Drawing Lewis diagrams | AP Chemistry | Khan Academy

In this video, we're going to think about constructing Lewis diagrams, which you've probably seen before. They're nice ways of visualizing how the atoms in a molecule are bonded to each other and what other lone pairs of valence electrons various atoms might have.

So, let's just start with an example, and then we'll come up with some rules for trying to draw these Lewis diagrams. The first example that we will look at is silicon tetrafluoride, and tetrafluoride is just a fancy way of saying four fluorines. So, tetrafluoride!

Now, the first step is to say, well, what are the electrons that are of interest to us? And if we're talking about the electrons that are likely to react, we're talking about the valence electrons. So, V. period E. period for short: valence electrons.

So, first, let's just think about how many total valence electrons are involved in silicon tetrafluoride. Well, to think about that, we could think about how many valence electrons silicon has and then how many valence electrons each of the fluorines has if they were just three atoms in neutral, and then multiply that times four because you have four fluorines.

So, let's get out our periodic table of elements. Then you can see here that silicon, its outer shell is the third shell, and in that third shell it has one, two, three, four valence electrons. So, silicon here has four valence electrons.

Then, to that, we're going to add the valence electrons from the four fluorines. A free neutral fluorine atom, its outer shell is the second shell, and in that outer shell it has one, two, three, four, five, six, seven electrons. So, each of these fluorines has seven valence electrons, but there are four of them.

So, one silicon tetrafluoride molecule is going to have four plus 28 valence electrons. So, this is going to be a total of 32. Now, the next step is to think about how might these be configured. As a general rule of thumb, we'd want to put the least electronegative atom that is not hydrogen at the center.

We've talked about this before, but you can even see from the periodic table of elements that fluorine is actually the most electronegative element. So, we would at least try to put silicon at the center and make fluorine a terminal atom, something on the outside.

So, let's try to do that. Let's put silicon in the center, and then we have to put the four fluorine someplace. Let's just put one fluorine there, one fluorine there, one fluorine there, and one fluorine there.

Now, the next step is, well, let's just say for simplicity that we just have single bonds between the silicon and each of the fluorines. So, let's do that: one bond, a bond, a bond, a bond.

Now, each of these covalent bonds, each of these lines in our Lewis diagram, they represent two electrons. For example, this one right over here that I'm doing in yellow represents two electrons that are shared by this fluorine and this silicon. This represents another two electrons that are shared between this fluorine and the silicon.

There's another two electrons that's shared between this fluorine and this silicon, and this is another two electrons shared between that fluorine and the silicon. So, so far, how many electrons have we accounted for? Well, each of these represents two electrons: so 2, 4, 6, 8 electrons.

If we subtract 8 from this, we are left with 24 electrons to account for, 24 valence electrons. So now our general rule of thumb would be to try to put those on those terminal atoms with the goal of getting those terminal atoms to having eight valence electrons.

In general, we try to get the octet rule for any atom except for hydrogen; hydrogen, you just need to get to two in that outer shell. But fluorine, you want to get it to eight. It already has two that it can share, so it needs six more.

So, let's add that: two, four, six. Let's do that again for this fluorine: two, four, six. Do it again for this fluorine: two, four, six. And then last but not least, for this fluorine: two, four, six.

Now, how many more electrons are now accounted for? Well, 6 in this fluorine, 6 in this fluorine, 6 in this fluorine, and 6 in this fluorine. So, 6 times four, we've now accounted for 24 more electrons. We've now used up all of the valence electrons.

Now, that's good because we want to account for all of the valence electrons. We want to represent them somehow in this Lewis diagram. The next thing to check for is how satisfied the various atoms are relative to the octet rule.

We've already seen that the fluorines are feeling pretty good. They each have six electrons that are not in a bond, and then they're able to share two electrons that are in a bond. So each of them can kind of feel like they have eight outer electrons, eight valence electrons hanging out with them.

Then the silicon is able to share in four bonds; each of those bonds have two electrons. So the silicon is also feeling good about the octet rule. I would feel very confident in this being the Lewis diagram, sometimes called the Lewis structure, for silicon tetrafluoride.

So, just to hit the point home on what we just did, I will give you these steps, but hopefully, you find them pretty intuitive. That's what I didn't want to show you from the beginning, but as you see, step one was to find the total number of valence electrons. We did that; that's the four from silicon and then the 28 from the fluorines.

It says add an electron for every negative charge, subtract an electron for every positive charge. We didn't have to do that in this example because it's a neutral molecule. Then it says decide the central atom, which should be the least electronegative, except for hydrogen. That's why we pick silicon because fluorine is the most electronegative atom.

Then we drew the bonds. We saw that the bonds accounted for eight electrons, and we subtracted those electrons from the total in step one, and that's just to keep track of the number of valence electrons that we are accounting for.

Then we had 24 left over. The next step says assign the valence electrons to the terminal atoms. That's where we assigned these extra lone pair electrons to the various fluorines, giving them an extra six each so that they were all able to fulfill the octet rule.

Then we subtracted that from the total, really just to account and to make sure that we're using all of our electrons. It says it right here: subtract the electrons from the total in step two. Then we saw that all of our electrons were accounted for.

But then in step four, it says if necessary, assign any leftover electrons to the central atom. We didn't have to do that in this example. If the central atom has an octet or exceeds an octet, you are usually done. In this case, it had an octet, so we felt done.

Finally, it says if a central atom does not have an octet, create multiple bonds. Once again, in this example, we were able to stay pretty simple with just single bonds. But in future examples, we're going to see where we might have to do some of these more nuanced steps.

So, I will leave you there, and I'll see you in the next example.