A) \[Be<B<C<N<O\]
B) \[B<Be<C<O<N\]
C) \[Be>B>C>N>O\]
D) \[B<Be<N<C<O\]
Correct Answer: B
Solution :
\[Be(2{{s}^{2}})\]and\[N(2{{s}^{2}}2{{p}^{3}})\]have completely filled and half-filled orbital, so Be and N have more stable electronic configurations and thus their IE. is more than B and O respectively. So, the correct order is \[B<Be<C<O<N.\]
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