question_answer

Why does \[L{{i}_{2}}C{{O}_{3}}\]  decompose at a lower temperature as compared to\[N{{a}_{2}}C{{O}_{3}}\]?

Answer:

                Upon heating, \[L{{i}_{2}}C{{O}_{3}}\] forms \[L{{i}_{2}}O\]  and \[C{{O}_{2}}\]. On the other hand, \[N{{a}_{2}}C{{O}_{3}}\] upon heating is expected to form\[N{{a}_{2}}O\]  and\[C{{O}_{2}}\]. Since the size of \[L{{i}^{+}}\] ion is smaller as compared to that of \[N{{a}^{+}}\] ion therefore, the crystal lattice of \[L{{i}_{2}}O\] is thermally more stable than that of \[N{{a}_{2}}O.\] \[L{{i}_{2}}C{{O}_{3}}\]  therefore, decomposes upon heating to form lithium oxide (stable oxide). \[N{{a}_{2}}C{{O}_{3}}\] does not decompose on heating since \[N{{a}_{2}}O\] is less stable.