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  • - [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.

- [Instructor] Let's do one more example

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B2 中高級

工作實例。二氟化氙的劉易斯圖(XeF₂)|AP化學|可汗學院。 (Worked example: Lewis diagram of xenon difluoride (XeF₂) | AP Chemistry | Khan Academy)

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    林宜悉 發佈於 2021 年 01 月 14 日
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