The process with negative entropy change is:

First, we recall the thermodynamic definition of entropy change for a chemical reaction:

$$\Delta S = \sum S_{\text{products}} \;-\; \sum S_{\text{reactants}}$$

If the disorder or randomness of the system decreases, $$\Delta S$$ becomes negative. A convenient qualitative guide is to look at the physical states and, in particular, at the number of moles of gaseous species on each side. Fewer moles of gas on the product side generally means a decrease in randomness and thus a negative entropy change.

Now we examine each option one by one.

Option A (Synthesis of ammonia):

$$\mathrm{N_2(g) + 3\,H_2(g) \;\longrightarrow\; 2\,NH_3(g)}$$

The reactant side has $$1 + 3 = 4$$ moles of gas, while the product side has only $$2$$ moles of gas. Since the number of gas molecules decreases, the randomness is reduced. Therefore

$$\Delta S = S_{2\,NH_3} - (S_{N_2} + 3S_{H_2}) \lt 0$$

So this process has a negative entropy change.

Option B (Dissolution of iodine in water): Solid iodine disperses into numerous hydrated species in the solution. The molecules move from an ordered crystal lattice to a more disordered aqueous phase, increasing randomness. Thus $$\Delta S \gt 0.$$

Option C (Dissociation of CaSO$$_4$$):

$$\mathrm{CaSO_4(s) \;\longrightarrow\; CaO(s) + SO_3(g)}$$

Here a gaseous product is created from an all-solid reactant. The appearance of any gas sharply increases disorder, so $$\Delta S \gt 0.$$

Option D (Sublimation of dry ice): Dry ice (solid CO$$_2$$) converts directly into gaseous CO$$_2$$. The solid-to-gas transition greatly enhances randomness, hence $$\Delta S \gt 0.$$

Among all the choices, only the synthesis of ammonia shows a decrease in the number of gas molecules and therefore a negative entropy change.

Hence, the correct answer is Option 1.