#ICl# has a higher boiling point than #Br_2#. What is the best explanation for this?

1 Answer
Oct 24, 2016

Answer:

Well, clearly, the intermolecular forces that operate in the interhalogen are of greater magnitude than those that operate in the bromine molecule,

Explanation:

The boiling point of bromine is #58.8# #""^@C#; the boiling point of #I-Cl# is #97.4# #""^@C#, so the difference is fairly dramatic. In the bromine molecule, however, only dispersion forces operate. Dispersion forces also operate in #I-Cl#, and these would be expected to be greater in that the iodine atom, has a larger more polarizable electron cloud. In addition, in #I-Cl# there would be some degree of bond polarity in that chlorine is certainly more electronegative than iodine, and would tend to polarize electron density towards itself.

We could represent the resultant dipole as #""^(delta+)I-Cl^(delta-)#, and these dipoles would tend to align in such a way that tends to maximize intermolecular interaction:

#""^(delta+)I-Cl^(delta-)cdots""^(delta+)I-Cl^(delta-)cdots""^(delta+)I-Cl^(delta-)#

No such intermolecular interaction is available for #Br-Br#, as this necessarily has a non-polar bond. And thus both bond polarity, and dispersion forces conspire to make the interhalogen less volatile.

So given all this, we know that olefins react in a certain way with diatomic halogens. What would you predict as the main product of reaction between propene and #ICl#?

#H_3C-CH=CH_2 + I-Cl rarr ??#