[jjkwest1]

I was checking my chemistry text and looked at a chart that compared the boiling points of different molecules. In the chart, it showed that hydrogen sulfide (H2S) had a higher boiling point than hydrogen chloride (HCl). I was wondering how this was so. Since they both have about the same London dispersion forces, isn't hydrogen chloride suppose to have the higher boiling point since it is more polar? I'm really confuse.

[vhpk]

If the molecules don't have hydrogen bond, what force determines the boiling point, and by what factors is it affected?

[jjkwest1]

Since there are no hydrogen bonds, doesn't the molecule that's the most polar determine the boiling point? In this case shouldn't HCl have a higher one?

[Black Mage]

Since there are no hydrogen bonds, doesn't the molecule that's the most polar determine the boiling point? In this case shouldn't HCl have a higher one?

In this case the electronegativities of Cl (2.58) and S (3.16) are close enough that the fact of the S attracting two hydrogen molecules gives it a slightly larger dipole moment (1.05 to .97). This seems like the reasonable explanation to me, but you should still ask your teacher.  

[jjkwest1]

How about Br₂ vs Hbr? Would Br₂ have the higher boiling point since its london dispersion forces are stronger even though it's a nonpolar molecule?

[Black Mage]

How about Br₂ vs Hbr? Would Br₂ have the higher boiling point since its london dispersion forces are stronger even though it's a nonpolar molecule?

In this case the small dipole of the HBr molecule is overcome by the fact that Br₂ has a stronger electron cloud (70 electrons) to the HBr electron cloud (36 electrons) thus the LDF in this case are stronger than the dipole-dipole interactions among HBr molecules.