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Lecture Comments (60)

1 answer

Last reply by: Professor Starkey
Fri May 6, 2016 12:41 AM

Post by likhitha muri on May 1 at 07:30:41 PM

Dr.Laurie you are an awsome and efficient teacher. Youu make things very easy to understand and remember.Thank you for being a good book for me.

1 answer

Last reply by: Professor Starkey
Tue Jan 12, 2016 12:58 AM

Post by Jinhai Zhang on January 9 at 03:02:05 PM

Prof. Starkey:
For the last example of this lecture what is the hybridization of N, is it should be sp3 since the first resonance structure contribute the most. or sp2? thank you for answering!

1 answer

Last reply by: Professor Starkey
Mon Sep 14, 2015 1:36 PM

Post by Ryan Rad on September 13, 2015

Just a quick question. on 21:16 would we be able to say that there is a third resonance structure with no double bonds and both oxygens negetive formal charge? I know it would be very unstable compared to rest but is it still a resonance? if not why not?

Thank you! :)

2 answers

Last reply by: Jason Smith
Sun Mar 29, 2015 1:45 PM

Post by Jason Smith on March 28, 2015

Hi professor. What's the difference between the formal charge of an atom and its oxidation state?

1 answer

Last reply by: Professor Starkey
Mon Feb 2, 2015 6:46 PM

Post by Saadman Elman on February 1, 2015

Hi professor, Starley, I signed up in last year. I just finished 1 year of General chemistry. Dr.Ow who taught general chemistry in was absolutely amazing! I got an A in both general chemistry classes through his help. Now i tutor General chemistry. I just listened to your first two lectures. Just got introduced to a lot of new stuff in your first lecture. Your second lecture was a very good review for me as your lecture was very explicit and you clarified it very nicely. To be honest, I am extremely excited about Organic chemistry. I just saw your rating in rating my I am really impressed. I will listen to all of your lectures  and will take a very good notes. We will keep in touch! :D

2 answers

Last reply by: Camille Fraser
Mon Dec 1, 2014 10:50 AM

Post by Camille Fraser on November 30, 2014

How can I just access a specific part of the lesson WITHOUT going through 40 min. of lecture. I have been on this same lesson since last night..over and over again..

3 answers

Last reply by: Professor Starkey
Sun Nov 9, 2014 10:11 PM

Post by Lalit Shorey on November 6, 2014

At 21:46 I don't understand why did we put all those lone pairs next to the oxygen? If the total was 24 valence electrons and 8 e- were used up leaving 16 electrons. Couldn't we just assign for it to have 16 electrons?

1 answer

Last reply by: Professor Starkey
Fri Oct 17, 2014 10:10 AM

Post by David Gonzalez on October 16, 2014

For the first formal charge example, CH3OH2, since Oxygen has a formal charge of +1, does this mean that the entire molecule also has a +1 charge? If so, is it considered an ion? Thank you!

1 answer

Last reply by: Professor Starkey
Tue Sep 9, 2014 11:55 PM

Post by Ahmed alkarkhi on September 8, 2014

Hey Professor nice and elegant lecture. However I do not know much about Pi and Sigma bonds and you have mentioned them quite a bit can you tell me place to get started (i.e  should I look up in the organic chem. lectures or what field should I start my search at). Thank you!

1 answer

Last reply by: Professor Starkey
Sat Aug 23, 2014 8:01 PM

Post by Torrey Poon on August 12, 2014

Dr. Starkey,

At 32:20, I don't understand why there is a negative charge in Example #1

1 answer

Last reply by: Professor Starkey
Sat Aug 23, 2014 6:53 PM

Post by David Gonzalez on August 9, 2014

Hi Professor Starkey, great lecture. Aside from helping us visualize how an organic molecule looks, what are some of the practical reasons of using Lewis structures? Thank you.

1 answer

Last reply by: Professor Starkey
Sat May 10, 2014 12:04 AM

Post by somia abdelgawad on May 7, 2014

can yo please tell me if we have to choose between a missing octete and the negative charge on the most electro negative ion which will be more stable resonace.

1 answer

Last reply by: Professor Starkey
Sun Feb 9, 2014 8:00 PM

Post by nneka igwemadu on February 9, 2014

What is the reason that you didn't move both lone pairs that were on the oxygen at the same time? at 39:33

1 answer

Last reply by: Professor Starkey
Sun Feb 9, 2014 7:58 PM

Post by nneka igwemadu on February 9, 2014

at 38:22 to start could the pid bond be added between the C and the N or is O more Electronegative and that is why you put the pie bond there?

1 answer

Last reply by: Professor Starkey
Sun Feb 9, 2014 2:07 PM

Post by nneka igwemadu on February 9, 2014

On determining resonance. when you are ready to move electron pairs. Do you move everything (electron pairs) that can be moved on a structure before you move on to draw the resonance structure? How do you know when to stop moving pairs? So if there are two areas to move e- pairs do you move them both then move on?

1 answer

Last reply by: Professor Starkey
Wed Nov 13, 2013 1:40 AM

Post by brian loui on November 11, 2013

For #2 of the "Rules for Estimating Stability of Resonance Structures", are the degrees of "separation of charge" determined by counting the bonds that separate the atom's with formal charge?

1 answer

Last reply by: Professor Starkey
Mon Oct 28, 2013 11:07 AM

Post by Razia Chowdhry on October 27, 2013

I just signed into
Thank you, I have a better understanding of resonance structures now.

2 answers

Last reply by: Julie Mohamed
Sun May 18, 2014 10:00 PM

Post by brandon oneal on September 29, 2013

How does carbon for H2C-HC=CH2 have 5 bonds?

1 answer

Last reply by: Professor Starkey
Fri Sep 20, 2013 1:48 PM

Post by Vinit Shanbhag on September 19, 2013

Great lecture.
Can you please elaborate on resonance and stability, does it have anything to do with the reduction of energy in electrons? thanks

1 answer

Last reply by: Professor Starkey
Sat Apr 20, 2013 7:44 PM

Post by Armaghan Shahid on April 20, 2013

you have helped so much with organic chemistry!!!
Thank you so much!

1 answer

Last reply by: Professor Starkey
Wed Aug 8, 2012 11:15 PM

Post by HEIDY PAZ on August 8, 2012

Great lecture!

1 answer

Last reply by: Professor Starkey
Tue Jul 17, 2012 9:54 AM

Post by vishal patel on July 13, 2012 has the best professors.

1 answer

Last reply by: Professor Starkey
Thu May 10, 2012 10:18 AM

Post by Lilian Comparini on May 6, 2012

I love your lectures, you are an amazing teacher. Thank you for being so great!!

2 answers

Last reply by: Professor Starkey
Wed Aug 17, 2011 4:36 PM

Post by alex koralewski on August 1, 2011

on the rules of estimating stabilty of resonance structures slide, in example one.. i was confused at first by the right example as why it was more stable, until i realized that lewis structure being all horizontal (no vertical atoms drawn) wrongly mislead me to believe that the center carbon wasn't octet-happy, since i thought it only had 3 bonds (double to the o and one to the hydrogen). that is, until i thought about it and realized that, had that hydrogen been drawn actually above/below the center carbon, i would have seen the carbon has an actual fourth bond and is indeed octet happy. i just wanted to make this note here in case anyone else gets confused by this as i did. you might want to consider re-drawing it with the vertical component for clarity's sake unless theres a reason its drawn in this (possibly) misleading way.. example #2/#3 have those vertical components of their lewis structures, but i guess hydrogen gets no vertical love normally :-P

4 answers

Last reply by: Professor Starkey
Wed Aug 17, 2011 4:37 PM

Post by Brianna Knoll on June 23, 2011

For this structure would it be possible to have a double bond between the oxygen that is attached to the hydrogen and the nitrogen. Which would mean three loan pairs on both the right side oxygens. In that case there would be more charges (on all three oxygens and the nitrogen) but they would cancel out. thanks

Lewis Structures & Resonance

Draw the Lewis Structure for:
Draw the Lewis Structure for: H2CO
  • Count valanece e
    1C x 4e = 4
    2H x 1e = 2
    1O x 6e = 6
  • Total e = 12
  • 6e used
  • O and C want complete octets
Draw the Lewis Structure for : (CH3NH3)+
  • Count valanece e
    1C x 4e = 4
    6H x 1e = 6
    1N x 5e = 5
  • Total e = 15
  • Subtract 1 for (+) charge so the final total = 14
  • 14e used
  • Now we need to assign charge to N
Draw a second resonance structure for:
Draw and rank the resonance structures from most stable to least stable:
Draw a second resonance structure for:

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.


Lewis Structures & Resonance

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Lewis Structures 0:08
    • How to Draw a Lewis Structure
    • Examples
  • Lewis Structures 6:25
    • Examples: Lewis Structure
    • Determining Formal Charges
    • Example: Determining Formal Charges for Carbon
    • Example: Determining Formal Charges for Oxygen
  • Lewis Structures 12:08
    • Typical, Stable Bonding Patterns: Hydrogen
    • Typical, Stable Bonding Patterns: Carbon
    • Typical, Stable Bonding Patterns: Nitrogen
    • Typical, Stable Bonding Patterns: Oxygen
    • Typical, Stable Bonding Patterns: Halogen
  • Lewis Structure Example 15:17
    • Drawing a Lewis Structure for Nitric Acid
  • Resonance 21:58
    • Definition of Resonance
    • Delocalization
    • Hybrid Structure
  • Rules for Estimating Stability of Resonance Structures 26:04
    • Rule Number 1: Complete Octets
    • Rule Number 2: Separation of Charge
    • Rule Number 3: Negative and Positive Charges
    • Rule Number 4: Equivalent
  • Looking for Resonance 32:09
    • Lone Pair Next to a p Bond
    • Vacancy Next to a p Bond
    • p Bond Between Two Different Elements
    • Other Type of Resonance: Benzene
  • Resonance Example 37:29
    • Draw and Rank Resonance Forms

Transcription: Lewis Structures & Resonance

Welcome back to Educator.0000

Let's talk about how to draw Lewis structures and how to draw resonance forms when resonance is involved.0001

There are several steps we will take in order to draw an accurate Lewis structure.0010

The first thing we need to do is to draw a skeleton.0014

This is going to show the connectivity of the atoms; how are they put together?0016

What we are showing here are the sigma(Σ) bonds; these are the single bonds that connect the atoms.0020

Of course, each atom must be connected by at least one bond to the overall structure.0026

The next thing we want to do is count the total # of valence electrons brought in by each of the atoms.0031

What we want to keep in mind is that the # of valence electrons equals the group # for that atom.0037

With the use of the periodic table, it will be pretty straight forward to figure out how many valence electrons are coming in.0050

Next, we will subtract the charge if it is a charged species.0056

Finally, we are going to fill in the missing electrons.0061

Now, the goal here, in filling in the electrons, is we want to fill in as many octets as possible.0064

We are going to follow the octet rule, and we are going to get as many atoms to have eight electrons around them as possible.0073

Now, there is one thing we really want to keep track here when we are adding in our electrons--is that we never will have more than eight electrons on carbon or nitrogen or oxygen.0082

OK?--those atoms along the second row of the periodic table have no way to expand and to accommodate any more than eight electrons.0098

So, we have to make sure that we never have, for example, five bonds to any of those atoms, or four bonds and a lone pair--anything that will ever total up to more than eight electrons.0106

OK, and finally, we are going to look at the atoms involved in the Lewis structure and determine the formal charges, if any are necessary, we will take a look at those.0116

If we see them, we will draw a nice, clear plus (+) charge or a minus (-) charge.0124

It is a real good idea to circle your charges so that they are very obvious to the observer and not to be mistaken as, maybe, an extraneous pencil mark on the page.0129

OK, so let's take a look at an example.0139

Here, we are given a condensed formula, and if we are asked to draw the Lewis structure, the first thing we will do is we will draw a skeleton.0142

We are going to show our connectivity--that chlorine is connected to the carbon.0149

OK?--and these two hydrogens--each hydrogen can only have one bond; so one is attached, both are attached to the carbon.0153

Then, that carbon is attached to the next carbon; and finally, that carbon is attached to a nitrogen.0162

There is the skeleton of the molecule; we have shown all the Σ bonds.0168

Before we fill in the remainder of the electrons, we need to figure out how many electrons are going to be in the Lewis structure in total.0174

We'll do that by counting up the valence electrons.0181

We have one chlorine; we have two carbons; we have one nitrogen; and we have two hydrogens.0183

With the use of the periodic table, we can see how many electrons each of these atoms brings in.0196

Chlorine is in Group 7, so it brings seven electrons.0200

Each carbon brings in four electrons, for a total of eight.0205

Nitrogen is in Group 5, so that will bring five electrons.0212

Each hydrogen brings along two electrons.0216

We are going to add these up, which will come to twenty-two total electrons.0219

That means that, in the end, that Lewis structure is going to contain twenty-two electrons as either single bonds or pi(Π) bonds, or as lone pairs.0225

Now, in drawing this skeleton--we've already used some electrons, right?0235

Every single one of these Σ bonds represents two electrons; so we have already shown two, four, six, eight, ten electrons.0239

We can show that ten electrons are already used, so we really have twelve electrons remaining to be filled in.0246

How are we going to decide where those electrons go?0259

Well, we are going to try and fill as many octets as possible.0261

We will start with this chlorine--this chlorine has just two electrons bonded to it, and so it wants a total of eight.0267

What we can do is we can add three lone pairs of electrons, so add another six electrons, and now this chlorine has a filled octet.0273

This carbon has two, four, six, eight electrons so that is a filled octet.0281

Remember, hydrogen can only have one bond so having just two electrons to a hydrogen fills its octet, if you will, because it is a filled valence shell.0285

But this carbon has only two bonds, so four electrons; so we could think about adding these as lone pairs. 0296

Then we come to this nitrogen--we find that this nitrogen also needs six electrons, so we can try adding lone pairs there as well.0308

But we are going to find out suddenly that we only had twelve electrons to add in, and so far we have added two, four, six, eight, ten, twelve.0314

These are the only electrons we have to play around with; and in this Lewis structure, we have not filled all our octets.0324

So, a better strategy, than just adding things as lone pairs, is to think about sharing the electrons between two atoms, and that way, you can count the electrons for both atoms.0331

For example, on this carbon, instead of showing this as a lone pair of electrons, we can share them as a Π bond between carbon and nitrogen.0341

In fact, we can do that with a second set of electron lone pair.0351

If we were to draw a triple bond between carbon and nitrogen, that gives this carbon four bonds, for a total of eight electrons.0356

It also gives this nitrogen eight electrons--three bonds and a lone pair--two, four, six, eight.0363

This would be the best Lewis structure for this molecule; it has as many filled-in octets as possible.0370

It turns out there is no formal charge in this molecule.0376

Let's look at an example that would have a formal charge, to see how we would handle that.0379

This next molecule, we could draw the backbone, if we wanted to the draw the Lewis structure for this.0387

We have a CH3 so that carbon has three hydrogens attached.0393

Then it is attached to an oxygen, and that oxygen has two hydrogens.0399

It does not matter the angles at which you draw these bonds; we are not trying to represent a three-dimensional molecule here.0403

We are just showing the connectivity so don't get too worried about whether you draw this hydrogen up or down or to the side, and so on.0410

OK, so there is our skeleton; let's count up our valence electrons.0419

We have one carbon, which brings four electrons.0423

We have one oxygen, which brings six electrons.0428

We have five hydrogens, and that brings five electrons.0433

So we have a total of fifteen electrons which should give us some concern because we have an odd # of electrons.0438

That is not very typical in a stable organic molecule because typically electrons come in pairs so it should be an even #.0447

But what is happening here, the way we are going to adjust that is, see that this has a charge.0454

Remember when we looked at the rules for drawing a Lewis structure, we said that we are going to count up the valence electrons, and then we are going to subtract the charge if necessary.0459

So we are going to subtract that +1 charge, which means instead of fifteen, we are only going to have fourteen electrons.0467

OK, does that make sense?0476

Well, what the positive charge is telling us is that we are deficient in electron; we are missing an electron; that is why we have that positive charge.0478

Instead of having the fifteen that we normally would have for neutral species, we only have fourteen.0484

OK, if this were negatively charged, that means we have an extra electron, so subtracting that charge means we would add an electron here.0489

Make sure this process makes sense to you rather than just following some rules blindly.0497

OK, so we have fourteen electrons to add to this Lewis structure; how many do we have in our structure already?0502

Two, four, six, eight, ten, twelve--we have already used twelve... already.0507

So there is only two electrons left to add to our structure; where do they go?0513

Well, there is only one atom that is missing its octet.0519

If we put a lone pair on this oxygen, we will now have the best Lewis structure possible.0523

The question now becomes: what do we do with this positive charge?--where do we put that?0529

In general chemistry, very often, we will just kind of put a bracket around this whole thing and put a + charge.0534

But in organic chemistry, we don't want to use that formalism because it is going to be very important to us in organic reaction mechanisms exactly where the charges reside.0540

So what we are going to do is we are going to determine the formal charge for each atom in the molecule.0552

Let's talk about the process for determining formal charges.0557

This is something that is for each atom; one by one, we are going to inspect each atom and determine whether or not it should be charged.0560

We are going to determine what is known as, what is described as the electron count for that atom.0567

Now, in a lot of textbooks, you will see a kind of complex calculation to be used for this formal charge.0572

But it's really very simple, and we should avoid using an equation if we can.0579

The electron count is simply counting up all the non-bonded electrons (in other words, all the lone pairs) and half of the bonded electrons.0584

That means, we are just going to count one electron for each bond.0592

We are going to compare that electron count with the valence electrons.0595

If we find that we are missing an electron, that means we have a + charge; and if we find that we have an extra electron, that means we have a - charge.0601

Let's look at an example; OK, so for example, let's look at carbon.0609

How many valence electrons does carbon bring in with it when it bonds with other atoms?0615

It brings in four; carbon wants four electrons to be stable and neutral; it still wants to have those four electrons.0621

Let's take a look at the carbon that is in this Lewis structure.0630

When we count up the electron count for this carbon, we are going to cut each of these covalent bonds in half and just see how many electrons are owned by that carbon.0633

We will count one, two, three, four, and so for this carbon--we see that it wants four, and it has four electrons.0644

That means it is a neutral atom; there is no charge for that carbon.0654

Now, let's take a look at this oxygen.0662

What is the valence for oxygen?--oxygen wants six electrons in order to be neutral.0665

And what is the electron count for the oxygen in this Lewis structure?0674

Well, we have two non-bonded electrons--those are owned completely by oxygen so we will count those completely--one, two; and then half of all our bonded electrons?three, four, five.0678

Oxygen has one, two, three, four, five electrons; so it wants six, but it has five electrons.0689

That means that oxygen is missing an electron.0696

It is one electron short, and if you are missing an electron, you are going to have a +1 charge.0702

Where do we locate this positive charge in the molecule?--we are going to put it right here on oxygen.0709

It doesn't matter where you put it (top, right, bottom, left).0714

Somewhere very close to the oxygen atom, indicating that in this molecule, it is the oxygen that is the most electron-deficient and has a +1 formal charge.0717

Let's take a look at some stable, some typical stable bonding patterns that you should really know by inspection, the more practice you have with Lewis structures.0731

For example, if we are looking at a hydrogen atom--what does a hydrogen atom usually look like when it is in a molecule?0740

Well, it is going to have just a single bond; there is going to be one bond and zero lone pairs.0747

When you have that bonding arrangement, the electron count for that hydrogen--we are going to cut that bond in half.0754

That electron count is just going to be one, and because this equals the valence, that would indicate we have a neutral hydrogen atom.0760

So any hydrogen that you attach to your Lewis structure by a single bond, you automatically know it is going to be neutral, and you don't even have to bother checking its formal charge.0770

How about a carbon?--what does a carbon typically look like to be neutral? 0778

Well, a neutral carbon will have four bonds--maybe, four single bonds, or combinations of triple bonds or double bonds.0781

But if we have four bonds and zero lone pairs, then the electron count for that carbon will be one, two, three, four--which equals the valence for that carbon.0788

So any carbon with four bonds, we know will be a neutral carbon just by inspection.0798

How about a nitrogen?--we just saw an example of nitrogen, and that example is pretty typical of a neutral nitrogen.0805

If we have a nitrogen with three bonds and one lone pair, that will be a neutral nitrogen.0812

Let's check the electron count to verify that.0819

That will have one, two, three, four, five electrons for its electron count--which is the same as the valence for nitrogens.0822

So any nitrogen with three bonds and one lone pair will be neutral.0830

How about oxygen, like the oxygen in water?--what we have are two bonds and two lone pairs. 0834

So we want two bonds and two lone pairs because that will give us one, two, three, four, five, six. 0842

That will give us six electrons for electron count, which is equal to oxygen's valence.0850

Finally, if we take a look at a halogen, we will represent a halogen with the letter, X, very frequently in organic chemistry--so something like fluorine, chlorine, bromine, iodine.0856

How does that typically bond in a molecule, in a Lewis structure?0866

Typically, we will have just one bond and three lone pairs just like the chlorine we saw on a recent example.0872

One bond and three lone pairs is a very stable-looking halide because it has two, four, six, seven electrons around it; and that is exactly what it wants according to its valence.0879

So just by a quick inspection, if we see this normal typical bonding pattern for any of these commonly encountered atoms, we know that it is going to be a neutral atom.0893

More importantly, if we see something that deviates from this normal pattern, we are alerted to the fact that there is going to be a formal charge.0903

Then we need to do our formal charge determination so that we can properly locate them.0910

Let's see another example--how about the Lewis structure for nitric acid, HNO3?0918

Well, HNO3 is very often the way we draw nitric acid, but a better shorthand or condensed notation would be HONO2--that better represents the actual connectivity of the atoms.0924

When we go to draw the skeleton for this, we might think that we do H-O-N-O-O.0937

Now, that would be a possibility for HONO2, but it gives us an arrangement where two oxygens are bound directly to one another.0947

This is called a peroxide--this group is called a peroxide; and this is really quite unstable, explosively so in fact; and so this is unusual.0955

While it might be possible that your structure is going to be a peroxide, we are going to look for an arrangement of atoms that avoids that unstable connectivity.0966

So instead, the Lewis structure, the correct connectivity has all three oxygens bonded to the central nitrogen; so that is actually the connectivity for the nitric acid.0978

OK, we can go ahead and add up our valence electrons.0994

Now, as we get more practice with this, we will eventually be able to skip this step and just try and fill in our Lewis structure by inspection--trying to fill every one, give everyone an octet.0999

But let's try this one more time--where hydrogen brings one electron, our nitrogen brings five, and our oxygens each have three so that gives us eighteen.1011

So we have twenty-four electrons total that will be in this system.1023

We have already shown two, four, six, eight electrons; so that means we have sixteen electrons left to distribute on this structure to try and fill as many octets as possible.1029

So, right here, I see this oxygen; I know oxygen likes to have two bonds and two lone pairs; so that would be a good idea--is add two lone pairs right here.1045

We could do that for this oxygen as well--we could have two bonds and two lone pairs to that oxygen; so that would give a good-looking oxygen.1054

This nitrogen now has a filled octet--it has four bonds, for a total of eight.1065

Our remaining six electrons can go on this oxygen to fill its octet.1070

So we've added two, four, six, eight, ten, twelve, fourteen, sixteen; so we now have a total of twenty-four electrons.1075

It is a good idea to check at the very end to make sure you have the correct # of electrons that you were expecting.1081

Because a lot of times, when we are shuffling around electrons or moving them around, it is very easy to maybe add an extra pair or lose a pair of electrons.1087

Then you have no chance of getting the proper Lewis structure.1094

This would be a Lewis structure for nitric acid, except when we take a look at it, we see that we have a few atoms with atypical bonding patterns.1099

For example, this nitrogen-- this nitrogen has four bonds; and usually, we like to see a nitrogen with three bonds and a lone pair; so we need to calculate the formal charge for that.1109

Nitrogen wants five electrons--that's its valence, but how many does it have when we count its electron count?1118

For the electron count, we count one electron for each bond--one, two, three, four; so it only has four.1128

What do we do when we have a missing electron?--that is a + charge.1136

So this nitrogen--every nitrogen with four bonds is going to be a positively charged nitrogen.1141

Where do we have another atom that has an unusual bonding pattern?1147

This oxygen, instead of having two bonds and two lone pairs, it has something different; so let's count this.1152

Let's remind ourselves: oxygen wants six.1159

What does it have here?--one, two, three, four, five, six, seven; this oxygen has seven electrons; so here we have a case where there is an extra electron.1165

And what charge would you associate with an extra electron?--a negative charge.1176

Any oxygen with one bond and three lone pairs is going to be an O-.1180

So this Lewis structure overall is a neutral structure, just like the formula told us.1185

There is no net charge, but it does, in fact, have formal charges within the structure.1192

That is why it is very important to recognize unusual bonding within a Lewis structure, so that you can be alert to finding these formal charges where they exist.1196

Now, there might be another Lewis structure that you could come up with for nitric acid that is equally good.1208

How about, instead of having the double bond to this top oxygen, we did the double bond to this bottom oxygen?1215

If we did that arrangement, and we filled in our lone pairs...1231

It will give us an N+ and an O-, and this would be a second Lewis structure.1239

When we have two Lewis structures that we can draw for a molecule, that means both Lewis structures are valid.1246

We call this relationship of having more than one Lewis structure--we call this resonance.1253

Resonance means we have more than one Lewis structure that we can draw, and we are going to talk a little bit more about resonance in just a second.1261

Let me show you one other possible Lewis structure that you might be considering.1268

Now what is very nice about this Lewis structure is it gets rid of all of our formal charges.1275

Right?--so that is very tempting, isn't it?--but, there is a problem with it.1282

This Lewis structure is impossible; what rule have I violated here?1285

That nitrogen has five bonds for a total of eight electrons.1291

Because it has more than eight electrons, I violated the octet rule, and this is impossible.1296

OK, so there are, in fact, two good Lewis structures that I can draw--valid Lewis structures I can draw for nitric acid, and these do represent resonance.1304

So let's talk a little bit more about resonance and what that means.1313

As we just saw with nitric acid, resonance means that more than one Lewis structure can be drawn.1321

When we compare those Lewis structures, what we see is we have movement or delocalization of Π electrons and non-bonded electrons in p orbitals.1326

We will see that these are in p orbitals when we look at some further examples of resonance.1337

One thing we are pointing out here is we are not moving Σ bonds; we are only moving Π bonds and lone pairs; the Σ bonds are staying the same.1345

The actual structure is a hybrid.1358

The fact that we can draw more than one Lewis structure for a given compound means that neither Lewis structure is an accurate representation of what the molecule really looks like.1362

So what we do is we try to draw as many Lewis structures as possible to show what characteristics the actual structure has.1373

The actual structure is described as a hybrid, and so for nitric acid, the hybrid looks something like this.1381

Remember, we had either a double bond to one oxygen or a negative charge to one oxygen.1390

The actual--the hybrid has a partial double bond spread out to both oxygens, and a ¥ä- on both, and a + charge on the nitrogen.1396

So this is the actual hybrid; this is what the structure really looks like.1407

But it is very difficult to work with a hybrid structure like this.1411

Instead, we just draw a discrete Lewis structure and work from there, recognizing that discrete Lewis structure is not a complete story.1414

OK, so what we are saying is we are not flipping back and forth between structures; it is not an equilibrium between one Lewis structure and the other.1424

Instead, it is called resonance, and we use this special arrow which has a line with an arrow on each end--indicates resonance.1435

It means that we have this special movement of just our Π and our non-bonded electrons.1445

Now, it turns out that all resonance forms may not contribute equally.1452

Although we might have two or three or even four resonance forms or even more--not all of them are equally good resonance forms, equally stable resonance forms.1456

So they don't all contribute equally necessarily.1467

We will take a look at that on the next slide.1470

Very commonly, we use curved arrows to show the electron movement from one resonance form to the next.1473

We usually start with the first resonance form and show arrows on how we would convert it to the next resonance form, and so on; so we only go from left to right.1479

We will see an example of that on the next slide as well.1488

Remember, because we are not moving our Σ bonds, no atoms are moving; there is no change in the bond length, no change in the bond angles.1492

Our atoms are fixed in space; it is just the lone pairs and the Π electrons that are shifting around, delocalizing in a continuous set of p orbitals.1499

No Σ bonds break; we already mentioned that.1513

Then, there is also going to be no change in overall charge; because we are not adding electrons, we are not removing electrons; the net charge has to remain the same.1516

But, of course, as we relocate electrons throughout the structure, our formal charges between atoms are most definitely going to be changing.1524

The key thing that we need to keep in mind is that resonance equals stability.1532

It is going to be very important to recognize resonance and be able to draw resonance forms and evaluate resonance forms.1543

Because when we find resonance occurring, that means we are delocalizing our electrons, spreading out electrons, and that is always a good thing.1548

That always means we are stabilized by resonance, or we are resonance-stabilized.1555

So how can we identify when resonance is going to be happening?1560

Actually, first, let's discuss the stability of various resonance forms and how we would estimate which form is more stable.1569

Here is an example of a molecule that has two different resonance forms.1576

Which of these is more stable and which contributes more to the overall hybrid?1580

Well, what we will find is the greater the number of covalent bonds, the greater the stability because more atoms will have complete octets.1585

So the number one thing we are always going to be looking for to evaluate a Lewis structure is, the more we have for complete octets, the better the Lewis structure.1593

That is always the first thing we want to take a look at.1602

When compare these two structures, we see that this carbon is missing an octet, while on the second structure every atom has a complete octet--a total of eight electrons around it.1606

So this is the more important; this is more stable.1620

We also say, we describe it as being a better contributor.1628

What does the actual hybrid look like?1634

The actual hybrid looks more like this structure on the right because it is the more stable structure, it is the better contributor, it is more important than the first.1635

Because this is missing an octet, this is less important.1651

OK, and finally, let's see if we can use curved arrows to get some practice shifting electrons around.1658

How would you say--what has changed in going from this structure on the left to this structure on the right?1665

Looks like this lone pair of electrons that used to be on oxygen is now being shared between the oxygen and carbon as a Π bond.1671

So what we do is we used a curved arrow, showing these two oxygens picking up and moving between the oxygen and the carbon.1677

This a really good habit to get into so we can keep track of our electrons--really important for bookkeeping electrons when we do curved arrows.1686

OK, another thing to look for is if we have a separation of charge.1694

If we could avoid formal charges, that would be a very good Lewis structure, so that is something to keep in mind.1698

If we compare this first structure to this next structure, those are both valid Lewis structures, but this is going to be the most important... because this one has formal charges.1705

How about comparing the second one to the third one, which of those is more important?--they both have formal charges.1721

You might think about bringing that + and ? closer together; that would be a good thing except something else is missing in this last structure that makes it not as good as the middle structure.1728

Looks like we have with this carbon again, a carbon with just three bonds; this is missing an octet.1740

This is the least important because octets are the most significant thing that we are looking for.1748

Again, covalent bonds are something you can look for; I lost a covalent bond in this structure so that tips me off to the fact that I must have lost an octet.1755

Let's do some curved arrow practice, and going from this first to second structure, how do we convert that?1765

This lone pair moved in to be a Π bond, and then this Π bond moved up to be a lone pair.1771

Notice our formal charges have... formal charges have been created as a result.1777

We can practice with those, double check if those charges are right.1784

How about going from this second structure to the third structure?1787

We've taken this Π bond and picked it up and moved it to be a lone pair on nitrogen.1790

We are moving Π bonds and we are moving lone pairs to do this resonance.1795

OK?--our third rule is that, all other things being equal, a structure with a negative charge on the more electronegative element will be more stable.1800

Similarly, if we have a positive charge that we are delocalizing, positive charges on the least electronegative will be more stable.1809

For example, when we look at this carbonyl--a C=O double bond is called a carbonyl--we can pick this lone pair up and move it to the oxygen to get a C+ and an O-.1817

Or we can pick this Π bond up and move it to the carbon to get a C- and an O+.1830

Now, one of these resonance delocalizations is quite significant, and the other is very unlikely.1836

Because oxygen is more electronegative than carbon, the oxygen prefers to have a negative charge, better handles a negative charge.1844

This is important resonance... and this resonance is unlikely.1852

Finally, resonance forms that are equivalent have no difference in stability and contribute equally.1866

Sometimes, we might compare two resonance forms and not find any significant difference in their structure.1873

For example, this guy is known as the allyl carbocation, and they both have one double bond and one carbon missing an octet--double bond, carbon missing an octet.1880

Because they have equal features of stability, they are going to be equal in energy, and therefore contribute equally to the resonance hybrid.1890

OK, let's show our curved arrows.1899

It looks like this Π bond is picked up and just jumped over to be on the other side, and as a result, this positive charge will now be on the end carbon.1901

Notice that with my arrows, I am never moving the charges; our curved arrows only move electrons so they are either going to start as a lone pair or they are going to start at a Π bond.1910

The charges will move as a result; so we will just calculate those charges on the new Lewis structure.1922

OK, how will we know when to expect resonance delocalization?--resonance stabilization?1929

Well, there is three main structural features that we can look for.1935

One possibility is to have an arrangement like this: where we have a lone pair that is next to a Π bond; so the lone pair is not on the same carbon as the Π bond; it is the next carbon over.1940

We call this being allylic so if we have an allylic lone pair, that tell us that resonance can occur; and the resonance that happens is as follows.1953

The lone pair becomes a Π bond; and the resonance can't be done yet because if all we did was share this as a bond, this carbon would now have five bonds to it and that would be impossible.1966

So what happens is as this lone pair comes in to be a Π bond, this Π bond picks up and becomes a lone pair.1978

We use those two arrows, and what do we get as a result?1984

We now have a double bond to the left-hand carbon... sorry, I had drawn this as a C-H.1989

We want to make sure that our atoms are not moving; and single bond now, with a lone pair on this carbon.1999

Now as a result, I think our formal charge has also moved; let's take a look at that.2007

What is the formal charge on this carbon?--carbon wants four; it has one, two, three, four, five.2011

Carbon has five; it only wants four; so it has an extra electron; this is where our negative charge is.2019

This is called allylic-type resonance, and we have that--every time there is a lone pair next to a Π bond, we will look for that.2026

OK, here is another possibility.2034

Now we have a vacancy next to a Π bond; we have an atom that is missing an octet.2036

This carbon has only six electrons around it; and when that is next to a Π bond, again, we call that allylic.2042

This is an example, in this case, of an allylic carbocation; a carbon with a positive charge is called a carbocation.2050

What we have in this vacancy is we have an empty p orbital.2060

A carbocation has an empty p orbital here, so it can interact with these two electrons in the p orbital right here.2068

What can happen is the Π bond can move over and fill that vacancy.2075

This carbon now has just three bonds; we took a bond away from it so that is where our carbocation is.2084

It has one, two, three electrons; it only has three electrons, but carbon wants four; so it is missing an electron.2089

OK, so we can have a vacancy next to a Π bond like an allylic carbocation.2096

One last thing to look for is having a Π bond between two different atoms, such as in a carbonyl; we call this carbonylic resonance; it is called a carbonyl when you have a C=O double bond.2100

OK, and every time we have something that looks like a carbonyl, we are going to pick up that Π bond, the electrons are going to move to the more electronegative atom.2110

We can pick up these two electrons and move them up to oxygen...2118

So there is just a single bond; and instead of just two lone pairs, now there is three lone pairs on that oxygen.2125

See how those arrows help us keep track of our electrons?--really good habit to get into.2130

Let's take a look at our formal charges: this carbon has just one, two, three electrons; it wants four; so it is missing an electron; this is a C+.2135

This oxygen has one, two, three, four, five, six, seven; oxygen has seven; it only wants six, so this is an O-; so that would be carbonyl-like resonance.2146

OK, now these are three patterns to look for in molecules to anticipate resonance.2158

There is another kind of resonance that exists, and this is something we will talk about a bit down the road--and it is looking at molecules such as benzene.2165

Benzene is the molecule that has a six-membered ring with three alternating Π bonds--we call these conjugated Π bonds...2175

Because they are alternating--we have double bond, single bond, double bond, single bond.2189

When you have three conjugated Π in a six-membered ring, it is a very special molecule, and it has something known as aromatic resonance; it is an example of an aromatic compound.2194

There are compounds related to benzene that have the same special stability.2205

What happens in benzene is you can take these Π bonds and move them around the ring.2209

This could shift over her, which causes this one to shift, which causes this one to shift; and you can draw the same molecule, but with the Π bonds located in different positions.2215

Now, does this look like it is resonance?--sure, we've kept all the atoms in their same position; we've simply moved around the Π electrons.2224

This is actually excellent resonance because since both of these are equivalent resonance forms, they contribute equally to the overall picture, and that really gives great resonance stabilization.2233

We will talk a lot about aromatic compounds in the future.2245

Let's try an example where we are asked to draw resonance forms, and then rank them in terms of their contribution.2250

I have given here, just as our starting compound, a skeleton of a molecule.2259

Let's see if we can fill in our lone pairs and Π bonds to give everyone a filled octet.2265

This oxygen has two bonds so let's put two lone pairs; that will make that oxygen neutral.2272

This oxygen wants two bonds and two lone pairs as well so we can make a Π bond here.2280

That also satisfies this carbon because now carbon has four bonds; I know carbon likes to have four bonds.2287

How about this nitrogen?--it has three bond right now; and if I added a lone pair, then this would be a neutral nitrogen as well.2292

So this is an example of a Lewis structure that is easy to complete just by inspection.2300

It is possible to fill the octet for every atom and have them be neutral so this, in fact, is a good Lewis structure for this.2305

Now are there any resonance forms?2314

What we need to do is we need to look for patterns in this molecule that would give rise to resonance.2316

For example, and there is several; for example, one thing I see is that I have a carbonyl.2323

Anytime I have a carbonyl, I have carbonyl-like resonance where that Π bond can move up and become a lone pair.2328

All carbonyls are resonance-stabilized; and so that means that this structure also has contributing to it an O-C+ structure.2336

OK, so this a second resonance form; what other possibilities are there?2351

Well, I have a vacancy; now I have a carbocation that is missing an octet--is there any way I can fill in that octet?2357

Perhaps I can use some of these neighboring lone pairs; if I use one of these lone pairs, it brings me back to the Lewis structure I already have.2365

But if I shared one of these lone pairs, that would give me a new Lewis structure.2372

I still have an O- up here... and this carbon now has four bonds so it is neutral.2387

But how about this oxygen?--that doesn't look like a typical oxygen.2395

Let's count: one, two, three, four, five; this oxygen has five, but it wants six; so it is missing an electron; so this is where our + charge has moved to.2398

Now, remember, this molecule started out neutral; so every Lewis structure, every resonance structure from here on out, has to have a zero net charge; and this still does, we have a + and -.2408

OK, there is actually one additional resonance form we can draw, one other Lewis structure that is possible for this.2421

Do you see any patterns that have not accounted for yet?--how about having a lone pair that is next to a Π bond?2427

This lone pair is allylic; every time we have an allylic lone pair, that can be resonance--where the lone pair comes in and the Π bond moves over.2435

That would give this structure, with an O- up top.2448

And where is our positive charge now?--our positive charge is on this nitrogen; one, two, three, four; nitrogen wants five; and this is a fourth resonance form.2452

Any other attempts at shifting around the lone pairs and the Π bonds will bring us back to one of these original structures that we drew.2463

These are all the possible resonance forms; now let's rank those resonance forms and decide who is the best contributor and the worst contributor and so on.2471

Of all these, I think this first one looks the best because it has no separation of charge, no formal charges.2482

This is number one--this is the best because it has no charges, it has filled octets (I should write that first actually), and it has no charges.2489

This would be our best Lewis structure; this is the Lewis structure we would typically draw, and this is the one that best represents the hybrid.2504

OK, but these other Lewis structures also exist; and so that means they are also contributing, to some extent, to the rest of the structure--to the actual hybrid.2511

OK, so between these remaining three, there is one that also stands out.2523

Remember, the most important thing we should look for in Lewis structures is to have filled octets.2530

In one of these structures, we are missing an octet--can you see it?2537

Right here, this carbon has only three bonds and so it does not have eight total electrons.2541

So this is going to be our worst because it is missing an octet... on carbon here, missing where that is.2547

OK, these last two structures have completely filled octets, so let's see what the difference is between them.2560

They both have an O- so that is the same; but here we have an N+, and here we have an O+.2568

Now, we take a look at the location of those formal charges and decide where a positive charge would rather be--which do you think is more stable, having an O+ or an N+?2575

Well, we know that oxygen is more electronegative than nitrogen; oxygen is the second most electronegative atom in the periodic table; so he is more electronegative than everyone, except for fluorine.2585

Because he is so electronegative, would he like to have a positive charge?--I don't think so.2595

This is... it is better to have a positive charge on nitrogen, so this is going to be our number two structure, and this is going to be our number three structure.2600

When we compare these two, they have filled octets...2609

OK, but the N+ is better than O+; and that is because nitrogen is more electropositive.2615

Or you could say that nitrogen is less electronegative--that says the same thing.2634

So in a case like this, we want to look for the differences in our Lewis structures, always looking for filled octets first.2639

And then, finding formal charges and trying to locate them on the best atom, depending on whether we are looking at a positive charge or a negative charge.2647

This concludes drawing Lewis structures and ranking resonance forms.2657

We will see you back soon at; thank you.2661