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Resonance

AP.Chem:
SAP‑4 (EU)
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SAP‑4.B (LO)
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SAP‑4.B.1 (EK)
Resonance arises when more than one valid Lewis structure can be drawn for a molecule or ion. The overall electronic structure of the molecule or ion is given by the weighted average of these resonance structures and is referred to as the resonance hybrid. Created by Sal Khan.

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  • leafers ultimate style avatar for user Nick Voelckers
    At until , Sal draws the resonance hybrid Lewis structure for the Nitrate anion. Contrary to his other Lewis structures for nitrate, he didn't add the valence electrons around the oxygen atoms. When drawing these kinds of structures, are the valence electrons included? If so, how are they drawn (do the dotted line covalent bonds take up a lone pair for each oxygen)?
    (8 votes)
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    • mr pants purple style avatar for user Ryan W
      He’s only showing you that rather than there being one double bond and two single bonds, there are in reality three equal bonds somewhere between a single and a double bond. That is what you need to be taking away from this video.
      (3 votes)
  • starky seed style avatar for user Runtian Du
    Does covalent bond have strict orientations?
    Is all the different resonance just different rotations of the same molecule since if you take one and rotate 60 degrees you would get the other ones.
    (9 votes)
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  • blobby green style avatar for user emilrichardwang
    Why would the oxygens share a 4th bond with nitrogen? Isn't nitrogen's electron affinity 0? How will this make the nitrate anion more stable in nature?
    (1 vote)
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    • leaf red style avatar for user Richard
      So it's important to remember what electron affinity is. It's the amount of energy released when a neutral atom (or molecule) gains an electron. Since it's defined as a release of energy, positive values indicate energy being released and negative values as energy being absorbed. And these values are energy measurements so they need energy units like eV or kJ; just saying 0 isn't meaningful.

      At any case, nitrogen's electron affinity isn't 0, instead it's -6.8 kJ/mol. Meaning that adding an electron to a neutral nitrogen atom requires adding 6.8 kJ of energy per mole of nitrogen atoms. And this means it is an unfavorable process because of neutral nitrogen's electron configuration which is: [He]2s^(2)2p^(3). This means it has a half-filled p subshell (a full p subshell holds 6 electrons) and is more stable as opposed to having an extra electron.

      However the nitrogen in nitrate isn't a neutral atom of nitrogen becoming an anion; instead it is bonding with three oxygen atoms to form a polyatomic anion. When nitrogen forms four bonds to the oxygen atoms, it will have eight valence electrons and complete it's octet (or fill its second electron shell). Having a complete octet is very stable and a favorable state to be in for an atom like nitrogen. So any unfavorable energetics nitrogen incurs from adding electrons to a half-filled p subshell are inconsequential compared to the stability achieved by a complete octet.

      Hope that helps.
      (3 votes)
  • piceratops ultimate style avatar for user Tanzz
    At , Sal allocates two valence electrons of Oxygen to Nitrogen to make another covalent bond. Aren't covalent bonds mean the bonds with mutual sharing of electrons from all the elements? Here, the Nitrogen isn't mutually sharing its one electron with Oxygen to make a double bond.
    (1 vote)
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    • leaf red style avatar for user Richard
      Well Sal only put those lone pairs on each of the oxygen atoms in the first place to use up all the valence electrons according to the method used to create Lewis structures. If we think of each atom's electron belonging to it then even before we form the double there's some unequal distribution in a sense. Each oxygen brings six valence electrons as individual atoms and the nitrogen atom brings five electrons. And before the double bond is formed each oxygen has 7 electrons and the nitrogen has only 3 electrons which means we've allocated some of the nitrogen's electrons to the oxygens already.

      Basically it's more correct to think about these valence electrons being more fluid and being able to move around the molecule more easily than being static and belonging to a single atom. So when we take that lone pair from oxygen, it's not really us taking oxygen's electrons since they never completely belonged to it in the first place. And remember the most important point is that it's not us creating these bonds, it's us trying to determine what type of bonding is actually occurring the molecule.

      So when we correctly identify the bond orders (single and double bonds) then these bonds are indeed covalent bonds as you stated where the electrons are being shared between the bonding atoms.

      Hope that helps.
      (3 votes)
  • aqualine sapling style avatar for user Erin Contreras
    Why did Sal subtract 6 for the valence e- ?
    (1 vote)
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  • leafers tree style avatar for user jaredthefuturephysican
    at , Sal makes the Oxygens look like there is 7 valence electrons by putting 3 lone pairs and 1 covalent bond connecting it to the nitrogen. I thought that it could only hold 6 electrons. Whats up with this?
    (1 vote)
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    • leaf red style avatar for user Richard
      I believe you mean valence electrons in your question since neutral oxygen would have eight electrons total. But we usually omit the core electrons and concern ourselves with the valence electrons with most chemistry questions.

      Three lone pairs (6) and a single covalent bond (2) would give oxygen eight valence electrons. Neutral oxygen has six valence electrons, but oxygen generally wants to follow the octet rule and obtain eight valence electrons. All the resonance structures of nitrate here has each oxygen with eight valence electrons. In total that would make oxygen feel like it has ten electrons in total.

      And still technically oxygen isn't strictly limited to having only ten, it can have more electrons. It's just that this isn't particularly stable and chemicals with oxygens with more than ten electrons would be short-lived.

      Hope that helps.
      (1 vote)
  • stelly blue style avatar for user mikespar18
    If Nitrogen only has three available binding sites and each oxygen requires two bonds to be stable, how can three oxygen bind to one nitrogen? Because oxygen would need to share six electrons with nitrogen, but nitrogen can only share three, right?
    (1 vote)
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    • leaf red style avatar for user Richard
      Nitrogen and oxygen are both second period elements and ideally want to follow the octet rule by having eight valence electrons. Atoms also generally want to have as little formal charge as possible since this is more stable. So the best Lewis structure will be one where the octets of the atoms are completed and the formal charges are as little as possible.

      In the case of nitrate we have the formula NO3^(-). This means we have 24 electrons to place in the Lewis structure; 5 from nitrogen, 18 from the oxygens, and an extra electron due to the negative charge. Nitrogen will be the central atom since it's the least electronegative of the two. The nitrogen will be connected to each oxygen by at least one single bond each using 6 electrons. The remaining 18 electrons will be placed as lone pairs on the terminal atoms (the oxygen atoms). Each oxygen will received three lone pairs or 6 electrons since that will complete the octet of each atom (and oxygen cannot exceed eight valence electrons) thus using up the remaining unplaced electrons. Each oxygen will have a complete octet; two electrons from the single bond to the nitrogen and six from the lone pairs. However the central nitrogen will be deficient two electrons and only have six from the three single bonds to the nitrogen atoms. Therefore one of the terminal oxygen's lone pairs can be moved and become a double bond with the nitrogen. This won't affect the oxygen's octet, but it will now give the nitrogen a complete octet; four electrons from the double bond and four more from the two remaining single bonds.

      Now if we were only concerned with completed octets this would be a perfectly acceptable Lewis structure, but we also have to consider formal charges. The nitrogen will have a +1 formal charge, the double bonded oxygen will have a 0 formal charge, and the single bonded oxygens will each have -1 formal charges. Summing these together this does give the original nitrate ion's charge, -1. However having atoms with permanent formal charges isn't the most stable, but we can alleviate this by resonance. This entails changing which oxygen atom has the double bond and spreading out the -1 formal charges to all the oxygens and therefore resulting in three resonance structures. The resonance hybrid which results from these resonance structures means that each nitrogen-oxygen bond has partial double bond character and the two -1 formal charges are spread out across the three oxygen atoms. But the nitrogen will retains its +1 formal charge.

      So the resonance hybrid, which is the most accurate Lewis structure for nitrate, finds a balance between fulfilling the octets of all the atoms and minimizing the formal charges of the atoms as much as possible.

      Hope that helps.
      (1 vote)

Video transcript

- [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.