question_answer2) The formation of the oxide ion\[{{O}^{2-}}(g)\]requires first an exothermic and then an exothermic step as shown below \[O(g)+{{e}^{-}}={{O}^{-}}(g);\] \[\Delta {{H}^{o}}=-142\,kJ\,mo{{l}^{-1}}\] \[{{O}^{-}}{{(g)}^{-}}+{{e}^{-}}={{O}^{2}}(g);\] \[\Delta {{H}^{o}}=844\,kJ\,mo{{l}^{-1}}\] This is because
A)
oxygen is more electronegative
doneclear
B)
oxygen has high electron affinity
doneclear
C)
\[{{O}^{-}}\]ion will tend to resist the addition of another electron
doneclear
D)
\[{{O}^{-}}\]ion has comparatively larger size than oxygen atom