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The formation of oxide ion $\mathrm{O}^{2-}(\mathrm{g})$, from oxygen atom requires first an exothermic and then an endothermic step as shown below
$\begin{aligned}
&\mathrm{O}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{O}^{-}(\mathrm{g}) ; \Delta H^{\circ}=-141 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
&\mathrm{O}^{-}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{O}^{2-}(\mathrm{g}) ; \Delta H^{\circ}=+780 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}$
Thus, process of formation of $\mathrm{O}^{2-}$ in gas phase is unfavourable even though $\mathrm{O}^{2-}$ is isoelectronic with neon. It is due to the fact that
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$\begin{aligned}
&\mathrm{O}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{O}^{-}(\mathrm{g}) ; \Delta H^{\circ}=-141 \mathrm{~kJ} \mathrm{~mol}^{-1} \\
&\mathrm{O}^{-}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{O}^{2-}(\mathrm{g}) ; \Delta H^{\circ}=+780 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}$
Thus, process of formation of $\mathrm{O}^{2-}$ in gas phase is unfavourable even though $\mathrm{O}^{2-}$ is isoelectronic with neon. It is due to the fact that
Solution:
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Verified Answer
The correct answer is:
electron repulsion outweighs the stability gained by achieving noble gas configuration
electron repulsion outweighs the stability gained by achieving noble gas configuration
$\mathrm{O}^{2-}$ has noble gas configuration and isoelectronic with neon but its formation is unfavourable due to strong electronic repulsion between the negatively charged $\mathrm{O}^{-}$ ion and the electron being added.
Thus, the electron repulsion will be more than the stability gained by achieving noble gas configuration.
Thus, the electron repulsion will be more than the stability gained by achieving noble gas configuration.
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