A) \[{{K}_{2}}O<N{{a}_{2}}O<A{{l}_{2}}{{O}_{3}}<MgO\]
B) \[A{{l}_{2}}{{O}_{3}}<MgO<N{{a}_{2}}O<{{K}_{2}}O\]
C) \[MgO<{{K}_{2}}O<A{{l}_{2}}{{O}_{3}}<N{{a}_{2}}O\]
D) \[N{{a}_{2}}O<{{K}_{2}}O<MgO<A{{l}_{2}}{{O}_{3}}\]
Correct Answer: B
Solution :
While moving from left to right in periodic table basic character of oxide of elements will decrease. \[\therefore \,\xrightarrow[\text{Increasing basic strength}]{A{{l}_{2}}{{O}_{3}}<MgO<N{{a}_{2}}O}\]and while descending in the group basic character of corresponding oxides increases. \[\therefore \,\xrightarrow[\text{Increasing basic strength}]{N{{a}_{2}}O<{{K}_{2}}O}\] Increasing basic strength \[\therefore \] Correct order is \[A{{l}_{2}}{{O}_{3}}<MgO<N{{a}_{2}}{{K}_{2}}O\]
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