Questions

# The formation of the oxide ion ${\mathrm{O}}^{2-}\left(\mathrm{g}\right)$ requires first an exothermic and then an endothermic step as shown below.$\begin{array}{l}\mathrm{O}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}={\mathrm{O}}^{-}\left(\mathrm{g}\right); \mathrm{\Delta }{H}^{\circ }=-142\mathrm{kJ}{\mathrm{mol}}^{-1}\\ {\mathrm{O}}^{-}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}={\mathrm{O}}^{2-}\left(\mathrm{g}\right); \mathrm{\Delta }{H}^{\circ }=844\mathrm{kJ}{\mathrm{mol}}^{-1}\end{array}$This is because

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a
oxygen is more electronegative
b
oxygen has high electron affinity
c
${\mathrm{O}}^{-}$ ion will tend to resist the addition of another electron
d
${\mathrm{O}}^{-}$ ion has comparatively larger size than oxygen atom.

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detailed solution

Correct option is C

The addition of negatively-charged electron to the negatively-charged species ${\mathrm{O}}^{-}$ requires an input of energy.

O− ion tends to resist the addition of another electron due to large repulsion of e−− e in small sized O− ion. Thus, energy is required to overcome this force of repulsion.

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