Examples of Hybridization

We have already discussed that whenever there are lone pairs they should be placed in equatorial positions. Now a question that may come to your mind that though the hybridization is sp3d, so the shape should be T.B.P. But when all the bonds are present the actual shape is TBP. But when instead of bond there are lone pairs in TBP the actual geometry is determined by the bonds not by the lone pairs. Here in ClF3 the bond present (2 in axial and 1 in equatorial) gives the impression of T shape.

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




 Now 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




 [Q l.p. are present in equatorial position and ultimate shape is due to the bonds that are formed]

11. PF2Br3 Total valence electrons : 40

Requirements : 5σ bonds

Hybridisation: sp3d

Shape : trigonal bipyramidal





 Here we see that fluorine is placed in axial position whereas bromine is placed in equatorial position. It is the more electronegative element that is placed in axial position and less electronegative element is placed in equatorial position. Fluorine, being more electronegative pulls away bonded electron towards itself more than that is done by bromine atom which results in decrease in bp – bp repulsion and hence it is placed in axial position.

In this context it can also be noted that in T.B.P. shape the bond lengths are not same. The equatorial bonds are smaller than axial bonds. But in square bipyramidal shape, all bond lengths are same.

12.  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.

13. NO2 Total valence electron: 18

Requirement : 2σ bonds + 1 lone pair

Hybridisation: sp2

Shape: angular



 14. 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.

                  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

15. CO2 Total valence electrons:16

Requirement: 2σ bonds

Hybridisation: sp

Shape: linear


O = C = O

16. Total valence electrons = 32

Requirement= 4σ bonds

Hybridisation: sp3

Shape: Tetrahedral



17. Total valence electron = 26

Requirement = 3σ bond + 1 lone pair

Hybridization: sp3

Shape: pyramidal



18. XeO2F2 Total valence electrons : 34

Requirement: 4σ bonds +1 lone pairs

Hybridization : sp3d

Shape: Distorted TBP (sea-saw geometry)



19. XeO3 Total valence electrons : 26

Requirement: 3σ bonds + 1 lone pair

Hybridization: sp3

Shape: Pyramidal



20. XeOF4 Total valence electrons : 42

Requirement: 5σ bonds + 1 lone pair

Hybridization: sp3d2

Shape: square pyramidal.





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