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

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/]()

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.

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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)#