In this example, we draw the Lewis structure for the hypoiodite ion. Our first step is to sum the valence electrons. Hypoiodite has one iodine with seven valence electrons and one oxygen with six valence electrons. Since hypoiodite has a one minus charge, we have to add one additional electron, so we have a total of 14 valence electrons. Next, we'll draw the structure with only single bonds. We'll start with one single bond between iodine and oxygen using two electrons; we'll have 12 electrons left. Our last step is to satisfy the octet rule by adding lone pairs of electrons. Both iodine and oxygen want six more electrons to satisfy the octet rule. Since we have 12 electrons left, we can satisfy both atoms by adding six electrons each. We've now used up all 14 electrons, and all of our atoms are satisfied, so we do not need to create multiple bonds. This is our Lewis structure for hypoiodite.