HYBRIDIZATION AND GEOMETRY SIMPLIFIED

PS:THIS FORMULA WILL ALWAYS GIVE YOU THE CORRECT ANSWER
Just get through wid all this u'll find inorganic petty easy....

SHORTEST FORMULAE FOR HYBRIDIZATION:

No of valence electrons + No. of atoms attached to central atom
2

CONDITIONS:
1. Don't count the multiple bonded attached atoms.
2. Don't coun't odd or unpaired electrons.

after putting the values in the formula if ya get:

2 then compound is sp hybridized
3 then compound is sp2 hybridized
4 then compound is sp3/dsp2 hybridized
5 then compound is sp3d/dsp3 hybridized
6 then compound is sp3d2/d2sp3 hybridized

TO DIFFERENTIATE b/w sp3 and dsp2 hybridization and so on.

1. Easiest way is to check whether d-orbital present or not.

2. If the compound undergoes reaction with strong ligand than dsp2, dsp3,d2sp3 hybridization occurs.

3.If the compound undergoes reaction with weak ligand than sp3, sp3d, sp3d2 hybridization occurs.

TO IDENTIFY STRONG LIGAND AND WEAK LIGAND:

ALL NITROGEN CONTAINING except nitrate ion are strong ligands.
CO is exceptionally strong ligand...

ALL STRONG LIGANDS CAUSE PAIRING OF UNPAIRED ELECTRON IN D-ORBITAL.

GEOMETRIES:
REQUIRED (R): the total no. of atoms which are necessary for perfect geometry of that hybridization or number of hybrid orbitals.

AVAILABLE(A): Atoms attached.

R-A=no. lone pair
1.sp hybridization:
R=2, A=2
R-A=0
linear geometry, 0 dipole moment.

R=2, A=1
R-A=1 lone pair

so dipole moment is not 0.

2.sp2 hybridization:

a. R=3
-A=3

0 = no. of lone pairs.
Geometry: trigonal planar and dipole moment =0

b. R=3
A=2

lone pair = 1
Geometry: V shape/angular

c. R=3
A=1

lone pair = 2
Geometry: Linear

3. sp3

a. R=4
A=4

lone pair=0
Geometry: Tetrahedral

b. R=4
A=3

lone pair = 1

Geometry: pyramidal

c. R=4
A=2

lone pair = 2

Geometry: V-shape

d. R=4
A=1

lone pair = 3

Geometry: Linear

4.sp3d

a. R=5
A=5

lone pair = 0

Geometry: Trigonal Bipyramidal

b. R=5
A=4

lone pair = 1

Geometry: see-saw

c. R=5
A=3

lone pair = 2

Geometry: T-shape

d. R=5
A=2

lone pair = 3

Geometry: Linear

LONE PAIR CAUSE MAXIMUM repulsion so they will be placed at equitorial postion (larger bond angle)

5. sp3d2:

a. R=6
A=6

lone pair = 0

Geometry: square bipyramidal

b. R=6
A=5

lone pair = 1

Geometry:square pyramidal

c. R=6
A=4

lone pair = 2

Geometry : square planar.

PRACTICE PROBLEMS:

1. BeCl2 = 2+2 = 2 =sp Linear
2

2. CO2 = 4/2 =2 =sp Linear

3.XeOF2 = 8/2 + 2/2 = sp3d T-shape

4. XeF6 = 8/2 + 6/2 = Sp3d3 Distorted octahedral or caped octahedrl