1. We need the element in each pair that releases more energy on gaining an electron, i.e. the one with more negative electron gain enthalpy (higher electron affinity in common language).

2. Compare each pair:

   (i) F vs Cl

   • Although fluorine is smaller, its very small size causes greater electron-electron repulsion in the compact $2p$ orbital.
   • Therefore, addition of an electron is slightly less favorable in F than in Cl.
   • Hence, Cl releases more energy than F on gaining an electron.

   So, from $(F, Cl)$ the correct choice is Cl.

3. (ii) S vs Se

   • For group 16 elements, electron gain enthalpy generally becomes less negative down the group.
   • Sulfur has a more negative electron gain enthalpy than selenium.
   • Hence, S releases more energy than Se on gaining an electron.

   So, from $(S, Se)$ the correct choice is S.

4. (iii) Li vs Na

   • For alkali metals, electron gain enthalpy is small, but generally becomes less negative down the group.
   • Lithium has more negative electron gain enthalpy than sodium.
   • Hence, Li releases more energy than Na on gaining an electron.

   So, from $(Li, Na)$ the correct choice is Li.

5. Combining the results: $\boxed{Cl,\ S,\ Li}$

6. Match with options:

   • A: $F, S, Li$ ❌
   • B: $Cl, Se, Na$ ❌
   • C: $Cl, S, Li$ ✅
   • D: $F, Se, Na$ ❌

Therefore, the correct answer is: $\boxed{\text{C}}$