6.05 Resonance & Formal Charge

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En provenance du cours de University of Kentucky

Chemistry

615 notes

University of Kentucky

Chemistry

615 notes

This course is designed to cover subjects in advanced high school chemistry courses, correlating to the standard topics as established by the American Chemical Society. This course is a precursor to the Advanced Chemistry Coursera course. Areas that are covered include atomic structure, periodic trends, compounds, reactions and stoichiometry, bonding, and thermochemistry.

À partir de la leçon

Week 6

This unit looks more in depth at molecular compounds to see how they are bonded together by looking at Lewis structures. These structures provide information about the types of bonds (single, double, or triple) as well as the connectivity of atoms. By knowing the Lewis structure, we can also predict the three-dimensional geometry of an individual molecule.

6.01 Lewis symbols 4:36

6.02 Covalent bonds 5:19

6.03 Electronegativity 11:21

6.04 Lewis Structures Part 1 13:11

6.04 Lewis Structures Part 2 18:49

6.04a Draw the Lewis structure for H2O 1:13

6.04b Draw the Lewis structure for H2CO 2:00

6.04c Draw the Lewis structure for CCL4 1:21

6.04d Draw the Lewis structure for NH3 1:16

6.05 Resonance & Formal Charge 16:27

6.05a Determine the formal charg on atoms in NH4+ 2:24

6.05b Determine the formal charge on atoms in H2CO 3:17

6.06 Exceptions to the Octet Rule 8:48

6.06a Draw the Lewis structure for I3- 1:16

6.06b Draw the Lewis structure for ClF4 1:41

6.06c Draw the Lewis structure for XeF4 1:23

6.07 VSEPR Part 1 11:01

6.07 VSEPR Part 217:07

6.07a Determine the electron pair and molecular geometries for I3- 1:42

6.07b Determine the electron pair and molecular geometries for NH3 1:22

6.07c Determine the electron pair and molecular geometries for ICl4- 1:57

6.07d Determine the electron pair and molecular geometries for PF5 1:52

6.07e Determine the electron pair and molecular geometries for XeF2 2:16

Rencontrer les enseignants

Dr. Allison Soult
Lecturer
Chemistry
Dr. Kim Woodrum
Sr. Lecturer
Chemistry

In this module we're going to talk about Resonance Structures and

determining Formal Charges.

By the end of this module, you should be able to draw resonance structures for

a molecule.

And determine the best structure from a set of

non-equivalent structures using formal charge.

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

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

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

maybe even three or four structures, and

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

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

So drawing these resonance structures is really a human invention.

It just helps us because,

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

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

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

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

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

those two oxygens.

And in reality when I look at a carbonate ion,

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

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

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

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

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

We will deal with having non equivalent structures, and

that's what we use formal charge for.

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

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

Here we show the double bond between the right oxygen and

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

actually show this arrow with the double head.

And so if we had a third resident structure,

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

Now, when we look at SO2, again, just like the carbonate,

we don't see the double bond and a single bond, we see an average of these two.

And that's an idea that's kind of hard to grasp, we're going to look at

an example that might be a little bit more familiar to you.

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

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

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

the other day the dog is a labrador.

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

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

Our electrons are not shifting back and forth.

We don't go from one structure to another,

even though we're showing that with an arrow.

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

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

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

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

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

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

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

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

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

And so the bond length here,

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

And this mixture is called our resonance hybrid.

So kind of like an average of these two structures.

So just like this is a hybrid dog breed.

Our molecule here is a hybrid, it's a mix of these two.

And we represent that by showing all the possible Lewis structures.

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

We have our Benzene rings here.

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

Lewis structure.

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

we draw it on either side.

And what we get is a two resident structure possibilities.

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

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

Everywhere where we see these intersections of these points, or

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

And we actually don't show the hydrogen.

Because the hydrogens are what we call implied hydrogens.

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

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

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

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

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

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

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

Because single bonds are going to be longer than double bonds, and

so what we see are 6 bonds that are longer than double bonds.

But shorter then single bonds.

So the key thing to remember here is that these are equivalent structures.

There's no difference between them.

And because of that, we frequently see Benzene ring written as.

The six carbon ring with a loop or an oval in the middle, indicating that

we have this resonance hybrid structure, that all of our bonds are absolutely

equivalent to one another, both in terms of bond link and bond energy.

So resident structures help us deal with equivalent structures, but

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

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

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

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

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

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

So let's look at an example.

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

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

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

structure is for this particular molecule.

So how do we calculate formal charge?

Well, it's done for each individual atom in a molecule.

And, we're going to look for a structure that minimizes our formal charge values,

so it minimizes the magnitude of them.

In other words, we want values that are as close to zero as possible.

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

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

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

And that’s not as good as a structure that has values all of zero.

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

so the same valence electrons we’ve been looking at for Lewis structures and

based on electron configurations.

We subtract the number of lone pair of electrons.

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

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

electrons around that particular atom in that particular Lewis structure.

Now a couple of things to keep in mind when we're looking at formal charges, and

some of these should sound a little bit

familiar when we assigning oxidation numbers.

So when we look at a neutral molecule we know that the sum of

the formal charges must be equal to zero for ions,

the sum of the formal charges must be equal to the charge on the ion.

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

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

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

And the answer is actually yes.

When we look at these possibilities,

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

Is the more electronegative atom.

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

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

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

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

Therefore it's going to be,

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

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

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

each atom in each of these structures.

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

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

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

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

And I end up with a formal charge of 0.

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

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

And minus half of the 4 electrons in the bonds.

And again, I get a formal charge of 0.

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

because the chlorines are exactly the same.

They're both single bonded,

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

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

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

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

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

And that looks pretty good.

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

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

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

I still have 4 electrons in the isolated atom.

I have no non-bonding electrons, and

I still have 8 bonding electrons around that.

So, I end up with a value of 0.

Now I look at oxygen, which has 6 valence.

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

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

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

because they have different bonding and different structural features.

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

it's actually going to look just like.

The chlorines in the first structure we looked at,

because it has the exact same structural features there.

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

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

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

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

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

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

We have to compare it to the alternatives.

Here we really only had two possible structures.

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

are not equivalent, that we have to decide among.

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

I would also have a resonance structure.

For this particular Lewis structure,

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

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

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

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

So now let's let you try one, and

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

So let's calculate the formal charges for

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

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

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

in this structure, and half of the bonding electrons.

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

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

minus half of my bonding electrons, which is 8.

And again I get a formal charge of 0.

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

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

So I get a formal charge of minus 1.

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

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

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

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

0 for all of these atoms.

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

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

Well, when we look at examples,

we see that there is more than one possible structure.

So when we looked at our COCl2 example.

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

went with the chlorine atom.

Sometime we have to worry about the connectivity of atoms,

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

Because their electronegativity values are the same,

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

because they have the same electronegativity.

So I might have to look at structures where,

sometimes carbon is the central atom.

And sometimes sulfur is the central atom.

And then I can find the formal charges and

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

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

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

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

So let's look at an example,

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

So this is our correct structure,

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

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

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

So, when I look at my formal charges.

I see that when I have nitrogen in the middle,

I get very high formal charges minus 2 plus 2 or minus 1.

Here, I'm minus 2 and plus 1.

And what we see is that nitrogen,

is not the least electronegative atom of the three.

So we would not expect it to go in the middle.

And so the formal charge just proves that to us as well.

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

with the oxygen and the carbon having the triple bond.

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

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

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

Because this is an ion,

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

I know that oxygen is the more electronegative atom,

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

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

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

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

In the isolated atom.

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

on the oxygen versus the carbon.

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

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

In the next module, we're going to look at EXCEPTIONS TO THE OCTET RULE.

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