Within each pair of elements F and Cl, S and Se, and Li and Na, respectively, the elements that release more energy…

🧠 "Releases more energy on electron gain" = more negative $\Delta_{eg}H$. Pick the more exothermic member of each pair Three independent pair comparisons. Get each right by knowing the small but important anomalies.

🗺️ Pair 1 — F vs Cl $\ce{F} = -328$, $\ce{Cl} = -349$. Cl is more negative. Cl wins. (F's $2p$ shell is too cramped.)

Pair 2 — S vs Se $\ce{S} = -200$, $\ce{Se} = -195$. S is more negative. S wins. (Standard Period 3 > Period 4 trend for Group 16, since Period 2 (O) is the anomaly here, not Period 3.)

Pair 3 — Li vs Na $\ce{Li} = -60$, $\ce{Na} = -53$. Li is slightly more negative. Li wins. (Despite both being alkali metals with low EGE magnitude, Li edges out Na because the smaller Li atom holds the new $2s^2$ electron pair more tightly.)

So the answer set is Cl, S, Li.

⚠️ The trap Students often pick (c) F, S, Li — they correctly remember "smaller atom wins for electron gain" and apply it everywhere. The S and Li picks are correct. But for halogens, the rule flips: F is too small, so Cl wins. Memorize: F < Cl in $|\Delta_{eg}H|$, but in Group 16 and Group 1, the smaller-period member usually still wins (S > Se, Li > Na).

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