I don't understand why the boiling point of $\ce{AsF3}$ is at 60.4 degrees Celsius and $\ce{AsF5}$ boils at -52.8 degrees Celsius. I understand that it has to do with the bondings, but I can't see what would cause such difference?

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    $\begingroup$ I guess AsF3 has a huge dipole moment, while AsF5 has none. $\endgroup$ May 27, 2017 at 16:40

1 Answer 1


This has to do with the dipole moment of those compounds. As the difference in electronegativity between As (2.18) and F (3.98) is very large, those bonds are highly polarized.

Using the VSEPR theory, $\ce{AsF3}$ is an $\ce{AX3E}$ molecule. As such, it has a pyramidal structure. Therefore, there is a separation of charge between $\ce{F^\delta-}$ and $\ce{As^\delta+}$ in the molecule, leading to a dipole moment. This implies dipole-dipole interactions between those molecules, which greatly increases the boiling temperature.

However, $\ce{AsF5}$ is an $\ce{AX5}$ molecule and has a trigonal bipyramidal structure. The centers of charges coincide and the net dipole moment is null. There is no dipole-dipole interaction between the molecules and the boiling point is much lower.


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