All of HYBRIDIZATION Explained in 8 Minutes

Hybridization is the process in which

atomic orbitals with different shapes

and energies mix to form a new set of

equivalent orbitals known as hybrid

orbitals which have the same shape and

energy. First we need to understand the

important rule of hybridization and

after that we will move towards the

types of hybridization. Always remember

that the number of hybrid orbitals

obtained is always equal to the number

of atomic orbitals intermixed. For

example, if 1 s and 3 p atomic orbitals

are involved in mixing, then the number

of hybrid orbitals obtained will be four

because four atomic orbitals are

involved in mixing. There are three

common types of hybridization based on

the mixing of s and p orbitals and they

are sp3, sp2 and sp hybridization.

First, let's explain sp3 hybridization

in detail. We can define sp3

hybridization as a type of hybridization

in which 1s and 3 p atomic orbitals are

intermixed to form four sp3 hybridized

orbitals. Let's take the example of

methane. The hybridization of carbon in

methane is sp3. But one important point

we need to clarify before going into

detail is how we can check the

hybridization of the central polyvalent

atom in a molecule. To check it, we need

to count the total steric number of the

polyvalent atom. The steric number

refers to the total number of bonded and

non-bonded electron pairs of that atom.

For example, if we apply this to

methane, it has four single bonds and no

lone pairs present on the carbon atom.

Its steric number will be four and sp3

hybridization involves 1 s and 3 p

orbitals. So in this way we can confirm

that the hybridization of carbon in

methane since steric number is four.

First let's have a look at the ground

state electronic configuration of

carbon. It has one electron in the 2px

orbital. One electron in the 2py orbital

and the 2pz orbital is empty. In the

excited state configuration of carbon,

one electron from the 2s orbital will be

promoted to the 2pz orbital. These four

orbitals will then overlap to form four

sp3 hybridized orbitals. If we look at

the shape of methane molecule, it has

four hybridized orbitals and each

orbital is overlapping with one s

orbital of hydrogen to complete four

carbon hydrogen bonds. In case the

overlapping is called sp3 and s orbital

because sp3 orbitals orbital of carbon

is overlapping with s orbital of

hydrogen to form sigma bond. The shape

of the methane molecule will be

tetrahedral and the bond angle is

109.5°. Let's take one more example of

sp3 hybridization to understand it more

deeply. Take the example of the ammonia

molecule. By applying the steric number

formula, we can check that nitrogen is

making three single bonds with each

hydrogen. So it means nitrogen has three

bonded electrons. Since nitrogen is a

member of group 5a to complete its five

veence electrons, it needs to have one

lone pair. Now let's calculate the total

steric number. There are three bond

pairs and one lone pair which will give

a total of four steric numbers. This

confirms that nitrogen in ammonia is sp3

hybridized. If we consider the

electronic configuration of nitrogen, it

has one electron each in the 2 p

orbitals. Keep in mind in this case

there is no need to excite an electron

from the 2s orbital to the p orbital

because the already present electrons

will repel the incoming electrons. We

need to excite electrons only in cases

where any 2p orbital is empty and an

electron can jump from the 2s orbital to

that p orbital. So in the case of

ammonia 1s and 3 p orbitals will overlap

to form four sp3 hybridized orbitals.

The shape of the ammonia molecule is

trional pyramidal and the bond angle

will be less than

109.5° due to the reason that the lone

pair will occupy more space than the

bond pairs causing the overall bond

angle to shrink. Now let's move towards

sp2 hybridization. We can define sp2

hybridization as the process in which 1

s and 2p atomic orbitals intermix to

form three sp2 hybridized orbitals. Here

the steric number will be three because

1 s and 2p orbitals are involved in

mixing. Some common examples of sp2

hybridization are boron trilouide and

the ethine molecule. First let's

understand the shape of boron

triloulloride based on hybridization.

Since boron is making three single bonds

with florine, its steric number will be

three. Confirming that the hybridization

state of boron in BF3 is sp2. The atomic

number of boron is five and its

electronic configuration shows that it

has one electron in the 2px orbital

while the 2py and 2pz orbitals are

empty. In this case, we need to excite

one electron from the 2s orbital to the

2py orbital which will be called the

excited state electronic configuration

of boron. Now the 2s orbital will

intermix with the 2p orbitals to form

three sp2 hybridized orbitals. Since the

2pz orbital is empty, it will not be

considered. If we look at the structure

of boron trilouide, three hybridized

orbitals of boron will form single bond

by headto-head overlapping of 2p orbital

of florine to make three bond between

boron and florine. The shape of boron

trilouide will be trional planer and its

bond angle will be 120°. In the case of

ethine, if we apply the steric number

formula, the result will also be three.

Here we need to understand one important

point that double and triple bonds are

also counted as one steric number. It

means if an atom forms a double or

triple bond with another atom, it must

be considered as one steric number.

Let's apply this to ethine. It has one

double bond and two single bonds with

hydrogen. So its steric number is three.

Confirming the sp2 hybridization of

carbon in the ethine molecule. Now let's

consider the electronic configuration of

carbon. Again, as we already discussed

in the case of methane, an electron from

the 2s orbital will be promoted to the

vacant 2pz orbital. During this process,

1 s and 2p orbitals will overlap to form

three hybridized orbitals. One thing we

need to consider here is that although

one electron is present in the 2pz

orbital, it is not involved in the

intermixing of atomic orbitals and will

be called an unhybridized orbital.

Unhybridized orbitals are those atomic

orbitals that do not take part in the

hybridization process and are involved

during the formation of a pi bond. The

geometry of the ethine molecule is

trional planer and the bond angle is

120°. Now we can move towards sp

hybridization. It can be defined as the

type of hybridization in which 1 s and 1

p atomic orbital intermix to form two sp

hybridized orbitals. In this case the

steric number will be two. Some common

examples of molecules having sp

hybridization are burillium dchloride

and ethine. First let's explain the

shape of burillium dchloride. To make it

clear, let's look at the structure of

burillium dchloride. Since burillium is

making two single bonds with chlorine,

its steric number will be two, which

confirms sp hybridization. If we look at

the ground state electronic

configuration of burillium, its atomic

number is four and the two p orbitals

are empty because all four electrons are

present in the 1 s and 2 s orbitals. In

this case, we again need to excite

electron and one electron from the 2s

orbital will be promoted to the 2px

orbital. In this way, these two atomic

orbitals will overlap to form two sp

hybridized orbitals. The shape of the

molecule will be linear and the bond

angle will be 180°. Now, let's take one

more example to make it clear. The

hybridization state of carbon in ethan

is sp, which is confirmed by the steric

number as well. Since carbon is forming

one triple bond with another carbon and

one single bond with hydrogen, the

overall steric number will be two, which

confirms its sp hybridization. Again

consider the ground state electronic

configuration of carbon. Here we again

need to promote an electron from the

ground state to the excited state. But

keep in mind only one s and one p atomic

orbital will overlap to form two sp

hybridized orbitals since the 2 py and 2

pz orbitals are not involved in

overlapping. They will be called

unhybridized orbitals and will only

participate in the formation of pi

bonds. If we look at the shape of the

ethn molecule based on hybridization, it

has a linear structure. One hybridized

orbital forms a single bond with

hydrogen and the second orbital forms a

sigma bond with the carbon atom. The

first pi bond is formed by the

overlapping of the unhybridized 2py

orbital of one carbon with another

carbon in ethn. And the second pi bond

is formed by the side to side

overlapping of the two pz orbitals. In

this way one sigma and two pi bonds are

formed between the two carbons in the

ethin molecule. The bond angle in the

case of sp hybridization will be 180°.

This concludes our hybridization topic

on bases of mixing of S and P orbitals.

If you have any question regarding this

topic, you can ask them in comments. And

if any other specific topics you want, I

should cover in next video, you also can

mention them in comment section. Thanks

for watching this complete. I hope it

was helpful for you in preparing exams.