Question

Among the following options, the correct trend in the electron gain enthalpy is

• A. $\text{I} \gt \text{Br} \gt \text{Cl} \gt \text{F}$
• B. $\text{F} \gt \text{Cl} \gt \text{Br} \gt \text{I}$
• C. $\text{Br} \gt \text{Cl} \gt \text{F} \gt \text{I}$
• D. $\text{Cl} \gt \text{F} \gt \text{Br} \gt \text{I}$

Hint:

This second- versus third-period size exception is a frequent favorite in competitive exams! It applies to both Group 17 ($\text{Cl} \gt \text{F}$) and Group 16 ($\text{S} \gt \text{O}$) due to the exceptionally high electron density of the compact second-period atoms (F and O).

Answer 1

The Correct Option is D

Step 1: Define electron gain enthalpy.
Electron gain enthalpy is the energy change when a gaseous atom accepts an electron. A more negative value means the atom holds the new electron more strongly. We compare magnitudes for the halogens F, Cl, Br, I.
Step 2: State the general down-group trend.
Going down a group, the atom gets larger, so the added electron is held less tightly and the value becomes less negative. This alone would suggest F is the most negative.
Step 3: Bring in the F vs Cl anomaly.
Fluorine is unusually small, so its \(2p\) shell is very compact and the incoming electron feels strong electron-electron repulsion. This makes fluorine release less energy than expected, so chlorine actually tops the list.
Step 4: Place chlorine first.
Because of that repulsion in tiny F, the order starts \(Cl > F\).
Step 5: Order the rest.
After Cl and F, the normal size trend takes over for the larger atoms, giving \(Br\) above \(I\). The full order is \(Cl > F > Br > I\).
Step 6: Select the option.
This is option 4.
\[ \boxed{Cl > F > Br > I} \]