Question

The electron gain enthalpy $(kJ/mol)$ of fluorine, chlorine, bromine and iodine, respectively, are:
(A) \[ - 333\], \[ - 349\], \[ - 325\] and \[ - 296\]
(B) \[ - 296\], \[ - 325\], \[ - 333\]and \[ - 349\]
(C) \[ - 349\],\[ - 333\], \[ - 325\] and \[ - 296\]
(D) \[ - 333\],\[ - 325\],\[ - 349\] and\[ - 296\]

Answer 1

Hint: As we know that fluorine, chlorine, bromine and iodine belong to the same group (\[{17^{th}}\]). Fluorine, chlorine, bromine and iodine are present at extreme right in the periodic table therefore, the size of fluorine is very small but as we go down the group the size increases.

Complete answer
The electron gain enthalpy is the tendency of elements to accept electrons in their outermost shell.
The halogens are present at extreme right in the periodic table so the size of the first element of halogen(fluorine) is very small so fluorine atoms cannot add an electron due to electron crowding in its outermost shell although it is a most electronegative element. So, the electron gain enthalpy is less negative.
As we go down the group in halogens, the chlorine atom is having large negative electron gain enthalpy due to more electronegativity as well as increased size.
As we further go down the group, the size of halogens increases and electronegativity decreases so the tendency to accept electrons in the outermost shell increases and the electron gain enthalpy becomes less negative.
Therefore, the correct order of the electron gain enthalpy is fluorine$(-333\, kJ/mol)$, chlorine$(-349\, kJ/mol)$, bromine$(-325\, kJ/mol)$ and iodine$(-296\, kJ/mol)$.

Hence, we conclude that the correct option is (A).

Note:
The electron gain enthalpy is measured in negative because when electrons are accepted by elements, energy is released.