The electron gain enthalpy (with negative sign) of fluorine is less than that of chlorine due to:

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

Which one of the following arrangements represents the correct order of least negative to most negative electron gain enthalpy for $C, Ca, Al, F$ and $O$?

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

Discuss the factors affecting electron gain enthalpy and the trend in its variation in the periodic table.

The formation of the oxide ion $O_{(g)}^{2-}$ requires first an exothermic and then an endothermic step as shown below. This is because
$O_{(g)} + e^{-} \rightarrow O_{(g)}^{-}; \Delta H^{o} = -142 \ kJ \ mol^{-1}$
$O_{(g)}^{-} + e^{-} \rightarrow O_{(g)}^{2-}; \Delta H^{o} = 844 \ kJ \ mol^{-1}$