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Hybridization

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.

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.

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.

The remaining two unhybridised p orbitals overlap with another unhybridised p orbital leading to the formation of triple bond as in HCCH.

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.

Compound and its Hybridization

Molecular geometry

1. NCl3 Total valence electrons = 26

Requirement = 3σ bonds + 1 lone pair

Hybridization = sp3

Shape = pyramidal

2. BBr3 Total 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. XeF2   Total valence electrons : 22

Requirements : 2σ bonds + 3 lone pairs

Hybridisation: sp3d

Shape : Linear

11.  Total valence electrons : 32

Requirement : 4σbonds

Hybridisation: sp3

Shape: tetrahedral

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. CO32– 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.

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

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. XeO2F2 Total 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

You can also refer to the following related links

JEE  Chemistry Syllabus

Reference books of Physical Chemistry

To read more, Buy study materials of Chemical Bonding comprising study notes, revision notes, video lectures, previous year solved questions etc. Also browse for more study materials on Chemistry here

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