CHAPTER 1 Stereochemistry and Bonding in Main Group Compounds: 19 ii) BrF 5 : The central atom in BrF 5 molecule is Br which has seven valence electrons (2, 8, 18, 7). Now because each fluorine atom needs one electron to complete its octet (2, 7), the Br atom uses its five valence electrons to create five bond pair domains, while two electrons are left as a lone pair domain. Now though the geometry for six electron pair domains is perfectly octahedral; but in this case, the non-bonding domain would require more space than the bonding domains. The question arises here is what position this non-bonding electron pair domain should occupy in an octahedral frame.
Now owing to the fact that all positions in an octahedral frame are equivalent, we can conclude that the lone pair can be placed at any of the six site. This, in turn, would result in greater lone-pair–bond-pair repulsion yielding the distortion of perfect octahedral geometry to a square- pyramidal one. Figure 15. Structure of BrF 5 molecule from the VSEPR model. iii) XeF 4 : The central atom in XeF 4 molecule is Xe which has eight valence electrons (2, 8, 18, 18, 8). Now because each fluorine atom needs one electron to complete its octet (2, 7), the Xe atom uses its four valence electrons to create four bond pair domains, while the four electrons are left as two lone pair domains.
Now though the geometry for six electron pair domains is perfectly octahedral; but in this case, the non-bonding domain would require more space than the bonding domains. The question arises here is what position this non-bonding electron pair domain should occupy in an octahedral frame. Now owing to the fact that lone-pair– lone-pair repulsion is highest, both non-bonding domains should be placed trans to each other in an octahedral frame to give a perfect square-planar geometry. Figure 16. Structure of XeF 4 molecule from the VSEPR model. Buy the complete book with TOC navigation, high resolution images and no watermark. Copyright © Mandeep Dalal