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PROFESSOR: OK, I want to
take 10 more seconds

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now the clicker slide.
 

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This is giving us one more
try on the vsper geometries,

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because it didn't go
so well on Wednesday.

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All right, excellent.
 

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So that is a very good job.
 

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Let's quickly go over why.
 

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We have p h 3.
 

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We told you that phosphorous
has 5 valence electrons plus 3

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from each of the hydrogens, so
we have a total of 8

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valence electrons.
 

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How many do we need to get full
valence shells everywhere?

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STUDENT: [INAUDIBLE]
 

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PROFESSOR: 14.
 

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So, we need 14 minus 8.
 

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That leaves us with 6
bonding electrons.

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And if we put that in our bond
here, we have 1, 2, 3 bonds,

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plus we have one lone
pair left over.

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So this is our Lewis
structure here.

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If these bonds were all
completely of equal distance

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apart, whether is was a lone
pair or bonding electrons,

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the angles would be 109 .
 

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5 degrees.
 

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But because there's this lone
pair here, it's pushing down on

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the other bonds, so we end up
with an angle of

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less than 109 .
 

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5 degrees.
 

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All right, so let's switch
over to notes for today.

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So we're going to finish
talking about molecular orbital

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theory, and then we'll switch
over to discussing bonding in

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larger molecules, even larger
than diatomic, so we'll move on

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to talking about valence bond
theory and hybridization.

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So, clearly you don't have your
notes in front of you yet, so

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you can just listen,
take it all in.

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What I'll do is I'll post the
notes filled in to the point

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where you actually get
your class notes today.

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So, this will be a little bit
more like a seminar to start

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with, and a little bit less
like a lecture in class.

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But let's go ahead and start
our discussion in terms of

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molecular orbital theory.
 

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So where we had left off with
was we'd fully discussed up to

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the point of considering
homonuclear diatomic molecules,

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so molecules that both
have the same nucleus.

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And where we had left off was
we were going to start one

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example of thinking about now
where we have a heteronuclear

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diatomic molecules, so two
different atoms in terms

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of forming the molecule.
 

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But first, I just want to
remind you when we're talking

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about molecular orbital theory,
this is treating electrons as

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waves, so what we're actually
able to do is either

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constructively or destructively
combine atomic orbitals to

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form molecular orbitals.
 

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So you should remember that any
time we combine 2 s orbitals,

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what we're going to find is if
we constructively interfere

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those two orbitals, we're going
to form a bonding orbital.

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And that's going to be lower
in energy than the two

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individual atomic orbitals.
 

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And we call that, for this
case, our sigma 2 s orbital.

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In contrast, if we have
destructive interference, what

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we're going to form is a sigma
2 s star, and what does

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the star designate?
 

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STUDENT: [INAUDIBLE]
 

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PROFESSOR: Anti-bonding, yup.
 

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So it's an
Anti-bonding orbital.

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It's going to be higher in
energy than the individual

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atomic orbitals.
 

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All right, great.
 

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So, I think we have these
molecular orbital energies

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down, so let's move on to
talking about more

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complex molecules.
 

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And to do this we're going
to introduce valence bond

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theory, and the idea of
hybridization of orbitals.

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So the idea behind valence
bond theory is very

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easy to understand.
 

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Essentially what you have is
bonds resulting from the

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pairing of unpaired electrons.
 

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So the simplest case we can
think of is with h 2 where we

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have two unpaired electrons,
each in a 1 s orbital

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of a separate h atom.
 

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And if we picture those
two coming together, we

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form the h 2 molecule.
 

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And again, we have the pairing
of the unpaired electrons, and

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we have two orbitals
coming together.

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So in molecular orbital theory,
what we did was we named

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orbitals based on
their symmetry.

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In valence bond theory, the
focus is on discussing the

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bonds, but it should look very
familiar to you, because

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there's two types of bonds that
we want to discuss here.

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We want to discuss sigma
bonds and pi bonds.

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So this is very similar to what
we saw in terms of sigma

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orbitals and pi orbitals.
 

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So in this first case
here, what we're seeing

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is a sigma bond.
 

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And a sigma bond forms any time
you have two orbitals coming

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together and interacting on
that internuclear axis.

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So we talk about a sigma bond
as being cylindrically

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symmetric about the bond axis,
and it's important to point out

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that it has no nodal plane
across this bond axis.

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This is in direct contrast
to when we're thinking

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about pi bonds.
 

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So pi bonds have electron
density both above and below

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the bond axis, but they
actually have a nodal plane at

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this z, this bond axis here.
 

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And remember for this class,
we always define z as the

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internuclear or the bond axis.
 

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So it might look like here, if
you don't understand about p

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orbitals, which I know all you
do, but if someone else was

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just looking and seeing,
it kind of looks like

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there's two bonds here.
 

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There's not two bonds, that's
one pi bond, and the reason is

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because it's 2 p orbitals
coming together, and remember p

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orbitals have electron density
above and below the axis, so

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when they come together, it
kind of looks like one bonds,

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but essentially what we
have here is one pi bond.

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So let's think about how we can
classify single and double and

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triple bonds, which is what
we're really used to dealing

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with in terms of these sigma
bonds and these pi bonds.

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So, if we take a look at what
a single bond is, and let me

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grab some molecules here.
 

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If we're talking about a single
bond, we're talking about 2

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orbitals overlapping in
the internuclear axis.

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So if we have a single bond
here, would you consider that

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a sigma bond or a pi bond?
 

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STUDENT: [INAUDIBLE]
 

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PROFESSOR: Right,
it's a sigma bond.

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Essentially what we're
seeing is overlapping

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in this z axis here.
 

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In contrast, if we talk about a
double bond, what we're now

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talking about is having
both a sigma bond and

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also one pi bond.
 

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And I apologize, I intended to
set this up right before class,

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but that didn't happen today.
 

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All right, so what we see here
is we have our sigma bond

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that's along the internuclear
axis here, but we also have a

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pi bond, because each of these
atoms now has electrons in it's

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in a p orbital, so we're going
to overlap of electron density

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above and below the bond.
 

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So that's exactly what our
definition of a pi bond

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is, so we have one sigma
bond, and one pi bond.

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00:07:20 --> 00:07:22
So now let's think
about a triple bond.

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A triple bond, again is going
to have one sigma bond on

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the internuclear axis.
 

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How many pi bonds
would you expect?

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STUDENT: [INAUDIBLE]
 

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PROFESSOR: Two, great.
 

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So, we're going to
see two pi bonds.

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The first one will be above and
below the bond axis is where

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we'll see the electron density,
and the second will be

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perpendicular to that, so it
will be a density in front of

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and behind the bond axis.
 

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So we can kind of flip it this
way -- this will be one pi

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00:07:48 --> 00:07:51
bond, this will be another
interacting between

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these p orbitals.
 

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All right, so that's really all
there is to thinking about

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valence bond theory in terms
of the most simple

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explanation here.
 

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But what we're going is we're
going to start trying to apply

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it to a molecule, and I
actually picked a molecule that

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it's not going to work for,
even though it would work even

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just at this level for
many, many molecules.

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And I picked looking at methane
so we could see if there are

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other factors that we're not
considering, that we need to

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maybe tweak our model a little
bit, and I think we'll find

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that we do if we take a look at
a polyatomic molecule,

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methane, so c h 4.
 

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So let's think about methane
using valence bond theory.

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So, using our simple valence
bond theory, what we would

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expect is that we want to pair
up any unpaired electrons in

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methane with unpaired electrons
from hydrogen and form bonds.

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But what we see we have
is that we only have two

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unpaired electrons here.
 

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Because we have paired set in a
2 s orbital, so all we're left

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essentially is two electrons
that are available for bonding.

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So this should immediately look
like a problem because we know,

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in fact, that methane is
tetravalent, and this is

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telling us it's only divalent.
 

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Essentially it would only
allow for us to bond

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to two hydrogen atoms.
 

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So if it did this, it now looks
like, from looking at the

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paired electrons that we have a
stable structure here, and our

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structure is not c h 4, it's a
stable structure of c h 2, and

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it will actually predict,
also, what this h c

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h bond angle it is.
 

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So according to this model
what is that bond angle?

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STUDENT: [INAUDIBLE]
 

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PROFESSOR: One more time.
 

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OK, I hear a mix.
 

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So, according to this model,
what we're seeing is a

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bond angle of 90 degrees.
 

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What do you know the
bond angle should be?

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It's 109 .
 

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5 is what we would expect
for methane because it's

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tetravalent, but here we're
just seeing something that's

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divalent, and they're both
in p orbitals that are

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perpendicular to each other.
 

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So what we're predicting is
a bond angle of 90 degrees.

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00:10:01 --> 00:10:05
This is totally wrong, this is
the wrong picture altogether.

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00:10:05 --> 00:10:07
If you had your notes, you
could do some fun scribbling

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00:10:07 --> 00:10:10
right now, so you
can do that at home.

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We're going to need to tweak
our explanation here, and take

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into account another factor,
and that factor is the fact

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that we know that we must have
four unpaired electrons in

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carbon if we're going
to form four bonds.

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00:10:24 --> 00:10:26
So the way that we can explain
this is through something

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00:10:26 --> 00:10:29
called electron promotion and
hybridization of

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00:10:29 --> 00:10:30
atomic orbitals.
 

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00:10:30 --> 00:10:34
So let's take a look at
what we mean by this.

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00:10:34 --> 00:10:39
So if we take our carbon atom
here, which has two electrons

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00:10:39 --> 00:10:44
in the 2 s orbital, and we
promote one of these electrons

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00:10:44 --> 00:10:49
into a 2 p orbital, what we see
now is that yes, we do, we

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00:10:49 --> 00:10:50
have four unpaired electrons.
 

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00:10:50 --> 00:10:54
So, looking at this, this might
not look so good for you.

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00:10:54 --> 00:10:59
What we're proposing here is
that you take a nice low energy

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00:10:59 --> 00:11:02
s electron and move it into
a higher energy p orbital.

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00:11:02 --> 00:11:05
And the truth is that yes, this
costs energy, we're going up

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00:11:05 --> 00:11:07
to a higher energy state.
 

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00:11:07 --> 00:11:10
But it doesn't actually cost as
much energy as you might think,

226
00:11:10 --> 00:11:13
because in this s orbital here
we have a paired electron

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situation where we're moving up
to a p orbital where the

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electron is no longer paired,
so it won't feel quite as much

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electron repulsion, but
nonetheless, this is

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going to cost us energy.
 

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So we'll have to think about
where that energy is going to

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come from and we'll see
that in just a minute.

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But let's assume that this is,
in fact, going to happen.

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So now what we have is
four unpaired electrons.

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That's great, but it's still
not quite the picture we need,

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00:11:36 --> 00:11:39
because actually, all the
electrons are not in equal

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00:11:39 --> 00:11:43
orbitals -- one's in an s
orbital, and 3 are in p.

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But what we need to remember
is the fact that we're

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00:11:45 --> 00:11:47
talking about electrons
which are waves.

240
00:11:47 --> 00:11:50
When we're talking about
orbitals, we're talking

241
00:11:50 --> 00:11:51
about wave functions.
 

242
00:11:51 --> 00:11:53
So we can actually
constructively and

243
00:11:53 --> 00:11:57
destructively combine these
waves, these atomic

244
00:11:57 --> 00:11:59
orbitals to make a hybrid.
 

245
00:11:59 --> 00:12:02
So if we go ahead and hybridize
our p orbitals and our s

246
00:12:02 --> 00:12:06
orbitals, we'll switch from
having these original

247
00:12:06 --> 00:12:10
orbitals to having something
called hybrid orbitals.

248
00:12:10 --> 00:12:13
And hybrid orbitals are all
going to be completely equal,

249
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and you'll notice that they're
higher in energy than the

250
00:12:16 --> 00:12:20
s orbital, and lower in
energy than the p orbital.

251
00:12:20 --> 00:12:23
That should make sense because
they come from combining s

252
00:12:23 --> 00:12:25
orbitals and p orbitals.
 

253
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And specifically, when we give
them a name it's very clear

254
00:12:28 --> 00:12:30
exactly which orbitals they
come from combining, we're

255
00:12:30 --> 00:12:34
calling these s p 3 orbitals --
that's because they come from

256
00:12:34 --> 00:12:38
combining 1 s orbital
and 3 p orbitals.

257
00:12:38 --> 00:12:41
You should never get the names
of hybrid orbitals wrong

258
00:12:41 --> 00:12:42
because it's very
straightforward.

259
00:12:42 --> 00:12:46
If it has 1 s and 3
p's, we call it s p 3.

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00:12:46 --> 00:12:49
Naming doesn't always make
sense in chemistry, so I

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00:12:49 --> 00:12:51
like to point out this is
a place where naming does

262
00:12:51 --> 00:12:52
make a lot of sense.
 

263
00:12:52 --> 00:12:56
All right, so let's consider
our methane situation now that

264
00:12:56 --> 00:12:58
we have our hybrid orbitals.
 

265
00:12:58 --> 00:13:01
So I want to mention also,
these are exactly equivalent,

266
00:13:01 --> 00:13:04
they're equivalent in energy,
they're equivalent in shape.

267
00:13:04 --> 00:13:06
The only thing that is
different about these orbitals

268
00:13:06 --> 00:13:08
is their orientation in space.
 

269
00:13:08 --> 00:13:10
So actually, first let's
take a look at how we

270
00:13:10 --> 00:13:11
got these orbitals.
 

271
00:13:11 --> 00:13:14
We got them from combining
again, 1 s orbital

272
00:13:14 --> 00:13:17
and the 3 p orbitals.
 

273
00:13:17 --> 00:13:20
If we hybridize these, what
we end up seeing are these

274
00:13:20 --> 00:13:22
four hybrid orbitals.
 

275
00:13:22 --> 00:13:24
You'll notice the shape is the
same, all that's different

276
00:13:24 --> 00:13:26
is their orientation.
 

277
00:13:26 --> 00:13:29
So essentially, each of these
orbitals come from linear

278
00:13:29 --> 00:13:33
combinations of all of the
original orbitals, and it's

279
00:13:33 --> 00:13:36
hard to picture exactly how
that happens, but one that you

280
00:13:36 --> 00:13:38
can at least start to get an
idea is if you think about

281
00:13:38 --> 00:13:42
combining the 2 s and the 2 p z
here, which is not quite

282
00:13:42 --> 00:13:44
accurate because of course,
we're combining all of them.

283
00:13:44 --> 00:13:46
But it would just give you a
little bit of an idea of shape.

284
00:13:46 --> 00:13:50
You can see if we combine the s
with the top lobe of the p,

285
00:13:50 --> 00:13:52
they're going to constructively
interfere because they

286
00:13:52 --> 00:13:53
have the same sign.
 

287
00:13:53 --> 00:13:57
So you see in the hybrid
orbital we actually have a

288
00:13:57 --> 00:14:01
larger lobe on top where they
constructively interfered.

289
00:14:01 --> 00:14:04
If you compare the s orbital
with the bottom lobe, these

290
00:14:04 --> 00:14:05
have a different sign
so they're going to

291
00:14:05 --> 00:14:07
destructively interfere.
 

292
00:14:07 --> 00:14:10
So what you see is actually a
diminished lobe on the back

293
00:14:10 --> 00:14:12
part of this s p 3 orbital.
 

294
00:14:12 --> 00:14:16
So s p 3 orbitals always
have one huge lobe and

295
00:14:16 --> 00:14:17
one really little lobe.
 

296
00:14:17 --> 00:14:20
A lot of times when people draw
them, they even only draw the

297
00:14:20 --> 00:14:22
big lobe just to keep their
paper looking nicer, but there

298
00:14:22 --> 00:14:27
is that little tiny lobe
on the other side.

299
00:14:27 --> 00:14:31
All right, so in terms of s p 3
hybrid orbitals, let's combine

300
00:14:31 --> 00:14:35
all four together on one axis,
because this is what's going to

301
00:14:35 --> 00:14:37
happen in an s p 3 carbon atom.
 

302
00:14:37 --> 00:14:40
So in this case, what would you
say that the angle is here?

303
00:14:40 --> 00:14:43
STUDENT: [INAUDIBLE]
 

304
00:14:43 --> 00:14:43
PROFESSOR: Right, great.
 

305
00:14:43 --> 00:14:45
So, we've achieved the angle
that we observed, which

306
00:14:45 --> 00:14:47
is good, which is a 109 .
 

307
00:14:47 --> 00:14:48
5.
 

308
00:14:48 --> 00:14:52
So we can think about now doing
bonding, and now we have

309
00:14:52 --> 00:14:56
four equal orbitals with
one electronic each.

310
00:14:56 --> 00:14:58
So we can bring in four
hydrogen atoms, which will

311
00:14:58 --> 00:15:01
each contribute another
unpaired electron.

312
00:15:01 --> 00:15:05
So now what we have
is four bonds.

313
00:15:05 --> 00:15:08
And we can think about where
we did get that energy for

314
00:15:08 --> 00:15:10
electron promotion that I
mentioned before where we

315
00:15:10 --> 00:15:13
moved the electron from
the 2 s to the 2 p.

316
00:15:13 --> 00:15:14
We get that from bonding.
 

317
00:15:14 --> 00:15:16
We're going to release a lot of
energy for bonding, it's going

318
00:15:16 --> 00:15:18
to more than make up for the
fact that we actually had

319
00:15:18 --> 00:15:22
to spend some energy to
promote that electron.

320
00:15:22 --> 00:15:26
So, we can think about now
how do we describe this bond

321
00:15:26 --> 00:15:28
in valence bond theory.
 

322
00:15:28 --> 00:15:30
So the way that you describe
a bond is you describe the

323
00:15:30 --> 00:15:33
orbitals that the bond
comes from, and also the

324
00:15:33 --> 00:15:34
symmetry of the bond.
 

325
00:15:34 --> 00:15:38
So would you expect this to be
a pi bond or a sigma bond here?

326
00:15:38 --> 00:15:41
STUDENT: [INAUDIBLE]
 

327
00:15:41 --> 00:15:43
PROFESSOR: OK, so I'm
hearing some mixed answers.

328
00:15:43 --> 00:15:45
It turns out that
it's a sigma bond.

329
00:15:45 --> 00:15:49
The reason that it's a sigma
bond is because the s p 3

330
00:15:49 --> 00:15:52
hybrid orbital is directly
interacting with the 1 s

331
00:15:52 --> 00:15:55
orbital of the hydrogen atom,
and that's going to happen on

332
00:15:55 --> 00:15:58
the internuclear axis, they're
just coming together.

333
00:15:58 --> 00:16:01
Any time two orbitals come
straight on together in that

334
00:16:01 --> 00:16:05
internuclear axis, you're
going to have a sigma bond.

335
00:16:05 --> 00:16:09
So if we go ahead and name this
bond, what we're going to name

336
00:16:09 --> 00:16:16
it is sigma, because that's the
-- basically the shape of

337
00:16:16 --> 00:16:19
the bond or that's how our
bond is coming together.

338
00:16:19 --> 00:16:21
And then we're going to name
the atomic orbitals that make

339
00:16:21 --> 00:16:26
it up, and it's being made up
of a carbon 2 s p 3 orbital,

340
00:16:26 --> 00:16:30
and a hydrogen 1 s orbital.
 

341
00:16:30 --> 00:16:32
All right, so let's think of
a case now that's getting a

342
00:16:32 --> 00:16:33
little bit more complicated.
 

343
00:16:33 --> 00:16:36
We were talking about
methane, which has

344
00:16:36 --> 00:16:37
only one central atom.
 

345
00:16:37 --> 00:16:40
We can also talk about
atoms that have two or

346
00:16:40 --> 00:16:41
more central atoms.
 

347
00:16:41 --> 00:16:44
So let's talk about ethane
now, which is c h 2.

348
00:16:44 --> 00:16:48
So let's take our carbon s p 3
hybridized carbon and just move

349
00:16:48 --> 00:16:53
it around here so we can make
the z inter- bonding

350
00:16:53 --> 00:16:56
axis between the two
carbons right here.

351
00:16:56 --> 00:16:58
So if we still have
an angle of a 109 .

352
00:16:58 --> 00:17:03
5 degrees, and again, we still
have four unpaired electrons

353
00:17:03 --> 00:17:06
available for bonding, we can
make one of those bonds with

354
00:17:06 --> 00:17:10
another s p 3 hybridized
carbon, so we're going to

355
00:17:10 --> 00:17:13
make up one pair here.
 

356
00:17:13 --> 00:17:16
If we think about that, that's
a sigma bond, right, they're

357
00:17:16 --> 00:17:22
coming together along
the nuclear axis.

358
00:17:22 --> 00:17:25
We also have six spots
available to form hydrogen

359
00:17:25 --> 00:17:28
bonds, so we can go
ahead and fill in those

360
00:17:28 --> 00:17:30
electrons as well.
 

361
00:17:30 --> 00:17:33
So in terms of thinking about
ethane, we actually have two

362
00:17:33 --> 00:17:38
bond types that we're going to
be describing just in terms of

363
00:17:38 --> 00:17:42
the carbon-carbon bond and
then the carbon h bonds.

364
00:17:42 --> 00:17:45
So let's talk about ethane
and how we would actually

365
00:17:45 --> 00:17:47
write these bonds.
 

366
00:17:47 --> 00:17:52
If we have the molecule ethane,
then what we're going to have

367
00:17:52 --> 00:17:55
first is our sigma bond that we
described between

368
00:17:55 --> 00:17:57
the two carbons.
 

369
00:17:57 --> 00:18:00
So it's going to be carbon,
and then what's the

370
00:18:00 --> 00:18:01
hybridization here?
 

371
00:18:01 --> 00:18:05
STUDENT: [INAUDIBLE]
 

372
00:18:05 --> 00:18:06
PROFESSOR: All right, start
again, what's the hybridization

373
00:18:06 --> 00:18:07
of the carbon atom?
 

374
00:18:07 --> 00:18:10
STUDENT: [INAUDIBLE]
 

375
00:18:10 --> 00:18:14
PROFESSOR: OK, so it's 2
s p 3, and our second

376
00:18:14 --> 00:18:17
carbon is also 2 s p 3.
 

377
00:18:17 --> 00:18:21
All right, so this is our
first type of bond here.

378
00:18:21 --> 00:18:24
Our second bond is going
to be between the carbon

379
00:18:24 --> 00:18:25
and the hydrogen atoms.
 

380
00:18:25 --> 00:18:28
Is that a sigma or a pi bond?
 

381
00:18:28 --> 00:18:28
STUDENT: [INAUDIBLE]
 

382
00:18:28 --> 00:18:30
PROFESSOR: Sigma, good.
 

383
00:18:30 --> 00:18:35
So again, our carbon is
going to be 2 s p 3.

384
00:18:35 --> 00:18:37
And what will our hydrogen be?
 

385
00:18:37 --> 00:18:40
1 s -- we don't have to
hybridize it, it already has

386
00:18:40 --> 00:18:43
only one unpaired electron
in a 1 s orbital.

387
00:18:43 --> 00:18:47
All right, so that's how
we describe ethane.

388
00:18:47 --> 00:18:49
We don't have to just stick
with carbon, we can think about

389
00:18:49 --> 00:18:52
describing other types of atoms
as well using this

390
00:18:52 --> 00:18:54
hybridization.
 

391
00:18:54 --> 00:18:57
For example, we can talk about
nitrogen, and nitrogen has five

392
00:18:57 --> 00:19:00
valence electrons shown here.
 

393
00:19:00 --> 00:19:03
Would you expect to see
electron promotion in nitrogen

394
00:19:03 --> 00:19:05
where we pull one of these 2 s
electrons into one of

395
00:19:05 --> 00:19:06
the 2 p orbitals?
 

396
00:19:06 --> 00:19:08
STUDENT: [INAUDIBLE]
 

397
00:19:08 --> 00:19:09
PROFESSOR: No, good.
 

398
00:19:09 --> 00:19:13
So, electron promotion does not
happen in terms of nitrogen,

399
00:19:13 --> 00:19:15
because it would not
increased our number of

400
00:19:15 --> 00:19:16
unpaired electrons.
 

401
00:19:16 --> 00:19:19
No matter what we do in terms
of promotion, we're always

402
00:19:19 --> 00:19:22
going to have three
unpaired electrons.

403
00:19:22 --> 00:19:25
We can still hybridize all
these orbitals, however, so we

404
00:19:25 --> 00:19:30
can still form four hybrid
orbitals, which are again,

405
00:19:30 --> 00:19:35
2 s p 3 hybrid orbitals.
 

406
00:19:35 --> 00:19:38
So if we take a look at
nitrogen here, what you'll

407
00:19:38 --> 00:19:40
notice is we have thre
available for bonding, and we

408
00:19:40 --> 00:19:42
already have our lone pair --
one of our orbitals is

409
00:19:42 --> 00:19:45
already filled up.
 

410
00:19:45 --> 00:19:49
So we can add three hydrogen
atoms here, and fill in our

411
00:19:49 --> 00:19:50
other orbitals right here.
 

412
00:19:50 --> 00:19:54
So if we do this and we form
the molecule ammonia, let's

413
00:19:54 --> 00:19:57
switch to a clicker question,
and have you tell me what the

414
00:19:57 --> 00:20:00
bond angle is going to be in
ammonia -- the h

415
00:20:00 --> 00:20:02
n h bond angle.
 

416
00:20:02 --> 00:20:05
Actually, let me draw it on the
board as you look -- actually,

417
00:20:05 --> 00:20:08
can you put the class notes on,
since you don't actually have

418
00:20:08 --> 00:20:10
your notes to refer to.
 

419
00:20:10 --> 00:20:11
So there's the
class notes there.

420
00:20:11 --> 00:20:17
All right, this should be a
pretty quick thing for you to

421
00:20:17 --> 00:20:32
figure out, so let's just
take 10 seconds on this.

422
00:20:32 --> 00:20:33
OK, great.
 

423
00:20:33 --> 00:20:36
Even thinking quickly, most
of you got it correct.

424
00:20:36 --> 00:20:39
So what we see is on
ammonia here, we know that

425
00:20:39 --> 00:20:41
it's less than a 109 .
 

426
00:20:41 --> 00:20:46
5, it's actually 107, so
it's less than a 109 .

427
00:20:46 --> 00:20:48
5, because of that lone
pair pushing down in

428
00:20:48 --> 00:20:50
the bonding electrons.
 

429
00:20:50 --> 00:20:52
And what is the shape,
for one more clicker

430
00:20:52 --> 00:20:59
question on ammonia?
 

431
00:20:59 --> 00:21:14
Let's take 10 seconds again,
this should be pretty quick.

432
00:21:14 --> 00:21:15
All right, pretty good.
 

433
00:21:15 --> 00:21:17
So, 70% of you.
 

434
00:21:17 --> 00:21:19
We'd like to get
this up higher.

435
00:21:19 --> 00:21:21
The shape is actually
trigonal pyramidal.

436
00:21:21 --> 00:21:23
And you need to just
remember your shapes.

437
00:21:23 --> 00:21:26
If they're not obvious to you
what they're called, you need

438
00:21:26 --> 00:21:28
to just study them
and learn them.

439
00:21:28 --> 00:21:31
So it's trigonal because we
have these three atoms that are

440
00:21:31 --> 00:21:35
bound to the central atom here,
and if you picture it, it's

441
00:21:35 --> 00:21:37
actually shaped like a pyramid.
 

442
00:21:37 --> 00:21:38
So it's trigonal pyramidal.
 

443
00:21:38 --> 00:21:42
That's what we call when we
have three bonding atoms

444
00:21:42 --> 00:21:46
and one lone pair.
 

445
00:21:46 --> 00:21:47
All right.
 

446
00:21:47 --> 00:21:50
So we can switch all the way
back to our notes here.

447
00:21:50 --> 00:21:53
And the last thing we can think
about is how do we name this

448
00:21:53 --> 00:21:57
n h bond, and again, we just
name it based on it symmetry.

449
00:21:57 --> 00:22:01
It's a sigma bond, and
it's going to be -- no.

450
00:22:01 --> 00:22:06
OK, it's going to be nitrogen 2
s p 3, because it's a nitrogen

451
00:22:06 --> 00:22:09
atom, and then hydrogen 1 s.
 

452
00:22:09 --> 00:22:11
So, I don't even have to worry
because you're not writing this

453
00:22:11 --> 00:22:14
down, so I can just fix it when
I post the notes and no one

454
00:22:14 --> 00:22:17
will ever know, except that
this is not OpenCourseWare.

455
00:22:17 --> 00:22:22
So let's switch to
thinking about oxygen

456
00:22:22 --> 00:22:23
hybridization here.
 

457
00:22:23 --> 00:22:27
So in oxygen we have a similar
situation where, in fact, we

458
00:22:27 --> 00:22:30
are not going to promote any of
the electrons because we have

459
00:22:30 --> 00:22:33
two lone pair electrons
no matter what we do.

460
00:22:33 --> 00:22:37
So when we hybridize our
orbitals, we're going to end up

461
00:22:37 --> 00:22:43
with again, four hybrid
orbitals, 4 s p 3 orbitals, and

462
00:22:43 --> 00:22:46
what we'll see is that two of
these are already going to be

463
00:22:46 --> 00:22:49
filled up with a paired
electrons, so we're only going

464
00:22:49 --> 00:22:51
to have 2 orbitals with
an unpaired electron

465
00:22:51 --> 00:22:53
available for bonding.
 

466
00:22:53 --> 00:22:56
So let's think about water
here as our simplest

467
00:22:56 --> 00:22:57
example with oxygen.
 

468
00:22:57 --> 00:23:03
So we can have our two hydrogen
atoms come in here, and what we

469
00:23:03 --> 00:23:06
will find is now that we have
all of our orbitals filled up

470
00:23:06 --> 00:23:10
-- so thinking about what this
angle is here, would you expect

471
00:23:10 --> 00:23:15
it to be less than or greater
than what we saw for

472
00:23:15 --> 00:23:16
ammonia before?
 

473
00:23:16 --> 00:23:17
STUDENT: Less than.
 

474
00:23:17 --> 00:23:18
PROFESSOR: Good, good, it's
going to be less than, and it's

475
00:23:18 --> 00:23:22
going to be less than because
now we have two lone pairs.

476
00:23:22 --> 00:23:24
So since we have two lone
pairs, we're going to be

477
00:23:24 --> 00:23:27
pushing down even further on
the bonding electrons, so we're

478
00:23:27 --> 00:23:29
going to smoosh those bonds
even closer together.

479
00:23:29 --> 00:23:31
The bond, it turns
out, is 104 .

480
00:23:31 --> 00:23:36
5 degrees, that h o h bond.
 

481
00:23:36 --> 00:23:40
So in terms of naming our o h
bond, good, it's right here.

482
00:23:40 --> 00:23:44
So it's going to be a sigma
bond, and we have oxygen

483
00:23:44 --> 00:23:51
2 s p 3 and hydrogen 1 s.
 

484
00:23:51 --> 00:23:54
And the geometry, which I
didn't ask you, is going to

485
00:23:54 --> 00:23:56
be bent for this molecule.
 

486
00:23:56 --> 00:23:59
All right, so that's s p 3
hybridization, but those aren't

487
00:23:59 --> 00:24:02
the only type of hybrid
orbitals that we can form.

488
00:24:02 --> 00:24:05
Let's take a look at what
happens if instead of combining

489
00:24:05 --> 00:24:07
all four orbitals, we just
combine three of those

490
00:24:07 --> 00:24:12
orbitals, and what we'll end up
with is s p 2 hybridization.

491
00:24:12 --> 00:24:16
So in s p 2 hybridization,
instead of combining all four,

492
00:24:16 --> 00:24:20
we're just combining two of the
p orbitals with the s orbital.

493
00:24:20 --> 00:24:22
So what we're going to
end up with now is

494
00:24:22 --> 00:24:24
three hybrid orbitals.
 

495
00:24:24 --> 00:24:27
And what happens to this last
p orbital is nothing at

496
00:24:27 --> 00:24:29
all, we just get it back.
 

497
00:24:29 --> 00:24:32
So we end up with 1 p orbital
completely untouched, and

498
00:24:32 --> 00:24:36
three hybrid s p 2 orbitals.
 

499
00:24:36 --> 00:24:39
So again, we can think
of an example here.

500
00:24:39 --> 00:24:44
So let's take boron, for
example, and this has --

501
00:24:44 --> 00:24:46
it starts off with three
valence electrons.

502
00:24:46 --> 00:24:49
Would you expect to see
electron promotion for boron?

503
00:24:49 --> 00:24:50
STUDENT: Yes.
 

504
00:24:50 --> 00:24:54
PROFESSOR: Yeah, absolutely. if
we move up one of our electrons

505
00:24:54 --> 00:24:56
into an empty p orbital, what
were going to see is now we

506
00:24:56 --> 00:24:57
have three unpaired electrons
that are ready for bonding.

507
00:24:57 --> 00:25:05
So, if we hybridize just these
three orbitals, what we're

508
00:25:05 --> 00:25:10
going to end up with is our
s p 2 hybrid orbitals.

509
00:25:10 --> 00:25:12
Again, the name is very
straightforward, it comes

510
00:25:12 --> 00:25:17
from 1 s and 2 p orbital,
so it will be s p 2.

511
00:25:17 --> 00:25:20
And again, you might be
thinking well, why didn't

512
00:25:20 --> 00:25:23
we actually hybridize
this 2 p y orbital.

513
00:25:23 --> 00:25:26
It doesn't actually have an
electron in it, so we don't

514
00:25:26 --> 00:25:28
have to worry about whether
it's very high in energy or

515
00:25:28 --> 00:25:30
not, we don't care that
it's high in energy.

516
00:25:30 --> 00:25:33
What we do care about is the
energy of our orbitals that

517
00:25:33 --> 00:25:37
have electrons in them, and if
we combined all four of the

518
00:25:37 --> 00:25:40
orbitals, then our hybrid
orbitals would have more p

519
00:25:40 --> 00:25:43
character to them, so they'd
actually be higher in energy.

520
00:25:43 --> 00:25:46
So if we don't have to
hybridize one of the p

521
00:25:46 --> 00:25:49
orbitals, we can actually end
up with a lower energy

522
00:25:49 --> 00:25:53
situation, because now these s
p 2 orbitals are 1/3 s

523
00:25:53 --> 00:25:59
character, and only 2/3 p
character, instead of 3/4.

524
00:25:59 --> 00:26:04
So we end up with 3 s p 2
hybrid orbitals, so we can

525
00:26:04 --> 00:26:09
think about what would happen
here in terms of bonding, and

526
00:26:09 --> 00:26:14
if we think about how to get
our bonds as far away as

527
00:26:14 --> 00:26:16
possible from each other, what
we're going to have is the

528
00:26:16 --> 00:26:18
trigonal planer situation.
 

529
00:26:18 --> 00:26:21
So if you picture, for
example, b h 3, it's

530
00:26:21 --> 00:26:23
going to look like this.
 

531
00:26:23 --> 00:26:26
All of our electrons are in our
bonds, we want to got them a

532
00:26:26 --> 00:26:29
120 degrees away from each
other, that's as far away

533
00:26:29 --> 00:26:31
as we can get them.
 

534
00:26:31 --> 00:26:33
Keep in mind we do have this
p orbital here and it's

535
00:26:33 --> 00:26:35
coming right out at us.
 

536
00:26:35 --> 00:26:38
And this p orbital is here, but
it's empty, it doesn't have any

537
00:26:38 --> 00:26:41
electrons in it, that's why we
don't have to worry about it in

538
00:26:41 --> 00:26:44
terms of getting our electrons
as far away from each

539
00:26:44 --> 00:26:45
other as possible.
 

540
00:26:45 --> 00:26:49
So what we'll have here is a
trigonal planar case, and you

541
00:26:49 --> 00:26:52
can see that we only have three
electrons that are set for

542
00:26:52 --> 00:26:56
bonding, so we'll add three
hydrogens, and for b h 3, we'll

543
00:26:56 --> 00:26:58
get a stable structure here.
 

544
00:26:58 --> 00:27:02
So, remember, boron was one of
those exceptions to our Lewis

545
00:27:02 --> 00:27:04
structure rules where it
was perfectly happy not

546
00:27:04 --> 00:27:05
having a full octet.
 

547
00:27:05 --> 00:27:09
So this can tell you why it's
so happy with only having

548
00:27:09 --> 00:27:10
six electrons around it.
 

549
00:27:10 --> 00:27:17
All right, so if we think about
b h bond here, again, it's the

550
00:27:17 --> 00:27:21
sigma bond, and we're going to
say it's a boron 2 s p 2 hybrid

551
00:27:21 --> 00:27:26
orbital interacting with
a hydrogen 1 s orbital.

552
00:27:26 --> 00:27:29
So let's take a look at another
case where we have s p 2

553
00:27:29 --> 00:27:33
hybridization, we can actually
also have it happen in carbon.

554
00:27:33 --> 00:27:35
So if we think about having
it happen in carbon, we're

555
00:27:35 --> 00:27:38
starting with the situation
where we've already promoted

556
00:27:38 --> 00:27:42
our electron into a 2 p orbital
here, and what we're going to

557
00:27:42 --> 00:27:46
do is just combine the s and
two of the p's, so we'll end up

558
00:27:46 --> 00:27:52
with electrons in one of each
three s p 2 hybrid orbitals.

559
00:27:52 --> 00:27:55
But unlike the case with boron
where we had an empty p

560
00:27:55 --> 00:27:59
orbital, we're actually going
to have an electron in the p

561
00:27:59 --> 00:28:02
orbital of carbon as well.
 

562
00:28:02 --> 00:28:06
So again, if we think about
that shape of that carbon atom,

563
00:28:06 --> 00:28:09
it's going to be trigonal
planar, it's going to have bond

564
00:28:09 --> 00:28:13
angles of 120 degrees, because
we have this set up of having

565
00:28:13 --> 00:28:16
three hybrid orbitals.
 

566
00:28:16 --> 00:28:18
So let's take a look at what
actually happens if we're

567
00:28:18 --> 00:28:22
talking about a carbon-carbon
double bond, such as in

568
00:28:22 --> 00:28:27
ethene, c 2 h 4, we're going
to have a double bond.

569
00:28:27 --> 00:28:30
If we have a double bond, we
know we need to have only one

570
00:28:30 --> 00:28:33
sigma bond, and we're also
going to have one pi bond.

571
00:28:33 --> 00:28:35
So it already should make sense
why we have that p orbital

572
00:28:35 --> 00:28:37
there, in order to form a pi
bond, we're going to

573
00:28:37 --> 00:28:38
need a p orbital.
 

574
00:28:38 --> 00:28:43
So if you picture this as our s
p 2 carbon atom where we have

575
00:28:43 --> 00:28:47
three hybrid orbitals, and then
one p y orbital coming

576
00:28:47 --> 00:28:49
right out at us.
 

577
00:28:49 --> 00:28:51
So again, we picture the
same thing as we pictured

578
00:28:51 --> 00:28:53
with the boron there.
 

579
00:28:53 --> 00:28:57
If we have, coming along this z
axis, another carbon atom, we

580
00:28:57 --> 00:29:00
can actually form one
bond between the two

581
00:29:00 --> 00:29:01
carbon atoms there.
 

582
00:29:01 --> 00:29:07
So if we picture how this
happens, what we have here if

583
00:29:07 --> 00:29:19
these are our 2 s p 2 carbon
atoms -- so here we have s p 2

584
00:29:19 --> 00:29:25
hybrid carbon, and here we have
s p 2 hybrid carbon atom.

585
00:29:25 --> 00:29:29
These 2 are going to come
together like this, and the

586
00:29:29 --> 00:29:31
first bond that we're going to
form is going to be a sigma

587
00:29:31 --> 00:29:32
bond, right, so we
see that here.

588
00:29:32 --> 00:29:36
If we're looking head on, we
see they form a sigma bond.

589
00:29:36 --> 00:29:38
We can also look at them coming
in from the side, and that's

590
00:29:38 --> 00:29:41
what I tried to depict here
where you can actually see

591
00:29:41 --> 00:29:43
in pink is the p orbital.
 

592
00:29:43 --> 00:29:46
So we can also show them coming
together this way, so now

593
00:29:46 --> 00:29:48
you're looking at it where you
can see the p orbital, and

594
00:29:48 --> 00:29:52
maybe just see well one
of the hydrogen atoms.

595
00:29:52 --> 00:29:57
So we can have four total
hydrogens bonding here, and we

596
00:29:57 --> 00:30:01
can think about how to describe
these carbon-carbon bonds.

597
00:30:01 --> 00:30:05
So in the first case of this
first bond here that I've put

598
00:30:05 --> 00:30:07
in a square, what type of a
bond is this, is

599
00:30:07 --> 00:30:08
the sigma or pi?
 

600
00:30:08 --> 00:30:10
STUDENT: Sigma.
 

601
00:30:10 --> 00:30:11
PROFESSOR: Yup,
it's a sigma bond.

602
00:30:11 --> 00:30:13
We're having two orbitals
coming together

603
00:30:13 --> 00:30:15
on the bond axis.
 

604
00:30:15 --> 00:30:19
So we'll call this sigma,
and it's between two s p

605
00:30:19 --> 00:30:21
2 hybrid carbon atoms.
 

606
00:30:21 --> 00:30:26
So it's stigma carbon
s p 2, carbon s p 2.

607
00:30:26 --> 00:30:29
What about this second bond
here where we're going to have

608
00:30:29 --> 00:30:32
interaction of 2 p orbitals,
is that sigma or pi?

609
00:30:32 --> 00:30:34
STUDENT: Pi.
 

610
00:30:34 --> 00:30:34
PROFESSOR: Pi, great.
 

611
00:30:34 --> 00:30:38
So our second bond is
going to be a pi bond.

612
00:30:38 --> 00:30:40
And again, this is between the
p orbitals, these are not

613
00:30:40 --> 00:30:43
hybrid orbitals, so when we
name this bond we're going to

614
00:30:43 --> 00:30:47
name it as a pi bond here,
because it's between two p

615
00:30:47 --> 00:30:51
orbitals, and it's going to be
between the carbon 2 p y

616
00:30:51 --> 00:30:54
orbital, and the other
carbon 2 p y orbital.

617
00:30:54 --> 00:30:57
Remember, we didn't hybridize
the 2 p y orbital, so that's

618
00:30:57 --> 00:31:00
what we have left over
to form these pi bonds.

619
00:31:00 --> 00:31:03
All right.
 

620
00:31:03 --> 00:31:07
So in addition to having these
two carbon bonds, we actually

621
00:31:07 --> 00:31:11
also have four carbon hydrogen
bonds in addition to our

622
00:31:11 --> 00:31:13
carbon-carbon bonds.
 

623
00:31:13 --> 00:31:17
So why don't you tell me what
the valence bond description

624
00:31:17 --> 00:31:19
would be of these
carbon hydrogen bonds?

625
00:31:19 --> 00:31:45
So let's take 10
seconds on that.

626
00:31:45 --> 00:31:45
OK, great.
 

627
00:31:45 --> 00:31:48
So most and you got it, so we
can switch to the notes and

628
00:31:48 --> 00:31:50
let's talk about this here.
 

629
00:31:50 --> 00:31:55
So in terms of the carbon
hydrogen bond, it's a sigma

630
00:31:55 --> 00:31:59
bond, because we define it --
any time we are bonding to an

631
00:31:59 --> 00:32:01
atom, we have to keep
redefining our bond axis

632
00:32:01 --> 00:32:04
to whatever two atoms
we're talking about.

633
00:32:04 --> 00:32:07
So it's along the bond axis and
it's between a carbon s p 2

634
00:32:07 --> 00:32:11
hybrid, and then the hydrogen
is just a 1 s orbital that

635
00:32:11 --> 00:32:12
we're combining here.
 

636
00:32:12 --> 00:32:17
So those are our three
types of bonds in ethene.

637
00:32:17 --> 00:32:19
One thing that I want to
mention that is really

638
00:32:19 --> 00:32:22
important is once you have
double bonds, what happens

639
00:32:22 --> 00:32:25
between those two atoms in the
molecule is they can no longer

640
00:32:25 --> 00:32:28
rotate in relation
to each other.

641
00:32:28 --> 00:32:29
So you can think
about why that is.

642
00:32:29 --> 00:32:33
When we have just a single bond
in them molecule, you have all

643
00:32:33 --> 00:32:36
the free rotation you want, you
can just spin it around,

644
00:32:36 --> 00:32:37
there's nothing
keeping it in place.

645
00:32:37 --> 00:32:41
But once you have a double bond
here, we have our pi bond,

646
00:32:41 --> 00:32:43
as well as our sigma bond.
 

647
00:32:43 --> 00:32:46
So there's electron density
above the bond and

648
00:32:46 --> 00:32:47
below the bond.
 

649
00:32:47 --> 00:32:51
So if I try to rotate my 2
atoms, you see that I have to

650
00:32:51 --> 00:32:54
break that pi bond, because
they need to be lined up

651
00:32:54 --> 00:32:56
so that the electron
density can overlap.

652
00:32:56 --> 00:33:00
So in order to rotate a double
bond, you have to actually

653
00:33:00 --> 00:33:02
break the pi bond, so
essentially what you're doing

654
00:33:02 --> 00:33:04
is breaking the double bond.
 

655
00:33:04 --> 00:33:07
So really, you can not ever
rotate a double bond, it makes

656
00:33:07 --> 00:33:09
your molecule very rigid.
 

657
00:33:09 --> 00:33:13
This is incredibly important
because if you picture having a

658
00:33:13 --> 00:33:16
double bond in a very large
molecule, you could have all

659
00:33:16 --> 00:33:19
sorts of other atoms off this
way and all sorts of other

660
00:33:19 --> 00:33:23
atoms off this way, and you can
picture the shape would be very

661
00:33:23 --> 00:33:25
different if you have
one confirmation versus

662
00:33:25 --> 00:33:27
another confirmation.
 

663
00:33:27 --> 00:33:31
So it's very important that
the double bond locks it in

664
00:33:31 --> 00:33:32
a particular conformation.
 

665
00:33:32 --> 00:33:36
This completely could change if
you were to flip from one to

666
00:33:36 --> 00:33:37
the other conformation
which can happen in

667
00:33:37 --> 00:33:39
chemical reactions.
 

668
00:33:39 --> 00:33:41
If you were to make that change
you would find that the

669
00:33:41 --> 00:33:44
molecule now has completely
different biological and

670
00:33:44 --> 00:33:46
chemical properties.
 

671
00:33:46 --> 00:33:48
So it's very important to be
keeping in mind that any time

672
00:33:48 --> 00:33:52
you see a double bond, you have
a pi bond there, so you're not

673
00:33:52 --> 00:33:56
going to see any rotation
around the bond axis.

674
00:33:56 --> 00:33:59
All right, so let's think of a
more complicated example of

675
00:33:59 --> 00:34:02
having a double bond, and maybe
a more interesting example, and

676
00:34:02 --> 00:34:03
this is talking about benzene.
 

677
00:34:03 --> 00:34:06
I think most and you have
talked a little bit about

678
00:34:06 --> 00:34:09
benzene over this past
week in recitation.

679
00:34:09 --> 00:34:13
Benzene is a ring that's
made up of six carbon atoms

680
00:34:13 --> 00:34:15
and six hydrogen atoms.
 

681
00:34:15 --> 00:34:17
So let's picture what this
looks like here, and we'll

682
00:34:17 --> 00:34:19
start with four and we'll
add in our last two.

683
00:34:19 --> 00:34:23
So essentially, we have two
ethene or ethylene molecules

684
00:34:23 --> 00:34:28
here to start with where these
blue are our 2 s p 2 hybrid

685
00:34:28 --> 00:34:32
orbitals, so you can see that
for each carbon atom, one is

686
00:34:32 --> 00:34:35
already used up binding
to another carbon atom.

687
00:34:35 --> 00:34:39
If we think about bringing in
those last two carbons, what

688
00:34:39 --> 00:34:43
you can see is that for every
carbon, two of its hybrid

689
00:34:43 --> 00:34:48
orbitals are being used to
bond to other carbons.

690
00:34:48 --> 00:34:53
So that leaves each carbon with
only one hybrid orbital left.

691
00:34:53 --> 00:34:56
And if we think about the six
hydrogens, now each of those

692
00:34:56 --> 00:34:59
are going to bind by combining
one of the carbon hybrid

693
00:34:59 --> 00:35:03
orbitals to a 1 s
orbital of hydrogen.

694
00:35:03 --> 00:35:07
So, if we think about what
bonds are in this molecule, we

695
00:35:07 --> 00:35:11
actually have six of these
sigma carbon s p 2,

696
00:35:11 --> 00:35:14
carbon s p 2 bonds.
 

697
00:35:14 --> 00:35:18
We also have carbon s p
2 hydrogen 1 s bonds.

698
00:35:18 --> 00:35:21
How many of those do we have?
 

699
00:35:21 --> 00:35:23
Yup, we also have six of
these, because we have six

700
00:35:23 --> 00:35:25
carbon hydrogen bonds.
 

701
00:35:25 --> 00:35:28
So that's two of our types of
bonds in benzene, and we have

702
00:35:28 --> 00:35:32
one type left, and that's going
to actually be the double bond

703
00:35:32 --> 00:35:35
or the pi bond that
forms between some

704
00:35:35 --> 00:35:37
of these p orbitals.
 

705
00:35:37 --> 00:35:41
So we can have one bond here
between this carbon's p orbital

706
00:35:41 --> 00:35:43
and this carbon's p orbital.
 

707
00:35:43 --> 00:35:46
So let's have a clicker
question here on how many

708
00:35:46 --> 00:35:50
total pi bonds do you
expect to see in benzene?

709
00:35:50 --> 00:35:56
Oh good, so it's left up --
the notes are left up on

710
00:35:56 --> 00:35:57
this screen right now.
 

711
00:35:57 --> 00:36:16
All right, so let's take
10 seconds on that.

712
00:36:16 --> 00:36:17
All right, great.
 

713
00:36:17 --> 00:36:20
So, most of you saw that what
we would expect to see is a

714
00:36:20 --> 00:36:22
three bond, some of
you thought six.

715
00:36:22 --> 00:36:25
So let's take a look at
why three is correct.

716
00:36:25 --> 00:36:30
So, what we end up having is
three of these pi -- 2 p y 2

717
00:36:30 --> 00:36:34
p y bonds, we can have one
between these two carbons here.

718
00:36:34 --> 00:36:36
Can we have one between
these two carbons here

719
00:36:36 --> 00:36:37
if we have one here?
 

720
00:36:37 --> 00:36:38
STUDENT: No.
 

721
00:36:38 --> 00:36:38
PROFESSOR: No, we can't.
 

722
00:36:38 --> 00:36:42
We're already using it up in
this pi bond here, so that

723
00:36:42 --> 00:36:45
means we're limited to only
two other spots on the

724
00:36:45 --> 00:36:47
molecule, so we have three.
 

725
00:36:47 --> 00:36:49
But, of course, what you saw in
recitation, and hopefully, what

726
00:36:49 --> 00:36:53
you can now think very quickly
by looking at this, is that

727
00:36:53 --> 00:36:56
this is not the only
configuration of pi bonds that

728
00:36:56 --> 00:36:58
we could have in benzene.
 

729
00:36:58 --> 00:37:00
There's absolutely no reason
I couldn't have switched it

730
00:37:00 --> 00:37:04
around and said that instead
the pi orbitals form between

731
00:37:04 --> 00:37:08
these atoms instead of those
first atoms I showed.

732
00:37:08 --> 00:37:11
So, let's look at this in a
more simple structure here

733
00:37:11 --> 00:37:15
where we have the two possible
forms of benzene, and the

734
00:37:15 --> 00:37:17
reality is is that it's going
to be some combination

735
00:37:17 --> 00:37:18
of the two.
 

736
00:37:18 --> 00:37:21
This is resonance, this is a
case of a resonance structure.

737
00:37:21 --> 00:37:25
So what we see is that those
six pi electrons are actually

738
00:37:25 --> 00:37:29
going to be de-localized around
all six of those atoms.

739
00:37:29 --> 00:37:33
So if you think about any one
of these carbon-carbon bonds,

740
00:37:33 --> 00:37:36
what type of a bond would
you expect that to be?

741
00:37:36 --> 00:37:38
What would the bond
order be for this bond?

742
00:37:38 --> 00:37:40
STUDENT: [INAUDIBLE]
 

743
00:37:40 --> 00:37:41
PROFESSOR: Yup, it's going
to be a 1 and 1/2 bond.

744
00:37:41 --> 00:37:45
It's a 1 and 1/2 because it's
halfway between a double

745
00:37:45 --> 00:37:46
bond and a single bond.
 

746
00:37:46 --> 00:37:49
So, of course, this is
resonance so we can go ahead

747
00:37:49 --> 00:37:51
and put our resonance notation
in there to indicate

748
00:37:51 --> 00:37:53
that benzene is a
resonance structure.

749
00:37:53 --> 00:37:55
All right.
 

750
00:37:55 --> 00:37:58
So let's quickly talk about our
last type of hybridization that

751
00:37:58 --> 00:38:01
we're going to discuss today,
which is s p hybridization.

752
00:38:01 --> 00:38:05
So s p hybridization comes now
from when we're combining

753
00:38:05 --> 00:38:08
an s orbital now with
only one p orbital.

754
00:38:08 --> 00:38:11
So let's take a look
at this with carbon.

755
00:38:11 --> 00:38:15
And if we hybridize these
orbitals in carbon, what we end

756
00:38:15 --> 00:38:19
up with is having two hybrid
orbitals, and then we're going

757
00:38:19 --> 00:38:22
to be left with two of our p
orbitals that are each going

758
00:38:22 --> 00:38:25
to have an electron
associated in them.

759
00:38:25 --> 00:38:28
So again, looking at the
shapes, now we're just

760
00:38:28 --> 00:38:32
combining two, we've got these
two equal hybrid orbitals plus

761
00:38:32 --> 00:38:34
these 2 p orbitals here.
 

762
00:38:34 --> 00:38:38
So let's take the case of
acetylene where we have two

763
00:38:38 --> 00:38:41
carbon atoms that are going to
be triple bonded to each other,

764
00:38:41 --> 00:38:45
each are bonded to a carbon
and then to one hydrogen.

765
00:38:45 --> 00:38:47
So this is a little bit
trickier to look at and see

766
00:38:47 --> 00:38:50
what it means, but essentially
we have two hybrid orbitals,

767
00:38:50 --> 00:38:53
which are shown in blue here,
and then we have one p orbital

768
00:38:53 --> 00:38:56
that's left alone that's going
up and down on the page.

769
00:38:56 --> 00:38:58
And then that second p
orbital's actually coming

770
00:38:58 --> 00:39:01
right out at you, it's coming
out of the screen at you.

771
00:39:01 --> 00:39:05
So, if we think about this z
bonding axis between the two

772
00:39:05 --> 00:39:09
carbon atoms, we can picture
overlap of those s p hybrid

773
00:39:09 --> 00:39:13
orbitals, and then we can also
picture bonding to hydrogen.

774
00:39:13 --> 00:39:19
So, in a settling, what
is the bond angle here?

775
00:39:19 --> 00:39:21
This is the easiest question
all day, what is the bond

776
00:39:21 --> 00:39:24
angle between all of these?
 

777
00:39:24 --> 00:39:26
Great, so it's 180 degrees.
 

778
00:39:26 --> 00:39:31
So, if we think about the bonds
that are forming -- oh I see

779
00:39:31 --> 00:39:34
our TAs are here, so you can
start handing them out, because

780
00:39:34 --> 00:39:37
we have two minutes left to go.
 

781
00:39:37 --> 00:39:44
So, as they're very quietly
handing out your class notes,

782
00:39:44 --> 00:39:47
let's think about what this
bond is here, this boxed

783
00:39:47 --> 00:39:50
bond, is it a pi bond
or a sigma bond?

784
00:39:50 --> 00:39:51
It's going to be a sigma bond.
 

785
00:39:51 --> 00:39:56
So, we have sigma 2
s p, carbon 2 s p.

786
00:39:56 --> 00:39:59
So they're two s p
bonds combining.

787
00:39:59 --> 00:40:01
Now let's think about this
first pi bond, which will

788
00:40:01 --> 00:40:04
be above and below
the bonding axis.

789
00:40:04 --> 00:40:07
Is this pi or sigma?
 

790
00:40:07 --> 00:40:08
This is pi.
 

791
00:40:08 --> 00:40:12
So we're talking about pi
carbon 2 p x, because it's

792
00:40:12 --> 00:40:17
the x axes combining
to carbon 2 p x.

793
00:40:17 --> 00:40:22
And the last bond that we have
here is a carbon-carbon bond,

794
00:40:22 --> 00:40:26
and this is our last p orbitals
that are coming together.

795
00:40:26 --> 00:40:29
These are the ones that are
coming right out at you, so

796
00:40:29 --> 00:40:32
this is going to be on
a second pi orbital.

797
00:40:32 --> 00:40:36
So this will be pi carbon
2 p y, carbon 2 p y.

798
00:40:36 --> 00:40:38
All right.
 

799
00:40:38 --> 00:40:42
So, we'll stop here today.
 

800
00:40:42 --> 00:40:44
Just stay in your seats for
another 30 seconds as they're

801
00:40:44 --> 00:40:46
handing out your notes.
 

802
00:40:46 --> 00:40:48
I want to mention that you're
going to get problem-set five

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is posted today, and I'll write
which ones you can already do

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so far, because you don't
have class on Monday.

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But remember that you do have
recitation on Tuesday, so that

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could be very helpful with the
problem-set, so be sure to go

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to recitation on Tuesday, and
have a great long weekend.

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