The formation of the oxide ion O^{2-}_{(g)} r | vedclass

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(C) The formation of the oxide ion $O^{2-}_{(g)}$ occurs in two steps.In the first step,an electron is added to a neutral oxygen atom,which is an exot

Vedclass · Accepted Answer

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

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(C) The formation of the oxide ion $O^{2-}_{(g)}$ occurs in two steps.In the first step,an electron is added to a neutral oxygen atom,which is an exothermic process $(\Delta H = -142 \, kJ \, mol^{-1})$ due to the attraction between the nucleus and the incoming electron.In the second step,an electron is added to the negatively charged $O^{-}$ ion.Since both the $O^{-}$ ion and the incoming electron are negatively charged,there is a strong electrostatic repulsion between them.To overcome this repulsion and force the electron into the ion,energy must be supplied,making the process endothermic $(\Delta H = 844 \, kJ \, mol^{-1})$.Therefore,the $O^{-}$ ion resists the addition of another electron.

The formation of the oxide ion $O^{2-}_{(g)}$ requires first an exothermic and then an endothermic step as shown below:
$O_{(g)} + e^{-} \to O^{-}_{(g)}; \Delta H = -142 \, kJ \, mol^{-1}$
$O^{-}_{(g)} + e^{-} \to O^{2-}_{(g)}; \Delta H = 844 \, kJ \, mol^{-1}$
This is because:

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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)}; \Delta H = -142 \, kJ \, mol^{-1}$$O^{-}_{(g)} + e^{-} \to O^{2-}_{(g)}; \Delta H = 844 \, kJ \, mol^{-1}$This is because: