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The formation of the oxide ion $O^{2-}_{(g)} $ requires first an exothermic and then an endothermic step, as shown below:

$O_{(g)} + e- \longrightarrow O^{-}_{(g)} ; \Delta H^0= -142 \;kJ mol^{-1} $ $O^{-}_{(g)} + e- \longrightarrow O^{2-}_{(g)} ; \Delta H^0= 844 \;kJ mol^{-1} $ This is because of which of the following reasons?

(A) $O^{-} $ ion is comparatively larger size than the oxygen atom
(A) $O^{-} $ ion is comparatively larger size than the oxygen atom
(C) Oxygen is more electrogenative
(D) Oxygen has high electron affinity
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Electron affinity is a measure of the attraction between the incoming electron and the nucleus - the stronger the attraction, the more energy is released.
The addition of second electron in this case to negatively charged ion requires more energy as it experiences more repulsion than attraction.
Therefore, $O^{-} $ ion will tend to resist the addition of another electron
answered Mar 15, 2014 by balaji.thirumalai

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