Hybridisation: Definition, Formula, Examples, Questions, C2, BF3, Water

Hybridization

Hybridization means mixing atomic orbitals to recombine into a new set of hybrid orbitals. Types of hybridization include $\mathrm{sp},{\mathrm{sp}}^2,{\mathrm{sp}}^3,{\mathrm{sp}}^3\mathrm{d}$, and ${\mathrm{sp}}^3{\mathrm{d}}^2$. These hybridizations correspond to different molecular geometries and bonding patterns. For instance, the sp³ hybridization gives a tetrahedral geometry, typical for molecules like methane, and CH₄, while the hybridization of sp² forms provides a trigonal planar geometry, seen for ethene, ${\mathrm{C}}_2{\mathrm{H}}_4$. Hybrid orbitals are formed by mixing the s- and p- orbitals of an atom (and sometimes the d orbitals).

The Salient Features And Conditions For Hybridization

1. Hybrid orbitals do not exist in isolated atoms. They are formed only in covalently bonded atoms.

2. Hybrid orbitals have shapes and orientations that are very different from those of the atomic orbitals in isolated atoms.

3. A set of hybrid orbitals is generated by combining atomic orbitals. The number of hybrid orbitals in a set is equal to the number of atomic orbitals that were combined to produce the set.

4. All orbitals in a set of hybrid orbitals are equivalent in shape and energy.

5. The type of hybrid orbitals formed in a bonded atom depends on its electron-pair geometry as predicted by the VSEPR theory.

6. Hybrid orbitals overlap to form σ bonds. Unhybridized orbitals overlap to form π bonds.

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Types Of Hybridisation

The hybridization can be of several types depending on the number of hybrid orbitals involved in the formation of molecules. The table given below describes all types of hybridization and their geometries.

How To Find Hybridisation

The hybridization depends upon sigma bonds and a lone pair of electrons.

Thus,

Hybridization = Number of sigma bonds + Number of lone pairs present on the central atom

For example, hybridization for NH₃ is sp³ and its molecular geometry is tetrahedral.

NH₃ has 3 sigma bonds and 1 lone pair, thus hybridization for NH₃:

3 sigma bonds + 1 lone pair = 4

Thus hybridization for NH₃ is sp³ and its geometry is tetrahedral.

sp Hybridization

This hybridization process involves mixing of the valence s orbital with one of the valence p orbitals to yield two equivalent sp hybrid orbitals that are oriented in a linear geometry as shown in the figure. The number of atomic orbitals combined always equals the number of hybrid orbitals formed. The p orbital is one orbital that can hold up to two electrons. The sp set is two equivalent orbitals that point 180^oC from each other. The two electrons that were originally in the s orbital are now distributed to the two sp orbitals, which are half filled.

${\mathbf{s}\mathbf{p}}^2$Hybridization

When 1 s-orbital and 2 p-orbitals are involved in the molecule formation then the equivalent set of orbitals are known as sp² hybrid orbitals. These hybrid orbitals arrange themselves at an angle of 120^oC as shown in the figure.

${\mathbf{s}\mathbf{p}}^3$Hybridization

When 1 s-orbital and 3 p-orbitals are involved in the molecule formation then the equivalent set of orbitals are known as sp³ hybrid orbitals. The bond angle between these hybrid orbitals is 109^oC as shown in the figure.

$s{p}^{3}d$ Hybridisation

When 1 s-orbital, 3 p-orbitals, and 1 d-orbital are involved in the molecule formation then the equivalent set of orbitals are known as sp³d hybrid orbitals. There are two kinds of bonds formed for sp³d hybridization, i.e., 2 axial bonds and 3 equatorial bonds. The angle between the axial bond and the equatorial plane is 90^oC while the bond angle between the equatorial bonds is 120^oC as shown in the figure given below:

$s{p}^3{d}^2$Hybridization

When 1 s-orbital, 3 p-orbitals and 2 d-orbitals are involved in the molecule formation then the equivalent set of orbitals are known as sp³d² hybrid orbitals. There are two kinds of bonds formed for sp³d² hybridisation, i.e., 2 axial bonds and 4 equatorial bonds. The angle between the axial bond and the equatorial plane is 90^oC while the bond angle between the equatorial bonds is 90^oC as shown in the figure given below:

${\mathrm{d}}^2{\mathrm{sp}}^3$ Hybridisation

When 2 d-orbital, 1 s-orbital and 3 p-orbitals are involved in the molecule formation then the equivalent set of orbitals are known as d²sp³ hybrid orbitals. There are two kinds of bonds formed for sp³d² hybridisation, i.e, 2 axial bonds and 4 equatorial bonds. The angle between the axial bond and the equatorial plane is 90^oC while the bond angle between the equatorial bonds is 90^oC as shown in the figure given below:

$s{p}^3{d}^3$ Hybridization

When 1 s-orbital, 3 p-orbitals and 3 d-orbitals are involved in molecule formation then the equivalent set of orbitals are known as sp³d³ hybrid orbitals. The sp³d³ hybridization has a pentagonal bipyramidal geometry i.e., five bonds in a plane, one bond above the plane and one below it.

Related Topics:

• Pi Bonds
• Vsepr Theory
• Molecular Geometry

Some Solved Examples

Example 1:

The type of hybridization and number of lone pair (s) of electrons of Xe in XeOF_4, respectively, are :

1) sp³d² and 1
2) sp³d and 2
3) sp³d²and 2
4) sp³d and 1

Solution

As we have learned in Hybridisation:

Hybridisation is sp³d²sp3 d2
It contains $5\sigma$ 5σ bond and $1\pi$ 1π bond.
It has 1 lone pair of electrons.

Hence, the correct answer is Option (1)

Example 2:

The correct statement about ICl₅ and ICl₄^-is:
1)Both are isostructural
2)ICl₅ is trigonal bipyramidal and ICl₄^-is tetrahedral
3)ICl₅ is square bipyramidal and ICl₄^- is tetrahedral
4)ICl₅ is square pyramidal and ICl₄^-is square planar.

Solution
ICl₅ is square bipyramidal and Cl₄⁻is square planar.
ICl₅:−

Square Pyramidal

Lone Pair =1

Bond Pair=5

hybridisation= sp³d²

ICl₄^- :-

Square planar

Lone Pairs = 2

Bond Pairs = 4

hybridisation = sp³d²

Hence, the correct answer is Option (4)

Example 3:

The orbitals undergoing Hybridization involve

1) Orbitals of the same atom with almost similar energies

2)Orbitals of different atoms but with equal energies

3)Orbitals of different atoms with different energies

4)Orbitals of the same atoms with exactly equal energies

Solution

The orbitals of the same atom having similar energies undergo hybridization to form hybrid orbitals which have the same energy.

Hence, the answer is option (1).

Example 4:

In which pair of species, both species have a similar geometry?

1)CO₂, SO₂

2)CO₂^3- and SO₃²⁻

3) SO₄^2- and ClO₄^-

4)PH₃ and BH₃

Solution:

The geometry of CO₂ is linear.

(O=C=O)

The geometry of SO₂ is V-shape

The geometry of CO₃^2- is trigonal planar.

The geometry of SO₃²⁻ is a pyramidal shape

The geometry of SO₄²⁻ is tetrahedral

The geometry of ClO₄⁻ is tetrahedral

The geometry of NH₃ is a pyramidal shape

The geometry of BH₃ is trigonal planar

Hence, the answer is option (3).

Example 5:

In ${\mathrm{SO}}_2,{\mathrm{NO}}_2^{-}$and ${\mathrm{N}}_3^{-}$the hybridizations at the central atom are respectively : [JEE Main 2025]

1) ${\mathrm{sp}}^2,{\mathrm{sp}}^2$ and sp

2) ${\mathrm{sp}}^2,\mathrm{sp}$ and sp

3) ${\mathrm{sp}}^2,{\mathrm{sp}}^2$ and ${\mathrm{sp}}^2$

4) $\mathrm{sp},{\mathrm{sp}}^2$ and sp

Solution:

${\mathrm{SO}}_2\Rightarrow 2\sigma$ bond +1 l.p. $\Rightarrow s{p}^2$ hybridisation

${\mathrm{NO}}_2^{-}\Rightarrow 2\sigma$ bond +1 l.p. $\Rightarrow s{p}^2$ hybridisation

${\mathrm{N}}_3^{-}\Rightarrow 2\sigma$ bond $\Rightarrow sp$ hybridisation

Hence, the correct answer is option (1).

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Summary

Hybridization describes mixing atomic orbitals to form hybrid orbitals, which allows for the determination of the shape and bonding properties of molecules. It was introduced by Linus Pauling, and it includes types such as sp, sp², sp³, sp³d, and sp³d²—each being associated with specific molecular geometries. For example, sp3 hybridization would give a tetrahedral shape and sp² results in a trigonal planar structure. In forming hybrid orbitals, the 's' and 'p' orbitals of an atom combine, and sometimes the 'd' orbitals, thereby forming orbitals that become equal in energy and shape.

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