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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Similar Questions
The formation of the $O^{2-}$ ion is first exothermic and then endothermic,as shown by the following reaction steps:
$O_{(g)} + e^- \to O^-_{(g)}; \Delta H^o = -142 \ kJ \ mol^{-1}$
$O^-_{(g)} + e^- \to O^{2-}_{(g)}; \Delta H^o = 844 \ kJ \ mol^{-1}$
This is due to:
$O_{(g)} + e^- \to O^-_{(g)}; \Delta H^o = -142 \ kJ \ mol^{-1}$
$O^-_{(g)} + e^- \to O^{2-}_{(g)}; \Delta H^o = 844 \ kJ \ mol^{-1}$
This is due to:
Easy
View SolutionWhich element in each of the following pairs has a more negative electron gain enthalpy?
$(i)$ $O$ or $F$ $(ii)$ $F$ or $Cl$
$(i)$ $O$ or $F$ $(ii)$ $F$ or $Cl$
Easy
View SolutionIn which of the following pairs,electron gain enthalpies of constituent elements are nearly the same or identical?
$A$. $Rb$ and $Cs$
$B$. $Na$ and $K$
$C$. $Ar$ and $Kr$
$D$. $I$ and $At$
Choose the correct answer from the options given below.
$A$. $Rb$ and $Cs$
$B$. $Na$ and $K$
$C$. $Ar$ and $Kr$
$D$. $I$ and $At$
Choose the correct answer from the options given below.
The electron affinity values of elements $A, B, C$ and $D$ are respectively $-135, -60, -200$ and $-348 \ kJ \ mol^{-1}$. The outer electronic configuration of element $B$ is
The formation of the oxide ion,$O^{2-}_{(g)}$ from an oxygen atom requires first an exothermic and then an endothermic step as shown below:
$O_{(g)} + e^- \to O^{-}_{(g)} ; \Delta_f H^o = -141 \ kJ \ mol^{-1}$
$O^{-}_{(g)} + e^- \to O^{2-}_{(g)} ; \Delta_f H^o = +780 \ kJ \ mol^{-1}$
Thus,the process of formation of $O^{2-}$ in the gas phase is unfavourable even though $O^{2-}$ is isoelectronic with neon. This is due to the fact that,
$O_{(g)} + e^- \to O^{-}_{(g)} ; \Delta_f H^o = -141 \ kJ \ mol^{-1}$
$O^{-}_{(g)} + e^- \to O^{2-}_{(g)} ; \Delta_f H^o = +780 \ kJ \ mol^{-1}$
Thus,the process of formation of $O^{2-}$ in the gas phase is unfavourable even though $O^{2-}$ is isoelectronic with neon. This is due to the fact that,
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