Practice English Speaking&Listening with: Lec 10 | MIT 3.091 Introduction to Solid State Chemistry

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A couple of comments about the test. You should have gotten your test

back yesterday. And here is the way things panned

out. We have a class average of about 77%. I said settle down.

Perhaps you think you have come to math or physics.

This is 3.091 and we have standards here. 77% is class average and the

standard deviation was 17%, so you can see where things lie.

50% is a pass. Congratulations,

a lot of people put in the effort and there is some learning going on.

While at the same time as I say congratulations,

I want to remind you that I think it comes as no surprise given the

lateness of Labor Day this year that this first test came rather early

and for many of you much of this material you had seen

in high school. Perhaps the data rate was a little

bit faster than you were accustomed to, but you had some familiarity.

Please don't, if you are up here, become complacent because a month

from now we will have another test. And I think you will wish you are

continuing to make the investment in time. If you are down here,

what I want you to do is talk to somebody. Talk to either your

recitation instructor, come in and talk to me.

We will provide tutorial assistance at no charge. If you want to

amplify the exposure to question and answer, all you have to do is

contact either Hillary or Lori at my office and we will arrange for

tutorial meetings. And maybe down here you got rattled

or maybe you weren't feeling too well that day.

Maybe you have to bone up on your test-taking skills.

I don't care what the score is down here at the moment.

As I told you, on the first day,

I do look at trends. And so let's make every effort to

take the steps necessary to master that material and move forward.

The good news is that we are going to be shifting gears very soon and

moving into subject matter unlike anything you have seen in high

school, and so you will get a fresh start. So I don't want anybody

feeling that if he or she is down here that it is hopeless at this

point. It is not. Quite the contrary.

And I have actually seen evidence over the years,

mercifully, just a small amount of evidence, but there are,

in a group like this, always a few students who come in with a really

strong background in high school. They do extremely well on the first

test and conclude that chemistry is in the bank and they can focus on

the other subjects. And little by little they start

creeping down here. It happens. The model solutions

will be posted, as I told you. And we pride ourselves in giving you

a rapid return, but there is a tradeoff.

And that is we are going to make some mistakes.

The mistakes are not indelible. You can come in and have your

grades adjusted. The first thing you do,

please talk to your recitation instructor. If the recitation

instructor feels that there was an oversight then the recitation

instructor will recommend that you come and see me and I will change

the grade. If it's just an addition error, the recitation instructor has

authority to make that adjustment. And you don't all have to go

stampeding down to my office right away. You can adjust those grades,

let's leave it until next week and we can take it at a leisurely pace.

But there is no time limit on it. In fact, you could be sitting three

years from now waiting before commencement and you can get that

grade adjusted. I can adjust the grade right up

until the moment you get your degree. After that the records are sealed.

The reason I tell you this is that I don't want people feeling that,

well, the paper is going to start to decompose or something if I don't

get to him by 5:00 today. Don't hurt yourself in the stampede

to my office. Congratulations again to those that did well and let's do

what we can to get everybody over to the right of that blue line.

And just to keep the rhythm going, we will have a test on Tuesday, the

weekly quiz. And I know it is only the third quiz,

but I like to keep the quiz number the same as the homework number.

So Q4 covers homework 4, otherwise there is some confusion later on.

Last day, Professor Ballinger introduced you to Lewis who taught

us that octet stability could be achieved via electron sharing as an

alternative to electron transfer. And he coined the term covalent

bonding. It means cooperative sharing of valence electrons.

That is what gets crunched down to covalent bonding.

Pauling came along later and told us that the electrons are not shared

equally, and this was important because there were problems in

computing the bond energies of heteronuclear molecules.

If you looked at heteronuclear molecules and you wanted to compute

the bond energy, you might start with the bond

energies of the two constituents. And you would find that the bond

energy of the heteronuclear molecule was nowhere on the average

of the two. And that did not make sense.

If I have one that's 400 kilojoules per mole and another that's 200

kilojoules per mole and I blend them, how do I get 500 kilojoules per mole?

That bothered people. And Pauling came along and said the

unequal sharing of electrons in a covalent bond can explain that extra

energy. And he coined the term "polar covalency."

And, in particular, he introduced a quantitative measure.

This is why he gets the Nobel prize and Lewis did not get the Nobel

prize, although Lewis did brilliant work, but Pauling's work was

quantitative. He introduced the concept of electronegativity which

was a measure, therefore, it is quantitative,

of the atom's ability to attract electrons within a covalent bond and

developed a scale of electronegativity.

That is shown here. And this looks a lot like the

average valence electron energy where the lowest values of

electronegativity are in the metallic zone and the highest values

of electronegativity are on the nonmetallic zone.

Think about it. If this is a measure of the ability of an

electron to be pulled in a covalent bond, so if I have H here and F here

and F is more electronegative than H, and so the electrons,

instead of being shared equally, I am going to show them over here as

if to say fluorine is hogging the electron.

It makes sense. If this is metallic,

it is a good electron donor. Well, if it is a good electron

donor in an electron transfer reaction, if the same element finds

itself in a covalent bond, it is going to be a good electron

donor, although it is not full transfer. Likewise,

the element that is a good electron acceptor in an electron transfer

reaction is going to be the element that is going to hog the electrons

in a covalent bond. So there is some consistency here.

And you see where the most electronegative elements are,

and fluorine is the most electronegative of the active

elements. I know neon has a higher yet electronegativity,

but normally it is inert. So how do we use this

electronegativity? Well, I mentioned to you that there

was this mystery. Let's go back and review one of

these. This is the homonuclear bond energy for hydrogen in pure hydrogen.

There we have perfect covalency. Both hydrogens are going to share

equally. We don't have two different species of hydrogen here,

so the electrons are perfectly shared. And this turns out to have

a bond strength of 435 kilojoules per mole. And,

likewise, if we look at fluorine in its diatomic molecule,

it is 160 kilojoules per mole. And you have the bond energy in HF.

Just so that we are clear, this is the bond in this compound.

As we see a little bit later, I could talk about the carbon-hydrogen

bond in methane where there is a plurality of bonds.

In this case, there is only the one bond but I just want to get the

formulas. And so this is the H-F bond in HF. It is nowhere on this

interval. And this is what puzzled people. It is 569 kilojoules per

mole. So there is something extra going on. And so what Pauling gave

us was a formula. He said if you take the XY bond

energy starting with the bond energies XX, and YY bond energies,

this is how to make sense of it. What Pauling taught us was that the

energy of the XY bond is equal to the average. We are going to start

with the average. That is a good place to start.

And he defined the average in terms of not the sum divided by two,

but he chose the geometric mean. So you take the square root of the

product. In fact, there are some very old books that

show this as the average, the sum divided by two arithmetic

mean, but the modern practice is to use the geometric mean.

And then he said there is a second term here, a contribution to the

unequal sharing. And to measure the intensity of

unequal sharing, we take the difference in

electronegativity, which I am going to use the Greek

symbol chi for. This is lower case chi.

I will take chi X minus chi Y. And the functionality goes as the

square of the difference in electronegativity.

And the prefactor is 96. when you want to get a result in

kilojoules per mole. Let's go through this calculation

for HF. If we look at 435 times 160, take the square root of that,

we will end up with 264 kilojoules per mole, which sensibly lies

between these two values. And then this second term,

if we go to the Periodic Table, we will find that the value here for

hydrogen is 2. and the value for fluorine is 3.

8. You plug all of this in and you get 344 kilojoules per mole,

which then sums to 608. Now, 608 is not 569.

But, remember, this is a very rough estimate.

But the beauty in this calculation is it shows that this concept of

electron sharing in an unequal manner is sound and it gets you into

the right ballpark. And so Pauling went further and

said this is the purely covalent component. This is purely covalent

because all it is is homonuclear bond energies.

Homonuclear means both elements in the molecule are the same.

There is no homoelemental. They say homonuclear. So this is

purely covalent. And what is this?

He said this is partial ionic character. Why did he choose that

term? Well, I like to think of it as I have this sort of sharing meter.

And I will give you the extremes. If I look at a homonuclear molecule

such as molecular hydrogen, this is perfect sharing.

If this is my sharing meter, this is my electron sharing meter.

It is right at 12:00. Now, if I get over here,

HF, I know that fluorine is hogging the electron, so it is not equal

sharing. It is over here. This is pure covalency. This is

polar covalency. And why did he call it polar

covalency? Because, if you look at this molecule,

you say that, well, the electrons are a little bit closer to the right

than to the left so the charge is not uniformly distributed.

The molecule is net neutral, but this right end is a little bit

more negative and the left end is a little bit more positive.

And in physics, when physicists want to say "a little bit of,

they write lower case delta. That is physics talk for a little bit of.

So this is a little bit negative and this is a little bit positive,

which means I could model this by drawing a dipole.

This is a dipole. It has a negative end and a

positive end and it has certain properties. If we put a dipole that

has the freedom to move, we put that dipole in an electric

field, it will align itself with the field lines. Because this is a

dipole, he chose the pole part of dipole to give us polar covalency.

And what is the ultimate? What is the extreme of unequal

sharing? It is electron transfer. So this is perfect sharing, unequal

sharing. And here the electrons are actually donated,

so this becomes F minus. This is denuded of an electron,

it is Na plus. And so, this one here, I am going to say let's bury

the needle on this one. This is the ultimate in unequal

sharing and this is ionic. This is ionic bonding.

So you can see that polar covalency is a tendency towards ionic bonding.

And that is why Pauling called this partial ionic character.

And he actually quantified it. He quantified it by this formula.

He said that the percent ionic character, and this is within a bond,

not for a compound, for a covalent bond.

He said that is going to equal one minus the exponential.

This is just base e, natural logarithm. E to the power of minus

one-quarter times the difference in electronegativity squared.

It is this thing again. You see the difference in electronegativity,

square it, multiply it by one-quarter and raise

that to the power e. Subtract that from one,

multiply it by 100 and you have something between zero and 100.

And I think there is some evidence of that here. This is right off

your Periodic Table. If you look up in the corner of

your Periodic Table on one of the sides, they have actually tabulated

this for you. I have rewritten the formula here. And there are two

different scales of ionic character. One by Pauling. Actually, there

are more than two. But they show these, too.

There is Pauling and there is Hannay and Smyth. Here is Pauling.

And we could do this for our compound here,

HF. For HF, just for grins and chuckles, let's do it.

The electronegativity of fluorine is 3.98 and the electronegativity of

H is 2.20 on this scale. And so, if you plug these values

into this formula, that will give you the percent ionic

character for the H-F bond. And I want to make sure that you

don't think that this is the percent ionic character of the compound,

so I am going to be really fastidious here and indicate that I

am talking about the H-F bond. If you plug into this formula here,

you end up with 56%. And that has already been tabulated up there in

the nomograph. You will see if you get --

I think the delta here is 1. 8, which is roughly 1.8. And there

you see the 56. Just to give you a sense of what

that means, if the 264 figure is pure covalency and the 344 is

partial ionic character, what fraction of the 608 is 344?

344, which is partial ionic character, divided by

608 times 100 is 57%. So it makes sense.

That is good. We can do all sorts of things with this.

So I indicated the molecule. There is another orthography that

is used to indicate dipole. You can use delta minus, delta plus.

Some people like to use an arrow, and the arrow points in the

direction of the more electronegative end.

And the way I remember that is I put a little cross here,

and then I make that the plus end. I don't know. You can have your

own mnemonic device to figure that out, but anyway.

Let's go on and look at another one. I think last day Professor

Ballinger did this one. He did some hybridization of

methane. Here is methane. Methane looks like this with the

sp3 hybridization. Let's look at that one.

Here we have a dipole. We only have the one bond so the actual HF

molecule is polar, it has a net dipole.

Whereas, H2 is nonpolar. How do I know its charge

distribution? It's perfectly symmetric. The electrons here are

shared equally. That is nonpolar.

And likewise for fluorine, F2. Let's look at methane. Methane

we can draw. It looks like this. We have carbon, sp3 hybridization.

CH4 is sp3 carbon hybridization. It allows four bonds to form instead

of just two. And so now I have hydrogens at the four corners of a

tetrahedron. Now I want to ask, what's the nature of the

carbon-hydrogen bond? And I look up and see that the

electronegativity of carbon is 2. 5 and the electronegativity of

hydrogen, we already know, is 2.20. So there is a difference

in electronegativity. And since carbon's

electronegativity is higher than that of hydrogen,

which you would expect from where carbon lies on the Periodic Table.

Think about it. See, if you know the Periodic Table

then you know where elements are relative to one another.

So, at any moment, you know in any bond which is the more

electropositive. Electropositive is down to the left

because that is where the metals live. So carbon is to the right of

hydrogen, this is more electronegative.

Let's look at this carbon-hydrogen bond.

What does that tell you? It tells you that, first of all,

there is polarity here. The electrons are not equally shared and

the carbon hogs the electrons a little bit more than the hydrogen,

so the carbon is electron-rich and the hydrogen is just a little bit

electron-deficient. Now, the question is --

Well, first of all, let's get that down.

Let's show that carbon-hydrogen is polar. It is a polar bond.

This is delta minus and this is delta plus. So I have a polar bond.

Here is the question. Is methane a polar molecule or a nonpolar

molecule? Let's look carefully. How do I answer that question? I

have to ask is there a net dipole? Let's look carefully here. All

four of these hydrogens are electron-deficient in the same

manner. Where is the center of net negative charge in this molecule?

The center of excess negative charge on all of the dipoles is at the very

center of the molecule. Where is the center of the excess

positive charge? And by here, I mean it's net

neutral, but the electron deficiency. But can you see that the centers of

electron deficiency lie on a sphere equidistant from the center?

So where is the center of the delta minus? The center of the delta

minus is the center of the molecule. So now that means if this is where

the center of positive excess charge lies and on top of it is the center

of negative excess charge there is no displacement of the charge so

there is no net dipole. And so, methane is nonpolar.

It is an aggregate of four polar bonds, but they are symmetry

arranged in space. Symmetric disposition of polar

bonds still results in a nonpolar molecule.

We can say that a nonpolar molecule you can have as a result of two

different conditions. Nonpolar molecules result from

either, it's one or the other. One is the trivial case,

homonuclear. If it is homonuclear they are all the same.

They all share. Homonuclear species.

For example, H2, N2, they are going to share the

electrons uniformly, equally. And you can have P4,

S8. All of these multiatomic moieties are nonpolar.

The second way to have something that is net nonpolar is to have

spatially symmetric disposition of polar bonds.

A good example of that is CH4, because the CH bond is polar but

symmetrically disposed in space. Now I want to look at the

energetics. I have not talked about it from a graphical standpoint.

We looked at it from an analytical standpoint.

Let's go back and look at it from an energetic standpoint.

There is too much talking in the room so I am asking you to stop now

in respect for your classmates. I do not like the rudeness. I want

it perfectly quiet in here. There is no compromise on that.

None. When you walk through that door it's an act of free will,

and I will not have anything other than complete silence.

It is not fair to classmates to have to listen to conversation.

Now let's look at the energetics

graphically now. I want to go back to our energy

level diagrams and see if I can rationalize energy level diagrams.

But now not for single atoms. I want to do energy level diagrams

for molecules. That is different.

Where am I going to get those energy levels from?

I get the energy levels from the Schršdinger equation.

The Schršdinger equation will give us the energy levels in molecules.

We saw the Schršdinger equation for atomic hydrogen,

but you can write it for more complex systems. And

it is horrible. But there is good news.

The Schršdinger equation is a linear equation.

And this is one of the times when you are actually going to be able to

use some of that math they teach you. What is one of the properties of a

linear equation? Linear equation means that

solutions are additive. For example, if I have an equation

that looks like this, f of x, y, z. I might as well do it as x,

y, z because we are talking about something that is going in three

space. Suppose f of x, y, z equals k1, that is my equation,

and it gives me a solution s1. And I have another equation f of x,

y, z. This is the same equation, the Schršdinger equation, only it

has different boundary conditions. It has k2 as the boundary

conditions. And when I solve it for k2 boundary conditions,

I get s2. Here is the power of linearity.

The power of linearity is if I come across f of x,

y, z equals k1 plus k2, if it is a linear equation,

I don't have to go and solve it all over again. Instead I can write

with impunity the solution will be s1 plus s2. And that saves a lot of

time. Linearity is the desirable way to go if you can manage.

What does this mean? Linearity of solutions, the elegant way of saying

it is superposition holds. And that is what people are going to

use. They are going to say, well, what is the solution? I said

here is a solution s1. What is the solution to the

Schršdinger equation? The solution of the Schršdinger

equation is psi, a wavefunction. What we are going

to say is that the wavefunction for molecular orbitals is going to be an

additive sum of the wavefunctions of atomic orbitals.

I am going to sum up the atomic orbitals that go into the molecular

orbital, and they are going to have some coefficients.

We are going to have coefficients on this, and they will add up.

And then we are going to construct molecular orbitals.

And the technique is called, watch because this is a six-letter

initialization, linear combination of atomic

orbitals into molecular orbital, LCAO-MO.

Linear combination of atomic orbitals into molecular orbitals.

And so we are going to use that in order to construct some energy level

diagrams. And so let's look at a few. Here is the first one.

I want to do one that is going to show that hydrogen,

H2 is favored over atomic hydrogen. Here is what we are going to do.

We are going to set up an energy level diagram. This

is E equals zero. N equals infinity.

And this is the ground state, n equals one. This is the 1s. And,

just to be clear, this is the 1s atomic orbital in atomic hydrogen.

And I am going to do the same thing. I am going to put another one over

here and put another one over here. Here are two hydrogens, and they

both look the same. And then over here is H2.

And, by using LCAO-MO, we can come up with the orbitals here.

And you are not going to calculate these. This would be given to you.

We would give you the energy levels and ask you to rationalize something

that we are going to tell you is true. What happens?

Well, we have to use two other ideas. Superposition.

We use the concept of conservation of states and the Aufbau Principle.

That includes Pauli Exclusion Principle and includes

Hund's Rule and so on. Let's look at electron filling first

of all. This has a single electron. And since it is Friday, I think I

will even give its quantum numbers, 1, 0, 0, ½. I only do this on

Fridays. And here is another one. It is over here. And its quantum

address is 1, 0, 0, ½. And you are saying wait a

minute, he has violated the Pauli Exclusion Principle.

He has two electrons here with the same set of quantum numbers.

But these are two separate hydrogen atoms. They are not part

of the same system. But now what we are going to do is

bring them in together sort of along the lines of this.

This cartoon shows the probability density function of 1s.

And this is kind of nice because you see it is not an abrupt

spherical surface. What you see is a scatterplot.

And you see that the dot intensity grows as you get closer and closer

to the center. So it tails off to infinity.

Now here are these two isolated hydrogen atoms.

Now they get close enough together that they start to sense that they

are becoming part of one system. And once they start to become part

of one system, Pauli Exclusion Principle kicks in.

You start to have to conserve states. We cannot have two 1s's

sitting at the same level. And ultimately we are going to get

to this state where we are going to form a bond. What happens is these

two levels displace. One is a little bit higher and one

is a little bit lower than they are in the atomic situation.

And so this lower level is called a bonding orbital,

and it is a bonding molecular orbital. And this one here,

because it is at a higher energy is called antibonding molecular orbital.

And these are s's. This is 1s atomic orbital.

This will be called lowercase sigma. And, since it came from 1s,

it will be called sigma 1s. And I am going to superscript it molecular

orbital, and this upper one, to indicate that it's antibonding,

has the asterisk. This is sigma star with the antibonding orbital

that came from 1s, and it is a molecular orbital.

This is the energy level diagram of H2. Now let's put the electrons in.

How do I do it? I put in the first one spin-up.

I put in the second one spin-down. I started with two electrons, one

each from two atomic hydrogens. And now I have to fill this

molecular hydrogen. The question now is,

is H2 stable or will it not form? And the answer lies clearly in the

energy difference. What we do is ask ourselves what is

the delta E? What is the energy change going from two atomic

hydrogens to one H2? That is going to equal the energy of

the electrons in H2 minus the energies of the electrons in H.

And, if this quantity is less than zero, the bonding is favorable.

We know it is favorable. We got it out of the tables,

it is 435 kilojoules per mole. That is pretty favorable. What is

happening is it turns out that putting the electrons in that lower

energy state is stabilizing that molecule

So that works. What about H2 plus?

What would that look like? H2 plus would take one electron out.

Would H2 plus be stable? According to this energy level

diagram, yes. H2 plus is stable and is observed in plasmas.

Let's go on. Let's do a few more. I want to ask about helium. Let's

start. Here is the zero. This is energy zero.

And here is 1s, atomic orbital for helium. Here is another 1s,

atomic orbital for helium. And, again, we will split. Here is the

antibonding and here is the bonding. And, in this case, I start with

helium - is 1s2. Each of the heliums has two

electrons. Now let's go and fill the molecular states for the

putative He2. I don't know if it exists yet or not.

We go one, two, three, four. Now,

He2, stable or not? Well, it is going to depend upon

whether this delta is greater than this delta. In other words,

if the rise in instability is greater than the fall in stability,

this is a positive delta E. And it turns out that is the case.

It turns out that the antibonding orbital is a little bit higher from

the atomic orbital level than the bonding orbital is lower.

And so the result is that the delta E, in this case,

delta E in going from the atomic states to the molecular states is

greater than zero. And so He2 unstable.

I would ask you to see how such energy level diagrams rationalize

what we know to be true. I don't expect you to start with one

of these and then predict what is going to happen.

I would say with the aid of an energy level diagram explain the

fact that helium is found as atomic gas and not molecular.

How about this one? What about He2 plus? What would that one look like?

For He2 plus there would be one electron missing.

And what do you see here? You have two electrons at much

lower energy. You have one electron at a higher

energy. And so, in fact, the delta here is negative.

And He2 plus is observed in helium plasmas at high pressure,

but not neutral He2. You don't see those because it's just not in the

cards. Let's do one more because I am having a ball.

I don't know about you, but this is so fun. Let's do one

more. Let's do lithium. Again, you know the drill.

Here is 1s atomic. But lithium has 2s1, so I need a 2s atomic orbital

here and likewise over here. Here is 2s atomic orbital. Here is

1s atomic orbital. This is lithium, lithium.

And this is all gas phase. These are single atoms. It has to

be gas phase. And now let's use the same principle.

Here is sigma star, sigma, sigma star and sigma.

These are the bonding and antibonding. The lower ones come

from 1s and the upper ones come from 2s. Let's look at the electron

filling in lithium, one, two, three. 1s2,

2s1 over here. The same thing, 1s2, 2s1. Because these lithiums

are independent. Now, what if they get close enough

together to start forming one system. Let's fill. What do we get?

One, two, three, four, five, six. What is the

conclusion? This indicates that by combining two lithiums,

the energy of the combined system is lower than the energies of the

atomic systems. Dilithium is stable.

And, in fact, this is the case for all the alkali metals.

When you go down the highway and you look up and see those sodium

vapor lamps, the sodium vapor exists as the dimer Na2.

This is good, dilithium is stable. So far we have looked only at the

formation of single bonds. Well, let's look at multiple bonds.

Let's try nitrogen. We know nitrogen bonds multiple bonds,

so let's look at nitrogen. How would that look?

Well, first of all, let's look at the Lewis structure.

Lewis structure for N2 is one, two, three, four, five. Here is the

second nitrogen, one, two, three, four,

five. Here is octet for the left, octet for the right. We see three

bonding pairs so this is a triple bond, indeed a multiple bond.

The bond is the electron pairs. This is homonuclear and so these

are nonpolar. They are perfectly shared between the two nitrogens.

And, furthermore, to put the background in front,

let's do the electronic structure. It is 1s2, 2s2, 2p3. That would be

one, two, three, four, Hund's Rule,

five, six, seven. Three unpaired electrons in nitrogen.

Oh, by the way. Look at this. I wasted your time here. I was

supposed to apologize to you. We learn nothing from examining

what is going on down here in n equals one shell.

It is filled here. It is filled here.

We learn nothing. Let's take this whole thing and

throw it away. So what is the lesson here?

If you want to understand whether compounds form or not look only at

the valence shell. It doesn't make any sense.

This is another example, sort of another thing that comes out of that

analysis. You can see that there is no value in studying inner shell

electrons to ask questions about bonding. What do these look like?

These are p-orbitals. And what does a p-orbital look like?

It is this dumbbell shape. That is what a p-orbital looks like.

And I have three of them. You know from the m quantum number there are

three. They are orthogonal. What does that look like? Here is

one, here is two and here is three. Put a second one next to it. One,

two, three. Excuse my drawing, but this is what you get. And the

convention is that when two atoms bond, they bond along the z-axis.

So the convention will be that the z is moving from left to right.

And so then the right-hand rule will kick in and require that you

start with the thumb as x and then you move to y and z.

And so if that happens this will be x into the board,

y will be the vertical and z will be moving to the right.

What we want to do is ask ourselves what happens when these two interact.

And here is what it looks like. I am going to do it first

pictorially and then I will write the algebra. What happens here?

We have two of them on the z-axis like this. These are both z-axis.

This is 2pz atomic orbital plus 2pz atomic orbital,

and they react to form something that smears in this manner.

It is going to smear in this manner and will look something like this

where I have dumbbell shapes with smearing. And this is now sigma of

2pz. And why did I call it a sigma? The sigma bond is characterized by

continuous electron density between nuclei.

We say there are no nodes, no dropouts, no holidays. No nodes.

You can start from one nucleus and go to the next nucleus,

and there are no zero planes, no nodes, nothing. And the same

thing that happens over here, what is this one going to look like?

Here you would have a 2s, here is a 2s, and when the two of them smear

we end with something that looks like this. Here is the sigma.

Again, we have continuous electron density from one nucleus to the

other. That gets us the first one. What are the other two going to do?

The other two have to do something different. The other two are going

to smear laterally. Let's look at 2py, which is

vertical. Here is a 2py atomic orbital, and it will react with a

second 2py atomic orbital. But it is not going to be able to

give a sigma. Because already I can see there is a

line of sight from one nucleus to the other with no electron density

whatsoever. These two are going to smear laterally.

They will smear to give you something like this where here are

the two nuclei. And these are called lobes.

This is a lobe. I have two lobes in a p-orbital.

And that zero point is the node. The node is the same thing you see

in a string. Point of absolute rest in a vibrating string we call a node.

This is the same idea. And how do you get from one side to

the other? You're vibrating. You're acting as a wave. That is

how you get from one side to the other. The two lobes smear and I

end up with something that looks like this. And so both of these

lobes together constitute a pi bond. And this will be called pi of 2py

molecular orbital. And so it has what?

In contrast to the sigma, it has a nodal plane containing both

nuclei. Let's take a look at some of these. First of all,

this is the two s orbitals in hydrogen, 1s plus 1s smearing to

give us this sigma molecular orbital.

And this one, just for completeness, is what the antibonding orbital

would look like. Where are we getting these again?

We are getting these by plotting the values from the Schrodinger

equation. You get these plots by taking the wave function times its

complex conjugate and operating on that. That is proportional to

probability. These functions are what are being plotted.

And there is a proportionality between that and where we expect the

electron to reside. That might be the contour of 99% of

the time. There is a 1% chance it is beyond, but the 99% surface is

spherical. Here is from your textbook. Same idea.

1s plus 1s gives you this oval ellipsoid which is the bonding,

and here are the antibonding, and then these are the energy levels

that I have been drawing for you. Now, from your book as well, this

is the pz's of the two atomic orbitals forming the

bonding orbital. You can see this is sort of p-like

but smeared. And then the antibonding which we don't care

about. And this is the lateral smearing. They call it x.

This is the vertical direction y. And just to give you a sense, if

you do a cross-cut of this, you cannot show it from this angle,

but it should really be sort of a dumbbell shape with the nodal plane

in between the two nuclei. Now let's go and do the energetics.

We can do the energetics on this one now. Let's look at it.

You would be given this filling sequence. What do you see here?

This is now for nitrogen. There is the 2s. Why are we not looking at

1s? Because it is not the valence shell. Forget about it.

We are only looking at 2s and 2p. Here is the atomic nitrogen, here

is the atomic nitrogen and these are the orbitals of molecular nitrogen.

This is just the ladder. Now we are going to put electrons on

the shelf here. Let's add them.

Nitrogen we start with 2s2, 2p3. One, two, three, four, five.

One, two, three, four, five. Five and five is ten.

Two, four, six, eight, ten. Look at that. When you are

done you have three electron pairs in bonding orbitals.

That is a good solid triple bond. What does it look like up here?

What it looks like up here, the simple Cartesian model of it is

these things smear this way. There is a sigma orbital.

And they are smearing this way. That is one pi orbital. There is

one sigma, one pi and there is a second pi, and that is how we are

getting the triple bond. N is one sigma and two pi.

There they are right here. You can see that on this one.

And so it gives us a whopping 946 kilojoules per mole.

That is a lot of energy because there is a really solid bond.

Here is oxygen and fluorine. And it is just a little bit

different. And I draw attention to it. We would tell you the filling

sequence, but I just want you to see that there are some subtle

differences here that in the case of oxygen and fluorine the p-orbitals,

the sigma and the pi flip-flop. See that? But we would tell you the

sequence. Let's go through it now. This is oxygen. Oxygen has six

valence electrons. Six plus six is 12.

Just as in nitrogen, we get two, four, six here,

but now we have two more electrons. And because of the way those

antibonding orbitals are stacked, the two electrons go one each into

those antibonding. At the end of the day you have some

unpaired electrons. You have two electrons in

antibonding to kind of offset the bonding. And so the double bond is

498. Remember the other one was 960. Now here is fluorine,

same skeleton, only fluorine has seven valence electrons.

Seven and seven is 14, so you can see that there is going

to be two sets in antibonding, three sets in bonding for a net of

one, giving us the single bond. And it is only 160 kilojoules per

mole. This energy level diagram helps us understand the relationship

between electron filling and bond strength. It is very nice.

And one last thing. Because these things are unpaired,

we have already seen how unpaired electrons play a role in the

Stern-Gerlach experiment. Here this leads to paramagnetism

because there is some mutual induction. Both of these are

spinning but they are spinning in the same level.

And so what happens is the result is paramagnetism.

Liquid oxygen is paramagnetic. And what can happen is that this

cartoon shows that if you put a magnetic field on liquid oxygen,

that liquid oxygen would be drawn towards the magnetic field.

And, in fact, if you take the jaws of a powerful magnet and pour liquid

oxygen in between the jaws, the oxygen will stop and will sit in

the jaws. Of course, it is evaporating like crazy because

it is so warm in the room. Whereas, liquid nitrogen just goes

zooming right through. This is a consequence,

again, of knowing the electron filling. Well,

let's see something really crazy with electronegativity.

This one I learned about in my own research about 20 years ago,

and it never ceases to amaze me. If you look at the

electronegativity difference for sodium iodide,

it is 1.73. And everybody would agree that is an ionic compound.

Sodium is a metal, iodine is a nonmetal. Look at cesium.

Cesium versus gold is the same delta electronegativity difference.

They are both metals, cesium and gold. When you melt them,

cesium melts at about 28 degrees centigrade, gold melts at 1063.

Liquid metal. Solid metals melt and become a liquid metal like mercury.

But when the concentration in the alloy is 50/50 by mole,

something magical happens. You have equal numbers of donors

and acceptors. Electron transfer occurs.

And you have this liquid alloy at 600 degrees centigrade,

and it turns into a molten salt. It just goes bang, clear and

colorless because salts are colorless. High band gap materials.

The electrical conductivity plummets about three orders of magnitude.

And, instead of being an electronic conductor, a liquid metal,

it is an ionic conductor. It is cesium auride.

And if you cool that it crystallizes in the form of salt.

You start with liquid metal one, liquid metal two,

you have the atom ratios proper, they mix, electron transfer occurs

and poof, it is clear and colorless. Sorcery. All right.

Have a nice weekend.

The Description of Lec 10 | MIT 3.091 Introduction to Solid State Chemistry