Correct order of electron affinity for $C$,$Si$,and $Ge$ is

The increasing order of electron affinity for the given electronic configurations of elements is:
$I$. $1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^5$ $(Cl)$
$II$. $1s^2 \ 2s^2 \ 2p^3$ $(N)$
$III$. $1s^2 \ 2s^2 \ 2p^5$ $(F)$
$IV$. $1s^2 \ 2s^2 \ 2p^6 \ 3s^2 \ 3p^1$ $(Al)$

The formation of oxide ion $O^{2-}$ from oxygen atom requires an exothermic reaction,followed by an endothermic step as shown below. The process of formation of $O^{2-}$ in gas phase is unfavorable $(\Delta H^{\ominus} = +ve)$,even though it has the stable configuration of the nearest noble gas neon. This is because:
$O_{(g)} + e^{-} \longrightarrow O^{-}_{(g)} ; \Delta H^{\ominus} = -141 \ kJ \ mol^{-1}$
$O^{-}_{(g)} + e^{-} \longrightarrow O^{2-}_{(g)} ; \Delta H^{\ominus} = +760 \ kJ \ mol^{-1}$

Which element in each of the following pairs has a more negative electron gain enthalpy?
$(i)$ $O$ or $F$ $(ii)$ $F$ or $Cl$

If the electron affinity of $Cl$ is $348 \ kJ/mol$,then how much energy is released to convert $1.0 \ g$ of $Cl$ into $Cl^-$ gas in $kJ$?

The formation of the oxide ion,$O^{2-}_{(g)}$,from oxygen atom requires first an exothermic and then an endothermic step as shown below:
$O_{(g)} + e^- \to O^{-}_{(g)} ; \Delta_f H^{\Theta} = -141 \ kJ \ mol^{-1}$
$O^{-}_{(g)} + e^- \to O^{2-}_{(g)} ; \Delta_f H^{\Theta} = +780 \ kJ \ mol^{-1}$
Thus,the process of formation of $O^{2-}$ in gas phase is unfavourable even though $O^{2-}$ is isoelectronic with neon. It is due to the fact that,