Hybridization

Table of Content

Important Characteristics of Hybridization?
Rule for Determination of total Number of Hybrid Orbitals
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The mixing or merging of dissimilar orbitals of similar energies to form new orbitals is known as hybridization and the new orbitals formed are known as hybrid orbitals.

Important Characteristics of Hybridization?

Orbitals belonging to the same atom or ion having similar energies get hybridized.
Number of hybrid orbitals is equal to the no. of orbitals taking part in hybridization.
The hybrid orbitals are always equivalent in energy and shape.
The hybrid orbitals form more stable bond than the pure atom orbital.
The reason hybridization takes place is to produce equivalent orbitals which give maximum symmetry.
It is not known whether actually hybridization takes place or not. It is a concept which explains the known behaviour of molecules.
The hybrid orbitals are directed in space in same preferred direction to have some stable arrangement and giving suitable geometry to the molecule.

Refer the following video for hybridization

Depending upon the different combination of s and p orbitals, these types of hybridization are known.

sp³ hybridization: In this case, one s and three p orbitals hybridise to form four sp³ hybrid orbitals. These four sp³ hybrid orbitals are oriented in a tetrahedral arrangement.

sp3 hybridization

sp² hybridization: In this case one s and two p orbitals mix together to form three sp² hybrid orbitals and are oriented in a trigonal planar geometry.

sp2 hybridization

The remaining p orbital if required form side ways overlapping with the other unhybridized p orbital of other C atom and leads to formation of p₂C = CH₂ bond as in H

sp hybridization: In this case, one s and one p orbital mix together to form two sp hybrid orbitals and are oriented in a linear shape.

sp hybridization

The remaining two unhybridised p orbitals overlap with another unhybridised p orbital leading to the formation of triple bond as in HC $\equiv$ CH.

Shape	Hybridisation
Linear	sp
Trigonal planar	sp²
Tetrahedral	sp³
Trigonal bipyramidal	sp³d
Octahedral	sp³d²
Pentagonal bipyramidal	sp³d³

Rule for Determination of total Number of Hybrid Orbitals

Detect the central atom along with the peripheral atoms.

Count the valence electrons of the central atom and the peripheral atoms.

Divide the above value by 8. Then the quotient gives the number of s bonds and the remainder gives the non-bonded electrons. So number of lone pair = non bonded electrons/2 .

The number of s bonds and the lone pair gives the total number of hybrid orbitals.

An Example Will Make This Method Clear:- SF₄ Central atom S, Peripheral atom F

∴ total number of valence electrons = 6 + (4 × 7) = 34

Now 34/8 = 4 2/8

∴ Number of hybrid orbitals = 4σ bonds + 1 lone pair)

So, five hybrid orbitals are necessary and hybridization mode is sp³d and it is trigonal bipyramidal (TBP).

Note:Whenever there are lone pairs in TBP geometry they should be placed in equatorial position so that repulsion is minimum.

1186_TBP geometry.JPG

Compound and its Hybridization

Molecular geometry

1. NCl₃ Total valence electrons = 26

Requirement = 3σ bonds + 1 lone pair

Hybridization = sp³

Shape = pyramidal

2. BBr₃Total valence electron = 24

Requirement = 3σ bonds

Hybridization = sp²

Shape = planar trigonal

3. SiCl₄ Total valence electrons = 32

Requirement = 4σ bonds

Hybridization = sp³

Shape = Tetrahedral

4. CI₄ Total valence electron = 32

Requirements = 4σ bonds

Hybridization = sp³

Shape = Tetrahedral

5. SF₆ Total valence electrons = 48

Requirement = 6σ bonds

Hybridization = sp³d²

Shape = octahedral/square bipyramidal

6. BeF₂ Total valence electrons : 16

Requirement : 2σ bonds

Hybridization : sp

Shape : Linear

F – Be – F

7. ClF₃ Total valence electrons : 28

Requirement: 3σ bonds + 2 lone pairs

Hybridization : sp³d

Shape : T – shaped

8. PF₅ Total valence electrons : 40

Requirement : 5σ bonds

Hybridization : sp³d

Shape : Trigonal bipyramidal (TBP)

9. XeF₄ Total valence electrons : 36

Requirement:4σ bonds+ 2 lone pairs

Hybridisation : sp³d

Shape : Square planar

Here three arrangements are possible out of which A and B are same. A and B can be inter converted by simple rotation of molecule. The basic difference of (B) and (C) is that in (B) the lone pair is present in the anti position which minimizes the repulsion which is not possible in structure (C) where the lone pairs are adjacent. So in a octahedral structure the lone pairs must be placed at the anti positions to minimize repulsion. So both structure (A) and (B) are correct.

10. XeF₂Total valence electrons : 22

Requirements : 2σ bonds + 3 lone pairs

Hybridisation: sp³d

Shape : Linear

11. Total valence electrons : 32

Requirement : 4σbonds

Hybridisation: sp³

Shape: tetrahedral

1061_PO3–4.JPG

Here all the structures drawn are resonating structures with O^– resonating with double bonded oxygen.

12. NO₂^– Total valence electron: 18

Requirement : 2σ bonds + 1 lone pair

Hybridisation: sp²

Shape: angular

13. CO₃^2– Total valence electrons: 24

Requirement = 3σ bonds

Hybrdisation = sp²

Shape: planar trigonal

But C has 4 valence electrons of these 3 form s bonds \ the rest will form a p bond.

287_4 valence electrons.JPG

In the structure one bond is a double bond and the other 2 are single. The position of the double bonds keeps changing in the figure. Since peripheral atoms are isovalent, so contribution of the resonanting structures are equal. Thus it is seen that none of the bonds are actually single or double. The actual state is

2127_Bond order.JPG

14. CO₂ Total valence electrons:16

Requirement: 2σ bonds

Hybridisation: sp

Shape: linear

O = C = O

15. Total valence electrons = 32

Requirement= 4σ bonds

Hybridisation: sp³

Shape: Tetrahedral

16. Total valence electron = 26

Requirement = 3σ bond + 1 lone pair

Hybridization: sp³

Shape: pyramidal

17. XeO₂F₂Total valence electrons : 34

Requirement: 4σ bonds +1 lone pairs

Hybridization : sp³d

Shape: Distorted TBP (sea-saw geometry)

18. XeO₃ Total valence electrons : 26

Requirement: 3σ bonds + 1 lone pair

Hybridization: sp³

Shape: Pyramidal

19. XeOF₄ Total valence electrons : 42

Requirement: 5σ bonds + 1 lone pair

Hybridization: sp³d²

Shape: square pyramidal.

20. PF2Br3 Total valence electrons : 40

Requirements : 5σ bonds

Hybridisation: sp3d

Shape : trigonal bipyramidal

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JEE Chemistry Syllabus

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