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5.111 Principles of Chemical Science, Fall 2005

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5.111 Principles of Chemical Science, Fall 2005 Transcript – Lecture 14

All right. Today we're going to start on a topic, one of the harder topics that we cover this semester. We're going to start talking about molecular orbital theory.

So this is a quantum mechanical description of the wave functions in a molecule. And we're going to use molecular orbital theory to look at the bonding in diatomic molecules. That is what we will do for the rest of today. We will look solely at diatomic molecules, and we will look mostly at homonuclear diatomic molecules.

At the end we will look at one heteronuclear diatomic molecule. And then on Monday we're actually going to use a different kind of approach to look at the chemical bonding in molecules that are larger than diatomic. So we're going to use molecular orbital theory for these diatomic molecules, and then on Monday we will go a little bit further.

We will look at bigger molecules. And we are going to use something else. We're going to use valence bond theory and hybridization ideas. But today we're going to deal with these diatomic molecules with the molecular orbital approach. All right? So here it goes. We're going to start talking about hydrogen, as we usually do when we start talking about chemical bonding. I am going to represent my hydrogen atom here by this nucleus here, a.

This is going to be hydrogen atom a. And I am going to represent the wave function of that hydrogen atom by a circle because, after all, the 1S wave function in hydrogen is spherically symmetric. I am going to represent it by a circle around the nucleus. So this is going to be 1Sa. And then I am going to, of course, have a second hydrogen atom represented here by this nucleus b and a circle around it because that wave function is spherically symmetric.

And what I am going to do in this molecular orbital theory approach is that I am going to take these two wave functions and I am going to bring them a distance of the bond length apart. In other words, I am going to take those two hydrogen atoms and I am going to put the two nuclei here, a and b, at a distance of a bond length apart.

And I am going to let these two wave functions overlap. And these are wave functions. And what we learned a few weeks ago is that waves constructively interfere or destructively interfere. What we're going to do first here is we're going to let these two waves constrictively interfere where they overlap.

When we do that, this part of the wave function here is going to constructively interfere with this part of this wave function. And then the sum of those two is going to look something like this where now there is this constructive interference between these two nuclei. There is a lot of wave function right here because we superimposed those two wave functions, each from one hydrogen atom. And so there is constructive interference here, there's a lot of wave function right there. And in so doing, what we're going to call this resultant wave function here, we are going to call that the molecular wave function or the molecular orbital.

And how we got it was simply a linear combination of atom orbitals. This is a linear combination of atomic orbitals. We just brought those two atoms close enough so that their wave functions overlapped, constructive interference, and the result is a molecular wave function. Now, we are going to give that molecular wave function a name. We're going to call it a sigma wave function. The sigma designates the symmetry of that wave function with respect to the bond axis. This is the bond axis right here. The sigma represents the fact that this is cylindrically symmetric about the molecular bond axis. That's what that sigma means.

And we are going to call this a sigma 1S orbital. It's a sigma 1S orbital because this was a linear combination of two 1S atomic orbitals. But now what I want to do is I want to remind you that the physically significant part of this wave function was actually the wave function squared.

Remember that? We talked about taking the wave function, squaring it. And that interpretation is a probability density. That was what was critical to our understanding of where the electrons are in a molecule. So that is what we're going to do here. We're going to take this sigma 1S molecular wave function now and we're going to square it. And that is then proportional to, or interpreted as a probability density. I will just abbreviate that by P, but this is a probability density, remember, not a probability. Well, if that is the case then I can multiply this out. Because sigma 1S right here, that sigma 1S is really equal to this linear combination of 1Sa, 1Sb.

That is sigma 1S. So let me take 1Sa plus 1Sb, multiply it by 1Sa plus 1Sb. That's what sigma 1S is. I'm just going to multiply this out. And when I do that I am going to get 1Sa squared plus 1Sb squared plus 2 times 1Sa, 1Sb.

That is my probability density. But now, to make this a little bit more real for you, let's plot it. I am going to plot the probability density as a function of the position along the bond axis. In this case here the nucleus a is here and the nucleus b is there.

So we have a molecule now. I am just going to plot the probability density as a function of the position along the internuclear axis. And what is it going to look like? Well, it's going to look like this. It is going to look something like that where the probability density is the highest right at nucleus a and the probability density is the highest right at nucleus b.

Because, after all, any wave function, an S wave function, the highest value of that wave function is right at r is equal to zero or right at the nucleus. So that is what that probability density looks like. And I have just been informed here that the Xeroxing on that particular figure did not turn out so well, but that is what it should read, if you cannot read it very well.

But the reason I plot this here is to contrast it with the following. Suppose I take just one hydrogen atom and the other hydrogen atom, but I take those wave functions and I just square them and then I add them up. Suppose I take a hydrogen atom with its wave function 1Sa.

I square it. And then I take another wave function for another hydrogen atom 1Sb. I square it. And I add those two up. In this case, what that is doing is taking two hydrogen atoms and just taking their wave function and squaring them. In this particular case, what I haven't done is I haven't allowed any interference to happen.

I haven't allowed the wave functions to constructively interfere. I did not add up these wave functions first before I squared them. I squared them and then I added them up. That is different than taking those wave functions, letting them constructively interfere and then squaring them. And you can see the difference is when I plot this sum and compare it to that sum.

So I am going to plot now this sum. When I do that it's going to look, well, pretty much like this. But then right at the center, right here, the probability density for this case, for two hydrogen atoms which are now nonbonded, they're nonbonded because I didn't let those two wave functions constructively interfere.

The result is that this probability density right in the center here, that probability density is less than what it would be if I let those two hydrogen atom wave functions constructively interfere. And that is the key in this chemical bond is this term right here.

It's this term that makes all of the difference. This is our constructive interference. This is the extra density that we get because those two wave functions have constructively interfered. That is what gives me this pile up of electron density right in the center here. This is the glue in this chemical bond. Make sense? OK.

All right. We have formed now a molecular wave function by this linear combination of atomic wave function. Now, what happens to the energy when we do that? Well, you can imagine what happens to the energy. The energy goes down. We are going to form a new molecular state. Let's draw an energy level diagram.

Energy going up here. And let me represent here, by these two lines, the energy levels of the separated hydrogen atoms. I am going to call this the 1S state on atom a, the 1S state on atom b. They're at the same energy. They are two hydrogen atoms. You cannot tell them apart. And each one of them has an electron.

Once we allow those two wave functions to constructively interfere, we're going to produce a new state. We're going to call that state sigma 1S. And that state is going to be lower in energy here than the two states from the individual hydrogen atoms. And that is good because if it is lower in energy, we've got these two hydrogen atoms stuck together.

We formed a chemical bond. These two hydrogen atoms are not going to fly apart. That's what we wanted to do. And, therefore, those two electrons here, one from each hydrogen atom are sitting now in this new molecular state that we are calling sigma 1S. And the wave function then that describes the electrons in that state is the sigma 1S molecular wave function. That's great. Everything looks cool, except now that these are waves. Not only can wave constructively interfere, but they can destructively interfere, too. So we have got to take care of that case now, the case of destructive interference. Let's do that.

In this case what is going to happen? Well, again, we have our 1Sa wave function and our 1Sb wave function. And now, when I am going to bring these two together to the equilibrium bond length ab here, what is going to happen is in destructive interference this wave function is going to cancel this part of the wave function.

When these overlap and I take this minus sign here or I am superimposing two wave functions with different sign, I am going to have destructive interference. And the result is that the molecular wave function is going to kind of look like this.

Right in the center here there is going to be a node. There is a node because this part of the wave function exactly cancelled this part of the wave function. Hey, there is no wave function right there because we have destructive interference. How are we going to label this wave function? Well, we are also going to call it a sigma wave function. It is sigma. That is the symmetry because it's cylindrically symmetric around the bond access.

But because it is a linear combination here that is a consequence of destructive interference, I am going to call this sigma star 1S. And I am going to call that sigma star 1S, which is an antibonding orbital or an antibonding wave function.

Now, how do I know that? Well, let's also calculate the probability density here for sigma 1S star. That is going to be interesting. We've got P, probability density, sigma 1S star, square it, so I have now a 1Sa minus 1Sb times 1Sa minus 1Sb.

Multiply that out so I get 1Sa quantity squared plus 1Sb quantity squared minus 2 times 1Sa times 1Sb. I am now going to plot this on my graph here. This is what I am plotting right on my graph of the probability density.

When I do that, well, way out here on one end of the molecule probability density is pretty low. Probability density goes to a maximum right at the nucleus and then it plummets. And right at the center here the probability density is zero because the wave function is zero.

Because we had destructive interference. I want you to notice something. Probability density here is zero. It is even lower than this case, which is just the sum of two hydrogen atoms, their wave functions squared. In the case when the two hydrogen atoms are not interacting, the probability density is some finite value.

But in the case of this antibonding state, the probability density is zero, goose egg, zip, nada, all of the other things. There is nothing there. This state here that we're forming is antibonding. It is not even nonbonding. Nonbonding you've got more electron density. Let me fix my microphone here. All right.

Nonbonding has more electron density than antibonding. This is antibonding against bonding. That is the kind of state we just formed by destructive interference. So where do you think, in an energy diagram, the sigma 1S star level is? Well, you can imagine where it is. It is increased in energy. It is up here, sigma 1S star. And it turns out that the amount of energy above the energy for the two separated hydrogen atoms that this state lies is about equal to the amount of energy below the separated atoms that this state lies. These two energies are about equivalent.

We now have a bonding state or a bonding orbital. We now have an antibonding state or an antibonding orbital. Now, when do we use that antibonding orbital? Well, we are going to see in a moment, but one case would be if we had an excited state of a hydrogen molecule. In an excited state what we could do is take one of these electrons.

And my eraser is kind of wet here so it is not erasing. Let me dry this. I could take one of these electrons here and populate it and put it in this state. That's the first excited state of H2, not H, H2. A second excited state would be, hey, this is permanent chalk once it gets wet. All right.

The second excited state would be taking that second electron and promoting it to this antibonding state. That would be the second excited state of molecular hydrogen. Now, I've got to tell you that if you have these kinds of excited states in molecular hydrogen, molecular hydrogen doesn't stick around. It is going to fall apart. But the ground state configuration of H2 has these two electrons here in the sigma 1S state.

I also just want to point out here what the significance physically is of the separation in energy between the hydrogen atom states and the molecular hydrogen state. Here is the significance. Remember we drew an energy of interaction for molecular hydrogen?

We talked about the general principles, the general attractions and repulsions that take place in every chemical bond, and the result was a curve that kind of looked like this where this is hydrogen, two separated hydrogen atoms, this is of course the equilibrium bond length of H2, the molecules bonded here everywhere where this energy is lower than the separated hydrogen atom limit.

Well, the physical significance here of this energy lowering of this state is, well, this is essentially delta E sub d, right? This is essentially the bond strength of molecular hydrogen. That is what that is physically. This energy from here to here, that's what it is going to take to pull the hydrogen atoms apart.

The bond energy. We took care of hydrogen. We have two molecular states, bonding and antibonding. For the ground state hydrogen, both electrons are in that bonding state. Now let's walk across the Periodic Table. Oh, one other thing.

If you were asked to write the electron configuration of molecular hydrogen in the ground state what would you do? Well, the electron configuration of molecular hydrogen then, the way we would write it is sigma 1S2, meaning two electrons in the sigma 1S state. And, just to be complete here, you could write this antibonding state as sigma 1S star with zero electrons in it.

That would be the electron configuration. Well, now what we are going to want to do is to write the electron configuration for the next molecular as we walk across the periodic table. The next diatomic molecule. Well, the next diatomic molecule is going to be He2. Where are the electrons in the case of He2? Well, in the case of helium, each helium atom brings in two electrons. Each helium atom has two electrons here in the 1S state, so I am now drawing this for helium.

And the way we draw this correlation diagram, as it is called, is following the Aufbau Principle that we had for filling atomic states. That is we take the states, we order them in terms of their energy, lowest energy at the bottom, next highest energy, and then we try to put electrons in those states one at a time starting at the bottom, starting at the lowest energy until we run out of electrons.

So that is what we're going to do for helium. What is going to happen? Two of these electrons are going to be in the bonding state and then two of these electrons are going to be in the antibonding state. And so the electron configuration here for He is (s1S)2 (s1S*)2. That's the electron configuration for helium dimer.

But this is interesting because, look, we have now, in this molecular state, two electrons that are at an energy that is lower than the two separated helium limit and two electrons that are higher. It seems like there wasn't an overall stabilization of this molecule because we have one state that is lower in energy and another state that is higher in energy by the same amount.

So the overall is zippo. And, indeed, that is the case. We've got a tool for counting or for telling how strong the bond is, and that tool is called the bond order.

The bond order is defined as one-half the number of bonding electrons minus the number of antibonding electrons. In the case of He2 we've got two bonding electrons and two antibody electrons.

In the case of He2 the bond order here is equal to zero. When you have a bond order equal to zero, what that means is that you have no bond. And, indeed, that is the case. He2 has no bond between them. Now, accurately speaking, He2 does have a bond between them.

He2 was discovered, so to speak, in 1993. And it was made under really very special conditions. And the bond strength of He2 is a whapping 0.01 kilojoules per mole. That is for He2. Whereas, for example, the bond strength in hydrogen is 432 kilojoules per mole.

So you can see how weak a bond that is. Effectively, this has no bond at all, practically speaking. And so molecular orbital theory really gets it right in the sense that it tells you there is no bond here. Great. All right.

Now what we're going to do is we're going to keep marching across the Periodic Table. Actually, we've got to go down one row now. And let's look at the electron configuration for lithium dimer. Now, if we're going to bring together two lithium atoms, lithium has two electrons that are in the 1S state and one electron each in a 2S state.

And let me put them in. Those are the electron configurations for the two separated lithium atoms. How do we write this electron configuration? Well, again, we follow the Aufbau Principle. We put two electrons here, two electrons here. But now, since we've got these 2S electrons, what we are going to have to do is to form molecular states, molecular orbitals between these 2S atomic orbitals. And the 2S atomic orbitals form these molecular states in the same way that the 1S atomic orbitals do. They're all spherically symmetric so they form these orbitals in absolutely the same way.

And so you get a sigma 2S bonding state and a sigma 2S star antibonding state. If you want to write the correlation diagram, as it is called, for Li2, you take all of lithium's electrons and you now put it in the lowest energy state, put it in the next highest energy state and put it in the next highest energy state. That is the correlation diagram.

We ran out of electrons. We had six electrons total. The electron configuration written here is this, (s1S)2 (s1S*)2 (s2S)2. What is the bond order here? Let's calculate it. We have two, four bonding electrons. We've got two antibonding electrons.

Four minus two divided by two is one. We've got a bond order of one. That means we've got a single bond. That bond strength is 105 kilojoules per mole. That is single bond. It's a weak single bond but that is a single bond. What about beryllium? Let's make Be2.

How did these two get up here? OK. These electrons are the electrons on the atom. The lithium atom has two 1S electrons and one 2S electron. And this is the energy ordering for the atomic states in lithium.

The molecular state for lithium here, the sigma 2S state, it is higher in energy than the sigma 1S star. This is the correct ordering of the energies of these states. I didn't clearly explain that to you. Does that make sense?

Yeah. OK. Now we will go to beryllium dimer. Now we've got two 2S electrons. Again, what do we do? We put two of the electrons in the lowest energy state, two in the next highest energy state, two in the next highest energy state and another two in the next highest energy state.

The electron configuration here of beryllium dimer (s1S)2 (s1S*)2 (s2S)2 (s2S*)2. The two is cut off here, but there is a two. What is the bond order? Zero. There is no bond here in Be2. Now, again, there is a very weak bond. It is 9 kilojoules per mole.

But, for all practical purposes, there is no bond between Be2. And this molecular orbital approach gets that right. Well, now boron. Boron has got a p electron in the atomic configuration. Now what we've got to do is look at the constructive and destructive interference between these p wave functions.

We've got to do that. We've got to make molecular states. We've got to let the p wave functions, the p atomic wave functions constructively and destructively interfere to make a new molecular wave function. You ready? Here is how we do it. Here is the 2px wave function or the 2py wave function. They look the same, right? One is just an x dimension. One is in the y dimension.

They've got a nodal plane right here. What we are going to do is take that wave function, overlap it with the same wave function, 2px or 2py from another atom. We're going to bring these two within the equilibrium bond length. We are going to let this part of the wave function overlap with this part of the wave function and this part of the wave function overlap with this.

And we are going to treat constructive interference first. And the result is a wave function that kind of looks like this. In the center right here we've got a lot of wave function because right here we've got constructive interference. In the center here we've got a lot of wave function. However, right along the internuclear axis we've got a nodal plane. Well, we've got a nodal plane there because we had a nodal plane in the atomic wave functions to begin with.

So that nodal plane is maintained here. I will try this again. OK. Now, this molecular wave function here, it is going to have a different name because it has a different symmetry. This wave function is not cylindrically symmetric around the bond axis.

There is wave function above the bond axis and wave function below the bond axis, but not on the sides here. It is not cylindrically symmetric so we have a new name for it. It is a pi wave function. And we are going to call it pi 2px or pi 2py. It's going to be a bonding wave function.

And, of course, since it is a bonding wave function, if we look at an energy level diagram, the pi 2px and pi 2py are going to be lower in energy than the atomic wave functions, just like this. And since the 2px and the 2py in the atomic states are at the same energy, the pi 2px and pi 2py, they are going to be at the same energy in the molecule.

Make sense? [LAUGHTER]

OK. Let's do B2. Boron has got one p electron. It also has some S electrons. The S states are down here. I didn't draw them in on this diagram. I am just treating right here the 2px electron.

What happens? Well, what happens is we are going to fill each one of these states with these p electrons. We are going to use Hunt's Rule again. Remember Hunt's Rule? It says that when you have states that are degenerate, that have the same energy the electrons fill such that one electron goes in each state before you can put a second electron in either one of those states.

So that is what we did. One electron has to go here. One electron has to go here. You cannot put two electrons in one of these states and leave the other one empty. That is not the lowest energy configuration. And you've got to put those electrons in with parallel spins, not opposite spins. That is not the lowest energy configuration. So that is the electron configuration here for B2.

(s1S)2 (s1S*)2 (s2S)2 (s2S*)2 and now (p2px)1 (p2py)1. Let's calculate the bond order here. The bond order is two, four, five, six bonding electrons. Two, four antibonding electrons. Six minus four over two is one. We've got a single bond 289 kilojoules per mole. That is a pretty good single bond.

A pretty strong single bond. That is B2. Let's do carbon. Well, carbon has got two 2p electrons. We follow the Aufbau Principle. C2. All four of those electrons then are going to go into these pi 2px and pi 2py states. Here is our electron configuration. We keep going.

Here is the (p2px)2 (p2py)2. Let's calculate the bond order. Number of bonding electrons, two, four, six, eight bonding electrons. Number of antibonding electrons, two, four. Eight minus four is four divided by two. Bond order is two. And we have a double bond. Let's look and see how strong that double bond is. 599 kilojoules per mole.

That's a pretty good double bond. Remember the double bond is shorter, the double bond is stronger? Our single bonds were at 300, 400 kilojoules. This is almost 600 kilojoules. That is a strong bond. And this molecular orbital picture tells you that you have got this double bond here by virtue of this bond order. Yes?

You don't, but I am just about to treat the 2pz. I will tell you what our convention is. Good question. Here comes the 2pz.

The 2pz atomic orbitals, by convention, we are going to put along the internuclear bond axis. In the case of the constructive interference of the 2pz electrons, in that case these 2p wave functions look like this. They were no longer coming in parallel to each other.

Now they're coming in head on. That is our convention. The 2pz is going to be along the bond axis. 3pz is going to be along the bond axis. We're going to bring these two atomic wave functions in, we're going to let this part overlap with that part, and the result is a wave function that looks like this.

Lots of electron density or wave function right in between the two nuclei. Here is a node. Here is a node. Well, these are nodes because there were nodes in the atomic wave functions to begin with. We are going to call this wave function the sigma 2pz. It is sigma because it is symmetric around the internuclear bond axis. So that is the symmetry. Call that sigma.

Now, what happens here? It turns out that this sigma state is a little higher in energy than the pi states in the next molecule. There are going to be some exceptions here in a moment. Nitrogen has one electron in each one of the p states.

Let's use the Aufbau Principle to fill them up. Well, there it is. Two here. Two here. Two there. The electron configuration is the 1S's and the (p2px)2 (p2py)2 and now (s2pz)2. Let's calculate the bond order in molecular nitrogen. Well, it's the number of bonding electrons.

Two, four, six, eight, ten minus the number of antibonding electrons, two, four, ten minus four is six divided by two, bond order three, and this means we've got a triple bond. Boy, this better be a whole lot stronger. And it is. If you look at the bond energy in molecular nitrogen, it is whapping 941 kilojoules per mole. We've got a triple bond. This is a strong bond.

This is short bond. Now, the next one is going to be molecular oxygen. We've taken care of all of the wave functions, the molecular wave functions that we can form from the 2p atomic wave functions using constructive interference. It is now time to form some molecular wave functions using destructive interference. Let's do that. Let's go back to 2px and 2py. They are the ones that come in parallel to each other. As we bring them in, and we're going do destructive interference, this part of the wave function is going to destructively interfere with this part. This part destructively interferes with that part.

The result is something that looks like this. We've got another node right along near the center. That is a node because of this destructive interference. We've got another node here where the plane of the molecular axis is. That is because these 2p wave functions had nodes in them to begin with. So that is what our wave function looks like. It has some wave function in this quadrant, in this quadrant, in this quadrant and this quadrant.

This is another high wave function because it is not symmetric around this bond axis. It is an antibody orbital. It is pi star because this was formed as a result of destructive interference. So we have a p*2px and a p*2py. Let's put those on an energy level diagram. If they are antibonding, the energies are above those of the atomic states.

Now, let's write the electron configuration for O2. Here it comes. There is the four 2p electrons for each oxygen atom. We are going to put two of them in the s2pz, two in the p2px and a p2py, and then we are going to follow Hunt's Rule.

And we are going to put one electron in the p*2px and one in the p*2py. We didn't put them both in the p*2px or both in the p*2py. And we put them in with parallel spins, not opposite spins. Thank you. You are going to ask me a question. Good question. I wanted you to know that.

Hey, here is a change. When we get to z=8 what happens? What happens is that they want to make this tough for you. What has happened is that this s2pz has dropped in energy compared to the p2px and p2py. At z=8 the energy ordering flips.

At z=7 for nitrogen p2px and p2py is lower in energy than s2pz. But for z=8, oxygen and above, the energy ordering flips. Sigma is lower in energy than the pi. That is something you have to know. It has to do with the fact that the s2ps has much more electron density in between the two nuclei.

That electron density is so high that when the nuclear charge is that of nitrogen plus 7e, the nuclear charge actually isn't large enough to support all that electron density. But when you get to z=8 it is and those energy levels flip. It is an exception.

I mean it is a change right there. It is one that you have to know. Here is the electron configuration for molecular nitrogen. If we go and we calculate here the bond order, what we will find is the bond order is two. That means we've got a double bond and that bond strength is 494 kilojoules per mole.

Now we have got to make one more molecular orbital, and that other molecular orbital is between the 2pz states and it is going to be destructive interference. When this part of the wave function overlaps with this part of the wave function, we're going to cancel some wave function. And the result is that we're going to have a node right here in the center. These are the two nuclei. Here is our nodal plane. We've got a nodal plane right here anywhere and one here from the atomic wave functions. This wave function is an antibonding wave function. It's s*2pz. And if you look at the energy level diagram, its energy, that antibonding state is higher in energy than sp2px and sp2py.

And so if we go to write the electron configuration for Ne2, well, all of the bonding and antibonding states are filled. The bond order in Ne2 is zero. There is no bond for all practical purposes in Ne2.