The entropy change can be calculated by using the expression \[\Delta S = \dfrac{{{q_{rev}}}}{T}\] . when water freezes in a glass beaker, choose the correct statement amongst the following.
A.\[\Delta S\] (system) decreases but \[\Delta S\] (surroundings) remains the same
B.\[\Delta S\] (system) increases but \[\Delta S\] (surroundings) decreases
C.\[\Delta S\] (system) decreases but \[\Delta S\] (surroundings) increases
D.\[\Delta S\] (system) decreases but \[\Delta S\] (surroundings) also decreases

Hint: Entropy is defined as the randomness or disorder Ness. More is the randomness, more will be the entropy. Freezing of water in a beaker is an exothermic reaction. Exothermic reaction means the releasing of heat. The heat releases from the system and adds to the surroundings.

Complete answer:
Freezing means conversion of the liquid into solid. When the temperature decreases, then only the liquids will solidify. Freezing of water in a beaker is an exothermic process. It means the heat releases from the beaker.
Thus, the entropy which is also defined as heat. System means beaker and surroundings means atmosphere. Entropy was represented by \[\Delta S\] . In freezing water in a beaker, the entropy of a system decreases leads to the increase in entropy of the surrounding.
Thus, \[\Delta S\] (system) decreases but \[\Delta S\] (surroundings) increases.

Thus, option C is the correct one.

Note:
In solids the atoms arrangement is very close and tight. Due to the less disorderness less will be the entropy. In liquids the atoms arrangement is loose compared to solids, leading to higher entropy. In gases, atoms are very far from each other. Thus, the gases have higher entropy compared to solids, liquids, and gases.