Among halogens, the correct order of amount of energy released in electron gain (electron gain enthalpy) is:

(i)$\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$

(ii) $\mathrm{F}<\mathrm{Cl}<\mathrm{Br}<\mathrm{I}$

(iii) $\mathrm{F}<\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$

(iv) $\mathrm{F}<\mathrm{Cl}<\mathrm{Br}<\mathrm{I}$

The correct option is C

(iii)

Explanation for correct option:

C. $\mathrm{F}<\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$

Electron gain enthalpy:

• "Electron gain enthalpy is the enthalpy change associated with an isolated gaseous atom when it gains an electron to form its corresponding anion."
• The reaction can be depicted as:

$\mathrm{X}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}\to {\mathrm{X}}^{-}\left(\mathrm{g}\right)$

• On moving down the group, the electron gain enthalpy becomes less negative as the atomic size, as well as nuclear charge increases but the effect of atomic size, is more prominent than that of nuclear charge.
• The electron gain enthalpy of $\mathrm{F}$ is less negative than that of $\mathrm{Cl}$ due to electron-electron repulsion in a small-sized $\mathrm{F}$ atom.

Thus, the trend for electron gain enthalpy in the case of halogens is $\mathrm{F}<\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$.

Explanation for incorrect options:

The trend for electron gain enthalpy in the case of halogens is $\mathrm{F}<\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$. Thus, options A, B, and D are incorrect.

Hence, the correct option is C i.e. iii. $\mathrm{F}<\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$.