The boiling points of ethylene, formaldehyde and dioxygen are $\pu{-103.7 ^\circ C}$, $\pu{-19 ^\circ C}$, and $\pu{−183 ^\circ C}$, respectively. I expect formaldehyde to have the highest boiling point of the three because of dipole moment mostly due to the carbon-oxgyen bond. However, I don't know why ethylene has a higher boiling point than dioxygen.

Different from this similar question, where the atoms involved are of very different size, molar mass, and polarizability, the structure and electronic structure of ethylene and dioxygen are more similar. They both have a single conformation, but they have different symmetry, which might have same bearing on the entropy change during phase transition. I know that both the $\ce{C-C}$ bond in ethylene and the $\ce{O-O}$ bond in dioxygen have a bond order of 2, but dioxygen is paramagnetic while ethylene is not.

The latent heat of vaporization is 6.8 kJ/mol for dioxygen and 13.7 kJ/mol for ethylene, so enthalpy seems a major contribution to the difference in boiling point.

What are the differences in intermolecular interactions in liquid $\ce{O2}$ compared to liquid $\ce{H2C=CH2}$ underlying the difference in normal boiling point?

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    $\begingroup$ Et tu Brute... Second answer you linked answers that. $\endgroup$ – Mithoron Mar 14 '20 at 16:15

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