To analyze the options provided, let's discuss the changes in entropy for the system and the surroundings separately during the phase change from liquid water (H₂O(l)) to water vapor (H₂O(g)) at 100°C and 1 atmosphere pressure. This is a typical boiling process.

First, let's consider the change in entropy, $$\Delta S_{\text{system}}$$, of the water itself (the system). When water boils and changes from a liquid to a gas, the molecules spread out significantly, occupying a larger volume and showing greater randomness in their motion. This transition from a more ordered phase (liquid) to a less ordered phase (gas) results in an increase in entropy. Thus, $$\Delta S_{\text{system}}>0$$.

Next, let's consider the change in entropy of the surroundings, $$\Delta S_{\text{surroundings}}$$. When the water boils at 100°C, it must absorb heat to do so. This heat is typically taken from the surroundings. As the surroundings lose heat, according to the formula for entropy change $$\Delta S = \frac{q}{T}$$, where $$q$$ is the heat transferred and $$T$$ is the temperature, the surroundings lose entropy if the process is happening at a constant temperature as typically assumed in such cases. Therefore, here $$\Delta S_{\text{surroundings}}<0$$ as heat loss by surroundings (negative $$q$$) results in a decrease in entropy.

Combining these insights, we find:

• The entropy of the system increases ($$\Delta S_{\text{system}}>0$$).
• The entropy of the surroundings decreases ($$\Delta S_{\text{surroundings}}<0$$).

Therefore, the correct answer is: Option B: $$\Delta S_{\text{system}}>0$$ and $$\Delta S_{\text{surroundings}}<0$$.