A) \[Be<B<N<O\]
B) \[B<Be<N<O\]
C) \[B<Be<O<N\]
D) \[O<N<B<Be\]
Correct Answer: C
Solution :
lionization energy generally increase from left to right in a period. Hence expected increasing order is Be, B, N, O but the actual order, found is: B < Be < O < N The reason is that Be and N, have comparatively stabler electronic configurations (\[n{{s}^{2}}\] and \[n{{p}^{3}}\]) than those of B and O (\[n{{p}^{1}}\] and \[n{{p}^{4}}\]). Hence, their ionisation energy becomes higher.
You need to login to perform this action.
You will be redirected in
3 sec
