Okay.
Good morning everybody.
Thanks for coming. I love it when people say good morning back.
We're going to change the dynamic here.
Okay,
cuz I know that's kind of like very stiff and proper.
I'm not, so the good news is,
maybe we'll have a little bit of fun.
The bad news is when you go to watch these videos back online, the ones with me or going to be all deleted out,
you know,
it's going to be just one big like beep and... [Laughters]
but that's okay.
We'll still have fun.
So we're going to pick up where Matt left off last time talking about and introducing the concept of molecular orbital theory.
We're going to get a little bit more complicated and work up to those those diatomic molecules that you learned way back in freshman chemistry.
So yes,
we're going to just keep rehashing things that you already know,
but maybe put a little bit of a deeper spin on it so that you have a better appreciation for how molecular orbital theory works
and that'll really help you next week when Matt comes back and takes you within the larger molecules
and in the second and a half of a quarter when I take over and we really start looking at transition metal complexes
things are going to going to go kind of haywire on the mo theory front.
Okay.
It's going to get a lot more complicated.
So it's really really important that you get a bulletproof understanding of these concepts with simple diatomics now
so that when we get to the crazy molecules, it's not quite so daunting.
So last time, okay, lots of animations here.
It's great last time, Matt introduced the... I'm sorry...Professor Law [Laughters]
Introduced the concept of molecular orbital theory to you again.
Now the important thing that I want you guys to always remember
is molecular orbital theory is a model that we use to describe the electronic structure of molecules.
It's one model. There are many other kinds of models.
Okay.
Do you want to think about it?
I like analogies.
So Legos are one way of building models of cars are spaceships or whatever.
Okay,
you can also go to the store and buy those little plastic models that look like a Ford Mustang that's a different type of model.
Okay.
Each model is a little bit different.
It makes different approximation.
So Mo theory is one type of model that we use to describe the electronic structure of molecules.
It's not the exact electronic structure.
We're not that smart.
Okay,
that's really what it comes down to we can solve the exact electronic structure for basically two cases.
The hydrogen atom, say we can solve the Schrodinger equation cuz there's one electron,
and we can solve the Schrodinger equation for the H2 cation so we can put two nuclei and remove one electron
solve the Schrodinger equation again. Anything else we can't solve the Schrodinger equation.
So we come up with these cheesy models to help us approximate.
What's great about them is they allow us to understand trends.
They often are pretty good until you get to ask me really fine detail questions,
but MO theory is one model, and now within that model of MO theory, the specific type of MO theory
that we're going to talk about is the linear combination of atomic orbitals (LCAO) approach.
Okay,
So when we build our molecular orbitals,
We're going to use the atomic orbitals that we're all familiar with: the s orbitals, the p orbitals, the d orbitals.
We can all draw these in our sleep. [Laughters]
And so it's what we call an easy basis.
Do you want to go back to the Lego model?
It's an easy set of bricks that we all understand, put together to build more complicated structures.
Okay.
So rule number one when we're doing this linear combination of atomic orbitals approach.
As Professor Law introduced last time when we say a LCAO sounds really complicated.
It just means we're taking orbitals and adding them together in different combinations
linear combination means you're adding them together.
If you start out with two Atomic orbitals,
like two s orbitals.
You're going to combine them in a linear fashion and you're going to get two molecular orbitals.
In this case.
We get the in phase combination that we saw last time that we call a sigma bond
and we get the out of phase combination that we call the sigma star or the anti-bonding combination.
We saw last time how these are only approximation of the picture.
Really.
What happens is you get a building up of electron density on this bond, the intranuclear separation between the two nuclei.
Okay,
the reason that we get a buildup of electron density there is because, remember, these orbitals are wavefunctions
When you add waves together so that they're in Phase.
They reinforce each other the amplitude of the wave gets larger.
And so that's what we call increasing electron density in between.
If you add two waves together and they're out of phase you get destructive interference,
ah we know wave right?
So if you add a positive wave together with a negative wave,
but they're exactly out of phase, they cancel each other out and get destructive interference.
You get a node.
That's what we call an anti-bonding orbital.
There's zero probability of the electron being in between the two nuclei.
Okay.
The math Professor Law went through it a little bit last time.
So,
the bonding orbital Matt went through the the math a little bit that describes this
the bonding orbital gets stabilized by less energy than the anti bonding orbital gets destabilized.
That's always true.
Go in a lot of molecular orbital diagrams,
you'll lose that information just because they get complicated and it's hard to draw it in
but you always have to remember that this bonding orbital is stabilized by less than the anti-bonding orbitals destabilize.
There's another consequence that was touched down briefly last time.
The bonding orbital is always smaller than the anti bonding orbital.
The anti bonding orbital is always bigger and more diffuse than the bonding orbital.
and both the energy effect and the size effect come out of the overlapping integral that S squared term [?]
that was introduced last time when Matt went through the math.
Okay,
so rule number two.
If the atomic orbitals are degenerate.
Their interaction is proportional to the overlapping integral.
Okay. Now,
Again,
qualitatively,
this should just make sense to you.
We would have a hydrogen atom over here and a hydrogen atom over here.
And we said how strong is the bond between them?
It's a pretty darn weak.
They're too far apart.
But if we bring them right next to each other at the distance that two hydrogen atoms form a bond.
0.8 Angstroms
well now all of a sudden the orbitals are in close contact you get a good overlap in the bond become strong.
Okay.
Again,
this is like I said,
I like analogies long-distance relationships,
right?
If you're close together,
it's easy to sticking out strong bond.
Everything's great.
You know,
you're far apart.
Well,
okay,
we know how long distance relationships usually work out that bond becomes weak,
all kinds of other things cause problems. But it's the overlapping integral.
The closer these orbitals are together.
The stronger the interaction.
So really what we're doing there is we're solving this sophisticated integral that relies on two-way functions
that you've seen the equations for before and if you ever tried to do something with them,
they gave you nightmares,
at least they gave me nightmares trying to do the integrals and everything.
That's why we like symmetry.
So we spent all that time teaching you point groups and symmetry operations because
group theory gives us the incredible power to predict when they overlap integral is exactly zero.
That's really the whole point of why we did symmetry.
Because you can now predict when the overlap integral between two orbitals is zero without actually solving the integral
without actually looking at the wave function for the orbitals and doing all kinds of crazy math.
If the two orbitals belong to the same irreducible representation
S is not zero.
If they belong to different irreducible representations,
S is zero all the time
You can't form bonds between orbitals that are parts of different irreducible representations.
It's fantastic - doesn't really matter that much for diatomics because you know more or less were talking about 8 [?] orbitals.
You can figure it out faster than you can do the symmetry,
but when you're talking about something like a porphyrin molecule.
And there's 25 atoms and each atom brings at least four orbitals and you're trying to sort all these things out
symmetry can make your life a lot easier and it's all based on this idea
about the orbitals belonging to the same irreducible representations.
We can look at these.
The idea of the overlapping integrals through some pictures.
Again, this is just what we were talking about right here.
So I actually don't like these graphs because they on the x-axis you have the internuclear separation and on the y-axis,
you have the overlapping integral and so you can see they start here at 0,
or you have an internuclear distance of 0, and that means the two nuclei right on top of each other.
Well,
this is showing you that the overlap integral.
That's the highest that should be the best,
right?
Clearly if the overlapping integrals = 1,
you're getting perfect overlap.
What's the problem?
Yeah,
right two positively charged nuclei are not going to want to be in the same spot.
Okay,
So you never get a diatomic molecule where the two nuclei are right on top of each other
because of the positive charged nuclei are going to repel each other.
So well,
it is true that the overlap maximizes at a distance of 0.
It's not a realistic picture of of what the bonding surface looks like.
Same thing happens with two parallel pi orbitals,
but then here we can see on the bottom half these two pictures show us
what happens when we have orbitals that belong to different representations.
Okay,
if we were to go through and do the symmetry on a diatomic molecule,
we would find that this I guess this is the px orbital and this is the pz orbital.
Okay, for a diatomic molecule those will always be parts of different representations.
We know because they're because they're different irreducible representations the overlapping integral = 0.  Pictorially, we can look at it.
We can say okay,
this is unshaded.
It's going to interact in a positive way here in a negative way.
There's going to cancel out.
So you can do it with pictures.
You can do it with this rule number three.
You can solve the integral if you're you know,
one of these overachiever types.
That's never been me.
So I like cartoons.
Okay.
Same thing here when we're talking about the overlapping integral
there are going to be cases and we're going to do is you're going to get into these next week and we'll really get into them
later in the quarter when we deal with transition metal complexes
where it's not exactly clear cut how the two orbitals are interacting so you can see in this case.
We have an S orbital coming in at an angle that's less than 90° to this p orbital.
Okay,
this is a special case.
It's not you can't get it with a diatomic.
Okay,
we'll see it with molecules that are threefold symmetric and so on
But in this case you get a positive interaction, you form a sigma bond
because this shaded s orbital interacts with this shaded lobe of the p orbital
much more strongly than it does with that unshaded orbital.
Oh,
sorry.
That was rule 3.
Okay, last one, rule 4.
rules 4 deals with the interaction between non-degenerate orbitals
The point of vocabulary when we talked about degenerate in quantum mechanics.
We mean they're of the same energy.
Okay,
I know there's some you know colloquial meanings of the word degenerate
but we're not talking about that. The definition of degenerate means the same energy.
So when we talk about non degenerate orbitals,
we're just talking about two orbitals that have different energies for instance carbon 2p orbital interacting with an oxygen to p orbital.
We know that carbon and oxygen have different electronegativities.
Electronegativities reflect the energy of the valence orbitals
Therefore the orbitals must be of different energies. That's going to impact the strength of their interaction.
And so what we can see here,
this is just done with two s orbitals not 2s orbitals,
but two s orbitals.
in this case the s orbital on atom 1 is higher in energy than the s orbital on atom 2
We represent that by remembering that energy is always plotted on the y-axis.
So we write this orbital higher in energy.
The bonding combination still is stabilized right?
It's lower in energy than the lowest atomic orbital.
The anti-bonding orbital is higher in energy than the highest energy atomic orbital.
You'll notice another thing happens or is represented by these cartoons.
The bond is polarized.
Correct?
we look at this bonding orbital.
We have unequal sizes for the two circles.
This is how molecular orbital theory deals with polarisation.
We know.
That if we take a hydrogen atom.
And we bonded to a fluorine atom.
To make a chap that the electrons are all going to be polarized to the fluorine side.
We all know how to assign dipoles.
Well,
this is how we describe it in molecular orbital speak.
The bonding orbital gets localized.
On the lower energy atom or the atom that contributes the lower energy orbital
the anti-bonding orbital gets polarized to the atom with the higher energy orbital.
so now if you are having flashbacks to Chem 51
you can assign different names,
right?
Atom 2 is going to be the nucleophile side.
Atom one is going to be the electrophilic side.
This anti-bonding orbital is going to be Lewis acidic or electrophilic side of the molecule
So there's something interesting down here. Orbitals with a difference in energy greater than 12 eV which is huge.
Have essentially no interaction.
That seems kind of weird.
So I don't I don't have all the numbers of the periodic table memorized but and there isn't even a periodic table in this room.
Fantastic.
But let's we know okay,
we know HF, we know lithium fluoride.
Let's drop down one of the last elements in the first column of the periodic table which is Rubidium
one of the most electropositive elements on the periodic table.
We know fluorine is one of the most electronegative.
So,
let's assume that there.
the Delta E between the orbitals on rubidium and on fluorine are 0
do you think rubidium fluoride exist?
That's a salt
It does because it's a salt, solid at room temperature.
So how does rubidium fluoride exist if we're telling you?
That there's zero interaction between the orbitals.
Ah! Ionic bonding.
I remember when I told you MO theory was just a model.
When you have models you make assumptions.
If you try and apply your model.
To a system where your assumptions aren't valid.
Your model is a piece of junk.
If you're talking about something like rubidium fluoride or lithium fluoride or anything of an ionic salt
MO theory is not the way that you want to describe it because
This bonding model was really developed to describe covalency not ionic bonding.
So you can still make molecules or compounds or salt when there's a big energy difference.
You just can't use Mo theory to accurately describe their electronic structure.
So the complicated part ends up in that gray area.
Okay,
where you have fairly large differences in Delta E,
but the compounds aren't quite ionic.
That's where we get the polar covalent bond and we describe.
The orbitals in this way so that they're polarized towards the more electronegative atom.
And so here's three cases.
The top case is one where we have the degenerate interaction.
Two atomic orbitals of the same energy interacting to form a sigma and sigma star.
From a covalency perspective.
That's when you get the strongest interaction.
The strongest covalent bond occurs between two atomic orbitals of the same energy.
We know that.
We know that a C-C single bond is stronger than other C-N bond or whatever
because it's a covalent bond and the orbitals are perfectly matched.
We have unequal energies,
but they're close.
You still get a strong interaction.
The bonding orbital will be localized on the lower energy atom the anti-bonding orbital on the higher energy atom.
This splitting between the bonding and antibonding is a little bit less than it is in the perfectly covalent case.
When we get to very unequal energies what we end up seeing is that this bonding orbital
only gets stabilized by a very small amount relative to the pure atomic orbital.
This is also going to be the most polarized case.
The bigger this difference in the energy, the bigger the polarisation of the bond towards one atom or the other.
Okay.
So those are your rules.
Now it's actually get in and make some.
Molecules using more than just s orbitals.
So O2 is the classic case for why molecular orbital theory rocks.
because if you use those in a really simplistic models that they teach you in chem 51,
Lewis dot structures and all the stuff you get the wrong electronic structure
and if you can't predict the electronic structure for a diatomic molecule well,
Just kidding. It actually, if you do it right it actually does work.
but this is kind of a simple
Dummies guide to Lewis structures that you would get for 02. You predict a double bond,
which is correct.
But you predict it all of the unpaired electrons are paired up.
And you can do a really really cool experiment with liquid oxygen
pour it through the poles of a magnet and the magnetic field will bend the stream of liquid oxygen.
Because oxygen has unpaired electrons, it's also blue in color.
Has anybody ever seen these this, like the oxygen experiments?
They're really cool.
Plus you have liquid oxygen and you drop a match into it really spectacular.
So okay Lewis dot structures don't get oxygen right?
I'm going to let you in on a secret.
How many of you guys are like organic chemists I could see you wearing a sweatshirt.
So.
Organic chemistry is great,
right?
You know that the bonding model they talk about in organic - sp2, sp3 hybridization all that.
It doesn't even get methane right,
I just want you to understand that happens.  MO theory does.
Okay,
so we're going to use this LCAO MO theory to see if we get a better electronic picture of the oxygen molecule.
I have a real complex when it comes to organic chemists.
They're smarter than me.
So,
you know,
I was trying to make fun of them.
[Laughter]
Okay,
so we know that we're dealing with oxygen. Oxygen is in the second row of the periodic table.
It's a p block element.
We count up the number of valence electrons. There are 6 and so we know that the 2s orbital is lower in energy than the 2p orbital.
For oxygen atom A, we put two electrons in the 2s the other four in the 2p.
same thing for oxygen atom B because it's two oxygen atoms.
The S and the p orbitals on each atom are at exactly the same energy.
This is a degenerate interaction.
And so now we can go through and we can build up the bonding and antibonding and whatever else types of orbital's based on.
What can we know about the shapes of these orbitals?
Now you can use symmetry.
Okay,
Okay, but again as easy as character tables are to use
for something like that oxygen where you're talking about the two s orbitals and a total of six p orbitals,
I'm pretty sure you can draw the cartoons faster than you can use the the character table.
So we take the s's.
We know that we're going to take linear combination.
So we do an in phase combination of the two s's to make a bonding orbital and out of phase combination to make the anti bonding orbital.
We can label these as sigma g and sigma star u or sigma.
u star however you whatever order you prefer.
This one is G because a homonuclear diatomic has an inversion center.
When it because it has an inversion center we can label this as a gerade orbital because it's symmetric with respect to inversion.
The ungerade orbital is unsymmetric with respect to inversion.
We can make a sigma Bond and a sigma star anti bonding orbital out of the pz orbitals.
By convention we put the z-axis along the bond.
So when the two lobes are shaded the same way facing each other for these p's to get the sigma orbitals,
It's gerade when they're out of phase.
It's Sigma star and it's ungerade.
Interesting point here.
And this is where you have to be careful when you're being rigorous and you're writing out your equations.
Okay.
Remember we're talking these are linear combinations.
This bonding combination between the s orbitals we would say is 2sa + 2sb.
They're added together.
This one we would say is -2sa + 2sb.
So you might want to think that anytime you add the orbitals together in a positive way you get a bonding orbital
and in the negative way to get an anti bonding orbital.
That's not true though.
In this case.
You'll notice that the phasing.
On the left pz is opposite of the phasing on the right pz so this is really 2pa - 2pb.
Where is the anti bonding orbital is to 2pa + 2pb on the z-axis.
Okay.
So the plus and minus doesn't tell you necessarily that it's a bonding or anti bonding orbital.
You have to draw out the shading to make sure you are getting it right.
Okay,
so there's the sigma [?].
We still have two orbitals left over on each atom.
It would be the px and the py and of course,
we know how to draw these.
It's the pi bonds between the two oxygen atoms.
These two are in the plane of the board, these two are coming out at you,
the shading on the lobes matches.
So these are the pi ungerade because again,
if you put an inversion center at the middle of that O-O Bond it unsymmetric with respect to inversion.
The anti bonding or pi * is now gerade.
So again,
Just because an orbital is gerade doesn't mean it's bonding
just because it's ungerade doesn't mean it's anti bonding.
You have to look at the actual orbital and decide because ungerade can be bonding or anti bonding
just depends on the actual shape of the orbital.
Okay,
we now have 12 electrons to put in. We fill from bottom to top. For now, we're always going to fill in lower energy orbitals before we fill put any electrons in a higher energy orbital
For now, we're always going to fill in lower energy orbitals before we fill put any electrons in a higher energy orbital
that will change later in the quarter just to make things confusing. So we get 10 electrons
In these two orbitals derived from the s's, the sigma derived from pz's,
the pi's derived from px's and py's and that leaves us two electrons to go into the pi star orbital.
So, bingo! Right away we predict that O2 is paramagnetic.
Got two unpaired electrons in the pi star orbitals.
The other thing that happens when you populate an anti bonding orbital is of course you break down the bond order.
So based on this we have four electrons in the pi bond,
but we have to subtract out the occupancy of the anti bond which drops the bond order.
You can see there's two different Energies.
Different splittings depending on whether you have a sigma bond or a pi bond,
all of the p orbital start out at the same energy.
That is an approximation that we may come back to later on.
But if we assume that all the p orbitals start out at the same energy, the sigma interaction is strong.
So that sigma bonding orbital comes way down much further than the pi bonding orbital again.
This is exactly in line with what you learned it in organic chemistry 51,
Sigma bonds are always stronger than pi bonds, significantly.
Okay,
and it goes back to rule number 3 or 2 or for whatever rule it was.
Why is it why is a sigma bond stronger because there's a stronger bigger overlapping integral
the overlap between two Sigma orbitals which are pointing directly at each other.
Is better than the overlap between two pi orbitals that are pointing parallel to one another.
Okay,
when we have this diagram you can write molecular orbital electron configurations.
Remember how much fun you had with this in freshman chemistry?
We can do the same thing and have just as much fun with molecules.
In this case.
We would write it as Sigma two Sigma * 2.
That's the sigma from the s, sigma star from the S Sigma 2 pi for pi * 2.
We have that filled Sigma from the pz's, the filled pi's from the px and py
and the half-filled pi * from the px and the py
so that's our molecular orbital electron configuration.
We can also calculate the bond order.
This is universal for any isolated bond.
We can calculate the bond order as the number of bonding electrons minus the number of anti bonding electrons and divide by 1/2
So if we just look at this diagram,
we have two electrons there in a bonding orbital, two here in a bonding orbital and four there in pi bonding orbital.
So that's a total of 8.
We have two orbitals in this Sigma star that comes from the S orbitals and two electrons in this pi *, that's four
1/2 * 4 = 2 so we calculate the bond order for O2 is exactly two.
Woohoo fun.
Right we're having a great time now.
So MO Theory is great.
It's wonderful way better than hybridization and all that stuff that you learned in organic chemistry.
We won't talk about all the problems that MO Theory has, yet we will get to them.
So,
you know.
We all know what really kind of sucks about chemistry right?
Started in freshman chemistry,
You'll be told kind of given out this simple model of how it all works are like,
okay,
I can remember that.
And then the other shoe drops.
Hey find out.
Well,
you know the model mostly works except when it doesn't and you have to make this other approximation.
Okay,
I can remember that one and then another one and you know,
it's all just we just keep piling on these things where you have to make your simple model more and more complicated
to reflect the real data that we collect in the lab.
So here's your first one.
Here's another representation of that Mo diagram that we just drew for O2
if somebody asks you to draw the mo diagram for fluorine, F2 and use the same one.
Perfect, works great.
If you look on the periodic table right here that's going to the right.
You start from oxygen and go to the left.
Okay,
what element?
You guys are all juniors and seniors chemistry majors at UC Irvine.
To the left of oxygen nitrogen.
Okay,
boy,
I didn't memorize periodic table either.
I figured that's what it's there for.
Okay, so what happens if we started oxygen we go one to the left.
We just are basically taking one electron away from each atom and making the N2 molecule.
So there's the mo diagram for that homonuclear diatomic.
You have really bad eyesight if probably looks the same to you.
If you don't have bad eyesight you'll notice.
It's somebody screwed up the ordering of those orbitals since when is a sigma bond weaker than a pi bond?
I know all of you organic chemistry out there saying yeah,
see you, looks like you're a model that sucks too!
So the question is what's going on and did we just draw this incorrectly?
The answer is no.
We do it this way first and then realized when we went and did experiments
that the model didn't fit the data so we had to refine the model.
And then we had to wave our hands and come up with an explanation why we were refining the model.
So what's the explanation? So here there's a lot of data on this slide.
Okay,
all we're looking at and it's all really small.
So I hope you got the notes last night.
We're going from dilithium to diberyllium, diboron, dicarbon, dinitrogen, dioxygen, difluoride and dineon.
Right across the second row of the periodic table.
And so if you look carefully we're just going to focus here for now on the sigma and pi orbitals that come from that the combination of two p orbitals
and you see for lithium beryllium boron carbon and nitrogen.
We all have this situation where for some reason the sigma orbital is higher in energy than the pi.
Which is another way of almost saying that the sigma bond is weaker than the pi bonds
which again we know from organic chemistry makes no sense whatsoever.
Once we get to oxygen.
The order corrects itself.
And the sigma is lower than the pi for oxygen fluorine and dineon even though dineon doesn't really hold together.
Okay,
What could be causing the problem here?
It turned out that the information is right on this slide.
[Student Question]
Um, yes or no?
So it's the diagram from the model that is modified to fit the experimental data.
Okay.
So these are the correct electron configurations based on experimental data.
But again,
the models been tweaked you get this crossover.
The sigma is lower or the pi is lower.
Okay,
so that's a really good observation.
Oxygen is the first combination that has electrons in the pi * and so maybe that's causing a problem.
I don't think that's what it is.
But that's a really good observation.
You'll notice there's something else changing.
As you go from left to right across the table.
Electronegativity is changing.
And so what's electronegativity? It's how tightly the nucleus holds onto the valence electrons.
That's a fair description, right?
Okay,
when in molecular orbital speak, in atomic orbital speak, that means the orbitals are lower in energy.
Oh,
you're just looking ahead on the notes,
aren't you?
So, effective nuclear charge, or Z_effective,
is one of the factors that control electronegativity.
Okay.
But look as we're going across look at what's happening to the sigma orbitals.
Sigma and sigma star that we derive from the S's.
They're dropping like a rock.
And so these are all tied together.
Right, remember for s orbitals,
there's actually a probability that the electron could be at the nucleus.
So what that means is the S orbitals really feel the effects of an increase in Z_effective.
It affects s orbitals more than it affects any other type of orbital because the S orbital has electron density at the nucleus.
The p has a node, where there's zero probability of an electron in a p orbital being at the nucleus.
So it has less of an influence.
These are dropping like a rock.
Okay.
So this is where things get a little screwy.
Or it's going to seem like they get a little screwy.
See that?
Symmetry label there.
We said that the bonding combination
between the 2s orbitals that forms Sigma Bond, we ascribed a symmetry label of Sigma G to it.
The sigma orbital comes from the combination of the pz's also is Sigma G.
So remember what the symmetry labels are?
It's the irreducible representation.
So these two Sigma orbitals have the exact same symmetry.
They belong to the same irreducible representation.
We said if we have two orbitals of the same symmetry,
they can interact they can mix that's what's going on.
Remember,
we also said that the strength of that mixing.
Is inversely proportional to the energy separation that is when the energy separation is small the mixing is significant.
When the energy separation is large.
The mixing is insignificant.
Hear the energies pretty small.
So what literally happens is because this Sigma bond is the same energy as this Sigma bond you can think of it as they repel each other.
This one is actually pushed down a little bit more than it normally would be.
This one gets pushed up higher than it normally would be.
Same thing here.
Same thing here.
Same thing here.
What happens when we get the oxygen while we've we've gone over the threshold.
This Sigma G combination is just too far away to interact significantly with that Sigma G.
And so this one does not get pushed up above the pi.
So we call that configuration interaction.
Or you can use another term that we've all used improperly before - hybridization.
It's another...this is what I like to say is hybridization done correctly.
You can hybridize orbitals when they have the same symmetry.
So this is one example where you basically have to put a fudge factor in to Mo theory to get the simple model to reproduce experimental data.
We're going to see all kinds of examples where this happens.
But for our purposes if you're asked for the diatomic Mo diagram for lithium beryllium boron carbon or nitrogen.
Just remember that Sigma is above pi.
You're asked for oxygen fluorine or neon, you get the traditional ordering where Sigma is below pi.
Now some of you that are thinking ahead or like well,
okay what happens if we do the CO molecule?
We have one from this side and one from that side of the line.
Without any added data, always assume this traditional ordering
another words is that your Sigma bonds are stronger than your pi.
You're given some other data.
Compound is this number of unpaired spins?
Okay,
where you have to flip the order of orbital's to get the right number of unpaired spins?
Okay,
then.
Then you can apply that this kind of fudge factor.
And so that's what happens with all of these situations where the model breaks down.
Okay,
and I know as soon as you worried about this cuz you're like,
well what if I'm asked question on a test and it's one of these weird ones where the model breaks down?
Well, the only way that we get it right is to go make a measurement and get a piece of data and realize we have to fudge the model or improve the model.
Okay,
so it's the same thing for you.
This is a case where I'm already telling you,
right?
If I ask on the test to 3 to do the diagram for diboron.
You have to remember that.
It's on the left side of the line.
Okay,
but for other cases you have to do the experiment chemistry is an experimental science.
Okay.
So rather than try and blitz through anything else,
I'm going to question.
Sorry.
[Student Question]
Oh great,
great question.
So what about the roads below so we only are talking about the second row of the periodic table.
What happens if we drop down to the third row in the fourth row?
So, first diatomic so that the third row are very very uncommon except for chlorine.
And, second, always follow the more traditional.
All right any other questions?
Okay,
have a great weekend.
Matt will be back on Monday and I will see you midway through the quarter.
