Drawing the Lewis Structure for BeF_{2}Viewing Notes:
Transcript: Hi, this is Dr. B. Let's do the Lewis structure for BeF2. On the periodic table, Beryllium is in group 2; 2 valence electrons. Fluorine has 7, but we have two Fluorines, for a total of 16 valence electrons. Put Beryllium here, it's the least electronegative, and the Fluorines can go on the outside. So we have 16 valence electrons. We'll put two here between to form chemical bonds. We've used 4. And then around the outside, 6, 8, 10, 12, 14, 16; and we've used all our valence electrons. You can see right away that Beryllium does not have 8 valence electrons, so it doesn't have an octet. The strange thing is, Beryllium doesn't really need 8. It's sort of an exception to the rule. So I could also draw Beryllium Fluoride a different way. Over here I'm using the same number of valence electrons, but by sharing, I've created double bonds between the Beryllium and the Fluorine, and that gives the Beryllium 8 while the Fluorine still has 8 as well. So everything has an octet on this side. But which one of these is the correct structure? To answer that, we can use formal charges. So let's calculate the formal charge on each atom. We'll start with this Fluorine right here. So Fluorine, on the periodic table, 7 valence electrons. Nonbonding, that's those guys right there, we have 4 of them. And then bonding, we also have 4. So we'll take 4 over 2; so seven minus four minus two gives us a positive 1. So the formal charge for Fluorine is +1. And this Fluorine over here has the same format, so it's also +1. For the Beryllium, we have 2 valence electrons; minus nonbonding, they're all involved in bonds; and then minus bonding, 8 over 2. Two minus eight over two gives us, actually, a minus two. So the formal charge on Beryllium is 2. On this molecule, let's do the Fluorine. So we have 7 minus 6 minus 2 over 2, that equals zero. So the formal charge is zero. The same for this one, since it's the same. And then for Beryllium, we have 2 minus zero minus the four bonding electrons over two, that also equals zero. So the formal charge here is zero. The most appropriate structure is going to be the structure that has a formal charge closest to zero. So in this case, they're all zero. That makes this the correct Lewis structure for BeF2, even though we didn't satisfy the octets. This structure here, we have the octets but the formal charges tell us that this isn't going to be the probable structure. That's the Lewis structure for BeF2. This is Dr. B., and thanks for watching. 
