Knowing the electron gain enthalpy values for $O \rightarrow O^{-}$ and $O \rightarrow O^{2-}$ as $-141 \ kJ \ mol^{-1}$ and $702 \ kJ \ mol^{-1}$ respectively,how can you account for the formation of a large number of oxides having $O^{2-}$ species and not $O^{-}$?
(Hint: Consider lattice energy factor in the formation of compounds).
(Hint: Consider lattice energy factor in the formation of compounds).
(N/A) The stability of an ionic compound is primarily determined by its lattice energy. $A$ higher lattice energy leads to greater stability of the compound.
Lattice energy is directly proportional to the product of the charges on the ions. When a metal reacts with oxygen,the lattice energy of an oxide containing $O^{2-}$ ions is significantly higher than that of an oxide containing $O^{-}$ ions.
Although the second electron gain enthalpy for oxygen is endothermic $(702 \ kJ \ mol^{-1})$,this energy is more than compensated for by the large amount of lattice energy released during the formation of the crystal lattice with $O^{2-}$ ions. Therefore,oxides containing $O^{2-}$ are more stable than those containing $O^{-}$.
Lattice energy is directly proportional to the product of the charges on the ions. When a metal reacts with oxygen,the lattice energy of an oxide containing $O^{2-}$ ions is significantly higher than that of an oxide containing $O^{-}$ ions.
Although the second electron gain enthalpy for oxygen is endothermic $(702 \ kJ \ mol^{-1})$,this energy is more than compensated for by the large amount of lattice energy released during the formation of the crystal lattice with $O^{2-}$ ions. Therefore,oxides containing $O^{2-}$ are more stable than those containing $O^{-}$.
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Similar Questions
Assertion $(A)$: Fluorine has a smaller negative electron gain enthalpy than chlorine.
Reason $(R)$: The electron-electron repulsion is higher in chlorine than in fluorine.
Reason $(R)$: The electron-electron repulsion is higher in chlorine than in fluorine.
What is the correct order of electron affinity for $B, C, N$,and $O$?
Medium
View SolutionThe formation of the oxide ion,$O^{2-}_{(g)}$,from oxygen atom requires first an exothermic and then an endothermic step as shown below:
$O_{(g)} + e^- \to O^{-}_{(g)} ; \Delta_f H^{\Theta} = -141 \ kJ \ mol^{-1}$
$O^{-}_{(g)} + e^- \to O^{2-}_{(g)} ; \Delta_f H^{\Theta} = +780 \ kJ \ mol^{-1}$
Thus,the process of formation of $O^{2-}$ in gas phase is unfavourable even though $O^{2-}$ is isoelectronic with neon. It is due to the fact that,
$O_{(g)} + e^- \to O^{-}_{(g)} ; \Delta_f H^{\Theta} = -141 \ kJ \ mol^{-1}$
$O^{-}_{(g)} + e^- \to O^{2-}_{(g)} ; \Delta_f H^{\Theta} = +780 \ kJ \ mol^{-1}$
Thus,the process of formation of $O^{2-}$ in gas phase is unfavourable even though $O^{2-}$ is isoelectronic with neon. It is due to the fact that,
Medium
View SolutionThe formation of the oxide ion $O^{2-}_{(g)}$ requires first an exothermic and then an endothermic step as shown below:
$O_{(g)} + e^{-} \to O^{-}_{(g)}; \Delta H = -142 \, kJ \, mol^{-1}$
$O^{-}_{(g)} + e^{-} \to O^{2-}_{(g)}; \Delta H = 844 \, kJ \, mol^{-1}$
This is because:
$O_{(g)} + e^{-} \to O^{-}_{(g)}; \Delta H = -142 \, kJ \, mol^{-1}$
$O^{-}_{(g)} + e^{-} \to O^{2-}_{(g)}; \Delta H = 844 \, kJ \, mol^{-1}$
This is because:
Which of the following represents an exothermic reaction?
Medium
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