A) \[B<Be<C<O<N\]
B) \[B<Be<C<N<O\]
C) \[Be<B<C<N<O\]
D) \[Be<B<C<O<N\]
Correct Answer: A
Solution :
[a] \[Be-1{{s}^{2}}2{{s}^{2}};\,\,B-1{{s}^{2}}2{{s}^{2}}2{{p}^{1}};\,\,C-1{{s}^{2}}2{{s}^{2}}2{{p}^{2}};\]\[N-1{{s}^{2}}2{{s}^{2}}2{{p}^{3}};\,\,O-1{{s}^{2}}2{{s}^{2}}2{{p}^{4}}.\] IP increases along the period. But IP of \[Be>B.\]Further IP of \[O<N\]because atoms with fully or partly filled orbitals are most stable and hence have high ionisation energy.
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