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

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Question

(A) The first step $O_{(g)} + e^- \rightarrow O^{-}_{(g)}$ is exothermic because energy is released when an electron is added to a neutral oxygen atom

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

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(A) The first step $O_{(g)} + e^- \rightarrow O^{-}_{(g)}$ is exothermic because energy is released when an electron is added to a neutral oxygen atom.However,the second step $O^{-}_{(g)} + e^- \rightarrow O^{2-}_{(g)}$ is endothermic.This is because the incoming electron experiences strong electrostatic repulsion from the already present negative charge on the $O^{-}$ ion.Therefore,energy must be supplied to overcome this inter-electronic repulsion to force the second electron into the $O^{-}$ ion.

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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:$O_{(g)} + e^- \rightarrow O^{-}_{(g)}$; $\Delta H_{eg} = -142 \ kJ \ mol^{-1}$$O^{-}_{(g)} + e^- \rightarrow O^{2-}_{(g)}$; $\Delta H_{eg} = +844 \ kJ \ mol^{-1}$This is because: