The absolute value of the electron gain enthalpy of halogens satisfies:

🧠 Cl wins over F because F's $2p$ shell is too cramped — adding a new electron there causes too much repulsion Halogens are the champs of electron gain. The expected order would be F > Cl > Br > I (smaller atom = stronger pull). But F breaks this rule. Its $2p$ subshell is so compact that the incoming electron faces unusually high inter-electronic repulsion. Cl has a roomier $3p$ — the new electron fits comfortably. So Cl is the most exothermic, F slightly less.

🗺️ Order by magnitude (kJ/mol) $\ce{Cl} = -349$ — biggest magnitude. $\ce{F} = -328$ — anomalously below Cl due to $2p$ repulsion. $\ce{Br} = -325$ — bigger atom, weaker pull. $\ce{I} = -295$ — largest atom, weakest pull.

Order of $|\Delta_{eg}H|$: Cl > F > Br > I.

⚠️ The trap Option (d) F > Cl > Br > I is the "smooth group trend" answer. It is wrong because of F's anomaly. Don't confuse electron gain enthalpy with electronegativity — F is still the most electronegative element, but it is NOT the most exothermic at gaining electrons. The Cl-beats-F fact is a high-yield exam memory.

$\boxed{\text{Answer: (c)}}$