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?

Answer:

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.