1. We need to identify processes in which entropy decreases.

Entropy, $S$, generally decreases when:

• disorder decreases,
• gas moles decrease,
• particles become more ordered,
• molecules get fixed on surfaces.

Now evaluate each process.

2. Option (A): Freezing of water to ice at $0^\circ$C

$\text{H}_2\text{O}(l) \to \text{H}_2\text{O}(s)$

Liquid water becomes solid ice, which is more ordered.

Therefore, $\Delta S < 0$

So, (A) is a process with decrease in entropy.

3. Option (B): Freezing of water to ice at $-10^\circ$C

Again, $\text{H}_2\text{O}(l) \to \text{H}_2\text{O}(s)$

Freezing always converts a less ordered liquid into a more ordered solid, so system entropy decreases.

Hence, $\Delta S < 0$

So, (B) also shows decrease in entropy.

4. Option (C):

$\text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g)$

Reactant gas moles $= 4$ moles of gas. Product gas moles $= 2$ moles of gas.

Decrease in number of gaseous molecules means less randomness.

Thus, $\Delta S < 0$

So, (C) is correct.

5. Option (D): Adsorption of $CO(g)$ on lead surface

In adsorption, gas molecules lose freedom of motion and become attached to a surface. This increases order and decreases randomness.

Therefore, $\Delta S < 0$

So, (D) is also a process with decrease in entropy.

6. Option (E): Dissolution of $NaCl$ in water

$\text{NaCl}(s) \to \text{Na}^+(aq) + \text{Cl}^-(aq)$

Dissolution usually increases disorder because the ionic crystal breaks into solvated ions. Hence entropy generally increases.

Therefore, $\Delta S > 0$

So, (E) does not correspond to decrease in entropy.

7. Collecting correct statements

Processes with entropy decrease are:

• (A)
• (B)
• (C)
• (D)

Thus the correct option is:

$\boxed{\text{C}}$

which corresponds to (A), (B), (C) and (D) only.