Explanation for high boiling point of Iodine Monobromide as compared to Iodine Monofluoride

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SUMMARY

The boiling point of Iodine Monobromide (IBr) is significantly higher than that of Iodine Monofluoride (IF), with IBr melting at 42°C compared to IF's -45°C. This disparity arises primarily from two factors: the greater molecular weight of IBr, which contributes to stronger van der Waals forces, and the geometric arrangement of the molecules. The comparable sizes of iodine and bromine in IBr lead to a more efficient packing in the crystal lattice, whereas the smaller fluorine atom in IF results in weaker intermolecular interactions due to less effective packing.

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neilparker62
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Apparently the melting point of IBr is 42 Celsius whereas that of IF is -45 Celsius. As Fluorine is highly electronegative, the dipole-dipole bonds in IF should be a lot stronger than in IBr. So I am wondering what forces are responsible for the huge disparity in melting points - the complete opposite of what one might expect based on bond polarity.
 
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Two things that come to mind:

1. Heavier molecules have more electrons, so typically heavier molecules have higher melting/boiling points (yes, this is assuming "all else kept equal", and you are perfectly right dipole moment could easily falsify this assumption).

2. Geometry - IBr is much more "well shaped" with I and Br being of comparable sizes. F is much smaller, so can be crystal lattice of a solid IF is not well packed and average distance between molecules is relatively high, which makes their interactions weaker.

These are just guesses of what is worth checking, not things that I would consider "the answer".
 
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