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  • - [Instructor] Let's see if we can draw the Lewis diagram

  • for a nitrate anion.

  • So a nitrate anion has one nitrogen and three oxygens,

  • and it has a negative charge.

  • I'll do that in another color.

  • It has a negative charge.

  • So pause this video and see if you can draw that,

  • the Lewis structure for a nitrate anion.

  • All right, well we've done this many times.

  • The first step is to just account

  • for the valence electrons.

  • Nitrogen has one, two, three, four, five valence electrons

  • in its outer shell, and in that second shell,

  • if it's a neutral, free nitrogen atom.

  • So we have five valence electrons there.

  • Oxygen has one, two, three, four, five,

  • six valence electrons.

  • But if you have three oxygens,

  • you're going to have six times three.

  • And so if you just add up the valence electrons,

  • if these were free, neutral atoms,

  • you would get five plus 18, which is 23 valence electrons.

  • Now the next thing we have to keep

  • in mind is this is an anion.

  • This has a negative one charge right over here.

  • So it's going to have one more extra electron,

  • one more extra valence electron than you would expect

  • if these were just free atoms that were neutral.

  • So let's add one valance electron here.

  • So that gets us to 24 valence electrons.

  • And then the next step is let's try

  • to actually draw this structure.

  • And the way we do it is we try

  • to pick the least electronegative atom

  • that is not hydrogen to be the central atom.

  • And in this case it is nitrogen.

  • It's to the left of oxygen in that second period.

  • So let's put nitrogen in the center,

  • right over there.

  • And around it let's put three oxygens.

  • So one, two, three oxygens.

  • Let's put a single bond between them.

  • And so so far we, and let me do that

  • in another color, so we can account for it better.

  • So I'll do them in purple.

  • So so far we have accounted for two, four,

  • six valence electrons.

  • So minus six valence electrons gets us

  • to 18 valence electrons.

  • The next step is we would try to allocate

  • as many of these as possible to our terminal atoms,

  • the oxygens over here.

  • Try to get them to a full octet.

  • So let's do that.

  • This, each of these oxygens,

  • they're already participating in one

  • of these covalent bonds,

  • so they already have two valence electrons hanging out.

  • So let's see if we can give them each another six,

  • to get to eight.

  • So two, four, six.

  • Two, four, six.

  • And then two, four, and six.

  • And so just like that

  • we have allocated 18 valence electrons,

  • six, 12, 18.

  • So minus 18 valence electrons.

  • And we are now left with no further valence electrons

  • to allocate.

  • But let's see how our atoms are doing.

  • We know the oxygens have a full octet,

  • but the nitrogen only has two, four,

  • six valance electrons hanging around.

  • It would be great if there was a Lewis structure

  • where we could have eight valence electrons

  • for that nitrogen.

  • Well one way to do that is to take one

  • of these lone pairs from one of the oxygens

  • and turn that into another covalent bond.

  • So let's do that.

  • So let me just erase this pair right over here,

  • and I'm just going to turn that

  • into another covalent bond.

  • And this is looking pretty good.

  • We have eight valence electrons around each

  • of the oxygens.

  • And now we have eight for the nitrogen,

  • two, four, six, eight.

  • And we have to remind ourselves

  • that this is an anion.

  • It has a negative one charge.

  • So to finish the Lewis diagram

  • we would just put that negative charge there.

  • And this is all well and good,

  • but if this was the only way that nitrate existed

  • when we observed nitrate anions in the world,

  • we would expect to see one shorter bond

  • and two longer bonds, and we would expect one

  • of the bonds to have a different energy

  • than the other two.

  • But in the real world we don't see that.

  • We see that all of the bonds actually have the same length,

  • and they actually have the same energy.

  • And so an interesting question is why is that?

  • And one thing that you might appreciate is,

  • when I took that lone pair

  • to create this covalent bond,

  • I could have done it with that top oxygen.

  • I could have done it with this bottom-left oxygen.

  • Or I could have done it with that bottom-right oxygen.

  • And so there's actually three valid Lewis structures

  • that we could have had.

  • Not only could we have had this Lewis structure,

  • we could have had this one,

  • and I'll draw it all in yellow to save us some time,

  • where you have this nitrogen.

  • It has a single bond with that top oxygen.

  • And so that top oxygen still has six electrons

  • in lone pairs.

  • And maybe it forms a double bond

  • with the bottom-left oxygen.

  • So this bottom-left oxygen only has two lone pairs.

  • One of them would have gone to form the double bond.

  • And then this oxygen would look the same.

  • So what I am drawing here is another valid Lewis structure.

  • Or the double bond might have formed

  • with this bottom-right oxygen,

  • so let me draw that.

  • So another valid Lewis structure could look like this.

  • So nitrogen bonded to that oxygen has three lone pairs.

  • This oxygen also has three lone pairs.

  • And now this one has the double bond

  • and only has two lone pairs.

  • And whenever we see a situation

  • where we have three valid Lewis structures,

  • we call this resonance.

  • Resonance.

  • Resonance.

  • And we'll put an arrow,

  • these two-way arrows between these structures.

  • And when you hear the word resonance,

  • it sometimes conjures up this image

  • that you're bouncing back, you're resonating

  • between these structures.

  • But that's actually not right.

  • What the right way to think about it is,

  • these different ways of visualizing the nitrate,

  • these contribute to a resonance hybrid,

  • which is really the true way

  • that the nitrate exists.

  • And so, if we wanted to draw a resonance hybrid,

  • it would look like this.

  • You have the nitrogen in the center.

  • You have your oxygens, one, two, three.

  • I can draw our first covalent bond like that.

  • And then you would show the bond

  • between nitrogen and each

  • of these oxygens are a hybrid between someplace

  • between a single bond and a double bond.

  • And so instead of just one of them having the double bond

  • and the other two having single bonds,

  • they're all somewhere in between.

  • So maybe you draw a dotted line,

  • something like that, to show what the reality is,

  • is that you actually have three bonds

  • that are someplace in between a single

  • and a double bond, because the electrons

  • in this molecule are delocalized throughout.

  • And of course you wanna make sure,

  • you always wanna make sure that people recognize

  • that this is a anion.

  • So this is the idea of resonance.

  • You have multiple valid Lewis structures.

  • They all contribute to a resonance hybrid,

  • which is actually what we observe.

  • We're not just bouncing between

  • these different structures.

  • The actual observation will be a hybrid of the three.

  • Now what we just drew here,

  • these three are all equivalent.

  • But in certain cases, we'll see this

  • in future videos, you don't have equivalent structures,

  • and some of them might contribute more

  • to the resonance hybrid than others.

  • But we'll see that in future videos.

- [Instructor] Let's see if we can draw the Lewis diagram

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B2 中上級

共鳴|分子・イオン性化合物の構造と性質|AP化学|カーンアカデミー (Resonance | Molecular and ionic compound structure and properties | AP Chemistry | Khan Academy)

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    林宜悉 に公開 2021 年 01 月 14 日
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