Question

How does the polarization of the carbonate ion make the thermal decomposition of CaCO3 more likely?

Answer 1

I think I see the confusion. You may not be seeing the intramolecular effects. The decomposition reaction is:

`"CaCO"_3(s) stackrel(Delta" ")(->) "CaO"(s) + "CO"_2(g)`

The effect of interest is the polarization towards the `"Ca"^(2+)` cation from the `"CO"_3^(2-)` anion, causing intramolecular bond weakening.

Yes, there is an ion-pairing attraction going on, but that is not going against the ability of the `"CaCO"_3` to decompose when you heat the compound (it is a `"Ca"^(2+)-"O"^(2-)` ion-pair interaction that will later hold the `"CaO"(s)` formula unit together, anyways). However, what you should also be looking at is how the polarization facilitates decomposition.

Since the calcium cation is highly-positively-charged, and is somewhat small, one might call it a "hard acid" (from Hard-Soft Acid/Base Theory), because it can capably concentrate negative charge density towards itself, and we call that great polarizing ability. We can also say that `"Ca"^(2+)` is very electropositive.

Thus, it pulls electron density from the `"CO"_3^(2-)`, specifically within the `"C"-"O"` bond (since the individual dipoles [that all cancel out] within `"CO"_3^(2-)` lie along the `"C"-"O"` bond, making it have the highest vector alignment). This unbalances the electron distribution and makes the `"C"-"O"` bond more ionic in character. That weakens the `"C"-"O"` bond, and facilitates breaking the bond.

That bond will be broken upon heating in order to perform the decomposition reaction.