1. Bonding – General Concepts
Boom

2. Chemical Bonds
bonds form when the potential energy of the system will be lowered
chemical bond – mutual electrical attraction between the nuclei and valence electrons of different atoms that binds the atoms together

3. Sodium Chloride Crystal Lattice
Ionic compounds form solids at ordinary temperatures.
Ionic compounds organize in a characteristic crystal lattice of alternating positive and negative ions.
Solid sodium chloride has a melting point ~ 800oC WOW!

4. Electronegativity: The ability of an atom in a molecule to attract shared electrons to itself.

5. Do we get it???? Let us see!
Order the following bonds according to their
Polarity (lowest to highest): H-H, O-H, Cl-H,
S-H and F-H. (Table on pg. 334)
H-H < S-H < Cl-H < O-H < F-H
POLARITY INCREASES

6. Ionic Bonds Electrons are transferred
Electronegativity differences are
generally greater than 1.7
The formation of ionic bonds is
always exothermic!

7. Determination of Ionic Character
Electronegativity difference is not the final determination of ionic character
Compounds are ionic if they conduct electricity in their molten state

8. Coulomb’s Law
“The energy of interaction between a pair of ions is proportional to the product of their charges, divided by the distance between their centers”

10. Sodium Chloride Crystal Lattice
ΔH……Again?

11. Estimate Hf for Sodium Chloride
Na(s) + ½ Cl2(g)  NaCl(s)
Lattice Energy
-786 kJ/mol
Ionization Energy for Na
495 kJ/mol
Electron Affinity for Cl
-349 kJ/mol
Bond energy of Cl2
239 kJ/mol
Enthalpy of sublimation for Na
109 kJ/mol
Na(s)  Na(g) kJ
Na(g)  Na+(g) + e kJ
½ Cl2(g)  Cl(g) ½(239 kJ)
Cl(g) + e-  Cl-(g) kJ
Na+(g) + Cl-(g)  NaCl(s) kJ
Na(s) + ½ Cl2(g)  NaCl(s) kJ/mol

12. Bond Length Diagram

13. Covalent Bonding Forces
Electron – electron
repulsive forces
Electron – proton
attractive forces
Proton – proton
repulsive forces

14. Covalent Bonds Polar-Covalent bonds Nonpolar-Covalent bonds
Electrons are unequally shared
Electronegativity difference between .3 and 1.7
Nonpolar-Covalent bonds
Electrons are equally shared
Electronegativity difference
of 0 to 0.3
Electrostatic
Potential Diagram

15. Polarity Dipole Moment uneven distribution of charge
a molecule with a imbalance in electrons has a dipole moment
polyatomic molecules have dipole moments if their polar bonds don’t cancel out

16. Practice
Determine whether each of the following bonds will be: ionic, polar covalent, OR nonpolar covalent
Cl and O
=0.5
polar covalent
Cl and Br
nonpolar covalent
S and H
=0.4
polar covalent
S and Cs
=1.8
ionic

17. Practice
For each of the following molecules, indicate which ones have a dipole moment.
HCl
Cl2
H: 2.1 < Cl: 3.0
Dipole moment - +
Cl: 3.0 = Cl: 3.0
No dipole moment

18. Practice CH4 C: 2.5 > H: 2.1 no dipole moment H2S+
4- + + + + +
2-

19. Ions: Electron Configurations and Sizes
nonmetals
sharing to make covalent bonds
taking electrons from metals to form anions
metals
giving electrons to nonmetals to form cations
Sizes of Ions
trends are most important
cation < parent atom
more protons attracting electrons in
anion > parent atom
more electrons without additional protons

20. “Lattice energy is more exothermic with more highly charged ions”

21. Sizes of Ions
isoelectronic ions: ions with the same number of electrons
Ex: S2- Cl- K+ Ca2+
electrons have greater attraction to nucleus as # of protons increases
so the size decreases as nuclear charge (Z) increases

22. Table of Ion Sizes

23. Example Give six ions that are isoelectronic with neon.
N3-, O2-, F-, Na+, Mg2+, Al3+
7 > 8 > 9 > 11 > 12 > 13
Place them in order of decreasing size.
(Largest to smallest)

24. Example
for each of the following groups, place atoms in order of decreasing size
Cu, Cu+, Cu2+
all have 29 p, lower #e- mean smaller ion
Cu > Cu+ > Cu2+
Ni2+, Pd2+, Pt2+
different amounts of e- and p-energy levels
Pt2+ > Pd2+ > Ni2+
O, O-, O2-
same #p, more electrons means larger ion
O2- > O- > O

25. SURPRISE!
Please close your books and notes and push them forward. Also, get out something to write with and….nothing else.

26. 1. The compound most likely to be ionic is:KF
CCl4
CO2
ICl
CS2

27. 2. Ranking the ions S2-, Ca2+, K+, Cl- from smallest
to largest gives the order as:
S2-, Cl-, K+, Ca2+
Ca2+, Cl-, K+,, S2-
K+, Ca2+, Cl-, S2-
Cl-, S2-, K+, Ca2+
Ca2+, K+, Cl-, S2-

28. 3. The type of bonding within a water molecule is:
ionic bonding
polar covalent bonding
nonpolar covalent bonding
metallic bonding
hydrogen bonding

29. 4. What type of bond would you expect in CsI?:
Ionic
Covalent
Hydrogen
Metallic
van der walls

30. PLEASE PASS YOUR RESPONSES RIGHT TO LEFT LAST PERSON…..BRING THEM UP.
THAT’S ALL FOLKS!
PLEASE PASS YOUR RESPONSES RIGHT TO LEFT
LAST PERSON…..BRING THEM UP.

31. The compound most likely to be ionic is:KF
CCl4
CO2
ICl
CS2
Answer: A Bonds formed from elements with
greater differences in electronegativity are most
likely ionic in nature. (remember trends!)

32. Ranking the ions S2-, Ca2+, K+, Cl- from smallest
to largest gives the order as:
S2-, Cl-, K+, Ca2+
Ca2+, Cl-, K+,, S2-
K+, Ca2+, Cl-, S2-
Cl-, S2-, K+, Ca2+
Ca2+, K+, Cl-, S2-
Answer E Note that all ions in this isoelectronic
series have an Ar electron configuration;
therefore the nuclear charge determines the size.

33. The type of bonding within a water molecule is:
ionic bonding
polar covalent bonding
nonpolar covalent bonding
metallic bonding
hydrogen bonding
Answer B Consider the nature of the bonding
between hydrogen and oxygen in water.

34. What type of bond would you expect in CsI?:
Ionic
Covalent
Hydrogen
Metallic
van der walls
Answer A You tell me why???

35. Bond Length and Energy
Bond
Bond type
Bond length
(pm)
Bond Energy
(kJ/mol)
C - C
Single
154
347
C = C
Double
134
614
C  C
Triple
120
839
C - O
143
358
C = O
123
745
C - N
305
C = N
138
615
C  N
116
891
Bonds between elements become shorter and stronger as multiplicity increases.

36. Bond Energy and Enthalpy
Energy required
Energy released
D = Bond energy per mole of bonds
Breaking bonds always requires energy
Breaking = endothermic (+DH)
Forming bonds always releases energy
Forming = exothermic (-DH)

37. Example
Calculate the change in energy for the following reaction: (use table 8.4 on pg 351 in your textbook)
H2(g) + F2(g)  2HF(g)
∆H = ∑ D (bonds broken) – ∑ D (bonds formed)
∆H = (DH-H + DF-F) – 2DH-F
∆H = (432 kJ kJ) – 2 (565 kJ)
∆H = -544 kJ

38. Example Using bond energies, calculate ∆H for the reaction:
CH4(g) + 2Cl2(g) + 2F2(g) 
CF2Cl2(g) + 2HF(g) + 2HCl(g)
∆H = ∑ D (bonds broken) – ∑ D (bonds formed)
∑ D (bonds broken) = 4 (413 kJ) + 2 (239 kJ) (154 kJ)
∑ D (bonds formed) = 2 (485 kJ) + 2 (339 kJ) (565 kJ) + 2 (427 kJ)
∆H = 2438 kJ – 3632 kJ = kJ

39. The Octet Rule
Combinations of elements tend to form so that each atom, by gaining, losing, or sharing electrons, has an octet of electrons in its highest occupied energy level.
Diatomic Fluorine

40. Formation of Water by the Octet Rule

41. Comments About the Octet Rule
2nd row elements C, N, O, F observe the octet rule.
2nd row elements B and Be often have fewer than 8 electrons around themselves - they are very reactive.
3rd row and heavier elements CAN exceed the octet rule using empty valence d orbitals.
Below are three examples of compounds containing elements not obeying the Octet Rule:

42. Lewis Structures
Shows how valence electrons are arranged among atoms in a molecule.
Reflects central idea that stability of a compound relates to noble gas electron configuration.

43. Completing a Lewis Structure -CH3Cl
Make carbon the central atom
Add up available valence electrons:
C = 4, H = (3)(1), Cl = 7 Total = 14
Join peripheral atoms
to the central atom
with electron pairs. H .. ..
Complete octets on
atoms other than
hydrogen with remaining
electrons H .. C .. Cl .. .. .. H

44. H H C Cl CH3Cl Example methyl chloride C: 1 x 4e- = 4e-
H: 3 x 1e- = 3e-
Cl: 1 x 7e- = 7e-
14 e-
Carbon is central
atom H
H C Cl
duet
octet

45. CH3Cl
Methyl Chloride

46. Multiple Covalent Bonds: Double bonds
Ethene
Two pairs of shared electrons

47. Multiple Covalent Bonds: Triple bonds
Ethyne
Three pairs of shared electrons

48. .. .. .. Example CH2O O H – C - H Formaldehyde C: 1 x 4e- = 4 e-
H: 2 x 1e- = 2 e-
O: 1 x 6e- = 6 e- 12 O
H – C - H

49. Resonance
Resonance is invoked when more than one valid Lewis structure can be written for a particular molecule.
Benzene, C6H6
The actual structure is an average of the resonance
structures.
The bond lengths in the ring are identical, and
between those of single and double bonds.

50. Resonance Bond Length and Bond Energy
Resonance bonds are shorter and stronger than single bonds.
Resonance bonds are longer and weaker than double
bonds.

51. Resonance in Ozone, O3 Neither structure is correct.
Oxygen bond lengths are identical, and intermediate to single and double bonds

52. Resonance in Polyatomic Ions
Resonance in a carbonate ion:
Resonance in an acetate ion:

53. Resonance in Polyatomic Ions
How about the nitrate ion(NO3-) ?:
How about the nitrite ion (NO2-) ?:
Lewis Structure
Nitrate ion
Lewis Structure
Nitrite ion

54. Resonance The lewis structure of which molecule requires
resonance structures?
MgCl2
PCl5
SiO2
SO2
OCl2
Answer: D In this case there is a single bond
between sulfur and one of the oxygens, and a
double bond between sulfur and the other oxygen.

55. Formal Charge
Helps you to compare various Lewis Structures and choose the best or most likely structure
FC = (# valence e-) – (#e- assigned to it)
Assigned:
all of the lone pairs belong to that atom
plus half of the electrons involved in bonding
Valence:
from periodic table

56. Formal Charge
Not REAL: but provides less extreme charges than oxidation numbers
sum of the FC on molecule must equal overall charge on molecule
atoms try to achieve a FC as close to zero as possible
any negative FC must be on most electronegative atom

57. Example SO42- sulfate ion 6e- + 4(6e-)= 30 e- + 2- = 32 electrons -1+2 -1 -1 -1

58. Example
NO2
nitrogen dioxide
17 electrons

59. Formal Charge (FC) Dinitrogen oxide, N2O has two double bonds. The
general structure is N=N=O. The formal charge
on the oxygen atom in this molecule is?
zero
Positive 1 (+1)
Positive 2 (+2)
Negative one (-1)
Negative two (-2)
Answer: A Use the formula! If necessary, draw
the lewis structure.

60. Localized Electron Bonding Model
Lewis structures are an application of the “Localized Electron Model”
L.E.M. assumes: That a molecule is composed of atoms that are bound together by sharing pairs of electrons using the atomic orbitals of the bound atoms.
L.E.M. says: Electron pairs can be thought of as “belonging” to pairs of atoms when bonding
Resonance points out a weakness in the Localized Electron Model.

61. METHANE
OXYGEN

62. Resonance in Ozone, O3

63. Localized Electron Bonding Model (LEM)
LEM is composed of three parts:
Description of the valence electrons
arrangement in the molecule (Lewis structure)
B. Prediction of the geometry of the molecule
(VSEPR)
C. Description of the type of atomic orbitals
used by the atoms (Chapter 9 – Hybridization)

64. VSEPR
Are you ready for 3-Dimensions?

65. O O Cl B Molecular Shapes
VSEPR Theory (Valence Shell Electron Pair Repulsion)
pairs of electrons are arranged as far apart from each other as possible.
Linear: 180o between bonds
Ex: O2
O O
2. Trigonal Planar: “flat triangle” – one central atom surrounded by three atoms with NO LONE PAIRS.
Ex: BCl3 B Cl
120o

66. C H N H 3. Tetrahedral: four atoms bonded to a central atom. 109.5o
4. Trigonal pyramidal: three atoms bonded to a central atom with one lone pair of electrons about the central atom. N H
107o

67. 5. Bent: a central atom bonded to two other atoms AND one or two LONE PAIRS.

68. VSEPR – Valence Shell Electron Pair Repulsion
Steric #
Overall Structure
Forms 2
Linear
AX2 3
Trigonal Planar
AX3, AX2E 4
Tetrahedral
AX4, AX3E, AX2E2 5
Trigonal bipyramidal
AX5, AX4E, AX3E2, AX2E3 6
Octahedral
AX6, AX5E, AX4E2
Steric # = X+E
A = central atom
X = atoms bonded to A
E = nonbonding electron pairs on A

69. VSEPR: Linear
AX2
CO2
Shape: Linear

70. VSEPR: Trigonal Planar
AX3
BF3
Shape: Trigonal Planar
AX2E
SnCl2
Shape: Bent

71. AX4 CCl4 AX3E PCl3 AX2E2 Cl2O Shape: Tetrahedral
VSEPR: Tetrahedral
AX4
CCl4
Shape: Tetrahedral
AX3E
PCl3
Shape: Trigonal Pyramidal
AX2E2
Cl2O
Shape: Bent

72. VSEPR: Trigonal Bi-pyramidal
AX5
PCl5
Shape: Trigonal Bipyramidal
AX4E
SF4
Shape: Seesaw
AX3E2
ClF3
Shape: T-shaped
AX2E3
I3-
Shape: Linear

73. AX6 SF6 AX5E BrF5 AX4E2 ICl4- Shape: Octahedral
VSEPR: Octahedral
AX6
SF6
Shape: Octahedral
AX5E
BrF5
Shape: Square Pyramidal
AX4E2
ICl4-
Shape: Square planar

74. The predicted geometry (shape) of PH3 according
VSEPR
The predicted geometry (shape) of PH3 according
to the VSEPR theory is?
linear
bent
Trigonal pyramidal
tetrahedral
Trigonal planar
Answer: C Easy one!

75. Covalent Bonding - Orbitals

76. Hybrids + = Poodle + Cocker Spaniel = Cockapoo Prius + Zeedonk =
Pidgpug
s orbital +
p orbital =
sp orbital

77. Hybridization - The Blending of Orbitals+ =
Poodle +
Cocker Spaniel =
Cockapoo + =
s orbital +
p orbital =
sp orbital

78. What Proof Exists for Hybridization?
We have studied electron configuration notation and
the sharing of electrons in the formation of covalent
bonds.
Lets look at a molecule of methane, CH4.
Methane is a simple natural gas. Its molecule has a
carbon atom at the center with four hydrogen atoms covalently bonded around it.

79. Carbon ground state configuration
What is the expected orbital notation of carbon
in its ground state?
Can you see a problem with this?
(Hint: How many unpaired electrons does this
carbon atom have available for bonding?)

80. Carbon’s Bonding Problem
You should conclude that carbon only has TWO
electrons available for bonding. That is not not enough!
How does carbon overcome this problem so that
it may form four bonds?

81. Carbon’s Empty Orbital
The first thought that chemists had was that carbon promotes one of its 2s electrons…
…to the empty 2p orbital.

82. However, they quickly recognized a problem with such
an arrangement…
Three of the carbon-hydrogen bonds would involve
an electron pair in which the carbon electron was a 2p, matched with the lone 1s electron from a hydrogen atom.

83. This would mean that three of the bonds in a methane
molecule would be identical, because they would involve
electron pairs of equal energy.
But what about the fourth bond…?

84. The fourth bond is between a 2s electron from the
carbon and the lone 1s hydrogen electron.
Such a bond would have slightly less energy than the other bonds in a methane molecule.

85. This bond would be slightly different in character than
the other three bonds in methane.
This difference would be measurable to a chemist
by determining the bond length and bond energy.
But is this what they observe?

86. The simple answer is, “No”.
Measurements show that
all four bonds in methane
are equal. Thus, we need
a new explanation for the
bonding in methane.
Chemists have proposed an explanation – they call it
Hybridization.
Hybridization is the combining of two or more atomic
orbitals of nearly equal energy within the same atom
into orbitals of equal energy.

87. In the case of methane, they call the hybridization
sp3, meaning that an s orbital is combined with three
p orbitals to create four equal hybrid orbitals.
These new orbitals have slightly MORE energy than
the 2s orbital…
… and slightly LESS energy than the 2p orbitals.

88. sp3 Hybrid Orbitals
Here is another way to look at the sp3 hybridization
and energy profile…

89. sp Hybrid Orbitals While sp3 is the hybridization observed in methane,
there are other types of hybridization that atoms
undergo.
These include sp hybridization, in which one s
orbital combines with a single p orbital.
This produces two hybrid orbitals, while leaving two normal p orbitals

90. sp2 Hybrid Orbitals
Another hybrid is the sp2, which combines two orbitals from a p sublevel with one orbital from an s sublevel.
One p orbital remains unchanged.

91. Hybridization Involving “d” Orbitals
Beginning with elements in the third row, “d” orbitals may also hybridize
dsp3 = five hybrid orbitals of equal energy
d 2sp3 = six hybrid orbitals of equal energy

92. Hybridization and Molecular Geometry
Forms
Overall Structure
Hybridization of “A”
AX2
Linear sp
AX3, AX2E
Trigonal Planar
sp2
AX4, AX3E, AX2E2
Tetrahedral
sp3
AX5, AX4E, AX3E2, AX2E3
Trigonal bipyramidal
dsp3
AX6, AX5E, AX4E2
Octahedral
d2sp3
A = central atom
X = atoms bonded to A
E = nonbonding electron pairs on A

93. “THINGS”

94. Sigma and Pi Bonds
Sigma () bonds exist in the region directly between two bonded atoms.
Pi () bonds exist in the region above and below a line drawn between two bonded atoms.
Single bond
1 sigma bond
Double Bond
1 sigma, 1 pi bond
Triple Bond
1 sigma, 2 pi bonds

95. Sigma and Pi Bonds Single Bonds
Ethane

96. Sigma and Pi Bonds: Double bonds
Ethene
1  bond

97. Sigma and Pi Bonds Triple Bonds
Ethyne
1  bond
1  bond