Sp2 hybridization Video Lecture - JEE

Voiceover: In an earlier video,
we saw that when carbon
is bonded to four atoms,
we have an SP3 hybridization
with a tetrahedral geometry
and an ideal bonding
over 109.5 degrees.
If you look at one of
the carbons in ethenes,
let's say this carbon right here,
we don't see the same geometry.
The geometry of the
atoms around this carbon
happens to be planar.
Actually, this entire molecule is planar.
You could think about
all this in a plane here.
And the bond angles are
close to 120 degrees.
Approximately, 120 degree bond angles
and this carbon that I've underlined here
is bonded to only three atoms.
A hydrogen, a hydrogen and a carbon
and so we must need a
different hybridization
for each of the carbon's presence
in the ethylene molecule.
We're gonna start with our
electron configurations
over here, the excited stage.
We have carbons four,
valence electron represented.
One, two, three and four.
In the video on SP3 hybridization,
we took all four of these
orbitals and combined them
to make four SP3 hybrid orbitals.
In this case, we only have a carbon bonded
to three atoms.
We only need three of our orbitals.
We're going to promote the S orbital.
We're gonna promote the S orbital up
and this time, we only
need two of the P orbitals.
We're gonna take one of the P's
and then another one of the P's here.
That is gonna leave one
of the our P orbitals
unhybridized.
Each one of these
orbitals has one electron
and it's like that.
This is no longer an S orbital.
This is an SP2 hybrid orbital.
This is no longer a P orbital.
This is an SP2 hybrid orbital
and same with this one,
an SP2 hybrid orbital.
We call this SP2 hybridization.
Let me go and write this up here.
and use a different color here.
This is SP2 hybridization
because we're using one S Orbital
and two P orbitals to form
our new hybrid orbitals.
This carbon right here is SP2 hybridized
and same with this carbon.
Notice that we left a P orbital untouched.
We have a P orbital
unhybridized like that.
In terms of the shape of
our new hybrid orbital,
let's go ahead and get
some more space down here.
We're taking one S orbital.
We know S orbitals are
shaped like spheres.
We're taking two P orbitals.
We know that a P orbital is shaped
like a dumbbell.
We're gonna take these orbitals
and hybridized them
to form three SP2 hybrid orbitals
and they have a bigger front lobe
and a smaller back lobe here like that.
Once again, when we draw the pictures,
we're going to ignore
the smaller back lobe.
This gives us our SP2 hybrid orbitals.
In terms of what percentage character,
we have three orbitals
that we're taking here
and one of them is an S orbital.
One out of three, gives us 33% S character
in our new hybrid SP2 orbital
and then we have two P orbitals.
Two out of three gives us 67% P character.
33% S character and 67% P character.
There's more S character
in an SP2 hybrid orbital
than an SP3 hybrid orbital
and since the electron
density in an S orbital
is closer to the nucleus.
We think about the electron density here
being closer to the nucleus
that means that we could think
about this lobe right here
being a little bit shorter
with the electron density being closer
to the nucleus and that's
gonna have an effect
on the length of the bonds
that we're gonna be forming.
Let's go ahead and draw the picture
of the ethylene molecule now.
We know that each of the
carbons in ethylenes.
Just going back up here
to emphasize the point.
Each of these carbons
here is SP2 hybridized.
That means each of those carbons
is going to have three SP2 hybrid orbitals
around it and once unhybridized P orbital.
Let's go ahead and draw that.
We have a carbon right here
and this is an SP2 hybridized orbitals.
We're gonna draw in.
There's one SP2 hybrid orbital.
Here's another SP2 hybrid orbital
and here's another one.
Then we go back up to here
and we can see that each
one of those orbitals.
Let me go ahead and mark this.
Each one of those SP2 hybrid orbitals
has one electron in it.
Each one of these
orbitals has one electron.
I go back down here and
I put in the one electron
in each one of my orbitals like that.
I know that each of those carbons
is going to have an
unhybridized P orbital here.
An unhybridized P orbital
with one electron too.
Let me go ahead and draw that in.
I'll go ahead and use a different color.
We have our unhybridized
P orbital like that
and there's one electron in
our unhybridized P orbital.
Each of the carbons was SP2 hybridized.
Let me go ahead and draw the dot structure
right here again so we
can take a look at it.
The dot structure for ethylene.
Let's do the other carbon now.
The carbon on the right
is also SP2 hybridized.
We can go ahead and draw
in an SP2 hybrid orbital
and there's one electron in that orbital
and then there's another one
with one electron and
then here's another one
with one electron.
This carbon being SP2 hybridized also has
an unhybridized P orbital
with one electron.
Go ahead and draw in that P orbital
with its one electron.
We also have some hydrogens.
We have some hydrogens
to think about here.
Each carbon is bonded to two hydrogens.
Let me go ahead and put in the hydrogens.
The hydrogen has a valance electron in
an unhybridized S orbital.
I'm going ahead and
putting in the S orbital
and the one valance electron from hydrogen
like this.
When we take a look at
what we've drawn here,
we can see some head
on overlap of orbitals,
which we know from our earlier video
is called a sigma bond.
Here's the head on overlap of orbitals.
That's a sigma bond.
here's another head on
overlap of orbitals.
The carbon carbon bond, here's also
a head on overlap of orbitals
and then we have these two over here.
We have a total of five sigma bonds
in our molecules.
Let me go ahead and write that over here.
There are five sigma bonds.
If I would try to find
those on my dot structure
this would be a sigma bond.
This would be a sigma bond.
One of these two is a sigma bond
and then these over here.
A total of five sigma bonds
and then we have a new type of bonding.
These unhybridized P orbitals can overlap
side by side.
Up here and down here.
We get side by side
overlap of our P orbitals
and this creates a pi bond.
A pi bond, let me go
ahead and write that here.
A pi bond is side by side overlap.
There is overlap above and
below this sigma bond here
and that's going to prevent free rotation.
When we're looking at
the example of ethane,
we have free rotation about the sigma bond
that connected the two carbons
but because of this pi bond here,
this pi bond is going to prevent rotations
so we don't get different confirmations
of the ethylene molecules.
No free rotation due to the pi bonds.
When you're looking at the dot structure,
one of these bonds is the pi bonds,
I'm just gonna say it's
this one right here.
If you have a double bond,
one of those bonds, the sigma bond
and one of those bonds is a pi bond.
We have a total of one pi bond
in the ethylene molecule.
If you're thinking about the distance
between the two carbons, let me go ahead
and use a different color for that.
The distance between this carbon
and this carbon.
It turns out to be
approximately 1.34 angstroms,
which is shorter than the distance
between the two carbons
in the ethane molecule.
Remember for ethane, the distance
was approximately 1.54 angstroms.
A double bond is shorter
than a single bond.
One way to think about that
is the increased S character.
This increased S character
means electron density
is closer to the nucleus
and that's going to make this lobe
a little bit shorter than before
and that's going to decrease the distance
between these two carbon atoms here.
1.34 angstroms.
Let's look at the dot structure again
and see how we can analyze this
using the concept of steric number.
Let me go ahead and
redraw the dot structure.
We have our carbon carbon double bond here
and our hydrogens like that.
If you're approaching this
situation using steric number
remember to find the hybridization.
We can use this concept.
Steric number is equal to
the number of sigma bonds
plus number of lone pairs of electrons.
If my goal was to find the steric number
for this carbon.
I count up my number of sigma bonds.
That's one, two and then
I know when I double bond
one of those is sigma
and one of those is pi.
One of those is a sigma bond.
A total of three sigma bonds.
I have zero lone pairs of electrons
around that carbon.
Three plus zero, gives me
a steric number of three.
I need three hybrid orbitals
and we've just seen in this video
that three SP2 hybrid orbitals form
if we're dealing with SP2 hybridization.
If we get a steric number of three,
you're gonna think
about SP2 hybridization.
One S orbital and two
P orbitals hybridizing.
That carbon is SP2 hybridized
and of course, this one is too.
Both of them are SP2 hybridized.
Let's do another example.
Let's do boron trifluoride.
BF3.
If you wanna draw the
dot structure of BF3,
you would have boron and then you would
surround it with your flourines here
and you would have an octet of electrons
around each flourine.
I go ahead and put those
in on my dot structure.
If your goal is to figure
out the hybridization
of this boron here.
What is the hybridization
stage of this boron?
Let's use the concept of steric number.
Once again, let's use steric number.
Find the hybridization of this boron.
Steric number is equal
to number of sigma bonds.
That's one, two, three.
Three sigma bonds plus
lone pairs of electrons.
That's zero.
Steric number of three tells us
this boron is SP2 hybridized.
This boron is gonna have
three SP2 hybrid orbitals
and one P orbital.
One unhybridized P orbital.
Let's go ahead and draw that.
We have a boron here
bonded to three flourines
and also it's going to have
an unhybrized P orbital.
Now, remember when you
are dealing with Boron,
it has one last valance electron
and carbon.
Carbon have four valance electrons.
Boron has only three.
When you're thinking about
the SP2 hybrid orbitals
that you create.
SP2 hybrid orbital, SP2, SP2
and then one unhybridized
P orbital right here.
Boron only has three valance electrons.
Let's go ahead and put in
those valance electrons.
One, two and three.
It doesn't have any electrons
in its unhybridized P orbital.
Over here when we look at the picture,
this has an empty orbital
and so boron can accept
a pair of electrons.
We're thinking about
its chemical behavior,
one of the things that BF3 can do,
the Boron can accept an electron pair
and function as a lewis acid.
That's one way in thinking
about how hybridizational
allows you to think about the structure
and how something might react.
This boron turns out to be SP2 hybridized.
This boron here is SP2 hybridized
and so we can also talk about the geometry
of the molecule.
It's planar.
Around this boron, it's planar
and so therefore, your bond
angles are 120 degrees.
If you have boron right here
and you're thinking about a circle.
A circle is 360 degrees.
If you divide a 360 by 3,
you get 120 degrees for
all of these bond angles.
In the next video, we'll
look at SP hybridization.