The formation of the oxide ion \(\mathrm{O}^{2-}(\mathrm{g})\) requires first an exothermic and then an…

The formation of the oxide ion \(\mathrm{O}^{2-}(\mathrm{g})\) requires first an exothermic and then an endothermic step as shown below. \(\mathrm{O}(\mathrm{g})+\mathrm{e}^{-}=\mathrm{O}^{-}(\mathrm{g}) ; \quad \Delta H^{\circ}=-142 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
\(\mathrm{O}^{-}(\mathrm{g})+\mathrm{e}^{-}=\mathrm{O}^{2-}(\mathrm{g}) ; \quad \Delta H^{\circ}=844 \mathrm{~kJ} \mathrm{~mol}^{-1}\)
This is because
  1. oxygen is more electronegative
  2. oxygen has high electron affinity
  3. \(\mathrm{O}^{-}\)ion will tend to resist the addition of another electron
  4. \(\mathrm{O}^{-}\)ion has comparatively larger size than oxygen atom.

Solution

The addition of negatively-charged electron to the negatively-charged species \(\mathrm{O}^{-}\)requires an input of energy. So , it will tend to resist the addition of another. *

Asked in: JEE-TOPICTESTS-CHEMISTRY

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