The correct order of C, N, 0 and F in terms of second ionisation potential is

We need to find the correct order of C, N, O, and F in terms of second ionisation potential (IE$$_2$$).

Write the electronic configurations of the singly charged cations.

The second ionization removes an electron from the singly charged cation:

C$$^+$$ (5 electrons): $$1s^2\,2s^2\,2p^1$$ — removing the lone $$2p$$ electron

N$$^+$$ (6 electrons): $$1s^2\,2s^2\,2p^2$$ — removing one of two $$2p$$ electrons

O$$^+$$ (7 electrons): $$1s^2\,2s^2\,2p^3$$ — removing from a half-filled, extra-stable $$2p$$ subshell

F$$^+$$ (8 electrons): $$1s^2\,2s^2\,2p^4$$ — removing one paired $$2p$$ electron

Apply trends and anomalies.

General trend: IE$$_2$$ increases with nuclear charge (left to right in period).

Anomaly: O$$^+$$ has the half-filled $$2p^3$$ configuration, which has extra exchange energy stabilization. Removing an electron from this configuration requires more energy than expected. This makes IE$$_2$$(O) anomalously high, even exceeding IE$$_2$$(F) despite F having a higher nuclear charge.

For F$$^+$$ ($$2p^4$$), the fourth electron is paired, making it relatively easier to remove compared to the half-filled O$$^+$$.

Determine the order.

Standard IE$$_2$$ values (kJ/mol): C (2353) < N (2856) < F (3374) < O (3388)

Order: $$\text{C} < \text{N} < \text{F} < \text{O}$$

The correct answer is Option C: C < N < F < O.