| The formation of the oxide ion \[{{O}^{2-}}(g),\] from oxygen atom requires first an exothermic and then an endothermic step as shown below, [NEET 2015 (Re)] |
| \[O(g)+{{e}^{-}}\to {{O}^{-}}(g);\,\,{{\Delta }_{f}}{{H}^{{}^\circ }}=-141\,\,KJ\,mo{{l}^{-1}}\] |
| \[{{O}^{-}}(g)+{{e}^{-}}\xrightarrow{\,}\,{{O}^{2-}}\,(g);\,{{\Delta }_{f}}{{H}^{o}}\]\[=\,\,+\,780\,\,kJ\,m\,o{{l}^{-1}}\] |
| 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) electron repulsion outweighs the stability gained by achieving noble gas configuration
B) \[{{O}^{-}}\] ion has comparatively smaller size than oxygen atom
C) Oxygen is more electronegative
D) addition of electron in oxygen result in large size of the ion
Correct Answer: A
Solution :
Since, electron repulsion predominate over the stability gained by achieving noble gas configuration. Hence, formation of
in gas phase is unfavourable.
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