• # question_answer 1) Why does $L{{i}_{2}}C{{O}_{3}}$ decompose at lower temperature whereas $N{{a}_{2}}C{{O}_{3}}$at higher temperature?

The $L{{i}^{+}}$on being smaller in size exerts a strong polarising action and distorts the electron cloud of the nearby oxygen atoms of the large sized $CO_{3}^{2-}$ion. It results in weakening of the C-O bond and strengthening of Li-O bond. This ultimately facilitates the decomposition of $L{{i}_{2}}C{{O}_{3}}$ into $L{{i}_{2}}O$ and$C{{O}_{2}}$ at lower temperature. The lattice energy of $L{{i}_{2}}O$is higher than the lattice energy of$L{{i}_{2}}C{{O}_{3}}$. This also favours the decomposition of$L{{i}_{2}}C{{O}_{3}}$. $N{{a}^{+}}$ion being bigger in size is not capable of exerting polarizing action on oxygen atoms of carbonate ions. Thus,$N{{a}_{2}}C{{O}_{3}}$is a stable compound.