CDS 8 Molecular Structures II
General Chemistry: Concept Development and Application
This course will cover the topics of a full year, two semester General Chemistry course. We will use a free on-line textbook, Concept Development Studies in Chemistry, available via Rice's Connexions project. The fundamental concepts in the course will be introduced via the Concept Development Approach developed at Rice University. In this approach, we will develop the concepts you need to know from experimental observations and scientific reasoning rather than simply telling you the concepts and then asking you to simply memorize or apply them. So why use this approach? One reason is that most of us are inductive learners, meaning that we like to make specific observations and then generalize from there. Many of the most significant concepts in Chemistry are counter-intuitive. When we see where those concepts come from, we can more readily accept them, explain them, and apply them. A second reason is that scientific reasoning in general and Chemistry reasoning in particular are inductive processes. This Concept Development approach illustrates those reasoning processes. A third reason is that this is simply more interesting! The structure and reactions of matter are fascinating puzzles to be solved by observation and reasoning. It is more fun intellectually when we can solve those puzzles together, rather than simply have the answers to the riddles revealed at the outset. Recommended Background: The class can be taken by someone with no prior experience in chemistry. However, some prior familiarity with the basics of chemistry is desirable as we will cover some elements only briefly. For example, a prior high school chemistry class would be helpful. Suggested Readings: Readings will be assigned from the on-line textbook "Concept Development Studies in Chemistry", available via Rice's Connexions project. In addition, we will suggest readings from any of the standard textbooks in General Chemistry. A particularly good free on-line resource is Dickerson, Gray, and Haight, "Chemical Principles, 3rd Edition". Links to these two texts will be available in the Introduction module.
This course and the accompanying textbook explain chemistry in a very simple, ground up way that I was able to easily understand.
Lectures are great and material is comprehensive. The only problem is there is a multiplicity of errors in the questions.
Bonding and Structures in Covalent Molecules
To understand the types of compounds which can be formed and the properties of those compounds, we have to understand how atoms bond together to form molecules. In this module, we develop a model for the bonding of non-metal atoms to non-metal atoms, called a covalent bond. The model can be used to predict which combinations of atoms are stable and which are unstable. Observations of the structures of the molecules lead to a model to understand molecular geometries and properties related to those geometries. From this, we build a foundation for understanding and predicting how molecular structure is related to molecular reactivity and function.
CDS 8 Molecular Structures I19:56
CDS 8 Molecular Structures II13:09
John Steven Hutchinson
Dean of Undergraduates and Professor of Chemistry
Following our discussion of isomers in the previous lecture, we're going to dig
into a real subtlety, having to do with molecular structures in this lecture.
This one's a bit more challenging, but it's extremely important, in a wide, a
wide variety of molecules. So to illustrate, we're going to take the
Lewis structure or structures for C6H6. Now there may be a variety of ways in
which this can be drawn, but we're going to draw it in one particular way.
We're going to consider first exactly how to work through our usual calculation
here. The number of available electrons is
equal to 6 for each carbon of ah,4 for each carbon atom plus 1 for each
hydrogen. So that's a total of 30 available
electrons for this atom or for this molecule.
The number needed to form octets for all of them is 6 for each carbon atom plus 2
for each hydrogen atom. So that's a total of 60 needed to satisfy
the octet rule for all of these atoms. The number shared therefore must be 30
and the number of bonds must be 15. There may be a variety of ways in which
we could do that, but it turns out in fact, the best way to do it is to draw a
ring of carbon atoms. And each carbon atom with six hydrogens
all together has a single hydrogen attached.
So we'll draw that structure. Looks a bit like the cyclohexane that we
drew in the previous lecture. Let's count up the number of electrons
we've drawn, and the number of bonds we've drawn.
We have one, two, three, four, five, six, seven, eight, nine, ten, 11, 12 of the 15
bonds all together. Notice as well, though, that none of the
carbon atoms in this structure have satisfied their valences.
They each have three bonds and they should all have four, and there are three
more bonds to go. With a little bit of inspection, what we
could discover is that if we applied double bonds in every other location in
this particular molecule, we'd have what we need.
We would have actually each carbon atom now has four bonds attached to it.
Let's count up all the electrons here and make sure they equal 30.
Two, four, six, eight, ten, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30.
So this structure works just fine. There is an interesting subtlety though.
Let's draw another carbon ring as follows.
Attach hydrogen to each one of those carbon atoms.
And notice that I made a choice along the line that you might not have noticed.
I put a double bond in this location on the left.
I could have put the double bond in that location on the right.
What happens if we do that? Well if we put the second double bond
there or the first double bond there, then the second double bond has to be two
away from it. And the third two away from both of
those. Notice these two structures are not the
same structure. The atoms which are bonded are not the
same atoms with the double bonds. So, that raises the question, which one
of these structures is correct? Is this like in the previous
consideration where when we control do, two different structures, that we were
drawing two different compounds with two different molecular properties.
And the answer is no. There's actually only a single compound
of this type. And these two structures are therefore
not isomers of each other. And the important distinction here is you
notice, we did not rearrange the atoms. We rearranged the electrons.
The bonding of the carbons to each other and with hydrogens are arranged in
exactly the same way in these two structures.
What has moved is not the atoms but rather the electrons between the atoms,
which have moved. That's a big difference.
So these are not resident structures at all.
Well, what is the appropriate structure? To try to understand that we need to look
at some experimental data. And it turns out the experimental data
tell us something interesting. All of the bond lengths, the
carbon-carbon bond lengths in this molecule are all the same.
They all have exactly the same length. Each one of these bonds here has a length
of 140 picometers. Well so what?
Well two conclusions, one is we already know the double bonds and single bonds
have different lengths. In either one of the structures that we
have drawn here, we should have three single bonds with longer length and three
double bonds with shorter length. And it would just be a difference of
which one of these structures gave us the right structure, but we didn't get that.
Experimentally, all of the bond lengths were exactly the same.
Furthermore, none of the bond lengths correspond to, the bond lengths
associated with either single bond, or double bond.
Notice these numbers here, a single bond is 154, a double bond is 134.
The actual bond length of the benzene molecule is part way in between those
two. Not exactly half but fairly close to the
middle, between the single bond length and the double bond lengths.
So we have a couple of puzzles to solve here.
Since the bond lengths are the same and they are neither single or double bond.
The indication is that neither one of these structures is correct, but rather
some other structure is correct and we refer to the structure as resonance
structures. These are two examples of resonance
structures. They have the same molecular formula.
They have the same molecular structure. What differs between the two, are the
arrangement of the electrons between the bonds.
And significantly, according to what we just argued a little while ago, the best,
most accurate structure that we can draw, for benzene in this particular case, is
what we will call a resonance hybrid of the two resonance structures.
It is neither this, nor this, but rather is a structure in which the, which
combines the properties of both of these. Meaning that everyone of these bonds is
somewhat double and somewhat single. How should we draw such a peculiar
structure. Conventionally chemists draw this in a
variety of ways. I will draw it in one.
We'll make our ring of carbon atoms here. We know that there's at least a single
bond between each of the carbon atoms. We also know that all of the bonds have
to be the same. And there have to be six more electrons
distributed in the bonding of this ring, because we know we're supposed to have a
total of let's see 15 bonds, and we have only 12 in this diagram.
The typical way in which this is shown is to draw a dotted line, indicating that
the second bond, the three double bond out there, are in fact delocalized around
the ring. So, we've got delocalization, and as I
noticed a little while, noted a little while ago, we refer to this as a
resonance hybrid. Meaning it is a combination of the two
structures that we drew on the previous page.
Since each of the bonds is partially single and partially double, then they
are all the same length and the bond length should be intermediate between a
single bond and a double bond. Well that's an interesting subtle
structure there. Let's look at another molecule that does
the same thing, to try to illustrate this property.
Nitrogen dioxide turns out to be an important example.
let's try to draw that one. This one actually has a couple of
subtleties in it, we'll focus on one of them.
So nitrogen dioxide, what I can tell you here is that the nitrogen is between the
two oxygens in this compound. One of the things we'll abserve is that
both of the bond lengths are exactly the same.
It might be then that what we would imagine is that these bond lengths are in
fact just single bond lengths, but let's do that calculation again.
The number of available electrons is 5 for the nitrogen for 6 for each Oxygen,
so that's 17. The number that we need is eight for the
nitrogen, plus 8 for each of the oxygens, so that's 24.
So, the number of shared electrons is 7, that's peculiar we've never seen an odd
number before. The number of bonds then needs to be
three or maybe four. So we have to figure out what we're
going to do to draw this structure. Let's go with three at first.
I need to add a second bond to one side or the other.
Let's draw it on the left side just because it's closer to me right now.
Then let's see we've used up six electrons, there are 11 electrons more to
go. We'll actually add four of them to that
oxygen to complete its octet. We'll add 6 of them to this oxygen to
complete its octet. And then we'll add one more to the
nitrogen because that's all we have left to add.
The nitrogen doesn't quite have its octet fulfilled.
But that's a consequence of the fact that we have an odd number of electrons.
The oxygens are both fine. This structure doesn't seem to be right.
Because what we noted a little while ago, is that both of the bond lengths are the
same. We should have gotten the same number.
We should have gotten bond links which in fact exactly corrrespond to a single bond
or a double bond. And not somewhere in between.
But you can tell immediately where we're going with this.
I made a random choice to put the double bond on the left, I could've made the
random choice to put the double bond on the right.
If I did, then I would wind up with structure which looks like that.
Once again, neither one of these structures is, in fact, correct.
Because both of them would predict, or either of them rather would predict, that
the bond lengths would be the different not the same.
So, if we make a comparison of the bond lengths to the actual bond lengths, we
notice that the NO bond length is somewhere between single and double.
And as a consequence what we can say is that there must be a residence hybrid.
These are referred to as resonance structures.
And the actual drawing should have a double bond delocalized across the
molecule, in such a way that we have the double bond delocalized to the two bond
lenghts come out to be the same. And we have a structure that we will
refer to as a resonance hybrid, once again.
Let's try another example, let's look at water.
Water seems to have been one of our simpler examples way back.
Let's think about it for a moment, and consider a couple of possibilities here.
Let's see. If we want to try to draw structures for
water, this is fairly easy, I won't do the calculation on it.
I might have drawn structures that look like, that.
Of course, when I started off with the original oxygen here, I could have drawn
it in such a way that the unshared unpaired electrons were adjacent to each
other rather then apart from each other. The question is, have I now drawn two
different structures, after all? These are clearly not isomers because the
molecular structure is the same. The question is are the astronomic
structure different? Have you moved the electrons around, at
all? So are these resonance structures?
And then answer is no. They're actually identical.
In this particular case all we have done is actually to draw the diagram
differently. We haven't actually changed the molecule
in any way. All we've shown in both of these diagram
is that the oxygen has two single bonds attached to hydrogen atoms, and to lone
pairs approaching, surrounding it, on different sides.
So when we interpret our Lewis Structures, here's what we can get out of
them. The Lewis Structures actually provides us
information on which atoms are bonded to which other atoms.
It tells us whether we're dealing with single, double or triple bonds.
It shows us whether we have lone pairs and what the numbers of lone pairs might
be. And it tells us if there's a presence of
resonance such as there was in benzyne and nitrogen dioxide.
What it does not tell us is the molecular geometry.
The implication here is that these two are geometrically different.
Like perhaps this is a linear molecule and this is a bent molecule.
In reality there's only one structure for water and it's neither linear or bent, at
least at a 90 degree angle. So at least directly, we do not get
information on molecular structures, out of these Lewis Stuctures.
Later on we will use the Lewis Stuctures and an additional theory, to make
predictions about molecular geometries. But the structures themselves do not
provide us that information. That concludes what we're going to talk
about with molecular structures. In the next lecture and concept
development study, we're going to take up the question of what does it actually
mean to share an electron pair.