A) \[C>N>O\]
B) \[O>N>C\]
C) \[C<N>O\]
D) \[C>N\sim O\]
Correct Answer: C
Solution :
Key Idea lonisation potential The amount of energy needed to take out the most loosely bound electron from a neutral isolated atom is called ionisation energy or ionisation potential. In a period from left to right ionisation potential of element generally increases. In a sub-group from top to bottom IP of elements decreases. Carbon, nitrogen and oxygen present in the group IV A, V A and VI A respectively. They present in the same period, ie, second period. Therefore, the IP of C, N and 0 must increase as \[C<N<O\] But the IP of nitrogen is greater than oxygen due to extra stability of half-filled p-sub-shell in nitrogen atom. Oxygen has \[2{{p}^{4}}\] electronic configuration in the outer most shell hence, its electron is more easily removed. So, the correct order of the first IP of N, 0, and C is \[C<N>O\]
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