CHEMISTRY COMMUNITY

4G.3 Which would you expect to have a higher molar entropy at T=0, single crystals of BF3 or COF2? Why?

The answer given in the solutions manual says that COF2 is the right answer because it is asymmetrical, with an O and two F instead of three F. I also said it was COF2, but I said that it was because this molecule had less electronegativity difference between its atoms to influence the molecular orientation away from randomness. Is this correct as well? If so, which factor, polarity or asymmetry, takes precedence in answering this sort of question, and why?

Even though BF3 and COF2 are both trigonal planar, COF2 has both oxygen and fluorine atoms occupying the same locations, as opposed to BF3 having only fluorine occupy these locations. It would be more likely for disorganization to occur within the COF2 molecule because of the different atoms rather than the identical fluorine atoms occupying sites around the boron atom. I think the electronegativity is a factor but not as important and the asymmetry.

Because they both have the same geometric shape, the orientation due to asymmetry is the biggest contributor to their different entropies, and the asymmetry just also contributes to polarity.

More the disorganization, more is the entropy which is why COF2 will have more entropy compared to BF3.