Practice English Speaking&Listening with: Lec 11 | MIT 3.091 Introduction to Solid State Chemistry

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The weekend is over. It's time to get back to business.

So, a couple of announcements: tomorrow we'll have quiz four based

on homework four. I'll be available from 2:15-3:15

under normal office hours. If you can't get it to see me at

that time, then just send a note and we'll arrange to meet at some other

time. And actually Lori has made a sign-up sheet,

sort of five-minute blocks so that if you do want to come by to look at

re-grades, you don't have to hang around waiting for a

long period of time. And, if you can't get them during

that time, there's no expiry. You can get the grade change at

some later date. And, we'll make sure we fix things

up. So, last day we were talking about covalent bonding,

looking at linear combinations of atomic orbitals into molecular

orbitals. And, we saw that there's two types of

molecular orbitals involving s and p-block elements that

are germane to us. We are not going to go into d-block

in any detail here. At the level of 3.091,

I think that s and p-block teaches you the rudiments.

And to go beyond that really requires some intensive quantum

mechanics. So, we're going to stay with s and

p-block. And we saw that if we had two atomic s-orbitals,

when they blended to form the molecular orbital,

it took on this sort of the ellipsoidal shape.

We called that a sigma orbital. And, the characteristic of the

sigma orbital is that you have continuous electron density from one

nucleus to the other. We said no holidays; constant

electronic density. We could then combine an s orbital

and a p orbital, as for example in the case of

hydrogen fluoride. In the case of s plus p,

will end up with an orbital that looks something like this.

And, it is also termed a sigma orbital because,

again, if you look from one nucleus to the other, there is continuous

electron density, no holidays. This sigma orbital is a little bit

in shape from that sigma orbital, but they're both sigmas. And then,

we saw what could happen with p orbitals. There's two options here.

The first is shown when the two p orbitals meet axially,

on end. When they meet axially on end, these two lobes smear to give

us the lobes shown in this cartoon: again, continuous electron density

from one nucleus to the other. That fits the definition of the

sigma orbital. And then, something only the p's can

do: when they smear laterally, these lobes smear to give a

molecular lobe. These lobes smear to give you a

molecular lobe. These two lobes collectively

comprise one molecular orbital. And, the characteristic here is,

in contrast to zero holidays, here you have zero density.

We have a nodal plane normal to the board. There is no

electron density. You can go clear line of sight from

one nucleus to the other with zero electron density.

And this is called a pi orbital. And, then we went on. We looked at

a couple of things like nitrogen, oxygen, and fluorine. But in the

case of nitrogen and oxygen we saw that in order to get the triple and

double bond, we needed a mix. We needed a mix of sigma orbital

and pi orbital. And, that's the template.

If you're going to make a single bond, you may be able to get away

with a sigma. If you're going to make a multiple bond,

the double bond or a triple bond, you're going to have to have a next

of sigma and pi. And I think that's the way we ended

the lecture last day, looking at energy level diagrams.

And let's see what else we have. Yeah, so we saw this. Here's the

end on end. This is a sigma orbital coming from two p atomic orbitals,

and remember the convention: the z axis is conventionally the axis

along which the two atoms bond? And then, this is the lateral

smearing to give us the pi orbital. And I think I showed you that if you

cut this, it looks this way. We went into this, and this,

and we just had a great time. So now what I want to do is I want to

show that you can extend this business of multiple bonds to

hybridized systems. Now, I know Professor Ballinger

last week talked about methane. And the way to rationalize methane

was to mix s and p atomic orbitals to make a hybridized orbital.

And a hybridized orbital that was constructed was the sp3 hybrid.

So, in order to get the CH4, CH4 was explained in terms of the sp3

hybrid orbital. As you recall, carbon in its

valence shell has six electrons in total, but four valence electrons.

And, the four valence electrons are depicted two in the s orbital,

and two in the p orbitals. And clearly that configuration doesn't

have the capacity to form four bonds. Somehow we've got to generate four

unpaired electrons. And, the sp3 was shown to take one

of the s orbitals. This is not a 1s. Let's say it's

one s orbital, and three p orbitals. Hence we get the term sp3.

And, these are equivalent. And so, now, by the Hund rule,

I've got four unpaired electrons ready to bond.

And since there's four of them, and they are equivalent in energy,

we disposed them equivalently in space. So, if I want to take a

central point and put four struts off that central point symmetrically

disposed in space, this will form the struts that go

from the center to the vertices of a tetrahedron. And this gives us the

109¡ bond angle. And there's a whole host of

compounds that are explained by this binding configuration.

But now, let's look at another one. But now let's look at another one.

How about this molecule: C2H4? It's known as ethylene, precursor

to polyethylene. And, let's first of all start with

the Lewis structure, two carbons here, and we've got

hydrogens. So, remember, I'm looking for octet

around the carbon. So: one, two, three,

four, for electrons around the left carbon. Let's bring in an electron

from hydrogen. That way, the hydrogen is

isoelectronic with helium. We'll bring an electron from the

other hydrogen. So, this is isoelectronic with

helium. And now, carbon has two, four,

six, and if it could pick up two electrons from the right carbon,

it's in business. And, the right-handed carbon mimics the

left-handed carbon. So, hydrogen each donates an

electron. And now, let's check. The left carbon has two,

four, six, eight. So, its octet's stable. The right carbon: two,

four, six, eight, its octet's stable. And, the way to explain this is

there is some kind of a double bond here. So, let's draw the double

bond. And for now we'll put the hydrogens. Later on,

we'll get to the point where when we see carbon, the strut will assume

that the default setting is if I don't write anything at the end of

the strut, assume that it means that there is hydrogen.

And, here's something to always look for: when you see a carbon,

check to see there is four struts off the carbon.

Start looking. There's probably something missing.

So now, how are we going to explain this? We've got a double bond.

We already know: double bond, there's a tip off there.

That means we need a Sigma and a pi. But with four struts 109¡ apart,

there's no way you can form a double bond. So, this isn't going to work.

The sp3 hybridization will not work. But maybe there's something else we

can do if we want to make a double bond. If we want to make a double

bond, what we could do, perhaps, is to start with that same

box structure. And, what's the gambit here?

What I want to do is I want to save an un-hybridized p orbital because

the p orbital is your friend because only with the p orbital do you have

the capacity to smear laterally. And, if you don't have the capacity

to smear laterally, you can't make a pi bond.

So, let's do something here, but not burn all the p orbitals.

So, how about doing this: instead of making four, what if we made three

and save one of the p's? So, this is p. It's un-hybridized,

and these are hybridized. So, I'm going to take one of an s,

the only s I've got, and I'm only going to take two of the p's.

So, I'll call this sp2 hybridized. So, I've got sp2 and p. Now, let's

go through the geometric rules. If I have three struts

symmetrically disposed, because these are equivalent

energies, if they are symmetrically disposed in space,

they are going to describe, first they're going to lie in a

plane. And they're going to be 120¡ apart

lying in a plane. That's what this tells me.

But I've got a fourth one out here. Where's the fourth one going to be?

It can be normal to the plane. So, I've got three sp2's lying in a

plane, and I've got an un-hybridized p --

-- out of the plane.

Hey, this is starting to look good because what I could do is start

bringing these together, and maybe make something work.

So, here's another sketch of what this looks like.

You can see the three sp2 orbitals lying in a plane,

and then above and below this rather bloated dumbbell shape,

p orbital. And, there's the top view, all right?

Now, here's two of them opposing one another. They are

opposing one another. So, if these are the carbons,

I could take these two sp2 orbitals, blend them, and I could make a sigma

orbital. And, now what? I've still got these

unhybridized p orbitals. I can bring those in, and they'll

smear top and bottom. And now I've got a pi orbital of

sigma, and a pi gives me my double bond. So, by hybridizing sp2,

I can give rise to the setting that will allow the formation

of the double bond. So, there it is.

There is the sigma is in the center here, sp2 plus sp2,

and then the pi's give me the double bond. And, remember,

the two lobes together constitute a single pi bond.

So, that's great. So we're just going to bring these

two in. So, we'll take sp2 plus sp2 axially, axially will give a sigma

bond. And then p plus p laterally will give the pi bond,

and together this is the double bond. And then, the other sp2's are

hanging off the end, and they each pick up a hydrogen.

So, I can draw this molecule instead of this format that I've

drawn here. I could really draw it in the following manner.

Here's carbon-carbon, and then show these as sort of,

I'm trying to give the idea that this is 120¡ angle.

And, I'm going to leave the hydrogens off because you're a

sophisticated audience. I've told you once; I don't have to

tell you again. So, there it is.

There's ethylene. And, just for grins and chuckles,

you might want to see how to do one more form of hybridization.

If you want to do a really good job of welding, you need a very,

very energy intensive fuel; use acetylene. Acetylene is C2H2.

It's known as acetylene. And, if you go through the Lewis

structure, you will conclude that you've got a triple bond there.

And, we already saw last day, the triple bond in nitrogen was

almost 1 megajoule per mole. So, there's a lot of energy in here.

So, you've got your acetylene torch and your oxygen feed,

I said you need four struts. So what's that? There's a hydrogen,

hydrogen, and a triple bond in the center. How are you going to make a

triple bond? Well, let's leave two unmixed p orbitals.

So instead of having one just up and down, I could have one up and down,

and one in and out of the board. And, I can form a double bond.

And, what's left over, you can go through and rationalize that this

would require sp hybridization, sp plus unmixed p orbitals. This p

smears with its neighbor to give a pi, this p smears with its neighbor

to give a pi, and this smears axially to give a sigma.

So, one sigma plus two pi's - triple bond.

That's the whole story. Cool. So, what are we learning

here? There is a paradigm here, and that is that covalent bonds, a

couple of characteristics that I want you to note.

First of all, they are saturated. Remember that I said that ionic

bonds are unsaturated. And you said, OK, well,

don't tell me what they aren't, tell me what they are. Well, what

they aren't is in contrast to what these are. So,

now I can finish the story. What does it mean that they are

saturated? Once you put an electron from the hydrogen and an electron

from the carbon into the orbital, that's the end of the story.

We've taken care of business here. There is no revisiting the

situation, unlike in the case of the ionic compounds where the positive

attracts the negative, but it has the capacity to track yet

another negative, yet another negative,

and the feed is felt attenuating of course, but it is palpable out to

infinity. So, we say that the ionic bonds are

unsaturated. The covalent bonds are saturated.

Once you've got two electrons in the orbital, move on.

The second thing is, what you're seeing here is that they

are directional. That is to say, in the case of sp3

they are 109¡ apart off the central carbon, that's the only hook on

which you can hang another atom. So, if you make methane, if you

make carbon tetrachloride, if you make freon, they are all

going to be the same shape molecule: tetrahedral, because the struts come

off the central carbon. Look at this. Anything that gives

me sp2 hybridization is going to be what? What's the shape of that

molecule? It lies in a plane. It's planar. That's what it means

to lie in a plane, doesn't it? And, look at the shape.

There are always 120¡ apart. So, this means that there is a

relationship between the directionality of the bond and the

shape of the molecule. And, why are the bonds directional

in the first place? Because of electronic structure.

This is the theme of 3. 91. It's going to be echoed over

and over again. The electronic structure dictates

everything. It's just dominoes after this, once you understand

electronic structure. We did all of this for one reason,

to achieve octet stability. If I peel away the board,

and I look in behind here, there's two words: octet stability.

That's the only thing that's going on here. This system will do

anything to achieve octet stability. It's savage; it just does

anything. All right, so now what I want to do

is I want to make the proposal that the electronic structure will

dictate the shape of all molecules. So, let's put up a grand system.

And, that's what I want to do. I want to put that up and work

through it. Oh, here's some more cartoons.

This is taken from another book. So, you see the pi bond here. And

here's the sigma bond from the two sp2's. And that the remaining sp2's,

and this is supposed to indicate that this is coming out at you.

But this does not give me the sense of 120¡. This is crummy art.

But really, this is really crummy. I'm taking that down. It's

horrible. Now, this is what I've put up.

This is the rules for determining molecular shape.

And, in our text, it's known as electron domain theory.

If your classmates down the hall in 5.111 and 5.112,

instead of electron domain, it's the same idea. But, they use a

different term. Instead of electron domain theory,

they call it VSEPR, which is called vesper so that they can pronounce it,

vesper which stands for Valence Shell Electron Pair Repulsion.

And, all of this will become clear in the next 15 minutes.

It'll start all making sense. So, here's a set of rules. I

generated these. This isn't in the book.

And, all of this will be posted by mid afternoon.

So, you can take it down if you want, or you can just take it off

the web in a little bit. All right, so I want to go through

and illustrate how this is used. And, so what I want to do is I want

to use this in order to deduce, to infer, the molecular architecture

of three molecules. So, let's start with this series,

SF6, BrF5, and IF4 minus. So, these are three molecules.

I would say, OK, tell me what the shapes of these molecules,

what will it be? So, let's start with SF6.

SF6 is interesting. We're going to have some of that in the class later

on. I'm going to teach diffusion. SF6 is a very large molecule, very

heavy. It's molecular weight is about 150 grams per mole.

Its density is five times that of air. But, its main claim to fame is

that it's chemically rather inert. And so, it's used extensively in the

metal casting industry. If you look under the hood of the

car, you see aluminum castings, magnesium castings for things like

water pump housing and so on, those castings were made under the

protection of a blanket of sulfur hexafluoride because as we already

know, aluminum and magnesium are really good metals.

And they will react violently with oxygen in the air.

So, when you're pouring molten aluminum, you've got to keep the

oxygen in the air away from it. And they use SF6.

You could say, well, we could use a liquid flux,

but what if some of the liquid flux gets entrained in the casting?

And the casting becomes weak, the wheel cracks, there is a collision,

maybe people lose their lives, people are sued; so it's better to

use a gas phase shield. OK, but the bad news is that the SF

bond, which you can calculate the energy for, right now using the

Pauling formula, it sits smack dab in the middle of,

just off visible. And so, this is a fantastic infrared

absorber. And this stuff goes up into the upper atmosphere,

and has greenhouse gas coefficient that is phenomenal.

And so, the industry is under pressure to get rid of this.

So, if somebody it here wants to do something good for the environment

and allow us to continue to make castings of aluminum,

magnesium, and reduce the mass of the vehicles, but do so in a way

that doesn't damage the upper atmosphere, this is the trade-off.

Somebody in this room is liable to be sitting in policy someday.

And, what's a simple solution? If I tell you that the greenhouse

gas coefficient of SF6 is 12, 00 times that of CO2, what would you

do? Say, well, let's just ban it.

All right, well that's dumb. OK, that's really dumb because now

we don't have aluminum and magnesium castings. And so,

now the mass of the automobiles go up. So, what was an MIT person

doing? An MIT person would say, well, let's figure out, what's the

tailpipe emission trade-off to cleaning the atmosphere by not

allowing SF6 to go into the atmosphere? See,

you've got to know some chemistry. Otherwise you just make dumb

decisions. All right, that having been said, I wonder what

the structure of this molecule is. OK, so first thing we do is Lewis

structure. Let's write the Lewis structure. So,

I'm going to put sulfur in here, and sulfur's got six valence

electrons. It's got six valence electrons, right?

Sulfur is 2s2, 2p4. And, so let's bring up a fluorine.

Fluorine's got seven valence electrons. So,

there is one, two, three, four, five, six, seven.

So now, fluorine by sharing one electron with sulfur becomes octet

stable. And, we can keep doing this. I'm not going to put all of these

down. I'm just going to put a single electron from each fluorine.

You will know that there's more like this. And here's another

fluorine. Here's another fluorine. Here's another fluorine. And

here's another fluorine. So, this is great. All the

fluorines are octet stable. Now, let's see what's going on with

sulfur. It's got two, four, six, eight, ten, 12: it's got

12 electrons. That's some octet with 12, isn't it?

OK, well, let me tell you something. With some of the very

electronegative elements that have very, very high average valence

electron energies, they can form. And the ones that

are in the dark blue are elements that, on occasion,

can form expanded octets. These are called expanded octets.

And, the expanded octet is an attempt. If octet stability is good,

and I have a huge appetite for electrons, I just might try to pull

in a few more. So, what happens?

And, let's look at the energy level diagram of that.

If I've got, here's my 2s, and here's my 2p, so this is one,

two, three, four, five, six. There's my six valence electrons,

right? But, here, what do I have? I need to be able to form six pairs

of bonds. So, I need to hybridize.

The best I can do is hybridize sp3. That will give me four. I need to

get more. You know where I'm going to go? I'm going to go to the

orbital store. And, where's the orbital store?

It's over here: d. You say, well, wait a minute, we've got to bring

down the d's, OK? We'll bring down the d's,

and here's how we do it. There's one, two, three,

if I grab two of these d's, and I mix s's, p's, and these two

d's, that can give me this configuration: sp3,

and I'm taking two of the d's, sp3 2d, and then I've got three d's

left here unmixed. So, now, Hund rule,

six electrons: one, two, three, four, five, six.

So, this is how one can hybridize in order to allow for expanded octet,

right? And, also account for the bonds that we are forming here.

So, this is called electron domain theory. Each one of these orbitals

here, each one of these orbitals that has electron occupancy around

the center atom, each one of these is termed an

electron domain. And, there's two types of electron

domains, those that involve bonds, and those that do not involve bonds.

So, the six that I've circled are bonding domains.

So, these are all bonding domains. And, for example, this one here,

which I'll put a square around its off to the side.

It's a nonbonding domain. It doesn't get involved in bonding.

But it's not even part of the central atom. So, we'll

see that in a second. All right, so I've got six electron

domains, and now I want to put up the skeletal structure.

So, if I have my central sulfur here, how can I put six domains

symmetrically in space. The way I do that is I make

essentially this triple cross that constitutes the coordinate system,

x, y, and z. This gives me six domains in symmetric disposition

around a central atom. And, in this case, so,

six electron domains implies octahedral skeleton.

Skeleton of what? It's the skeleton of the molecule.

It's where the electrons can reside. And, why is this called octahedral?

You look at this, if you were to link these up in the plane,

so you link up four in the plane and then build a pyramid to the top,

can you see that the pyramid has four faces to it?

So, four faces on the up, four faces on the down; four plus

four is eight, so octahedral, eight-faced.

But, I look at that and I see the six struts. And I do this cognitive

discordance. What I'm saying: it's octahedral but I'm seeing six.

You just get used to it. So, this is octahedral.

And clearly we can put the fluorines at the ends of the struts.

So, there's SF6. That's the shape of the molecule.

That's the shape of the molecule because we have six bonding domains.

All the domains are engaged in bonding. So, they're all the same.

There's no distinction between them. We have an octahedral skeleton,

and in this case, we have an octahedral molecule.

Let's see, what else can we say about it? Polar or nonpolar?

What do you think? As soon as you see fluorine, it's just like in the

Manchurian Candidate. When someone says fluorine,

you go, ah, electronegative, most electronegative, right?

So, bingo, this is really super electronegative.

So, we know this is a very polar bond. But, as I showed you last day,

there is six symmetrically disposed polar bonds. So,

it's a nonpolar molecule. Terrific. All right, let's go on.

This is fun. Let's do the next one. Let's do bromine pentafluoride.

OK, let's look at BrF5. So, according to this,

let's go back to our rules here. Write the Lewis structure.

And that will give us electron distribution. So,

here's bromine. And it's got one, two, three, four, five, six, seven.

And, each fluorine is going to come up, same as last time.

Each fluorine will share one electron with bromine.

So, that will give fluorine its octet. It's two,

four, six, eight, and then I'm going to do the same.

There is the fluorine, one, two, three, another fluorine,

another fluorine, another fluorine, so, I've got one, two, three, four,

five. And then, I've got a pair of electrons that

are hanging around. So, bromine has shared one each of

five of its electrons with the fluorine. And then,

there is a pair here that is unshared.

OK, so first of all, let's count the number of electron

domains around the central bromine. So, I've got one, two, three, four,

five, six electron domains. But, we're not panicking now,

because we know that expanded octets exist, and it's OK.

All right, but there's a difference here. There's a difference.

So first of all, six electron domains, so that means,

again, octahedral skeleton. So, the electrons and the bonds are

going to be in this kind of a formation again with the

bromine at the center. But, there's one little distinction

here. And, that is that they are not all bonding domains.

You see, this pair over here is not involved in any bonding.

So, I've got five bonding domains, and one nonbonding domain. OK, so

the five are in an oval, and the one is in the rectangle here.

All right, and sometimes if you look in some old chemistry books,

nonbonding is one of these strange, it's clear it's a late comer to the

language. Sometimes people call these lone pairs.

They're lone pairs. Both electrons come from the

central atom. And, they haven't gotten picked up for

any kind of duty. So, they are alone.

They are lone pairs. All right, so now let's take a look

at the shape of the molecule. The shape of the molecule,

it turns out that it doesn't really matter. There is a symmetry here,

right? It doesn't matter if I have the five fluorines in the plane and

above, or the five fluorines underneath, and the lone pair up

here. Those are symmetric. One's just upside-down from the

other. There's nothing distinguishable.

So, just to get this thing over with, I'm going to put the fluorines

in the plane. OK, so this is the xy plane.

And then, the z direction is up here, and then this is the

nonbonding pair is off to one side here.

So, you don't see the electrons. If you try to detect to the shape

of this molecule, you simply see the atoms.

So, you'll see bromine in the center, and you'll see five

fluorines about it. And, what's going to be the shape

of this thing? Well, four of the fluorines lie in

a plane. And then, the fifth fluorine lies above the

plane dead center over the bromine. So, I can pitch a tent here.

And, what am I going to have? I'm going to have something that's

square based and a pyramid. So, the molecule is square

pyramidal molecule. The structure is still octahedral

for the electrons. But the molecule: square pyramidal

molecule. What else can we say about this? Oh,

how about, is this polar or nonpolar? It's polar. You can see there's

asymmetry here. There's extra electrons down on the

bottom the way I've drawn it. So that means that I could model is

as a dipole with the bottom part of it negative and the top part of it

positive, net neutral, or we could draw it this way.

I think we were going to put an arrow pointing in the direction of

the negative end of the dipole. So, that's good. Wow, we are

clicking. There's a piece of paper down here, it's got more information

on it. Let's get this one. So, let's do one more. It's so

much fun. That's do IF4, tetrafluoroiodate.

It's a complex ion. All right, so we're going to do the

same thing. We'll write the Lewis structure, and iodine one,

two, three, four, five, six, seven. OK, so here's fluorine one,

two, three, four, five, six, seven. So, by sharing one electron with

iodine, fluorine becomes octet stable. So, this one will do that

as well. You know, I've got enough room here.

I think I'm going to go for the gold. I'm going to put all the electrons

here. So, there's fluorine over here. Here's fluorine.

So, there's eight there. And, let's put another fluorine up

here. I'm going to move this just a little bit, raise this,

put this off to the side, IF4 minus, I'll put a fluorine here.

One, two, three, four, five, six, seven.

So, there's the four fluorines. And now, let's look at the

electrons around iodine. I gave one, two, three, four,

five, six, seven. But, you see, this thing's net negative. Where

did the net negative come from? How does something become net

negative? It acquires an electron through electron transfer.

This is really cool, because this is covalent bonding within something

that is engaged in electron transfer. So, to make this thing,

this complex ion net negative means that I had to throw one more

electron in here into the mix. And, where is it going to go?

Where do electrons live? They live in orbitals. And,

there is one orbital that's just waiting for it right here.

So, I'm going to put this in electron in here.

So now, what do I have? I've got one, two, three,

four bonding orbitals, four orbitals where I have one electron from the

iodine, and one electron from the fluorine. So let's write that down.

Four bonding domains, and I'm still talking about the central iodine.

Here's two of its electrons that had no place to go.

So that's one lone pair. Here's the seventh electron,

no place to go. And then, the net negative charge is an electron.

And, it's sitting in this. So, there's two nonbonding domains, for

a total of six. So, this means six electron domains.

So that means octahedral skeleton again. Now, question.

Where do I put the fluorines? Well, you say, put them at the end

somewhere. So, there's two possibilities now unlike

the other case. See, I've got this possibility.

This is door number one. I put the fluorines in the plane.

OK, door number two is the same electron structure.

I'll put the fluorines above and below, and then I'll put the

electron pairs next to one another. These are distinguishable,

agreed? So, what's it tell you? It says number three is maximized

separation between domains. Number four: give more space to

nonbonding domains. Let's think about the physics

behind that. Why? Electrons that are in nonbonding

domains are sitting there with no interaction with a nucleus,

whereas the ones that are in bonding domains, they are on this axis.

See, I've got two electrons sitting in here. But,

they are scrunched in, in between nuclei, whereas these two

are just hanging out there, just brandishing this negative

charge, you see? And here's another negative charge.

So, what is it going to try to do? It's going to try to get as far

apart as possible. So, that's why that rule is there.

Number four: so which of these two configurations keeps these farther

apart from one another, door number one or door number two?

Number one. One's up here, and one's up here.

These two are too close together. So, this is not what happens. And

in fact, this is the structure of tetrafluoroiodate.

So, what do we see? All we see is the atoms,

the iodine, the four fluorines; they lie in a plane.

And, what's the top view of this thing? It's just a simple,

regular cross. So, you have fluorines at four corners of a

square, iodine in the center. This is called square planar.

This is a square planar molecule. Polar or nonpolar?

It's nonpolar. This is cool. Look, it's net negative, but it's

nonpolar. So, that means that even though it has a

net negative charge, if I get really far away,

how do I model that thing? How do I model it? When I get

really far away, this is what it looks like: point

charge negative, see, way out here.

I don't care what the internal structure looks like.

I don't care about the electrons. This is all symmetric anyways.

It's very, very good. All right, so let's look at one

more. This is too much fun. All right, so what's the punch line

of this one here? You see, we had three,

SF6, BrF5, IF4 minus. They all had octahedral skeletons,

but the molecular architecture was different because of the relative

number of bonding and nonbonding domains.

But, they all started with octahedral skeleton.

Now I'm going to do something else. Now, I'm going to compare this

series that have the same formula. So, let's do three tetrafluorides:

CF4, SF4, and tetrafluoroiodate. See, they all have, it's F4. And,

I'm going to show you three different things.

Well, what's this one? You know this one in your head.

This is tetrahedral, right, this one here? Just write it by

inspection, boom, boom, boom, boom, that's done.

Finished, polar, nonpolar, fluorine, strongly electronegative,

but symmetric bonds, nonpolar. And this one, we just did.

So, if your memory is good, you know that this one's square


OK, now, so let's do SF4. What's SF4 look like? What does it

look like? It looks something like this. We start with the Lewis

structure. That's what it says up there. So, sulfur: one,

two, three, four, five, six. And fluorines: I'm going to put F,

F, and F. Actually, I know what's coming. I'm going to put the F over

here. So, there is seven fluorine electrons from fluorine,

and then one from the sulfur makes eight. So, we have a bonding domain

here, octet stability, and likewise, the same will happen

here, here, and here. So, what do we have?

Two, four, six, eight, and two of the electrons from the

sulfur are not involved in bonding. So, they constitute a nonbonding

pair. So, I've got two, four, six, eight, ten. I've got ten

electrons, or let's say, five electron domains.

There's five electron domains here. That's different. See, here,

there's four electron domains. Here, there's six electron domains,

four ED, and they're all bonding, and four bonding.

Here is six ED, and four bonding. And, this is five ED, five electron

domains, four bonding. Gee, they're all tetrafluoride,

and I've just proved that if you have four fluorines, you've

got four bonds. This is really consistent.

It's not terribly deep, but anyways, OK, so here we are.

Here we are. No, it's important that at the end of the day,

these things rationalize. OK, so we've got four bonding, and one

nonbonding: one nonbonding. So, if I have a central atom here,

how do I put five struts off it equally spaced?

So, that's going to mean one, two, three in the plane, one above,

and one below. That's five, OK? So, this is called,

the five struts, trigonal bipyramid. And, how do they get that? If you

look at the plane, you can join the three in the plane.

This is 120¡, agreed, if I'm looking down on this plane?

And now, what I can do is make a pyramid here. But,

this is a pyramid that has three faces. So, I've got,

this is called a trigonal pyramid. This is a trigonal pyramid, but

this would have above and below. So, it's a trigonal bipyramid.

That's a fancy name for something that's got five struts sticking out

of a simple atom, OK, trigonal bipyramid.

So, now, what are we going to do? We're going to, now, put the

fluorines. So, we've got two choices,

again. We've got sulfur in the center, one, two,

three, four, five. So, here's one possibility is to

put the nonbonding electrons in the plane with the other fluorines.

This is called the equatorial position. If you think of this as

the globe, there's North Pole, South Pole, and these three lie at

the equator, so this is called the equatorial position.

This is equatorial. So, that's one possibility. The other

possibility is to put the lone pair either at the North Pole

or the South Pole. And, if we use the same idea as

before, in other words, we try to minimize the interaction

between the unpaired electrons, excuse me, the nonbonding electrons,

and other atoms, which one is going to give this less contact with its

neighbors, door number one or door number two? See,

this one has actually your book even shows a little cartoon for this.

See, if you put it at equatorial position, it's got two nearest

fluorines, OK, whereas if you put it at a polar

location, it's got three nearest fluorines.

So, if we use that, this is called polar.

By that analysis, door number one should be minimal interaction.

And in fact that was one of the things that we were told to do.

See, place nonbonding domains, this is rule number five,

nonbonding domains at equatorial positions in a trigonal bipyramid.

That's what you say if someone asks you to say something into a

microphone. Just say that. And, so, that means the molecule is

going to look like this. And this molecule is called see-saw,

or teeter-totter because you don't see the nonbonding pair here.

Your characterization apparatus would find fluorine,

sulfur, fluorine, and then these two are underneath like

a sawhorse at 120¡. And, is this polar or nonpolar?

This is polar because I've got electrons over here.

So, I can model this as a dipole, or in this fashion. OK, so, this

has been really good. What else do we have here?

So, so far I've been doing this with heteronuclear molecules.

But, we can do this with homonuclear molecules. So,

I want to show you two conditions. Look at sp3, and we'll look at sp2.

All right, and I'm going to work with carbon alone,

no hydrogen. So, sp3, I start with a central carbon,

sp3 hybridized, and carbon goes everywhere. So,

each carbon has four nearest neighbors, and they're all at 109¡,

and they all have the same strut length, the same bond length,

one, two, three, four, and so on, carbon here, carbon up here, and

these are going in and out of the board.

But, I think you can see that one thing is certain.

There will be regular arrays in three dimensions,

regular arrays, regular 3D array. And, if you look at this from a

distance, what you will find is this is forming the diamond cubic

crystal structure. In this instance,

carbon is hybridized sp3. And, the consequence of that is

diamond cubic. Now, if carbon,

instead, hybridizes as sp2, we know that we have three bonds

identical lying in the plane. So, let's do that one. So, here's

carbon; carbon, carbon, carbon. And then, this one has one,

two, and three over to here. OK, and what are we seeing here?

We are seeing a hexagonal pattern, aren't we? This is just three of

the bonds. But I said, look, remember, you've got to keep

me honest here. Look at this carbon.

How many sticks off of the carbon? One, two, three. I told you, we'd

better have four. And, I've only got three.

So, what happened? Where's the last stick?

Well, I've taken this configuration over here, s and p,

and I've mixed them to get sp2. That means there must be some p

orbital here somewhere. And, where is it? Well,

it's sitting right here. There's a p orbital here,

and there's a p orbital here. And, there's a p orbital here,

and there's a p orbital here, and a p orbital here,

and p orbital here, and so on, and so on,

and so on. And, it turns out that the bond length is such that these p

orbitals are able to bond cooperatively, OK?

And so, they form p bonds, are cooperative. And, they're so

cooperative that the electrons are not associated with any particular

bond. We say that the electrons are delocalized. And,

we'll see a little bit more of that later, but we might as well learn it

now. They are delocalized. This is the structure of graphite.

This is the structure of graphite. And, what do we have here?

We have strong covalent bonds in the plane, and then we have these

weak, delocalized bonds normal to the plane so that the planes can

slide over one another, which is what gives graphite its

lubricity. That's why you squirt graphite powder into a lock in the

wintertime to give it some lubrication. So,

these glide over one another, and furthermore, it's not only

lubricity. You get electronic conduction. And,

graphite has a mildly decent electronic conductivity,

unlike diamond where all of the electrons are tightly bound,

very tightly bound. And by the way,

these bonds are very, very strong. The planes,

the bonds in the plane of graphite are extremely strong.

But, plane to plane, they glide. OK, so you can think of these

almost diamond-like rafts in terms of the bond strength.

But, these are flat, and these are three-dimensional.

The other thing that happens is, we have really, really tight bonds.

What's these optical properties here? This one is transparent to visible

light. Why? Because the electrons are tightly bound.

So, you've got this really weak photon, visible light,

two electron volt photon comes in, and it can't excite anything. But

the bonds are so strong that it's refracted. So,

one of the properties: this has a very high refractive index,

whereas here, electronic conduction means it's got very,

very low excitation, and hence, graphite is an absorber.

It appears black because any light that shines on it is absorbed.

See, everything comes from the electronic properties.

You can explain that its lubricity, and you can explain its electronic

conduction. And, you can explain its absorption of

photons. Now, I mentioned electronics here,

and the impact on photonics, you see? High index of refraction,

this is socially a very dangerous quantity because what happens is

this is what makes diamonds so precious. It's not just that it's

rare. It's how it functions in society. It takes a small amount of

diamond. You're just sitting; you're just minding your own

business. You just went out, you just want to have a drink,

you're sitting there, there's some gal sitting over in the corner,

this dimly lit café, she's got a diamond stud on.

That diamond stud takes a tiny, tiny amount of lumens, and thanks to

this insidious property, concentrates them and it shoots them

across, catches your eye, and then the trouble starts.

So, this sp3 hybridized carbon is very dangerous owing to this

property. Now, I draw your attention to the

periodic table. You know, it may happen that some

of you are going to feel compelled to make a long term commitment to

another person, and initiate that commitment with

the presentation of a stone, a stone typically made of diamond.

Well, I draw your attention to the fact that at room temperature,

the stable form of carbon is graphite. Now,

I want you to think about this. If you want to present a ring that

contains the stone that's symbolic of eternal love,

doesn't it make sense to present the stable form, not the

metastable form? I mean, I was just thinking about

this. So, my point is that this is the stable form.

I've done my part to explain it in terms of sp2 hybridization.

I leave it up to you to explain it to your sweetheart.

OK, have a nice day.

The Description of Lec 11 | MIT 3.091 Introduction to Solid State Chemistry