1. Catherine MacGowan Armstrong Atlantic State University Chapter 9 Chemical Bonding I: The Lewis Model © 2013 Pearson Education, Inc. Lecture Presentation

2.  It is a measure of the ability of an atom in a bond to attract electrons to itself.  This attraction or pulling of electrons causes a separation of charge within the bond.  Dipole moment is formed.  Symbol:   The greater the difference, the more POLAR the bond. What Is Electronegativity (  )?

3. Electronegativity and Bond Polarity If the difference in electronegativity between bonded atoms is: – Zero (0), then the bond is pure covalent. Equal sharing of electrons between atoms – 0.1 to 0.4, then the bond is nonpolar covalent. – 0.5 to 1.9, then the bond is polar covalent. – ≥2.0, then the bond is ionic. Transfer of electrons between atoms 100% 00.42.04.0 4%51% Percent ionic character Electronegativity difference

4. | SectionChapter | Periodic Trends: Electronegativity 4 Periodicity of the Elements 1 4 Increasing electronegativity

5. Electronegativity and Bond Types

6. Illustration of Main Types of Intramolecular Bonds

7. Ionic Bonding Occurs between a metal and a nonmetal element.Occurs between a metal and a nonmetal element. Transfer of electron(s) from one element to another.Transfer of electron(s) from one element to another. Cation/anion formationCation/anion formation Strength of bond can be calculated using Coulomb’s law.Strength of bond can be calculated using Coulomb’s law. E = kQ1Q2E = kQ1Q2 RR RR Ionic bonding results in the attraction of the between the two charges (cation + anion).Ionic bonding results in the attraction of the between the two charges (cation + anion). Q2Q1

8. What Is Covalent Bonding? The bond arises from the mutual attraction of two nuclei for the same electrons. The bond arises from the mutual attraction of two nuclei for the same electrons. – Electron sharing results. The bond is a balance between: The bond is a balance between: – Attractive forces (p+ to e-) (p+ to e-) – Repulsive forces (e- to e-) (e- to e-) (p+ to p+) (p+ to p+) Bond can result from a “head-to-head” overlap of atomic orbitals on neighboring atoms. Bond can result from a “head-to-head” overlap of atomic orbitals on neighboring atoms. Occurs between nonmetal elements Occurs between nonmetal elements

9. Sigma Bond Formation by Orbital Overlap

12. Measured by the energy required to break a bond Bond Bond dissociation enthalpy (kJ/mol) H - H436 C - C346 C=C602 C = C 835 N = N945 The GREATER the number of bonds (bond order), the HIGHER the bond strength and the SHORTER the bond. The GREATER the number of bonds (bond order), the HIGHER the bond strength and the SHORTER the bond. Bond Energies: Strength

13. Decide on the central atom; never H. Therefore, N is central. Count the number of valence electrons each atom is bringing. H = 1 but you have 3 H atoms so, 3 x 1 = 3 electrons N = 5 Total = 3 + 5 = 8 electrons Eight valence electrons need to be accounted for in the Lewis structure such that H has 2 electrons or 1 bonding pair and N has a total of 8 valence electrons surrounding it. 1 LONE PAIR – 3 BOND PAIRS and 1 LONE PAIR H N H H HN H H NH 3 : Ammonia’s Lewis Dot Structure

14. Central atom = B (6 electrons surrounding max.) – The B atom has a share in only 6 electrons or 3 pairs. – B atom in many molecules is electron deficient Total valence electrons in molecule = 24 B atom is an exception to octet “rule.” Boron Trifluoride, BF 3

15. Why? According to quantum mechanics (quantum numbers) elements in the 3rd period can have more than 8 valence electrons if center because “3d” predicted by the quantum number ( l ) Step 1: Center atom is S Step 2: Count number valence electrons F: 7 x 4 = 28 F: 7 x 4 = 28 S: 6 x 1 = 6 S: 6 x 1 = 6 Total valence electrons = 34 Step 3: Make bonds F only single bonds F only single bonds S must have 8 valence electrons S must have 8 valence electrons BUT can have > 8 valence electrons if center atom BUT can have > 8 valence electrons if center atom *Usually nonmetal elements of periods 3 and greater SF 4 : Exception to “Octet Rule”

16. Central atom = C Total valence electrons = 16 Place lone pairs on outer atoms. NOTE NOTE: For carbon and oxygen to have 8 valence electrons, need to form double bond Carbon Dioxide, CO 2

17. Central atom is oxygen. O – O – O Total valence electrons for this molecule is 18. Oxygen needs to form a double bond so that it has 8 electrons around it. This leads to the following structures; both satisfy the octet guidelines. These equivalent structures are called RESONANCE STRUCTURES. These equivalent structures are called RESONANCE STRUCTURES. The true electronic structure is a HYBRID of the two. The true electronic structure is a HYBRID of the two. Ozone, O 3

18. Total number of electrons: 24 – N atom brings in 5 electrons – O atom brings in 6 each You have 3 oxygen atoms so 6 x 3 = 18 electrons – Anion = 1 electron The N atom and O atom must have 8 electrons surrounding each of them. Three legitimate Lewis dot structures for the nitrate ion can be drawn. Three legitimate Lewis dot structures for the nitrate ion can be drawn. – Resonance Nitrate ion, NO 3 -

19. CO SeOF 2 NO 2 − H 3 PO 4 SO 3 2− P 2 H 4 Practice Problems: Lewis Structures

20. CO (10 e - ) : c = o : SeOF 2 (26 e - ) NO 2 − (18 e - ) Answers:

21. H 3 PO 4 (32 e - ) SO 3 2− (26 e - ) P 2 H 4 (14 e - ) Answers:

22. H 3 PO 4 (32 e - ) SO 3 2− (26 e - ) P 2 H 4 (14 e - ) Answers:

23. Problem: Predict the most stable structure: ONC - or OCN - or NOC -............. : O :: N :: C : : O :: C :: N : : N :: O :: C : : O :: N :: C : : O :: C :: N : : N :: O :: C : (A) (B) (C) Formal charge: (# valence electrons) - (# of bonds) - (# nonbonding electrons) (A) (B) (C) O6 – 2 - 4 = 06 – 2 - 4 = 06 – 4 - 0 = +2 N5 – 4 - 0 = +1 5 – 2 - 4 = -1 5 – 2 - 4 = -1 C4 – 2 - 4 = -2 4 - 2 - 0 = 0 4 – 2 - 4 = -2 NOTE: Structure B has the combination with the lowest formal charge, and it has the negative formal charge on one of the more electronegative atoms.

24. Sigma Bond Formation by Orbital Overlap

25. Measured by the energy required to break a bond Bond Bond dissociation enthalpy (kJ/mol) H - H436 C - C346 C=C602 C = C 835 N = N945 The GREATER the number of bonds (bond order), the HIGHER the bond strength and the SHORTER the bond. The GREATER the number of bonds (bond order), the HIGHER the bond strength and the SHORTER the bond. Bond Energies: Strength

26. Net energy is equal to the ∆H rxn. ∆H rxn can be calculated using the following equation: Breaking bonds REQUIRES energy (use + value of bond energies). Making bonds RELEASES energy (use the - value of bond energies). Breakage of weak bonds and the formation of stronger bonds: exothermic reaction occurs. Breakage of strong bonds and the formation of weak bonds: endothermic reaction occurs. Using Bond Dissociation Enthalpies

27. Problem: Find the  H rxn for CH 4 (g) + Cl 2 (g)  HCl(g) + CH 3 Cl(g) Average Bond Energies Given:H – H436 kJ/molCl - Cl242 kJ/mol H – Cl432 kJ/molH – C414 kJ/mol Cl – Cl339 kJ/mol ∆H rxn =  energy required to break bonds +  energy evolved when bonds are made ∆H rxn = {(4 mole x 414 kJ/mol) + (1 mol x (242 kJ/mol)} + {(1mol x - 432 kJ/mol) + (3 mol x - 414 kJ/mol) + (1 mole x -339 kJ/mol )} = 1898 kJ + (-2013 kJ) = - 115 kJ This reaction is exothermic!