# Hybridization

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.

**Types of Hybridization**

It is based on the types of orbitals mixed together and can be classified as sp, sp^{2}, sp^{3}, sp^{3}d, sp^{3}d^{2}.

**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 180^{ο} and each sp orbital has 50% s character and 50% p character.

Example: BeF_{2}

Be^{4 }= 1s^{2} 2s^{2 }

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.

F-Be-Fe

Other examples: BeCl_{2}, BeBr_{2}, BeH_{2} and all compounds of carbon contain triple bond like C_{2}H_{2}.

**sp**^{2} Hybridization

^{2}Hybridization

In this type of hybridization, one s and two p orbitals mix together to form three new sp^{2} hybrid orbitals of same energy. They form triangular planar geometry with an angle of 120^{ο} and each sp^{2 }orbital has 33.33% s-character and 66.66% p-character.

Example: BF_{3}

B^{5} = 1s^{2} 2s^{2} 2p^{1 }

In the electronic configuration of Boron, there is one unpaired electron but boron is forming three bonds with fluorine in BF_{3}. 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 sp^{2} orbitals of the same energy which make three new bonds with three fluorine atoms.

**sp**^{3} Hybridization

^{3}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 109^{ο} 28’. Each sp3 hybrid orbital has 25% s character and 75% p character.

Example: Methane (CH_{4})

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

C^{6} = 1s^{2} 2s^{2} 2p^{2}

^{}

**sp**^{3}d Hybridization

^{3}d Hybridization

In sp^{3}d hybridization, one s, three p and one d orbitals mix together to from five sp^{3}d orbitals of same energy. They form trigonal bipyramidal geometry and three hybrid orbital which are on horizontal plane maintain an angle of 120^{ο} to each other are known as equatorial orbitals. The other two orbitals lie on vertical plane at 90^{ο} plane of equatorial orbital known as axial orbitals.

Example: PCl_{5}

**sp**^{3}d^{2} Hybridization

^{3}d

^{2}Hybridization

When one s, three p and two d orbitals mix together and results six sp^{3}d^{2} hybrid orbitals of equivalent energy, is known as sp^{3}d^{2 }hybridization. These orbitals are arranged in octahedral geometry by making 90^{ο} to each other.

Example: SF_{6}

**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 = sp^{2}

1s + 3p = sp^{3}

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 → sp^{3}

Z= 5 → sp^{3}d

Z=6 → sp^{3}d^{2}

For example: SO_{3}

Z=3, sp^{2}, Triangular planar

H_{2}O

Z= 2 sigma bond + 2 lone pair = 4

sp^{3}, tetrahedral

NH_{3}

Z = 3 sigma bonds + 1 lone pair = 4

sp^{3}, 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

Examples:

**Hybridization of Nitrogen in NH _{3}**

Z = ½ (5+0-0+3) = 4 → sp^{3}

**Carbon in CO _{3}^{2-}**

Z = ½ (4+2-0+0) = 3 → sp^{2}

**Practice Problems to Find Hybridization**

- Si in SiF
_{4} - S in SF
_{4} - S in H
_{2}S - Xe in XeO
_{2}F_{2} - Xe in XeF
_{4}

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Aqsa RafiqV easy method ..good inspiration

Editorial StaffThank you Aqsa

M.AshrafIn sp hyb.angle 180. Not 120. 2 Trick. See your formula and example NH3

Editorial StaffThank 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.

XMCPLEnjoyed looking through this, very good stuff, thankyou .

Editorial StaffThank you