Exceptions to the octet rule | AP Chemistry | Khan Academy

In this video, we're going to start talking about exceptions to the octet rule, which we've talked about in many other videos. The octet rule is this notion that atoms tend to react in ways that they're able to have a full outer shell; they're able to have eight valence electrons.

Now, we've already talked about some exceptions, things like hydrogen. Its outer shell is that first shell, which gets full with two electrons, so it's trying to get to that duet rule. But as we'll see, there are other exceptions. Boron and aluminum, for example, can form stable molecules where the boron or the aluminum only have six valence electrons, not eight.

There are exceptions in the other direction as you get to the third period and beyond. We'll actually see atoms that can maintain more than eight valence electrons, and we're actually going to see an example of that with xenon. So let's just go into a few examples. Given what I've told you, see if you can come up with the Lewis diagram for aluminum hydride.

So, aluminum hydride has one aluminum and three hydrogen. See if you can draw the Lewis diagram for that.

All right, now let's do this together. The first thing you want to do is account for all of the valence electrons. Aluminum's outer shell is the third shell—here in the third period—and it has one, two, three valence electrons. Then we have three hydrogens, and each hydrogen has one valence electron.

So, you add all of this up together: three plus three is equal to six valence electrons in aluminum hydride. Now, the next step after that is to try to draw the structure with some covalent bonds. We don't want to make hydrogen our central atom; that would be very atypical.

So, let's put aluminum in the center, and then we're going to have three hydrogens: one, two, and three. Now let's put some covalent bonds in here. Let's see how many valence electrons have we now accounted for.

This is two in this covalent bond; another two gets us to four, another two gets us to six. So, we have just accounted for all six valence electrons—so we have no more valence electrons to play with. Let's think about how the various atoms are doing.

So, the hydrogens are all meeting their duet rule—these two electrons in this bond are hanging around hydrogen and around the aluminum. But from hydrogen's point of view, it has a full duet—and that hydrogen as well, and that hydrogen as well. But notice the aluminum over here; it has two, four, six electrons—valence electrons—around it, and so it's not a full octet.

But aluminum hydride is actually something that has been observed. Let's think about another example: let's think about xenon pentafluoride cation. So, positively charged ion here. Pause this video and see if you could draw the Lewis diagram for this.

All right, now let's do this together. If any of this seems unfamiliar, I encourage you to watch the video on the introduction to drawing Lewis diagrams. But what we'd want to do here is first think about our valence electrons.

So, xenon, right over here, is actually a noble gas; it already has a full octet in its outer shell, so it has eight valence electrons. So, xenon has eight valence electrons, and then fluorine—we've seen this multiple times—has one, two, three, four, five, six, seven valence electrons. But there's five of them, so five times seven… I'm going to be drawing a lot of electrons in this.

So, this gives us a total of eight plus thirty-five, which is forty-three valence electrons. But we have to be careful; this is a cation. It is a positively charged molecule; it has a positive one charge. So, we have to take one electron away because of that.

So, let's take away one valence electron to get that cation, and so we are left with forty-two valence electrons. The next step is to try to draw its structure with some basic single covalent bonds.

Xenon would be our preferred central atom because fluorine is more electronegative—it's actually the most electronegative element. So, let's put a xenon in the middle, and then let's put some fluorines around it—five of them to be specific: one, two, three, four… I'm having trouble writing an F… four and then five fluorines.

Now let me make five covalent bonds: one, two, three, four, five. So, just like that, I have accounted for ten valence electrons because you have two valence electrons in each of these covalent bonds: two, four, six, eight, ten.

So, let me subtract ten valence electrons, and then we are left with thirty-two valence electrons. Now the next step is to try to allocate some more of these valence electrons to the terminal atoms so that they get to a full octet.

So, let me do that to the fluorines. Each of these fluorines already are participating in a covalent bond, so they already have two valence electrons hanging out with them. So, let's give them each six more. So, let's give that fluorine six, and that fluorine gets six, and that fluorine gets six valence electrons, and that fluorine gets six valence electrons, and then last but not least, this fluorine gets six valence electrons.

So, I've just given away six valence electrons to each of five fluorine atoms, so that is thirty valence electrons that I have just allocated. And then, what does that leave me with? That leaves me with two valence electrons that have gone unallocated.

The only place to now put them is on the xenon. As I said, things that are lower down in that periodic table of elements, especially as we get below the third period, these can defy the octet rule. Xenon already has ten valence electrons, and I'm just… I'm about to allocate it two more to it, just like that.

So, you allocate those two more, and then we have allocated all of our valence electrons. I want to make sure I remind myself and everyone that this is a cation, so I have to put that plus charge just like this. But this is something that has been observed where you can actually have a central atom like this that goes beyond an octet number of valence electrons.

In this case, it has two, four, six, eight, ten, twelve valence electrons. Now, an interesting question is how do these atoms that are in the third period or beyond handle more than eight valence electrons? It is a matter of debate, but some chemists believe that it's possible because they're able to place their electrons in their empty valence d orbitals. But once again, this is controversial in the chemistry community.