Observing the trend of boiling points of the compounds listed, choose the appropriate terms to fit into the blanks: \begin{array}{lr} \text{Compound}& \text{boiling point}\\\hline \ce{H2Te} & −2\ ^{\circ}\mathrm{C}\\ \ce{H2Se} & −41.5\ ^{\circ}\mathrm{C}\\ \ce{H2S} & −60.7\ ^{\circ}\mathrm{C}\\ \ce{H2O} & 100\ ^{\circ}\mathrm{C}\\ \end{array} The ___ the compound, the stronger the __ present between molecules. The stronger the ___, the higher the boiling point of the compound?
More polar, smaller, London dispersion forces, larger, less polar, dipole-dipole and ion-dipole forces

I noticed that the electronegativity for each compound increases from top to bottom, so higher electronegativity = higher boiling point?

I've also tried putting more polar the compound, stronger the dipole-dipole & ion forces. No luck, I still got it wrong.

  • $\begingroup$ You are certainly on the right track. Think about what the higher electronegativity means to the bonds in the molecule; this will give you the first word to insert, that you have correct. I would further think you are right by assigning dipole-dipole and ion-dipole forces to the second gap. The third gap uses in my opinion the more general term for these forces. $\endgroup$ – Martin - マーチン Nov 28 '16 at 4:16

I think it is wrong to say there should be a trend as We study the trends of individual atoms not molecule and compound itself . Boiling points depend on the intermoleculer force which depends on the polarity of whole molecule which in turns depends on the constituting species , so there should not be specific trend . It is only the matter of comparison between the species in molecule , moreover there are not single type of intermolecular forces which bond molecules such as dispersion , metallic or hydrogen , all have different strenght . I have not elaborated it but if you want further help i m on it . Sorry for errors i am not strong at that.


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