- Dot Structures Questions
- Drawing dot structures
- Formal charge and dot structures
- Resonance and dot structures
- VSEPR for 2 electron clouds
- VSEPR for 3 electron clouds
- VSEPR for 4 electron clouds
- VSEPR for 5 electron clouds (part 1)
- VSEPR for 5 electron clouds (part 2)
- VSEPR for 6 electron clouds
Resonance and dot structures
Drawing resonance structures and resonance hybrid using example of nitrate anion. Created by Jay.
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- At4:00, aren't the three dot structures the same, just rotated? Why are they considered different?(29 votes)
- I wondered the same thing initially. I think that an important thing to consider is that the diagrams are only the-same-but-rotated if you don't care which oxygen atom is which. If you labeled the oxygen atoms, then it wouldn't be the same. So in real life, if you were somehow able to hold the molecule still and look at just one oxygen atom, the three structures would not be the same (it could have either a single or double bond to the nitrogen).(29 votes)
- while drawing the resonance hybrid structure, why does one bond have 4 dots while the other has 5. shouldn't all of the bonds have six dots(7 votes)
- He is just using dotted lines to represent the partial bonds.
The dots do not represent electrons.
The dotted lines could have 3 or 6 or 10 dots. It wouldn't matter.(25 votes)
- I can't find in either the chemistry or organic chemistry a good explanation for "resonance structure". It is kind of clear from the several drawings of molecular structures where there are several equivalent configurations where the electrons may be found, and where the pi bounds are, but when you call it resonance it should have to do with some kind of oscillatory process, as if the molecule would swing back and forth between two or more states. But then there should be a physically measurable evidence for that. Is this covered anywhere here in more detail?(6 votes)
- i think this page might provide the solution to your query http://www.chem.ucla.edu/harding/tutorials/resonance/draw_res_str.html(8 votes)
- If resonance do not actually exist then why they are only made on paper ?
Why not only resonance hybrid structures are directly made ?(4 votes)
- Resonance structures show different possible placements on electrons. In reality, the real molecule is a hybrid of all these resonance structures and we draw them so it can help us understand why molecules behave and react the way they do.(12 votes)
- Which molecule is More stable :
A compound in which resonance occurs OR
a compound in which resonance do not occur..
Please help. Than you.(4 votes)
- A molecule with resonance will be more stabile since charges are delocalized.(6 votes)
- what will be the oxidation no of N in NO3^- ?(3 votes)
- The oxygen atoms each have a oxidation state of -2 and there are three such oxygens. Therefore, the total is -6. The entire molecule is only -1. The difference is the oxidation state of nitrogen, which is +5.(7 votes)
- What's happening with the orbitals when electrons are delocalized? If a sigma bond is a head on overlap of lobes but a pi bond is an side overlap, then how are resonance electrons being shared?(4 votes)
- Excellent question! This can be best answered by the idea of molecular orbitals. Just as two atomic orbitals with p-character can overlap to form a π bonding orbital, multiple bonding orbitals can overlap to form systems that cover large parts of the molecule. A good example is benzene: if benzene did just have 3 π bonds with no delocalization, all the electrons would be cramped together, hovering above and below the three sides of the hexagonal ring that have the double bond. But, since every carbon has electrons in an atomic orbital that has p character, in reality, the electrons are hovering in big circular "halos" above and below the ring -- with more room to swish and swirl around in, not all cramped up (like they would be if benzene really just had 3 double bonds), and therefore much more stable.(1 vote)
- At5:20, I get confused. When you de-localize the electrons, you added 4 electrons in a line connecting the first Oxygen to Nitrogen. Why does that Oxygen electron not need 8 total electrons anymore? Also, its seems that you added 14 de-localized electrons...why is it 18 total electrons to complete the octets? And doesn't adding those de-localized electrons between the Oxygens and Nitrogen give Nitrogen more than 8 electrons?(3 votes)
- When he draws in the delocalized electrons, it's not literally showing that 14 of them are added — those dots represent the idea of delocalized electrons generally, not individual electrons. No electrons are added, but rather some of them are no longer bound to or paired with any one atom. The octets of each atom are still satisfied — you can think about it as if those electrons in the structure not associated with any one atom are spending enough time near each oxygen to keep all of them satisfied.(2 votes)
- Can oxygen ever have a single bond in any compound ? I mean shouldn't it have 2 lone pairs and share the third pair in a double bond ? Even if it does have a single bond , shouldn't it be a coordinate bond as none of oxygen's original six electrons are shared ?
P.S. I dont mean single bond as in the resonancee structures of no3. I mean can oxygen be connected with a single bond in any compound ?(2 votes)
- How about any alcohol which will have a carbon bonded to an oxygen bonded to a hydrogen(C-O-H) but otherwise, no, oxygen cannot have only 1 single bond because it needs to follow the octet rule(3 votes)
- Does formal Charge still apply to Resonance Dot Structures? If so, how do you calculate the formal charges in the depicted dot structures for NO3-?(2 votes)
- Yes, each atom in a dot structure has a formal charge.
In the nitrate ion, the N atom is +1 and the two single-bonded O atoms are each -1.(3 votes)
Now that we know how to draw dot structures, let's apply our rules to the nitrate anion. And we're going to see that we can draw a few different dot structures for this anion. And we're going to call those resonance structures of each other. But first, we need to calculate the total number of valence electrons. And so nitrogen is in Group 5 in the period table, therefore, five valence electrons. Oxygen is in Group 6, therefore, six valence electrons for each oxygen. I have three of them. So 6 times 3 is 18 valence electrons, plus the 5 from the nitrogen gives me 23. And I have a negative charge. This is an anion here. So we have to add one electron to that. So 23 plus 1 gives us a total of 24 valence electrons that we need to represent in our dot structure. So we know that nitrogen is going to go in the center, because oxygen is more electronegative. So nitrogen goes in the center. Nitrogen is bonded to three oxygens. So I can go ahead and put them in there like that. And let's see. How many valence electrons have we represented so far? 2, 4, and 6. Therefore, 24 minus 6 gives us 18 valence electrons left over. We're going to put those leftover valence electrons on our terminal atoms, which are our oxygens. And oxygen's going to follow the octet role. Currently, each oxygen has two valence electrons around it, the ones in magenta. So if each oxygen has two, each oxygen needs six more to complete the octet. And so I go ahead and put six more valence electrons on each one of my oxygens. Now each oxygen is surrounded by eight electrons. So the oxygens are happy. We added a total of six valence electrons to three oxygens. So 6 times 3 is 18. So we've used up all of the electrons that we need to represent. And so this dot structure, so far, it has all of our valence electrons here. Oxygen has an octet. So oxygen is happy. But nitrogen does not have an octet. If you look at the electrons in magenta, there are only six electrons around the nitrogen. And so the nitrogen wants to get to an octet. And there are a couple of different ways that we could give nitrogen an octet. For example, we could take a lone pair of electrons from this top oxygen here and move them into here to share those electrons between that top oxygen and that nitrogen. So let's go ahead and draw that resulting dot structure. So we would have our nitrogen now with a double bond to our top oxygen. Our top oxygen had three lone pairs of electrons. But now it has only two, because electrons in green moved in to form a double bond. This nitrogen is bonded to an oxygen on the bottom left and an oxygen on the bottom right here. So this is a valid dot structure. We followed our steps. And we'll go ahead and put this in brackets and put a negative charge outside of our brackets like that. So that's one possible dot structure. But we didn't have to take a lone pair of electrons from the top oxygen. We could've taken a lone pair of electrons from the oxygen on the bottom left here. So if those electrons in blue moved in here, we could have drawn another dot structure which would have been equally valid. We could have shown this oxygen on the bottom left now bonded to this nitrogen, and it used to have three lone pairs. Now it has only two. And now this top oxygen is still a single bond with three lone pairs around it. And this bottom right oxygen is still a single bond with three lone pairs around it. So this is a valid dot structure as well. So let's go ahead and put our brackets with a negative charge. And then, of course, we could have taken a lone pair of electrons from the oxygen on the bottom right. So I could have moved these in here to form a double bond. And so now, we would have our nitrogen double bonded to an oxygen on the bottom right. The oxygen on the bottom right now has only two lone pairs of electrons. The oxygen at the top, single bond with three lone pairs. And then the same situation for this oxygen on the bottom left. And so this is, once again, another possible dot structure. And so these are considered to be resonance structures of each other. And the way to represent that would be this double-headed resonance arrow here. And I think when students first see resonance structures, the name implies that, in this case, the ion is resonating back and forth between these three different possible, equally valid dot structures. And that's not quite what's going on here. Each of these dot structures is an attempt to represent the structure of the ion. But they're really not the best way of doing that. You need to think about combining these three dot structures in a resonance hybrid of each other. And so let's go ahead and draw just a simple representation of a way of thinking about a resonance hybrid. So if I combined all three of my dot structures here into one picture, I had a double bond to one oxygen in each of my three resonance structures here. And so the top oxygen had a double bond in one of them, the bottom left in the middle one, and then the bottom right in the third one. So, in reality, if we take a hybrid of all those things, we could think about the electrons being delocalized or spread out among all three of our oxygens. And so instead of giving our top nitrogen-oxygen, instead of making that a double bond, we can just show some electrons being delocalized in that area, so stronger than a single bond, but not as strong as a double bond. And we could do the same thing between this nitrogen and this oxygen. So the electrons are delocalized a little bit here. It's not a double bond. It's not a single bond. And the same idea for this nitrogen-oxygen in here. And one way we know that the ion looks more like this hybrid is because of bond length. When the ion is measured in terms of the bond length, all the nitrogen and oxygen bonds are the same length. And of course, if we thought about one of these resonance structures as being the true picture of the ion-- let's say this one, for example-- that wouldn't be the case for this ion, because this double bond here, we know that would be shorter than one of these single nitrogen-oxygen bonds. And so it's actually more of a hybrid with the electrons delocalized throughout. And that's the idea of resonance structures here.