0:00
sp2 hybridization explained. In this
0:02
lecture, I will explain sp2
0:05
hybridization in complete detail with
0:07
examples. After watching this video, you
0:09
will fully understand how sp2
0:11
hybridization works in different
0:13
molecules. So, let us begin. First, we
0:15
need to understand how we can define sp2
0:18
hybridization. The type of hybridization
0:20
in which 1 s and 2p atomic orbitals mix
0:24
together and form three sp2 hybridized
0:26
orbitals is called sp2 hybridization.
0:29
Now before we move further, there is one
0:32
important rule about hybridization that
0:33
you must always keep in mind. The rule
0:35
is that the number of atomic orbitals
0:37
that mix together will always be equal
0:39
to the number of hybrid orbitals that
0:41
are formed. For example, here in sp2
0:44
hybridization, one s orbital and two p
0:47
orbitals are mixing together. That means
0:49
three atomic orbitals are mixing. So
0:51
according to this rule, three hybrid
0:53
orbitals will always be formed. And you
0:55
need to keep in mind that this rule
0:56
applies to every type of hybridization.
0:59
Now I will tell you about the shapes of
1:00
these orbitals. As we know the shape of
1:03
S orbital is spherical and the shape of
1:05
P orbital is dumbbell. When we look at
1:07
this through a diagram, one S orbital
1:10
which has spherical shape mixes with two
1:12
P orbitals which are PX orbital and PY
1:15
orbital. These three orbitals mix
1:17
together and form three sp2 hybridized
1:20
orbitals. If we look at the overall
1:22
shape of these sp2 hybrid orbitals,
1:24
their arrangement is trional planar. To
1:27
understand sp2 hybridization in complete
1:30
detail, we will cover three examples. In
1:32
the first example, we will understand
1:34
sp2 hybridization in BF3. In the second
1:38
example, we will see sp2 hybridization
1:41
in ethine molecule. And in the third
1:43
example, we will cover sp2 hybridization
1:46
in benzene. Let's start with the first
1:48
example. If we look at the BF3 molecule
1:52
as it is clear from the formula, the
1:54
central atom here is boron and three
1:56
florine atoms have formed bonds with
1:58
boron. Now before explaining the
2:00
electronic configuration, I will share a
2:02
very useful trick with you. Using this
2:04
trick, you can find the hybridization of
2:06
any specific atom in any molecule very
2:09
quickly. The trick is this. First write
2:11
the expanded structure of the molecule.
2:14
Expanded structure means open all the
2:16
bonds and write them separately. Then
2:18
count all those bonds. The number you
2:20
get will be the steric number of that
2:22
atom. And the hybridization type in
2:24
which the number of hybrid orbitals
2:26
equals that steric number will be the
2:28
hybridization of that atom. Let's apply
2:30
this trick on boron in BF3. When we
2:33
write the expanded structure of BF3, it
2:36
becomes clear that boron is forming
2:38
three bonds, one bond with each florine
2:40
atom. So the steric number of boron is
2:43
three. Now sp2 hybridization gives us
2:46
three hybrid orbitals. Therefore the
2:49
hybridization of boron in BF3 is sp2.
2:54
Now look at the electronic configuration
2:55
of boron. The atomic number of boron is
2:58
five. This means boron has five
3:01
electrons. The ground state electronic
3:03
configuration of boron is 1 s2 2 s2 2
3:12
in this ground state. Boron has only one
3:14
unpaired electron. But boron needs to
3:17
form three bonds with three florine
3:19
atoms. So boron needs three unpaired
3:22
electrons. To get three unpaired
3:24
electrons, one electron from the 2s
3:26
orbital gets excited. This electron
3:28
moves from 2s orbital and goes into the
3:30
empty 2py orbital. Now the excited state
3:33
electronic configuration of boron
3:35
becomes 1 s2 2 s1 2px1 2py1 2pz 0. Here
3:44
always remember one important point. The
3:46
electron will always be excited from the
3:48
2s orbital only. The electron from 1 s
3:51
orbital will never get excited. The
3:53
reason is that 1 s is not the valence
3:55
shell of boron. Only valence shell
3:58
electrons participate in hybridization
4:00
and bonding. Now hybridization takes
4:02
place. The s orbital and the 2p orbitals
4:05
which are 2px and 2py mix together.
4:08
According to the rule of hybridization,
4:10
three sp2 hybrid orbitals are formed.
4:13
These three sp2 orbitals form three
4:15
bonds with three florine atoms. The
4:17
shape of BF3 molecule is trional planar
4:20
and the bond angle in BF3 is 120°. Now
4:24
let's move to the second example that is
4:27
the ethine molecule. In eene the central
4:30
atom is carbon. Let us apply the same
4:32
trick here. First write the expanded
4:35
structure of aine. When we open the
4:37
structure of ethine it becomes clear
4:39
that each carbon is forming a double
4:41
bond with the other carbon and along
4:42
with that each carbon also has two
4:45
hydrogen atoms attached to it. Now if we
4:47
count the characters of carbon we have
4:49
two single bonds with two hydrogen atoms
4:51
and one double bond with the other
4:53
carbon. These make a total of three
4:55
characters for carbon. So the steric
4:57
number of carbon is three. And since sp2
5:00
hybridization gives three hybrid
5:01
orbitals, the hybridization of carbon in
5:04
ethine is sp2. Now let's look at the
5:07
electronic configuration of carbon. The
5:09
atomic number of carbon is six. The
5:12
ground state electronic configuration of
5:13
carbon is 1 s2 2 s2 2 px1 2 p y1 2 pz0.
5:20
In this ground state, carbon already has
5:23
two unpaired electrons in two px and 2py
5:26
orbitals. But for sp2 hybridization,
5:29
carbon needs three unpaired electrons.
5:31
So one electron from 2s orbital gets
5:34
excited and moves into the empty 2pz
5:36
orbital. Now the excited state
5:38
electronic configuration of carbon
5:40
becomes 1 s2 2 s1 2 px1 2 p y1 2 pz1.
5:47
Now hybridization takes place. The 2 S
5:50
orbital and two px and two py orbitals
5:54
mix together to form 3 spp2 hybrid
5:57
orbitals. The 2 pz orbital does not take
6:00
part in hybridization. It remains
6:02
unhybridized and forms the pi bond in
6:05
the double bond of ethine. The shape
6:07
around each carbon atom in ethine is
6:09
trional planar and the bond angle is
6:12
120°. Now let's move towards the third
6:15
example that is benzene. In benzene, the
6:18
carbon atom shows sp2 hybridization. Let
6:21
us apply the trick here as well. Write
6:23
the expanded structure of benzene. In
6:25
benzene, each carbon atom has one
6:27
hydrogen atom attached to it. Along with
6:29
that, each carbon is connected to two
6:31
other carbon atoms. One with a single
6:34
bond and one with a double bond. So, if
6:36
we count the characters of carbon in
6:37
benzene, there is one bond with
6:39
hydrogen, one single bond with adjacent
6:42
carbon and one double bond with the
6:44
other adjacent carbon. This gives us a
6:46
total of three characters. So the steric
6:48
number is three and therefore the
6:50
hybridization of carbon in benzene is
6:53
sp2. Now the ground state electronic
6:55
configuration of carbon is the same as
6:57
we discussed in athen. It is 1 s2 2 s2
7:01
2px1 2p y1 2pz 0. And just like in a one
7:08
electron from 2s orbital gets excited
7:10
and moves into the empty 2pz orbital.
7:12
The excited state configuration becomes
7:15
1 s2 2 s1 2 px1 2 p y1 2 pz1. After
7:21
excitation, the same hybridization
7:23
process takes place. The 2 s 2 px and
7:26
2py orbitals mix together to form 3 s p2
7:31
hybrid orbitals. The two pz orbital
7:34
remains unhybridized and participates in
7:36
the deoized pi bonding system of
7:38
benzene. The shape around each carbon
7:40
atom in benzene is also trigonal planar
7:44
and the bond angle is 120°. Now at the
7:47
end, let me give you a quick summary of
7:49
this entire lecture. Sp2 hybridization
7:52
occurs when 1 s orbital and 2 p orbitals
7:55
mix together to form three sp2 hybrid
7:58
orbitals. The shape of sp2 hybridized
8:01
molecules is trional planer with a bond
8:04
angle of 120°. We also learned an
8:07
important trick. Write the expanded
8:08
structure of any molecule. Count the
8:10
total bonds or characters of the central
8:13
atom and that number will tell you the
8:15
hybridization. If the number is three,
8:17
the hybridization will be sp2. We
8:19
studied three examples. In BF3, boron
8:23
underos excitation from ground state and
8:26
then sp2 hybridization occurs. In athen
8:29
and benzene carbon also underos
8:31
excitation from 2s to 2pz and then sep
8:36
hybridization takes place. In all three
8:38
cases the shape is tragonal planar and
8:41
bond angle is 120°. Remember the rule of
8:44
hybridization. The number of orbitals
8:46
that go in will always equal the number
8:48
of hybrid orbitals that come out. This
8:51
concept is very important for your
8:53
exams. Practice these examples and you
8:55
will master sp2 hybridization
