# Why is sodium carbonate more thermally stable than aluminium carbonate

Sodium carbonate will not decompose in the presence of heat. This is because sodium is a very reactive metal, and hence forms very strong bond with the carbonate ion. In contrast, aluminium is not as reactive as sodium. Hence the bond by aluminium is weaker, and aluminium carbonate will decompose when heated.

But sodium ion has a charge of +1, while aluminium ion has a charge of +3. If that's the case, wouldn't the ionic bond formed by aluminium be stronger? Hence, shouldn't aluminium carbonate be more stable to heat?

So does the reactivity of a metal, or the charge on an ion, determine the thermal stability of an ionic compound?

Thank you :)

Borek
Mentor
Try to write decomposition reactions for both carbonates and think about the stability of the products.

Thermal stability depends on the reactivity of the metal. Taking aluminium and sodium for example. Even though aluminium has greater ionic charge, it has 3 valence electrons while sodium only has 1. Hence, sodium is able to form ions more easily.

Borek
Mentor
Thermal stability depends on the reactivity of the metal. Taking aluminium and sodium for example. Even though aluminium has greater ionic charge, it has 3 valence electrons while sodium only has 1. Hence, sodium is able to form ions more easily.

And? I don't see how it is related to the original question. We start with ions and we end with ions.

And so the ability to form ions dictates the reactivity which then affects thermal stability.

The thermal stability is related to the metal reactivity. It is because a more reactive metal has higher tendency to obtain a stable octect/duplet outermost shell electrons to form a stable ion. So energy needed for it to return to an unstable state (i.e. a metal atom) is higher.
On the otherhand, melting point and boiling point is related to the size of the ion. The smaller the size of an ion, the higher the melting point the compound has.
As the reactivity of sodium is higher than that of aluminium, therefore it has a higher thermal stability. However, as the atmoic size decrease across the period, aluminum has a smaller atomic radius than sodium and hence it has a stronger ionic bond and melting point than sodium

Borek
Mentor
The thermal stability is related to the metal reactivity.

Thermal stability of what? What you wrote later suggest you mean "thermal stability of a cation" but it makes no sense, as cation never (unless in exotic conditions) exist separately, and we were discussing here stability of a salt. And the stability of a salt doesn't depend on the properties of a metal alone.

Besides, I have no idea how to define "thermal stability of a cation". The only thing that can happen to the cation when it is heated is it can get ionized even more than it already is ionized, but it won't ever get back to the neutral state.

Of the ionic compound. What I think is that the thermal stability of a compound is mainly related to its metallic ions, i.e. the higher the the reactivity series the metallic ions at, the higher the thermal stability its ionic compound is.
Also cation can return to its neutral state when heated. e.g CuS(
s) + O2(g) →Δ Cu(s)+ SO2(g)

Borek
Mentor
Of the ionic compound. What I think is that the thermal stability of a compound is mainly related to its metallic ions, i.e. the higher the the reactivity series the metallic ions at, the higher the thermal stability its ionic compound is.

No, the reactivity of a cation is not enough to predict stability of an ionic compound. Different salts of a same cation have different stabilities, so the identity of the anion plays a role as well. Decomposition of CaCO3 is trivial, decomposition of CaSiO3 not so.

Also cation can return to its neutral state when heated. e.g CuS(s) + O2(g) →Cu(s)+ SO2(g)

No, it is not a cation returning to a neutral state on heating, it is a metathesis reaction. Entirely different situation. Not to mention the fact I wouldn't call CuS ionic, with electronegativity difference around 0.7 it is mostly covalent.