Question:

Which is the correct order of second ionization potential of \(\mathrm{C, N, O}\) and \(\mathrm{F}\) in the following ?

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Second ionization energy is the energy to remove an electron from the monopositive ion, so compare the electronic configurations of C+, N+, O+, F+.
Updated On: Oct 31, 2026
  • \(\mathrm{O} \gt \mathrm{N} \gt \mathrm{F} \gt \mathrm{C}\)
  • \(\mathrm{O} \gt \mathrm{F} \gt \mathrm{N} \gt \mathrm{C}\)
  • \(\mathrm{F} \gt \mathrm{O} \gt \mathrm{N} \gt \mathrm{C}\)
  • \(\mathrm{C} \gt \mathrm{N} \gt \mathrm{O} \gt \mathrm{F}\)
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The Correct Option is B

Solution and Explanation

IE2 refers to \(\mathrm{M^+ \rightarrow M^{2+} + e^-}\).
Configurations of the ions: \(\mathrm{C^+}: 2s^2 2p^1\), \(\mathrm{N^+}: 2s^2 2p^2\), \(\mathrm{O^+}: 2s^2 2p^3\), \(\mathrm{F^+}: 2s^2 2p^4\).
\(\mathrm{O^+}\) has a stable half-filled \(2p^3\) set, so removing an electron from it is the hardest: O has the highest IE2.
Next, F+ (\(2p^4\)) has the highest effective nuclear charge among the remaining ions, so F comes second, then N+ (\(2p^2\)), and C+ (\(2p^1\), removal gives stable \(2s^2\)) is the easiest.
Order: \(\mathrm{O} \gt \mathrm{F} \gt \mathrm{N} \gt \mathrm{C}\) (about 3388, 3374, 2945, 2352 kJ/mol).
Answer: (2) \(\mathrm{O} \gt \mathrm{F} \gt \mathrm{N} \gt \mathrm{C}\)
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