Determine the electron pair and molecular geometries for ICl4- To determine the geometries of a particular molecule I first need to determine the Lewis structure. And before I can do that, I have to know the number of valence electrons. So, I'm going to count the number of valence electrons I have. I have five halogen atoms, so they each have seven electrons, seven valence electrons. And then I have a minus sign with a minus one charge, that means I'm going to have one additional electrons, so I have a total of 36 electrons in my Lewis structure. I'm going to start by drawing my skeletal structure with all single bonds. Putting I in the middle because it is the lesser electro-negative atom of those two. And now I'm going to start filling in the remaining electrons. I see I've used eight. I had 36 to begin with. So I have 28 electrons left. So I first complete the octets on my terminal atoms. And when I've done that, what I see is that I have used up 32 of my electrons. I have four electrons remaining, and so I'm going to actually put those last two on my central atom on the iodine. Now I can look at the general form, for this molecule which will be AX4 for four bonding groups. E2 for two non-bonding groups. And so now I can say well, my electron geometry, depends on the total number of groups which is six. So my electron geometry will be octahedral. And my molecular geometry, because I have four bonding groups and two non-bonding groups, is going to be square planar.