Question

The electron affinity of halogens are $F - 322,Cl - 347,Br - 324,I - 295kJmo{l^{ - 1}}$ . The higher value of $Cl$ 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-orbital in $Cl$.

Answer 1

The size of the $Cl$ atom is greater than the size of the $F$ atom as there is an addition of another shell in the chlorine atom. Due to this increase in size of the chlorine atom, the electrons are dispersed throughout the increased volume of the chlorine atom and thus there is a weaker electron-electron repulsion in it.

Complete answer:
The electron affinity is the amount of energy released when an electron is added to the valence shell of a neutral isolated gaseous atom to convert it into a negatively charged ion. It is an exothermic reaction and with the loss of the electron enthalpy, the anion attains a lower energy state and high stability. This is the reason that the higher the value of electron affinity, the lower is its numerical value due to the presence of a negative sign in front of its value which also signifies that it is an exothermic reaction.
Now, coming to the case of $Cl$ and $F$, the higher the interelectronic repulsion in the atom’s valence shell, the higher it opposes the incoming electron, thus making it difficult for the electron to get attached with the atom. Thus, fluorine has a lower value of electron affinity than the chlorine atom because of its high interelectronic repulsions and comparatively lower interelectronic repulsions in the chlorine atom.

Thus, the correct option is A. weaker electron-electron repulsion in $Cl$.

Note:
The value of electron affinity decreases down the group and increases across a period. This order is decided without applying the negative sign in front of the numerical value of electron affinity or electron enthalpy.