How to Write Lewis Structures

How to write Lewis structures for N2, H2O, and BF3.

TRANSCRIPT:
Today we’re going to write Lewis structures for N2, H2O, and BF3. And I have a couple of rules here that we have to follow. Each element must have an octet of electrons surrounding it, except for boron, that’s the only exception. And the total number of electrons drawn, which we’re gonna symbolize electrons with either dots or lines, must match the total number of valence electrons that you calculate from the periodic table. So I’m gonna make some space for myself here. Let’s start with N2. So first we have to calculate the number of valence electrons that N2 has. So what I usually do is I follow the numbers that are written here. If you don’t have these numbers, you could just write in 1, 2, 3, 4, 5, 6, 7, 8. And then I read it off from the periodic table, so we go to N, I see that we have 5 valence electrons here, ignore the 1 in front of it. So we have 5 from one N, and then we have another 5, so that’s going to be 10 in total, valence electrons. To write the Lewis structure, I like to write in the elements first, and remember, one of our rules was that the number of drawn electrons has to match the number of electrons that we got from the periodic table. So what I like to do first is I like to give each element an octet of electrons ,which is what you see here. They have 2 electrons on each side, and whenever you see this, you can write in just a line to symbolize they’re sharing the electrons. And a line is going to account for 2 electrons, even though you saw 1, 2, 3, 4 like that before I erased it. These are actually just shared electrons, so it’s really just 2 electrons.
So now let’s count how many valence electrons we have. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14. That is way too much. We want 10. So instead, we can take out a pair of electrons and make them share it. And that’s still gonna count as an octet, because remember we had 4 sides, and we had 2 electrons on each side. Well, we have 1, 2, 3, 4 for this element. 1, 2, 3, and 4 for this element so we’re still good. Now let’s count how many electrons we have again (1-12). Okay, looks like we just have to get rid of 1 more pair, and we should be good. And that’s the Lewis structure for N2.
Okay, now H2O. We have 1 valence electron for hydrogen, and we have two hydrogens. For oxygen over here, looks like we have 6. That’s gonna be 8 valence electrons in total so let’s start out by giving each element an octet. But, for hydrogen, you do not need to have an octet. I forgot to write hydrogen and helium as an exception to the rule because they’re just in the first row which means that they can only hold 2 electrons. Don’t forget that for hydrogen and helium. And then remember here, we can replace this with a line, replace these 4 with a line. So let’s count how many valence electrons we have. And we’re good.
Last one, we have 3. I’m gonna write that over here. So we have B, F, F, and F. Remember boron is an exception to the octet rule. So boron actually, it can just have 6 valence electrons surrounding it, not 8. So what we can do is, these fluorines need an octet so I’m just gonna surround the fluorines with it. Boron can have just 6 valence electrons so I see this group of 4 here, let’s erase that – make a line. This group of 4, make a line. This group of 4, that’s a really messy group of 4 – but, okay. So boron fulfills its rule because it’s an exception, it only needs 6. Fluorines have an octet. Now let’s count the number of valence electrons – oh, and I forgot to calculate the amount that you need. Boron is over here, it has 3 valence electrons, then you have 3 fluorines, each of them have 7, and that totals to 24 valence electrons. So let’s count (1-24). We’re all good to go.