${ }_8 O \rightarrow 1 s ^2 \quad 2 s ^2 \quad 2 p ^4$

${ }_9 F \rightarrow 1 s ^2 \quad 2 s ^2 \quad 2 p ^5$

${ }_7 N \rightarrow 1 s ^2 \quad 2 s ^2 \quad 2 p ^3$

${ }_6 C \rightarrow 1 s ^2 \quad 2 s ^2 \quad 2 p ^2$

$2 nd\,\,IE \rightarrow O ^{-} \rightarrow 1 s ^2 \quad 2 s ^2 \quad 2 p ^3$

$F ^{-} \rightarrow 1 s ^2 \quad 2 s ^2 \quad 2 p ^4$

$N ^{-} \rightarrow 1 s ^2 \quad 2 s ^2 \quad 2 p ^2$

$C ^{-} \longrightarrow 1 s ^2 \quad 2 s ^2 \quad 2 p ^1$

Oxygen having half filled configuration is most stable and will require high energy to remove an $e ^{-}$. Followed by fluorine being the most electronegative element, then followed by $N$ which is less electronegative than $F$ but more than $C$.

$\therefore$ Order of second ionization enthalpy is $O\, >\, F\, >\, N\, >\, C$