A) \[Li<Be<B<C<N<O<F<Ne\]
B) \[Li<B<Be<C<O<N<F<Ne\]
C) \[Li>Be>B>C>N>O>F>Ne\]
D) \[Li>B>C>Be>O>N>F>Ne\]
Correct Answer: B
Solution :
The first lonisation Energy (IE) increases in a period. Thus, the first IE of the elements of second period should be as follows: \[Li<Be<B<C<N<O<F<Ne\] But in practice, the elements do not follow the above order. The first IE of these element is \[Li<B<Be<C<O<N<F<Ne\] The lower IE of B than that of Be is because in \[B(1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}),\]electron is to be removed from 2p which is easy while in\[Be(1{{s}^{2}},2{{s}^{2}}),\]electron is to be removed from 2s which is difficult. The low IE of is lower than that of N because of the half-filled \[2p\]orbitals in\[N(1{{s}^{2}},2{{s}^{2}},2{{p}^{3}})\],
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