In comparison with alkali metals,the electron affinity of halogens is

$Na^{-} \xleftarrow{x} Na \xrightarrow{5 \ eV} Na^{+}$
Predict the value of $X$ in $eV$.

The formation of the $O^{2-}$ ion is first exothermic and then endothermic,as shown by the following reaction steps:
$O_{(g)} + e^- \to O^-_{(g)}; \Delta H^o = -142 \ kJ \ mol^{-1}$
$O^-_{(g)} + e^- \to O^{2-}_{(g)}; \Delta H^o = 844 \ kJ \ mol^{-1}$
This is due to:

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

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

Explain why the electron gain enthalpy of fluorine is less negative than that of chlorine.