Solution:
The correct order of second ionisation potential of C,N,O,F is
O>F>N>C
The electronic configuration of N,O,F, and C are as follows
Nitrogen (N)⟶[He]2s2,2p3
Oxygen (O)⟶[He]2s2,2p4
Fluorine (F)⟶[He]2s2,2p5
Carbon (C)⟶[He]2s2,2p2
After removing one electron the configuration will be :
Nitrogen (N+)⟶[He]2s2,2p3
Oxygen (O+)⟶[He]2s2,2p3
Fluorine (F+)⟶[He]2s2,2p4
Carbon (C+)⟶[He]2s2,2p1
In nitrogen after removing one more electron, only one electron will be left in outermost p-subshell.
In oxygen, after removing one more electron the atom will become unstable as it will loose its half-filled stable configuration and will be left with only 2 electrons in valence p-subshell. So, have very high IP (Ionisation potential). In fluorine after removal of one more electron, it will occupy stable half-filled configuration and at last in carbon, after removal of one electron it will have ideal gas configuration. Hence, electron will be easily removed. So, it will have lowest second ionisation potential.
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