A) Li
B) Na
C) Rb
D) K
Correct Answer: A
Solution :
\[L{{i}^{+}}\] being smallest possesses strong positive field and, thus, combines with small anion \[{{O}^{2-}}\]to form stable \[L{{i}_{2}}O\]compound. The\[N{{a}^{+}},{{K}^{+}}\]and Rb"1 being relatively larger exert less strong positive field around them and thus react with larger oxygen anion, i.e., \[O_{2}^{2-}\]and \[O_{2}^{1-}\] to form stable oxides.
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