# How would you order the intermolecular force for the molecules F_2, HF, and H_2? How would you assess the intermolecular force?

Mar 6, 2017

We would propose the order ${F}_{2}$, $F - C l$, and $H - F$, from the least intermolecular force on the left, and the most intermolecular force on the right.

#### Explanation:

$H - F$ has the strongest intermolecular force, on account of the phenomenon of hydrogen bonding. Where $H$ is bound to a strongly electronegative element, charge separation occurs and this so-called hydrogen-bonding operates intermolecularly as the dipoles line up, i.e. ${\text{^(+delta)H-F^(delta-)...""^(+delta)H-F^(delta-)......}}^{+ \delta} H - {F}^{\delta -} \ldots$. And thus $H F$ has the (relatively) high normal boiling point of $19.5$ ""^@C. Water, where hydrogen bonding also operates as a potent intermolecular force, has an even higher boiling point.

For $F - C l$, there is some (little) polarity in the $F - C l$ bond, which we could represent as ""^(-delta)F-Cl^(delta+). That the chlorine atom in this interhalogen molecule is large, and that its electron cloud is potentially polarizable, also contributes to the magnitude of the intermolecular force by reason on dispersion forces. But compared to $H F$, $F - C l$ has a much lower normal boiling point of $- 100.1$ ""^@C, which reflects that hydrogen bonding (which is here absent) is the most potent intermolecular force.

And now we go to $F - F$, which is a homonuclear diatomic molecule, with no intermolecular forces save dispersion forces. Its normal boiling point, $- 188.5$ ""^@C certainly reflects the degree of intermolecular interaction with respect to $F - C l$, and $H - F$.

In all these examples, the normal boiling point is the best metric for comparing the degree of intermolecular force. The normal boiling points should really have been provided with the question. You are required to interpret data not to remember them.