Sp Hybridisation Video Lecture - Chemistry Class 11 - NEET

Voiceover: The video on SP3 hybridization,
we saw a carbon is bonded to four atoms
and in the video in SP2 hybridization,
we saw that carbon
is bonded to three atoms
and in this video,
we're gonna look at the
type of hybridization
that's present when carbon
is bonded to two atoms.
If I look at this carbon right here
and the ethyne or the acetylene molecule.
This carbon is bonded to a hydrogen
and it's also bonded to another carbon.
We have carbon bonded to only two atoms
and the shape of the acetylene molecule
has been determined to be linear.
We have a linear geometry.
We also have a bond angle here.
These bond angles are 180 degrees
and so we must have a
different hybridization
for this carbon.
We have a different geometry,
a different bond angle
and a different number
of atoms that this carbon is bonded to.
To find our new type of hybridization,
we look at our electronic configuration
already in the excited stage.
We have carbons, four valence electrons
represented here in the excited stage.
One, two, three and four
and we're looking for two hybrid orbitals
since carbon is bonded to two atoms.
We're going to take an S orbital.
We're gonna promote an S orbital
in terms of energy and
we're going to demote
a P orbital, only one P orbital this time.
We have an S orbital with one electron.
A P orbital with one electron.
That's gonna leave behind two P orbitals.
Each one of those P orbitals
has one electron in it.
We have carbons, four valence electrons
but this is no longer an S orbital
because we're going to hybridized it
with a P orbital to make
an SP hybrid orbital.
This is no longer a P orbital
because we're going to hybridize it
to form our SP hybrid orbital.
This is called SP hybridization.
This is SP hybridization
because our new hybrid orbitals
came from one S orbital
and one P orbital like that.
This carbon right here is SP hybridized
since it bonded to two atoms
and this carbon right here
is also SP hybridized.
Let's think about the shape
of our new SP hybrid orbitals.
Let's get a little bit of room down here.
Once again, we know an S orbital shaped
like a sphere.
We took one S orbital and
we took one P orbital,
which is shaped like a dumbbell
and we hybridized these
two orbitals together
to give us two new hybrid orbitals.
Two SP hybrid orbitals.
Let me go ahead and draw in
an SP hybrid orbital here
and once again, we're going
to ignore the small lobe.
We're going to ignore the small lobe.
When we draw our picture,
only think about this
bigger frontal lobe here.
When I think about the
percentage of S character.
We use one S orbital and one P orbital.
That means it's 50% S character
and 50% P character
and this is more S character
than in the previous videos.
On the video on SP3 hybridization,
we're talking about 25% S character
The video on SP2 hybridization,
we talked about 33% S character
and then for these hybrid orbitals,
we have even more S character,
up to 50% and since the electron density
for an S orbital is
increased electron density
closer to the nucleus for an S orbital
than for a P orbital,
that means that this lobe
here has an increased
electron density closer to the nucleus,
which is one way to think
about why these bonds
get shorter as you
increase in S character.
In general, as you
increase in S character,
you're going to get shorter bonds
because you have smaller
hybrid orbitals here.
Let's go back up here to this picture
of acetylene.
I wanna see if we can draw it.
We now know that both of these carbons
in a acetylene are SP hybridized.
Let's go back down here
and let me draw up a dot
structure really fast.
We have acetylene here.
We have carbon triple
bonded to another carbon.
We know each of those
carbons as SP hybridized.
If each of those carbons is SP hybridized,
each carbon has two SP hybrid orbitals.
We go ahead and draw in
one SP hybrid orbital.
Again, I'm ignoring the smaller back lobe
and here's our other SP
hybrid orbital on this carbon.
Let me go back and look
at our diagram again.
Each carbon, we use red for this.
Each SP hybridized
carbon has an SP orbital
with one valence electron in it
and we put that in here
and then there's another one.
There's another SP hybrid orbital
with one valence electron in it.
I'm gonna go ahead and put
the other electron over here
in this hybrid orbital.
Also, notice, if you're
dealing with an SP hybridized
carbon, you also have two P orbitals.
Two unhybridized P orbitals.
Each P orbital with a valence electron.
Let me go back down here
and I'm gonna draw in.
here's one P orbital.
There's one P orbital
with one valence electron
and there here's another
P orbital right here
with another one of
those valence electrons.
Now, we have our picture
of an SP hybridized carbon.
Let's say that was this carbon over here
on the left.
Now, let's go ahead
and draw in this carbon
on the right.
This carbon on the right
is also SP hybridized.
Therefore, this carbon on the right
has an SP hybrid orbital
with one valence electron in here
and then another SP hybrid orbital
with one valence electron here.
This carbon, it's SP hybridized.
Once again, go back up
here to this diagram.
This carbon is also going to have
a P orbital with a valence electron
and another P orbital with
another electron in it.
Go back down to here
and we draw in those P orbital.
Here is one P orbital.
Here is one P orbital
with one valence electron
and there here's another P orbital
with an electron in here like that.
Finally, we have to add in hydrogen.
We have a hydrogen on either side here.
We know that hydrogen
has one valence electron
in an unhybridized S orbital.
Over here, we have a hydrogen
with one valence electron
in unhybridized S orbital.
Now, we can finally analyze the bonding
that's present.
We know that if we have
head on overlap of orbitals
like right in here.
That's a sigma bond.
There's one sigma bond.
Here's a head on overlap of orbitals
between our two carbons.
That's a sigma bond.
Then finally, we have a head on overlap
of orbitals here.
That's another sigma bond.
With a total of three sigma bonds
in the acetylene molecule.
The video on SP2 hybridization,
we saw how to make a pi bond.
We had this side by side
overlap of orbitals.
Here we have one pi bond.
We have interaction above and below.
That's one pi bonds
and then we have another pi bond here.
We have side by side
overlap of orbitals here
as well and so we have
two pi bonds present.
We have two pi bonds present
in the acetylene molecules.
Let's look at our dot structure again.
We saw the bond between this carbon
and this hydrogen was a sigma bond.
We saw there was one sigma bond
between our two carbons.
I'm just gonna pick the
one on the middle here.
Say that's a signal bond
and then this bond over here
we said was a sigma bond.
There's our three sigma bonds
and then we have a triple bond presence.
There were two pi bonds also present.
Two of these are pi bonds here.
A total of two pi bonds
and three sigma bonds
for the acetylene molecule here.
Remember pi bonds prevent free rotation.
We can't rotate about the sigma bond
between the two carbons
because of the pi bonds.
There's no free rotation
for our triple bond.
We have a linear shape.
Let me go ahead and
draw that line in here.
You see there's linear geometry
for this molecule like that.
Also in terms of bond length.
The distance between these two carbons.
The distance between this carbon
and this carbon, let me circle them.
The distance between these two carbons
turns out to be
approximately 1.20 angstroms.
An even shorter bond length
than in our previous videos.
Once again, that's due to
the increased S character.
Increased S character gives
you these smaller orbitals
and that's one way to think
about the shorter bond distance
and the triple bond
compared to a double bond
or a single bond.
That's a lot that we've covered here.
Let's go ahead and draw the dot structure
one more time and analyze
it using steric number.
We have our triple bonds.
If we're doing steric number to find out
the hybridization state,
we know to do steric number,
you take the number of sigma bonds.
Let's say our goal was to figure out
the steric number for this carbon.
The number of sigma bonds.
I know this is a sigma bond.
I know on a triple bond,
I have one sigma bond and two pi bonds.
There are two sigma bonds here
and zero lone pairs of electrons.
Two plus zero gives me two.
I need two hybrid orbitals.
Which you make from one S orbital
and one P orbital.
If you get a steric number of two,
you think SP hybridization.
This carbon is SP hybridized
and so is this carbon as well.
That's how to think about
it using steric number.
Once again, a linear
geometry with a bond angles
of 180 degrees.
Let's do one more example
using steric number
to analyze the molecule.
Let's do carbon dioxide.
If we wanted to figure
out the hybridization
of the carbon there.
Let's go ahead and do that.
Using steric number.
The hybridization of this carbon.
The steric number is equal to the number
of sigma bonds.
If I focus in on the double bond
between one of these
oxygens in this carbon,
I know that one of these
bonds is a sigma bond.
from our previous video.
I have one sigma bond here
and then for this other
double bond on the right,
I know that one of them is a sigma bond.
I have two sigma bonds here
and zero lone pairs of electrons
around the carbon.
Two plus zero gives me
a steric number of two.
I need two hybrid orbitals for that carbon
and of course, that must mean this carbon
is SP hybridized.
This carbon here is SP hybridized as well
and therefore, we know that
this is a linear molecule
with a bond angle of 180 degrees.
Once again, steric
number is just a nice way
of analyzing the hybridization
and also the geometry of the molecule.
In the next video, we'll look at a couple
of examples of organic molecules
in different hybridization states.