 16, 2017 at 14:01
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    $\begingroup$ It's not about crystal structure, really. The thing is that you have to consider the products - in this case it forms $\ce{MO}$ and that's very favourable for $\ce{Mg}$ and not as favourable for $\ce{Ba}$. (That's because of, again, lattice energy.) $\endgroup$ Feb 16, 2017 at 14:11
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    $\begingroup$ Why wouldn't it matter ? Clearly, some crystal structures are more stable than others. So, energy required to break the ionic lattice and separate the ions into its constituent atoms would be obviously high. Okay, could it be because crystal structures would affect both, lattice energy and thermal stability? $\endgroup$
    – Arishta
    Feb 16, 2017 at 14:18
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    $\begingroup$ The factors would probably depend on the precise route of thermal decomposition. Essentially if thermal decomp. is a reaction $\ce{A -> B + C +}\cdots$, to find how favourable this reaction is, you would need to not only look at the stability of the reactant $\ce{A}$ (which is indeed measured by the lattice energy) but also the stability of the products $\ce{B + C + }\cdots$. The exact factors would depend on exactly what $\ce{B}$ and $\ce{C}$ etc. are. $\endgroup$ Feb 16, 2017 at 14:36

When you supply energy to an ionic compound to separate it into its component ions (e.g. $\ce{MgCO3}$ to $\ce{Mg^2+}$ and $\ce{CO3^2-}$) it is called the lattice energy. The same amount of energy is released when this same ionic compound is formed ($\ce{MgCO3}$).

You can also heat this compound ($\ce{MgCO3}$) until it breaks down to $\ce{MgO}$ and $\ce{CO2}$. And that is called thermal decomposition. I don't know what determines whether the heat supplied will break down the ionic compound or separate it into its component ions but that is the difference between the two processes.


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