Question

The formation of the oxide ion, $O^{2-}_{(g)}$, from oxygen atom requires first an exothermic and then an endothermic step as shown below:
$O_{\left(g\right)} + e^{-} \to O^{-}_{\left(g\right)} ; \Delta H^{\circ} = -141\,kJ\,mol^{-1}$
$O^{-}_{\left(g\right)} + e^{-} \to O^{2-}_{\left(g\right)} ; \Delta H^{\circ } = +780\,kJ\,mol^{-1}$
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:
$O ( g )+ e ^{-} \rightarrow O ^{-}( g ) ; \Delta_{ f } H ^{\ominus}=-141 kJmol ^{-1}$
$O ^{-}( g )+ e ^{-} \rightarrow O ^{2-}( g ) ; \Delta_{ f } H ^{\ominus}=+780 kJ mol ^{-1}$
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 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.