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CATHERINE DRENNAN: If you
haven't clicked in yet,
please pay attention to
the clicker question,
and I'll give you a little
bit more time to click in.
And I'll give you a
10-second warning.
All right.
Let's just go ahead and
take 10 more seconds.
All right.
Does someone want to say
how they got this one right?
Anyone willing to say
how they got it right?
We have a American
Chemical Society,
whatever these are called that
hang around your neck and clip
things to them.
Yes.
There's the word.
No?
Oh, there we go.
AUDIENCE: Thank you.
So this molecule
is the phosphorus
has five valence electrons and
then each hydrogen has three
so it's per total of eight.
So it has three bonding
and then one lone pair,
which makes it a tetrahedral,
but then the lone pair
pushed the other bond, so
it's less than 109.5 degrees.
CATHERINE DRENNAN: Excellent.
Yeah.
So I see that some people
just decided that there
were three things bound to it.
And so then they decided
either 120 or less than 120,
but you really need to
do the Lewis structure
and see how many loan
pairs there are first.
So once you do the
Lewis structure,
then you figure out
the parent geometry,
Sn forces sum of
tetrahedral system.
All right.
So more practice
on these coming up.
All right.
So we're continuing on now
with molecular orbital theory,
and we started the course
talking about atomic structure
and then we talked about atomic
orbitals or wave functions.
And then we moved on to bonding,
that atoms can come together
and bond and talked about
the structure of molecules,
and now we're going
to molecular orbitals.
And then we'll also talk
more about the structure
of molecules based on
those molecular orbitals
on Wednesday.
So we're really coming
all the way around,
we're using a lot of the
ideas that you've seen before,
but now we're applying
them to molecules.
And then, to me, this is
really an exciting part.
I love getting up
to the molecules
and talking about
structures and molecules
and how orbitals really play a
role in those properties that
molecules have.
And then, to me, the really
exciting-- I like reactions.
So after we finish
with the structures,
we're going to talk more
about how molecules react.
And so on Friday,
we're going to be
starting that,
reactions of molecules,
and getting into thermodynamics.
So we're sort of winding our way
away from orbitals for a while.
We will come back to d orbitals
around Thanksgiving time,
but we'll have a long
time before that where
we're talking about reactions.
So we're about to
do a transition
in the type of material.
But before we do
that, more orbitals.
But these are super cool because
these are molecular orbitals.
So we're going to talk
today about MO theory, MO
for molecular orbitals.
So molecular orbital
theory presents the idea
that these valence
electrons are really
going to be delocalized
around these molecules
and not just sitting
on individual atoms.
So to think about this
electron do localization,
we need to think about
molecular orbitals.
Molecular orbitals,
as we've learned,
another word for orbital
is also wave functions.
Wave functions are orbitals.
Orbitals are wave functions.
You need to consider
the wavelike properties
of electrons to think about
where the electrons are going
to be, what is their
probability density,
how are they going
to be arranged
with respect to the nucleus.
And so when we take
atomic orbitals
and we bring them together
as atoms come together
to form bonds, atomic
orbitals come together
to form molecular orbitals.
So we're going to be adding
superimposing, atomic orbitals
to form these
molecular orbitals.
And this is called a linear
combination of atomic orbitals,
or LCAO.
And so we're going to bring
those atomic orbitals together
and create molecular orbitals.
And we're going to create two
types of molecular orbitals.
We're going to create
bonding and antibonding.
And some basic math
principles apply here,
and that is that you can
create N molecular orbitals
from N atomic orbitals.
All right.
So that's really the basis
of molecular orbital theory,
and now let's apply it to
our friends, the s orbitals.
All right.
So we're going to think about
really simple molecules,
bringing together two atoms that
are identical with each other
and what happens to their s
orbitals when this happens.
So first we'll talk
about bonding orbitals.
So bonding orbitals,
again, arise
from this linear combination
of atomic orbitals, the LCAO.
And if it's a
bonding orbital, it's
going to arise from
constructive interference.
So we talked before about the
properties of waves, and one
of the great
properties of waves is
that they can constructively
interfere, destructively
interfere.
And orbitals are wave functions
so they can constructively
interfere and
destructively interfere.
Bonding orbitals
are generated by
the constructive interference.
So let's look at
two atomic orbitals.
And so here we have an orbital.
The nucleus is in the middle.
It's a little dot.
It's a nucleus.
And these two atomic orbitals
are going to come together.
There's going to
be a bonding event.
And so we have a 1s
orbital from atom a
and a 1s orbital from
atom b, and they're
going to come
together, and they're
going to form a
molecular orbital,
an ab molecular orbital,
and this is called sigma 1s.
So a sigma orbital is
symmetric around the bond axis,
so the bond axis here would
be just a vertical, a line
between these two nuclei here.
And so this molecular orbital
is symmetric around that bond
axis.
There are no nodal planes for
something that is symmetric.
There are no nodal planes
for our s orbitals,
and so there's none for the
molecular orbital either.
And we can also write
this as 1sa plus 1sb.
So the atomic orbital from 1sa,
the atomic orbital 1sb coming
together to form
sigma orbital 1s.
That is a bonding
orbital because it's
constructive interference.
It's a bonding MO,
or molecular orbital.
So now let's consider
the wavelike properties
and think about these atomic
orbitals coming together
as waves in what is happening.
So here we have the
same equation up here.
We're bringing
together 1sa and 1sb,
but now let's think about
this as a wave function.
So there is an
amplitude associated
with the wave function
for 1sa, and there
is an amplitude associated
with the wave function for 1sb.
These wave functions
come together.
Here is one nuclei,
here's the other nuclei.
And for a bonding
orbital, it will
be constructive interference,
and so the amplitude
where these atomic
wave functions overlap
will be increased when you
have constructive interference.
So our sigma 1s
now has increased
amplitude between
these two nuclei
due to that constructive
interference.
So an increased amplitude
between these two nuclei,
again, this is the bonding
axis-- here is one nuclei, here
is another-- so an
increased amplitude here
corresponds to an increased or
enhanced probability density.
Remember our wave function
squared is probability density.
It suggests the
likelihood of finding
an electron in a
certain region of volume
and if it's enhanced by this
constructive interference.
So if that probability
density is enhanced,
you have a greater
chance of finding
an electron between these
two nuclei, which will
be attracted by both nuclei.
So now why don't
you tell me what
you think is going to
happen to an electron that
is in this region of
constructive interference,
this increased area of
probability density.
OK, 10 more seconds.
Interesting.
OK.
So we should have probably
not put up the answer there
and re-pole.
Yes.
So here, if we have
an electron that's
attracted to both
nuclei, then we
want to think about
whether that's
going to be lower
or higher in energy
than something in
an atomic orbital.
So if it's attracted
to both, it's
going to be more
stably bound to those,
It will be harder to remove
that electron, which means
that it's lower down in energy.
And so we should look at
that, and we're going to.
So the answer is it should
be lower energy, more stable,
harder to remove that electron.
It's in a sweet spot.
It has two
positively-charged nuclei,
and it's hanging out
right in the middle.
It's very happy.
It's going to be more stable,
and that means lower in energy.
All right.
So let's take a look at that.
So the electron is
lower in energy,
and the bonding
orbital energy is also
going to be decreased compared
to the atomic orbitals,
and that has to be true if
that's where the electron is.
So let's look at
the atomic orbital
from a and the atomic
orbital from b.
And now, I'm going to put the
bonding orbital at a lower
energy level.
First, I was going to
put the electrons on,
and now, I'm going to put the
orbital at a lower energy.
So remember, this is
increasing energy here,
so the atomic orbitals
would be up here,
whereas the molecular
orbital is down here.
The molecular bonding orbital
will be lower in energy.
And now we'll put
our electrons there.
So we have one electron
up here and one here,
and so when they
come together, we're
going to have two electrons
in this molecular orbital.
So when you have
these two electrons
and they both occupy
the bonding orbital,
and this is the case
for H2, each H atom
is bringing one electron,
H2 has these two electrons,
and that's going to
make H2 more stable.
And we saw that before that
to associate the H2 bonds,
you have to add energy
into the system.
H2 is more stable
than free H plus H.
So when you bring
atomic orbitals together
and you have constructive
interference, an increased
probability of electrons
between those two nuclei,
that's a sweet spot.
Those electrons are going
to be very happy there,
and that will result in a more
stable, lower energy structure.
That's bonding.
But whenever there is a
positive event like this,
there's always a negative
event because that's
just how life works.
So we have bonding
orbitals, but we also
have antibonding orbitals.
So antibonding arise from
the linear combination
of atomic orbitals,
LCAO, through
destructive interference.
So here these are going to
be destructively interfering,
and that will generate
a molecular orbital
that is an antibonding orbital.
So here are our
little nuclei again,
and this antibonding orbital
is called sigma 1s star.
So we can write an
equation for this
as 1sa minus 1sb
equals sigma 1s star.
That is an antibonding
molecular orbital.
And let's just think about
how this kind of shape
arises considering the
wavelike properties
of these atomic orbitals.
So I'm going to now move
this equation up to the top.
And now, I have my
1sa here, my 1sb here,
and now it's destructive
interference.
So we have opposite phase
of the wave function.
And we'll put up a
wave function there.
Now, the next one has
the opposite phase,
so they're going to
destructively interfere.
There is overlap
over here, but when
you have destructive
interference,
then the amplitude
is going to decrease.
So now, when we consider
this destructive interference
between these two
orbitals, you see that you
have what arises between them.
Instead of enhanced probability
of finding an electron,
you actually get a node.
So you have decreased
amplitude translates
to decreased probability density
between these two nuclei-- one
here, one here-- and
that results in a node
between the two nuclei.
So in the antibonding
orbital, there
is a much lower
probability that it
will be in this sweet spot
between the two nuclei
that have this nice,
positive charge for it's
little negative
charge, so there's
pretty much no shot at being
in that nice, sweet spot.
And so what that
ends up meaning is
that an electron and
an antibonding orbital
is pretty much excluded
from that internuclear
region, that region
between those two nuclei,
and that results in
a molecular orbital
that has higher energy
than the atomic orbital.
There's just no chance of
being in that wonderful spot.
It's really very sad
for the poor electron
that has to occupy an
antibonding orbital.
So now, let's put this
on our energy scale.
So we'll go back to
our energy scale.
And we saw before that
when we had 1sa and 1sb
and you had a sigma
1s, a bonding orbital,
that was lower in energy.
Electrons that occupy it
are more stable compared
to their positions in the
atomic orbital, but we also
now have an antibonding orbital
from destructive interference
between the wave functions
of the atomic orbitals,
and that's higher in energy.
And so that's up here.
So this is what our
diagram is going
to look like that brings
two atomic orbitals together
to create two
molecular orbitals.
So the antibonding
orbital up here
is raised in energy
by the same amount
that the bonding
energy is lowered.
And so that gives
rise to this diagram.
And importantly,
as I mentioned, we
have N atomic orbitals
forming N molecular orbitals.
So if we have two
atomic orbitals,
we generate two
molecular orbitals,
one is bonding, lower in energy,
and one is antibonding, higher
in energy.
All right.
So let's take a look
at some examples.
So always start with hydrogen.
So hydrogen has how many
electrons?
One hydrogen, hydrogen atom.
AUDIENCE: One.
CATHERINE DRENNAN: One.
So we have two hydrogen
atoms, and so we
have two 1s orbitals, 1sa, 1sb.
And now where do we want
to put our electrons?
In the highest energy
possible, lower energy?
Where do we want to put them?
AUDIENCE: Lower.
CATHERINE DRENNAN: Always
start with lower energy.
So we're going to
put them down here.
And so this is now
the MO diagram for H2.
And we can write the
electron configuration
for the MO diagram, which you'll
note is a different electron
configuration than you were
writing when you were looking
at the periodic table because
now we're not writing it
in terms of 1s 2, 2s
2, we're writing it
in terms of molecular orbitals.
And there are some of
these on the problem set,
and I tried to indicate
example to make
sure you know what kind
of electron configurations
we're talking about.
So for MO diagrams, when it
says electron configuration
we're talking about where the
electrons are with respect
to the molecular orbitals.
So the answer to this
would be sigma 1s 2.
There are two electrons in the
molecular orbital, sigma 1s.
Good.
So that means that you can do
the same thing for dihelium,
and that's a clicker question.
So let's just take
10 more seconds.
OK, you can vote.
All right.
So does someone want to
tell me-- that is correct--
what's number two for?
For hydrogen. What about number
three, what was wrong there?
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: OK,
everyone's doing really well.
Yeah.
So all of the electrons
were parallel in there,
and what does that violate?
What would be true in that case?
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: Yeah.
And so they'd have the same
four principle quantum numbers.
This is not allowed.
So that one's not good.
And what's wrong with four?
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: Yeah,
star was on the bottom.
So they were flipped around.
Antibonding was lower in energy.
Great.
So we'll just put
those in over here,
and so we had two electrons.
Helium brings two.
So two went into
the bonding, and two
went into the antibonding
with opposite spins.
And we have the
bonding, lower energy;
antibonding, higher energy.
All right.
So then we can put in the
electron configuration here,
and we have sigma 1s
2, sigma 1s star 2,
so that's the electron
configuration.
Now interestingly,
you see, you have two
at lower energy and
two at higher energy,
so there's no net loss
or gain in energy of H2
compared to just
two helium atoms.
And so that raises the
question, does helium 2 exist?
So what would molecular
orbital theory
tell you about whether it
exists, and it would actually
predict that it does not exist.
And the way that molecular
orbital theory gives you
these predictions is through
the calculation of something
called bond order.
So bond order is
half of the number
of bonding electrons minus
the number of antibonding
electrons.
So let's just write out what the
bond order for helium would be.
So helium 2, our
dihelium molecule.
So we have a bond
order equals 1/2.
There's always 1/2.
And now, the number of
bonding electrons, so
how many bonding electrons
do we have for helium?
AUDIENCE: Two.
CATHERINE DRENNAN: Two.
How many antibonding electrons
do we have for dihelium?
AUDIENCE: Two
CATHERINE DRENNAN: Two.
And can someone do
this math for me?
AUDIENCE: Zero.
CATHERINE DRENNAN: Zero.
Right.
So that would
suggest a bond order
of 0, which means no bond.
And let's just compare
that to hydrogen, H2,
which should have a
bond order equal to 1/2.
It's always a 1/2.
In the hydrogen, how many
bonding electrons did we have?
AUDIENCE: Two.
CATHERINE DRENNAN: We had two.
How many antibonding
electrons did we have?
AUDIENCE: Zero.
CATHERINE DRENNAN: Zero.
And the math?
Bond order is?
AUDIENCE: 1.
CATHERINE DRENNAN: 1.
So that means 1 bond
or a single bond.
So MO theory would predict that
dihelium, has no bond, i.e.
it's not really a
molecule without a bond,
but H2 should exist.
So let's take a look at
what experiment tells us,
and it does exist, but
only really not that much.
So it was not discovered until
1993, which for some of you
might seem like quite
a long time ago,
but since we've
been mostly talking
about discoveries that
were made in the 1800s,
it took a long time
before someone could
prove that dihelium existed.
And if we look at the bond
association energy for H2,
0.01, and compare
that to H2, 432,
dihelium really doesn't
exist very much.
0 is a much better approximation
of its bond than 1 would be.
It really is not a
very good molecule.
So I would call this a win
for molecular orbital theory.
It correctly predicts
that our helium 2 is not
going to be a very
good molecule,
but H2 will be a good
molecule, and that works.
All right.
Now, let's consider 2s.
So 2s orbitals are
analogous to 1s
except that you have to
remember that they're bigger.
So we have our 1s and our 2s,
but for the purposes of this,
it doesn't really matter.
So let's look at a diagram
now that has 1s and 2s.
So we have lithium.
So how many electrons
does lithium have?
AUDIENCE: Three.
CATHERINE DRENNAN: Yep.
So we'll put on
lithium, dilithium.
We'll put on lithiums has
two in 1s and one in 2s,
and so we have one lithium
here and one lithium over here.
And our 1s orbitals are
going to be lower energy,
so they're down here.
Our 2s orbitals are
higher in energy.
So that goes for both the atomic
and the molecular orbitals
that are generated.
So bringing together 1s with
1s, we get sigma 1s and also
sigma star 1s.
And so we can start
putting our electrons in.
We're going to start with the
lowest energy and move on up.
So we have four electrons
that we need to put in,
so we fill up everything here.
And now we go up to our 2s.
The 2s orbitals will generate
signal 2s and sigma 2s star.
We have two electrons, so
they both can go down here.
So here is what our
MO diagram looks like.
We can write out the
electron configuration again,
based on this diagram.
So we have two electrons in
sigma 1s, so we have a 2 there.
We have two electrons in
sigma 1s star, our antibonding
orbital, and we have two
electrons in sigma 2s.
And now we can calculate
the bond order, which
is another clicker question.
OK, just 10 more seconds.
Excellent.
84%.
So if we do out the math
here, we always have our 1/2.
We have four bonding electrons.
We have two down
here and two up here.
We have two antibonding
electrons in our sigma 1s star,
and so that gives us
a bond order of 1.
And in fact, the
dissociation energy
does suggest there is a bond.
It's not a great bond.
It's 105 kilojoules per mole,
not necessarily enough to power
a starship, but still
this molecule does exist.
All right.
So let's look at beryllium
now, do another example.
So how many electrons is
beryllium going to have?
It will have four.
So we'll put those up.
Now, we're going to start with
our lowest energy orbital,
and we're going to put some
in the antibonding as well.
And then we'll do this,
and then we'll do that.
And we filled up everything.
And so we can write
out our configuration.
We have two in our
sigma 1s orbital.
We have two in
our sigma 1s star.
We have two in our sigma 2s,
and two in our sigma 2s star
antibonding orbital there.
And now there are
two different ways
that I can calculate
the bond order here,
and we'll do both of them and
show you that they come out
the same way.
One way that we
can calculate this
is to consider all electrons.
So if we consider all
electrons, our bond order,
and that's often just said
B.O., is 1/2 of our bonding.
And now we can count up
both the 1s and the 2s.
So how many bonding
electrons do we have?
We should have four, so we have
two here, two here, so four.
How many antibonding?
Also four, so that
suggest a bond order of 0.
Or we could just consider
our valence electrons,
so that would be
the electrons in 2s,
and if we do that,
bond order equals 1/2.
It's always a 1/2, so how
many valence electrons
do we have in bonding orbitals?
Two.
How many in
antibonding orbitals?
Two.
And that gives the answer of 0.
So it should always work unless
you do something very strange.
You should be able to do it both
ways and get the same answer.
So if this is a
complicated problem,
you might want to only
consider the valence electrons.
In fact, on a test,
you may only be
asked to draw the
molecular orbital diagrams
for the valence
electrons, and you don't
have to do the other ones.
So that should work both ways.
And we get a bond order
of 0, and in fact,
the dissociation energy is
only 9 kilojoules per mole.
It's a little stronger
than dihelium,
but this is an
exceedingly weak molecule.
So when you have the same
number of electrons and bonding
as antibonding, it doesn't
lead to a very strong molecule.
All right.
That's the orbitals.
Now, it's time for molecules
that have p orbitals as well.
So in this case, now I have my
p orbital and another p orbital.
And we have our
nuclei in the middle,
and we're going to
bring these together,
and we will have
constructive interference.
And so here we're
bringing together
2p x of a and 2p xb
or 2p ya and 2p yb.
So just x and y we're
considering right now.
And if we bring those together
with constructive interference,
then we're going to form
a bonding orbital that
has enhanced probability
density in both cases
and nodal plane
along the bond axis
because we had nodal planes
along there to begin with.
So if we think about this
and we have both of these,
we're bringing them
together and they're
going to interfere
constructively, enhance
probability density here,
enhance probability density
down here, but we still have
our nodal plane because we
started out with one.
And so if we have a nodal plane,
this cannot be a sigma orbital.
It has to be a pi orbital,
because sigma orbitals are not
going to have a nodal plane
along the bonding access.
All right.
So we could generate pi
2px or pi 2py this way.
So a pi orbital is
a molecular orbital
that has a nodal plane through
the bond axis or maybe I
should say along the bond axis,
so here is our nodal plane
right through the bond axis.
We can also have
antibonding, which
means destructive interference.
So now, I'm going
to be subtracting
one of these from
one of these, and I'm
going to get something
that looks like this,
and it will be pi 2px
star or pi 2py star,
and it will have
two nodal planes.
So let's look at this.
So this is destructive
interference.
I'm subtracting one of
these from one of these,
so the phase has to change.
So I'm going to change
the phase on one of them
and then bring them together.
Now, we're not going to have
that awesome, constructive
interference increased
probability density.
These are negatively
interacting with each other.
And this generates a nodal
plane between the molecules.
They really look much
more like this now,
so we have still our nodal
plane through the bond.
We had that before.
We're always going to have that.
But now we have an additional
nodal plane between the nuclei.
So in one case, we have enhanced
density, probability density
again.
And the other case,
in antibonding,
we have another nodal plane.
All right.
So now, let's look
at what happens
to the energies of
these pi orbitals.
And the diagrams that
I'm about to show you,
we're only talking
about px and py now.
We have for the
moment forgotten 2pz.
So these diagrams are
rated I for incomplete.
Warning to the viewer,
people come to me
and go where are the 2pz?
Yes, they're not
in these diagrams,
but when you're asked
for a complete diagram,
you will always have to
put those orbitals in.
And in fact, completing
these diagrams
could be a question
you get later.
But for now, we're
going to have 2pz here,
but it's not forming a molecular
orbital in this diagram.
This diagram is,
thus, incomplete,
but we're going to start simple
and build more complicated.
So only 2p orbitals first, and
then we'll add the third one.
Because this compound doesn't
actually need that orbital,
so we're good for now.
All right.
So we have moved on to
the first element that has
a p electron in a p orbital.
We have boron, and
we have two of them.
So now I dropped off
1s to simplify this.
Now, we just have our
valence electrons,
so it's a good thing we can
calculate bond order just using
our valence electrons.
All right.
So we have two in 2s
and one in 2px or 2py.
I could have put
it in either place.
So let's put in
where they would go.
So we have 2s orbitals.
They'll go into bonding
sigma 2s first, then
into antibonding, sigma 2s star,
and now we have p electrons,
and we're going to put
them into our pi orbitals,
our molecular orbitals.
I'll put one in, and where am
I going to put the other one?
Am I going to put it
next to it or over here?
What do you think?
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: Yeah.
So we're going to do that
because, again, when you're
going to sit on a bus, you want
to have if they're degenerate
in energy levels
as they are here,
you're always going to put
one electron in each orbital
of the same energy first
before you pair them up,
and they'll have parallel spins.
So we're reviewing
things we learned before.
I love doing that.
So now, let's see what our
electron configuration is,
and this is just the valence
electron configuration.
We're simplifying.
We're not going to
consider the 1s orbital,
and we can write this down.
So we have two
electrons in sigma 2s,
and we have two electrons in
sigma 2s star, our antibonding
orbital, and we have one in
pi 2px and one in pi 2py.
And I can put a
1 or not put a 1.
If I don't put anything,
for a 1, 1 is assumed.
And we can calculate
our bond order as well.
So we have 1/2, and
again, we're just
using our valence
electrons, but that's OK.
We have four now who are
bonding, two down here, two
up here, these are
bonding orbitals,
and we have two that
are antibonding.
And notice for our
pi orbitals, this
is what we saw before
bonding are lower in energy,
antibonding are
higher in energy.
With the bonding orbitals, we
had constructive interference,
enhanced probability of the
electrons near the nuclei,
and so that's lower in energy.
But in our antibonding
ones, we have a nodal plane
in between our nuclei.
So we don't have any
probability that electrons
are right in between there
because there's a nodal plane,
so those are higher in energy.
So here is our B2 diagram.
So now let's try the same
thing for carbon, C2,
and that is a clicker question.
All right.
Let's just take 10 more seconds.
So let's take a look
at that over here.
The easiest thing to do
to answer this question
was to fill in the
diagram in your handout.
And so if you did that, you
would have put two down here
and you would have
put two up here.
Then you would have put
one here, one there,
another one there,
and another one there.
So now we have used these
up, and so our configuration
is sigma 2s 2, sigma star
2, pi 2px 2, pi 2py 2.
And the bond order is 1/2.
There are six bonding
electrons-- 1, 2, 3, 4, 5, 6--
and two antibonding
electrons, and so that adds up
to a bond order of 2.
And so sometimes
on a test, they'll
be a simple question what is
the bond order, but to get there
you have to draw your whole
molecular orbital diagram
and figure out how many bonding
and how many antibonding,
so these are not really
that fast questions.
And it's nice.
Sometimes we give you
like a little space,
and you see this whole little
molecular orbital diagram
fit in there to
answer the question.
All right.
So let's just compare these
two diagrams for a minute.
So in both cases, we had 2s
orbitals, two atomic orbitals
for 2s, and they both generated
two molecular orbitals,
a bonding and an antibonding.
The bonding is lower in
energy, and the antibonding
is higher in energy.
We also had two 2px
atomic orbitals.
They generated 2
pi 2 px orbitals,
one bonding, one antibonding.
And the same for our two
atomic orbitals for 2 py.
We had two of those, and they
generated one lower energy
bonding, pi, bonding and
one pi star antibonding.
So you always have N
atomic orbitals generating
N molecular orbitals.
So the stability of
the resulting molecules
in these cases depend on
how many of the electrons
are bonding, how many
are in energy lower
as a result of formation
of the molecule,
and how many are at higher
energy as a result of formation
of the molecule.
And if the net result are more
electrons in lower energy, more
electrons in bonding
orbitals, then that molecule
is more stable.
If there's a very
slight or no difference,
then that's not a
very stable molecule.
So now let's just
compare these two
and think about which of these
is going to be more stable.
So we have our
configurations again.
So in the case of B2,
how many electrons
are in lower energy
or bonding orbitals?
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: Yeah,
we have four-- 1, 2, 3, 4.
How many in higher?
AUDIENCE: Two.
CATHERINE DRENNAN: Two.
Up here.
For carbon, we had six-- 1,
2, 3, 4, 5, 6-- two in higher.
And so the bond order here was
1, the bond order here was 2.
Which is more stable?
Higher dissociation energy,
which one do you think?
AUDIENCE: [INAUDIBLE].
CATHERINE DRENNAN: Carbon.
Right.
Has a bond order of two.
It has more electrons in
lower energy orbitals.
So it cannot really well
out of this bonding deal,
and the dissociation energy for
B2, 289, whereas, for C2, 599.
So when molecules come together
such that more of the electrons
are in lower energy
or bonding orbitals,
you form a nice,
stable molecule.
When molecules come
together such that more
of their electrons are an
antibonding or higher energy,
that's not a happy molecule.
So I'll just and with one
way to think about this.
In this cartoon
molecular, break up lines,
sometimes two atoms just
have an incompatible number
of valence electrons.
And there are just too many--
just here this molecule saying,
I'm sorry.
Too many of your electrons
are in my antibonding regions.
I don't know how many
times we've all heard that,
but it's time to dissociate, but
our atomic orbitals, well, they
can still be friends.
OK.
See you on Wednesday.
Take a look at the
clicker question.
All right.
Let's just take 10 more seconds.
All right.
Let's just go through this one.
So this is a review of
where we were last time.
So the correct answer is 1.
So sigma orbitals are
cylindrically symmetrical.
Let's quiet down a minute.
You can hear the answer.
So this one is true.
The second one is not true.
A bond order of zero
doesn't mean that you just
have antibonding orbitals.
Whenever you bring together
two atomic orbitals,
you have to make two
molecular orbitals.
So it isn't that sometimes
you make bonding orbitals
and sometimes you make
antibonding orbitals.
Every time you bring
together two atomic orbitals,
you make two molecular orbitals,
one that's lower in energy,
and that's the bonding
orbital, and one
that's higher in energy, and
that's the antibonding orbital.
So that is not what that means.
A bond order of
zero means that you
have equal numbers of electrons
in your bonding and antibonding
orbitals.
So there's no net stabilization
due to the formation
of these bonds.
So here bonding occurs
when you bring together
two atomic orbitals to make
two molecular orbitals that
are both of lower energy.
No.
Every time you make the two
orbitals, one is lower energy,
one is higher that energy.
You can't make two that
are both lower in energy.
And bond order of 1 means
constructive interference
is generated at one
bonding orbital.
That's not what a
bond order of 1 is.
And again, every time you
generate a bonding orbital,
you generate an
antibonding orbital.
And 1 means that you have
twice as many electrons
in your bonding orbitals
as antibonding orbitals
because the formula is
1/2 the number of bonding
minus antibonding.
But that's good, and it's
important to remember
that sigma orbitals are
symmetric around the bond axis.
All right.
So we had these diagrams
for boron and carbon,
just talking about the
interactions of the px and py,
and so we had
forgotten about our pz.
And you can't do that on a test.
You get into trouble,
so I'm always
telling people for
these two handouts,
you must include the
molecular orbitals that
are derived from p to z.
So on a test, you need to put
them even if they're empty.
Even if they don't
have anything in them,
they need to be part of your
molecular orbital diagram.
We didn't have
them in the diagram
because we hadn't
talked about them yet,
so now we're going
to talk about them.
So two pz orbitals, again,
this is on Monday's handout.
You have this linear
combination of atomic orbitals.
And all are p orbitals.
They all look the
same as each other,
they're just different
in orientation.
You have one along x,
one along y, one along z,
but they're the same.
So now we're going to bring
our two pz orbitals together,
and we're going to do it
along the bonding axis.
So we've defined this as the
bonding axis in the class,
so we'll bring them
together, and they'll
be constructive interference
with our bonding orbitals.
There's always constructive
interference that
generates bonding orbitals.
And so we're going to create a
enhanced amplitude as the wave
functions come
together, and it's
going to be
cylindrically symmetric.
So what type of orbital do
you think this is going to be,
sigma or pi?
AUDIENCE: Sigma.
CATHERINE DRENNAN:
It'll be sigma
because it's
cylindrically symmetric.
So we do not have any bonding
plane along the bond axis,
and it's symmetric around.
And we have enhanced
probability density,
and we have the wave
function squared
enhanced probability of having
an electron between the two
nuclei.
And so this is a sigma 2pz.
So p orbitals can form
sigma molecular orbitals.
So we do have nodes
passing through our nuclei.
Here are our nuclei again.
We do have them.
They were here before
in our p orbitals.
There's the nodal plane in our
p orbitals, but we do not have,
in this case, a node
along the bond axis.
So that is our bonding.
So whenever you generate
a bonding orbital, which
is going to be
lower energy, we're
going to have our increased
amplitude between the nuclei,
again, our increased probability
density and therefore,
lower energy.
So whenever you have
constructive interference
generating a molecular
orbital of lower energy,
you got to create
something of higher energy.
That's just how life works.
So we also are going
to have antibonding
orbitals, which are generated
by destructive interference.
And again, these
orbitals can be thought
about as wave functions,
and a property of waves
is that they constructively
interfere and destructively
interfere.
So now we can subtract
our two orbitals,
which we're going
to switch the sign
and they're going
to be out of phase.
So they'll
destructively interfere,
and that's going
to look like this.
So now, you're going to generate
a nodal plane between the two
nuclei, but it's still
symmetric around the bond axis.
So this is a sigma 2pz star,
so it's an antibonding orbital.
So again, it's
sigma, so it's still
cylindrically symmetrical with
no nodal plane along that bond
axis, but you do have a new
nodal plane that's generated.
So nodes pass, again,
through the nuclei,
but also now between
these two orbitals.
So we have a new
nodal plane that's
generated that's
between these nuclei,
and that's a result of
destructive interference.
Generates that nodal plane.
You have decreased
probability density
for an electron
being found there.
And so that poor electron is
shut out of that sweet spot.
The electrons like to be
between those two nuclei
where they have the two positive
charges of the nuclei and then
their little negative charge,
and they can sit right there
and be very happy in
a low-energy state.
But here there's really
lower probability density,
lower likelihood the
electron will be found here,
and that generates
a molecular orbital
that's antibonding
or higher in energy.
All right.
So now, we have to go
back to our MO diagrams
and figure out where to put
these new molecular orbitals
onto our nice diagrams.
And it's not as simple as
it was before because where
we put these new sigma 2pz
molecular orbitals depends
on what z is.
So it depends on the value of z.
So if z is less than
the magic number of 8,
then we have our pi 2px and 2py
orbitals are lower in energy
than our sigma 2pz
molecular orbital.
But if we are equal
to or greater than 8,
then the sigma 2pz
orbital is lower energy
than the pi 2px and 2py.
So less than 8, pi is first.
It's lower in energy as you
go up your energy scale.
And if z is equal to or
greater than 8, pi is second
and sigma is first.
So how are you going
to remember this?
There could be many different
ways one can remember it,
but I'll tell you
how I remember it.
And my life revolves around
my daughter and my dog.
And so at Thanksgiving, we
always have the question,
can I eat pie first?
So if you are
under the age of 8,
you always want your pie first.
So if z is less than
8, pi comes first.
Pi is lower in energy
and sigma is higher.
However, when you mature
to the grand age of 9,
say, or 10, if
you were a kid, 10
is like the oldest
you can possibly
imagine being, very mature.
And you can eat your
Thanksgiving dinner
and wait for pie.
So that is how I
would remember it,
under 8 pi is first,
equal to 8 or greater,
you can wait till after
dinner to have your pie,
pi comes second.
Note that the ordering of
the antibonding orbitals
is the same, so all
you have to remember
is down here depends on z,
is pi first or is pi second?
So let's take a
look at an example.
Let's look at our friend
molecular oxygen that has a z
equal to 8.
So oxygen is at the
old, mature age of 8,
and so it's going to
have its sigma 2pz first.
It can wait for its pi
orbitals until later.
So let's start putting
in our electrons.
So we have each oxygen making
up molecular oxygen, or O2.
Brings two 2s orbitals to the
Thanksgiving dinner table,
and two of them
go down in energy
into the bonding sigma
2s orbital and two
go into our antibonding,
sigma star 2s orbital.
Now, we have four
electrons in our atomic pz
orbitals, four
from each molecule,
so we need to put
all of those in.
So we always start with
the lowest energy orbital.
So we'll put two in
there, then we'll go up.
We have two more
here, two more here.
We'll put them in singly
with their spins parallel,
and then we'll pair them up
in the lowest energy orbitals.
And then we have two
more left, so we're
going to have to put those up
in our antibonding orbitals.
So they go into pi 2px and
pi 2py star orbitals up here.
All right.
So now we can calculate
the bond order for oxygen,
and that's a clicker question.
All right.
Let's just take 10 more seconds.
All right.
So let's take a look over here.
So bond order, again,
is 1/2 the number
of bonding electrons minus
the number of antibonding
electrons.
And we have eight bonding--
1, 2, 3, 4, 5, 6, 7, 8.
And we have four
antibonding-- 1, 2, 3, 4.
So that gives us
a bond order of 2.
And the bond order
equation is one
that you do have to memorize.
That will not be given
to you on an exam.
And with a bond order of 2,
we have a pretty big number
for dissociation energy.
Again, that's energy you have
to put into a bond to break it,
to dissociate it, and that
means it's a pretty strong bond
if it's a big number.
If you need a lot of energy,
that's a strong bond.
Another thing that you
can see from this diagram
is that O2 is a biradical.
It has two lone pair electrons,
so two unpaired electrons,
which also makes
it paramagnetic,
or attracted to
a magnetic field.
So whenever you have
unpaired electrons,
you will have a
paramagnetic species,
and diamagnetic means
they're all paired.
And so there are some questions
on problem sets and on exams,
so you need to know the
definitions of those.
All right.
So we talked about this when
we were doing Lewis structures,
if you recall.
And we drew a beautiful Lewis
structure of molecular oxygen
that had two lone pairs on
each oxygen and a double bond.
But I told you that
that was not really
a complete description
of molecular oxygen,
that molecular oxygen
was actually a biradical,
but you would not get a hint of
that from the Lewis structure.
So if you draw the Lewis
structure as a biradical,
then you have a
single electron here
and a single electron
there and a single bond.
But we also see
the bond orders 2,
so this one describes
the biradical nature,
but doesn't really
describe that double bond
character that it has.
So neither of these Lewis
structures really completely
describe molecular
oxygen. And we
need our molecular orbital
diagram to really help
us understand the properties
of molecular oxygen,
that it's pretty
strong bond between it,
it has double bond character,
but it also is very reactive.
It's a biradical.
So this is just a
bizarre molecule.
And really that is why this
diagram right here really tells
us why our life, our
planet is what it is,
right here, that describes it.
It's because of
oxygen that we have
the life forms that we did.
Life was very different
on this planet
before molecular
oxygen came about.
You did have microbes that lived
anaerobically without oxygen.
But when oxygen came,
everything changed.
And so if I say, can
you explain life to me?
You can draw this,
and there it is.
This explains life as
we know it because life
as we know it exists because of
this crazy molecule that is O2,
nothing else really like it.
It is an amazing molecule
that allows us to live.
So there you go.
I was going to say you don't
learn anything in chemistry.
This diagram I
just explains life
as we know it to
you, right there.
All right.
It's not just oxygen, there's a
few other elements that, yeah,
are pretty important,
and one of them
is nitrogen. We wouldn't
really be much anywhere
without nitrogen either.
Oxygen is, no, nitrogen
is pretty special too.
This also helps.
Maybe these two
molecular orbital
diagrams really sum
up life as we know it.
All right.
So let's look at
molecular nitrogen, N2.
So we have two electrons
in our 2s orbitals,
so we're going to
bring them together.
We're going to put two in the
lower energy orbital, our sigma
2s, and two in our
antibonding orbital.
And then we have three
in our p orbitals
from this nitrogen, three
electrons from this nitrogen,
and so we'll put them down here.
We'll put in 2, 3, 4, 5, 6.
So we didn't need to use any
of these antibonding orbitals,
and this is a z
less than 8 case.
So here we have our
pi orbitals first
because it is z less than 8.
So let's look at
the bond order here.
And the bond order is 1/2
of 8-- 1, 2, 3, 4, 5, 6, 7,
8-- eight bonding
electrons and just
these two antibonding electrons,
so 1/2 of 8 minus 2 is 3.
And you have a really big number
for your dissociation energy,
941.
This is a very stable
molecule, and you
can draw the Lewis structure of
this without much difficulty.
This one works quite well,
and you get a triple bond
and two lone pairs.
So this would be a
diamagnetic molecule,
no unpaired electrons,
no radicals here.
But this is a
crazy, strong bond.
It's really hard
to split nitrogen
because of this triple bond.
And molecular orbital
theory tells you
it should be its triple bond.
It should be really,
really strong interactions
between these nitrogens.
And so this is a
very hard thing.
We want to do is
industrially, and we'll
talk more about this when we
get a chemical equilibrium.
How do you split
the nitrogen bond.
This is actually, currently,
a big area of research,
how you break that bond because
we need nitrogen for life.
And so how do you split it?
There's lots of nitrogen
N2 in the atmosphere,
but we need it here,
and we need it usable,
so we need to break
that bond to use it.
So we'll come back to this
in chemical equilibrium
and how people are able to
split the nitrogen bond.
So forget man of steel.
We should have man of
nitrogen. That's strong.
Nitrogen is strong.
Forget steel.
Man of nitrogen.
One more thing to do before
we move on to today's handout
and this is really
fast, you can also
be asked to draw molecular
orbital diagrams where
both atoms are not the same.
So if you were asked
to do that, you
can use the following rules.
If z is less than 8 for
both atoms, pi is first.
If z is not less than
8 for both atoms,
it's hard to know
what to do, so you
don't need to know what to do.
And that's it.
So we might tell you
something about it
and then you can do
it, but otherwise, just
worry about things
when z is less than 8.
So that's molecular
orbital theory.
