Question

Among halogens, the correct order of amount of energy released in electron gain is?

• A. $\mathrm{F}>\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$
• B. $\mathrm{F}<\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$
• C. $\mathrm{F}>\mathrm{Br}>\mathrm{Cl}>\mathrm{I}$
• D. $\mathrm{I}>\mathrm{Cl}>\mathrm{Br}>\mathrm{F}$

Answer 1

The correct option is B

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

Explanation for correct option:

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

Electron gain enthalpy:

• “It is the amount of energy released when an electron is added to an isolated gaseous atom.”
• Electron gain enthalpy is the enthalpy change when a gaseous atom gains an electron and results in its respective ion.
• The reaction is as shown:
• $\mathrm{X}\left(\mathrm{g}\right)+{\mathrm{e}}^{-}\to {\mathrm{X}}^{-}\left(\mathrm{g}\right)$

Trend for electron gain enthalpy:

On going down the group, the electron gain enthalpy becomes less negative as both atomic size as well as nuclear charge increase but the effect of atomic size is dominant over the nuclear charge. This results in decreasing the force of attraction between the nucleus and the added electron and thus enthalpy becomes less negative.

In halogens, the trend will be as follows:

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

The electron gain enthalpy of F is less negative than that of Cl due to electron-electron repulsion in a small-sized F atom.

Explanation for incorrect options:

The correct order of electron gain enthalpy is $\mathrm{F}<\mathrm{Cl}>\mathrm{Br}>\mathrm{I}$. Thus, options A,C and D stand incorrect.

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