Q.The formation of the oxide ion, O^2- (g), from oxygen atom requires first an exothermic and then an endothermic step as shown below:
O

Concept: Electron Affinity and Electrostatic Repulsion

The first electron addition to O(g) is exothermic because the incoming electron is attracted to the nucleus and occupies a stable orbital. However, adding a second electron to O⁻(g) forces it into an already negatively charged species. The small, compact O⁻ ion (with eight electrons in a confined $n=2$ shell) creates intense electrostatic repulsion between the incoming electron and the existing negative charge. …

When we add electrons to an atom, we expect energy to be released if the atom "wants" that electron—that is, if the electron experiences a net attraction to the nucleus. The first electron affinity of oxygen is indeed exothermic ($-141 \, \text{kJ mol}^{-1}$), confirming that a neutral oxygen atom readily accepts one electron to form $\mathrm{O}^-$. But the second step, adding another electron to the already-negative $\mathrm{O}^-$ ion, costs $+780 \, \text{kJ mol}^{-1}$. Why does nature resist this step so strongly, even though the resulting $\mathrm{O}^{2-}$ ion is isoelectronic with neon and has a filled valence shell?

The answer lies in the balance between two competing factors: the attraction of the added electron to the nucleus, and the repulsion between the incoming electron and the electrons already present.

Step-by-step reasoning

1. First electron addition is exothermic.

   A neutral oxygen atom has six valence electrons and a nuclear charge of $+8$. When the first electron approaches, it feels the effective nuclear charge (partially shielded by inner electrons) and is drawn into the $2p$ subshell. The atom gains stability, releasing $141 \, \text{kJ mol}^{-1}$.

2. Second electron addition faces a negative ion.

   Now we attempt to add a second electron to $\mathrm{O}^-$, which already carries a net charge of $-1$. The incoming electron must overcome the electrostatic repulsion from this negative charge. The ion is small (oxygen is a second-period element with a compact electron cloud), so the electron density is high and the repulsion is intense.

3. Electron–electron repulsion dominates.

   Although adding the second electron does complete the octet and make $\mathrm{O}^{2-}$ isoelectronic with neon, the energy cost of forcing two negative charges so close together in a small volume far exceeds the stabilization from achieving the noble-gas configuration. The repulsion term wins, making $\Delta H^\circ = +780 \, \text{kJ mol}^{-1}$ strongly positive.

4. Why the process is unfavorable in the gas phase.

   In isolation (gas phase), there is no external stabilization—no lattice energy, no solvation—to offset this huge endothermic step. The $\mathrm{O}^{2-}$ ion is thermodynamically unstable in the gas phase. It exists in ionic solids like $\mathrm{MgO}$ only because the lattice energy (the attraction between $\mathrm{Mg}^{2+}$ and $\mathrm{O}^{2-}$) more than compensates for the cost of forming $\mathrm{O}^{2-}$.

Evaluating the options

(A) Oxygen is more electronegative.

Electronegativity measures an atom's ability to attract electrons in a bond. High electronegativity would favor electron addition, not oppose it. This does not explain the endothermic second step. …