how  to  find  hybridisation  state  of  any  molecule

Hybridisation of a molecule or ion can be predicted by making use of the following formula:

Number of Hybrid orbitals (H)=

   1/2( No. of valence electrons in valence shell of central atom  the atom+ No. of   monovalent  atoms    surrounding - Charge   on cation + Charge on anion)

 

 For example,

  BeF2:  H=1/2[2 + 2 -0 + 0] = 2 ; Hybridisation : sp

   CH4:  H = 1/2[4 + 4 – 0 +0] =4  ; Hybridisation : sp3

  BF3:  H= 1/2[3 + 3 -0 + 0] = 3  ; Hybridisation : sp2

  SF6:  H= 1/2[6+ 6 - 0 + 0] = 6  ;  Hybridisation: sp3d2

  PCl5H=1/2[5+5 – 0 + 0] = 5 ; Hybridisation: sp3d

   NH4+H= 1/2[ 5+ 4 – 1 +0] = 4 ; Hybridisation =sp3

  CO32-: H=1/2[4+ 0 -0 +2] = 3 ; Hybridisation = sp2

  NO3-:  H=1/2[5 + 0 -0+1] =3; Hybridisation = sp2

  The above formula cannot be used for predicting hybridization state of any complex ion.

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he general formula is

 

(m + v – a + c)/2

 

Where

 

m = No. of monovalent  atoms surrounding the central atom

 

v = No. of valence electrons in central atom

 

c and a = cationic or anionic charge (if any)

 

The number that you get will give total no. of hybrid  orbitals. You can figure out the hybridization thereafter.

 

If  the answer is

2 = sp

 

3 = sp2

 

4 = sp3

 

5 = sp3d

 

6 = sp3d2

 

7 = sp3d3

 

Example,

 

XeF4

 

m = 4

v = 8

 

No. of hybrid orbitals  = (4+8)/2 = 6

 

So six orbitals. Hence hybridization = sp3d2

 

  • 6

Gud answer.

But in XeF4 how does the valancey bcms 8?

Do u mean valancy of Xe  = 8?

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Also please giv an answer for:   [Co(NH3)6]3+

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