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### Question 9.23

Posted: Fri Jan 27, 2017 11:56 am
Question 9.23 in the home work asks, "Which would you expect to have a higher molar entropy at T=0, single crystals of $BF_{3}$ or of $COF_{2}$? Why?

The solution manual answer says "$COF_{2}$. $COF_{2}$ and $BF_{3}$ are both trigonal planar molecules, but it would be possible for the molecule to be disordered with the fluorine and oxygen atoms occupying the same locations. Because all the groups attached to boron are identical, such disorder is not possible."

I understand the shape and structure of both molecules, but I don't understand what it means when it says that fluorine and oxygen atoms could occupy the same locations and why that leads to disorder.

### Re: Question 9.23

Posted: Fri Jan 27, 2017 5:22 pm
Entropy is the measure of disorder in a system. That being said, when we describe the randomness/disorder for the fluorines in BF3, it is essentially 0 because no matter how we arrange the fluorine atoms it will have the same end arrangement. COF2 however, while having a trigonal planar shape, has two different types of elements around the carbon. Therefore, there are different arrangements that it can be in and will have a greater total entropy than BF3.