Is the formula given for hybridisation i.e. (total valenceelectrons)/8 always applicable? If not what are its limitations and how dowe solve it then

Solution:

Hybridisation of a molecule or an ion can be predicted by making use of the following formula.
Number of Hybrid orbitals (H)=n
=1/2[group no central atom +no of atoms attached to central metal atom (like H , Cl, F.except O) + magnitude of negative charge - magnitude of positive charge]
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
PCl5: H=1/2[5+5 – 0 + 0] = 5 ; Hybridisation: sp3d
NH4 +: H= 1/2[ 5+ 4 – 1 +0] = 4 ; Hybridisation =sp3
CO3 2-: 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 hybridisation state of complex ions.

Dear student,

Hybridisation of a molecule or an ion can be predicted by making use of the following formula.
Number of Hybrid orbitals (H)=n
=1/2[group no central atom +no of atoms attached to central metal atom (like H , Cl, F.except O) + magnitude of negative charge - magnitude of positive charge]

Regards

Sumit