1. Chemical Bonding I: Basic Concepts
Chapter 9
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2. Topics:
Lewis Dot symbols
The ionic bond
Lattice energy of ionic compound
Electronegativity
Writing Lewis structure
Formal charge and Lewis structure
Thee concept of resonance
Bond energy

3. Lewis Dot symbols
• keep track of valence electrons as dots
Examples: Na • and :C:
• Elements in a group:
– have similar Lewis symbols
– have same number of valence electrons

4. Valence electrons are the outer shell electrons of an
atom. The valence electrons are the electrons that
participate in chemical bonding.
Group
# of valence e-
e- configuration 1A 1
ns1 2A 2
ns2 3A 3
ns2np1 4A 4
ns2np2 5A 5
ns2np3 6A 6
ns2np4 7A 7
ns2np5
9.1

5. Lewis Dot Symbols for the Representative Elements &
Noble Gases
9.1

6. Group practice problem of chap9:Q1
What are the correct Lewis dot symbols of
(A)F,F-, (B)S,S2-, (C) O,O2-, (D) N,N3-?
Group practice problem of chap9:Q1

7. The Ionic Bond - Chemical Bonds Two general types:
• Ionic bonds---- between ions of opposite charges
• Covalent bonds---- sharing of electrons
The Ionic Bond Li + F
Li+ -
1s22s1
1s22s22p5
1s2
1s22s22p6
[He]
[Ne]
Li+ + e-
e- +
Li+ +
9.2

8. The Octet Rule
Atoms tend to gain, lose, or share electrons until they
are surrounded by 8 valence electrons.
An octet = 8 electrons
= 4 pairs of electrons
= noble gas arrangement
= very stable
Exception: H = 2 electrons only
Example:
Na [Ne]3s1 loses 1 e----> Na+ [Ne]
Cl [Ne]3s23p5 gains 1 e- --> Cl- [Ar]

9. Electrostatic (Lattice) Energy
Lattice energy (E) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions.
Elattice
LiF(s) Li+(g) + F-(g)
E = k
Q+Q- r
Q+ is the charge on the cation
Q- is the charge on the anion
r is the distance between the ions
cmpd
lattice energy
MgF2
MgO
2957
3938
Q= +2,-1
Q= +2,-2
Lattice energy (E) increases as Q increases and/or
as r decreases.
LiF
LiCl
1036
853
r F- < r Cl-
9.3

10. • Elattice increases as Charge on ions increases
Lattice Energy
• Elattice increases as Charge on ions increases
• Elattice decreases as inter-ionic distances increase
(charge is much more important than distance)
Size increase
E decrease
Q1=1;Q2=1
Q1=2;Q2=1
9.3

11. Practice problem of chap9:Q2
Which of the following is a correct order of lattice energy?
NaCl < AlCl3 < MgCl2 (b) LiF < LiCl < LiBr
(c) Na2O < MgO < Al2O3 (d) Ga2O3 < CaO < K2O
(e) LiCl < NaCl < KCl
Lattice Energy
• Elattice increases as Charge on ions increases
• Elattice decreases as inter-ionic distances increase
(charge is much more important than distance)
Practice problem of chap9:Q2

12. Born-Haber Cycle for Determining Lattice Energy
Lattice energy (E) is the energy required to completely separate one mole of a solid ionic compound into gaseous ions.
For an ionic compound the lattice enthalpy is the heat
energy released when one mole of solid in its standard state
is formed from its ions in the gaseous state.
Li+(g) + F-(g) LiF(s) ΔH0=- Elattice
This value cannot be determined directly and so we make
use of changes for which data are available and link them
together with an enthalpy cycle.
This enthalpy cycle is based on the formation of the compound
from its elements in their standard states.
Hess’s Law

13. Consider the reaction: Li(s) +1/2 F2(g) LiF(s)
Li(s) Li(g) DH01 =155.2 kJ/mol
½ F2(g) F(g) DH02 =75.3 kJ/mol
Li(g) Li+(g) +e- DH03=520 kJ/mol
F(g) + e F-(g) DH04=-328 kJ/mol
Li+(g) + F-(g) LiF(s) DH05=?
Li(s) + ½ F2(g) LiF(s) DH0overall = KJ/mol
DHoverall = DH1 + DH2 + DH3 + DH4 + DH5 o
DH05=?=-1017KJ/mol
Lattice energy of LiF is 1017KJ/mol
*Lattice energy is always a positive quantity.

14. Why should two atoms share electrons?
A covalent bond is a chemical bond in which two or more electrons are shared by two atoms.
Why should two atoms share electrons?
7e-
7e-
8e-
8e- F F + F
Lewis structure of F2
lone pairs F
single covalent bond
single covalent bond F
9.4

15. Lewis structure of water
single covalent bonds
2e-
8e-
2e- H + O + H O H or
Double bond – two atoms share two pairs of electrons
8e-
8e-
8e-
double bonds O C or O C
double bonds
Triple bond – two atoms share three pairs of electrons
triple bond
8e- N
8e- or N
triple bond
9.4

16. Lengths of Covalent Bonds
Bond Lengths
Triple bond < Double Bond < Single Bond
Bond strength:
triple > double> single
9.4

17. 9.4

18. Bond polarity
Polar covalent bond or polar bond is a covalent bond with greater electron density around one of the two atoms
e- poor
e- rich
electron rich
region H F
electron poor
region F H d+ d-
Electron density is pulled more tightly around the F end of the molecule.
Nonpolar bond: two identical atoms form a bond & equally share the bond’s electron pair
9.5

19. Electron Affinity - measurable
Electronegativity is the ability of an atom to attract toward itself the electrons in a chemical bond.
Electron Affinity - measurable
X (g) + e X-(g)
Electronegativity - relative, F is highest
• the term used to describe the relative attraction of an atom for the electrons in a bond
• a mathematical average of IE and EA
• Atom of the bond with the greater
electronegativity will carry the partial
negative charge
9.5

20. Practice problem of chap9:Q4
Which of the following is a polar covalent bond?
The HN bond in NH3.
(b) The SiSi bond in Cl3SiSiCl3.
(c) The CaF bond in CaF2.
(d) The OO bond in O2.
(e) The NN bond in H2NNH2.
Practice problem of chap9:Q4

21. The Electronegativities of Common Elements
9.5

22. Variation of Electronegativity with Atomic Number
9.5

23. Electronegativity:
• The most electronegative element: F (4.0)
• Next most electronegative element:
O (3.5)
• Next two most electronegative:
N (3.0) and Cl (3.0)
• Least electronegative element:
Cs (0.7)
Also
H (2.1)
C (2.5)
Na (0.9)

24. Classification of bonds by difference in electronegativity
Bond Type
Covalent
 2
Ionic
0 < and <2
Polar Covalent
Increasing difference in electronegativity
Covalent
share e-
Polar Covalent
partial transfer of e-
Ionic
transfer e-
H-H
HCl
NaCl
9.5

25. Practice problem of chap9:Q11
Classify the following bonds as ionic, polar covalent,
or covalent: The bond in CsCl; the bond in H2S; and
the NN bond in H2NNH2.
Cs – 0.7
Cl – 3.0
3.0 – 0.7 = 2.3
Ionic
H – 2.1
S – 2.5
2.5 – 2.1 = 0.4
Polar Covalent
N – 3.0
N – 3.0
3.0 – 3.0 = 0
Covalent
Practice problem of chap9:Q11
9.5

26. Writing Lewis Structures
Draw skeletal structure of compound showing what atoms are bonded to each other. Put least electronegative element in the center. Arrange atoms & connect with single bonds
– Single bond = 2 electrons
2. Count total number of valence e-. Add 1 for each negative charge. Subtract 1 for each positive charge.
3. Complete an octet for atoms bonded to the central atom except hydrogen (H=2). Then add leftover electrons to central atom. Electrons belonging to the central or surrounding atoms must be shown as lone pairs if they are not involved in bonding.
4. If the central atom has fewer than 8 electrons, try adding double and triple bonds on central atom as needed.
9.6

27. 5 + (3 x 7) = 26 valence electrons
Write the Lewis structure of nitrogen trifluoride (NF3).
Step 1 – N is less electronegative than F, put N in center
Step 2 – Count valence electrons N - 5 (2s22p3) and F - 7 (2s22p5)
5 + (3 x 7) = 26 valence electrons
Step 3 – Draw single bonds between N and F atoms and complete
octets on N and F atoms.
Step 4 - Check, are # of e- in structure equal to number of valence e- ?
3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons F N
9.6

28. 4 + (3 x 6) + 2 = 24 valence electrons
Write the Lewis structure of the carbonate ion (CO32-).
Step 1 – C is less electronegative than O, put C in center
Step 2 – Count valence electrons C - 4 (2s22p2) and O - 6 (2s22p4)
-2 charge – 2e-
4 + (3 x 6) + 2 = 24 valence electrons
Step 3 – Draw single bonds between C and O atoms and complete
octet on C and O atoms.
Step 4 - Check, are # of e- in structure equal to number of valence e- ?
3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons
Step 5 - Too few electrons, form double bond and re-check # of e-
2 single bonds (2x2) = 4
1 double bond = 4
8 lone pairs (8x2) = 16
Total = 24 O C
9.6

30. Formal Charge
• Another way (other than Oxidation Number) to keep track of electron density
• Big difference:
– Oxidation number divides shared electrons unequally
– Formal charge divides shared electrons equally
An atom’s formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure.

31. An atom’s formal charge is the difference between the number of valence electrons in an isolated atom and the number of electrons assigned to that atom in a Lewis structure.
formal charge on an atom in a Lewis structure =
total number of valence electrons in the free atom -
total number of nonbonding electrons - 1 2
total number of bonding electrons ( )
FC = [ Val. e-] -[(nonbonded e-) + 1/2(bonded e-)]
The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion.
• Remember: formal charges are NOT real charges
9.7

32. Practice problem of chap9:Q12

33. Formal Charge and Lewis Structures
For neutral molecules, a Lewis structure in which there are no formal charges is preferable to one in which formal charges are present.
Lewis structures with large formal charges are less plausible than those with small formal charges.
Among Lewis structures having similar distributions of formal charges, the most plausible structure is the one in which negative formal charges are placed on the more electronegative atoms.
Which is the most likely Lewis structure for CH2O? H C O -1 +1 H C O
9.7

34. ( ) - -1 +1 H C O = 1 2 = 4 - 2 - ½ x 6 = -1 = 6 - 2 - ½ x 6 = +1
C – 4 e-
O – 6 e-
2H – 2x1 e-
12 e-
2 single bonds (2x2) = 4
1 double bond = 4
2 lone pairs (2x2) = 4
Total = 12 H C O
formal charge on an atom in a Lewis structure = 1 2
total number of bonding electrons ( )
total number of valence electrons in the free atom -
total number of nonbonding electrons
formal charge on C
= 4 -
2 -
½ x 6 = -1
formal charge on O
= 6 -
2 -
½ x 6 = +1
9.7

35. ( ) - H C O = 1 2 = 4 - 0 - ½ x 8 = 0 = 6 - 4 - ½ x 4 = 0 C – 4 e-H C O
C – 4 e-
O – 6 e-
2H – 2x1 e-
12 e-
2 single bonds (2x2) = 4
1 double bond = 4
2 lone pairs (2x2) = 4
Total = 12
formal charge on an atom in a Lewis structure = 1 2
total number of bonding electrons ( )
total number of valence electrons in the free atom -
total number of nonbonding electrons
formal charge on C
= 4 -
0 -
½ x 8 = 0
formal charge on O
= 6 -
4 -
½ x 4 = 0
9.7

36. Resonance Structures
• Why necessary?
• Lewis structures are not adequate
• More than one Lewis Structure is needed to represent the real molecule
A resonance structure is one of two or more Lewis structures for a single molecule that cannot be represented accurately by only one Lewis structure.
Example: experimentally, ozone has two identical bonds whereas the Lewis Structure requires one single (longer) and one double bond (shorter). O + - O + -

37. Resonance in Benzene: C6H6
Resonance Structures:
• connect different resonance structures with <-->
• actual molecule is not depicted by resonance structures
• actual molecule does not go between resonance structures
• Molecules that can be depicted by resonance structures are more stable than expected
Resonance in Benzene: C6H6
9.8

38. Practice problem of chap9:Q12, 7
What are the resonance structures of the
carbonate (CO32-) ion? O C - O C - O C -
Which of the following has resonance structure?
NH4+ (b) CO2 (c) AlI3
(d) C6H6 (e) H2O
Practice problem of chap9:Q12, 7
9.8

40. Exceptions to the Octet Rule
The Incomplete Octet (too few electrons on central atom)
Be – 2e-
2H – 2x1e-
4e-
BeH2 H Be
B – 3e-
3F – 3x7e-
24e-
3 single bonds (3x2) = 6
9 lone pairs (9x2) = 18
Total = 24 F B
BF3
9.9

41. Exceptions to the Octet Rule
Odd-Electron Molecules
Result: unpaired electron
N – 5e-
O – 6e-
11e- NO N O
The Expanded Octet (too many e- on central atom; central atom with principal quantum number n > 2) S F
S – 6e-
6F – 42e-
48e-
6 single bonds (6x2) = 12
18 lone pairs (18x2) = 36
Total = 48
SF6
9.9

42. Practice problem of chap9:Q3,9
Which of the following is correct concerning the lone pairs on the underlined atoms in compounds?
ICl, 3 (b) H2S, 6 (c) CH4, 4
(d) CaH2, 2 (e) SCl2, 4
Which of the following obeys the octet rule?
BF3 (b) SF4 (c) NO
(d) PF5 (e) NO3- F B
Practice problem of chap9:Q3,9

43. Single bond < Double bond < Triple bond
Bond energy
The enthalpy change required to break a particular bond in one mole of gaseous molecules is the bond energy.
Bond Energy
H2 (g)
H (g) +
DH0 = kJ
Cl2 (g)
Cl (g) +
DH0 = kJ
HCl (g)
H (g) +
Cl (g)
DH0 = kJ
O2 (g)
O (g) +
DH0 = kJ O
N2 (g)
N (g) +
DH0 = kJ N
Bond Energies
Single bond < Double bond < Triple bond
9.10

44. Average bond energy in polyatomic molecules
H2O (g)
H (g) +
OH (g)
DH0 = 502 kJ
OH (g)
H (g) +
O (g)
DH0 = 427 kJ
Average OH bond energy = 2
= 464 kJ
9.10

45. Bond Energies (BE) and Enthalpy changes in reactions
Imagine reaction proceeding by breaking all bonds in the reactants and then using the gaseous atoms to form all the bonds in the products.
DH0 = total energy input – total energy released
= SBE(reactants) – SBE(products)
Endothermic
reaction
Exothermic
reaction
9.10

46. H2 (g) + Cl2 (g) HCl (g)
2H2 (g) + O2 (g) H2O (g)
9.10

47. Use bond energies to calculate the enthalpy change for:
H2 (g) + F2 (g) HF (g)
DH0 = SBE(reactants) – SBE(products)
Type of bonds broken
Number of bonds broken
Bond energy (kJ/mol)
Energy change (kJ) H 1
436.4 F 1
156.9
Type of bonds formed
Number of bonds formed
Bond energy (kJ/mol)
Energy change (kJ) H F 2
568.2
1136.4
DH0 = – 2 x = kJ
Practice problem of chap9:Q9
9.10

48. O C (A) (B) (C) (D) -2 O C -2 O O H Ca H S O C - N N Cl S Cl F F -2 -2-3 N
Cl S Cl -1 F F

49. Cl S Cl
Cl S Cl H Cl C
Cl S Cl
Cl S Cl Cl C H Cl C H Cl C H

50. Cl N Cl O C O Cl N Cl Cl Cl O C O Cl N Cl Cl N Cl O C O Cl Cl C O N C H N C H N C H N C H

51. O C -2 O C -2 O C -2 O C -2