Q. The formation of the oxide ion $O^{2 -}$(g) requires first an exothermic and then an endothermic step as shown below; $O(g) + e^{-} = O^{-}(g);\ \mathrm{\Delta}H{^\circ} = - 142\ kJmo1^{- 1}$ $O(g)^{-} + e^{-} = O^{2 -}(g);\mathrm{\Delta}H{^\circ} = 844\ kJmo1^{- 1}$ This is because
A) Oxygen is more electronegative
B) Oxygen has high electron affinity
C) $O^{-}$ion will tend to resist the addition of another electron
D) $O^{-}$has comparatively larger size than oxygen atom
Explanation
(c)
$O^{-}(g) + e^{-} \rightarrow O^{2 -}(g),\ \mathrm{\Delta}H{^\circ} = 844\ kJmol^{- 1}$
This process is unfavorable in the gas phase because the resulting increase in electron-electron repulsion overweighs the stability gained by achieving the noble gas configuration.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.