1. CH 10: Molecular Geometry & chemical bonding theory
Vanessa N. Prasad-Permaul
Valencia Community College
CHM 1045

2. Molecular Shapes: VSEPR
Molecular Geometry: the general shape of a molecule, as determined by the relative positions of the atomic nuclei.
Valence-Shell Electron-Pair Repulsion Theory: predicts the shapes of molecules and ions by assuming that the valence-shell electron pairs are arranged about each atom so that the electron pairs are kept as far away from one another as possible, thus minimizing electron-pair repulsions.

3. Molecular Shapes: VSEPR
Two Electron Groups: Electron groups point in opposite directions.
LINEAR ARRANGEMENT

4. Molecular Shapes: VSEPR
Three Electron Groups: Electron groups lie in the same plane and point to the corners of an equilateral triangle.
TRIGONAL PLANAR ARRANGEMENT
Trigonal planar geometry or bent (angular)

5. Molecular Shapes: VSEPR
Four Electron Groups: Electron groups point to the corners of a regular tetrahedron.
TETRAHEDRAL ARRANGEMENT

6. Molecular Shapes: VSEPR
The approximate shape of molecules is given by Valence-Shell Electron-Pair Repulsion (VSEPR).
Step #1: Count the total electron groups.
Step #2: Arrange electron groups to maximize separation.
Groups are collections of bond pairs between two atoms or a lone pair.
Groups do not compete equally for space:
Lone Pair > Triple Bond > Double Bond > Single Bond

7. Molecular Shapes: VSEPR
Table of the Molecular Geometry
and Examples of Certain Compounds

8. ¨ ¨ ¨ ¨ EXAMPLE 10.1 : Predict the geometry of the following
Molecular Shapes: VSEPR
EXAMPLE 10.1 : Predict the geometry of the following
Molecules or ions, using the VSEPR method:
BeCl B. NO C. SiCl4
Be = 2 electrons Cl = 7 electrons each = 14
Total of 16 electrons
:Cl:Be:Cl:
2 sets of electron pairs, 0 lone pairs  linear

9. ¨ ¨ ¨ ¨ B. NO2- N = 5 electrons, O = 6 electrons each = 12
Molecular Shapes: VSEPR
B. NO N = 5 electrons, O = 6 electrons each = 12
Total of 17 electrons plus 1 electrons = 18 e-
:O:N::O:
3 sets of electron pairs, 1 lone pair  trigonal planar arrangement and a bent geometry

10. Molecular Shapes: VSEPR
C. SiCl4
Si = 4 electrons, Cl = 7 electrons each = 28 electrons
Total of 32 electrons
4 electron pairs, 0 lone pairs tetrahedral

11. EXERCISE 10.1 : Use the VSEPR method to predict the
Molecular Shapes: VSEPR
EXERCISE 10.1 : Use the VSEPR method to predict the
geometry of the following ion and molecules:
ClO3-
OF2
SiF4

12. Molecular Shapes: VSEPR
Five Electron Groups: Electron groups point to the corners of a trigonal bipyramid.

13. Molecular Shapes: VSEPR
Six Electron Groups: Electron groups point to the corners of a regular octahedron.

14. Molecular Shapes: VSEPR
EXAMPLE 10.2 : What do you expect for the geometry of tellurium tetrachloride?
Tellurium has 6 electron pairs in its valence shell
Cl = 7 electrons each = 28 electrons
Total of 34 electrons
There are 4 bonding pairs and 1 lone pair of electrons 
Trigonal bipyramidal arrangement and a seesaw geometry

15. Molecular Shapes: VSEPR
EXERCISE 10.2 : According to the VSEPR model, what molecular geometry would you predict for iodine trichloride?

16. Dipole Moment & Molecular Geometry
Dipole Moment: a quantitative measure of the degree of charge separation in a molecule.
 -
H Cl
If a molecule has a dipole moment it is polar
A compound could contain polar bonds but the molecule could be non-polar because there is no dipole moment
Bond dipoles can cancel each other out
O C O

17. Based on electronegativity differences between atoms in a molecule
Dipole moment
Based on electronegativity differences between atoms in a molecule
The most electronegative atom is partially negative
The less electronegative atom is partially positive
The dipole moment is the average of all dipoles in the molecule
Exception (C-H bonds do not have a dipole)

18. Molecular Shapes: VSEPR
EXAMPLE 10.3 : Each of the following molecules has a non-zero dipole moment. Select the molecular geometry that is consistent with this information. Explain.
SO2: linear or bent
BENT because in linear geometry, the dipole moment would be zero. There are 2 electron pairs and 1 lone pair which is a trigonal planar arrangement giving a bent geometry.

19. PH3: trigonal planar or trigonal pyramidal
Molecular Shapes: VSEPR
PH3: trigonal planar or trigonal pyramidal
In a trigonal planar geometry, the dipole moment is zero, therefore PH3 has a trigonal pyramidal geometry. Tetrahedral arrangement; 3 electron pairs and one lone pair.

20. Molecular Shapes: VSEPR
EXERCISE 10.3 : Bromine trifluoride BrF3, has a nonzero dipole moment. Indicate the correct arrangement and molecular geometry is consistent with this information.

21. Molecular Shapes: VSEPR
EXERCISE 10.4 : Which of the following would be expected to have a dipole moment of zero on the basis of symmetry. Explain.
SOCl2
SiF4
OF2

22. 3. The greater the amount of orbital overlap, the stronger the bond.
Valence Bond Theory
1. Covalent bonds are formed by overlapping of atomic orbitals, each of which contains one electron of opposite spin.
2. Each of the bonded atoms maintains its own atomic orbitals, but the electron pair in the overlapping orbitals is shared by both atoms.
3. The greater the amount of orbital overlap, the stronger the bond.

23. According to Valence Bond Theory:
An orbital on one atom comes to occupy a portion of the same region of space as an orbital on the other atom. The two orbitals are said to overlap.
The total number of electrons in both orbitals is no more than two.

24. Combined hybrid orbitals
Valence Bond Theory
Combined hybrid orbitals
Hybrid orbitals
# of orbitals
VSEPR geometry

25. Valence Bond Theory
sp hybrid:

26. Valence Bond Theory
sp2 hybrid:

27. Valence Bond Theory
sp2 hybrid (π bond):

28. Valence Bond Theory
sp3 hybrid:

29. Valence Bond Theory
sp3d hybrid:

30. Valence Bond Theory
sp3d2 hybrid:

31. Steps to obtain the bonding description about an atom in a molecule:
Valence Bond Theory
Steps to obtain the bonding description about an atom in a molecule:
Write the Lewis electron-dot formula
From the Lewis formula, use the VSEPR model to obtain arrangement
From the geometry, s what type of hybrid orbitals are required
Assign valence electrons to the hybrid oritals of this atom one at a time, pairing them only when necessary
Form the bonds by overlapping singly occupied orbitals of other atoms

32. EXAMPLE 10.4: Describe the bonding in H2O according to
Valence Bond Theory
EXAMPLE 10.4: Describe the bonding in H2O according to
valence bond theory.
Using the VESPR theory, notice there are 3 sets of electron pairs, and two lone pairs giving a tetrahedral arrangement and a bent geometry

33. Valence Bond Theory
EXERCISE 10.5: Using hybrid orbitals, describe the bonding in NH3 according to valence bond theory.

34. EXAMPLE 10.5: Describe the bonding in XeF4 using hybrid orbitals.
Valence Bond Theory
EXAMPLE 10.5: Describe the bonding in XeF4 using hybrid orbitals.
4 single bonds, 2 lone pairs octahedral arrangement and square planar geometry.

35. EXERCISE 10.6: Describe the bonding in PCl5 using hybrid orbitals.
Valence Bond Theory
EXERCISE 10.6: Describe the bonding in PCl5 using hybrid orbitals.

36. Molecular Orbital Theory
Additive and subtractive combination of p orbitals leads to the formation of both sigma and pi orbitals.

37. Valence Bond Theory
EXAMPLE 10.6: Describe the bonding on a given N atom in dinitrogen difluoride, N2F2, using valence bond theory. 
:F N N F: 

38. Valence Bond Theory
EXERCISE 10.7: Describe the bonding on the carbon atom in carbon dioxide using valence bond theory.

39. Valence Bond Theory
EXERCISE 10.8: Dinitrogen difluoride exists as cis and trans isomers. Write structural formulas for these isomers and explain using valence bond theory why they exist.

40. Molecular Orbital Theory
The molecular orbital (MO) model provides a better explanation of chemical and physical properties than the valence bond (VB) model.
Atomic Orbital: Probability of finding the electron within a given region of space in an atom.
Molecular Orbital: Probability of finding the electron within a given region of space in a molecule.

41. Molecular Orbital Theory
Additive combination of orbitals (s) is lower in energy than two isolated 1s orbitals and is called a bonding molecular orbital.

42. Molecular Orbital Theory
Subtractive combination of orbitals (s*) is higher in energy than two isolated 1s orbitals and is called an antibonding molecular orbital.

43. Molecular Orbital Theory
Bond Order is the number of electron pairs shared between atoms.
Bond Order is obtained by subtracting the number of antibonding electrons from the number of bonding electrons and dividing by 2.
BO = Bonding electrons – antibonding electrons 2

44. *2p *2p 2p 2p *2s 2s *1s 1s B2
*2p *2p 2p 2p *2s 2s *1s 1s
B has 5 electrons
So B2 has 10 elec
Core electrons don’t count toward BO

45. *2p *2p 2p 2p *2s 2s *1s 1s C2
*2p *2p 2p 2p *2s 2s *1s 1s
Carbon has 6 electrons so C2 has 12 electrons

46. Molecular Orbital Theory
Molecular Orbital Diagram for H2:

47. Molecular Orbital Theory
Molecular Orbital Diagrams for H2– and He2:

49. Molecular Orbital Theory
Second-Row MO Energy Level Diagrams:

50. Valence Bond Theory
EXAMPLE 10.7: Give the orbital diagram of the molecule. Is the molecular substance diamagnetic or paramagnetic? What is the electron configuration? What is the bond order?

51. Valence Bond Theory
EXERCISE 10.9: The C2 molecule exists in the vapor phase over carbon at high temperature. Describe the molecular orbital structure of this molecule (orbital diagram and electron configuration). Is the substance paramagnetic or diamagnetic? What is the bond order?

52. Valence Bond Theory
EXAMPLE 10.8: Write the orbital diagram for dinitrogen monoxide (nitric oxide). What is the bond order for NO?

53. Valence Bond Theory
EXERCISE 10.10: Give the orbital diagram and electron configuration for the carbon monoxide molecule. What is the bond order and is this molecule paramagnetic or diamagnetic?

54. Molecular Orbital Theory
MO Diagrams Can Predict Magnetic Properties:

55. Example 1: VSEPR
Draw the Lewis electron-dot structure and predict the shapes of the following molecules or ions:
O3 H3O+ XeF2
PF6– XeOF4 AlH4–
BF4– SiCl4 ICl4– AlCl3

56. Example 2: Molecular Orbital Theory
The B2 and C2 molecules have MO diagrams similar to N2. What MOs are occupied in B2 and C2, and what is the bond order in each? Would any of these be paramagnetic?

57. Example 3: Dipole moment
Draw the dipole moment for the following molecules, are they polar?
HCl NH3 CHCl3
H2O SF6 CCl4
CO2 CH2Cl2

58. Hybridization Easy Way