What is electron gain enthalpy?
(N/A) Electron gain enthalpy $(\Delta_{eg}H)$ is defined as the change in enthalpy when an electron is added to a neutral gaseous atom $(X)_{(g)}$ to convert it into a negative ion $(X)_{(g)}^{-}$.
The process is represented as: $X_{(g)} + e^{-} \rightarrow X_{(g)}^{-}; \Delta_{eg}H$.
The value of $\Delta_{eg}H$ can be either negative or positive depending on the element.
The process is represented as: $X_{(g)} + e^{-} \rightarrow X_{(g)}^{-}; \Delta_{eg}H$.
The value of $\Delta_{eg}H$ can be either negative or positive depending on the element.
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Similar Questions
The correct order of electron affinity is
Which elements have the most positive and most negative electron gain enthalpy,respectively?
Medium
View SolutionThe correct pair of electron affinity order is
The formation of the oxide ion $O^{2-}_{(g)}$ requires first an exothermic and then an endothermic step as shown below:
$O_{(g)} + e^{-} \to O^{-}_{(g)}; \Delta H = -142 \, kJ \, mol^{-1}$
$O^{-}_{(g)} + e^{-} \to O^{2-}_{(g)}; \Delta H = 844 \, kJ \, mol^{-1}$
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 transformations requires the least amount of energy?
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