A) N
B) O
C) Be
D) Li
Correct Answer: D
Solution :
After losing one electron, these have, following configurations: \[{{\,}_{7}}{{N}^{+}}=1{{s}^{2}},2{{s}^{2}}2{{p}^{2}}\] \[{{\,}_{8}}{{O}^{+}}=1{{s}^{2}},2{{s}^{2}}2{{p}^{3}}\] \[{{\,}_{4}}B{{e}^{+}}=1{{s}_{2}},2{{s}^{1}}\] \[{{\,}_{3}}L{{i}^{+}}=1{{s}^{2}}\] As \[\text{L}{{\text{i}}^{\text{+}}}\]has the most stable electronic cofigurtion, its ionisation energy (i.e., \[\text{I}{{\text{E}}_{\text{2}}}\,\text{Li}\] of Li) will be the highest.
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