Which is the correct order of second ionization potential of C, N, O and F in the following?

The second ionization potential is the energy required to remove an electron from a positively charged ion (after the first electron has been removed). We need to compare this for Carbon (C), Nitrogen (N), Oxygen (O), and Fluorine (F).

First, recall the electron configurations of the neutral atoms:

• Carbon (C, atomic number 6): $$1s^2 2s^2 2p^2$$
• Nitrogen (N, atomic number 7): $$1s^2 2s^2 2p^3$$
• Oxygen (O, atomic number 8): $$1s^2 2s^2 2p^4$$
• Fluorine (F, atomic number 9): $$1s^2 2s^2 2p^5$$

After the first ionization (removing one electron), the ions are:

• C⁺: Remove one electron from $$2p^2$$ → $$1s^2 2s^2 2p^1$$
• N⁺: Remove one electron from $$2p^3$$ → $$1s^2 2s^2 2p^2$$
• O⁺: Remove one electron from $$2p^4$$ → $$1s^2 2s^2 2p^3$$
• F⁺: Remove one electron from $$2p^5$$ → $$1s^2 2s^2 2p^4$$

The second ionization energy involves removing an electron from these ions:

• From C⁺ ($$2p^1$$) → C²⁺ ($$1s^2 2s^2$$)
• From N⁺ ($$2p^2$$) → N²⁺ ($$1s^2 2s^2 2p^1$$)
• From O⁺ ($$2p^3$$) → O²⁺ ($$1s^2 2s^2 2p^2$$)
• From F⁺ ($$2p^4$$) → F²⁺ ($$1s^2 2s^2 2p^3$$)

Ionization energy depends on the stability of the electron configuration and the effective nuclear charge. Half-filled or fully filled orbitals are more stable, requiring more energy to remove an electron.

Analyzing each:

• O⁺: Has a half-filled $$2p^3$$ configuration, which is very stable. Removing an electron breaks this stability, so the second ionization energy is high.
• F⁺: Has $$2p^4$$. Removing an electron leads to $$2p^3$$ (half-filled), which is stable. This makes removal easier, but fluorine has a high nuclear charge (Z=9), so the energy is still relatively high, though less than oxygen's.
• N⁺: Has $$2p^2$$. Removing an electron breaks this to $$2p^1$$, which is not particularly stable. Nitrogen has a lower nuclear charge (Z=7) than fluorine and oxygen, so its second ionization energy is lower.
• C⁺: Has $$2p^1$$. Removing an electron leaves $$2s^2$$ (stable), but carbon has the lowest nuclear charge (Z=6), and the electron is in a higher-energy p orbital, so removal is easiest.

Based on stability and nuclear charge, the order of second ionization potential is: Oxygen (highest) > Fluorine > Nitrogen > Carbon (lowest).

Comparing with the options:

• A: O > N > F > C → Incorrect, as F should be greater than N.
• B: O > F > N > C → Matches our order.
• C: F > O > N > C → Incorrect, as O should be greater than F.
• D: C > N > O > F → Incorrect, as C has the lowest ionization energy.

Hence, the correct answer is Option B.