Worked example: Lewis diagram of xenon difluoride (XeF₂) | AP Chemistry | Khan Academy

Let's do one more example of constructing a Lewis diagram that might be a little bit interesting. So let's say we want to construct the Lewis structure or Lewis diagram for xenon difluoride. So pause this video and have a go at that.

All right, now let's work through this together. So first step, we just have to account for the valence electrons. Xenon, right over here, it is a noble gas. It has eight valence electrons: one, two, three, four, five, six, seven, eight in that fifth shell. It's in the fifth period, so it has eight valence electrons.

Then fluorine, we have looked at fluorine multiple times. We know that it has seven valence electrons: one, two, three, four, five, six, seven in that second shell. We have two of these fluorines, so two times seven. And then this gives us a total of eight plus 14 valence electrons, which gets us to 22 valence electrons in total.

Now the next step, and we've done this multiple times in multiple videos now, is we would try to draw the structure with some single covalent bonds. We would put xenon as our central atom because it is less electronegative than fluorine. So let's put a xenon there and let's put two fluorines on either side. So fluorine there and a fluorine there.

And let's set up some single covalent bonds. So how many of our valence electrons have we now accounted for? Well, two in that bond and then two in that bond. So we've accounted for four. So, minus four valence electrons, we now have a total of 18 valence electrons.

Now the next step is we want to allocate them to our terminal atoms and try to get them to a full octet. Each of these fluorines already have two valence electrons that they are sharing. So we need to give each of them six more: two, four, six, two, four, six. So I've just allocated 12 more valence electrons.

So, minus 12 valence electrons means that we still have six valence electrons left to allocate. And there's only one place where we can allocate those leftover six valence electrons, and that's at the central atom, at the xenon. So let's do that: two, four, and six.

And there you have it, we have the Lewis diagram, the Lewis structure for xenon difluoride. Now what's interesting here is our fluorines, they have an octet of valence electrons. But what's going on with xenon? Xenon has two, four, six, eight, ten valence electrons hanging around. So this is one of those examples of an exception to the octet rule, where we go beyond eight valence electrons, which is possible for elements in the third or higher period.