The formation of the oxide ion O^2– (g), from the oxygen atom requires first an exothermic and then an endothermic step as shown below, 
$\mathrm{O}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}\to {\mathrm{O}}^{-};$ ${∆}_{\mathrm{f}}{\mathrm{H}}^0=–141$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
${\mathrm{O}}^{-}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}\to {\mathrm{O}}^{2-}\left(\mathrm{g}\right);$ ${∆}_{\mathrm{f}}{\mathrm{H}}^0=+780$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$

Thus, the process of formation of O^2– in the gas phase is unfavorable even though O^2– is isoelectronic with neon. It is due to the fact that:

The value of electron gain enthalpy of Na⁺, if IE₁ of Na = 5.1 eV, is:

1. +10.2 eV 

2. –5.1 eV

3. –10.2 eV 

4. +2.55 eV

The correct order of increasing electron affinity for the elements, O, S, F and Cl is:

Which one of the following arrangements represents the correct order of electron gain enthalpy (with negative sign) of the given atomic species?

1.  F < Cl < O < S

2.  S < O < Cl < F

3.  O < S < F < Cl

4.  Cl < F < S < O