Explain why "Li"_2"CO"_3 decomposes at a lower temperature, whereas "Na"_2"CO"_3 decomposes at a higher temperature?

1 Answer
Dec 31, 2016

First, write out what the decomposition reaction is; it may help:

${\text{M"_2"CO"_3(s) -> "M"_2"O"(s) + "CO}}_{2} \left(g\right)$

Now, recall what ${\text{CO}}_{3}^{2 -}$ looks like:

Clearly, it has a negative charge. The alkali metals are known as hard acids (from HSAB theory), because:

• They have high charge density (relative to same-row elements to their right).
• Their typical ionic form is a cation, ${\text{M}}^{+}$.
• They polarize electron density towards themselves, because it is ${\delta}^{-}$ while the hard acids are ${\delta}^{+}$.
• Their ionic radii are quite small.

So, when they are near negative ions, they polarize the ${\delta}^{-}$ end of the $\text{C"-"O}$ bond so that the electron distribution is more negative on one side than the other.

This weakens the $\text{C"-"O}$ bond. Therefore, the stronger the polarization is, the easier it is for ${\text{CO}}_{2}$ to form in the decomposition process.

Note that $\text{Na}$ is larger than $\text{Li}$, so ${\text{Na}}^{+}$ is larger than ${\text{Li}}^{+}$ as well. Here's a flow chart of how I would go about thinking this out:

$\text{Na"^(+): "Larger cationic radius}$

$\to \text{Less tightly-packed charge density}$ $\left(\text{softer hard acid}\right)$

$\to \text{More polarizable by carbonate ion}$

$\to \text{Less polarizing towards carbonate ion}$

$\to \text{C"-"O}$ $\text{bond on carbonate ion less weakened}$ $\left(\text{electrons more evenly shared}\right)$

$\to \text{The sodium carbonate solid is therefore}$ $\text{harder to decompose}$

$\to \textcolor{b l u e}{{\text{Higher decomposition temperature for Na"_2"CO}}_{3}}$