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

May 5, 2017

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

${\text{CaCO"_3(s) stackrel(Delta" ")(->) "CaO"(s) + "CO}}_{2} \left(g\right)$

The effect of interest is the polarization towards the ${\text{Ca}}^{2 +}$ cation from the ${\text{CO}}_{3}^{2 -}$ anion, causing intramolecular bond weakening.

Yes, there is an ion-pairing attraction going on, but that is not going against the ability of the ${\text{CaCO}}_{3}$ to decompose when you heat the compound (it is a ${\text{Ca"^(2+)-"O}}^{2 -}$ ion-pair interaction that will later hold the $\text{CaO} \left(s\right)$ 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 ${\text{Ca}}^{2 +}$ is very electropositive.

Thus, it pulls electron density from the ${\text{CO}}_{3}^{2 -}$, specifically within the $\text{C"-"O}$ bond (since the individual dipoles [that all cancel out] within ${\text{CO}}_{3}^{2 -}$ lie along the $\text{C"-"O}$ bond, making it have the highest vector alignment). This unbalances the electron distribution and makes the $\text{C"-"O}$ bond more ionic in character. That weakens the $\text{C"-"O}$ bond, and facilitates breaking the bond.

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