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

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(C) The formation of $O^-_{(g)}$ from $O_{(g)}$ is exothermic because energy is released when an electron is added to a neutral oxygen atom.However,th

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$O^-$ ion will lead to resisting the addition of another electron.

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(C) The formation of $O^-_{(g)}$ from $O_{(g)}$ is exothermic because energy is released when an electron is added to a neutral oxygen atom.However,the addition of a second electron to the $O^-_{(g)}$ ion is endothermic.This is because the incoming electron experiences strong electrostatic repulsion from the already present negative charge on the $O^-_{(g)}$ ion.Therefore,energy must be supplied to overcome this inter-electronic repulsion,making the process endothermic.

Column-$I$	Column-$II$
$A$. National Botanical Research Institute	$I$. Carolus Linnaeus
$B$. Indian Botanical Garden	$II$. Taxon
$C$. Binomial Nomenclature	$III$. Howrah (India)
$D$. $A$ unit of Classification	$IV$. Lucknow

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