The formation of the oxide ion O^2-(g) requires first an exothermic and then an endothermic step as shown below: O(g) + e^- → O^-(g);\ H^= -142\ kJ mo - Chemistry MCQ AIEEE 2004 2026 | ExamTest.live

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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^{-} \to O^{-} (g); Δ H^{\circ} = - 142 kJ mol^{- 1} O^{-} (g) + e^{-} \to O^{2 -} (g); Δ H^{\circ} = 844 kJ mol^{- 1} The second step is endothermic because:

Solution & Step-by-step Explanation

The addition of the first electron to an oxygen atom is exothermic because of the attractive force of the nucleus. However, the addition of a second electron to a negatively charged O^{-} ion involves strong electrostatic repulsion between the ion and the incoming electron. This repulsion must be overcome by supplying energy, making the second step highly endothermic.

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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^{-} \to O^{-} (g); Δ H^{\circ} = - 142 kJ mol^{- 1} O^{-} (g) + e^{-} \to O^{2 -} (g); Δ H^{\circ} = 844 kJ mol^{- 1} The second step is endothermic because:

Oxygen is more electronegative

O^{-} ion has comparatively larger size than oxygen atom

O^{-} ion will tend to resist the addition of another electron

Oxygen has high electron affinity

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