To determine the correct order of the second ionization potentials for carbon (C), nitrogen (N), oxygen (O), and fluorine (F), we need to consider how the removal of the second electron affects the electronic structure of each element's ion.

Ionization potential (IP) is the energy required to remove an electron from a gaseous atom or ion. The second ionization potential specifically refers to the energy needed to remove a second electron after the first has already been removed. Generally, the second ionization potential is higher than the first because the remaining electrons experience a higher effective nuclear charge after one electron is removed.

Let’s consider the electronic configurations of the singly ionized ions of these elements:

• $$C^+: 1s^2 2s^2 2p^1$$
• $$N^+: 1s^2 2s^2 2p^2$$
• $$O^+: 1s^2 2s^2 2p^3$$
• $$F^+: 1s^2 2s^2 2p^4$$

Second ionization potential will involve removing an electron from the configurations as shown above. Here are some considerations:

• Removing a second electron from a half-filled or fully filled p-orbital (as in $$N^+$$ and $$F^+$$) requires more energy because of the increased stability of half-filled and fully filled configurations.
• In the case of $$O^+$$, removing one electron from the 2p orbitals will lead to a half-filled stable configuration (similar to $$N$$), which will make this ion less stable than $$N^+$$ initially, hence requiring lesser energy to remove an electron compared to $$N^+$$.
• $$C^+$$, having only one electron in the 2p orbitals, will have the least stable arrangement among these; hence it will likely have the lowest second ionization potential.

Based on these considerations:

1. $$F^+$$, having to remove an electron from a near half-filled p-orbital (2p4 to 2p3), will involve high energy.
2. $$N^+$$, with a half-filled p-orbital configuration, will have a next higher second ionization potential.
3. $$O^+$$, removal leads to a half-filled configuration which is energetically favorable, thus requiring less energy compared to $$N^+$$.
4. $$C^+$$ will require the least energy to remove the second electron.

Thus, the correct order of increasing second ionization potential is:

Option C: $$C > O > N > F.$$

This suggests that carbon will require the lowest energy to lose its second electron, followed by oxygen, then nitrogen, and fluorine needing the highest energy to lose its second electron among these elements.