2. The smaller the atom, the higher the electronegativity
Types of Bonds
A. Electronegativity - The ability of an atom to attract electrons to itself in a bond
1. Periodic Trends (link to size)
Metals – Low Electronegativity
Non- Metals – High Electroneg
The smaller the atom, the higher the electronegativity

3. Ionic vs. Molecular Compounds
Ionic Compounds
Molecules
Metal + Non-Metal
Stealing of Electrons
Called Salts
Clumps of Ions
No prefixes, may need Roman #
Two Non-metals
Sharing of electrons (Covalent or Polar Cov.)
Separate Molecules
Polar and Non-Polar Molecules (H2O vs CH4)
Prefixes

5. Types of Bonds 3. Types of bonds Electroneg. Difference Ionic >2
Polar Cov. 0.2 – 1.9
Covalent <0.2
Example: Li - F

6. Which of the following form predominantly ionic, covalent, or polar covalent bonds?
B-Cl P-H
Na-F P-Cl
C-Cl O-H

7. L. Dot for Ionic Compounds
Why do Ionics steal? To gain an Octet
Draw Lewis Dot Pictures for:
NaCl
CaCl2
BaO
Li2O

8. NaCl(s)  Na+(g) + Cl-(g)
Lattice Energy – Energy required to convert a mole of an ionic solid to its gaseous ions
NaCl(s)  Na+(g) + Cl-(g)
Increases as distance (d) decreases
Increases with increasing charge (Q)
U = k Q1Q2 d

10. Melting Point and Charge
MgCl2 MgO
(778 oC) (2800 oC)
CaCl2 CaO
(772 oC) (2528 oC)

11. Which would have a higher lattice energy?
NaCl or KCl
CaBr2 or Ca3N2
NaCl or NaBr
CaI2 or CaO

12. Old School Lewis Dots
Molecular Compounds
CH4 CO2 C2H4
H2O HCN C2H2

14. Michael Faraday's Benzene Sample (1825)

15. The Lone Pear(Pair) rides again!

16. Lewis Dots Rules 1. Sum all valence electrons, including charges
2. Single Bonds
3. Outer atoms get an octet except H
4. Center gets rest even if it violates the octet
5. Double/triple bonds if center atom still does not have an octet

17. Lewis Dots
NH3
NCl3
SF6
CO2
HCN
ClF5

18. Lewis Dots
You try:
SF4
H2SO4
KrF4
Cl2O
NH2CH3

19. CH3CH2CHCH2
CHCCH2NCl2
Lewis Dots

20. Lewis Dots
CN-
ICl4-
BrO3-
NO+

21. Lewis Dots
You try:
CO32-
IBr4-
BF4-
SO42- 21

22. Resonance Structures Warm-Up: O3
Definition – When a molecule can exist in more than one arrangement of electrons
1. Atoms remain static
2. Only the electrons move

23. Resonance Structures Examples NO2- CHO2- HNO3
Which needs resonance, SO3 or SO32-
Order the species in the previous problem from shortest to longest bond length.
Resonance Structures

24. Benzene
SO2
SO22-
Resonance Structures 24

25. Less Than an Octet Hydrogen Only gets 2 Beryllium, Boron, and Aluminum
BeCl2 BF3 AlF3

26. More Than an Octet
Just follow the rules and you will be able to draw these
Ex: AsF6-

28. Single < Double < Triple
Strengths of Covalent Bonds
Single < Double < Triple
Bond
Strength (kJ/mole)
Bond Length
(Å)
C-C
348
1.54
C=C
614
1.34
C = C
839
1.20
N = N
941
1.10

29. CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g)
Calculating Enthalpies of Reaction
Hrxn =  Hbroken – made
Calculate the heat of reaction for the following reaction.
CH4(g) + Cl2(g)  CH3Cl(g) + HCl(g)

30. H-CH3 + Cl-Cl  Cl-CH3 + H-Cl
Let’s look at bonds broken and made
H-CH3 + Cl-Cl  Cl-CH3 + H-Cl
Bonds broken: One mol C-H, One mol Cl-Cl
Bonds made: One mol C-Cl, One mol H-Cl

31. Hrxn =  Hbroken – made
Hrxn = [1(C-H) + 1(Cl-Cl)] – [1(C-Cl) + 1(H-Cl)]
Hrxn = [413 kJ kJ] – [328 kJ kJ]
Hrxn = -104 kJ (Exothermic reaction)

33. Calculate the H for the following reaction:

34. Bonds Broken Bonds Made
6 C-H C=O
1 C-C O-H
7/2 O2

35. Hrxn =  Hbroken – made
Hrxn = [6C-H + 1C-C + 7/2O2] – [4C=O + 6O-H]
Hrxn = [6(413) + 1(348) + 7/2(495)]
– [4(799) + 6(463)]
Hrxn = kJ (Exothermic reaction)

36. Calculate the H for the following reaction:

37. Bonds Broken Bonds Made
4 N-H N=N
1 N-N H-H
Hrxn =  Hbroken – made
Hrxn = [4N-H + 1 N-N] – [1 N=N + 2 H-H]
Hrxn = [4(391) + 1(163)] – [1(941) + 2(436)]
Hrxn = -86 kJ (Exothermic reaction)

38. Double Bonds
Calculate the enthalpy change for the following reaction. Be sure to always break the multiple bond and remake a single C-C bond.
Br Br
| |
H-C=C-H Br2  H-C-C-H
| | | |
H H H H
(ANS: -93 kJ)

39. Double Bonds Calculate the enthalpy change for the following reaction.
H H
| |
H-C=C-H H2  H-C-C-H
H H
(ANS: kJ)

40. CH3Cl(g) + H2O(g)  CH3OH(g) + HCl(g)
If the enthalpy change for the following reaction is 0 kJ, calculate the C-Cl bond energy.
CH3Cl(g) + H2O(g)  CH3OH(g) + HCl(g)
Bond energy
H-Cl 430 kJ/mol
C- O 360 kJ/mol
O-H 460 kJ/mol
(ANS: 330 kJ/mol)

41. C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
If the enthalpy change for the following reaction is kJ, calculate the bond energy of C=O.
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
Bond energy
C-C 345 kJ/mol
C-H 415 kJ/mol
O kJ/mol
O-H 460 kJ/mol (800 kJ)

42. Suffix Organic Naming Suffix Class Characteristic -ane Alkane
All single bonds
-ene
Alkene
Double bond(s)
-yne
Alkynes
Triple bond(s) 42

43. IUPAC Names of Alkanes Suffix = -ane # of C atoms Name & Formula 1
Methane CH4 6
Hexane C6H14 2
Ethane C2H6 7
Heptane C7H16 3
Propane C3H8 8
Octane C8H18 4
Butane C4H10 9
Nonane C9H20 5
Pentane C5H12 10
Decane C10H22 43

44. Name the following and write the chemical formula44

46. 14. [Ca]2+2[F]- (8 ve- around F)
BaF2 CsCl Li3N Al2O3
20 a) KF has a larger LE because F- is much smaller than Cl-
b) Na – Cl ~ 2.8 A K-F ~ 2.7A
a) (i) Increases with charge (ii) decreases with d
b) KBr < NaF < MgO < ScN
24 a) Ca2+ is smaller than Ba2+, higher LE
b) NaCl is smallest pair, highest LE
c) BaO has highest charges, highest LE

47. 34. a) Draw LD of H2O2 and O2
b) O2 has a double bond, shorter bond length
38. a) O b) Al c) Cl d) F
40. a) O-F < C-F < Be-F
b) S-Br < C-P < O-Cl
c) C-S < N-O < B-F

48. Bond Length SO2 < SO3 < SO32-
52. CO2 (no resonance needed)
54. Bond Length NO+ < NO2- <NO3-
66. a) kJ b) 20 kJ c) 5 kJ
68. a) kJ b) kJ c) kJ
70. a) kJ b) kJ
a) C2H3Cl3O2 b) Same

49. 62.
100. In2S (I) [Kr]5s24d10
InS (II) [Kr]5s14d10
In2S3 (III) [Kr]4d10
In(III) is smallest (least mutual electron repul)
In(III) has the highest lattice energy
a) C2H3Cl3O2 b) C2H3Cl3O2
c) Structure CCl3CH(OH)2