Key idea: For entropy changes, count moles of dissolved species and consider how ions order the surrounding water. Creating free ions from a neutral solid usually lowers ΔS° because each ion binds nearby water molecules into a more ordered hydration shell, overriding the increase from breaking up the solid lattice. Conversely, when ions combine to form a neutral species, those ordered water molecules are released, raising the entropy of the system.

(A) $\ce{Ga(l)->Ga(s)}$: liquid → solid is a freezing process, which reduces molecular disorder, so $\Delta S^\circ<0$.

(B) $\ce{C60(s)->60C(s,graphite)}$: converts one highly ordered molecular solid into 60 separate graphite units. Although the mole count rises, both are rigid, low-entropy solids, and breaking strong covalent cage bonds into another solid network gives only a small, often near-zero or even negative, entropy change (experimentally $\Delta S^\circ<0$ for this transformation because graphite's ordered layered lattice offsets the gain from separation).

(C) $\ce{CaCO3(s)->Ca^{2+}(aq)+CO3^{2-}(aq)}$: dissolving a salt into ions seems to increase disorder, but the two ions are small and highly charged, so they strongly organize surrounding water molecules into rigid hydration shells. This ordering effect dominates, making $\Delta S^\circ<0$ for this dissolution (a classic case where intuition about "more particles = more entropy" fails).

(D) $\ce{NH4+(aq)+OH-(aq)->NH3(aq)+H2O(l)}$: here two separately hydrated ions combine into two neutral species. The tightly held hydration water around $\ce{NH4+}$ and $\ce{OH-}$ is released as those ions disappear, freeing water molecules into the bulk solvent. This net release of ordered water increases the overall entropy, so $\Delta S^\circ>0$.

Answer: D