# How to explain the periodic trends in boiling points in groups?

Observing the trend of boiling points of the compounds listed, choose the appropriate terms to fit into the blanks: $$\begin{array}{lr} \hline \text{Compound} & \text{b.p.}/\pu{°C}\\ \hline \ce{H2Te} & −2 \\ \ce{H2Se} & −41.5 \\ \ce{H2S} & −60.7 \\ \ce{H2O} & 100 \\ \hline \end{array}$$ 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 does higher electronegativity implies higher boiling point?

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

• 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. – Martin - マーチン Nov 28 '16 at 4:16
• The LARGER the compound, the stronger the LONDON DISPERSION FORCES present between molecules. The stronger the LONDON DISPERSION FORCES, the higher the boiling point of the compound. – Daddy Ratchet May 24 at 3:21

## 1 Answer

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 intermolecular force which depends on the polarity of the whole molecule, which in turn depends on the constituting species, so there should not be specific trend.

It is only the matter of comparison between the species in a molecule. Moreover, there is not a single type of intermolecular forces which bonds molecules such as dispersion, metallic or hydrogen: all have different strength.