Answer
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Hint: Electron affinity is the energy released when an electron is gained by a gaseous atom to form a negatively charged ion. Halogens have a strong tendency to gain one electron in their valence shell to attain the nearest noble gas configuration. Fluorine belongs to the second period and is smaller in size than chlorine which belongs to the third period.
Complete step by step solution:
Electron affinity of fluorine is less than that of chlorine. This is due to the reason explained below:
Fluorine has seven electrons in 2p-subshell whereas chlorine has seven electrons in its 3p-subshell. 3p-subshell is relatively larger than 2p-subshell. Therefore, repulsion among the electrons will be more in the 2p-shell of fluorine than 3p-subshell in chlorine.
Due to the smaller size and thus, the greater electron-electron repulsions, fluorine will not accept an incoming electron with the same as chlorine. As a result, a lesser amount of energy released is when one electron is added into the 2p-subshell of F (g) to form ${{F}^{-}}$(g) ion.
Since, energy released during the formation of $C{{l}^{-}}$ (g) ion is more than that released in case of ${{F}^{-}}$(g) ion, the electron affinity of chlorine is more than that of fluorine.
\[F(g)+{{e}^{-}}\to {{F}^{-}}(g);\,\Delta {{E}_{eg}}=-333kJmo{{l}^{-1}}\]
\[Cl(g)+{{e}^{-}}\to C{{l}^{-}}(g);\,\Delta {{E}_{eg}}=-349kJmo{{l}^{-1}}\]
Note: We generally compare the electron affinities of atoms by their electronegativity values. A more electronegative element has higher tendency to attract electrons, i.e. higher electron affinity. So the basis of electronegativities of fluorine and chlorine, we may expect fluorine to have more electron affinity than chlorine. But the size of fluorine is so small that there is no room to accommodate one extra electron easily.
Complete step by step solution:
Electron affinity of fluorine is less than that of chlorine. This is due to the reason explained below:
Fluorine has seven electrons in 2p-subshell whereas chlorine has seven electrons in its 3p-subshell. 3p-subshell is relatively larger than 2p-subshell. Therefore, repulsion among the electrons will be more in the 2p-shell of fluorine than 3p-subshell in chlorine.
Due to the smaller size and thus, the greater electron-electron repulsions, fluorine will not accept an incoming electron with the same as chlorine. As a result, a lesser amount of energy released is when one electron is added into the 2p-subshell of F (g) to form ${{F}^{-}}$(g) ion.
Since, energy released during the formation of $C{{l}^{-}}$ (g) ion is more than that released in case of ${{F}^{-}}$(g) ion, the electron affinity of chlorine is more than that of fluorine.
\[F(g)+{{e}^{-}}\to {{F}^{-}}(g);\,\Delta {{E}_{eg}}=-333kJmo{{l}^{-1}}\]
\[Cl(g)+{{e}^{-}}\to C{{l}^{-}}(g);\,\Delta {{E}_{eg}}=-349kJmo{{l}^{-1}}\]
Note: We generally compare the electron affinities of atoms by their electronegativity values. A more electronegative element has higher tendency to attract electrons, i.e. higher electron affinity. So the basis of electronegativities of fluorine and chlorine, we may expect fluorine to have more electron affinity than chlorine. But the size of fluorine is so small that there is no room to accommodate one extra electron easily.
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