Determine the electron pair and molecular geometries for I3 minus. So, in order to find a geometry for a molecule, I first need to know the Lewis structure. And so I need to determine the number of valence electrons. Here we have three atoms, all iodine, each with seven valence electrons. And, I have a minus 1 charge, which means I'm going to add one additional electron. So, I have a total of 22 electrons. Now I'm going to draw my skeletal structure with all single bonds, and that only uses up four electrons, so I have 18 left. I'm going to go through and fill in the octet on my terminal atoms. And after I've done that, I've used 16 electrons, so I still have six electrons remaining. And so, all of those left over electrons will go on my central iodine atom. When I look at the general form of this molecule, I see that I have A for my central atom, X2 representing two bonding groups, and E3, representing three non-bonding groups. Now I can say that my electron geometry, because I have five electron groups between the bonding and non-bonding, is trigonal, bipyramidal. And that's what it will always be, when there are five groups around that central atom. And our molecular geometry, will be linear.