CHEMISTRY COMMUNITY

COF2 will have a higher molar entropy than BF3.

They would both be trigonal planar molecules. The answer says that since O and F will disorder the COF2 by occupying the same location.

Is my understanding right that the molecule that has a distorted composition will have a higher entropy than the molecule that is non polar (undistorted)? Or is there any further explanation?

I would appreciate your reply.

That could be a possible explanation, but it would best to draw out lewis structures of COF2 and BF3. By drawing out the structures, you can see that the elements can be in different positions, and whichever molecule has the most displacements, it will have higher entropy.

Taravat_Lakzian_1C wrote:That could be a possible explanation, but it would best to draw out lewis structures of COF2 and BF3. By drawing out the structures, you can see that the elements can be in different positions, and whichever molecule has the most displacements, it will have higher entropy.

So it is because, the more arrangements of the molecule structure there are, the higher the degeneracy W is, thus the higher entropy.