Class 11 Chemistry (India)
- Ionic bonds
- Ionic bonds: Reaction of metals & Non-metals
- Covalent bonds
- Single and multiple covalent bonds
- Metallic bonds
- Drawing Lewis diagrams
- Predicting bond type (metals vs. nonmetals)
- Worked example: Lewis diagram of formaldehyde (CH₂O)
- Worked example: Lewis diagram of the cyanide ion (CN⁻)
- Exceptions to the octet rule
- Worked example: Lewis diagram of xenon difluoride (XeF₂)
- Lewis diagrams
In some molecules, the central atom exceeds the octet rule (is surrounded by more than eight electrons). See an example of a molecule that violates the octet rule (XeF₂) and learn how to draw its Lewis diagram in this video. Created by Sal Khan.
Want to join the conversation?
- wait Xe can have up to 12 valence electrons, but in this drawing, it has only 10? Won't the valence shell remain incomplete then?(2 votes)
- Well, technically it normally holds 8 valence electrons because the first shell holds 2 electrons, and the second shell holds 8 electrons, which for Xe would be it's valence electrons.
If you watch the video Sal explains how sometimes atoms can remain stable & hold "extra" or "less" valence electrons then they are actually supposed too.
So in this case, "Xe" can hold 2 extra valence electrons, and still remain stable.
Hope this helps,
- Convenient Colleague(4 votes)
- Why didn't he put a dubble or tripple bond between Xe and Fe instead of adding it to Xe?(3 votes)
- F in general never forms double bonds, also I don't think that would help the issue here, F can only have 8 valence electrons in total around it, the rest have to go onto Xe.(2 votes)
- Hi thanks for great explanation.
Just curious, would Xe than have an electron configuration of
1S²2S²2P⁸ ??(1 vote)
- Not really. First of all neutral xenon has 54 electrons (same as its atomic number and number of protons) and here you only have 12 electrons. Second, xenon makes use of its 4th and 5th electron shells and here you're only going up to the 2nd. Third, you have 8 electrons in the 2p subshell which can only hold a maximum of 6 electrons.
Using noble gas configuration xenon should have an electron configuration of: [Kr]5s^(2)4d^(10)5p^(6).
Hope that helps.(1 vote)
- I answered this one on the "exception to the octet rule" video, but again this isn't really a chemistry question.(1 vote)
- [Instructor] Let's do one more example of constructing a Lewis diagram that might be a little bit interesting. So let's say we wanted to construct the Lewis structure or Lewis diagram for xenon difluoride. So pause this video and have a go of 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. And 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. And 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 and 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. And 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 wanna 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. So 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 left over six valence electrons and that's at the central atom. At the xenon. So let's do that. So, 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, 10 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.