The formation of the oxide ion O2−(g) requires first an exothermic and then an endothermic step as shown below.O(g)+e−=O−(g); ΔH∘=−142kJmol−1O−(g)+e−=O2−(g); ΔH∘=844kJmol−1This is because - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

A
oxygen is more electronegative
B
oxygen has high electron affinity
C
$O^{-}$ ion will tend to resist the addition of another electron
D
$O^{-}$ ion has comparatively larger size than oxygen atom.

Solution:

The addition of negatively-charged electron to the negatively-charged species $O^{-}$ 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.

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

Solution:

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