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

Concept: Electron gain enthalpy ($\Delta_{eg}H$) is defined as the enthalpy change that occurs when an electron is added to an isolated, gaseous atom to form a univalent negative ion. A more negative value indicates a greater release of energy and a stronger affinity of the atom for an incoming electron. In general periodic chemistry trends, when comparing the magnitudes (absolute values) of the exothermic electron gain enthalpies for halogens:

• As a general rule, electron gain enthalpy becomes less negative down a group because the atomic radius increases, placing the valence shell further from the positive nucleus.

• However, a significant

anomaly occurs between the second-period ($n=2$) and third-period ($n=3$) elements within the same group due to shifts in local electron density.

Step 1: Explaining the anomaly between Chlorine and Fluorine.
Based purely on atomic size trends, one might expect Fluorine (F) to have a more negative electron gain enthalpy than Chlorine (Cl). However, the actual experimental magnitude trend is $\text{Cl} \gt \text{F}$. Let us explore the underlying physical cause:

• Fluorine belongs to the second period and has an extremely compact atomic volume. Its valence electrons are crowded together into a small $2p$ subshell.

• When an external electron approaches a fluorine atom, it experiences severe inter-electronic repulsion from the dense cloud of existing valence electrons. This repulsion offsets some of the nuclear attraction.

• Chlorine belongs to the third period and has a significantly larger atomic radius with a more spacious $3p$ subshell. The added electron spreads out comfortably with minimal inter-electronic repulsion, allowing a greater net release of energy.
Therefore, Chlorine possesses a more negative electron gain enthalpy value than Fluorine: \[ |\Delta_{eg}H_{\text{Cl}}| \gt |\Delta_{eg}H_{\text{F}}| \]

Step 2: Ordering the remaining halogens down Group 17.
Moving further down from Chlorine to Bromine (Br) and Iodine (I), the atomic size increases substantially as new primary quantum shells are added. The valence shell moves further away from the nucleus, and the shielding effect from core electrons increases. As a result, the effective nuclear charge felt by an incoming electron decreases steadily, causing the electron gain enthalpy to become predictably less negative down the group: \[ |\Delta_{eg}H_{\text{Cl}}| \gt |\Delta_{eg}H_{\text{F}}| \gt |\Delta_{eg}H_{\text{Br}}| \gt |\Delta_{eg}H_{\text{I}}| \]

Step 3: Matching with experimental values.
Let us verify this trend using standard IUPAC experimental data for halogen electron gain enthalpies:

• Chlorine (Cl): $-349 \text{ kJ mol}^{-1}$

• Fluorine (F): $-328 \text{ kJ mol}^{-1}$

• Bromine (Br): $-325 \text{ kJ mol}^{-1}$

• Iodine (I): $-295 \text{ kJ mol}^{-1}$
Arranging these absolute energy magnitudes in descending order yields: \[ \text{Cl} \gt \text{F} \gt \text{Br} \gt \text{I} \] This corresponds exactly to the trend listed in Option (4).