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The formation of the oxide ion $0^{2-} (g)$ requires first on exothermic and then an endothermic step. $O_{(g)} +e^- \to O^{-} _{(g)}; \Delta r^{\circ} =-142\;kj mol^{-1}$$O_{(g)} +e^- \to O^{2-}_{(g)}; \Delta r^{\circ} =+844\;kj mol^{-1}$This is because of

$\begin{array}{1 1} \text{Oxygen is more electropositive} \\ \text{Oxygen has high electron affinity} \\ \text{$O^-$ ion will tend to resist the addition of another electron} \\ \text{$O^-$ ion has comparability larger size than oxygen atom.} \end{array} $

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Solution :
When an electron is added to negatively charged ion, it experiences more repulsion rather than a attraction .
Hence addition of second electron usually requires energy .
As a result, second electron affinity value are positive i.e endothermic
Answer : $ \text{$O^-$ ion will tend to resist the addition of another electron}$
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