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

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 .

9.1-Lewis dot symbols .

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 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.

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.

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 

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.

9.2 The ionic bond .

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

Q: Use Lewis dot symbols to show the formation of Li2O 2- 2 Li + O 1s22s1 1s22s22p4 1s2 1s22s22p6 [He] [Ne] 2e- + O 2- 2 Li 2Li+ + 2e- O 2- 2Li+ + 2Li+

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

K F

K F

K F

K F

K F

+ _ K F

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

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

9.4 The covalent bond .

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

Covalent Bonds

Chlorine forms a covalent bond with itself Cl2

How will two chlorine atoms react? Cl Cl

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

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?

Cl Cl

Cl Cl

Cl Cl

Cl Cl

Cl Cl octet

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

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

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

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

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

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

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

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

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

Oxygen is also one of the diatomic molecules

How will two oxygen atoms bond?

Each atom has two unpaired electrons

O

O

O

O

O

O

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

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

O

O O

O O

O O

Both electron pairs are shared.

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

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

O O two bonding pairs, making a double bond

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

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

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

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

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

9.5 Electronegativity .

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

The Electronegativities of Common Elements

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

Variation of Electronegativity with Atomic Number

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-

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

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 =2.5-2.5=0 covalent d) CsCl =3-1=2 Ionic e) H2S =2.5-2.1=0.4 Polar covalent f) N-N =3-3=0 Covalent g)Si-Cl in Cl3SiSiCl3=3-1.2=.1.8

9.6 Writing Lewis structure.

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.

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

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).

(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

9.7 Formal charge and Lewis structure.

( ) - 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.

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

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

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.+1 B.0 C.-1 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

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

(b) (a) Which is the most likely Lewis structure for CH2O? -1 +1 H C O H 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.

(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 = 20 - 12 = 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

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

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 = 16 - 8 = 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+

(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 = 24 - 18 = 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

(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 = 32 - 24 = 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

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.

9.8 The concept of resonance.

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-

(Ex7) Write the Lewis structure of nitric acid (HNO3). Step 1 – No. of valence electrons = 1 + 5 + (3x6) = 24 valence electrons Step 2 – No. of stable state electrons = 2 + 8 + (3x8) = 34 e Step 3 – No. of shared electrons = 34 - 24 = 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

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 = 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) -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

9.9 Exceptions to the octet rule.

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-

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-

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-

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

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