Question

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

Answer 1

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

`"M"_2"CO"_3(s) -> "M"_2"O"(s) + "CO"_2(g)`

Now, recall what `"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, `"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 `"C"-"O"` bond so that the electron distribution is more negative on one side than the other.

This weakens the `"C"-"O"` bond. Therefore, the stronger the polarization is, the easier it is for `"CO"_2` to form in the decomposition process.

Note that `"Na"` is larger than `"Li"`, so `"Na"^(+)` is larger than `"Li"^(+)` as well. Here's a flow chart of how I would go about thinking this out:

`"Na"^(+): "Larger cationic radius"`

`-> "Less tightly-packed charge density"` `("softer hard acid")`

`-> "More polarizable by carbonate ion"`

`-> "Less polarizing towards carbonate ion"`

`-> "C"-"O"` `"bond on carbonate ion less weakened"` `("electrons more evenly shared")`

`-> "The sodium carbonate solid is therefore"` `"harder to decompose"`

`-> color(blue)("Higher decomposition temperature for Na"_2"CO"_3)`