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Worked example: Lewis diagram of xenon difluoride (XeF₂)

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.

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  • blobby green style avatar for user Nancy Salman
    Why didn't he put a dubble or tripple bond between Xe and Fe instead of adding it to Xe?
    (6 votes)
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  • blobby green style avatar for user trisha.lodha
    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)
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    • hopper cool style avatar for user Iron Programming
      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
      (5 votes)
  • area 52 yellow style avatar for user Marco Pan
    I'm a chemist and I never heard of noble gases forming covalent bonds. Are these examples with Xe just rare cases from the literature, or are they actually stable compounds commonly observed? Thanks
    (P.S. I'm here to review all the fundamentals which I completely forgot, and this whole course is being great, thanks to Sal, and Richard for all the great answers to common questions)
    (2 votes)
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    • leaf red style avatar for user Richard
      The noble gases are some of my favorite elements because of their unusual inertness.

      Earlier chemists in the 19th century weren’t even aware of their existence because elements were discovered primarily based on their reactivity. It was only in the late 19th century that William Ramsay discovered the noble gases by identifying a discrepancy in the density of nitrogen gas obtained by the decomposition of nitrogen compounds and the density of the same gas obtained by isolating it from the atmosphere. His colleague Lord Rayleigh (yeah the Rayleigh from Rayleigh scattering) though that the air containing decomposed nitrogen also contained unknown lighter substances, but Ramsey suspected that the atmospheric nitrogen was contaminated by a heavier gas. In due course, he found that he could separate the atmospheric nitrogen into nitrogen and another gas, which was far less reactive; thus he discovered argon (from the Greek for ‘lazy’) in 1894.

      And so even though Ramsey discovered more noble gases and how abundant, especially argon, they are on Earth we were content to just label them as inert gases which did not react. But then in 1962 chemists successfully were able to make compounds from xenon and fluorine, specifically xenon difluoride. And apparently it wasn’t very difficult to do so, they simply combined xenon and fluorine gas and let sunlight initiate and run the reaction. Generally, chemists paired noble gases with the more reactive chemicals like fluorine and oxygen and achieved success. Since then we’ve been able to create stable noble gas compounds with krypton and radon too. Usually these compounds are stable as low temperatures and decompose into their starting materials upon heating.

      Hope that helps.
      (4 votes)
  • blobby green style avatar for user Cole Bender
    Why isn't there a triple bond formed between Xe and each F? If I'm not mistaken there are enough electrons to form two triple bonds
    (2 votes)
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    • leafers seed style avatar for user San
      You are correct in that there are enough electrons for triple bonds, but it does not result in the most favorable Lewis structure according to formal charges. (Sal doesn't go over formal charge until the next lesson.)

      The formal charge is a tool to keep track of the "charge" of each atom in a molecule. It also helps determine the most likely Lewis structure. It is calculated with the following formula:

      formal charge = valence electrons - [lone pair electrons + 1/2 * bonding electrons]


      FC = VE - [LPE + 1/2(BE)]

      Rules for arranging electrons for most favorable structure:
      1. Keep formal charges as close to 0
      2. Keep negative formal charges on the most electronegative atoms
      3. Keep like charges (+ and +, or - and -) away from adjacent atoms

      * 4. All the individual formal charges should add up to the overall charge of the molecule

      For the XeF2 structure with single bonds in the above video, here are the formal charges:

      FC = VE - [LPE + 1/2(BE)]

      F: 7 - [6 + 1/2*(2)] = 0
      Xe: 8 - [6 + 1/2*(4)] = 0

      The formal charges of 0 is a clue that this Lewis structure is the most favorable.

      Let's try a XeF2 Lewis structure with two triple bonds, with 4 LPE on each F and 2 LPE on the Xe.

      FC = VE - [LPE + 1/2(BE)]

      F: 7 - [2 + 1/2*(6)] = +2
      Xe: 8 - [6 + 1/2*(12)] = -4

      The formal charges aren't as close to 0 compared to those the other Lewis Structures. Moreover, despite F being more electronegative than Xe, Xe has the negative formal charge. Thus, the triple bond Lewis structure is not as likely.

      Hope this helped!


      Source: chem.libretexts & KhanAcademy
      (2 votes)
  • primosaur ultimate style avatar for user Marvyn Greco
    why are there double bonds, how do they work again?
    (1 vote)
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  • blobby green style avatar for user Hwang Juhee
    Hi thanks for great explanation.
    Just curious, would Xe than have an electron configuration of
    1S²2S²2P⁸ ??
    (1 vote)
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    • leaf red style avatar for user Richard
      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.
      (2 votes)
  • marcimus purple style avatar for user l
    since Xe is a noble gas and is very inert wouldnt it hold to its electrons more than F and thus be more electronegative?
    (1 vote)
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  • blobby green style avatar for user woolamel
    Would this molecule be polar?
    (1 vote)
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    • blobby green style avatar for user DerpingDerp202
      So, when you're trying to figure out if XeF₂ (Xenon Difluoride) is polar, you need to look at its shape and the pull of the atoms on the electrons. XeF₂ has a linear shape because of how the electrons are arranged around the xenon atom, with the two fluorine atoms on opposite sides. Fluorine is more electronegative, meaning it pulls the electrons towards itself more than xenon does. But since the fluorine atoms are directly opposite each other, their pulls cancel out. Even though the Xe-F bonds are polar, the whole molecule ends up being nonpolar. So, XeF₂ isn’t a polar molecule.
      (1 vote)

Video transcript

- [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.