Explain why the electron gain enthalpy of fluorine is less negative than that of chlorine.
(N/A) The electron gain enthalpy of fluorine is less negative than that of chlorine due to its small atomic size.
In fluorine,the incoming electron experiences significant inter-electronic repulsion because the $2p$ orbitals are compact.
In contrast,chlorine has a larger atomic size and the incoming electron enters the $3p$ orbital,which has more space available.
This results in less inter-electronic repulsion in chlorine,making the process more exothermic and leading to a more negative electron gain enthalpy compared to fluorine.
In fluorine,the incoming electron experiences significant inter-electronic repulsion because the $2p$ orbitals are compact.
In contrast,chlorine has a larger atomic size and the incoming electron enters the $3p$ orbital,which has more space available.
This results in less inter-electronic repulsion in chlorine,making the process more exothermic and leading to a more negative electron gain enthalpy compared to fluorine.
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Which one of the following statements is incorrect?
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View SolutionWhich of the given atoms has the greatest electron affinity?
The formation of oxide ion $O^{2-}$ from oxygen atom requires an exothermic reaction,followed by an endothermic step as shown below. The process of formation of $O^{2-}$ in gas phase is unfavorable $(\Delta H^{\ominus} = +ve)$,even though it has the stable configuration of the nearest noble gas neon. This is because:
$O_{(g)} + e^{-} \longrightarrow O^{-}_{(g)} ; \Delta H^{\ominus} = -141 \ kJ \ mol^{-1}$
$O^{-}_{(g)} + e^{-} \longrightarrow O^{2-}_{(g)} ; \Delta H^{\ominus} = +760 \ kJ \ mol^{-1}$
$O_{(g)} + e^{-} \longrightarrow O^{-}_{(g)} ; \Delta H^{\ominus} = -141 \ kJ \ mol^{-1}$
$O^{-}_{(g)} + e^{-} \longrightarrow O^{2-}_{(g)} ; \Delta H^{\ominus} = +760 \ kJ \ mol^{-1}$
The electron affinity for the inert gases is
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