Which of the following is correct regarding the electron affinity of halogens?

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:
$O_{(g)} + e^{-} \longrightarrow O^{-}_{(g)} ; \Delta H^{\ominus} = -141 \ kJ \ mol^{-1}$
$O^{-}_{(g)} + e^{-} \longrightarrow O^{2-}_{(g)} ; \Delta H^{\ominus} = +760 \ kJ \ mol^{-1}$

Electron affinity is the

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,

Which electron affinity process would liberate the most energy?

The correct order of electron gain enthalpies of $Cl$,$F$,$Te$,and $Po$ is .....