Question

The formation of the oxide ion, $O_{(g)}^{2-}$, from oxygen atom requires first an exothermic and then an endothermic step as shown below:
$O_{\left(g\right)}+e^{-}\to O_{\left(g\right)}^{-};\Delta H^{\circ }=-141kJmo{l}^{-1}$
$O_{\left(g\right)}^{-}+e^{-}\to O_{\left(g\right)}^{2-};\Delta H^{\circ }=+780kJmo{l}^{-1}$
Thus, process of formation of $O^{2-}$ in gas phase is unfavourable even though $O^{2-}$ is isoelectronic with neon. It is due to the fact that,

• A. oxygen is more electronegative
• B. addition of electron in oxygen results in larger size of the ion
• C. electron repulsion outweighs the stability gained by achieving noble gas configuration
• D. $O^{-}$ ion has comparatively smaller size than oxygen atom.

Answer 1

Answer: (C)

The formation of the oxide ion, $O^{2-}(g)$, from oxygen atom requires first an exothermic and then an endothermic step as shown below:
$O(g)+e^{-}\to O^{-}(g);{\Delta }_f{H}^{\ominus }=-141kJmo{l}^{-1}$
$O^{-}(g)+e^{-}\to O^{2-}(g);{\Delta }_f{H}^{\ominus }=+780kJmo{l}^{-1}$
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 electron repulsion outweighs the stability gained by achieving noble gas configuration.
When an electron is added to $O^{-}$anion, there is strong electrostatic repulsion between the two negative charges. Due to this, the second electron gain enthalpy of oxygen is positive.