0:00
sp3 hybridization explained. In this
0:03
lecture, you will learn sp3
0:04
hybridization in the simplest way
0:06
possible. After watching this video, you
0:09
will completely understand how sp3
0:11
hybridization works in different
0:12
molecules. First of all, you need to
0:14
understand what is hybridization and how
0:17
we can define it. Actually,
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hybridization is a process in which
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different atomic orbitals with different
0:22
shape and energy intermix to form new
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set of orbitals. Having same shape and
0:27
energy is called hybridization. and
0:29
orbitals obtained are called hybrid
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orbitals. In simple words, when atomic
0:33
orbitals mix together, they form new
0:35
orbitals called hybrid orbitals. Now,
0:37
let's move towards types of
0:38
hybridization. When S and P orbitals mix
0:41
together, three types of hybridization
0:43
can occur. First is sp hybridization.
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Second is sp2 hybridization and third is
0:50
sp3 hybridization. In this lecture, we
0:54
will understand sp3 hybridization in
0:57
detail. First of all, we need to define
0:59
sp3 hybridization. The type of
1:01
hybridization in which 1 s and 3 p
1:04
atomic orbitals intermix to form four
1:07
sp3 hybridized orbitals is called sp3
1:10
hybridization. This means one s orbital
1:13
and 3 p orbitals combine to make four
1:15
new sp3 orbitals. To understand sp3
1:19
hybridization completely, we will study
1:21
three important examples. In the first
1:24
example, we will see sp3 hybridization
1:27
in methane. In the second example, we
1:29
will understand sp3 hybridization in
1:32
ammonia. And in the third example, we
1:34
will learn sp3 hybridization in water
1:37
molecule. These three examples will make
1:39
the concept crystal clear. Let's start
1:41
with first example which is sp3
1:43
hybridization in methane. Let us start
1:45
with methane molecule. The formula of
1:48
methane is CH4. In methane, the central
1:51
atom is carbon. carbon shows sp3
1:54
hybridization in methane molecule. Now
1:57
let me show you the electronic
1:59
configuration of carbon. The atomic
2:01
number of carbon is 6. So it means
2:03
carbon has six electrons. The ground
2:06
state electronic configuration of carbon
2:08
is 1 s2 2 s2 2 p2. In this ground state
2:13
carbon has only two unpaired electrons
2:15
in two p orbitals. But in methane carbon
2:18
forms four bonds with four hydrogen
2:20
atoms. So carbon needs four unpaired
2:22
electrons. So to get four unpaired
2:25
electrons, carbon gets excited. When
2:27
carbon gets excited, one electron from
2:30
2s orbital jumps to the empty 2p
2:32
orbital. Now the excited state
2:34
electronic configuration of carbon
2:36
becomes 1 s2 2 s1 2 p3. In the excited
2:40
state, carbon has four unpaired
2:42
electrons. One electron is in 2 s
2:45
orbital and three electrons are in two p
2:48
orbitals. Now hybridization takes place.
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When an electron moves from ground state
2:52
to excited state, the orbitals that mix
2:55
together give us the hybridization
2:56
state. Here in excited state, 1 2
2:59
orbital and 32p orbitals are present
3:02
with unpaired electrons. These four
3:04
orbitals mix together. So 1 s orbital
3:06
and 3p orbitals undergo hybridization.
3:09
Therefore, the hybridization of carbon
3:11
is sp3. After hybridization, four sp3
3:15
hybrid orbitals are formed. All four sp3
3:18
orbitals have same shape and same
3:20
energy. These four sp3 orbitals form
3:23
four bonds with four hydrogen atoms. The
3:26
shape of methane molecule is
3:27
tetrahedral. The bond angle in methane
3:32
Now let us understand the second example
3:35
that is ammonia molecule. The formula of
3:37
ammonia is NH3. In ammonia the central
3:40
atom is nitrogen. Nitrogen also shows
3:43
sp3 hybridization. The atomic number of
3:46
nitrogen is 7. So it means nitrogen has
3:49
seven electrons. The ground state
3:51
electronic configuration of nitrogen is
3:53
1 s2 2 s2 2 p3. In this ground state,
3:57
nitrogen already has three unpaired
3:59
electrons in two p orbitals, which means
4:02
one electron in each 2p orbital. Now
4:04
here is an important point. In ammonia,
4:07
nitrogen forms three bonds with three
4:09
hydrogen atoms. Nitrogen already has
4:11
three unpaired electrons in ground
4:13
state. So nitrogen does not need to get
4:15
excited. There is no excitation in case
4:17
of nitrogen. But the question is why no
4:20
excitation takes place. The answer is
4:22
simple because nitrogen already has
4:24
enough unpaired electrons to form bonds.
4:27
Also all three 2p orbitals are already
4:29
occupied. There is no empty 2p orbital
4:32
available. So electron cannot jump to 2p
4:35
orbital. Therefore nitrogen remains in
4:37
ground state only. Even though there is
4:39
no excitation, hybridization still
4:42
occurs. In nitrogen 12s orbital and 32p
4:45
orbitals undergo hybridization. So the
4:48
hybridization state will be sp3. After
4:51
hybridization, four sp3 hybrid orbitals
4:54
are formed. Out of these four sp3
4:57
orbitals, three orbitals have one
4:59
unpaired electron each. These three
5:01
orbitals form three bonds with three
5:03
hydrogen atoms. The fourth sp3 orbital
5:06
contains a lone pair of electrons. This
5:08
lone pair does not participate in
5:10
bonding. The shape of ammonia molecule
5:12
is trional parameal. It is not
5:14
tetrahedral like methane because one
5:16
position is occupied by lone pair. The
5:18
bond angle in ammonia is approximately
5:20
107°. Now let us discuss the third
5:23
example which is water molecule. The
5:26
formula of water is H2O. In water the
5:29
central atom is oxygen. Oxygen also
5:32
underos sp3 hybridization. The atomic
5:35
number of oxygen is 8. So oxygen has
5:38
eight electrons. The ground state
5:40
electronic configuration of oxygen is 1
5:43
s2 2 s2 2 p4. In this ground state,
5:47
oxygen has two unpaired electrons in two
5:50
p orbitals. The remaining two electrons
5:52
in two p orbitals are paired. In water,
5:55
oxygen forms two bonds with two hydrogen
5:58
atoms. Oxygen already has two unpaired
6:00
electrons. So oxygen does not require
6:03
excitation. Oxygen remains in ground
6:05
state. So again the question is why does
6:08
oxygen not get excited? The reason is
6:10
that all three 2p orbitals are already
6:12
filled. There is no empty 2p orbital
6:15
available for electron to jump. If an
6:17
electron tries to jump, it needs to go
6:19
to 3s orbital which requires very high
6:22
energy. So excitation does not occur in
6:24
oxygen. However, hybridization takes
6:27
place even without excitation. 1 2
6:30
orbital and 32p orbitals of oxygen
6:32
undergo hybridization. So the
6:35
hybridization will be sp3. 4 sp3 hybrid
6:39
orbitals are produced. Out of four sp3
6:43
orbitals, only two orbitals contain
6:45
unpaired electrons. These two orbitals
6:47
form two bonds with two hydrogen atoms.
6:50
The remaining two sp3 orbitals contain
6:53
lone pairs of electrons. The shape of
6:55
water molecule is bent or v-shaped. This
6:58
is because two positions are occupied by
7:00
two lone pairs. The bond angle in water
7:03
is approximately 104.5°.
7:06
Now at the end, let's have a look at the
7:08
summary of all three examples. I will
7:10
summarize all three examples for you. In
7:12
methane, carbon underos excitation from
7:15
ground state to excited state. One
7:17
electron jumps from 2s to 2p orbital.
7:20
Then sp3 hybridization occurs. Methane
7:23
has tetrahedral shape. In ammonia,
7:25
nitrogen does not undergo excitation. It
7:28
already has three unpaired electrons.
7:30
Nitrogen directly underos sp3
7:33
hybridization in ground state. Ammonia
7:35
has trional parameal shape due to one
7:38
lone pair. In water, oxygen does not
7:41
undergo excitation. It already has two
7:43
unpaired electrons. Oxygen directly
7:46
underos sp3 hybridization in ground
7:48
state. Water has bent shape due to two
7:50
lone pairs. In all three molecules, sp3
7:54
hybridization occurs. But the process is
7:56
slightly different in each case. In
7:58
methane, excitation happens first. In
8:01
ammonia and water, no excitation is
8:03
needed. Therefore, sp3 hybridization is
8:06
the mixing of one s orbital and three p
8:09
orbitals to form four sp3 hybrid
8:11
orbitals. This concept is very important
8:14
for understanding molecular geometry and
