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Hybridization is defined as the intermixing of atomic orbitals of same or nearly same energy to give new hybrid orbitals of equivalent energy. Hybridization is the responsible for the geometry of the molecule.

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Types of Hybridization

It is based on the types of orbitals mixed together and can be classified as sp, sp2, sp3, sp3d, sp3d2.

sp Hybridization

In sp hybridization, one s and one p orbital of the same shell mix together to form two new orbitals of same energy. They forms linear geometry with an angle of 180ο and each sp orbital has 50% s character and 50% p character.

sp Hybridization

Example: BeF2

Be4 = 1s2 2s2

If we see the electronic configuration of Be, there is only two valence electrons present in 2s orbital which are paired, so how it makes two bonds with fluorine. We know that p orbital has no electron so one of 2s electron will shift to the 2p orbital, after that Be will be able to form two bonds with fluorine.



Other examples: BeCl2, BeBr2, BeH2 and all compounds of carbon contain triple bond like C2H2.

sp2 Hybridization

In this type of hybridization, one s and two p orbitals mix together to form three new sp2 hybrid orbitals of same energy. They form triangular planar geometry with an angle of 120ο and each sp2 orbital has 33.33% s-character and 66.66% p-character.

sp2 Hybridization

Example: BF3

B5 = 1s2 2s2 2p1

In the electronic configuration of Boron, there is one unpaired electron but boron is forming three bonds with fluorine in BF3. One of the electrons from 2s orbital will be shifted to 2p vacant orbital. After that it can have one s and two p orbitals which get mix together and form three new sp2 orbitals of the same energy which make three new bonds with three fluorine atoms.

sp2 Hybridization

sp3 Hybridization

When one s and three p orbital mix together to get four sp3 orbitals of equal energy, is said to be sp3 hybridization. They form tetrahedral geometry with an angle of 109ο 28’. Each sp3 hybrid orbital has 25% s character and 75% p character.

Example: Methane (CH4)

All four bonds of methane are equivalent in all respects which have same bond length and bond energy.

C6 = 1s2 2s2 2p2

sp2 Hybridization

Molecular Orbital structure of methane


sp3d Hybridization

In sp3d hybridization, one s, three p and one d orbitals mix together to from five sp3d orbitals of same energy. They form trigonal bipyramidal geometry and three hybrid orbital which are on horizontal plane maintain an angle of 120ο to each other are known as equatorial orbitals. The other two orbitals lie on vertical plane at 90ο plane of equatorial orbital known as axial orbitals.

Example: PCl5



sp3d2 Hybridization

When one s, three p and two d orbitals mix together and results six sp3d2 hybrid orbitals of equivalent energy, is known as sp3d2 hybridization. These orbitals are arranged in octahedral geometry by making 90ο to each other.

Example: SF6



Key points

1. Intermixing of atomic orbitals of the same or nearly same energy to give new orbitals of the same energy.

2. Half filled, empty, fully filled orbitals will go into hybridization.

3. The numbers of hybrid orbitals are always equal to number of intermixing orbitals.

4. Concept of hybridization is only for sigma bond and lone pair where as pi bond is not relevant with hybridization.

5. Hybrid orbitals are named after parent orbitals.

1s + 1p = sp

1s+2p = sp2

1s + 3p = sp3

6. The shapes of hybrid orbitals are same.

7. The hybrid orbitals orientate such that the repulsion between them is minimum.

How to find hybridization?

There are two methods to find out hybridization of atoms:

1. By analyzing Structure

If structure is provided, hybridization can be calculated on the basis of number of sigma bonds and lone pairs.

Z = No. of sigma bonds + lone pair of central atom

Z= 2 → sp

Z= 3 → sp2

Z = 4 → sp3

Z= 5 → sp3d

Z=6 → sp3d2

For example: SO3


Z=3, sp2, Triangular planar



Z= 2 sigma bond + 2 lone pair = 4

sp3, tetrahedral



Z = 3 sigma bonds + 1 lone pair = 4

sp3, tetrahedral

2. Super Trick to find the hybridization:

By using the simple formula, hybridization of any molecule or ion can be determined easily.

Z = ½ (No. of valence of electron on central atom + (-ve charge) – (+ve charge) + No. of monovalent atoms

Monovalent atoms: H, F, Cl, Br, I


Hybridization of Nitrogen in NH3

Z = ½ (5+0-0+3) = 4 → sp3

Carbon in CO32-

Z = ½ (4+2-0+0) = 3 → sp2

Practice Problems to Find Hybridization

  1. Si in SiF4
  2. S in SF4
  3. S in H2S
  4. Xe in XeO2F2
  5. Xe in XeF4
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6 thoughts on “Hybridization”

    1. Thank you for the comment. In sp hybridization, it was typographical mistake, which is corrected and I do not see anything wrong in super trick formula and NH3 calculation. There are 5 valence electrons of nitrogen and 3 monovalent atoms so 5+3= 8 and divided by 2, we get 4, it means Nitrogen has sp3 hybridization in NH3.

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