The formation of the oxide ion ${\mathrm{O}}^2-\left(\mathrm{g}\right)$ requires first an exothermic and then an endothermic step shown below

$\mathrm{O}\left(\mathrm{g}\right)+e^{-}={\mathrm{O}}^{-}\left(\mathrm{g}\right);\mathrm{\Delta }{H}_{e\mathrm{g}}=-142{\mathrm{kJmol}}^{-1}$

${\mathrm{O}}^{-}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}={\mathrm{O}}^2-\left(g\right);\mathrm{\Delta }{H}_{e\mathrm{f}}=+844\mathrm{kJ}{\mathrm{mol}}^{-1}$

This is because:

${\mathrm{O}}^{-}$ion will tend to resist the addition of another electron of account of same charge,

Oxygen has high electron affinity 

Oxygen is more electronegative

${\mathrm{O}}^{-}$ ion baa comparatively larger size than oxygen atom.