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

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(A) The formation of $O_{(g)}^{2-}$ involves the addition of an electron to a negatively charged $O_{(g)}^{-}$ ion.Since both the incoming electron an

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

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(A) The formation of $O_{(g)}^{2-}$ involves the addition of an electron to a negatively charged $O_{(g)}^{-}$ ion.Since both the incoming electron and the $O_{(g)}^{-}$ ion are negatively charged,there is a strong electrostatic repulsion between them.To overcome this repulsion and force the electron into the $O_{(g)}^{-}$ ion,energy must be supplied to the system,making the process endothermic.Therefore,the $O^{-}$ ion resists the addition of another electron.

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

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