A) \[2NaOH\xrightarrow{\Delta }N{{a}_{2}}O+{{H}_{2}}O\]
B) \[2LiOH\xrightarrow{\Delta }L{{i}_{2}}O+{{H}_{2}}O\]
C) \[MgC{{O}_{3}}\xrightarrow{\Delta }MgO+C{{O}_{2}}\]
D) All of the above
Correct Answer: A
Solution :
Key Idea The thermal stability of group I hydroxides increases with increasing size of cation. \[LiOH\] decomposes readily but other alkali hydroxides are most stable. \[2NaOH\xrightarrow{\Delta }N{{a}_{2}}O+{{H}_{2}}O\] This reaction is not feasible because sodium hydroxide is thermally stable. \[2LiOH\xrightarrow{\Delta }L{{i}_{2}}O+{{H}_{2}}O\] \[LiOH\] is unstable towards heating due to small size of Li atom. \[MgC{{O}_{3}}\] gives magnesium oxide and \[C{{O}_{2}}\] on heating. \[MgC{{O}_{3}}\xrightarrow{\Delta }MgO+C{{O}_{2}}\]
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