18 A Textbook of Inorganic Chemistry – Volume I iv) I 3 : The central atom in I 3 ion is I which has seven valence electrons (2, 8, 18, 18, 7). Now because one iodine atom needs one electron to complete its octet (2, 8, 18, 18, 7) while the iodide ion needs zero electrons (2, 8, 18, 18, 8), the I atom uses its one valence electrons to create two bond pair domains, while six electrons are left as three lone pair domains.
Now though the geometry for five electron pair domains is trigonal bipyramidal with equatorial and axial bond angles of 120° and 180°, respectively; but in this case, the non- bonding domain would require more space than the bonding domains, and therefore need to be placed on the equatorial position. This, in turn, would result in a perfectly linear geometry with a normal bond angle of 180°. Figure 13. Structure of I 3 from the VSEPR model. 5. Six electron-pair domains: i) SF 6 : The central atom in SF 6 molecule is S which has six valence electrons (2, 8, 6).
Now because each fluorine atom needs one electron to complete its octet (2, 7), the S atom uses its all six valence electrons in sharing to create six bond pair domains only. Hence the geometry for minimum repulsion will be perfect octahedral and the normal angle between two any adjacent bonds will be 90°. Figure 14. Structure of SF 6 molecule from the VSEPR model. Buy the complete book with TOC navigation, high resolution images and no watermark. Copyright © Mandeep Dalal