The formation of the $O^{2-}$ ion is first exothermic and then endothermic,as shown by the following reaction steps:
$O_{(g)} + e^- \to O^-_{(g)}; \Delta H^o = -142 \ kJ \ mol^{-1}$
$O^-_{(g)} + e^- \to O^{2-}_{(g)}; \Delta H^o = 844 \ kJ \ mol^{-1}$
This is due to:
$O_{(g)} + e^- \to O^-_{(g)}; \Delta H^o = -142 \ kJ \ mol^{-1}$
$O^-_{(g)} + e^- \to O^{2-}_{(g)}; \Delta H^o = 844 \ kJ \ mol^{-1}$
This is due to:
- AOxygen is highly electronegative.
- BThe size of the $O^-$ ion is larger than the $O$ atom.
- C$O^-$ ion resists the addition of another electron.
- DOxygen has a high electron affinity.
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Similar Questions
The formation of the oxide ion $O_{(g)}^{2-}$ requires first an exothermic and then an endothermic step as shown below. This is because
$O_{(g)} + e^{-} \rightarrow O_{(g)}^{-}; \Delta H^{o} = -142 \ kJ \ mol^{-1}$
$O_{(g)}^{-} + e^{-} \rightarrow O_{(g)}^{2-}; \Delta H^{o} = 844 \ kJ \ mol^{-1}$
$O_{(g)} + e^{-} \rightarrow O_{(g)}^{-}; \Delta H^{o} = -142 \ kJ \ mol^{-1}$
$O_{(g)}^{-} + e^{-} \rightarrow O_{(g)}^{2-}; \Delta H^{o} = 844 \ kJ \ mol^{-1}$
MediumAIEEE 2004
View SolutionConsider the following conversions:
$(i) \, O_{(g)} + e^- \to O_{(g)}^{-}, \Delta H_1$
$(ii) \, F_{(g)} + e^- \to F_{(g)}^{-}, \Delta H_2$
$(iii) \, Cl_{(g)} + e^- \to Cl_{(g)}^{-}, \Delta H_3$
$(iv) \, O_{(g)}^{-} + e^- \to O_{(g)}^{2-}, \Delta H_4$
Based on the given information,which statement is incorrect?
$(i) \, O_{(g)} + e^- \to O_{(g)}^{-}, \Delta H_1$
$(ii) \, F_{(g)} + e^- \to F_{(g)}^{-}, \Delta H_2$
$(iii) \, Cl_{(g)} + e^- \to Cl_{(g)}^{-}, \Delta H_3$
$(iv) \, O_{(g)}^{-} + e^- \to O_{(g)}^{2-}, \Delta H_4$
Based on the given information,which statement is incorrect?
Difficult
View SolutionElectron affinity is the lowest for
Easy
View SolutionIn which of the following processes is the highest energy absorbed?
Medium
View SolutionThe electron affinity values for the halogens show the following trend:
Medium
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