Transcript - Lecture 15

-- talking about molecular orbital theory.

And remember what the key idea was there? The key idea was to take the wave functions, the atomic wave functions, the wave functions on the atoms that are going to form this chemical bond.

Take those wave functions and let them overlap.

Let them constructively and destructively interfere. After all, they are waves. They are wave functions. They can constructively and destructively interfere. And we saw last time that when these atomic wave functions constructively interfered, the result was a molecular wave function that was a bonding wave function.

Consequently, it had a bonding state.

For example, in the case of oxygen, we had the 2s wave functions on each of the isolated oxygen atoms. When they overlap they can constructively interfere to form a sigma 2s bonding wave function.

And consequently a sigma 2s bonding state.

And there are two electrons in that. And you will remember that we labeled the molecular states with respect to the symmetry of the wave function around the molecular axis. That wave function was sigma if that symmetry was cylindrical around the molecular axis.

It was pi here in this case for molecular oxygen when the wave function was not cylindrically symmetric around the bond axis.

And we also saw what happened when these two wave functions come together. They overlap and they destructively interfere. When they destructively interfere, we form what we called an antibonding wave function.

We gave it a designation star.

In the case here of the 2s electrons or wave functions in oxygen, this formed the sigma 2s star wave function, the antibonding wave function. Its energy is higher than that of the atomic states for the two separated oxygen atoms.

The bonding state is lower in energy.

The antibonding state is higher in energy. And so this was for the 2s electrons. Below this lower energy, because they're more strongly bond, are the 1s atomic states and the molecular states formed by the constructive and destructive interference with the 1s wave function.

I didn't draw that. 2s then is higher in energy.

And finally here are the 2p states that are coming together in molecular oxygen forming a sigma 2pz and then two molecular states, pi 2py and a pi 2px.

And then in molecular oxygen we also saw that there was one electron each here in the pi star 2px and pi star 2py.

Now, I just wanted to say a few words here about the physical significance of these energies here.

This is the molecular orbital in the case of molecular oxygen here. This is the molecular state that is highest occupied. What we mean by highest occupied is that it is the state that is the highest lying in energy.

It has an electron in it that is the highest lying in energy.

What this binding energy means is, or what this energy means is that if I put in energy to pull this electron, the one that is highest lying in energy off, the energy from here to 02+ plus an electron, this physically is the ionization energy of 02.

Because the first ionization energy is always the energy required to take off the most weakly bond electron or the electron in the most highest lying energy state.

So that is the ionization energy of 02. That turns out to be about 12.01 electron volts. That is what that physically means.

In contrast here, if I were to pull an electron off of an oxygen atom, these are the oxygen atom states, so I have 0+ plus an electron, well, this is the ionization energy of oxygen.

This is a little bit larger than it is for molecular 02. This is 13.58 eV.

And you can see that here. These 2p states are lower lying in energy than the pi states in the molecular oxygen. This isn't in your notes, OK? And, likewise, if I were to look at the electron affinity of oxygen.

Now I am just going to draw here the p*2px, p*2py. And here are my atomic states. Here are the other two pi. Here is the sigma. Here is oxygen. If I were to look at the electron affinity of oxygen, molecular oxygen, so way up here I have an 02 molecule plus an electron.

Putting in an electron in, I would have to put it in that state. And so the energy difference here, from here to here is minus the electron affinity, which is, in the case of molecular oxygen, minus 0.045 eV.

I just wanted to give you a feeling for what these energies actually mean. We are not plotting them as a function of energy here.

But they physically mean something in terms of, for example, the ionization energy of molecular oxygen or the ionization energy of atomic oxygen.

You can see the relative orders of the energies of all of these different states. This kind of diagram that we have been drawing here sometimes is called a molecular orbital energy diagram.

Sometimes it is called the correlation diagram.

It is called the correlation diagram because it shows you, by the means of these kinds of dotted lines here, how the atomic states correlate to the molecular states. That is why it is called the correlation diagram, again.

It shows you how those atomic states correlate to the molecular states.

Questions? OK. If not then I want to proceed. There was one page on the lecture last time that I had not finished, so I want to talk about that right now.

And that was drawing the molecular orbital diagram, or the correlation diagram for a heteronuclear diatomic molecule.

And that is CO. In the case of the heteronuclear diatomic molecule, we proceed in the same way as we do for a homonuclear diatomic molecule.

That is we had the states of the individual atoms out here, and now I am just showing you the 2p states. 2s are down here. 1s is down here lower in energy, but I haven't drawn it all.

And what you also see is that the 2p states on the oxygen atom here are lower in energy than the 2p states on the carbon.

And that's right. Remember we talked about the fact that as the nuclear charge gets larger those atomic states are more strongly bond because the Coulomb interaction is greater?

So the atomic states of oxygen are lower in energy than those of carbon.

If you were asked to draw this on a homework problem or on an exam, you would have to show that the oxygen atomic states are lower in energy than those for carbon. But we proceed much in the same way.

That is the 2pz states here overlap.

Remember those are the states that, by definition, we are putting parallel or along the bond axis? So they overlap to form the s2pz wave function for the molecular.

And then the 2px and 2py are overlapping, those are the ones coming in like this, parallel to each other, to form the p2px and p2py.

And likewise you have the antibonding state. These wave functions can destructively interfere resulting in the antibonding p*2px, p*2py and s*2pz.

Now we've got all the states there, and now we proceed using the Aufbau Principle. We start populating them, lowest energy first until we use up all the electrons.

We have got six electrons, so we put two of them in that lowest energy state and then the other four here in these pi states.

And that then is the correlation diagram, or just the top part of it for this molecule CO. And if I had to write the electron configuration, here it is.

Here is the (s1s)2 (s1s*)2 (s2s)2 (s2s*)2 (s2pz)2 (p2px)2 (p2py)2.

Now what is the bond order? Remember that was a measure here of the strength of the bond. The bond order is one-half times the number of bonding electrons minus antibonding electrons.

Well, in this case it is three, right? Because we have two, four, six, eight, ten bonding electrons.

We have two, four antibonding electrons. Ten minus four is six divided by two is three. This is a triple bond. And the binding energy here is 162 kilojoules. This is the strongest bond known, this triple bond here between carbon and the oxygen.

Yes?

Most negative, right.

No, the lowest energy is always the more negative. Great. There is one other thing I want to point out here. Remember I talked about the relative ordering of sigma and pi, how in the case when z=8 the sigma level drops below the pi level?

Well, that is true.

But here we have CO where one of the atoms has z=8 and one of them is less than z=8. What do we do in that case? In that case we are going to put the sigma lower than the pi. We are going to follow the z=8 or greater rule when you have a choice there.

That is the heteronuclear diatomic molecule.

Now what we are going to do is leave molecular orbital theory. We are going to start talking about larger molecules than diatomic molecules. And we have another approach for these larger molecules.

And that is what is called a valence bond method along with a hybridization.

And I will explain that to you. You can use the molecular orbital approach for larger molecules. We are not going to do that. It doesn't always work for larger molecules. Let me show you just an example here when it doesn't work.

And that is the case of methane.

Here I have just drawn a carbon atom. I show you the 2px and the 2pz wave functions for a carbon. And in the molecular orbital approach what did we do? Well, what we did was let the wave function from one atom overlap with the wave function from the other to form a new molecular wave function.

And so that is what we would do if we started with a molecular orbital approach for methane.

We would let the 1s wave function overlap with the 2pz on the carbon, and over here the 1s wave function overlap with the 2pz on the carbon. But if we did that we are now stuck.

We are stuck because we don't have another electron up here to overlap.

We do not have another wave function to overlap with the hydrogen. The electron configuration of carbon is 2s, two electrons in it, and then one in one of the 2p wave orbitals, the other one in the other 2p state.

If you use this molecular orbital approach strictly on methane, what it would predict is that we could only make two bonds to carbon.

And it would say methane is really CH2 and the bond angle here is 90 degrees.

So molecular orbital theory, at least on methane, doesn't work, although it does work for some other polyatomic molecules. That was just an example here of how it doesn't work.

Now we are going to take an approach that will work to describe the chemical bonding in methane, and that is this hybridization coupled with what we call the valence bond approach.

Here is the important part. Here is the atomic state for carbon. Notice two 2s electrons and then one in each of the 2p states.

The first thing that we are going to do is we are going to do an electron promotion.

We are going to take this electron and put it up there. That is electron promotion. That costs energy. Where did the energy come from? I am going to tell you soon where the energy comes from. We're just going to do it now, and then we will go back and look at where the energy comes from.

What we are then going to do is draw out here each one of these atomic wave functions.

I am drawing this atomic wave function, this one, this one and this one on the next slide. That is what this is. Here is the 2s. That is spherical. Here is the 2pz we looked at before. Here is the 2ps.

Here is the 2py. Those are the four atomic wave functions.

Now what I am going to do, and this is all on carbon, we haven't formed any bonds yet, is I am going to take these four wave functions. They are all on the same carbon, but I am going to let those four wave functions constructively and destructively interfere.

Now I am taking the wave functions on the same atom and letting them constructively and destructively interfere.

In molecular orbital theory, we took the wave functions on different atoms and let them constructively and destructively interfere.

This is going to be taking the wave functions on the same atom and letting them constructively and destructively interfere. When I do this, what you are going to get is something that looks like this.

You are going to get four different wave functions that we are going to call the sp3 wave functions or the sp3 orbitals.

More precisely, they are the four sp3 hybrid wave functions or hybrid orbitals.

Because they are this linear combination, the overlap of these four different wave functions on the same carbon atom to make four equivalent hybrid wave functions that we call sp3 wave functions.

They all look the same. They are just oriented in space a little differently. To understand that, and it is not clear, I know, from looking at this, how that gets like this.

But here is one simple way to look at one of these.

Take this 2s wave function and let it constructively and destructively interfere with the 2pz wave function, so this sphere is essentially on top of here. What you notice on the 2pz wave function is that this part of the wave function has got a positive amplitude.

Remember we talked about the positive sign on one of these lobes meaning it had a positive amplitude? And this part of the wave function, that has got a negative sign, it has a negative amplitude.

But the 2s wave function here has a positive amplitude everywhere. If we put the 2s on top of here, you can see that right here we are going to have two positives, amplitudes of two waves that are positive.

They are going to constructively interfere and the result is above here, right here, if you just take these two and look at this you are going to have a lot of wave function right up there.

However, below this plane where we have a positive wave function at the same part where we have a negative wave function, well, you know what, this is destructive interference. And the result is that the wave function here is going to be pretty small.

It is still going to be a little negative but it is going to be pretty small.

That is probably the simplest just example to look at so that you understand how I get from here to here. So we are going to form these four sp3 wave functions. Here is a picture of that. When I take these four wave functions.

I have just separated them out in space, but now I am really just going to put them on top of each other because they really all have the same origin here.

When I do that my wave function kind of looks like this where these lobes here are the positive parts. They are my sp3 wave function. And in this picture, I left out the little negative lobes that would be sticking out here.

I left those out because it is just too hard to draw this.

And what I want you to notice here is that this angle from here to here is 109.5 degrees. This is a little bit better picture that I didn't draw of those sp3 wave functions.

In this case you can see a little bit of the negative wave function from this positive lobe over here, and you can kind of see the little negative here and the little negative there.

That is a better picture than the one I drew. We have got this kind of configuration here with an angle of 109.5 from this lobe to that lobe or this lobe to that lobe or this lobe to that lobe.

And you can understand that 109.5 degrees if I do the following, that is if I now draw lines from the outer most point of each lobe.

What you see is that I have a tetrahedron now with carbon in the center. And each one of these vertices here of this tetrahedron is at the outer most points of these wave functions.

The angle in a perfect tetrahedron is 109.5 degrees.

That is where that 109.5 degrees comes from. That is a tetrahedral configuration around that carbon. What did we do here? We took an electron.

We promoted it. When then took these four wave functions and let them constructively and destructively interfere.

The result now is four new wave functions corresponding to four new states. Recalling these states sp3 states. There is one electron in each one of those sp3 states. Where do these energies lie?

Well, they lie in between the 2s and the 2p.

They are not as high in energy as the 2p but the certainly are higher than the 2s. And, overall, you can see we are still going to have to need some energy input into the system to make this happen.

And I haven't told you yet where that energy comes from.

Here we have this carbon. And we have just let those wave functions on the carbon itself constructively and destructively interfere. Now we are going to make bonds.

Now we are going to take these hybrid wave functions and let them overlap with a wave function, an atomic function from another atom.

Now we are going to start to make the bonds. Here it goes. We are bringing in a hydrogen, making a bond, bringing in a hydrogen, making a bond, bringing in a hydrogen, bringing in another hydrogen, and now we've got methane.

What did we do to do this? What we did is we took that sp3 wave function on carbon and then brought in the hydrogen atom with its 1s wave function.

And we let it overlap. Now there is constructive interference right in here, or destructive interference.

And the result is now a new bond. A result is this constructive interference between this sp3 wave function and the 1s wave function. We are going to call that a sigma bond. It is sigma because that wave function is symmetric along the CH axis.

It is cylindrically symmetric along the CH axis.

So this is a new bond. That bond is the sigma bond composed of the C 2sp3 and the H 1s. And so we've got two electrons now in each one of those sp3 states.

Now the question is where did this energy come from for that hybridization? Remember we had to promote an electron to do this trick here? So where did this come from? Well, where it comes from is the bond formation.

When you form a carbon-hydrogen bond, you are going to release some energy. It is lower energy.

You are going to release some energy. And that energy from the hybridization comes from that bond formation.

In other words, when you make the CH bond, that is going to be an exothermic process, you are going to release energy, but because you had to do this electron promotion the amount of energy you are going to release when you make that CH bond is not all that you could have released if you didn't have to do this hybridization, if you didn't have to do this electron promotion.

So that energy, to promote the electron, comes from the actual CH bond formation.

Now, here is my carbon with just the sp3 wave functions. I haven't gotten that bonded to anything in this picture.

What I am going to do is I am going to rotate this around, and it will be for a reason.

I just rotated it around. And one electron in each one of the sp3 states. Now what I am going to do is I am going to bring in another sp3 hybridized carbon along this axis here. That is why I turned this around.

Here comes the other sp3 hybridized carbon.

And I am going to let these two sp3 hybrid wave functions overlap so that now one of my sp3 states has two electrons in it. What am I doing right here? I am letting those two sp3 wave functions overlap, constructive interference.

I am going to make another bond, a sigma bond because this wave function is cylindrically symmetric around that axis.

That is a sigma bond. It is a sigma bond composed of the carbon sp3 hybrid wave function and another carbon sp3 hybrid wave function.

I just formed a carbon-carbon bond. I just brought in some hydrogens.

In each case, I now have overlap between the 1s wave function and the sp3 wave function of the carbon, and the molecule that I have is ethane.

And this angle here is 109 degrees, right? It is 109 degrees because you had that tetrahedral configuration around the carbon.

Because that is how the 2s wave functions and the three 2p wave functions around the carbon constructively and destructively interfered to give you that tetrahedral shape for the sp3 wave functions.

Oh, I thought we were going to do this more closely. Sorry. All right. We made these CH bonds here. They are all sigma bonds also.

That is this sp3 hybridization for carbon, but other atoms can also undergo this sp3 hybridization.

And one of those atoms here is hydrogen. Here is the electron configuration for nitrogen. In this case, we have got two electrons in the 2s state and then one in each one of the 2p states.

What we are going to do here is allow now these four wave functions to constructively and destructively interfere, do what we call hybridize.

And the result will be, again, four sp3 wave functions.

Except the difference is that because we have one more electron in nitrogen, one of these sp3 states has already got two electrons in it. It is not going to be available for bonding in a moment.

In nitrogen, the electron arrangement, or the wave function arrangement is also tetrahedral, sp3 to this corner, sp3, sp3, sp3.

Except that one of these sp3 wave functions really is kind of two sp3 wave functions.

There are two electrons here in this state. This is the lone pair on the nitrogen, and then sp3 hybridized nitrogen atom. This is the lone pair pointed at one of these vertices, but then each one of these lobes just represents one electron.

So they are going to be available for bonding.

What are we going to do here? We are going to bring in some hydrogens. Here is a hydrogen. Here is a hydrogen coming in. Here is a hydrogen. We have got ammonia. What did we do to form these nitrogen-hydrogen bonds? Well, we took that sp3 wave function on nitrogen and brought in the hydrogen, let that 1s wave function constructively and destructively interfere with that sp3 wave function.

The result is a sigma bond.

It is sigma. It is composed of the N 2sp3 and the hydrogen 1s wave function. Now, the angle right in here, the hydrogen-nitrogen-hydrogen angle here, I want you to see that it is actually a little smaller than the ideal tetrahedral angle.

It is no longer 109.5 degrees.

It is 107 degrees. And the reason for that is because of this lone pair out here. This lone pair has a very repulsive interaction with the electrons in these nitrogen-hydrogen bonds here bringing that angle in a little bit more close.

They are being repelled.

And so that will generally be the case. If you've got a lone pair of electrons somewhere, that will tend to repel, everything will be repelled from it. And the consequence is that this bond angle here is going to be a little smaller than a tetrahedral angle of 109.5 degrees.

But now here is a really important point.

You are often asked for what is the geometry around some atom? In this case, you might be asked for what is the geometry around nitrogen? When you are asked such a question, what it means is they are asking you, what is the geometry of the atoms around the nitrogen? Not the electrons.

The distinction is that the geometry of the electrons around this atom is about tetrahedral.

But usually one isn't interested in that. Usually one is interested in what is the geometry of the other atoms around the nitrogen. And so, in this case here, the geometry of the other atoms is a trigonal pyramidal geometry.

It is not tetrahedral.

If you are asked for the shape of the nitrogen molecule, the answer is a trigonal pyramid, not a tetrahedron. That is really important. That trips up a lot of people. So the lone pairs pointing out here, the electron geometry is approximately tetetrahedral, but that is not what we are usually interested in.

We are interested in where the atoms are, so the geometry is a trigonal pyramid.

That is the sp3 hybridization on ammonia or on nitrogen. Oxygen can also undergo this sp3 hybridization. Here is the electron configuration for an oxygen atom, two electrons in the 2s and four in the 2p states.

What we are going to here again is we are going to let these wave functions constructively and destructively interfere.

When we do that, again, we are going to get four new wave functions, the sp3 hybrid wave functions. But in the case of oxygen, because we have one more electron than that of nitrogen, we are going to have two electrons in that sp3 state and two electrons in the other sp3 state.

And then one electron each in the sp3 states.

What is the electron arrangement around oxygen going to look like? Well, it is going to look like this where, again, each one of these sp3 wave functions is pointed to, is there a question over here that I can help somebody with? OK.

Each one of these wave functions is pointed to a corner of this tetrahedron.

But, again, one of these wave functions represents two electrons, this lone pair, another one represents two electrons, that lone pair, and then only these two sp3 wave functions are going to be available for overlap with another atom.

Let's bring another atom in.

Hydrogen is our atom of choice today. We have just made the water molecule. What did we do? We took that sp3 wave function on oxygen and brought in a hydrogen, overlap them and formed a sigma bond.

A sigma bond between oxygen sp3 and hydrogen 1s. That is a sigma bond.

Now, notice this angle here. It is 104.5 degrees. It is even smaller than 107 degrees. And 107 was smaller than the tetrahedral angle of 109.5.

Why? Well, because we got these two lone pairs here of electrons. They're sticking out there. They really are repulsive to all the atoms around them.

And so all the other atoms, these two hydrogens here really try to get away from these two lone pairs.

And so this angle here, the hydrogen-oxygen-hydrogen angle is even smaller than that in ammonia and much smaller than that in an ideal tetrahedron. It is 104.5 degrees.

If you were asked for the geometry of this molecule, what would you say? Sorry? Bent.

All right, it is bent. It is also planer, right? Hey, this is a no-brainer. If you've got three atoms, the molecule is planer. Three points to find a plane. And then the question is whether or not those three atoms are linear, in a line, or whether they are bent?

And so, in this case, for the water molecule, this is a bent planer arrangement.

That is the atom arrangement. That is not the electron arrangement. The electron arrangement is approximately tetrahedral. But, again, usually you're going to be asked for the atomic arrangement.

The geometry around the oxygen, well, this is a planer bent molecule with the two lone pairs sticking out here.

And, hey, you know what? I want you to remember these two lone pairs on water. Very important because I think Wednesday we are going to talk about something called hydrogen bonding.

And these two lone pairs play a very critical role in hydrogen bonding. We have just covered this sp3 hybridization.

There are other ways in which the s wave functions and the p wave functions can constructively and destructively interfere.

And to illustrate that, I am going to start with this atom boron, one less electron than carbon. So a different kind of a hybridization now.

In the case of boron, we are also going to have to do this electron promotion.

We just move that 2s electron up to a 2p state, so now we've got three wave functions, 2s, 2p, 2px, 2py. We've got three populated states, and we are going to let them hybridize.

And when we let them hybridize, what we are going to get are three what we call sp2 wave functions each with one electron.

Let me try to explain that. These are the wave functions that we've got after this electron promotion, one wave function for that one electron in the 2s state on boron, it's spherically symmetric.

One 2px.

One electron represented by the 2py wave function, here it is. What we are going to do now is to let these three wave functions, not four like in sp3 hybridization, but these three wave functions constructively and destructively interfere.

And when we do that we are going to get what we call sp2 wave functions.

Or, sp2 hybrid wave functions or hybrid orbitals. They all look the same. They are just oriented differently. These three wave functions all lie in one plane.

The sp2 all lies in one plane, unlike sp3 hybrid wave functions. And so what I am going to do is take these three wave functions now.

Since they all have the same origin, I am going to plot them on the same graph, and they are going to look like this.

Again, I am going to leave out this little negative lobe here, little negative part of the wave function. Just for clarity in this drawing, I just show you the positive part of the wave function. And the positive parts of those wave functions lie in one plane.

And the angle between these wave functions is 120 degrees, each one of them.

This is the sp2 hybridization on boron, one electron in each one of the sp2 states. I am going to bring in a hydrogen now, bring in another hydrogen and bring in another hydrogen. Hey, I've got BH3.

So what did we do? Well, with each one of these hydrogens we made a sigma bond.

It is sigma because it is cylindrically symmetric around that axis. It is a sigma bond composed of boron 2sp2 overlapping with a hydrogen 1s bond. This molecule here is planer because these hybrid wave functions all lie in the same plane.

We call this trigonal planer because there is an angle of 120 degrees between each hydrogen-boron-hydrogen, or these hydrogen-boron-hydrogen angles are all 120 degrees.

It is trigonal planer. Well, carbon, it can also undergo this sp2 hybridization.

It can undergo sp3, but it can also undergo sp2 hybridization. We have to do an electron promotion again. We did that.

We are going to let them hybridize to sp2. The result is three sp2 wave functions and one of the 2p wave functions is left alone. We are going to label it 2py. That one is not going to participate in this hybridization.

A picture is here are the individual atomic wave functions on carbon, and I am going to let just these three hybridize to make three sp2 hybrid orbits, or three sp2 hybrid wave functions.

And they have identically the same shapes as they did for boron.

And then this guy hasn't been touched. It is still the atomic wave function, 2py. Yes?

Because you've got different kinds of environments.

Some environments allow these three wave functions to constructively and destructively appear. Others will allow all the four to constructively and destructively be interfered.

Thermodynamics, OK? Whether you're going to form here ethane or ethylene is a function of a lot of other parameters.

No, until we get to thermodynamics and then you are. We will get to that, OK? OK. You are quite welcome. That is all right. Because I know this is a little bit abstract, but it is actually the case.

Which one you're going to form, ethane or ethylene depends on what other constituents you've got around.

It is going to depend on the relative pressures of your carbon and your hydrogen. In this case here, because we have the same origin, I am now going to put these three wave functions on top of each other.

Plot them in the same graph.

They are going to look like they did in the case of boron. I think we will pick it up there on Wednesday. See you then.

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