Which would you expect to have a higher molar entropy at \(T=0\), single crystals of \(\mathrm{BF}_{3}\) or of \(\mathrm{COF}_{2}\) ? Why?

The third law of thermodynamics holds significant implications for the physical properties of substances, especially as they approach absolute zero temperature. This law states that as the temperature of a perfect crystalline substance approaches absolute zero, its entropy (a measure of disorder) also approaches zero. However, no substance can ever reach absolute zero due to the infinite steps required to remove all thermal energy. At absolute zero, a substance is envisioned to have only one possible microstate, which means it is perfectly ordered.
Understanding this law is crucial when comparing molar entropy—the amount of entropy per mole of a substance—of different materials. Although theoretical calculations often assume perfect crystals, in reality, substances have imperfect crystalline structures and never reach zero entropy at absolute zero. Instead, they will have a non-zero entropy value due to their molecular complexity and the number of accessible microstates.
To contextualize, during the analysis of compounds such as \(\mathrm{BF}_{3}\) and \(\mathrm{COF}_{2}\), we would not be able to directly apply the third law without considering real-world deviations that arise due to molecular imperfections. Hence, for comparing molar entropy at very low temperatures, just above absolute zero, it's critical to consider the complexity of molecular structure, among other factors.

Microstates are fundamental to the concept of entropy, representing the different ways in which a system can be arranged while maintaining the same macroscopic properties, such as energy, volume, and number of particles. A microstate includes all the different positions and momenta of each particle in the system. The greater the number of microstates, the higher the entropy, because there is more 'disorder' or 'randomness'.
In dealing with the molar entropy of substances like \(\mathrm{BF}_{3}\) and \(\mathrm{COF}_{2}\), the comparison hinges on their available microstates. Owing to its symmetrical structure, \(\mathrm{BF}_{3}\) would have fewer microstates compared to \(\mathrm{COF}_{2}\), because the latter has a more complex shape and more atoms. This complexity allows for a larger number of ways that the molecules can be arranged, ultimately leading to a higher entropy at a given temperature.

Entropy and Temperature

As temperature increases, the number of accessible microstates grows because the molecules have more energy and can therefore take on more varied configurations. This complements entropy's dependency on temperature, illustrating how these concepts are intertwined and how microstates play a critical role in the thermal behavior of materials.

Molecular structure complexity can greatly influence a substance's physical properties, including its molar entropy. The complexity is determined by many factors: the number of atoms, the variety of elements involved, the arrangement of these atoms in the molecule, and the degrees of freedom such as vibrations and rotations that the molecule exhibits.
Consider the case of \(\mathrm{BF}_{3}\) vs. \(\mathrm{COF}_{2}\): the former has a simple, symmetrical trigonal planar shape with identical atoms bonded to boron. In contrast, \(\mathrm{COF}_{2}\) has a more complicated geometry, different elements, and thus a larger variety of chemical bonds, leading to additional vibrational and rotational states.

Vibrational and Rotational Modes

With more vibrational and rotational modes available due to the differing bond types and lengths, \(\mathrm{COF}_{2}\) inherently possesses a greater number of potential microstates even at low temperatures. This increased number of states contributes to more chaos and randomness, which is directly associated with higher entropy. Consequently, molecules with a high degree of structural complexity will generally exhibit greater molar entropy than their simpler counterparts, provided that all other conditions are similar.