For which of the following processes, $$\Delta S$$ is negative?

First, let us recall the basic thermodynamic idea of entropy. Entropy ($$S$$) is a quantitative measure of disorder or randomness. Whenever the randomness of a system increases, $$\Delta S = S_{\text{final}}-S_{\text{initial}}$$ becomes positive, and whenever the randomness decreases, $$\Delta S$$ becomes negative.

We now examine each of the four given processes one by one, always asking, “Does the disorder go up or down?”

Process A : $$\text{C (diamond)} \rightarrow \text{C (graphite)}$$

Diamond has a rigid three-dimensional covalent lattice, while graphite consists of planar sheets that can slide over one another. The atomic arrangement in graphite is therefore much less ordered than in diamond, i.e., more random. So

$$\Delta S_{\text{A}} > 0$$

Hence $$\Delta S$$ is positive, not negative.

Process B : $$\text{N}_2$$(g, 1 atm) $$\rightarrow$$ $$\text{N}_2$$(g, 5 atm)

This is an isothermal compression of an ideal gas from an initial pressure $$P_1 = 1\;\text{atm}$$ to a final pressure $$P_2 = 5\;\text{atm}$$. At constant temperature, the entropy change of an ideal gas can be written in two equivalent ways:

$$\boxed{\displaystyle \Delta S = nR\ln\!\left(\frac{V_2}{V_1}\right)} \qquad\text{or}\qquad \boxed{\displaystyle \Delta S = -\,nR\ln\!\left(\frac{P_2}{P_1}\right)}$$

We use the second form because pressures are given. Substituting $$P_1 = 1\;\text{atm}$$ and $$P_2 = 5\;\text{atm}$$, we obtain

$$\Delta S_{\text{B}} = -\,nR\ln\!\left(\frac{P_2}{P_1}\right) = -\,nR\ln\!\left(\frac{5}{1}\right) = -\,nR\ln 5$$

Since $$\ln 5$$ is positive, the entire right-hand side is negative:

$$\Delta S_{\text{B}} < 0$$

Thus, the entropy decreases during this compression; we have a negative $$\Delta S$$ here.

Process C : $$\text{N}_2$$(g, 273 K) $$\rightarrow$$ $$\text{N}_2$$(g, 300 K) (pressure unchanged)

Heating a gas at constant pressure increases the average molecular speed and the number of accessible microstates. Therefore randomness rises:

$$\Delta S_{\text{C}} > 0$$

Process D : $$\text{H}_2(g) \rightarrow 2\text{H}(g)$$

One diatomic molecule breaks into two monatomic species, doubling the number of independent particles. The positional and translational disorder increases markedly, giving

$$\Delta S_{\text{D}} > 0$$

Out of the four situations, only Process B shows a decrease in entropy. Hence, the process for which $$\Delta S$$ is negative corresponds to Option B.

Hence, the correct answer is Option 2.