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

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The melting point of IBr is significantly higher at 42°C compared to IF, which melts at -45°C. This discrepancy raises questions about the forces influencing melting points, especially given that fluorine's high electronegativity suggests stronger dipole-dipole interactions in IF. Two primary factors are considered: Firstly, heavier molecules generally have higher melting and boiling points due to increased electron count, although this can be influenced by dipole moments. Secondly, the molecular geometry plays a crucial role; IBr has a more favorable shape with iodine and bromine being similar in size, allowing for better packing in the crystal lattice. In contrast, the smaller fluorine atom in IF may lead to less effective packing and weaker intermolecular interactions. These factors highlight the complexity behind melting point disparities beyond simple bond polarity.
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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