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The s-block Elements — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryThe s-block Elements3 Marks
Question
Why is $\mathrm{Li}_2 \mathrm{CO}_3$ decomposed at a lower temperature whereas $\mathrm{Na}_2 \mathrm{CO}_3$ at higher temperature?
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Answer

$\mathrm{Li}_2 \mathrm{CO}_3$ decomposes on heating because the $\mathrm{Li}^{+}$ion exerts a strong polarising action and distorts the electron cloud of the nearby oxygen atom of the large $\mathrm{CO}_3^{2-}$ ion. This results in the weakening of the $\mathrm{C}-\mathrm{O}$ bond and strengthening of the $\mathrm{Li}-\mathrm{O}$ bond. This ultimately facilitates the decomposition of $\mathrm{Li}_2 \mathrm{CO}_3$ into $\mathrm{Li}_2 \mathrm{O}$ and $\mathrm{CO}_2$. The lattice energy of $\mathrm{Li}_2 \mathrm{O}$ is higher than the lattice energy of $\mathrm{Li}_2 \mathrm{CO}_3$. This also favours decomposition of $\mathrm{Li}_2 \mathrm{CO}_3 . \mathrm{Na}^{+}$ion being bigger in size, the lattice energy of $\mathrm{Na}_2 \mathrm{O}$ is less stable than that of $\mathrm{Na}_2 \mathrm{CO}_3$. Therefore, $\mathrm{Na}_2 \mathrm{CO}_3$, does not decompose on heating.

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