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:

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

Now, recall what "CO"_3^(2-) looks like:

![http://upload.wikimedia.org/](useruploads.socratic.org)

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.

![www.chemguide.co.uk)

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)