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Last time we were talking about the general interactions that
are present in a chemical bond. The nuclear-nuclear repulsion.
The electron-nuclear attractions.
The electron-electron repulsions.
And we talked about how the competition of those three
interactions are what gave rise to this interaction potential
that we drew. It is the competition between
those interactions that determines what the bond length
of a molecule is. It is the competition between
those interactions that determines what the well depth
is, where the bond strength is in a chemical bond.
Well, today what we're going to do is move on and talk about a
specific model for chemical bonding.
And the model we're going to talk about is that of Lewis
diagrams. Now, interestingly,
this is a model that is non-quantum mechanical.
This was invented 20 years before quantum mechanics,
by G. N.
Lewis. Somehow G.
N. Lewis recognized,
and I don't know how he recognized --
That's genius, I guess.
But he recognized that the key to chemical bonding was a
sharing or an overlap of two electrons, one from each atom.
And he represented these ideas using these Lewis diagrams in
which he took the valence electrons, the total number of
valence electrons in the two atoms that were forming a bond.
And he redistributed them so that one electron from each atom
that was making this bond shared an electron and the rest of the
valence electrons were distributed amongst those two
atoms such that each atom had an octet of electrons around them.
Each atom had a rare gas electron configuration.
And that's the basis of this idea of Lewis diagrams.
The idea of the Octet Rule. There's no chalk today.
I've got a piece. Okay, the Octet Rule.
For example, if you have a fluorine atom
with its seven valence electrons here coming together with
another fluorine atom with its seven valence electrons to make
a fluorine molecule. Well, the idea is that each one
of these extra electrons now on the fluorine,
or these unpaired electrons, are shared between the two
fluorines. The result is now,
given the sharing, that this fluorine has an octet
of electrons around it and this fluorine now has an octet of
electrons around it. They have this octet
configuration which is the lower energy filled-shell
configuration. Now, that's the case,
of course, for all the elements except for hydrogen.
In the case of hydrogen, we're coming in with one atom.
Say it's combining or reacting with a chlorine atom.
Well, in this case, this electron from the hydrogen
is going to be shared with this electron from the chlorine.
And now, again, this chlorine has this octet
around it. But the hydrogen,
of course, is just two electrons around it.
Its inner gas configuration is that of helium,
and so it is satisfied with just two electrons around it.
With this sharing idea, Lewis could really understand
the stoichiometry of many compounds.
He understood why hydrogen gas was H two and not a
hydrogen atom. He understood why nitrogen gas
was N two and not a nitrogen atom.
He could understand why oxygen was O two and not an
oxygen atom. Now, these Lewis diagrams have
absolutely nothing to do with orbitals or wave functions,
which we actually know to be a more accurate description of
chemical bonding. We're going to get to them.
Lewis diagrams work 90% of the time to tell you where the
electrons are in a molecule. It works so well and it is easy
that chemists have used these Lewis diagrams before quantum
mechanics and have continued to use them after quantum
mechanics. They use them because it's easy
and it works. It allows you to understand
where the electrons are in a chemical bond,
on which atoms the electrons are centered,
where the lone pairs are, where those electrons that are
in the molecule that actually don't participate in the
chemical bond. For example,
here in HCl, these two electrons,
these two electrons, these two electrons,
they aren't participating in the chemical bond.
We call these electrons here lone pairs.
That's an important definition that we're going to use.
So we're going to look at this system.
Again, it's easy. The alternative is solving the
Schrödinger equation, which you don't want to do if
you don't have to do, because that's hard.
You cannot do that quickly. So that's the value here.
It's going to give us a framework, these Lewis diagrams,
to kind of organize a lot of chemical reactions.
To kind of organize our ideas of bonding and what will react
with what. Let's start doing this.
And to that we're going to have a set of rules here.
And let me put this front screen down.
Actually, on the side screen, could you put my rules up
there?
All right. The basic idea in Lewis
structures is that we're going to take the total number of
valence electrons and the two atoms that are going to form
this bond. We are going to share electrons
between the two atoms, and then take the rest of the
valence electrons and distribute them amongst the two atoms in
that molecule such that each atom has an octet configuration.
To draw a Lewis diagram you're going to have to know what the
valence electrons are. In particular,
how many valence electrons each atom has.
This you have to know really off the top of your head.
And this is just a quick list of the number of valence
electrons. Carbon has four,
nitrogen has five, oxygen has six,
etc. Let's, as an example,
illustrate how to draw these Lewis diagrams.
Here are our set of rules. We are going to go through them
step-by-step. As you get better at this,
you won't have to go through these rules in such detail.
But for now, when we're just learning how to
do it, I think this is a very effective way.
So let's start. Let's take an easy one.
We want to draw the Lewis structure of methane.
The first thing you have to do is you have to figure out what
the skeletal structure is. You have to kind of figure out,
as a first pass, what is bonded to what.
And now I'm going to give you some hints.
First of all, hydrogen, if you have a
hydrogen atom. That's always what we call a
terminal atom. Terminal means that it is only
bonded to one other atom. Chlorine atoms also are usually
only terminal atoms. Hydrogen always is.
Fluorine is 99% of the time only a terminal atom.
In the case of methane, which you know has four
hydrogens and one carbon, if the hydrogens have to be
terminal, well, then there is only one choice
here. You put the carbon in the
center. So that's our skeletal
structure. Number two here on our Lewis
diagrams, it says count the total number of valence
electrons. So let's do that.
Hydrogen has one valence electron, but there are four
hydrogens. Carbon has four valence
electrons. The total number of valence
electrons in our molecule here is eight, and I am going to draw
those eight valence electrons. Number three here in our rules
says count the total number of electrons that you would have to
have if each of the atoms had an inner gas configuration.
These are the number that we've got, eight.
And now we are calculating how many we would like to have in
the optimal situation if each atom has an inner gas
configuration. Let's do that.
Well, hydrogen wants two. And we have four hydrogens,
so four times two. Carbon wants eight.
If each of these atoms had an inner gas configuration we would
need 16 electrons. Step four says to calculate the
number of bonding electrons, the way you do that is by
taking the total number of electrons that you would have to
have if each atom had a noble gas configuration and subtract
from it the total number of electrons that you actually do
have. So we're going to take 16,
subtract from that eight. We're going to have eight
bonding electrons. The way I'm going to represent
the eight bonding electrons, or bonding electrons in general
is that I am going to draw a red square around my electrons that
are bonding electrons. Up here these are the valence
electrons that I've got. It says eight of those valence
electrons are bonding electrons, so I am going to draw a box.
Those are my bonding electrons. Step five says to assign two
bonding electrons to each bond. Well, let's do that,
assign two bonding electrons to each bond.
Well, if we put two there, two there, two there and two
there, then we've used up all of our valence electrons and all of
our bonding electrons in that structure.
Number six says if you've got bonding electrons that remain,
well, the answer is no, we used them all up.
And number seven says do we have any valence electrons left?
The answer is no, we don't.
We've used them all up. So we are done.
And you can see that. The hydrogen has an inner gas
configuration around it, the carbon has an inner gas
configuration, an octet of electrons around
it. Hey, we're done with methane.
But you could have done that anyway, right?
Yes? Yeah.
Here is another one. Oh, one other concept here.
In this particular structure, carbon is the central atom.
What I mean by a central atom is an atom to which everything
else is bonded. Sometimes we're going to ask
you to write a Lewis structure, and we are going to tell you
what atom is the central atom so you know to put it in the center
and everything else gets connected to it.
Let's do a higher one. Here is this molecule.
It is going to be hydrogen cyanide, HCN.
The first thing you have to do, according to our rule number
one, is to draw a skeletal structure.
You have a choice here of how to draw that skeletal structure.
You can put the nitrogen in the center or the carbon in the
center. Hydrogen is not going to go in
the center because we know the hydrogen is terminal always.
What you often do is a first pass when you're drawing the
skeletal structure, is that you put the element
with the lowest ionization energy in the middle.
That is what our rule number one said.
Element with lowest ionization energy goes in the middle.
That often is true, but not always.
This is a first pass. I am going to show you a little
bit later some cases where this won't be true and how we are
going to determine that the lowest ionization element
doesn't go in the center. We're going to have a way to
figure this out. But for right now,
if you had no other choices, that's what you would do,
you'd write a structure with the lowest ionization energy
element in the center. Okay, so we did that.
Number two, calculate the total number of valence electrons in
this molecule. Well, hydrogen brings one
valence electron into the party, carbon brings four and nitrogen
brings five. We have a total of ten valence
electrons. I am going to draw those ten
valence electrons. While you are learning to do
this, you might also want to draw ten valence electrons like
this. Next step, calculate the total
number of electrons each atom would have to have if,
in an ideal world, they had a noble gas
configuration without sharing. Well, the total number of
electrons that we would have to have is two for the hydrogen,
eight for the carbon and eight for the nitrogen.
A total number of 18. Step four says we calculate the
number of bonding electrons. To do that we subtract the
total number that we would have to have for an inner gas
configuration, 18, from the total number that
we actually do have, which is ten.
The result is eight bonding electrons.
I am going to now draw this red square around the valence
electrons that are going to be bonding electrons.
Here it goes, we've got eight bonding
electrons and two electrons left out here.
We will see that in a moment, where they go.
Number five, assign two bonding electrons to
each bond. We'll put two there,
we put two there, that's great.
We've now used up four bonding electrons.
We've got four more bonding electrons left.
Our next step is, do we have any bonding
electrons left? We sure do.
We've got four of them. In that case you start to make
double bonds or triple bonds. A double bond is one in which
four electrons are shared. A triple bond is one in which
six electrons are shared. In general, which atoms do you
make double bonds or triple bonds between?
Well, in general, double bonds form between
carbon, nitrogen, oxygen and sulfur.
Triple bonds usually restrict it to carbon,
nitrogen and oxygen. In this case,
we've got a carbon and a nitrogen here,
we've got four extra bonding electrons.
Well, what we are going to do is we are going to put these in
between the carbon and the nitrogen.
We're going to have a triple bond right there.
We took care of all our bonding electrons now.
Step seven, do we have any valence electrons remaining?
We sure do. We've got two valence electrons
remaining. What do we do with them?
Well, we assign them as lone pairs.
We look to see which atoms don't have an inner gas
configuration. Well, hydrogen does.
The carbon does now. It's got eight electrons around
it. But the nitrogen doesn't.
It only has six. So we're going to put those two
extra valence electrons, as a lone pair,
here on the nitrogen. And now the nitrogen has an
octet around it. These two electrons are called
lone pairs. They don't participate in the
chemical bonding. We did that one.
Next one, thionyl chloride. It is Cl two SO.
How do you do this one? Well, as a first guess,
the halogens usually are terminal atoms so they are not
bonded to anything else. That's usually the case.
Not always, especially for chlorine.
We're going to put the chlorines out on the ends
somewhere. So then do we put oxygen in the
center or sulfur in the center? Well, we're going to pick the
lowest ionization energy element and will put that in the center.
That is going to be sulfur. So we will try this as a
skeletal structure. Count the total number of
valence electrons. Chlorine brings seven valence
electrons into the party, and there are two chlorine
atoms. Sulfur brings six.
Oxygen brings six. We've got a total number of 26
valence electrons. Let me draw them all out.
Step number three is calculate the total number of electrons
that we would have to have for each atom to have an inner gas
configuration. We've got four atoms here.
Each one of them has got eight. We're going to have 32.
That's what we would need for an inner gas configuration for
each atom. Let's calculate the total
number of bonding electrons. Well, that's 32 minus the
number of valence electrons. That's six.
Of those 26 valence electrons, let me draw my bonding box
around six of them. Next step is to assign those
bonding electrons. That's going to be easy.
Put two there, two there and two there.
That means we've just used up our six bonding electrons.
Do we have any more left? No, there aren't going to be
any double bonds here. However, we've got lots of
valence electrons left. We've got 20 of them.
We're going to have to make lone pairs with those 20 excess
valence electrons. We're going to have to make ten
lone pairs so that each atom has an octet around it.
So let's do that. We're going to put six of them
around the oxygen, so now the oxygen has an octet,
another six, there they go,
another six around the chlorine, there they go,
another six around this chlorine, there they go.
And, finally, we've got two left.
Well, that's going to have to go onto sulfur to make an octet.
There it goes. We are done.
We've used up all of our valence electrons and we have an
octet around each one of the atoms in that molecule.
So that's the basic procedure for writing these Lewis
diagrams. But now, I talked about putting
the lowest ionization energy element in the center sometimes.
And I talked about you've got try this skeletal structure
first. How do we really know if we've
drawn the right Lewis diagram? Well, the way we're going to
tell is by this concept called Formal Charge.
Formal charge is a measure of the extent to which an atom has
gained or lost an electron when it has formed a chemical bond.
So compared to when it is an atom, the formal charge tells
you how much of an electron that has been gained by it or lost by
it. I will make that a little more
real in just a moment. How do you assign a formal
charge? Well, we've got a recipe for
this. That formal charge is the
following. You count up the number of
valence electrons on the free atom.
We're going to call that V. We're going to subtract from it
the number of electrons that are lone pairs on that atom in the
molecule, so V minus L. And then the third step is
you're going to subtract one-half of the number of
bonding electrons that atom has. Or, one-half the number of
electrons that it shares in that molecule.
So it's V minus L minus S over 2.
S is the number of electrons it shares.
That's the formal charge.
Let's illustrate this. Let's take this ion,
CN minus. I already drew the Lewis
structure here, so now what we're going to do
is we're going to calculate the formal charge for carbon and the
formal charge for nitrogen within this Lewis structure.
Formal charges depend on the Lewis structure.
Ready? Let's do it.
For carbon, V is 4. We have four valence electrons.
In this structure L is 2. L is the number of lone pair
electrons. Look at in this structure.
Here are these electrons, the lone pair that don't
participate in the bond. We have two lone pair
electrons. It's minus 2.
And then we have to subtract from that minus one-half the
number of shared electrons. Well, the number of shared
electrons in the case of the carbon is 6.
We've got a triple bond here. That's 6 over 2,
that's 3. The formal charge on carbon in
this Lewis structure is minus 1. What we often do is put a minus
1 somewhere near that carbon and circle it.
That is the formal charge on the carbon.
What that means is that the carbon has gained one more
electron than it usually has as an atom.
It's a little bit more negatively charged in this
molecule than it is in the atom. Let's do nitrogen.
Nitrogen, valence electrons five.
Then minus the number of lone pairs.
Well, that's 2. Here is that lone pair.
And then it's minus one-half the number of shared electrons
or bonding electrons. Well, that's 3.
The result is a formal charge of 0.
And so what we would usually do is we'd write a zero with a
circle around it near that nitrogen.
So the formal charge on carbon is minus 1, the formal charge on
nitrogen is 0 in this structure, but now you've got to check to
make sure you did everything right.
And here is the check. The check is that the total
charges here, minus 1 plus 0 is minus 1,
the sum of the formal charges has to equal the overall charge
on the molecule. Minus 1 plus 0 is minus 1.
That does equal the overall charge on the molecule.
So we did everything right. That is an important thing to
check. That's how you calculate formal
charge. Now, how do you use it?
Well, first of all, before I tell you how you use
it, let me just emphasize that formal charge is not oxidation
number. Oxidation number does not
depend on the Lewis structure. Formal charge depends on the
Lewis structure. You're going to talk about
oxidation number with Professor Drennan.
If you don't know what it is, that is okay.
You will learn about it in a few weeks.
But if you do know what it is, I just want to emphasize that
formal charge is not the oxidation number.
How do we use formal charge to make sure that we've got the
correct Lewis structure? That's what you do.
[LAUGHTER] You use the formal charge.
What you do is you realize that structures with lower absolute
values of formal charge are the more stable structures.
Structures which have formal charges on the atoms that are
lower are more stable. Structures where the number of
electrons that the atom has gained or lost in forming a
molecule is smaller are the stable structures.
This is really important because you're going to often
have a situation where you've got two different skeletal
structures. You've obeyed all of the rules
for drawing the Lewis diagrams of those two different skeletal
structures. Those two different
arrangements of atoms. But only one of those
structures is going to be correct.
And the one that is correct is going to have the lower set of
absolute values of the formal charge.
So let's illustrate this. Let's take this ion.
It's the thiocyanate ion, NCS minus,
as I have written it here. Now, if you were just told
you've got an ion that has nitrogen, carbon and sulfur in
it, you might say, well, how do I start?
Well, you could start by putting the element with the
lowest ionization energy in the center.
That would be a good place to start.
You should do that. But what we're going to do is
we're going to look at all three possibilities here.
We're going to look at a Lewis structure where the carbon is in
the middle, where the sulfur is in the middle,
and now where the nitrogen is in the middle.
And we drew these Lewis structures.
I've already drawn them here for you according to our rules
posted here on these walls. But only one of these
structures is correct. And we're going to use the idea
of formal charges to determine which one is correct,
which one has the lowest energy.
If we take this structure and we calculate the formal charge
for nitrogen, we will find it is minus 1.
The formal charge on the carbon here is 0.
The formal charge on the sulfur here is 0.
So I already did these calculations for you.
If you sum up the formal charges we get minus 1.
The overall charge on this ion is minus 1.
We did it right. We did everything right.
You should check your formal charges like this.
Let's do the same thing for this structure.
We calculate the formal charge on carbon is minus 2.
The formal charge on sulfur is 2.
The formal charge on nitrogen is minus 1.
Let's sum these up. When we do that we get a minus
1. It indeed matches the overall
charge on the ion. Let's calculate the formal
charges for each atom here. Sulfur is 0.
Nitrogen is 1. Carbon is minus 2.
Overall charge minus 1. We did that right.
Now, which one of these structures is the lowest energy
structure? Ready?
I'm going to ask you to raise your hand or to scream.
Is it structure number one? Yes.
Is it structure number two? No.
Is it structure number three? No.
You're right. It's structure number one.
It has the lowest set of formal charges.
Minus 1, 0, 0. Only one atom here has a formal
charge other than zero. These two structures have
formal charges that are minus 2 or 2.
I don't think that there is a structure on earth,
well, maybe there is, that has a formal charge of
minus 2 or 2 that will make it the lowest energy structure.
If you've got a structure with a minus 2 or 2,
that's very, very likely not the lowest
energy structure. This is a powerful tool to know
if you have drawn your skeleton correctly, so remember this.
It's really important. On an exam, we're going to tell
you to draw the Lewis structure with the lowest energy.
You're going to have to try several of them to see which one
has the lowest set of absolute values for the formal charge.
Question?
Number of valence electrons on the carbon is 4 minus the number
of lone pairs on the carbon is 4.
That's 2. 4, is that your question?
Yeah. Other questions?
Okay. Here is another use of the
formal charge. This is also determining the
skeletal structure here, but you will see a problem in
just a moment. Here is another special case.
We have CH three NHO minus.
How do you draw, in this case here,
that skeletal structure? Could I have some lights on the
board here?
CH3 here, this is what we call a methyl group.
It is always terminal since the hydrogens are always terminal.
So the structure looks like this.
And the carbon, then, is bonded to the next
element over. If you see a structure like
this, where you've got lots of different elements,
the first pass, just take these elements in a
row. The CH3, this methyl group is
bonded to the nitrogen. The next one is hydrogen,
but we know we're not going to put the hydrogen between the
nitrogen and the oxygen because the hydrogen is always terminal.
So the hydrogen is either going to go here, or the hydrogen is
going to go here. And that's the question that
we're going to ask here, on this slide,
is this the correct structure, or is this the correct
structure? I drew the skeletal structures,
I drew the Lewis diagrams. Which one is right?
Well, if I go and calculate the formal charges here,
I would find that the formal charge on oxygen is minus 1 and
that the formal charge on all other atoms is 0.
In this structure, if I go calculate the formal
charge on the nitrogen in this structure is minus 1.
And then the formal charges on all the other atoms are 0.
We've got a problem here. We've got two different
structures, and they have absolutely the same set of
formal charges in terms of the absolute magnitudes.
We have only one atom that has a minus 1 formal charge,
only one atom that has one more electron that it usually does in
its unbonded state, in its atomic state.
So which one of these is correct?
Well, it turns out that this structure is correct.
It is correct because of this. It is correct because we put
the formal charge on the most electronegative atom which is
oxygen. Oxygen is more electronegative
than nitrogen. That is the other key here,
is that if you've got two structures, same sets of formal
charges, and you have a choice of putting the minus formal
charge on an atom, you put it on the most
electronegative atom. That is going to be the lowest
energy structure. In general here we talked about
the electronegativities. The electronegativity of
fluorine is greater than oxygen, is greater than nitrogen,
is greater than carbon. That is something that you will
need to know. So we've seen how this formal
charge now is useful in figuring out whether or not we had the
right skeletal structure to begin with.
Now, here is another --
-- utility of formal charge. Let's look at this structure
NO3 minus. Well, I know ahead of time that
we're going to put this nitrogen in the center.
If you just had NO3 minus and you didn't know anything,
you would have to play around with some structures and look at
formal charges, but I have already done that
playing around, so I know to put nitrogen in
the center. But let's do the Lewis diagram
for this. We count up the valence
electrons. We've got six for every one of
the oxygens and five for the nitrogen.
And now, this one is important, we added an extra electron when
we counted up the valence electrons.
That extra valence electron comes because this is an O3
minus. We have an extra electron on
NO3, so we have to count it in our valence electrons.
If this was NO3 plus, we would subtract one right
here. But this is NO3 minus,
so we've got to add one. That's 24 valence electrons.
Let me draw them all out. Number three.
Let's calculate the number of electrons that we would have to
have if each atom had a noble gas configuration.
That would be 32 because we've got four atoms.
Let's calculate the number of bonding electrons.
Well, that's 32 minus 24 or eight.
So, of those 24 valence electrons, eight of them are
bonding electrons. Let's assign them.
Well, here is two, here is two,
there is two. That's great.
We used up six. Do we have any remaining?
We sure do. We've got two,
so we're going to have to make one of these a double bond.
So we've got one double bond there between nitrogen and
oxygen. Now, we've got a lot of
remaining valence electrons. We've got 16 that are
remaining. We're going to distribute them
around the oxygens in order to make an octet around each atom.
There we go. We did that.
And so now we've used up all of our valence electrons.
Let's calculate the formal charges.
The formal charge on the singly bonded oxygen is minus 1.
I put a minus 1 there. I'm going to put a minus 1
there. The formal charge on the doubly
bonded oxygen right in here, that's 0.
I'm going to put a 0 there. The formal charge on the
nitrogen, well, that's 1.
I'm going to put a 1 there. If I sum up the formal charges,
I get a minus 1. That does agree with the
overall charge on this molecule, NO3 minus.
So we did everything right. But now you see something a
little strange here, right?
And that is, for whatever reason,
I picked this nitrogen and oxygen to make a double bond.
But I could have also made the double bond between this
nitrogen and oxygen. And if I had calculated the
formal charges, I would have identically the
same set of formal charges. Or, I could have picked this
oxygen and nitrogen to put in the double bond.
If I calculated the formal charges, I'd have the same set
of formal charges. So we've got three different
structures, all which have the same set of formal charges.
What are we going to do here? Which one of these is correct?
Well, when you've got this situation here,
this is a hint that you've got what we call a resonance
structure. So if you have the same
skeletal arrangement and you've just got three different
structures, and the electrons are in different places,
and the same set of formal charges exist,
this is a hint that you have something called a resonance
structure. What is a resonance structure?
Well, here is the bottom line. In NO3 minus,
if you look at the structure of that ion, you would not find
that one of the nitrogen-oxygen bonds was a double bond and that
the other two were single bonds. A double bond is stronger than
a single bond. A double bond is shorter than a
single bond. So if you do the appropriate
experiment, you can tell whether or not you have a shorter or a
longer bond. You can tell whether you have a
weaker or a stronger bond. The bottom line is,
that when you look at this molecule experimentally,
you do not see one bond that is shorter and stronger than the
other two nitrogen-oxygen bonds. What you find is that all three
bonds are equivalent. All three bonds have the same
strength and the same length. You find that the length of the
nitrogen-oxygen bonds are a little bit shorter than those of
a single bond and you find that the nitrogen-oxygen bonds are
uniformly, all three of them, a little bit stronger than a
nitrogen-oxygen single bond. But not quite a double bond.
What you find is that all of these three bonds is like a bond
and a third. So what is happening here?
What is happening here is that this extra pair of electrons,
instead of being localized on just one of the nitrogen-oxygen
bonds, those extra pair of electrons are actually
delocalized over all three bonds.
And we call that a resonance hybrid.
We denote that resonance hybrid in this way, with three
different structures with double-headed arrows.
This does not mean, this is important,
that the structure is flickering between here,
here and here. That's not what is happening.
Rather, these two extra electrons are delocalized over
all three bonds such that the bond is like a bond and a third
rather than one bond or two bonds.
This is called a resonance structure, a resonance hybrid.
And you will be clued in that you've got this kind of
resonance hybrid if you're able to write three structures with
the same set of formal charges. You will know to do this.
Okay. Have a nice cool weekend.
See you Wednesday.