SPEAKER: The following content is provided under a Creative
Commons License.
Your support will help MIT OpenCourseWare continue to
offer high-quality educational resources for free.
To make a donation or view additional materials from
hundreds of MIT courses visit MIT OpenCourseWare at
ocw.mit.edu.
PROFESSOR: OK.
OK.
Settle down.
So the weekend doesn't begin until after the 3.091 lecture.
OK.
A couple of announcements.
Tuesday, Quiz 3 based on Homework 3.
The final exam has been scheduled.
In fact, all the finals are scheduled now, so please
consult the Registrar's listing.
The celebration of celebrations, the 3.091 final
exam, will be Tuesday, 15th of December in the morning, 9:00
am to 12:00 noon over in the Johnson Athletic Center.
So I urge you to go through the final exam schedule and
then make your travel plans when you know what your last
obligation is.
And book them soon, because we've got about a quarter of a
million students in the Boston area and the semesters all
come to a close within a very narrow time window, and
everybody's trying to get through a wormhole called
Logan Airport.
So you want to be ready.
Do not try to leave town before you've met all of your
obligations.
You can't leave before you've finished all of your finals.
But you know what they are now, so call your travel agent
or make your internet booking.
And an announcement for Professor Paul's section.
Owing to the holiday on Monday, he's going to have
office hours from noon to 1:30 today.
So if you're in Professor Paul's section, you'd like to
catch up with him, noon till 1:30 today.
All right.
Let's get to the lesson.
The last day we started looking at octet stability,
and we looked at octet stability and what it means in
terms of shell filling and some sweet spot in energy with
respect to reactivity.
And a filled shell leaves us with this electron
configuration, ns2np6 for n greater than 1.
In the case of n equals 1, there's just ns2 for helium.
Otherwise, ns2np6.
2 plus 6 is 8.
There's the octet stability.
It's observed trivially in noble gases.
And then we saw that it can also be
observed in certain ions.
And how do we get to octet stability?
Electron transfer.
And here's the prototypical reaction.
I like this.
This is very iconic.
It puts it in broad terms. The marriage of an electron donor
with an electron acceptor leads to the formation of a
cation and an anion, thanks to electron transfer from the
donor to the acceptor.
It doesn't end there, though.
You have cations and anions in the gas phase and they're
attracted to one another by coulombic forces, or
electrostatic forces, and that leads to ionic bonding.
And today I want to go deeper into it and study the
energetics of ion-pair formation.
So let's do it.
And what I'm going to do is study this with reference to
an example of sodium chloride.
So we will go through the energetics of sodium chloride,
which we know is going to exist as
Na plus and Cl minus--
sodium being the electron donor and chlorine being the
electron acceptor.
So what I'm going to do is show you a plot of energy as a
function of separation.
So you're going to have to be pluralistic
here in your ideas.
So, for example, if I wrote this word by itself you don't
know if I'm saying lead or I'm talking about the metal lead.
The only way you know is in context.
And so you have to know in context here.
So r is not the radius of the atom.
In this case it's the symbol.
And I'm using this symbol not to confuse you.
This is what professionals use.
So if you go to the text and the literature you'll see r.
This is the interionic separation, and it's
determined nucleus to nucleus.
So I'm going to put a sodium ion at the origin.
This is the origin.
Energy is on the ordinate.
So I'm going to model this as a hard sphere.
So this is sodium ion, Na plus, and it's
got some finite radius.
And its radius, I'm going to call it r-plus.
That's the radius of the cation.
And next to sodium I'm going to put the chloride anion.
And it's bigger.
It's bigger.
It's not to scale, but it's bigger.
So this is chloride, the Cl minus, and it has
its radius, rCL minus.
And then the interionic separation is measured from
the sodium nucleus to the chloride nucleus, and it's
given the symbol r0.
That's the interionic separation.
And we could we write that r0, in fact, is strictly r-plus
plus r-minus because we're using a hard sphere model.
So there's no shrinkage when the two touch.
So that makes sense.
Now what I want to do is calculate the energy here.
The only energy that we have here is electrostatic.
So we're going to start and say, imagine if we had these
two ions separated by infinite distance and we brought them
to a separation of r0.
How much electrostatic energy would be stored there?
So I can write that as E, and I'm going to call it
attractive.
There's a force that attracts these two ions together, and
that's given by Coulomb's Law.
So that's the product q1 q2 over 4 pi
epsilon0r, in general.
I'm making this as a function of r and then we're going to
figure out how to get to r0.
Let's put a little ledge in here.
So q1, I'm going to make that the sodium just
for grins and chuckles.
So q1 is equal to the z, the valence on the sodium, times
the charge, the elementary charge.
So the charge on the sodium is plus 1-- so it's plus 1e--
and q2 is going to equal the charge on the
chloride times e.
And in this case the chloride ion is negative 1, so q2 is
minus e, q1 is plus e.
Now if instead I were doing magnesium oxide, magnesium
would be plus 2, oxide would be minus 2.
So that's where the charge on the ion comes into play.
So we can put that in here and will make this z-plus times e,
z-minus times e.
So there's q1 q2 over 4 pi epsilon0 r.
And in this case, for sodium chloride, this is plus 1 times
minus 1 is minus 1 e squared over 4 pi epsilon0 r.
So this is a function of r.
This e is 1/r.
That's the hyperbole, and I can draw that here.
So it's going to look something like this.
All right?
It's going to be a right like this.
Good.
So that's the attractive force.
Now how do we avoid the ions just blending together?
Why do they stop at r0?
What puts the brakes on the coulombic force?
Ah!
For that we have to look at the fine structure of sodium.
So let's look at the fine structure of sodium.
So sodium is net charge plus 1.
But sodium more properly, if I go to the next level of
structure, is really 11 protons in the nucleus and
it's got electrons around it--
in total, 10 electrons.
So it's net charge is plus 1, all right?
Now if I'm way over here and I look at sodium I just see
something of charge plus 1.
And, in fact, I could model sodium as just a point charge,
a plus 1, and do all the electrostatics and I would be
perfectly accurate in my estimation.
But as I get closer I see there's fine structure.
It's like so many other things in life.
You know, from a distance it looks really good, and you get
up close you go, eww.
[LAUGHTER]
PROFESSOR: I'm not going to mention any names, all right?
But it's Friday.
Be careful.
So as I get up closer I realize that it's plus 1, but
the plus 1 is plus 11 mediated by minus 10.
So when things start getting close together this exterior
of negative charge becomes manifest, palpable.
So let's look at the chlorine.
The chlorine's coming, sees plus 1, plus 1.
It starts getting closer and closer.
And now what happens?
The negative electronic configuration on the outside
of sodium is interacting with the negative electronic
configuration on chlorine, and the result is you have
electron-electron repulsion.
And we've got an equation for that, and that's given by E
repulsion is equal to some constant b over r to the n.
And this n is called the Born exponent.
It's not the quantum number.
So r today means interionic separation,
n means Born exponent.
And you have to determine it by experiment and the value
lies between 6 and 12.
And we have to determine b by experiment.
So let's say this thing's got a-- pick a
number in the middle.
Say it's 8.
So let's plot r to the eighth.
So that's going to hug the abscissa, and it's going to
come in really, really close.
And then somewhere around this distance here, where the
electron-electron repulsion starts to be felt,
this thing takes off.
But 1 over r to the 8 goes way, way up fast. So instead
of being a gentle curve it's more like a
hockey stick shape here.
And so now the net energy is the sum of the negative
attractive energy and the positive repulsive energy.
And if we sum the two what are we going to get?
Well, way out here 1 over r to the 8 is negligible.
And the net value--
so E net, in red--
is essentially equal to E attractive.
And at very, very low values of r, r to the 8 dominates r.
So we have this as the net value.
And the two sum--
and I can't quite do it because of the way this is
drawn, but I think you can see.
These two eventually equilibrate and
go through a minimum.
So I'm going to cheat a little bit here and I'm going to add
these two in such a way as to go through a minimum value of
energy at r equals r0, which is the sum of E attractive and
E repulsive.
So we can sum those two and let's see what we get.
At r equals r0, E net is equal to its minimum value.
So how do you find a minimum in a function?
You take the derivative.
I'm going to put some math to work here.
So we'll take dE by dr and set it equal to 0.
And the value of r at which it equals 0 is termed r0.
OK?
So we'll just go through and take the
derivative of that thing.
And I'm not going to go through all the math but just
show you the set-up here.
What's the other thing that we know?
dE by dr represents what?
That's force.
So I could say that at r equals r0 the net force is 0,
which is what you'd expect.
Because if the net force isn't 0 it's going to either push
the ions farther apart or draw them closer together.
So this is mathematics imitating reality.
What a concept.
Math working for you instead of you working for math.
So what do I have?
I have everything in these equations except
the value of b.
I'm assuming we know the value of n from experiment.
We don't know the value of b.
And so you can solve to get the value of b, and once
you've got that you can put everything together and give
an expression for the energy of the system at r equals r0.
When you plug everything in you get this: z-plus, which is
the net charge on the cation, times z-minus, which is the
net charge on anion, divided by 4 pi epsilon0 r0.
1 minus 1/n, where n is the Born exponent.
And this is valid at r equals r0 only.
If you're not at r equals r0 then you can get the value of
b and then put it into this expression, where E net will
equal E attractive plus E repulsive.
So there it is.
And so this represents--
this is the energy of a single ionic bond, because that's all
the energy that's there.
It's the single ionic bond.
And the second thing that we realize is plus times minus is
net minus, so this means that it's negative quantity,
as it should be.
It has to be a negative quantity.
All right.
So what do we have here?
What we've seen by going through this derivation is the
recognition that the ionic bond is electrostatic
attraction mediated by electronic repulsion.
It's the balance of the two.
And those words sound so good to me that I'm going
to write them down.
Electrostatic attraction mediated--
another lovely word--
mediated by electronic repulsion.
So that's how you get to the final setting here of the
interionic separation.
So what are the characteristics?
What does this lead to in terms of
characteristics of this bond?
Characteristics of the ionic bond.
First of all, it's omnidirectional.
This is a concept based on the fact that the electric field
radiates in all directions uniformly.
So the negative field coming from the chloride ion is
uniform in all directions.
There's no preferred direction.
Omnidirectional.
OK?
E field--
oh, I better not say that.
Electric field, not the energy field, radiates in all
directions uniformly.
And that's going to have consequences.
I'm not just telling you this because we
like cataloguing things.
This isn't a bookkeeping class.
So we're going to come back, use this fact.
And the second thing is that the bond is unsaturated, which
is a chemical way of saying that a given ion can bond to
more than one other ion.
In other types of bonds that's not the case.
A given atom can only bond once and then it's done.
Whereas in this case the ion can bond to a plurality of
other ions.
So ions bond to more than one.
OK?
Plurality of bonds is formed.
They're polygamous, if you like.
So what does that mean?
That means that here is what happens.
We've got the blues as the sodiums, and for any given
sodium it forms bonds without limit until the number of
bonds is stopped by physical limitations--
not because the E field was saturated.
It's unsaturated.
You just can't jam any more chlorides physically around
the sodium.
That's why the sodium is only bonding to the number of
chlorides that it bonds to.
There's no intrinsic limitation.
So what happens when you get to this situation where you
have omnidirectional forces, unsaturated bonds, and ions
that you can model as hard spheres of constant radius?
All the sodiums have the same radius, all the chlorides have
the same radius.
You make a 3-dimensional ordered array.
So you can make an infinite atomic ordered array, which we
use the simple Anglo-Saxon word last day
to describe: crystal.
You form a crystal.
And as a result ionics have to be solid at room temperature,
because if you've got thousands and thousands of
atoms together in one aggregate they're
not going to float.
They're going to settle.
Put another way, the strength of the bond, the amount of
energy in here, is so great that the thermal energy of the
room isn't great enough to disrupt this bond.
It's a combination of unsaturated, omnidirectional,
and high energy.
So we form solids at room temperature.
OK.
Now I want to show the energetics of that one because
this is good.
You know, I promised you I wouldn't do derivations, so
I'm not going in detail on this.
I'm giving you just enough so that I can introduce the
characters here.
You know, how else am I going to
introduce the Born exponent?
Am I'm just going to say, there's this exponent n, the
Born exponent.
We're going to introduce it in context.
So now you know what the energetics are.
So now I want to prove to you energetically along this line
that crystals will form.
So let's imagine--
we're going to do this thought experiment.
We're going to take three ion-pairs of sodium chloride.
So here's three ion-pairs of sodium chloride.
And I want to compare these three ion-pairs.
So this is an ion gas.
And the distance between ion-pairs is great enough that
one pair doesn't affect the other.
The electrostatics are only strong within the pair.
So we'll just label this infinity with quotation
marks around it.
They're very far apart.
When a physicist says they're very far apart, very is code
for infinity.
So this one doesn't interact with this one, which doesn't
interact with this one.
And I want to compare the energetic state of the ion
dispersion to what would happen if I were to put all of
those in a single line.
Plus, minus, plus, minus, plus, minus.
So then this has a certain energy state, it's the energy
of the ion line.
And I want to show you that there's an energy decrease in
collecting all of these and ordering them into a line.
So there's more energy in a line dance
than in ballroom dancing.
That's what we're going to say ultimately.
OK?
So let's compare the energies.
That's all we're going to do.
So what's the energy of this?
Well, we're not going to do 3 versus 3 Let's think big.
It's Friday.
So let's take Avogadro's number of pairs, shall we?
So the energy of the ion dispersion would then equal--
that's the energy of one pair, and they're infinite distance
apart, so there's nothing to be gained by putting them in
the same chamber.
So it's just going to be N Avogadro, if that's the
number, times the energy evaluated at r equals r0.
So we're done.
And we know what that is.
I don't have to rewrite it for you.
OK.
So now what I have to do is get an estimate of the energy
of the line and show you that the line is at a
lower energy state.
Well, let's see.
Jocelyn, take the top one and the middle one, please, but
not the bottom one.
Thanks.
OK.
So now let's look at the energy of a line.
So we're going to do-- here's my line.
I'm going to get the colored chalk.
Green and blue.
You know, some people think being a professor is so cool
because you get to travel, you get to research, and so on.
It's the colored chalk.
[LAUGHTER]
PROFESSOR: It's the colored chalk.
All right.
So there's a sodium.
And on each side of the sodium we'll put a chloride.
All right?
And I'm going to just keep going this way.
Here's a sodium and here's a chloride.
All right.
And now what I'm going to do, I'm going to start here at
this sodium and I'm going to count the energy,
electrostatic energy, that's in this system.
So we're going to count to the left, we're going to count to
the right, and then we're going to multiply it by the
total number of ion-pairs.
And I know that the ones on the end aren't the same, but
the number of ends over N Avogadro, that's peanuts.
The edge effects are negligible because there's
such a giant middle.
That's how you model this stuff.
You don't obsess over the fact that the last ion doesn't see
anything on that side.
You just do it and forget about it, because you know you
wouldn't do it for 5.
This would be a big problem.
But if you have Avogadro's number, who cares?
It's called risk assessment.
All right.
So let's look at the energetics here.
So what we've got is the energy of the ion line.
OK?
So let's start with this central one.
And separated by distance r0 is the chloride, and there's
an attractive energy here.
So that's going to equal minus e squared over
4 pi epsilon0 r0.
Now I'm going to keep going, because the field is
unsaturated and goes in all directions.
So this sodium, that's a distance 2r0 away.
It's got a repulsive force exerted on this sodium.
So let's add that.
So that's going to be plus.
A repulsive force raises the energy of a system.
That's e squared over 4pi epsilon0 times 2r0.
And let's keep going.
So now let's go 3r0 away.
So 3 times r0, that gets me out to the chloride over here.
Let's put 3r0.
Now that will take me from the center of the sodium to the
center of the next chloride.
And that's going to be attractive.
So that'll be minus e squared over 4pi epsilon0 times 3r0
plus, et cetera.
So you see how this goes.
So you go all the way out, you add them all up, and you go
the other way.
And so on and so forth.
So that's how the derivation goes.
You might say, hey wait a minute.
What happened to the Born exponent and the repulsive
energy term?
Well, where's the repulsive energy term going to be felt?
It's called electron-electron repulsion.
Axiomatically, the electrons here are
nowhere near these electrons.
See, you only have to count it for the nearest neighbor.
So we can patch that in at the end.
And we do.
I haven't forgotten.
But I'm not going to spend a whole day trying to derive
this thing.
I'll show you how it starts to evolve.
At some point you end up with something
that looks like this.
e squared over--
in fact, I'll put the minus sign up here.
Minus e squared over 4 pi epsilon0 r0.
And you're going to double it, because you're going to go one
side and the other side, and you're going to get a series
that looks like this: 1 minus 1/2 plus 1/3 minus 1/4 plus
blah, blah, blah.
Yeah.
OK.
So what does this look like?
I've really broken this into two pieces.
So this coefficient out in front here, you should now be
able to repeat this in your sleep: e squared over 4 pi
epsilon0 r.
This is electrostatics, isn't it?
Electrostatics.
This is the consequence of Coulomb's law.
What's this second term here?
What's this all about?
Geometry.
This is dictated by atomic arrangement.
So I could calculate this if I took, instead of a line, what
if I put them in a sheet subject to the constraints of
those sizes and plus 1 and minus 1?
So what would be?
I'd start at the sodium and count how many chlorides?
If I'm on a plane there'd be one, two, three, four.
And then how many sodiums?
Well, they'd be on the backside
of each of the chlorides.
And I'd add them all up in 2-space and I'd end up with
another coefficient here.
All right?
And we compress all of this into a coefficient which we
call the Madelung constant.
And it's a function of the atomic arrangements.
So different crystal structures have different
Madelung constants.
It's named after a German professor, Madelung.
In 1910, he published calculations for the energy of
a system of point charges--
just abstract theoretical paper.
And then about 10 years later another German professor by
the name of Paul Ewald--
he did his PhD for Sommerfeld--
he published a paper in which he actually made the
calculation for ion crystals, and he came
up with this constant.
And to show you the class of the guy, instead of naming the
constant after himself he named it after Madelung.
Now that's class.
So Madelung did the first calculation so he gets named.
So now what we're going to do is we're going to multiply by
the N Avogadro, because I've got N Avogadro of these
things, and we're going to put in the Born exponent
patch and so on.
And here's what the final expression looks
like for the line.
There are a few algebraic tricks that I'm not willing to
do in class because I don't think that's a profitable use
of our time in a chemistry class.
But if you want to try the derivation I have it in full
and we can compare notes.
So once we get the patch in it's going to look like this.
It'll be minus.
There'll be the Madelung constant times
N Avogadro e squared--
and this is already assuming it's plus 1, minus 1.
If this were magnesium oxide there'd be a 4 in here.
4 pi epsilon0 r0 times 1 minus 1/n, where
n is the Born exponent.
So compare this to this one here.
What's the only difference?
The only difference is the Madelung constant, right?
It's the only difference.
So E of the pair dispersion is really equal to E of the line
divided by the Madelung constant.
So what I'm trying to prove to you is that E line is more
negative than E of the dispersion.
So it all hinges on the magnitude of M.
If M is greater than 1 we win.
If M is less than 1 I've just proved to you that water runs
uphill, so that's a bad day.
All right?
So let's calculate the value of M.
And you can go to your algebra books.
And you've got this series natural log-- and engineers
write natural log "ln." I know the mathematicians write
"log." Uh-uh.
Engineers--
uh.
That's 1 plus x.
OK?
Natural log, 1 plus x.
You can expand this as x minus x squared over 2 plus x cubed
over 3, dah, dah, dah, dah.
You know, look that one up.
Set x equal to 1, which is what we've got, right?
Because we've got 1 minus 1/2, 1/3, dah, dah, dah, dah.
Go through it and you'll get the value that M, according to
this, will give you 2 times the natural logarithm of 2,
which is 1.386--
which is greater than 1.
And so we're golden.
That means that the energy of the line is lower, more
negative, than the energy of the dispersion.
So I'm going to do this pictorially.
Let's make an energy level diagram.
And the energy level diagram will look like this.
So up here the energy is 0 and everything is negative.
So if I put this as minus 1 unit.
All right?
These are all negative values increasing in this direction.
So this is the dispersion of ion-pairs.
So all we've done is take a cation and
put it to the anion.
We've just seen that if we do the calculation for the line
it's 1.386 times whatever this is.
So that gives us-- this is for the ion line, which I'm going
to take the liberty of calling a 1-dimensional crystal.
It's a 1-dimensional ordered array.
And what I can do is go to the 3-dimensional array, start at
the lower right-hand corner with that chloride.
Calculate the distance to each of the nearest neighbor
sodiums. Go through the geometry.
The next nearest neighbor chlorides, the next nearest
neighbor sodiums. And you'll build an infinite series and
you'll evaluate it.
And in three dimensions it's even lower.
It's 1.7476.
So this is for the 3-dimensional crystal.
This is for the ionic crystal.
All right?
3-D crystal.
So what this is showing is that the system keeps making
more and more bonds.
Why does it make bonds?
Because the more nearest neighbors it has the lower the
energy goes.
So making a 3-dimensional crystal is
energetically favored.
Now there are different Madelung constants for
different crystal structures.
You say, well wait a minute.
How do you get different crystal structures?
Suppose instead of sodium it's potassium.
What's the only difference?
Potassium is plus 1.
What's the difference?
Size.
Potassium is larger.
They're not going to pack quite the same.
And so depending on the relative ion sizes--
I mean what if I have something like silver iodide?
Iodide is huge.
Silver is so small it'll fit into the interstices between
touching iodines.
So it's going to have a different crystal structure,
and the different crystal structure will give us a
different Madelung constant.
And there it is.
So we've come a long way with that little assumption of
octet stability.
So now let's take a look at what the properties are of
these things.
They're solid at room temperature because we've got
strong bonds.
High melting points.
Bonding is related to melting point.
Now think point is dictated by bonding, because now you're
comparing thermal energy versus the cohesive energy of
the crystal.
So tightly bonded substances melt at high temperatures,
weakly bonded substances melt at low temperatures.
Transparent to visible light.
How do I know that?
Because I'm the professor.
No.
How do I know that?
How do we think about it when someone says to you is
something transparent, in this case to visible light.
Well, what I do is I say, here's the solid.
This is the ionic solid and here is visible light, h nu.
And I'm going to write 2 to 3 electron volts per photon.
And what happens when I want to decide whether this is
transparent to visible light?
Now look at the modeling here.
Photon is a squiggle.
I don't know if that's Cartesian space.
This is like a crystal, right?
I'm speaking California.
It's like a crystal.
So now if I go inside I want to make the energy diagram.
So what's the energy diagram look like?
All right?
So now the question is how does--
if this is the energy diagram of the crystal and I make this
the energy of the photon of visible light, I'm going to
compare how much energy the photon has versus how much
energy it takes to excite electrons all the
way to a new state.
Because if you don't excite them all the
way to a new state--
they can't go part way, so nothing happens.
And if nothing happens the photon goes through and that's
transparent.
Now what do I know about the binding energy and the energy
level diagram of Na plus?
Well, it's isoelectronic with neon.
And I know that neon has an average valence electron
energy of about 20 electron volts.
So my guess is the visible light is
not going to do anything.
And so it just passes right on through.
See, we got all that.
Electrical insulator.
How do I know that?
Well, all that glitters is not gold, but it
must have free electrons.
And these electrons are all bound, and
they're tightly bound.
Hard and brittle.
If it's going to be ductile the atoms need to be able to
slide over one another.
Well, there's no way these can slide over one another because
to slide over one another requires that at some point
the two sodiums are going to be nearest neighbors, and the
repulsive forces are so high the crystal fractures.
So if you try to deform an ionic solid you will get it
moving in accordance to the elasticity.
So force will be proportional to the extension--
Hooke's Law-- but if you try to plastically deform
it, you shear it.
Soluble in water.
We'll come back to that later.
Melt to form ionic liquids.
And good for electrolytic extraction of metals.
I showed you magnesium last day.
Today I'll talk a little bit about aluminum.
But that comes later.
OK.
So where do we find elements that are going
to form ionic solids?
Well, you go back to the beginning of the lecture.
What do you look for?
You've got to find the box of really good electron donors
and the box of really good electron acceptors.
That's where you go.
So the good electron donors are at the left side and the
good electron acceptors are at the right side.
So if you take sodium plus chlorine
you get sodium chloride.
If you get calcium plus fluorine, calcium fluoride.
Magnesium plus oxygen, and so on.
OK?
Yeah, this shows.
Yeah.
Aluminum plus oxygen, yeah.
There's Max Born.
Max Born, he got a Nobel Prize.
And so did Fritz Haber, but we're going to
come to him in a minute.
All right.
So now I want to do this energetic calculation one more
way, because right now we've been operating with ion gas
but sodium isn't found normally in the
form of an ion gas.
So let's do something that starts with
elements found in nature.
So we want to form an ionic crystal from elements in their
natural state.
And what's going to happen is en route we're going to be
able to define a few more terms. So that's going to make
everybody happy because we get more definitions.
And this is called the Born-Haber Cycle, named after
Born and Haber.
And what we're going to do is we're going to--
pardon me, there's a C in there-- we're going to invoke
Hess's Law.
And Hess's Law is sort of like Kirchhoff's law
for electric circuits.
Hess's law says that the energy of a chemical change is
path independent.
So energy change in a chemical reaction is path independent.
It's sort of like potential energy in Newtonian mechanics.
It doesn't matter if you take the elevator to the top of the
Hancock Tower or if you walk up the stairs, the change in
potential energy is the same when you express it from the
top of the Hancock Tower, although you might be somewhat
more exhausted having walked the steps instead of taking
the elevator.
But you don't get any credit in terms of
the potential energy.
So let's use Hess's Law in order to describe the
formation of sodium chloride.
So I'm going to start with sodium as it's found in
nature, and I'm going to talk about room temperature.
So sodium is a solid at room temperature, and I'm going to
react it with chlorine gas.
Chlorine's a gas at room temperature and it's a
diatomic molecule.
And we're going to react it to form sodium chloride, which is
a solid and a crystal.
Now you might say, well isn't that kind of redundant?
No, because later on I'm going to teach you about a form of
solid matter that does not consist of atoms
in a regular array--
disordered solids.
So we're specifying, I want to form crystal and solid sodium
chloride, because that's the reaction that would occur if
we were to do it in the lab.
And what have we calculated so far?
What we've calculated so far is this.
We've calculated chloride ion in the gas phase plus sodium
ion in the gas phase reacting to form the crystal.
And we've called this the energy of crystallization.
That's this Madelung stuff.
This is the Madelung energy here.
All right?
I'll even put M here.
Madelung energy.
And why am I using H?
Because that's what the books use.
H is enthalpy, and for condensed matter the
difference between enthalpy and energy
doesn't amount to much.
So H, just for the record, is enthalpy.
And we've been operating with E as energy.
It's almost equal to E, which is energy
for condensed matter.
For the gas phase it gets hairy.
OK.
So I want to get us from sodium solid, chloride solid
over to here.
So how am I going to do that?
Well, first of all, I know how to make sodium ion gas.
I start with sodium gas and then by ionization I make the
electron plus sodium ion.
So this is called the ionization energy, isn't it?
This is the ionization energy.
Sodium gas goes to that.
And now how do I get sodium gas from sodium?
Well that's just called sublimation.
So this I'm going to need delta H of sublimation.
Sublimation is the conversion of solid to vapor.
And you can look that up on the Periodic Table.
I'll show you how to get that in a second.
And now how do I get chloride gas?
Well chloride gas is going to start with atomic chlorine
gas, but instead of losing an electron I've got to acquire
an electron.
And this action of adding an electron is sort of an inverse
ionization, and this is called electron affinity.
And there are tables of electron affinity.
So each element has the ability to lose an electron,
it has the ability to gain an electron.
Losing an electron is ionization energy, acquiring
an electron is electron affinity.
And just as with ionization energies, if you have multiple
electrons you have a first electron affinity, second
electron affinity, and so on.
And now how do I get to atomic chlorine?
I've got to dissociate diatomic chlorine.
So this is called dissociation.
So that's the whole thing.
And I'm going to now put some numbers on here.
Let's see.
I'm going to call sublimation step one, dissociation step
two, ionization I've got here is step three, electron
affinity is step four, and crystallization or
Madelung is step five.
And so working off of Hess's law we can say that the total
energy required for the formation of the crystal--
delta H for the reaction.
What's the reaction?
The reaction of sodium plus chlorine
to make sodium chloride--
is going to be the sum of all of the constituent components.
The sum of all the delta's H.
Not delta H's.
delta's H, like attorneys general.
All right?
So now let's add these up.
So we go number one.
Number one I can look up on the Periodic Table.
Where is it?
There's Fritz Haber.
So that's given here.
If you look on the Periodic Table that'll
give you the number.
And for sodium it's 108 kilojoules per mole.
Number two, get that from tables.
It's 122.
Number three is just the first ionization energy.
You look it up on the Periodic Table.
First ionization energy of sodium is about 5.3 electron
volts, which turns out to be 496 kilojoules per mole.
But look, these are all positive energies and we're
trying to make a net negative energy.
So these three steps are all raising the
energy of the system.
Finally, acquiring an electron by chlorine is going to
decrease the energy of the system, because chlorine is a
good electron acceptor.
So that's minus 349.
But watch this people.
The energy in forming the crystal from the discrete
ion-pairs is 787 kilojoules per mole, which gives us a net
value of minus 410 kilojoules per mole.
So what this is showing you is what the relative
contributions are of the different
components of that thing.
So here it is in graphical form.
There's the vaporization of sodium.
This is the dissociation of chlorine.
This is the ionization of sodium.
All positive energies.
And now electron affinity.
And look at this contribution from the Madelung energy.
So when things crystallize a lot of heat's given off.
In fact, we can use that in cooling and moderating climate
if we're clever about it.
All right.
Now just to show you what the different values are here that
you're going to go in and get the lattice energies, you need
to know the various r values, you see.
The r0 is simply going to be the r-plus and the r-minus.
So lithium fluoride, there it is.
There's the lattice energy and it's based on the combination
of lithium cation and fluoride anion.
So they've gone through and calculated these values.
So there's sodium chloride is 787.
And, you know, you even get things like the boiling
points, melting points.
So sodium chloride, for example, melts at about 800
degrees Celsius.
Now if you take magnesium oxide, magnesium oxide is,
look, 3,700 versus 700.
And the melting point of magnesium oxide is
2,800 degrees C.
Look at aluminum.
Aluminum plus oxygen, look at the end binding energy there.
It's phenomenal, which means it might be useful for--
I'm standing underneath the shuttle here.
This is the tiles underneath the shuttle, and they're made
of aluminum oxide because the Madelung energy is so high so
it's got the thermal shock resistance.
So now you know how to go and design things for thermal
ablation resistance.
All you need is this table.
That's all.
No, you need a little more than that, but this is a good
place to start.
If you don't understand this table I don't want you working
on the project.
OK?
What else?
All this shows you-- yeah, we're going skip that.
All right.
So now we've got a few minutes here.
What I want to do is last day I talked about magnesium,
today I want to talk about aluminum.
And it is also made in an electrochemical process.
In this case the electrodes are horizontal.
We feed aluminum oxide in and we pass current--
and huge currents.
This thing typically runs at 300,000, 400,000 amperes and
about 4 volts.
So with the cathode we are running-- remember, we're
running nature in reverse.
Instead of the electron donor that aluminum is, we're
shoving electrons onto aluminum ion and converting it
back to aluminum.
Unfortunately, on the anode side we have to use carbon,
and the carbon itself is consumed.
So we consume about a half a ton of carbon to
make a ton of aluminum.
So aluminum smelters generate a lot of greenhouse gases.
So you can see this is like a drafting pencil: it's
constantly being fed.
And to give you a sense of scale this is probably about
10 feet across, and this gap is about 2 inches, and this is
about, I don't know, a foot and a half deep.
It's going to get about 1,000 degrees Centigrade--
liquid aluminum and liquid salt.
So this is what a smelter looks like.
What's the sound of electric current?
Yeah.
The only sound you hear is the fans on the top that keep the
place clean.
There's the busbars that are bringing in the current.
And all of these various posts are these things.
So all of the magic is occurring
below the floor here.
It was invented simultaneously in the United States by
Charles Martin Hall and in France by Paul Heroult.
In the same year they filed patents independently and
eventually crossed license when they collided at the
World Court.
So this is what happens.
We dissolve aluminum oxide in a molten fluoride called
cryolite, which originally came from Greenland, make
liquid aluminum carbon dioxide.
Now if we wanted to make this truly green, we want to
eliminate greenhouse gas emissions, you need to find an
inert anode.
So on an inert anode aluminum oxide would be converted into
aluminum and oxygen.
So not only would you not produce greenhouse gases but
you'd produce tonnage oxygen, which is marketable.
So some of the work that goes on in my lab is in advanced
materials with a view to trying to find an inert anode
that would then make this process very, very clean and
justify substituting aluminum for steel in cars.
By the way, when you have the field, even though it's a DC
current it's a divergent field.
And this is me in a magnesium smelter just in Utah.
There is the busbar for the magnesium cell.
So I'm about, oh, two meters away from
the edge of the busbars.
The magnetic field is so high I've got one, two, three,
four, five paper clips standing against gravity.
And I asked them for paper clips and there
was no office there.
They managed to find a few.
I wanted to see how many would go.
I'm willing to bet I could probably have about seven or
eight paper clips up before the gravity would cause them
to collapse.
That's the intensity of magnetic
fields in these smelters.
So when you drive up to the smelter for the tour you park
a fair distance away, you leave your wallet with the
credit cards in it, because this is the biggest bulk
demagnetizer you can imagine.
If you've got a watch that's got hands that move they're
going to be going like this.
[GESTURING]
PROFESSOR: Yeah.
This is the shuttle.
This is forged aluminum wheels.
The shuttle lands on Centerline racing wheels.
They're forged at a place in California.
They're made of aluminum alloy.
And I don't know if you can read on the side here, it's a
little bit light, but these are special tires.
They're made by Michelin
So, anyways, the whole message here is learn the lessons here
in 3.091 and then we can work together to make metal in an
environmentally acceptable way.
And just before I send you on your way for the weekend I
thought I'd tell a little joke related to
the uncertainty principle.
So the joke goes like this.
Heisenberg is racing down the Autobahn and he gets pulled
over by the state trooper who comes to the car and says,
where's the fire buddy?
Do you know how fast you were going?
And Heisenberg looks at him he says, no, but I
know where I am.
[LAUGHTER]
PROFESSOR: All right.
Get out of here.
Have a good weekend.