Compounds of hypervalent halogens model considerations

If the hypervalent halogen atoms are assumed to form a 2c,2e bond to each ligating atom, the central atoms in CIF3 and BrFs are surrounded by 10 valence electrons, viz. three bond pairs and two lone pairs. The central atoms in XF5, X = Cl, Br or I, are surrounded by 12 electrons, viz. five bond pairs and one lone pair. Finally the iodine atom in IF7 is surrounded by 14 electrons, or seven bond pairs. 

According to the VSEPR model the five electron pairs in CLF3 or BrF3 should arrange themselves in a trigonal bipyramidal manner with the two lone pairs in equatorial positions. Since the electron lone pairs require more space around the central atom they are expected to deform the molecule is such a way that the valence angles spanned by one axial and one equatorial F atom are reduced to values below 90 . This is indeed what is observed. 

The equatorial bonds in the trifluorides may be may be described as 2c,2e bonds, and the two axial bonds as a 3c,4e bonding system. This description is consistent with the observation that the equatorial bonds are shorter than the axial. 

Problem 18.1 Draw a Lewis structure of the pentafluorides and discuss structure and bonding in terms of the VSEPR and MO models. 

The pentagonal bipyramidal geometry in IF is difficult to rationalize in terms of the VSEPR model since the smallest FIF valence angles could be larger than 72° if the coordination polyhedron was a distorted capped octahedron [3]. See Fig. 18.3. 

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