1. CHAPTER 9A Chemical Bonding I: Basic Concepts
Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

2. Chapter 9 9.1 Lewis dot symbols . 9.2 The ionic bond .
9.4 The covalent bond .
9.5 Electronegativity .
9.6 Writing Lewis structure .
9.7 Formal charge and Lewis structure .
9.8 The concept of resonance .
9.9 Exceptions to the octet rule .

3. 9.1-Lewis dot symbols .

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

5. 9.1 Lewis Dot Symbols نظام النقط للويس
قاعدة الثمانية
The Octet Rule: in forming chemical bonds, atoms usually gain, lose or share electrons until they have 8 in the outer shell to reach the same electronic configuration of the noble gasses (ns2 np6).
Lewis Dot Representation: In the representation of an atom, the valence electrons of an atom are represented by dots.
There are two main types of chemical bonds: ionic bond and covalent bond.

6. Lewis Dot Symbols for the Representative Elements &
Lewis dot symbol consists of the symbol of an element and one dot for each valence electron in an atom of the element.
Lewis Dot Symbols for the Representative Elements &
Noble Gases
valence electrons represented by dots ((نقط.

7. Lewis dot structures for atoms and ions
1-The Lewis dot symbol for the chloride ion is
First, write the electron configuration of atom or ion
2-The Lewis dot symbol for the lead atom is 
Then, determine the valence electrons
3-The Lewis dot symbol for the calcium ion is 
Finally, draw the Lewis dot structure
Ca2+
4-The Lewis dot symbol for the S2- ion is 

8. nonmetals + metalloide
Types of Bonds
Types of Atoms
Type of Bond
Bond Characteristic
metals + nonmetals
Ionic
electrons
transferred
nonmetals + Nonmetals
nonmetals + metalloide
Covalent
shared
Non-metals have high electronegativity Metals have low electronegativity.

9. 9.2 The ionic bond .

10. The Ionic Bond is the electrostatic force that holds ions together in an ionic compound.
Ionic compounds are compounds consist of (metal + nonmetal) that forms only ionic bonds. Li + F
Li+ -
وتتكون الرابطة الأیونیة من القوة الكھروستاتیكیة التي تعمل على ربط الأیونات المتخالفة في الشحنة مع بعضھا البعض.
1s22s1
1s22s22p5
1s2
1s22s22p6
[He]
[Ne]
Other Examples:
CaO
Li2O
Mg3N2 Li
Li+ + e-
e- + F - F -
Li+ +
Li+
LiF

11. Q: Use Lewis dot symbols to show the formation of Li2O2-
2 Li + O
1s22s1
1s22s22p4
1s2
1s22s22p6
[He]
[Ne]
2e- + O 2-
2 Li
2Li e- O 2-
2Li+ +
2Li+

12. F K In an IONIC bond, electrons are lost or gained,
resulting in the formation of IONS
in ionic compounds. F K

13. K F

14. K F

15. K F

16. K F

17. K F

18. + _ K F

19. K F _ + The compound potassium fluoride
consists of potassium (K+) ions
and fluoride (F-) ions

20. K F _ + The ionic bond is the attraction between the positive K+ ion
and the negative F- ion

23. 9.4 The covalent bond .

24. Why should two atoms share electrons?
A covalent bond is a chemical bond between (nonmetal + nonmetal) or (metalloide + nonmetal) in which two or more electrons are shared by two atoms.
A covalent compounds are compounds that contain only covalent bonds.
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
lone pair is a valence electron pair without bonding or sharing with other atoms

25. Covalent Bonds

26. Chlorine
forms a
covalent
bond
with
itself
Cl2

27. How
will
two
chlorine
atoms
react? Cl Cl

28. Cl Cl Each chlorine atom wants to
gain one electron to achieve an octet

29. Cl Cl do to achieve an octet? What’s the solution – what can they
Neither atom will give up an electron –
chlorine is highly electronegative.
What’s the solution – what can they
do to achieve an octet?

30. Cl Cl

31. Cl Cl

32. Cl Cl

33. Cl Cl

34. Cl Cl
octet

35. Cl Cl octet circle the electrons for each atom that completes
their octets

36. Cl Cl The octet is achieved by each atom sharing the
electron pair in the middle
circle the electrons for
each atom that completes
their octets

37. Cl Cl The octet is achieved by each atom sharing the
electron pair in the middle
circle the electrons for
each atom that completes
their octets

38. Cl Cl This is the bonding pair circle the electrons for
each atom that completes
their octets

39. Cl Cl It is a single bonding pair circle the electrons for
each atom that completes
their octets

40. Cl Cl It is called a SINGLE BOND circle the electrons for
each atom that completes
their octets

41. Cl Cl Single bonds are abbreviated with a dash
circle the electrons for
each atom that completes
their octets

42. Cl Cl Cl2 This is the chlorine molecule, circle the electrons for
each atom that completes
their octets

43. 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
8e-
triple bond N
8e- or N
triple bond
9.4

44. Oxygen is also one of the diatomic molecules

45. How will two oxygen atoms bond?

46. Each atom has two unpaired electrons

47. O

48. O

49. O

50. O

51. O

52. O

53. O Oxygen atoms are highly electronegative.
So both atoms want to gain two electrons.

54. O Oxygen atoms are highly electronegative.
So both atoms want to gain two electrons.

55. O

56. O O

57. O O

58. O O

59. Both electron pairs are shared.

60. O O
6 valence electrons
plus 2 shared electrons
= full octet

61. O O
6 valence electrons
plus 2 shared electrons
= full octet

62. O O
two bonding pairs,
making a double bond

63. O O =
For convenience, the double bond
can be shown as two dashes.

64. This is the oxygen molecule,=
this is so cool!!
This is the oxygen molecule, O2

65. Triple bond < Double Bond < Single Bond
Lengths of Covalent Bonds is the distance between the nucleis of two covalently bonded atoms in a molecule.
Bond Lengths
Triple bond < Double Bond < Single Bond

66. Chemical bond is the force that holds the atoms together in a molecule, and the intermolecular force operates between molecules. The intermolecular forces in the ionic compounds is stronger than the forces in the covalent compounds.
Ionic and Covalent
Intermolecular Forces
Attractive Forces Between ATOMS
Attractive Forces Between Molecules

67. Polar covalent bond or polar bond is a covalent bond with greater electron density around one of the two atoms
Polar covalent bond
covalent bond H F
electron rich
region
electron poor
region
e- poor
e- rich F H d+ d-
Polar covalent bond:
Both have High EN
Different atoms
So: significant difference EN
Covalent bond:
Both have High EN
Almost the same atoms
So: no significant difference EN

68. 9.5 Electronegativity .

69. Electron Affinity - measurable, Cl is highest
Electronegativity is the ability of an atom to attract toward itself the electrons in a chemical bond.
Electron Affinity - measurable, Cl is highest
X (g) + e X-(g)
Electronegativity - relative, F is highest
Electronegativity doesn’t have a unit

70. The Electronegativities of Common Elements

71. Cs Ga As K Which of these elements has the greater Electronegativity?
Increasing Electronegativity
Decreasing Electronegativity

72. Variation of Electronegativity with Atomic Number

73. 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-

74. 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

75. Worked Example 9.2
Classify the following bonds as ionic, polar covalent, or covalent
a) HCl =3-2.1=0.9
Polar covalent
b) KF =4-0.8=3.2
Ionic
c) C-C = =0
covalent
d) CsCl =3-1=2
Ionic
e) H2S = =0.4
Polar covalent
f) N-N =3-3=0
Covalent
g)Si-Cl in Cl3SiSiCl3=3-1.2=.1.8

76. 9.6 Writing Lewis structure.

77. Writing Lewis Structures
Draw skeletal structure of compound showing what atoms are bonded to each other. Put least electronegative element in the center.
Count total number of valence e-. Add 1 for each negative charge. Subtract 1 for each positive charge.
Complete an octet for all atoms except hydrogen
If structure contains too many electrons, form double and triple bonds on central atom as needed.

78. كتابة أشكال لويس
1- يمثل الإلكترون بنقطة والرابطة التساهمية بخط ويعرف الزوج من الإلكترونات غير المرتبطة بالزوج الحر.
2- نوجد مجموع إلكترونات التكافؤ للذرات المكونة للجزئ وليكن رمزه a (عدد إلكترونات التكافؤ لعناصر المجموعات الرئيسية يساوي رقم المجموعة) إذا كان المركب عبارة عن أيون:
الأيون ذو شحنة سالبة تضاف قيمة الشحنة لمجموع إلكترونات التكافؤ.
الأيون ذو شحنة موجبة تطرح قيمة الشحنة من مجموع إلكترونات التكافؤ.
3- نوجد عدد الإلكترونات اللازم للوصول إلى مدار التكافؤ المشابه للغاز الخامل لكل ذرة في الجزئ وليكن رمزه b. (العدد للهيدروجين 2 ولبقية الذرات 8).
4- نوجد عدد الإلكترونات المشتركة في تكوين الروابط وذلك بطرح الخطوة 2 من 3 وليكن رمزه c.
5- نوجد عدد الروابط التساهمية وليكن رمزه d وهو يساوي c/2.
6- نوجد عدد الإلكترونات الحرة (غير المرتبطة) وليكن رمزه e
e = a - c

79. 7- تكتب رموز العناصر وترسم الروابط التساهمية بين كل ذرتين فإذا لم ينتهي المحسوب في الخطوة 5 يتم إضافة روابط مضاعفة إلى أن ينتهي عدد الروابط. (لاحظي أن ذرة الهيدروجين لا تكون أكثر من رابطة) وأخيراً توزع الإلكترونات المفردة على الذرات بحيث يكون حول كل ذرة ثمانية إلكترونات (ما عدا ذرة الهيدروجين حولها إلكترونين).
8- تحسب الشحنة الرسمية لكل ذرة بتطبيق العلاقة
الشحنة الرسمية = رقم المجموعة – (عدد الروابط + عدد الإلكترونات الحرة).

80. (Ex1) Write the Lewis structure of nitrogen trifluoride (NF3).
Step 1 – No. of valence electrons = 5 + (3x7) = 26 valence electrons
Step 2 – No. of stable state electrons = 8 + (3x8) = 32 e
Step 3 – No. of shared electrons = 32 – 26 = 6 e
Step 4 - No. of bonds = 6e ÷ 2 = 3 bonds
Step 5 - No. of non-bonded electrons = 26 – 6 = 20e (10 lone pairs)
3 single bonds (3x2) + 10 lone pairs (10x2) = 26 valence electrons
N is less electronegative than F, put N in center F N

81. 9.7 Formal charge and Lewis structure.

82. ( ) - Formal Charge 1 = 2 Formal Charge of atom =
Formal charge: show how the charge distributed in a molecule
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.
عدد الالكترونات المكونة للرابطة والتي ساهمت بها كلتا الذرتين المكونتين للرابطة
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
عدد الالكترونات المكونة للرابطة والتي ساهمت بها الذرة المراد حساب الشحنة لها
Formal Charge of atom =
No. of valence electrons – total number of electrons (bonding+non-bonding)
The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion.

83. O - - O C O 2- O O C O Example 9.6 p385:
Write the formal charges for the carbonate ion (CO32-). 6
– 4
–(½x4) O
Solution: - - O C O 2- O 6
– 6
–(½x2) -1 6
– 6
–(½x2) -1 O C O
The sum of the formal charges of the atoms in a molecule or ion must equal the charge on the molecule or ion.
-1 -1 = -2 √
4 – 0 –(½x8) = 0

84. The molecule is neutral
Calculate the Formal Charge of Nitrogen triflouride? (continue of EX1) F N
Formal Charge of atom =
No. of valence electrons – total number of electrons (bonding+non-bonding)
Formal charge of (F) atoms = 7 – (1+6) = 0
Formal charge of (N) atoms = 5 – (3+2) = 0
The total charge of the molecule = 0
The molecule is neutral

85. 1- How many lone pair around C atom in CH4?
A. 0 B. 1 C. 3 D. 2
2-What is the formal charge on the oxygen atom in N2O
(the atomic order is N-N-O)?  A B C D.-2
3-In the Lewis structure of the carbonate ions CO3-2,that satisfies the octet rule, the formal charge on the central carbon atom is
A. +2 B. +1 C. 0 D.-1
4-In the Lewis structure of the sulfite ions SO3-2,that satisfies the octet rule, the formal charge on the central sulfur atom is
A. +2 B. +1 C. 0 D.-1

86. Formal Charge and Lewis Structures
If there is more than one possible structure for a molecule!!!
Formal charges help to decide which structure is the correct one, by applying the following:
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.
للجزيئات المتعادلة تركيب لويس الأكثر تفضيلا هو
عندما تكون الشحنة الرسمية لكل ذرة مساوية للصفر.
تركيب لويس الأكثر تفضيلا هو عندما تكون الشحنات الرسمية صغيرة
عندما تحتوي تراكيب لويس على نفس توزيع الشحنة الرسمية فإن تركيب لويس الأكثر تفضيلا
هو عندما تكون الذرة ذات السالبية الكهربية العالية في الجزيء ذات شحنة رسمية سالبة

87. (b) (a) Which is the most likely Lewis structure for CH2O? -1 +1 H C OH C O H C O 4
– 2
–(½x6) -1 6
– 2
–(½x6) +1 6
– 4
–(½x4) 4
– 0
–(½x8)
Structure (b) has no formal charges → Thus: it is the most likely structure of formaldehyde.

88. (Ex2) Write the Lewis structure of formaldehyde (CH2O).
Step 1 – No. of valence electrons = 4+(2x1)+6 = 12 valence electrons
Step 2 – No. of stable state electrons = 8+(2x2)+8 = 20 e
Step 3 – No. of shared electrons = = 8 e
Step 4 - No. of bonds = 8e ÷ 2 = 4 bonds
Step 5 - No. of non-bonded electrons = 12 – 8 = 4e (2 lone pairs) -1 +1 H C O H C O
unacceptable
acceptable
F. C. of (H) atoms = 1 – 1 = 0
F. C. of (H) atoms = 1 – 1 = 0
F. C. of (C) atoms = 4 – 5 = -1
F. C. of (C) atoms = 4 – 4 = 0
F. C. of (O) atoms = 6 – 5 = +1
F. C. of (O) atoms = 6 – 6 = 0

89. The molecule is neutral
(Ex3) Write the Lewis structure of carbon disulphide (CS2).
Step 1 – No. of valence electrons = 4 + (2x6) = 16 valence electrons
Step 2 – No. of stable state electrons = 8 + (2x8) = 24 e
Step 3 – No. of shared electrons = 24 – 16 = 8 e
Step 4 - No. of bonds = 8e ÷ 2 = 4 bonds
Step 5 - No. of non-bonded electrons = 16 – 8 = 8e (4 lone pairs)
C is less electronegative than S, put C in the center
S = C = S
Formal charge of (S) atoms = 6 – (2+4) = 0
Formal charge of (C) atoms = 4 – (4+0) = 0
The total charge of the molecule = 0
The molecule is neutral

90. The total charge of the molecule = 1+
(Ex4) Write the Lewis structure of amonium ion (NH4+).
Step 1 – No. of valence electrons = 5 + (4x1) - 1 = 8 valence electrons
Step 2 – No. of stable state electrons = 8 + (4x2) = 16 e
Step 3 – No. of shared electrons = = 8 e
Step 4 - No. of bonds = 8e ÷ 2 = 4 bonds
Step 5 - No. of non-bonded electrons = 8 – 8 = 0e (no lone pairs)
Formal charge of (H) atoms = 1 – (1+0) = 0
Formal charge of (N) atoms = 5 – (4+0) = +1
The total charge of the molecule = 1+

91. (Ex5) Write the Lewis structure of Ozone (O3).
Step 1 – No. of valence electrons = (3x6) = 18 valence electrons
Step 2 – No. of stable state electrons = (3x8) = 24 e
Step 3 – No. of shared electrons = = 6 e
Step 4 - No. of bonds = 6e ÷ 2 = 3 bonds
Step 5 - No. of non-bonded electrons = 18 – 6 = 12e (6 lone pairs)
Resonance Structures
Formal charge of (O=) atoms = 6 – (2+4) = 0
Formal charge of (-O=) atoms = 6 – (3+2) = +1
Formal charge of (O-) atoms = 6 – (1+6) = -1
The total charge of the molecule = 0
The molecule is neutral

92. (Ex6) Write the Lewis structure of carbonate ion (CO32-).
Step 1 – No. of valence electrons = 4 + (3x6) +2 = 24 valence electrons
Step 2 – No. of stable state electrons = 8 + (3x8) = 32 e
Step 3 – No. of shared electrons = = 8 e
Step 4 - No. of bonds = 8e ÷ 2 = 4 bonds
Step 5 - No. of non-bonded electrons = 24 – 8 = 16e (8 lone pairs)
Formal charge of (C) atoms = 4 – (4+0) = 0
Resonance Structures
Formal charge of (O=) atoms = 6 – (2+4) = 0
Formal charge of (O-) atoms = 6 – (1+6) = -1
Formal charge of (O-) atoms = 6 – (1+6) = -1
The total charge of the molecule = -2

93. 1-The electron Lewis dot structure for HCN shows
three single bonds and 10 lone pairs
one single bond,one triple bond,and 1 lone pair.
two single bonds,and double bond,and 9 lone pair.
two single bonds,and 1 lone pair.
2-The electron dot structure for H2O2 shows
a total of 84 electrons dots.
two single bonds, one double bond, and 9 lone pair.
one single bonds, and two double bond, and one lone pair.
three single bonds, and 4 lone pairs.
3-The electron dot structure for AsCl3 shows 
two single bonds, one double bond, and 9 lone pairs.
three single bonds and 10 lone pairs
one single bond, two double bonds, and 8 lone pairs.
three single bonds and one lone pair.

94. 9.8 The concept of resonance.

95. Resonance Structure of Ozone O3
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.
resonance structure differ only in the arrangement of electrons, not the arrangement of atoms O + - O + -
Resonance Structure of Ozone O3 O C - O C - O C -
Resonance Structure of Carbonate ion CO32-

96. (Ex7) Write the Lewis structure of nitric acid (HNO3).
Step 1 – No. of valence electrons = (3x6) = 24 valence electrons
Step 2 – No. of stable state electrons = (3x8) = 34 e
Step 3 – No. of shared electrons = = 10 e
Step 4 - No. of bonds = 10e ÷ 2 = 5 bonds
Step 5 - No. of non-bonded electrons = 24 – 10 = 14e (7 lone pairs)
Resonance Structures
Formal charge of (N) atoms = 5 – 4 = +1
The total charge of the molecule = 0
Formal charge of (O1) atoms = 6 – 6 = 0
Formal charge of (O2) atoms = 6 – 7 = -1
Formal charge of (O3) atoms = 6 – 6 = 0
The molecule is neutral
Formal charge of (H) atoms = 1 – 1 = 0

97. More stable Not acceptable
(Ex8) Write the Lewis structure of dinitrogen oxide (N2O).
Step 1 – No. of valence electrons = (2x5)+6 = 16 valence electrons
Step 2 – No. of stable state electrons = (2x8)+8 = 24 e
Step 3 – No. of shared electrons = = 8 e
Step 4 - No. of bonds = 8e ÷ 2 = 4 bonds
Step 5 - No. of non-bonded electrons = 16 – 8 = 8e (4 lone pairs) -1 +1 +1 -1 -2 +1 +1
More stable
Not acceptable
because the –ve charge is on the more electronegative oxygen atom
resonance structures for the molecule N2O

98. 9.9 Exceptions to the octet rule.

99. Exceptions to the Octet Rule
1-Incomplete octet: the number of electrons surrounding the central atom in stable molecule is less than 8.
Examples:
AlI3
BeF2
and
Be – 2e-
2H – 2x1e-
4e-
BeH2 H Be
Be is only surrounded by 4 e-
B – 3e-
3F – 3x7e-
24e-
BF3 F B
3 single bonds (3x2) = 6
9 lone pairs (9x2) = 18
Total = 24
B is only surrounded by 6 e-

100. Exceptions to the Octet Rule
2-The Expanded Octet
3rd period and beyond (4th, 5th, 6th, 7th ) may form molecules in which the central atom is surrounded by more than 8 electrons (expanded octet)
Examples: Xe F
4 single bonds (4x2) = 8
14 lone pairs (14x2) = 28
Total = 36
Xe – 8e-
4F – 28e-
36e-
XeF4
Xenon is surrounded by 12 e- P F
P – 5e-
5F – 35e-
40e-
5 single bonds (5x2) = 10
15 lone pairs (15x2) = 30
Total = 40
PF5
Phosphorus is surrounded by 10 e-

101. sulfur is surrounded by 12 e-
S – 6e-
6F – 42e-
48e-
6 single bonds (6x2) = 12
18 lone pairs (18x2) = 36
Total = 48
SF6
sulfur is surrounded by 12 e-

102. Exceptions to the Octet Rule 3-Odd-Electrons Molecules
Odd-electrons molecules = RADICALS
Radical: atom has one electron alone
Examples:
N – 5e-
O – 6e-
11e- NO N O
Total electrons
N – 5e-
2O – 12e-
17e-
O = N + - O -
NO2

103. Problems
9.16 – 9.18 – 9.30 – 9.36 – 9.74
9.44 – 9.46 – 9.48
9.52 – 9.56 – 9.64 – 9.80