A) \[F<N<C<O\]
B) \[C<N<O<F\]
C) \[C<O<N<F\]
D) \[F<O<N<C\]
Correct Answer: C
Solution :
: In a period, the value of ionisation enthalpy increases from left to right with irregularity where the atoms have somewhat stable configurations. In nitrogen (second period) 2p- orbitals are half filled leading to higher value of ionisation energy than oxygen. Hence, the order is \[\underset{(1086)}{\mathop{C}}\,<\underset{(1314)}{\mathop{O}}\,<\underset{(1402)}{\mathop{N}}\,<\underset{(1681)}{\mathop{F}}\,\,\,\,\,\,\,kj\,mo{{l}^{-1}}\]
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