1. Molecular shape: VSEPR:
Valence Shel Electron Pair Repulsion model.
This model is based on an arrangement that minimizes the repulsion of shared & unshared pairs of e- around the central atom.
This repulsion forces results in atoms existing at fixed angles to each other.
Bond angle: Is the angle formed by any two terminal atoms
& the central atom.

2. The bonding atoms will seek the maximum separation
Facts:
- Shared e- pairs repel one another.
- Lone pairs occupy a slightly larger orbital than shared e-.
Examples:
I – Molecules than contain no lone pairs of e- in the central atom:
1) BeCl2
8-2 = 6 lone pairs
(3 pairs in each Cl)
Cl – Be –Cl
The bonding atoms will seek the maximum separation
Molecules with:
2 pairs of bonding e-
and “0” lone pairs
Bond angle = 180o
Geometry = Linear

3. Be --- 1s2 2s2 (Ground state configuration)
Geometry = Linear
It has 2 identical covalent bonds.
BeCl2
Bond angles = 180o
Be --- 1s2 2s2 (Ground state configuration)
Be --- 1s2 2s1 2p1 (Excited atom configuration)
Beryllium atom undergoes hybridization:
1 atomic orbital “s”
1 atomic orbitals “p”
2 hybrid orbitals (sp)
(Linear)
Be --- 1s2 2(sp)1 2(sp)1

4. Hybridization: Hybrid orbitals: Result from combining two or more
atomic orbitals.
# of hybrid orbitals = Total number of atomic orbitals combined.

5. The bonding atoms will seek the maximum separation
Examples:
I – Molecules than contain no lone pairs of e- in the central atom:
2) AlCl3
12-3 = 9 lone pairs
(3 pairs in each Cl)
Cl – Al –Cl Cl Cl Al
Cl Cl
Bond angle = 120o
The bonding atoms will seek the maximum separation
Geometry = Trigonal planar
Molecules with:
3 pairs of bonding e- and “0” lone pairs

6. Al --- 1s2 2s2 2p6 3s2 3p1 (Ground state configuration)
Geometry = Trigonal planar
It has 3 identical covalent bonds.
AlCl3
Bond angles = 120o
Al --- 1s2 2s2 2p6 3s2 3p1 (Ground state configuration)
Al --- 1s2 2s2 2p6 3s1 3p1 3p1 (Excited atom configuration)
Aluminum atom undergoes hybridization:
1 atomic orbital “s”
2 atomic orbitals “p”
3 hybrid orbitals (sp2)
(Trigonal planar)
Al --- 1s2 2s2 2p6 3(sp2)1 3(sp2)1 3(sp2)1

7. The bonding atoms will seek the maximum separation
Examples:
I – Molecules than contain no lone pairs of e- in the central atom:
3) CH4
4-4 = 0 lone pairs H
H – C –H H C
H H H
109.5o
Bond angle = 109.5o
The bonding atoms will seek the maximum separation
Geometry = Tetrahedral
Molecules with:
4 pairs of bonding e- and “0” lone pairs

8. C --- 1s2 2s2 2p2 (Ground state configuration)
C --- 1s2 2s1 2p1 2p1 2p1 (Excited atom configuration)
It can form 4 covalent bonds; but not identical.
Carbon atom undergoes hybridization:
1 atomic orbital “s”
3 atomic orbitals “p”
4 hybrid orbitals (sp3)
(Tetrahedral)
C --- 1s2 2(sp3)1 2(sp3)1 2(sp3)1 2(sp3)1
Bond angle = 109.5o

9. The bonding atoms will seek the maximum separation
Examples:
II – Molecules than contain one lone pairs of e- in the central atom:
1) PH3
4-3 = 1 lone pairs
(in P atom)
H – P –H H P
H H H
The lone pair occupies larger orbital, so it pushes the bonds down.
The bonding atoms will seek the maximum separation
Bond angle = 107.3o
Geometry = Trigonal pyramidal
Molecules with:
3 pairs of bonding e- and one lone pairs

10. P --- 1s2 2s2 2p6 3s2 3p13p1 3p1 (Ground state configuration)
Geometry = Trigonal pyramidal
PH3
Bond angles = 107.3o
P --- 1s2 2s2 2p6 3s2 3p13p1 3p1 (Ground state configuration)
Phosphorus atom undergoes hybridization:
1 atomic orbital “s”
3 atomic orbitals “p”
4 hybrid orbitals (sp3)
(Trigonal pyramidal)
P --- 1s2 2s2 2p6 3(sp3)2 3(sp3)1 3(sp3)1 3(sp3)1

11. The bonding atoms will seek the maximum separation
Examples:
III – Molecules than contain two lone pairs of e- in the central atom:
1) H2O
4-2 = 2 lone pairs
(in O atom)
H – O –H O
H H
The two lone pair occupy larger orbitals, so it pushes the bonds down.
The bonding atoms will seek the maximum separation
Bond angle = 104.5o
Geometry = Bent
Molecules with:
2 pairs of bonding e- and two lone pairs

12. O --- 1s2 2s2 2p2 2p1 2p1(Ground state configuration)
Geometry = Bent
H2O
Bond angles = 104.5o
O --- 1s2 2s2 2p2 2p1 2p1(Ground state configuration)
Oxygen atom undergoes hybridization:
1 atomic orbital “s”
3 atomic orbitals “p”
4 hybrid orbitals (sp3)
(Bent)
O --- 1s2 2(sp3)2 2(sp3)2 2(sp3)1 2(sp3)1

13. The bonding atoms will seek the maximum separation
Examples:
IV – Molecules than contain no lone pairs of e- in the central atom,
which are exceptions of the Octet Rule:
1) NbBr5
20-5 = 15 lone pairs
(three in each Br atom) Br Br Br Nb Br Br
Bond angle = 90o /120o
The bonding atoms will seek the maximum separation
Geometry = Trigonal bipyramidal
Molecules with:
5 pairs of bonding e- and “0” lone pairs

14. Nb --- [Kr] 5d3 4s2 (Ground state configuration)
Geometry = Trigonal bipyramidal
NbBr5
Bond angles = 90o/120o
Nb --- [Kr] 5d3 4s2 (Ground state configuration)
Nb --- [Kr] 5d1 4s1 4p1 4p1 4p (Excited atom configuration)
Phosphorus atom undergoes hybridization:
1 atomic orbital “s”
3 atomic orbitals “p”
1 atomic orbital “d”
5 hybrid orbitals (sp3d)
(Trigonal bipyramidal)
Nb --- [Kr] 5(sp3d)1 5(sp3d)1 5(sp3d)1 5(sp3d)1 5(sp3d)1

15. The bonding atoms will seek the maximum separation
Examples:
IV – Molecules than contain no lone pairs of e- in the central atom,
which are exceptions of the Octet Rule:
2) SF6 F
24-6 = 18 lone pairs
(three in each F atom) F F S F F F
The bonding atoms will seek the maximum separation
Bond angle = 90o
Geometry = Octahedral
Molecules with:
6 pairs of bonding e- and “0” lone pairs

16. S --- 1s2 2s2 2p6 3s2 3p4 (Ground state configuration)
Geometry = Octahedral
SF6
Bond angles = 90o
S --- 1s2 2s2 2p6 3s2 3p4 (Ground state configuration)
S --- 1s2 2s2 2p6 3s1 3p1 3p1 3p1 3d1 3d1 (Excited atom configuration)
Sulfur atom undergoes hybridization:
1 atomic orbital “s”
3 atomic orbitals “p”
2 atomic orbitals “d”
6 hybrid orbitals (sp3d2)
(Octahedral)
S – 1s2 2s2 2p6 3(sp3d2)1 3(sp3d2)1 3(sp3d2)1 3(sp3d2)1 3(sp3d2)1 3(sp3d2)1

17. SUMMARY: Example: BeCl2 2 Linear 180o sp 3 AlCl3 Trigonal planar 120o
# of shared pairs (single bonds)
Lone pairs
Geometry
Bond Angle
Hybridization
Example:
BeCl2 2
Linear
180o sp 3
AlCl3
Trigonal planar
120o
sp2
109.5o
sp3
CH4 4
Tetrahedral
Trigonal pyramidal
PH3 3 1
107.3o
sp3
Bent
104.5o
sp3
H2O 2 2
Trigonal bipyramidal
90o/120o
sp3d
NbBr5 5 6
SF6
Octahedral
90o
sp3d2

18. valence e- = 8 e- bonding pairs = = 4 pairs 4-3 = 1 lone pairs
Example:
Determine the molecule geometry, bond angle & type of hybridization of PH3
valence e- =
5 + 3(1) =
8 e-
1) PH3
bonding pairs =
= 4 pairs
𝟖 𝟐
H – P –H H
4-3 = 1 lone pairs
(in P atom) P
H H H
3 single covalent bonds.
1 lone pairs of e- in the
central atom.
Geometry = Trigonal pyramidal
Bond angle = 107.3o
Hybridization = sp3

19. Classification of covalent bonds:
All single covalent bonds are very strong bonds, named sigma (σ).
Any double covalent bond contains one sigma (σ) & one pi (π) bonds.
Any triple covalent bond contains one sigma (σ) & two pi (π) bonds.
Sigma (σ) covalent bonds are stronger than pi (π) covalent bonds.

20. Exercise: Determine the molecular geometry, bond angle, & type of
hybridization for the following:
1) BF ) SCl ) CF ) NH4) ) BeF2