0:19

So, when we look at resonance structures what we see is that.

Â For some molecules or ions, we cannot draw one structure that best represents that

Â particular molecule, and so we actually have to draw a couple of structures, or

Â maybe even three or four structures, and

Â what we see is that we don't actually see all of those structures in nature.

Â what we actually see is kind of a hybrid, or an average of those structures.

Â So drawing these resonance structures is really a human invention.

Â It just helps us because,

Â its really hard to draw an average of four different structures.

Â So we just draw all four and know that those are resonance structures.

Â So the example we looked at in the previous unit, we looked at carbonate ion

Â and we said we drew the double bond to the oxygen going up, but there was really.

Â No particular reason why we couldn't have shown the double bond between either of

Â those two oxygens.

Â And in reality when I look at a carbonate ion,

Â what I'm going to see is not a double bond and two single bonds, I'm going to see

Â three bonds that are kind of a mixture, an average, of a double and two single bonds.

Â 1:21

So let's look at an example here with SO2.

Â We have two possible equivalent structures, and that's a key thing to

Â remember about resonant structures, that these are equivalent to one another.

Â We will deal with having non equivalent structures, and

Â that's what we use formal charge for.

Â Where we have two structures that are equivalent, we have to show them both.

Â And so here, we show the double bond, between the left oxygen and the sulfur.

Â Here we show the double bond between the right oxygen and

Â the sulfur, and the way we indicate these are resident structures is we

Â actually show this arrow with the double head.

Â And so if we had a third resident structure,

Â we could actually draw another arrow and draw that third structure as well.

Â 2:21

So if we look at two breeds of dogs, here we have a poodle and a labrador.

Â And if we bring these two dogs together, what we end up with is a labradoodle.

Â We don't see a scenario where one day the dog is a poodle, and

Â the other day the dog is a labrador.

Â It is a labradoodle, it is a mix between them.

Â And the same thing is happening when we look at our molecules.

Â Our electrons are not shifting back and forth.

Â We don't go from one structure to another,

Â even though we're showing that with an arrow.

Â What we're saying is that these two structures best represent the Lewis

Â structure or the structure of the bonds, within this molecule which is

Â actually ozone and that we can't really draw that hybrid structure.

Â We can try to, we can show these dash lines to kind of indicate.

Â That it is a hybrid between a single and double bond.

Â However, it's not always this easy to draw the symbol of a molecule, and

Â so trying to draw an average can be very complicated for some larger molecules.

Â So it's easier to show the multiple resident structures but knowing.

Â That when we look at the molecule, what we actually see, is a cross between them.

Â And so the bond length here,

Â will be somewhere between the bond length of a single bond, and a double bond.

Â 3:55

Here we can look at another example of a molecule, that has resonant structures.

Â We have our Benzene rings here.

Â We know that Benzene has alternating double bonds when we draw our

Â Lewis structure.

Â But we don't know which set of carbons are involved in those double bonds, and so

Â we draw it on either side.

Â And what we get is a two resident structure possibilities.

Â Now, we can actually show this a little more simplified,

Â which is the way we shorthand things a lot of times in organic chemistry.

Â Everywhere where we see these intersections of these points, or

Â if we had a line with no atom on it, that would represent a carbon atom.

Â And we actually don't show the hydrogen.

Â Because the hydrogens are what we call implied hydrogens.

Â We know that each carbon has to have four bonds, and that if

Â there are not four bonds, those remaining spots are actually filled up by hydrogens.

Â When we start looking at larger organic molecules, simplifying the structure in

Â this way makes our lives a lot easier, and a lot easier to draw.

Â Now just like with the carbonate and the SO2, I don't see three single bonds and

Â three d, double bonds when I look at these molecules.

Â What I see is, I have 6 bonds that are the exact same length.

Â 6:01

So resident structures help us deal with equivalent structures, but

Â we also have to look at how we deal with non-equivalent structures.

Â And so formal charge is a way we can actually kind of

Â keep up with our electrons, a little bookkeeping going on,

Â to see how many electrons are assigned to an atom in an isolated atom.

Â Verses how many electrons are in an atom in a particular Lewis structure.

Â And so we use this to evaluate non-equivalent Lewis structures.

Â So let's look at an example.

Â Here are two non-equivalent Lewis structures, and we actually want to be

Â able to do is, to assign a formal charge to each atom in each of these structures.

Â And from that we're going to be able to decide what the best Lewis

Â structure is for this particular molecule.

Â 7:07

So, for example, if we had a set of values of minus 2, 0,

Â plus 2, that is a not as good of a structure,

Â as something that has a set of Lewis structures of minus 1, 0, plus 1.

Â And thatâ€™s not as good as a structure that has values all of zero.

Â So what we do is we take the number of valence electrons in their isolated atom,

Â so the same valence electrons weâ€™ve been looking at for Lewis structures and

Â based on electron configurations.

Â We subtract the number of lone pair of electrons.

Â Or the number of non-bonding electrons in that, assigned to

Â that particular atom in the structure, and then we subtract half of the bonding

Â electrons around that particular atom in that particular Lewis structure.

Â 8:13

So say we get down and we figured out to the two possible structures.

Â And what we see are having formal charges of say 0,0 minus 1, and minus 10,0.

Â So we say, well, they're basically the same values, does it really matter?

Â And the answer is actually yes.

Â When we look at these possibilities,

Â what we have to do is we have to figure out which atom.

Â Is the more electronegative atom.

Â And it's going to be better able to handle that negative formal charge.

Â Because when we have a negative formal charge, what we're saying is

Â that there are more electrons assigned to that atom in the compound,

Â than it normally has when it's in an isolated atom.

Â Therefore it's going to be,

Â the more electronegative atom is going to be better able to handle that

Â extra electron density, that's assigned to in that particular structure.

Â 9:07

So now let's go back to our examples, and let's calculate the formal charges for

Â each atom in each of these structures.

Â So I'm going to look at my carbon first,

Â in carbon I have 4 valance electrons in the isolated atom.

Â I don't have any non bonding electrons assigned to carbon in this structure.

Â And for half of my bonding, I have 8 bonding electrons.

Â And I end up with a formal charge of 0.

Â For oxygen, I know that I have 6 electrons in the isolated atom,

Â minus the 4 that are at non-bonding electrons assigned to oxygen.

Â And minus half of the 4 electrons in the bonds.

Â And again, I get a formal charge of 0.

Â For chlorine, I only need to do this calculation once,

Â because the chlorines are exactly the same.

Â They're both single bonded,

Â they both have the same number of non-bonding electrons assigned to them.

Â So I know that chlorine has 7 valence electrons minus 6 electrons

Â non-bonding assigned to it in this particular molecule, minus half

Â of the bonding electrons, and I also get 4 more charges of 0 for my chlorine atoms.

Â So, now I have a set of formal charges of 0, 0, 0.

Â And that looks pretty good.

Â I can't get any closer to 0 then that.

Â But for comparison we want to look at the formal charges of our other structure.

Â So for carbon, I see that it's going to look exactly the same.

Â I still have 4 electrons in the isolated atom.

Â I have no non-bonding electrons, and

Â I still have 8 bonding electrons around that.

Â So, I end up with a value of 0.

Â Now I look at oxygen, which has 6 valence.

Â Here it has 6 non-bonding electrons, minus half of the bonding,

Â so I have 6 minus 6 minus 1, gives me minus 1.

Â Now, I have to look at my chlorine atom separately,

Â because they have different bonding and different structural features.

Â So I'm going to look at the chlorine that has the single bond first, and

Â it's actually going to look just like.

Â The chlorines in the first structure we looked at,

Â because it has the exact same structural features there.

Â When I look at the chlorine with a double bond, I still start with 7 electrons.

Â I subtract off the 4 that are non-bonding in the chlorine, minus half.

Â Of the 4 that are bonding, and what I get is the formal charge of plus 1.

Â So looking at this I have values of 0 minus 1, 0 and plus 1.

Â Together they really aren't that bad of a set of formal charges, and

Â to determine the best structure we can't just look at one set of formal charges.

Â We have to compare it to the alternatives.

Â Here we really only had two possible structures.

Â But for some molecules, we'll actually have more structures that

Â are not equivalent, that we have to decide among.

Â Now, if it were to be that this were the better of the two structures,

Â I would also have a resonance structure.

Â For this particular Lewis structure,

Â because I can put the double bond between this carbon and chlorine.

Â Because all my values are zero, I've got the values as close to zero as possible.

Â I see that this is going to be the better Lewis structure, and

Â this is equivalent to what I'll actually see in nature.

Â 12:20

So now let's let you try one, and

Â calculate the formal charges for each atom in this given structure.

Â So let's calculate the formal charges for

Â sulfur, carbon, and nitrogen so we can see why this answer is correct.

Â When I look at sulfur, I know that it has 6 electrons in the isolated atom.

Â I subtract out the four non-bonding electrons assigned to it

Â in this structure, and half of the bonding electrons.

Â And when I do, I get a formal charge of 0.

Â For carbon, 4 electrons in the isolated atom, minus 0 non-bonding electrons,

Â minus half of my bonding electrons, which is 8.

Â And again I get a formal charge of 0.

Â For nitrogen I have 5 electrons in the isolating atom, minus 4 non-bonding,

Â minus half of my bonding, and I see that that's going to be 4.

Â So I get a formal charge of minus 1.

Â So, when I look at the sum of those values, I get minus 1, and

Â because the charge on my ion is minus 01 this is exactly what I would expect.

Â So, then, my formal charges must add up to minus 1, therefore,

Â it's not possible to give any structure that has a set of formal charges of 0, 0,

Â 0 for all of these atoms.

Â Somewhere there has to be something with a charge, so that the sum equals minus 1.

Â 13:45

So why do we use formal charges in the first place?

Â Well, when we look at examples,

Â we see that there is more than one possible structure.

Â So when we looked at our COCl2 example.

Â We didn't know whether the double bond, went with the oxygen, or

Â went with the chlorine atom.

Â Sometime we have to worry about the connectivity of atoms,

Â particularly when I'm looking at, say, carbon and sulfur.

Â Because their electronegativity values are the same,

Â I don't know which one is necessarily going to be the central atom,

Â because they have the same electronegativity.

Â So I might have to look at structures where,

Â sometimes carbon is the central atom.

Â And sometimes sulfur is the central atom.

Â And then I can find the formal charges and

Â decide, which of those structures is the best one for that particular compound.

Â It also helps me to determine the most probable structure, so I can take into

Â account where the double bond is going, what the central atom is, and this is

Â going to be the one that gives me the one that's most likely to be found in nature.

Â So let's look at an example,

Â where we determine the preferred Lewis structure for the cyanate ion.

Â So this is our correct structure,

Â because when we look at the formal charges we minimize those values.

Â So let's look at how we calculate the formal charges for each of these

Â structures, so we can see how we decided that, that was in fact the best one.

Â 15:24

When we look at the structure up at the top with the carbon in the middle, but

Â with the oxygen and the carbon having the triple bond.

Â We see that we also get minus 2 and plus 1,

Â also not a great set given that I have these two values here for

Â answers two and three, where I have 0, 0, minus 1 and 0, minus 1 0.

Â Because this is an ion,

Â I know that I'm going to have to have one that has values not all equal to 0.

Â I know that oxygen is the more electronegative atom,

Â which means it has more attraction to those electrons in a covalent bond,

Â and that means it's better able to handle that extra electron density.

Â Because, when I have a formal charge of minus 1, remember that,

Â that means I have more electrons assigned to that atom than it has.

Â In the isolated atom.

Â And so, I have to choose between having the formal charge of minus 1

Â on the oxygen versus the carbon.

Â And I choose to put it on the oxygen, because it's the more electronegative of

Â those two atoms, and it's better able to handle the negative charge.

Â