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PROFESSOR: OK, let's get started here.
Go ahead and take 10 more seconds on the clicker
question, which probably looks all too familiar at this
point, if you went to recitation yesterday.
All right, and let's see how we do here.
OK.
So, let's talk about this for one second.
So what we're asking here, if we can settle down and listen
up, is which equations can be used if we're talking about
converting wavelength to energy for an electron.
Remember, the key word here is electron.
This might look familiar to the first part of problem one
on the exam, and problem one on the exam is what tended to
be the huge problem on the exam.
I think over 2/3 of you decided on the exam to use
this first equation, e equals h c over wavelength.
So I just want to reiterate one more time, why can we not
use this equation if we're talking about an electron?
C. OK, good, good, I'm hearing it.
So the answer is c.
What you need to do is you need to ask yourself if you're
trying to convert from wavelength to energy for an
electron, and you are tempted, because we are all tempted to
use this equation, and if you were tempted, say, does an
electron travel at the speed of light?
And if the answer is no, an electron does not travel at
the speed of light, light travels at the speed of light,
then you want to stay away from using this equation.
And I know how tempting it is to do that, but we have other
equations we can use -- the DeBroglie wavelength, and this
is just a combination of energy equals 1/2 m v squared,
and the definition of momentum, so we can combine
those things to get it.
You might be wondering why I'm telling you this now, you've
already -- if you've lost points on that, lost the
points on it, and what I'm saying to you is if there are
parts of exam 1 that you did not do well on, you will have
a chance to show us again that you now understand that
material on the final.
One quarter of the final is going to be exam 1 material,
and what that means is when we look at your grade at the end
of the semester, and we take a look at what you got on exam
1, and you're right at that borderline, and we say well,
what happened, did they understand more at the end of
the semester, did the concepts kind of
solidify over the semester?
And if they did and if you showed us that they did, then
you're going to get bumped up into that next grade category.
So keep that in mind as you're reviewing the exam, sometimes
if things don't go as well as you want them to, the
temptation is just to put that exam away forever and ever.
But the reality is that new material builds on that
material, and specifically exam 1 a, question 1 a that
deals with converting wavelength to
energy for an electron.
I really want you guys know this and to understand it, so
I can guarantee you that you will see this on the final.
Specifically, question 1, part a.
You will see something very, very similar
to this on the final.
If you are thinking about 1 thing to go back and study on
exam 1, 1 a is a really good choice for that.
This is important to me, so you're going to
see it on the final.
So if you have friends that aren't here, you might want to
mention it to them, or maybe not, maybe this is your reward
for coming to class, which is fine with me as well.
All right.
So I want to talk a little bit about exam 1.
I know most you picked up your examine in recitation.
If you didn't, any extra exams can be picked up in the
Chemistry Education office, that's room 2204.
So, the class average for the exam was a 68%, which is
actually a strong, solid average for an exam 1 grade in
the fall semester of 511-1.
What we typically see is something right in this range,
either ranging from the 50's for an exam 1 average to
occasionally getting into the 70's, but most commonly what
we've seen for exam 1 averages is 60, 61 -- those low 60's.
So in many ways, seeing this 68 here, this is a great sign
that we are off to a good start for this semester.
And I do want to address, because I know many of you,
this is only your second exam at MIT, and perhaps you've
never gotten an exam back that didn't start with a 90 or
start with an 80 in terms of the grades.
So one thing you need to keep in mind is don't just look at
the number grade.
The reason that we give you these letters grade categories
is that you can understand what it actually means, what
your exam score actually says in terms of how we perceive
you as understanding the material.
So, for example, and this is the same categories that were
shared in recitation, so I apologize for repeating, but I
know sometimes when you get an exam back, no more information
comes into your head except obsessing over the exam, so
I'm just going to say it one more time, and that is between
88 and 100, so that's 20% of you got an A. This is just
absolutely fantastic, you really nailed this very hard
material and these hard questions on the exam where
you had to both use equations and solve problems, but also
understand the concept in order to get yourself started
on solving the problem.
The same with the B, the B range was between 69 and 87 --
anywhere in between those ranges, you've got a B, some
sort of B on the exam.
So again, if you're in the A or the B category here, this
is really something to be proud of, you really earned
these grades.
You know these exams, our 511-1 exams, we're not giving
you points here, there are no give me, easy points, you
earned every single one of these points.
So, A and B here, these are refrigerator-worthy grades,
hang those up in your dorm.
This is something to feel good about.
All right.
So, for those of you that got between a 51 and a 68, this is
somewhere in the C range.
For some people, they feel comfortable being in the C
range, other people really do not like being in this range.
We understand that, there is plenty of room up there with
the A's and the B's.
You are welcome to come up to these higher ranges starting
with the next exam.
And what I want to tell you if you are in the C range, and
this is not a place that you want to be in, anyone that's
got below the class average, so below a 68 -- or a 68 or
below, is eligible for free tutoring, and I put the
website on the front page of your notes.
This means you get a one-on-one tutor paid for by
the Chemistry Department to help you if it's concepts
you're not quite up on, if it's exam strategy that you
need to work on more.
Whatever it is that you need to work on, we want to help
you get there.
So, if you have a grade that you're not happy with, that
you're feeling upset or discouraged about, please, I'm
happy to talk to all of you about your grades
individually.
You can come talk to me, bring your exam, and we'll go over
what the strategy should be in terms of you
succeeding on the next exam.
You can do the same thing with all of your TAs are more than
happy to meet with each and every one of you.
And then in addition to that, we can set you up with a tutor
if you are in the C range or below, in terms
of this first exam.
All right.
So 44 to 50, this is going to be in the D range.
And then anything below a 44 is going to be
failing on this exam.
And also keep in mind, for those of you that are
freshman, you need at least a C to pass the class.
So, if you did get a D or an F on the first exam, you are
going to need to really evaluate why that happened and
make some changes, and we're absolutely here
to help you do that.
So the real key is identifying where the problem is -- is it
with understanding the concepts, are you in a study
group that's dragging you along but you're not
understanding?
Do you kind of panic when you get in the exam?
There are all sorts of scenarios we can talk about
and we want to talk about them with you.
Seriously, even if we had a huge range in this exam from
17 to 100, if you're sitting there and you're the 17, and
actually there's more than 1 so don't feel alone, if you're
a 17 or you're a 20, it's not time to give up, it's not time
to drop the class and say I'm no good at
chemistry, I can't do this.
You still can, this is your first couple of exams,
certainly your first in this class, potentially one of your
first at MIT, so there's tons of room to improve
from here on out.
This is only 100 points out of 750.
So, the same thing goes if you did really well, you still
have 650 other points that you need to deal with.
So, make sure you don't just rest on your high score from
this first exam.
So, OK, so that's pretty much what I wanted to say about the
exam, and in terms of there's tons of resources if things
didn't work out quite as you wanted.
If you feel upset in any way, please come and talk to me.
We want you to love chemistry and feel good about your
ability to do it.
Nobody get into MIT by mistake, so you all deserve to
be sitting here, and you all can pass this class and do
well in it, so we can help you get there no matter what.
You all absolutely can do this.
And then one more time, to reiterate, in case anyone
missed it, 1 a, make sure you understand that, I feel like
that's important.
And actually all of 1 --
I really feel like the photoelectric effect is
important for understanding all of these energy concepts.
So, as you go on in this class, make sure you don't go
on before you go back and make sure you
understand that problem.
All right, so let's move on to material for exam 2 now, and
we're already three lectures into exam 2 material.
And I do want to say that in terms of 511-1, what tends to
happen is the exam scores go up and up and up, in terms of
as we go from exam 1, to exam 2, to exam 3.
One of these reasons is we are building on material, the
other reason is you'll be shocked at how much better you
are at taking an exam just a few weeks from now.
So this will be on, starting with the Lewis structures, so
go back in your notes -- if this doesn't sound familiar,
if you spent too much time -- or not too much time, spent a
lot of time studying exam 1 and didn't move on here.
Today we're going to talk about the breakdown of the
octet rule.
Cases where we don't have eight electrons around our
Lewis structures, then we'll move on to
talking about ionic bonds.
We had already talked about covalent bonds, and then we
talked about Lewis structures, which describe the electron
configuration in covalent bonds.
So now let's think about the other extreme of ionic bonds,
and then we'll talk about polar covalent bonds to end,
if we get there or will start with that in class on Monday.
Also, public service announcement for all of you,
voter registration in Massachusetts, which is where
we are, is on Monday, the deadline if you want to
register to vote.
There's some websites up there that can guide you through
registering and also can guide you through, if you need an
absentee ballot for your home state.
And I actually saw, and I saw a 5.111 student manning,
there's some booths around MIT that will register you or get
you an absentee ballot.
So, the deadline's coming soon, so patriotic duty, I
need to remind you of that as your chemistry teacher --
chemistry issues are important in politics as well.
So make sure you get registered to vote.
I just remembered one more announcement, too, that I did
want to mention, some of you may have friends in 511-2 and
have heard their class average for exam 1.
And I want to tell you, this happens every year, their
average was 15 points higher than our average.
Last year, their average was 15 points
higher than our average.
This is for exam 1.
This is what tends to happen to 511-2 grades as
the exam goes on.
This is what happens to 511-1.
You guys are in a good spot.
Also, I want to point out that what's not important is just
that number grade, but also the letter that goes with it.
So, for example, if you got a 69 in this class on this exam,
that's a B minus.
If you got a 69 on your exam in 511-2, that's a D, you
didn't pass the exam.
So keep that in mind when your friend might have gotten a
higher number grade than you and you know you understand
the similar material just as well.
Similarly, an 80 in this class on the exam was a B plus, a
very high B. An 80 in that class is going to be a C. So,
just don't worry so much about exactly where that average
lies, you really want to think about what the
letter grade means.
OK, I've said enough.
I just --
I hate to see people discouraged, and I know that a
few people have been feeling discouraged, so that's my
long-winded explanation of exam 1 grades.
All right.
So, let's move on with life though, so talking about the
breakdown of the octet rule.
The first example where we're going to see a breakdown is
any time we have an odd number of valence electrons.
This is probably the easiest to explain and to think about,
because if we have an odd number that means that we
can't have our octet rule, because our octet rule works
by pairing electrons.
And if we have an odd number, we automatically have an odd
electron out.
So, if we look at an example, the methyl radical, we can
first think about how we draw the Lewis structure -- we draw
the skeletal structure here.
And then what we're going to do is add up our valence
electrons -- we have 3 times 1 for the hydrogen atoms, carbon
has 4 valence electrons, so we have a total of 7.
If we want to fill all of our valence shells in each of
these atoms, we're going to need a total of 14 electrons.
So, what we see we're left with is that we
have 7 bonding electrons.
So we can fill in 6 of those straightforward here, because
we know that we need to make 3 different bonds.
And now we're left over with 1 electron, we
can't make a bond.
So, what we'll do is carbon does not have an octet yet.
We can't get it one, but we can do the best we can and
help it out with adding that extra electron onto the carbon
atom, so that at least we're getting as close as possible
to filling our octets.
This is what we call a radical species or a free radical.
Free radical or radical species is essentially any
type of a molecule that has this unpaired electron on one
of the atoms. This might look really strange, we're used to
seeing octets.
But you'll realize, if you calculate the formal charge on
this molecule, that it's not the worst
situation ever for carbon.
At least it's formal charge is zero, even if it doesn't have
-- it would rather have an extra bond
and have a full octet.
But it's not the worst scenario that we can imagine.
But still, radicals tend to be incredibly reactive because
they do want to fill that octet.
So, what happens when you have a radical is it tends to react
with the first thing that it runs into, especially highly
reactive radicals that are not stabilized in some other way,
which you'll tend to talk about it organic chemistry --
how you can stabilize radicals.
So the term free radical should sound familiar to you,
whether you've heard it in chemistry before, or you
haven't heard it in chemistry, but maybe have heard it, I
don't know, commercials for facial
products or other things.
People like to talk about free radicals, and they're sort of
the hero that gets rid of free radicals, which are
antioxidants.
So you hear in a lot of different creams or products
or vitamins that they have antioxidants in them, which
get rid of free radicals.
The reason you would want to get rid of free radicals is
that free radicals can damage DNA, so
they're incredibly reactive.
It makes sense that if they hit a strand of DNA, they're
going to react with the DNA, you end up breaking the
strands of DNA and causing DNA damage.
So, this is actually what happens in aging because we
have a lot of free radicals in our body.
We can introduce them artificially, for example,
cigarette smoke has a lot of really dangerous free radicals
that get into the cells in your lungs, which damage your
lung DNA, which can cause lung cancer.
But also, all of us are living and breathing, which means
we're having metabolism go on in our body, which means that
as we use oxygen and as we metabolize our food, we are
actually producing free radicals as well.
So it's kind of a paradox because we need them because
they are a natural by-product of these important processes,
but then they can go on and damage cells, which is what
kind of is causing aging and can lead to cancer.
We have enzymes in our body that repair damage that is
done by free radicals, that will put the strands of DNA
back together.
And we also have antioxidants in our body.
So, you might know that, for example, very brightly colored
fruit is full of antioxidants, they're full of chemicals that
will neutralize free radicals.
Lots of vitamins are also antioxidants, so we have
vitamin A on the top there and vitamin E.
So, the most common thing we think of when we think of free
radicals is very reactive, bad for your
body, causes DNA damage.
But the reality is that free radicals are also
essential for life.
So this is kind of interesting to think about.
And, for example, certain enzymes or proteins actually
use free radicals in order to carry out the reactions that
they carry out in your body.
So, for example, this is a picture or a snapshot of a
protein, this is a crystal structure of ribonucleotide
reductase is what it's called.
It's an enzyme that catalyzes the reaction of an essential
step in both DNA synthesis and also DNA repair, and it
requires having radicals within its active site in
order to carry out the chemistry.
So, this is kind of a neat paradox, because radicals
damage DNA, but in order to repair your DNA, you need
certain enzymes, and those enzymes require different
types of free radicals.
So, free radicals are definitely very interesting,
and once we get -- or hopefully you will get into
organic chemistry at some point and get to really think
about what they do in terms of a radical mechanism.
We can think about radicals that are also more stable, so
let's do another example with the molecule nitric acid.
So we can again, draw the skeleton here, and just by
looking at it we might not know it's a radical, but as we
start to count valence electrons, we should be able
to figure it out very quickly, because what we have is 11
valence electrons.
We need 16 electrons to have full octets.
So, we're left with 5 bonding electrons.
We put a double bond in between our nitrogen and our
oxygen, so what we're left over with is this single
bonding electron, and we'll put that on the nitrogen here.
And I'll explain why we put it on the nitrogen and not the
oxygen in just a minute.
But what we find is then once we fill in the rest of the
valence electrons in terms of lone pairs, this is the
structure that we get.
And if you add up all of the formal charges on the nitrogen
and on the oxygen, what you'll see is they're both 0.
So if you happen to try drawing this structure and you
put the lone pair on oxygen and then you figured out the
formal charge and saw that you had a split charge, a plus 1
and a minus 1, the first thing you might want to try is
putting it on the other atom, and once you did that you'd
see that you had a better structure
with no formal charge.
I have to mention what nitric oxide does, because it's a
very interesting molecule.
Don't get it confused with nitrous oxide, which is happy
gas, that's n o 2.
This is nitric oxide, and it's actually much more interesting
than nitrous oxide.
It's a signaling molecule in your body, it's one of the
very few signaling molecules that is a gas, and obviously,
it's also a radical.
What happens with n o is that it's produced in the
endothelium of your blood vessels, so the inner lining
of your blood vessels, and it signals for smooth muscle that
line your blood vessels to relax, which causes
vasodilation , and by vasodilation, I just mean a
widening of the blood vessels.
So, n o signals for your blood vessels to get wider and allow
more blood to flow through.
And if you think about what consequences this could have,
in terms of places where they have high altitude, so they
have lower oxygen levels, do you think that they produce
more or less and n o their body?
More?
Yeah, it turns out they do produce more.
The reason they produce more is that they want to have more
blood flowing through their veins so that they can get
more oxygenated blood into different parts of their body.
N o is also a target in the pharmaceutical industry.
A very famous one that became famous I guess over 10 years
ago now, and this is from a drug that actually targets one
of n o's receptors, and this drug has the net effect of
vasodilation or widening of blood vessels in a certain
area in the body.
So this is viagra, some of you may be familiar, I think
everyone's heard of viagra.
Now you know how viagra works.
Viagra breaks down, or it inhibits the breakdown of n
o's binding partner in just certain areas, not everywhere
in your body.
So, in those areas, what happens is you get more n o
signaling, you get more vasodilation, you get
increased blood flow.
So that's a little bit of pharmacology
for you here today.
All right, so let's talk about one more example in terms of
the breakdown of the octet rule with radicals.
Let's think about molecular oxygen.
So let's go ahead and quickly draw this Lewis structure.
We have o 2.
The second thing we need to do is
figure out valence electrons.
6 plus 6, so we would expect to see 12.
For a complete octet we would need 8 electrons each, so 16.
So in terms of bonding electrons, what we have is 4
bonding electrons.
So, we can go ahead and fill those in as a double bond
between the two oxygens.
So, what we end up having left, and this would be step
six then because five was just filling in that, is 12 minus
4, so we have 8 lone pair electrons left.
So we can just fill it in to our oxygens like this.
All right, so using everything we've learned about Lewis
structures, we here have the structure of molecular oxygen.
And I just want to point out for anyone that gets confused,
when we talk about oxygen as an atom, that's o, but
molecular oxygen is actually o 2, the same for molecular
hydrogen, for example.
All right, so let's look at what the actual Lewis
structure is for molecular oxygen, and it turns out that
actually we don't have a double bond, we have a single
bond, and we have two radicals.
And any time we have two radicals, we talk about what's
called a biradical.
And while using this exception to the Lewis structure rule,
to the octet rule for odd numbers of valence electrons
can clue us into the fact that we have a radical, there's
really no way for us to use Lewis structures to predict
when we have a biradical, right, because we would just
predict that we would get this Lewis structure here.
So, when I first introduced Lewis structures, I said these
are great, they're really easy to use and they work about 90%
of the time.
This falls into that 10% that Lewis structures
don't work for us.
It turns out, in order to understand that this is the
electron configuration for o 2, we need to use something
called molecular orbital theory, and just wait till
next Wednesday and we will tell you what that is, and we
will, in fact, use it for oxygen.
But until that point, I'll just tell you that molecular
orbital theory takes into account quantum mechanics,
which Lewis theory does not.
So that's why, in fact, there are those 10% of cases that
Lewis structures don't work for.
All right, the second case of exceptions to the octet rule
are when we have octet deficient molecules.
So basically, this means we're going to have a molecule
that's stable, even though it doesn't have a complete octet.
And these tend to happen in group 13 molecules, and
actually happen almost exclusively in group 13
molecules, specifically with boron and aluminum.
So, any time you see a Lewis structure with boron or
aluminum, you want to just remember that I should look
out to make sure that these might have an incomplete
octet, so look out for that when you see those atoms.
So, let's look at b f 3 as our example here.
And what we see for b f 3 is the number of valence
electrons that we have are 24, because the valence number of
electrons for boron is 3, and then 3
times 7 for each fluorine.
For total filled octets we need 32, so that means we need
8 bonding electrons.
So, let's assign two to each bond here, and then we're
going to have two extra bonding electrons, so let's
just arbitrarily pick a fluorine to give
a double bond to.
And then we can fill in the lone pair electrons, we have
16 left over.
So thinking about what the formal charge is, if we want
to figure out the formal charge for the boron here,
what we're talking about is the valence number for boron,
which is 3, minus 0 because there are no lone pairs, minus
1/2 of 8 because there are eight shared electrons.
We get a formal charge of minus 1.
What is our formal charge since we learned this on
Monday for thinking about the double
bonded fluorine in boron?
So, look at your notes and look at the fluorine that has
a double bond with it, and I want you to go ahead and tell
me what that formal charge should be.
All right, let's take 10 more seconds on that.
OK, so 49%.
So, let's go look back at the notes, we'll talk about why
about 50% of you are right, and 50% need to review, which
I totally understand you haven't had time to do yet,
your formal charge rules from Monday's class, there were
other things going on.
But let's talk about how we figure out formal charge.
Formal charge is just the number of valence electrons
you have. So fluorine has 7.
You should be able to look at a periodic table and see that
fluorine has seven.
What we subtract from that is the number of lone pair
electrons, and there are four lone pair electrons on this
double bonded fluorine, so it's minus 4.
Then we subtract 1/2 of the shared electrons.
Well we have a double bond with boron here, so we have a
total of 4 shared electrons.
And when we do the subtraction here, what we end up with is a
formal charge plus 1 on the double bonded fluorine.
Without even doing a calculation, what do you think
that the formal charge should be on you
single bonded fluorines?
Good.
OK, it should be 0 and it is 0.
The reason it's zero in terms of calculating it is 7 minus 6
lone pair electrons minus 1/2 half of 2 shared
electrons is 0.
The reason that you all told me, I think, and I hope, is
that you know that the formal charge on individual atoms has
to equal the total charge on the molecule.
So if we already have a minus 1 and a plus 1, and we know we
have no charge in the molecule, and we only have one
type of atom left to talk about, that formal charge had
better be 0.
OK.
So this looks pretty good in terms of a Lewis structure, we
figured out our formal charges.
These also look pretty good, too, we don't have too much
charge separation.
But what actually it turns out is that if you experimentally
look at what type of bonds you have, it turns out that all
three of the b f bonds are equal in length, and they all
have a length that would correspond to a single bond.
So, experimentally, we know we have to throw out this Lewis
structure here, we have some more information, let's think
about how this could happen.
So this could happen, for example, is if we take this
two of the electrons that are in the b f double bond and we
put it right on to the fluorine here, so now we have
all single bonds.
And let's think about what the formal charge situation would
be in this case here.
What happens here is now we would have a formal charge of
0 on the boron, we'd have a formal charge of 0 on all of
the fluorine molecules as well.
So, it turns out that actually looking at formal charge, even
though the first case didn't look too bad, this case
actually looks a lot better.
We have absolutely no formal charge separation whatsoever.
It turns out again, boron and aluminum, those are the two
that you want to look out for.
They can be perfectly happy without a full octet, they're
perfectly happy with 6 instead of 8 in terms of electrons in
their valence shell.
So that is our exception the number two.
We have one more exception and this is a valence shell
expansion, and this can be the hardest to look out for,
students tend to forget to look for this one, but it's
very important as well, because there are a lot of
structures that are affected for this .
And this is only applicable if we're talking about a central
atom that has an n value or a principle quantum number
that's equal to or greater than three.
What happens when we have n that's equal to or greater to
three, is that now, in addition to s orbitals and p
orbitals, what else do we have available to us?
D orbitals, great.
So what we see is we have some empty d orbitals, which means
that we can have more than eight electrons that fit
around that central atom.
If you're looking to see if this is going to happen, do
you think this would happen with a large or
small central atom?
So think of it in terms of just fitting.
We've got to fit more than 8 electrons around here.
Yeah, so it's going to be, we need to have a large central
atom in order for this to take place.
Literally, we just need to fit everything around is probably
the easiest way to think about it.
And what happens is it also tends to have small atoms that
it's bonded to.
Again, just think of it in terms of all fitting in there.
So, let's take an example p c l 5.
The first example is the more straightforward example,
because let's start to draw the Lewis structure, and what
we see is that phosphorous has five chlorines around it.
So we already know if we want to form five bonds we've
broken our octet rule.
But let's go through and figure this out and see how
that happens.
What we know is we need 40 valence electrons,
we have those --
5 from the phosphorous, and we have 7 from each of the
chlorine atoms. If we were to fill out all of those octets,
that would be 48 electrons.
So what we end up with when we do our Lewis structure
calculation is that we only have 8 bonding electrons
available to us.
So we can fill those in between the phosphorous and
the chlorine, those 8 bonding electrons.
So, this is obviously a problem.
To make 5 p c l bonds we need 10 shared electrons, and we
know that that's the situation because it's called p c l 5
and not p c l 4, so we can go right ahead and add in that
extra electron pair.
So we've used up 10 for bonding, so that means what we
have left is 30 lone pair electrons, and I would not
recommend filling all of these in your notes right now, you
can go back and do that, but just know the rest end up
filling up the octets for all of the chlorines.
So, in this first case where you actually need to make more
than for bonds, you will immediately know you need to
use this exception to the Lewis structure octet rule,
but sometimes it won't be as obvious.
So, let's look at c r o 4, the 2 minus version here, so a
chromate ion, and if we draw the skeletal structure, we
have four things that the chromate needs to bond to.
So, let's do the Lewis structure again.
When we figure out the valence electrons, we have total, we
have 6 from the chromium, we have 6 from each of the
different oxygens, and where did this 2 come from?
Yup, the negative charge.
So, remember, we have 2 extra electrons hanging out in our
molecule, so we need to include those.
We have a total of 32.
40 are needed to fill up octets.
So again, we have 8 bonding electrons available, so we can
go ahead and fill these in between each of the bonds.
What happens is that we then have 24 lone pair electrons
left, and we can fill those in like this.
And the problem comes now when we figure
out the formal charge.
So, when we do that what we find is that the chromium has
a formal charge of plus 1, and that each of the oxygens has a
total charge of minus 1.
So we actually have a bit of charge separation here.
Without even doing a calculation, what is the total
charge of these that are added up?
OK, it's minus 2, that's right.
We know that the total charge of each of the formal charges
has to add up to minus 2, because that's the charge in
our molecule.
We can also just calculate it -- the chromate gives us a
plus 2, then we have 4 times minus 1 for each of the
oxygens, so we have a minus 2.
So, we have some charge separation here, and in some
cases, if we're not at n equals 3 or higher, there's
really nothing we can do about it, this would be the best
structure we can do.
But since we have these d orbitals available, we can use
them, and it turns out that experimentally this is what's
found, that the length and the strength are not single bonds,
but they're actually something between a single bond and a
double bond.
So how do we get a 1 and 1/2 bond, for example, what's the
term that let's us do that?
Resonance.
That's right.
So that's exactly what's happening here.
So, if we went ahead and drew this structure here where we
have now two double bonds and two single bonds, that would
be in resonance with another structure where we have two
double bonds instead to these two oxygens, and now, single
bonds to these two oxygens.
We can actually also have several other resonance
structures as well.
Remember, the definition of a resonance structure is where
all the atoms stay the same, but what we can do is move
around the electrons -- we're moving around those extra two
electrons that can be in double bonds.
So, why don't you tell me how many other resonance
structures you would expect to see for this chromate ion?
All right, let's take 10 more seconds on this.
All right.
This is good.
I know this is a real split response, but the right answer
is the one that is indicated in the graph
here that it's four.
This takes a little bit of time to get used to thinking
about all the different Lewis structures you can have. So,
you guys should all go back home if you can't see it
immediately right now and try drawing out those four other
Lewis structures, for chromate,
there are four others.
You'll probably get a chance to literally do this example
in recitation where you draw out all four, but it's even
better to make sure you understand it before you get
to that point.
So, we can go back to the class notes.
So it turns out there's four other Lewis structures, so
basically just think about all the other different
combinations where you can have single and double bonds,
and when you draw those out, you end up with four.
So, for every single one of these Lewis structures, we
could figure out what the formal charges are, and what
we would find is that it's 0 on the chromium, it's 0 for
the double bonded oxygens, and it's going to be negative 1
for the single bonded oxygens.
So, what you can see is that in this situation, we end up
having less formal charge separation, and that's what
we're looking for, that's the more stable structure.
So any time you can have an expanded octet -- an expanded
valence shell, where you have n is equal to or greater than
3, and by expanding and adding more electrons into that
valence shell, you lower the charge separation,
you want to do that.
I also want to point out, I basically said there's 6
different ways we can draw this in terms of drawing all
the resonance structures.
You might be wondering if you have to figure out the formal
charge for each structure individually, and the answer
is no, you can pick any single structure and the formal
charges will work out the same.
So, for example, if you pick this structure and your friend
picks this structure, you'll both get the right answer that
there's just the negative 1 on the oxygens and no other
formal charges in the molecule.
All right.
So those are the end of our exceptions to the octet rule
for Lewis structures, that's everything we're going to say
about Lewis structures.
And remember, that when we talk about Lewis structures,
what they tell us is the electron configuration in
covalent bonds, so that valence shell electron
configuration.
So we talked a lot about covalent bonds before we got
into Lewis structures, and then how to represent covalent
bonds by Lewis structures.
So now I'll say a little bit about ionic bonds, which are
the other extreme, and when you have an ionic bond, what
you have now is a complete transfer of either one or many
electrons between two atoms. So the key word for covalent
bond was electron sharing, the key word for ionic bonds is
electron transfer.
And the bonding between the two atoms ends up resulting
from an attraction that we're very familiar with, which is
the Coulomb or the electrostatic attraction
between the negatively charged and the
positively charged ions.
So let's take an example.
The easiest one to think about is where we have a negative 1
and a positive 1 ion.
So this is salt, n a c l -- actually lots of things are
call salt, but this is what we think of a table salt.
So, let's think about what we have to do if we want the form
sodium chloride from the neutral sodium and chlorine
atoms. So, the first thing that we're going to need to do
is we need to convert sodium into sodium plus.
What does this process look like to you?
Is this one of those periodic trends, perhaps?
Can anyone name what we're looking at here?
Exactly, ionization energy.
So, if we're going to talk about the energy difference
here, what we're going to be talking about is the
ionization energy, or the energy it takes to rip off an
electron from sodium in order to form the sodium plus ion.
So, we can just put right here, that's 494
kilojoules per mole.
The next thing that we want to look at is chlorine, so in
terms of chlorine we need to go to chlorine minus, so we
actually need to add an electron.
This is actually the reverse of one of the periodic trends
we talked about.
Which trend is that this is the reverse of?
Electron affinity, right.
Because if we go backwards we're saying how badly does
chlorine want to grab an electron?
Chlorine wants to do this very badly, and it turns out the
electron affinity for chlorine is huge, it's 349 kilojoules
per mole, but remember, we're going in reverse, so we need
to talk about it as negative 349 kilojoules per mole.
So if we talk about the sum of what's happening here, what we
need to do is think about going from the neutrals to the
ions, so we can just add those two energies together, and
what we end up with is plus 145 kilojoules per mole, in
order to go from neutral sodium in
chlorine to the ions.
So, the problem here is that we have to actually put energy
into our system, so this doesn't seem favorable, right.
What's favorable is when we actually get energy out and
our energy gets lower, but what we're saying here is that
we actually need to put in energy.
So another way to say this is this process
actually requires energy.
It does not emit energy, it does not give off excess
energy, it requires energy.
So, we need to think about how can we solve this problem in
terms of thinking about ionic bonds, and the answer is
Coulomb attraction.
So there's one more force that we need to talk about, and
that is when we talk about the attraction between the
negatively and the positively charged ions, such that we
form sodium chloride.
So this process here has a delta energy, a change in
energy of negative 589 kilojoules per mole.
So that's huge, we're giving off a lot of energy by this
attraction.
So if we add up the net energy for all of this process, all
we need to do is add negative 589 to plus 145.
So what we end up getting is the net energy change is going
to be negative 444 kilojoules per mole, so you can see that,
in fact, it is very favorable for neutral sodium and neutral
chloride to form sodium chloride in an ionic bond.
And the net increase then, is a decrease in energy.
So, I just gave you the number in terms of what that Coulomb
potential would be in attraction, but we can I
easily calculate it as well using this equation here where
the energy is equal to the charge on each of the ions,
and this is just multiplied by the value of charge for an
electron divided by 4 pi epsilon nought times r, are r
is just the distance in terms of the bond length we could
talk about.
So, let's calculate and make sure that I didn't tell you a
false number here.
Let's say we do the calculation with the bond
length that we've looked up, which is 2 .
3 6 angstroms for the bond length
between sodium and chloride.
So we should be able to figure out the Coulombic
attraction for this.
So, if we talk about the energy of attraction, we need
to multiply plus 1, that's the charge on the sodium, times
minus 1, the charge on the chlorine, times the charge in
an electron, 1 .
6 0 2 times 10 the negative 19 Coulombs, and that's all
divided by 4 pi, and then I've written out epsilon nought in
your notes, so I won't write it on the board.
And then r, so r is going to be 2 .
3 6 and times -- what is angstrom, everyone?
Yup, 10 to the negative 10.
So 10 to the negative 10 meters.
So, if we do this calculation here, what we end up with is
negative 9.774 times 10 to the negative 19 joules.
So that's what we have in terms of our energy.
That does not look the same as what we saw -- yup, do you
have a question?
STUDENT: [INAUDIBLE]
PROFESSOR: OK.
Luckily, although, I did not write it in my own notes, I
did it when I put in my calculator, thank you.
So you need to square this value here and then you should
get this value right here, negative 9.77.
All right, so what we need to do though is convert from
joules into kilojoules per mole, because that's what we
were using.
So if we multiply that number there by kilojoules per mole
-- or excuse me, first kilojoules per joule, so we
have 1,000 joules in every kilojoule.
And then we multiply that by Avagadro's number, 6.022 times
10 to the 23 per mole.
What we end up with is negative 589
kilojoules per mole.
So this is that same Coulombic attraction that we saw in the
first place.
So, notice that you will naturally get out a negative
charge here, remember negative means an attractive force in
this case, because you have the plus and
the minus 1 in here.
So we should be able to easily do that calculation, and what
we end up getting matches up with what I just told you,
luckily, and thank you for catching the square, that's an
important part in getting the right answer.
So, experimentally then, what we find is that the change in
energy for this reaction is negative 444
kilojoules per mole.
If we look experimentally what we see, it's actually a little
bit different, it's negative 411 kilojoules per mole.
So, in terms of this class, this is the method that we're
going to use, and we're going to say this gets us close
enough such that we can make comparisons and have a
meaningful conversations about different types of ionic bonds
and the attraction between them.
But let's think about where this discrepancy comes from,
and before I do that I want to point out, one term we use a
lot is change in energy for a reaction where, for example,
you break a bond.
Remember that the negative of the change in energy is what's
called delta e sub d.
We first saw this when we first introduced the idea of
covalent bonds.
Do you remember what this term here means, delta e sub d?
A little bit and some no's, which this was pre-exam, I
understand, you still need to review those notes, it's
dissociation energy.
So you get a negative energy out by breaking the bond.
The dissociation energy means how much energy that bond is
worth in terms of strength, so it's the opposite of the
energy you get out of breaking the bond -- or excuse me, the
energy that you get out of forming the bond.
It's the amount of energy you need to put in to break the
bond is dissociation energy.
It takes this much energy to dissociate your
bond, excuse me.
All right.
So, let's take a look here at our predictions, so I just put
them both ways so we don't get confused.
The dissociation energy is 444.
The change in energy for forming the
bond is negative 444.
We made the following approximations, which explain
why, in fact, we got a different experimental energy,
if we look at that.
The first thing is that we ignored any repulsive
interactions.
If you think about salt, it's not just two single atoms that
you are talking about.
It's actually in a whole network or whole lattice of
other molecules, so you actually have some other
chlorines around that are going to be having repulsive
interactions with our chlorine that we're talking about.
We're going to ignore those, make the approximation that
those don't matter, at this point, in these calculations.
And the result for that is that we end up with a larger
dissociation energy than the experimental value.
That's because the bond is going to be a little bit more
broken than it was in our calculation, because we do
have these repulsive interactions.
The other thing that we did is that we treated both sodium
and the chlorine as point charges.
And this is what actually allowed us to make this
calculation and calculate the Coulomb potential so easily,
we just treated them as if they're point charges.
We're ignoring quantum mechanics in this -- this is
sort of the class where we ignore quantum mechanics, we
ignored it for Lewis structures,
we're ignoring it here.
We will be back to paying a lot of attention to quantum
mechanics in lecture 14 when we talk about MO theory, but
for now, these are approximations, these are
models where we don't take it into consideration.
And I think you'll agree that we come reasonably close such
that we'll be able to make comparisons between different
kinds of ionic bonds.
All right.
So, the last thing I want to introduce today is talking
about polar covalent bonds.
We've now covered the two extremes.
One extreme is complete total electron sharing -- if we have
a perfectly covalent bond, we have perfect sharing.
The other is electron transfer in terms of ionic bonds.
So when we talk about a polar covalent bond, what we're now
talking about is an unequal sharing of electrons between
two atoms.
So, this is essentially something we've seen before,
we just never formally talked about what we would call it.
This is any time you have a bond forming between two
non-metals that have different electronegativities, so, for
example, hydrogen choride, h c l.
The electronegativity for hydrogen is 2.2, for
chlorine it's 3.2.
And in general, what we say is we consider a difference in
terms of a first approximation if the difference in
electronegativity is more than 0.
5, so this is on the Pauling electronegativity scale.
So what we end up having is we sort of have a kind of, and
what we call it is a partial negative charge on the
chlorine, and a partial positive
charge in the hydrogen.
The reason we have that is because the chlorine's more
electronegative, it wants to pull more of that shared
electron density to itself.
If it has more electron density, it's going to have a
little bit of a negative charge and the hydrogen's
going to be left with a little bit of a positive charge.
So, we can compare this, for example to, molecular hydrogen
where they're going to have that complete sharing, so
there's not going to be a delta plus or a delta minus,
delta is going to be equal to zero on each of the atoms.
They are completely sharing their electrons.
And we can also explain this in another way by talking
about a dipole moment where we have a charged distribution
that results in this dipole, this electric dipole.
And we talk about this using the term mu, which is a
measurement of what the dipole is.
A dipole is always written in terms of writing an arrow from
the positive charge to the negative charge.
In chemistry, we are always incredibly interested in what
the electrons are doing, so we tend to pay attention to them
in terms of arrows.
Oh, the electrons are going over to the chlorine, so we're
going to draw our arrow toward the chlorine atom.
So, we measure this here, so mu is equal to q times r, the
distance between the two.
And q, that charge is just equal to the partial negative
or the partial positive times the charge on the electron.
So this is measured in Coulomb meters, you won't ever see a
measurement of electronegativity in Coulomb
meters -- we tend to talk about it in terms of debye or
1 d, or sometimes there's no units at all, so the d is just
assumed, and it's because 1 debye is just equal to this
very tiny number of Coulomb meters and it's a lot easier
to work with debye's here.
So, when we talk about polar molecules, we can actually
extend our idea of talking about polar bonds to talking
about polar molecules.
So, actually let's start with that on Monday.
So everyone have a great weekend.