Identify the two steps

Formation of $O^{2-}(g)$ occurs in two successive electron-gain steps:

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

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

Interpret the signs of enthalpy change

• First step is exothermic: energy is released when neutral oxygen gains one electron.
• Second step is endothermic: energy must be supplied to add an electron to the already negatively charged ion $O^-$.

Reason for the second step being endothermic

In the second step, an incoming electron is being added to an ion that already has a negative charge.

Hence, the incoming electron experiences electrostatic repulsion from $O^-$. Therefore, $O^-$ resists the addition of another electron, and energy is required.

Check the options

A: $O^-$ ion will tend to resist the addition of another electron
This is correct. The second electron is added against electron-electron repulsion, so the process is endothermic.

B: Oxygen has high electron affinity
This explains why the first electron addition is exothermic, but not why the second is endothermic.

C: Oxygen is more electronegative
Electronegativity concerns attraction of bonding electrons, not directly the reason for the positive second electron gain enthalpy.

D: $O^-$ ion has comparatively larger size than oxygen atom
Though $O^-$ is larger than $O$, this is not the main reason for the endothermic second step. The key factor is repulsion due to negative charge.

Conclusion

The correct option is:

$\boxed{A}$