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

First Step (Exothermic):

$$O(g) + e^- \rightarrow O^-(g), \quad \Delta H^\circ = -142 \text{ kJ mol}^{-1}$$

Second Step (Endothermic):

$$O^-(g) + e^- \rightarrow O^{2-}(g), \quad \Delta H^\circ = +844 \text{ kJ mol}^{-1}$$

When we attempt to add a second electron to the already negatively charged $$O^-$$ ion, a strong inter-electronic repulsion occurs between the incoming electron and the existing negative charge of the ion.

To overcome this powerful electrostatic repulsion and force the second electron into the valence shell, a significant amount of external energy must be supplied. This is why the second electron gain enthalpy is heavily endothermic.

[C]: $$O^-$$ ion will tend to resist the addition of another electron.