1. Ch. 8 Bonding and
Molecular Structure
Ch. 9 Orbital Hybridization
The content of Ch. 8 – 9 will be
taught through both this PPT and
LM 8-9.1
Text references:
8.1, 8.2, 8.4-8.9, 9.1-9.2

2. The rest of this PPT…
• Focuses on substances with covalent bonds
– Molecules
– Polyatomic ions
• For any molecule or PAion, you will learn to…
–
–
–
–
–
draw the Lewis Dot Structure (LDS)
identify the 3D molecular geometry
determine the polarity
Identify hybridization and sigma/pi bonding
(These are all components of LM 8-9.1)

3. Chapter 8
LM 8-9.1 Part I
Lewis Dot Structures

4. How to draw LDS for molecules and ions:
1.
2.
3.
4.
5.
6.
7.
Write formula for compound.
Write symbol for least electronegative atom in center
and write symbols for all other atoms around it.
Add valence e- from all atoms to find out total # of e- you
must assign to the compound.
Draw one line (single bond) from central atom to all
outer atoms. (1 line = 2 e-) Subtract 2e- for each line
you’ve drawn from the total # valence e-.
Give all OUTER atoms their complete “octet.” Subtract
this value from total.
If there are ANY remaining e-, assign them in pairs
around the central atom.
If there are no more e- but the central atoms still needs
more to complete the octet, remove a lone pair from an
outer atom and turn this into a double bond. Repeat if
you need to form a triple bond.
Ex. #1
Ex. #2

5. Example #1
• Draw LDS for nitrogen trifluoride:
Formula = NF3
• ••
• •F
•
••
N
• ••
F•
••
•F•
•• ••
Total valence e- = 5 + 3(7) = 26eRemaining e- = 26 – 6 = 20eRemaining e- = 20 – 18 = 2eClick this link and go to Lewis Dot Structures for more info.

6. Example #2
• Draw LDS for carbon dioxide:
Formula = CO2
O=C=O
Since C did not have enough e- to complete the octet,
each oxygen must donate a lone pair of e- to make a
double bond.

7. Electron Deficient Atoms
• Elements from Groups I, II, & III will not
hold 8 e- when bonded
valence e• Group I  has 21e- in bonding
• Group II  has 4 2 valence ee- in bonding
• Group III  has 6 3 valence ee- in bonding

8. Expanded Octet
• Some atoms (usually group V and higher)
can POTENTIALLY take on more than 8 e- in
bonding
– Group IV – VIII must always have at least
8 e-, however some will take more
• Ex. AsF5
• As has 5 fluorine atoms
bonded to it and thus is
sharing 10 total e- with
the outer atoms
F
F
As F
F
F
Click on this link and go to Expanded Octet for more info.

9. Resonance Structures
• When there are equivalent LDS for a molecule
or ion we say that the structure experiences
resonance.
• Applies to a symmetric molecule that contains
at least one double bond
• The double bond could be in multiple places
and still have the same structure.
Click this link and go to Resonance for more info.

10. Why Resonance Structures?
• Electrons in molecules w/double bonds are often delocalized
between two or more atoms.
• Electrons in a single Lewis structure are assigned to specific
atoms-a single Lewis structure is insufficient to show electron
delocalization.
• Composite of resonance forms more accurately depicts
electron distribution. Molecule or ion doesn’t actually switch
between resonance structures. The structure behaves as a
blend or composite of all possible structures
• Proof of resonance?
– Bond lengths in resonance structures shorter than single
bonds/longer than double bonds
– Bonds are stronger than single bonds/weaker than double
bonds

11. Resonance Structures
• Draw all resonance structures for
ozone (O3)

12. Chapter 8
LM 8-9.1 Part II
Molecular Geometry
Getting a 3D perspective of molecules…
For a preview of what’s to come, click on
this link and click on VSEPR Model.

13. Applying VSEPR to Balloons
• Valence Shell Electron Pair Repulsion =
model for predicting molecular
geometry (3D arrangement of BP and
LP)
• Watch what your instructor can do
w/balloons…

14. VSEPR
• Valence Shell Electron Pair Repulsion model for
predicting molecular geometry
• Model describes the 3D arrangement of atoms in
molecule based on # of lone pairs (LP) and bond
pairs (BP) of e- on a central atom.
• A molecule adopts the geometry that minimizes
the repulsive force among the e- pairs on central
atom
• By minimizing repulsion around central atom you
get the most stable geometry of a molecule
(minimize repulsion to maximize stability)

15. For the next slide, draw LDS for:
• BF3, O3
• CH4, NH3, H2O
• PCl5, SF4, ClF3, XeF2
• What, does 5each grouping have in common? What
SF6 BrF , XeF4
changes as you go from one grouping to the next? How
might this effect the placement of e- or “geometry”
according to VSEPR theory?

16. Applying VSEPR Theory
• Lowest energy arrangement when there are:
Linear
180o
– 2BP, 0LP =
120o
– 3BP, 0LP = Trigonal Planar
– 4BP, 0LP = Tetrahedral
109.5o
– 5BP, 0LP = Trigonal bipyramidal
120o, 90o
– 6BP, 0LP =
Octahedral
90o
• Bond angle = angle formed
when 3 atoms bond together

17. Linear Shapes
Examples;
Examples;
BeH2
HF
CO2
HCl
HCN
H2
Images from http://intro.chem.okstate.edu/1314F97/Chapter9/VSEPR.html

18. Trigonal Planer Geometry
Examples;
BF3
CO32COCl2
Images from http://intro.chem.okstate.edu/1314F97/Chapter9/VSEPR.html

19. Tetrahedral Geometry
Examples;
CH4
CCl4
SO42-
Images from http://intro.chem.okstate.edu/1314F97/Chapter9/VSEPR.html

20. Trigonal Bipyramidal Geometry
Examples;
PF5
PCl5
Images from http://intro.chem.okstate.edu/1314F97/Chapter9/VSEPR.html

21. Octahedral Geometry
Examples;
SF6
SiF63-
Images from http://intro.chem.okstate.edu/1314F97/Chapter9/VSEPR.html

22. Applying VSEPR Theory (cont’d)
• Lowest energy arrangement when there are lone
pairs??? Leads to molecular geometry
• LP’s push harder on surrounding BP’s
– LP can spread out more in nonbonding orbital than BP
can (BP more localized/constrained in space)
– LP’s make bond angles slightly smaller than expected
– The bond angle tends to decrease as # of LP
increases
• Steric # (total # of e- pairs) determines e- pair
arrangement to minimize repulsion; #BP vs. LP
determines name of molecular geometry (as
opposed to electronic geometry); LP influence
bond angles
• See how this works…try link above

23. Steric # = 3 (3 Total Pair of E-)
• 3BP, 0LP
• 2BP, 1LP
Trigonal Planar
Bent
<120o
Examples;
NO2SO2
O3

24. Steric # = 4 (4 Total Pair of E-)
• 4BP, 0LP
Tetrahedral
• 3BP, 1LP
Trigonal
pyramidal <109.5o
• 2BP, 2LP
Bent
<<109.5o
Examples;
NH3
NF3
PCl3
Examples;
H2O
ClO2
OF2

25. Steric # = 5 (5 Total Pair of E-)
• 5BP, 0LP
Trigonal bipyramidal
• 4BP, 1LP
Seesaw
<90o
Examples;
SF4
<120o
BrF4+

26. Steric # = 5 (contd)
• 3BP, 2LP
T-Shaped
Examples;
ClF3
BrF3
SeO32-
<90o
• 2BP, 3LP
180o
Linear
Examples;
XeF2
ICl2I3 -

27. Steric # = 6 (6 Total Pair of E-)
• 6BP, 0LP
Octahedral
• 5BP, 1LP
<90o
4BP, 2LP
Examples;
BrF5
IF5
Square
Pyramidal
90o
Examples;
XeF4
ICl4-
Square
Planar

28. What if there’s >1 central atom?
1
2
3
4
• Number each atom in the chain (this one has 4
central atoms)
• Count the number of BP and LP on each central
atom
• Identify the geometry of each central atom (there’s
no geometry for the whole structure)
• C1: 4BP, 0LP  tetrahedral
• C2, 3, 4: 3BP, 0LP  trigonal planar (each one)

29. Chapter 8
Lab 8-9.1 Part III
Molecular Polarity
To be polar or not to be polar, that is the
question!

30. Bond Polarity vs. Molecular Polarity
• Bond polarity – refers to
distribution of e- cloud
around 2 atoms in a bond
– Polar if unequal distribution
(dipole-one end of bond
slightly pos and the other
slightly neg) (has a dipole
moment)
– Nonpolar if equal
distribution (has no dipole
moment)
• Determined by difference
in EN values of atoms
More info on polarity at this link and select Partial Charges and Bond Dipoles

31. Chemical Bond Formation
• Chemical bond – results when a chem
rxn occurs between 2 or more atoms and
the valence e- reorganize so that a net
attractive force occurs between them
– Ionic Bonds
• Transfer of valence e• Electrostatic attractn of opp. charged ions
• Usu. btwn M + NM, M + PAion or PAion + NM
– Covalent Bonds
• Sharing of valence e• Electrostatic attractn of e- of one atom to nucleus
of another
• Usu. btwn NM + NM

32. Covalent Bond Formation

33. Electronegativity (EN)
• Definition: a value that represents the
relative ability of an atom to attract etowards itself in a bond
– EN is NOT an energy
– Scale ranges from 0 - 4
• What this means:
larger EN = atom has greater attraction for
e- in a bond

34. Back

35. Electronegativity (EN)
• Trend:
– Down Group = decrease
– L to R in Period = increase
• Explain Trend:
– Down Group = inc shielding e- = decrease pull
on own valence e- = decrease pull on any ein bond
– L to R in Period = inc Z = increase pull on own
valence e- = increase pull on any e- in bond

36. Electronegativity (EN)
• Deviations:
Group 8: Most Grp 8 elements tend to have
EN = 0 b/c they have full outer en levels.
This means they tend not to form bonds like
other elements. Some Grp 8 elements do
have EN values, though, b/c they are large
enough to form bonds.

37. Is bond ionic or polar?
• Difference in EN values between two atoms will
determine if bond is Ionic or Covalent (found in
Ch. 8.7)
• On a continuum of ∆EN:
100% covalent
0
0.4
Nonpolar
0% ionic
covalent
0% covalent
1.7
Polar
covalent
4
Ionic
100% ionic
• Can also use distance on PT as indicator of bond
type: the closer two atoms are, the more covalent
the bond; the farther apart they are, the more ionic
the bond

38. Is bond ionic or polar?

39. Bond Polarity vs. Molecular Polarity
• Molecular polarity – related to distribution
of e- cloud around entire molecule
– Polar molecule – experiences unequal
distribution of e- cloud (aka: dipole – one end
of molecule slightly + and other end slightly -)
(has a dipole moment)
– Nonpolar molecule – experiences equal
distribution of e- cloud (no dipole moment)
• Determined by symmetry of molecule
(geometry and arrangement of LP)
• LP tend to cause asymmetry in structure
which results in a shift in the e- cloud

40. What is a dipole moment?
• A substance possesses a dipole moment if its
centers of positive and negative charge
do not coincide and cancel each other out.
µ=exd
—
+
(expressed in Debye units)
(you won’t need to use this)
polar
• The larger the dipole moment, the more polar is the bond or
molecule (due to a greater asymmetry in the distribution of
electrons around the bond or molecule)
• The greater the ∆EN, the larger the dipole moment for a
bond.

41. Why is water polar?
• H—O bonds (polar bonds) are not
“balanced” around the structure (the pull
of each bond does not cancel the other)
• Lone pairs cause
asymmetry (imbalance)
• E- cloud of molecule
is shifted
such that there
is greater
neg charge
near O and
greater pos
charge by H

42. General Rule for Molecular Polarity
• If all outer atoms are same and structure is
symmetrical (dipole moments of bonds “cancel”
each other out b/c same magnitude but opposite
direction)  molecule is NP (no overall dipole
moment)
– If 0 LP then NP (if all outer atoms same)
• If structure is asymmetrically arranged (has LP or
different outer atoms) molecule is Polar (has dipole
moment)
– If has LP* then polar
– *Exception: Square planar (4 BP, 2LP) and Linear (2 BP,
3 LP) are NP even though they have LP
– LP’s cancel each other out

43. Attractive forces in & btwn molecules
• Intramolecular forces = attractive forces
within the molecule (bonds)
• Intermolecular forces (IMF) = attractive
forces between molecules (sticky factors)
• Properties of covalent compounds are
attributed to their IMF
– Ex. Polar cmpds tend to have higher MP and
BP due to stronger IMF

44. Chapter 9
Lab 8-9.1 Part IV
Hybridization
The “real story” behind bonding…
For a preview of what’s to come,
click on this link and select
Hybridization.

45. Draw the LDS of CH4:
What is the e- configuration for C? for H?
Carbon = 2s2 2p2
s
s
s
s
px py pz
H = 1s1
s
Is there really room for each H to share e- with Carbon?

46. Is there really room for Hydrogen to share 4 ewith Carbon?
Carbon = 2s2 2p2
s
s
s
s
px py pz
H = 1s1
s
No! Carbon must make room for 2 more e- so
it promotes an e- (moves it to a p orbital)
s
px py pz

47. Here’s what it would look like so far:
s
p
p
p
H = 1s1
Carbon = 2s2 2p2

48. And then if Hydrogen bonds with Carbon:
Remember that each p orbital is on an X, Y or Z axis at 90 o
angles from one another.
Also remember that the s orbital is smaller in radius than
the p orbitals.

49. Evidence that something
“different” is happening…
• Each C-H bond in methane has the same
length (109 pm)
• Each H-C-H bond angle is the same (109.5 o)
• How is this possible if :
– the s and p orbitals are different sizes from one
another (this would lead to bonds having
different lengths)?
– The s and p orbitals are at 90o not at 109.5o
angles

50. Then answer is….Hybridization
• All of the s, p atomic orbitals mix to create new
hybrid orbitals
• # of new hyrbrid orbitals = # of atomic orbitals
that were mixed
EX. If s and p atomic orbitals mix get 2 new hybrid
orbitals, each called sp hybrid
• Each hybrid orbital is exactly like the other and
the geometry is based on these (not atomic
orbitals)

51. More on Hybridization…
• Identify the hybridization if the following atomic
orbitals are mixed:
– s + p + p sp2 hybridization
– s + p + p + p + d sp3d hybridization
• What geometry could you expect from the
above hybridization?
3 hybrids (3BP or 2BP, 1LP)
5 hybrids (5BP or 4BP, 1LP, or 3BP, 2
LP or 2BP, 3LP)
Hybrids can contain LP or BP. # BP & LP determines
geometry. Even if you know hybridization, must know
what each orbital contains to determine geometry.

52. BeF2
http://www.mikeblaber.org/oldwine/chm1045/notes/Geometry/Hybrid/Geom05.htm

53. BH3

54. NH3
H2O

55. Sulfur difluoride
1.
2.
3.
4.
Draw the LDS.
Identify the geometry. Bent
Is the molecule polar? Yes, it is polar
What is the hybridization of the central
atom? 2BP+2LP = 4 hybrids = sp3 hybridization
••S••
F
F

56. Xenon tetrafluoride
1.
2.
3.
4.
Draw the LDS.
Identify the geometry. Squar Planar
Is the molecule polar? No, it is not polar
What is the hybridization of the central
atom? 4BP+2LP = 6 hybrids = sp3d2 hybridization
F
F
••
Xe
••
F
F

57. Sigma and Pi Bonding
• Draw LDS and Label all sigma and pi bonds
in each structure:
• Ethane (C2H6)
• Ethene (Ethylene) (C2H4)
• Ethyne (Acetylene) (C2H2)
Go to this link for more info about this section.

58. Sigma and Pi Bonding
• Ethane (C2H6)

59. Sigma and Pi Bonding
• Ethene (Ethylene) (C2H4)

60. Sigma and Pi Bonding
• Ethyne (Acetylene) (C2H2)

62. Sigma and Pi Bonding
• Sigma Bonding (σ) – head to head overlap
of hybrid or unhybridized orbitals
*All single bonds consist of σ
bonds.
*One of the double bonds is a σ
bond.
• Pi Bonding (π) – side to side overlap of
unhybridized p bonds (contains 2 e- total)
*The second and third bond
of a double/triple bond are π
bonds.
*Can only occur when central
atom is sp or sp2 hybridized

63. Benzene (C6H6)
b/c of resonance
we write…

64. Covalent Bond Characteristics
• Bond Order: The number of bonds between
two atoms
What’s the BO in a resonance structure?
• C—C
• C==C
•C C
BO = 1
BO = 2
BO = 3
• Bond Energy: Energy a bond must absorb to
break (sign?)
• Bond Length: Avg distance between the
centers of two nuclei (Next slide…)
• How are they all related?

65. Determining Bond Length

66. Factors Affecting Bond Length
Larger atoms = longer bonds

67. How are they all related?
___BL = ___BE indirect
___BO = ___BE
___BO = ___BL
direct
indirect

68. Using Bond Energies to find ∆Hrxn
• Draw the dot structures of the reactants
and the products
• Determine the energy needed to BREAK
all the bonds in the reactants
(Endothermic, positive value)
• Determine the energy change to MAKE all
the bonds in the products (Exothermic,
negative value)
• Add them together to get an approximate
value for the ∆H of the reaction

69. Using Bond Energies to find ∆Hrxn
• Determine the ∆H for the combustion of
methane gas. (use BE values on next slide)
• What is the ∆Hrxn when you use ∆Hf
values?

71. CH4
+
2O2

CO2
+
2H2O
Notice these are pos energies b/c
bonds are breaking!!!
BEreactants = 4(C-H) + 2(O=O) = 4(413kJ) + 2(498 kJ)
BEreactants = 2648 kJ
BEproducts = 2(C=O) + 4(H-O) = 2(-745 kJ) + 4(-463 kJ)
Notice these are neg energies b/c
BEproducts = -3342 kJ
bonds are being made!!!
∆H Rxn = BEproducts + BEreactants = -694 kJ

72. More links
• Click this link for a site that provides an
overview of this information
• The first 2min 15sec of this video gives
a great visual representation of the big
picture (where this unit fits into all of
chemistry)
• A song about ionic and covalent bonds