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

$\begin{array}{l}\mathrm{O}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}={\mathrm{O}}^{-}\left(\mathrm{g}\right);\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{\Delta }{H}^{\circ }=-142\mathrm{kJ}{\mathrm{mol}}^{-1}\\ {\mathrm{O}}^{-}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}={\mathrm{O}}^{2-}\left(\mathrm{g}\right);\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{\Delta }{H}^{\circ }=844\mathrm{kJ}{\mathrm{mol}}^{-1}\end{array}$

This is because

The addition of negatively-charged electron to the negatively-charged species ${\mathrm{O}}^{-}$ requires an input of energy.

O− ion tends to resist the addition of another electron due to large repulsion of e−− e− in small sized O− ion. Thus, energy is required to overcome this force of repulsion.