Choose the correct order of second IE of O, C, N and F.

To determine the correct order of the second ionization energy (IE₂) for carbon (C), nitrogen (N), oxygen (O), and fluorine (F), we must look at the electronic configurations of their respective univalent cations (mono-positive ions) from which the second electron is removed.

The first ionization process removes one electron from the neutral atom to form a univalent cation:
C (Z = 6): 1s² 2s² 2p² ⇒ C⁺: 1s² 2s² 2p¹
N (Z = 7): 1s² 2s² 2p³ ⇒ N⁺: 1s² 2s² 2p²
O (Z = 8): 1s² 2s² 2p⁴ ⇒ O⁺: 1s² 2s² 2p³
F (Z = 9): 1s² 2s² 2p⁵ ⇒ F⁺: 1s² 2s² 2p⁴

The second ionization energy (IE₂) is the energy required to remove an electron from these cations:

1. Carbon (C⁺) has the configuration 1s² 2s² 2p¹. It has the lowest nuclear charge (Z = 6) among the four and only one electron in its outer 2p orbital, making it the easiest to remove. Therefore, it has the lowest IE₂.

2. Nitrogen (N⁺) has the configuration 1s² 2s² 2p². With an increased nuclear charge (Z = 7), the outer electrons are held more tightly than in carbon, resulting in a higher IE₂ than carbon.

3. Oxygen (O⁺) has the configuration 1s² 2s² 2p³. The 2p subshell is exactly half-filled, which provides extra stability due to symmetrical distribution of electron density and high exchange energy. Removing an electron from this highly stable half-filled configuration requires an exceptionally large amount of energy.

4. Fluorine (F⁺) has the configuration 1s² 2s² 2p⁴. Although fluorine has a higher nuclear charge (Z = 9) than oxygen, removing an electron from F⁺ destroys a non-half-filled configuration to yield a stable, half-filled 2p³ configuration. Because O⁺ starts with an already stable half-filled configuration, its IE₂ is higher than that of F⁺.

Comparing the values, the order of second ionization energy is:
C < N < F < O