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At the end of last hour, we had just gotten to the point
of having developed the molecular orbital energy level
diagram for the BH three molecule, this trigonal planar
entity. And I had not had time last
hour to show you what these molecular orbitals look like in
their calculated form. And so, I am going to start off
today by doing that, and then will proceed to answer
a quadrangle of questions that we can attack regarding diatomic
molecules using molecular orbital theory.
Remember that the BH three MO problem is a seven
orbital problem. And so, we will go through and
just look at these seven molecular orbitals in ascending
energy from the lowest energy on up.
This one is the lowest energy molecular orbital for BH three
as calculated using a modern quantum chemical package.
And what you can see is that since we are indicating the
phase of the wave function by color, there is only one phase
to be seen. This, of course,
is a characteristic property of an s orbital.
Remember that? And what we have here,
at the center of the molecule, is the 2s orbital on boron.
And in this molecular orbital theory, effectively what is
happening is that 2s orbital on boron is reaching out and
simultaneously overlapping with the three 1s orbitals on each of
the hydrogens. And the way that we have
orthogonalized it and normalized it amounts to a single molecular
orbital that can, if you will,
house a pair of electrons in this orbital that is
simultaneously bonding between boron and the three hydrogens.
You have one electron pair in here.
Two of the six electrons in the valance shell of this molecule
are in this lowest lying, most tightly held,
most electronegative molecular orbital.
Let's look now at the next one.
You should be thinking in your mind just what was the next
highest lying orbital that we had, and you will realize that
this is an orbital formed from one of the boron p orbitals and
a linear combination of two of the hydrogen wave functions.
And let's see if you can recognize it based on what we
did last time. Maybe I can reorient it to help
in that regard. Here is a molecular orbital
that can house a pair of electrons, one spin-up and one
spin-down, and that is simultaneously bonding from that
boron to this hydrogen, up here.
And then there is a nodal surface, here,
that is coincident with the nodal surfaces of the boron 2px
orbital, using the same coordinate system that we were
using last time. And so, the wave function
changes sign as you pass through the boron.
And then the opposite phase lobe of the boron 2px orbital is
able to overlap with these two hydrogens, here.
If you will remember, when I wrote down the
normalization constants for this molecular orbital,
we had double the coefficient on the hydrogen up here as
compared with the two hydrogens down here.
And they were opposite in sign. And that leads to a really nice
overlap, as you can see here. This is BH bonding here,
and BH bonding, down here.
And it involves that boron 2px orbital.
One way you can think about this molecular orbital theory is
that this 2px orbital on the central atom is reaching out
with its intrinsic plus-minus phase combination that is
intrinsic to a 2px orbital simultaneously reaching out and
bonding with all three hydrogens in the only way that it can,
being a 2px orbital. And so, we created that linear
combination with that idea in mind.
And then, if we go up one more in energy, strike that.
Not one up more in energy, but to the other orbital in the
molecule that is at the same energy.
What we will find is that we are involving,
now, the 2py orbital. And it will reach out and
interact with hydrogens in the only way it can,
given its intrinsic nodal properties as a 2py orbital.
Remember that a 2py orbital has the x,z-plane as a nodal
surface. And there is a hydrogen up here
on that x,z-plane, namely the one that lies on the
plus x-axis. And because it lies on the
intrinsic nodal surface of this 2py orbital, it can contribute
nothing to this molecular orbital.
All we are seeing up here at the top is an arbitrarily sized
sphere, just to show the location of that nucleus,
the hydrogen nucleus that is on the plus x-axis.
And then here, what you see is the 2py orbital
oriented along y and overlapping simultaneously with a plus 1s
wave function here and minus 1s wave function here so that you
see a beautiful bonding molecular orbital that is at the
same energy at the one we just looked at.
And so, as you step up in energy, looking at the energy
level diagram for BH three from last time,
that is what you see. We have now six electrons in
the molecule, two in this orbital and two
each in the two orbitals we just looked at that provide the three
BH bonds that we represent in the valance bond theory as
electron pair bonds. But this is a totally different
way of looking at the electronic structure that is very appearing
because it just takes advantage of the atomic orbital properties
that you would calculate for your central atom.
Now, let's go up one more. The very next one,
I will skip showing to you for purposes of saving time,
here, but you know that the next orbital up is simply the
2pz orbital on the boron, which is our lowest unoccupied
molecular orbital, as well as being an atomic
orbital in this system and responsible for the Lewis acid
characteristics of this species. And one up above that,
one higher in energy than the LUMO is the orbital that I did
show you last time, the out-of-phase combination of
the boron 2s with the three hydrogens all with the same
sign. It is sort of a round nodal
surface intersection each of the BH bonds.
And then, finally, orbitals number six and seven
are the antibonding counterparts to the ones that I just showed
you involving boron's 2px and 2py wave functions.
Here is what these look like. And this one,
you will see, involves the 2px orbital.
Here is our x-axis pointing up. And the blue lobe of 2px,
the positive lobe here, as you can see,
is squished out and down. These pictures can be a little
bit more complicated than what we are used to looking at from
just sketching it out on the board or the qualitative
pictures that you sometimes see in your textbook.
But what we do find here that is the important feature is that
as we go from the 1s contribution on this hydrogen
over here to the positive lobe of 2px we are going through a
nodal surface. And that is antibonding in
character. And you can see that this
deformation of this lobe of the p orbital, of course,
looks like a high energy type of phenomena.
And indeed it is. This is one of the two highest
energy orbitals in the energy level diagram for the BH three
molecule. And then, similarly,
here is the other lobe of the 2px orbital, the minus lobe.
And, as you try to go from it to the adjacent hydrogens,
remember, in the bonding one we were allowing this to overlap
with the same sign on the hydrogens.
Now we have the opposite sign on the hydrogens,
so that we have a beautiful antibonding node indicated here
and here between the minus phase lobe of 2px and the adjacent
hydrogens. So, this antibonding orbital
has lots of internuclear nodes, nodes that appear between
nuclei and indicate that we are not getting overlapped and
bonding, but rather we are getting a high frequency orbital
that is likewise high in energy. Finally, I will show you the
counterpart to this one, that involves 2py.
It is a little easier to understand from inspecting it.
And it also evinces the internuclear nodes that you
would expect for an antibonding molecular orbital.
And I am orienting it so that you will be able to understand
it with reference to our coordinate system as introduced
last time. And so, you have the boron and
the hydrogen along x, up here.
Here is our 2py. It is sort of bending away from
these two hydrogens whose wave functions have opposite sign to
the lobes that are directed into their vicinity.
You can see that we have here a nodal surface between boron and
this hydrogen wherein the bonding counterpart that we
looked at had the same sign and a nice bond.
And then over here, once again, the mirror image.
So, this is one of our final orbitals in looking at the BH
three energy level diagram.
It is the case that if you start to put electrons into
antibonding orbitals, they will cancel the bonding
properties of the bonding counterparts.
And we will explore that in quite a bit more detail here
shortly. Let me leave that up there for
now and tell you what quadrangle of problems it is that I hope to
answer for you today using more MO theory ideas.
First of these will be why He two is so stable?
And so, today's lecture is predominantly devoted to
diatomic molecules. And He two will be the
first of these. Let's make this unstable.
Let's do stable here in just a moment, but first let's say,
next, why does O two have unpaired electrons?
Certainly, if we draw the valance bond picture for O two,
like that, it does not give us any
indication that this diatomic molecule would have unpaired
electrons. And so, we are going to see how
MO theory can shed some light on this issue.
And then, the final two questions that I will fold into
one line here given the way I am using up space on this board is,
why is N two so stable --
-- and CO poison?
MO theory can shed very clear light on all of these questions.
And so, that is what we will work on for the rest of the
hour. In MO theory,
there are a couple things you should keep in mind when trying
to grasp all the subtleties of a given energy level diagram with
which you may be confronted in your textbook or on an exam or
just through reading papers in the Journal of the American
Chemical Society, wherever you may find these.
You should keep in mind that, number one, interaction is
strong. That is, when atomic orbitals
interact to form molecular orbitals, the AO interaction is
strong --
-- when there is good spatial overlap.
Overlap is a central concept in molecular orbital theory.
The idea is that if you have atoms that are far apart from
each other in space, then their wave functions have
dropped off exponentially and are not interacting much and
there is not good overlap. But when orbitals are close in
space and if their orbitals are directed in space toward one
another, then there may be overlap that leads to good
bonding. And also I can echo this first
part, AO interaction is strong when, and say two,
AOs are close in energy. The interacting orbitals are
close in energy.
When I talk about where atomic orbitals are in energy,
I am going to really want you to think about your periodic
table and the properties that appear with periodicity and are
organized and collated in the periodic table.
We are going to be talking about properties like
electronegativity as it relates to the energy.
And I will get to that a little bit on the next board,
but let's first make a small MO diagram here.
And this will be a diagram that we may use both for H two
or He two. This is the simplest of all MO
diagrams and one with which you should be quite familiar.
We have a bonding molecular orbital at low-end energy.
And notice that if this is a 1s orbital over here and this is a
1s orbital over here for H two or He two
systems, then when these come together and bond,
here are your two nuclei, --
-- and here is a probability density isosurface that I am
drawing around that shows in-phase combination of those
two 1s orbitals merging with each other, giving good overlap,
and the interacting AOs being of the same energy.
Because the H two and He two molecules have such
symmetry we have a good bonding orbital.
And then, up here, we must find there to be an
internuclear node in that corresponding antibonding
orbital. And so, it will look something
like this. And, therefore,
I am going to associate with this high-lying orbital an
asterisk to indicate the antibonding character,
the presence of this internuclear node,
the change of sign as we go from one nucleus to the other,
the absence of bonding character.
Here we have the buildup of electron density in the
internuclear region, electrons being stabilized by
being held simultaneously by more than one nucleus that is
positively charged. And here, the exact inverse of
that, constructive interference and destructive interference in
the H two or He two systems.
And then we can ask, how many electrons do we have
to put into the diagram? If it is the H two case,
then, to figure out the bond order, --
-- what we do is populate the diagram according to the same
kind of Aufbau principle that you would for generating the
energy level diagram for an atom, and taking into account
things like Hund's rule. If it is H two,
we have two electrons that we can put in, which in here with
opposing spin, like that.
That is our bonding molecular orbital.
We have two bonding electrons. Zero antibonding electrons
divided by two equals a bond order of one.
We have an H-H single bond in the case of the H two
molecule. But if we have He two,
then in the He two system, we have two more electrons,
and the only place they can go is up here in the antibonding
molecular orbital. We have two bonding minus two
antibonding over two is equal to zero for a bond order of zero.
And so, the idea is that if two He atoms collide with each
other, when they do so, their orbitals may overlap and
give rise to bonding. But, simultaneously,
you get antibonding. And the net bond order is zero,
and so He two is not bound, but these atoms just
bounce off each other. There is no stabilization.
And you can certainly understand that if we were in
the middle somewhere, as we would be if we had He two
plus, missing one electron up here,
then our bond order would be one-half.
It is going to get a lot more complicated than this,
but this you have to know because this is a really nice
starting point for thinking about bonding in all kinds of
scenarios where you may need to consider it.
And so, let's look at some periodic properties.
This will be with reference, again, to energy.
And I am interested in how some of the properties of the atoms
vary as we go across the periodic table,
lithium to beryllium to boron to carbon to nitrogen to oxygen
and over to fluorine. And I will stop there.
We won't worry about any more of the noble gases today.
As we go from left to right, one of the things that you know
from your study of the periodic table is that electronegativity
increases.
This increases from left to right across the Periodic Table
in concert with increasing Z.
That is, the increasing atomic number.
That is, the increasing number of positively charged protons in
our nucleus, as we go across. Each time we are adding more
protons to that nucleus. The nucleus is becoming more
and more positively charged as we go across.
And an important consequence of that has to do with the fact
that s orbitals don't have a node at the nucleus.
Whereas, p orbitals do. Since these elements are in the
so-called p block of our periodic table,
well, lithium in the s block, but we will consider the
properties all the way across. And the energy issue up here
that I am going to be concerned with is the energy of the atomic
2s versus 2p orbitals as we go from left to right across the
periodic table, systematically adding to the
number of protons in our nucleus in a way that does not affect 2s
orbitals the same way that it affects 2p orbitals.
It turns out that by the time we get over to fluorine,
where we have essentially our most electronegative element,
the 2s orbital has sunk to a very low energy.
And the 2p orbital has gone down in energy as well,
but just not by nearly as much. And so, it is nearly an
isotonic function as we go across.
2p is decreasing in energy in response to this increased Z,
but not as much as 2s is decreasing in energy,
having to do with the fact that 2s electrons see much more of
the nucleus than 2p electrons do because 2s electrons don't have
a node there. They spend a lot more time in
close to the nucleus than do your 2p electrons.
This has a really important influence on that set of
diatomic molecule MO theory problems, eight orbital
problems, that we call homo or heteronuclear diatomic molecules
of these elements. And that has to do with the
fact that in between nitrogen and oxygen, a fundamental change
takes place in the energy level diagram.
And that change, that results from the atomic
structure of the elements, ends up with the result that
the MO diagram is much easier to draw for oxygen,
O two, and fluorine,
F two, than it is for the earlier
ones. And it is much easier to
understand for O two and F two than it is for N
two, C two, B two, Be two, Li two.
So, let's just see what that means.
First of all, when we are going to draw an
energy level diagram for a diatomic molecule,
if it is a homonuclear diatomic molecule, meaning the two atoms
that are going to be bonded together are the same,
that means that the atomic orbitals for the contributing
atoms are the same. Here, on the left,
what I am drawing is, let's say, an F or an O atom.
We have the 2s and the 2p, x-y-z.
That is four of the eight orbitals in our problem.
And then, over here, we have the other identical F
or O atom, with its 2p x-y-z and 2s orbitals at exactly the same
energy as for the counterpart, over here.
So, this is another F or O atom.
And then we are going to ask, how can these atomic orbitals
combine to give molecular orbitals in these diatomic
molecules? This is unlike the BH three
problem because there is no atom at the center of gravity
of our system, but the ideas are the same.
And the ideas are very much the same as what we have here for H
two, where also we could make molecular orbitals despite
the fact that there was no atom at the center of our system.
We can pretend that there is, and we can use that pretend
atom to identify what linear combinations we should make.
A simpler way to do it, at this point,
would be to say that our molecule has the two nuclei
oriented along the z-axis. And then you have x and y
perpendicular to z. And so, if we take this choice
of coordinate system that the molecule is oriented along z,
the two nuclei lie one on plus z, one on minus z.
Then we can immediately classify our atomic orbitals as
being either sigma or pi with respect to the z-axis.
I will say AO is sigma or pi with respect to z,
which is the molecular axis. And remember what that means in
terms of the nodal properties. If something is sigma with
respect to z, that means if you view it down
z it will look like a cylinder. It will look like a sphere,
actually. It will be cylindrically
symmetric about the z-axis. If it is sigma then,
if you are viewing down z, you see that.
That could be either an s orbital or it could be a pz
orbital. They would look the same if you
were viewing right down z. And then pi,
viewing down z, would look like this.
And that would be a px or a py orbital perpendicular to the
z-axis. And this would be an s orbital
or a pz orbital. What I can say right away is
that we can separate out the molecular orbitals that we are
going to form from these atomic orbitals into their sigma or pi
characteristics relative to the z-axis.
And because we are in the O or F atom case, very far on the
right-hand side the periodic table, we have a large gap.
This gap is large.
We can make the approximation that this 2s orbital here,
which could in principle form a sigma bond with this 2pz orbital
over here, that would be an interaction looking like that,
-- -- that that does not occur.
Because remember here I said you get strong interaction when
the interacting AOs are close in energy.
When you are talking about these very electronegative
elements, the s manifold of orbitals is very well separated
in energy from the p manifold. That leads to simplicity in the
case of the diagram that we are about to draw.
Because we say that this interacts with this in a sigma
fashion, giving rise to bonding and antibonding interactions
that look exactly like what we have drawn over there.
I will draw the bonding orbital here.
Actually, the antibonding orbital here and the bonding
orbital here. And I will show the parentage
of this molecular orbitals in this way.
And you will recognize that this is our in-phase combination
and this is our out-of-phase combination, analogous to the H
two problem. This is like the H two
problem built into the F two or the O two
problem. And next what we find is that
we can also make a sigma bond by interaction of pz with pz.
This is going to be a nice strong directed bonding
interaction that will look like this.
You see we have overlap here in the center, a nice strong
bonding interaction in the center with 2pz orbitals that
are directed at each other. And this is a sigma bond
because you view that down z, it is going to look like a
round thing. You are not going to see any
nodal surfaces. And then it has up here a
corresponding antibonding orbital right there.
So far I have --
-- two pair-wise combinations that give us two sigma bonding
orbitals and two sigma star antibonding orbitals.
So, that is pretty straightforward.
And, once again, I can say that we have the H
two problem built into this energy level diagram twice,
with the s interactions, with the pz interactions.
The only thing I have left is to do my px and py interactions.
And we can make a pair of pi bonds and a pair of pi
anti-bonds that corresponds to those pi bonds.
This one is pi. This one is pi star.
And what do they look like? Well, they look like
side-to-side bonds between pairs of p orbitals.
This is just like what I showed you for the ethylene molecule.
The pi bond, in fact, has this nice overlap
on one side and up here on the other.
There is your pi bond. And you have one of those that
lie in the x,z-plane and one of those that lie in the y,z-plane.
You have two of them, and they are at equal energy
here. And then up here,
what does the pi star look like?
Well, the same, except we turn around one of
the p orbitals, introducing a nodal surface,
like that. And we have two of those.
And here is our nodal surface. Now it is an internuclear
between the nuclei nodal surface, indicating antibonding
character there. That is the easy case because s
and p are very well separated. What you can see,
though, is that type of interaction of a 2s on our left
atom with a pz on the right atom, that might start to become
important as you are over here in this part of the diagram,
where s and p are pretty close together in energy.
And that is what happens, in fact.
And that is what happens as soon as you go to the left of
oxygen and to nitrogen or anything lighter than nitrogen.
Suddenly, you cannot ignore interactions of s on one side
with p on the other.
And so, I am going to draw here the diagram that you would use,
say, for N two. And, for simplicity,
I am just going to draw the middle part.
This part here in the middle is that which corresponds to the
molecule. This one is what you would use
for O two or F two.
And let's just go ahead and draw that diagram for the
molecule. This one would be this set of
energy levels corresponding to, for example, N two.
And we have increasing energy
on the vertical axis. We are going to find that
again, we have a sigma and a sigma star.
And then, the difference comes right here where we find that as
we go up in energy we next get to our pi bonds.
And then, the next sigma orbital is slightly higher in
energy than the pi system rather than slightly lower in energy.
And then we will encounter up here our pi star.
And then, finally, that highest in energy,
sigma star. This switch,
this sigma going up relative to that pi is really a consequence
of all of the sigma orbitals being responsive to the value of
the electronegativity of the 2s orbital of the atom.
You can really almost consider the pi system independent from
the sigma. These four sigma orbitals,
sigma, sigma star, sigma, sigma star,
I can call them one sigma, two, three, four,
just to show you that we have, in ascending order,
four orbitals of sigma symmetry.
That whole manifold, when you can start mixing a
little bit of very low energy s into it, sinks down relative to
pi. In the case over here,
we are basically pulling down the four sigma orbitals relative
to pi. And here, because our s
orbitals are higher in energy relative to our p orbitals,
they go up a little bit and bumps three sigma up over pi.
And what it means is that these orbitals are not so simple
anymore. Any one of them is a linear
combination of four atomic orbitals.
This orbital here, one sigma, will not look like
it looks in the H two problem, anymore.
This one sigma will have some p mixed into it.
And so, one way we can represent that through a simple
drawing is like this.
You see the difference between that sigma bonding orbital and
the one up there? We have some extra little nodes
in here that are an inherent characteristic of the
contribution of 2pz orbitals into this.
This is a linear combination of the 2s and the 2pz orbitals on
the two atoms we have. There are four atomic orbitals
that go in with different coefficients to these different
parts of the sigma manifold. And then up here,
the other bonding orbital, we can draw it this way.
And this molecular orbital, this one here,
three sigma, has diminished bonding
character between the two nuclei and enhanced character on the
outside, the part of the p orbital that is contributing to
it that points away from the other atom.
And so, while this has bonding character, it also has what we
might think of as lone pair characteristics.
In MO theory, some orbitals are not that
simple. They might be at the same time
a little bit bonding and a little bit lone pair in
character. Let's look at a couple of
orbitals like that, briefly.
And here I have done this for the C two molecule.
I have done the calculation for C two,
so now we are over here on the left-hand side of that dividing
line where s and p are a big closer together.
And I am opening up this orbital, this one down here.
It is a little more complicated than the way I drew it on the
board, but what you should be able to see is --
Think back, now, to the first orbital I showed
you today, which was the lowest lying orbital in the BH three
molecule. You remember,
no matter which way we turned it, we only saw blue,
which was the positive wave function.
It looked like an s orbital, everywhere.
It had the same sign everywhere, and you could not
see any nodal properties. Well, that would be true also
for this one down here in the case of O two or F two
because it is not mixing in p character.
Because the atomic s and p orbitals are so far away in
energy. But now, when they are closer
in energy, this orbital down here, the s over here sees the p
over here and they mix in, and vice versa.
And, if I turn this around, what you are going to see are
these little negative lobes here that I am indicating that are
mixing in. But this is hugely bonding
orbital, still, because look at how much of the
orbital is centered in the region between the nuclei.
This is a very strongly bonding orbital in this system.
Let's now visualize the three sigma, the third sigma symmetry
orbital, as we ascend in energy. And it will look hopefully
something like what I drew here. There it is.
You see this is a sigma orbital, because as we look down
the z-axis, it looks just round. But it certainly has a lot of p
orbital character because as we go from outside the molecule
through the nucleus, we get a change in sign.
And then there is overlap in here, but the amount of the
overlap in here is small compared to the amount of this
lobe out here that is just nonbonding in character because
it is pointing out into space. It is like a lone pair of
electrons at the same time as it is a bonding orbital.
If this molecule actually had enough electrons to fill up
through three sigma and no higher, than you would recognize
that this orbital that we are looking at would be the highest
occupied molecular orbital. And it would be responsible for
the basicity and the nucleophilicity of the molecule
-- -- because any Lewis acid that
came in would want to come in on the z-axis to interact with this
lone pair electron density out there.
So, that is how our three sigma orbital now looks when we are on
the left side of that red dotted line to the lighter elements.
I will leave that up there for a moment while we consider a
couple of interesting points. One of the things that you will
often be asked to do is to write down the configuration.
And I want to finish these two questions here of O two,
why does it have unpaired electrons?
In the case of O two, we are over on this diagram,
we have 12 valance electrons. We can go two,
four, six, eight, ten, and then 11 and 12,
two electrons up here in the pi star manifold.
We are completely filled up through pi.
And then, we have a half-filled pi star with 12 electrons
populating the O two diagram.
Let's represent that this way. We have our one sigma with two
electrons in. We have our second sigma,
which is antibonding with two electrons.
Then we have our third sigma with two electrons.
And then, as we go up in energy, the next is our first pi
orbital bonding. That accommodates four
electrons. And then we have next our
second pi type orbital. It is antibonding and has two
electrons in it. And then all the other orbitals
are empty, so I won't indicate them.
This is the configuration for the O two molecule.
That means, we come back over here, by Hund's rule of maximum
multiplicity, we have two spin-up electrons
in the O two molecule with all these other orbitals
down here being filled. That configuration that I wrote
down for the O two molecule over there is a very shorthand way of
representing the MO energy level diagram --
-- because I started here at the lowest energy,
and I went up in energy like this.
And what you can see is sigma star cancels sigma,
so no bonding there. We have here three electron
pairs in bonding orbitals, sigma and 2pi,
but we have two electrons that are unpaired electrons,
responsible for the paramagnetism of the O two
molecule, that cancel some of the bonding
down here. When we want to write that down
in terms of our bond order convention, we have six bonding
electrons minus two antibonding electrons over two,
which is going to be equal to a bond order of two.
That is our bond order for O two.
While the bond order is what we expect to find based on that
valance bond representation of O two up there,
what we find out is that the representation of O two up there
in the valance bond terminology would not tell us that the
molecule should actually have two electrons that are not
paired in the molecule, --
-- that it should have paramagnetism due to the lack of
pairing up of all the electrons, spin pairing in the molecule.
And it has to do with this high symmetry of the O two molecule.
And it cannot distort from high
symmetry because it has just two atoms.
It has very few degrees of freedom.
It cannot bend or something to get those electrons to pair up,
and so they are unpaired. And that is why dioxegen has
unpaired electrons. And then, if we go to nitrogen,
the order that we write these things down in is a little
different because the energy level diagram is different
because now s and p are closer together.
And what we have is one sigma with two electrons,
or two sigma star with two electrons.
And then, next, we have our one pi with its
four electrons. And then next,
we have our three sigma with its two electrons.
And that is it. We have two less electrons in
the system than we do in O two.
That is N two. And our bond order now is
three. That is in accord with our
Lewis structure of N two, but it is a triple bond.
And it is a triple bond in a molecule that is completely
non-polar. And it is one of the strongest
chemical bonds known. That triple bond is worth about
226 kilocalories per mole. And it makes N two a
very fascinating molecule. And, of course,
it is the major constituent of our atmosphere.
And it is very inert, due to the stability associated
with having this diagram populated up all through the
bonding orbitals, and then no more antibonding
electrons present in the system. So, that is pretty interesting.
And given the time, I think I will leave you after
the following. I just want to show you one
more picture as a prelude to where I will start next time.
And that has to do with what happens when the two atoms that
are bonding are different. Let's look at the three sigma
orbital for the carbon monoxide molecule.
It would look like it does for N two.
It is the same number of electrons.
But look here. It is very asymmetric.
And that has to do with the fact that carbon and oxygen have
very different electronegativity.
And so, this highest occupied molecular orbital for carbon
monoxide, again, an eight orbital problem,
and again, a ten electron system, is one where we have a
larger coefficient on the lone pair piece that is on the carbon
atom of the molecule. And I will start off with this
next time, but that is one of two reasons why CO is a poison.
Have a nice weekend.