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 :
| \[Be-1{{s}^{2}}2{{s}^{2}};\] |
| \[B-1{{s}^{2}}2{{s}^{2}}2{{p}^{1}};\] |
| \[C-1{{s}^{2}}2{{s}^{2}}2{{p}^{2}};\] |
| \[N-1{{s}^{2}}2{{s}^{2}}2{{p}^{3}};\] |
| \[O-1{{s}^{2}}2{{s}^{2}}2{{p}^{4}}.\] |
| 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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