(A) The electron affinity of $Cl$ $(349 \ kJ \ mol^{-1})$ is higher than that of $F$ $(322 \ kJ \ mol^{-1})$. This is because the $2p$ subshell of $F$

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Weaker electron-electron repulsion in $Cl$

(A) The electron affinity of $Cl$ $(349 \ kJ \ mol^{-1})$ is higher than that of $F$ $(322 \ kJ \ mol^{-1})$. This is because the $2p$ subshell of $F$ is very small,leading to strong inter-electronic repulsion when an incoming electron is added. In contrast,the $3p$ subshell of $Cl$ is larger,which results in weaker electron-electron repulsion,making the addition of an electron more favorable.

Class 11 Chemistry Classification of Elements and Periodicity in Properties Electron affinity

Easy

The electron affinities of halogens are $F = 322$,$Cl = 349$,$Br = 324$,$I = 295 \ kJ \ mol^{-1}$. The higher value for $Cl$ as compared to that of $F$ is due to

A
Weaker electron-electron repulsion in $Cl$
B
Higher atomic radius of $F$
C
Smaller electronegativity of $F$
D
More vacant $p-$subshell in $Cl$

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Chapter

Classification of Elements and Periodicity in Properties

Subtopic

Electron affinity

Explain the following :
$(1)$ $Pb^{4+}$ acts as an oxidizing agent but $Sn^{2+}$ acts as a reducing agent.
$(2)$ Electron gain enthalpy of chlorine is more negative as compared to fluorine.

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Explain the trends in electron gain enthalpy across periods and down groups.

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The electron affinity values for the halogens show the following trend:

Medium

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The second electron affinity of an element is ...

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The formation of the oxide ion $O_{(g)}^{2-}$ requires first an exothermic and then an endothermic step as shown below. This is because
$O_{(g)} + e^{-} \rightarrow O_{(g)}^{-}; \Delta H^{o} = -142 \ kJ \ mol^{-1}$
$O_{(g)}^{-} + e^{-} \rightarrow O_{(g)}^{2-}; \Delta H^{o} = 844 \ kJ \ mol^{-1}$

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The electron affinities of halogens are $F = 322$,$Cl = 349$,$Br = 324$,$I = 295 \ kJ \ mol^{-1}$. The higher value for $Cl$ as compared to that of $F$ is due to

Similar Questions

Explain the following :$(1)$ $Pb^{4+}$ acts as an oxidizing agent but $Sn^{2+}$ acts as a reducing agent.$(2)$ Electron gain enthalpy of chlorine is more negative as compared to fluorine.

Explain the trends in electron gain enthalpy across periods and down groups.

The electron affinity values for the halogens show the following trend:

The second electron affinity of an element is ...

The formation of the oxide ion $O_{(g)}^{2-}$ requires first an exothermic and then an endothermic step as shown below. This is because$O_{(g)} + e^{-} \rightarrow O_{(g)}^{-}; \Delta H^{o} = -142 \ kJ \ mol^{-1}$$O_{(g)}^{-} + e^{-} \rightarrow O_{(g)}^{2-}; \Delta H^{o} = 844 \ kJ \ mol^{-1}$

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Explain the following :
$(1)$ $Pb^{4+}$ acts as an oxidizing agent but $Sn^{2+}$ acts as a reducing agent.
$(2)$ Electron gain enthalpy of chlorine is more negative as compared to fluorine.

The formation of the oxide ion $O_{(g)}^{2-}$ requires first an exothermic and then an endothermic step as shown below. This is because
$O_{(g)} + e^{-} \rightarrow O_{(g)}^{-}; \Delta H^{o} = -142 \ kJ \ mol^{-1}$
$O_{(g)}^{-} + e^{-} \rightarrow O_{(g)}^{2-}; \Delta H^{o} = 844 \ kJ \ mol^{-1}$