Chapter 10: Chemical Bonding I: Basic Concepts

1. Petrucci • Harwood • Herring • Madura GENERAL Ninth Edition CHEMISTRY Principles and Modern Applications Chapter 10: Chemical Bonding I:Basic Concepts General Chemistry: Chapter 10

2. 10-1 Lewis Theory: An Overview • Valence e- play a fundamental role in chemical bonding. • e- transfer leads to ionic bonds (metal-non metals) • Sharing of e- leads to covalent bonds (two non-metals) • e- are transferred of shared to give each atom a noble gas configuration • the octet. General Chemistry: Chapter 10

3. •• •• •• •• •• As Bi P Sb N • • • • • • • • • • • • • • • •• •• •• Al Ar • Se •• I • •• • • •• • • •• • •• Lewis Symbols • A chemical symbol represents the nucleus and the coree- (inner shell) • Dots around the symbol represent valence e-. • Si • • • General Chemistry: Chapter 10

4. Lewis Structure (Dot and cross diagram) • A lewis structure is a combination of Lewis symbols that represents either the transfer or the sharing of electrons in a chemical bond.

5. EXAMPLE 10-2 • •• 2+ 2- •• Ba O Ba O •• • • • •• •• •• Writing Lewis Structures of Ionic Compounds. Write Lewis structures for the following compounds: (a) BaO; (b) MgCl2 ; (c) aluminum oxide. BaO Note the use of the “fishhook” arrow to denote a single electron movement. A “double headed” arrow means that two electrons move. General Chemistry: Chapter 10

6. 10-2 Covalent Bonding: An Introduction General Chemistry: Chapter 10

7. + H H - •• N H Cl •• •• H N H •• •• H H Coordinate Covalent Bonds A covalent bond in which a single atom contributes both of the electrons to a shared pair. Cl H • Note • the “double headed” arrow showing that two electrons move. • Any atom with electron pairs (non-metals: O, S, P) can form • coordinate covalent bonds. General Chemistry: Chapter 10

8. • • • •• •• C O O •• •• • •• •• • • • •• •• C O O C O O •• •• •• •• • Multiple Covalent Bonds • • • O C O • • • • • • • • • • • • • General Chemistry: Chapter 10

9. • • N N •• •• • • • N N N N •• •• •• •• • Multiple Covalent Bonds • • •• N N •• • • • • General Chemistry: Chapter 10

10. Paramagnetism of OxygenFailure of the Lewis Structure Model Explain by: Molecular Orbital Theory General Chemistry: Chapter 10

11. Practice Examples

12. 10-3 Polar Covalent Bonds and Electrostatic Potential Maps A covalent bond in which electrons are not shared equally between two atoms General Chemistry: Chapter 10

13. Electronegativity (EN) • It is an atoms ability to compete for electrons with other atoms to which it is bonded. • The lower the EN, the more metallic the element is • The higher the EN, the more nonmetallic the element is General Chemistry: Chapter 10

14. Percent Ionic Character Electronegativity difference, DEN = the absolute value of the difference in EN values of the bonded atoms DEN >1.5 (Ionic bond) 0.3<DEN <1.5 (polar covalent bond) DEN<0.3 (covalent bond) General Chemistry: Chapter 10

15. Practice Questions

16. 10-4 Writing Lewis Structures • All the valence e- of atoms must appear. • Usually, the e- are paired. • Usually, each atom requires an octet. • H only requires 2 e-. • Multiple bonds may be needed. • Readily formed by C, N, O, S, and P. General Chemistry: Chapter 10

17. Skeletal Structure • Identify central and terminal atoms.

18. Skeletal Structure • Hydrogen atoms are always terminal atoms. • Central atoms are generally those with the lowest electronegativity. • O is only central in peroxo (-O-O-) or a hydroxy go (-O-H) • Carbon atoms are always central atoms (organic molecules) • Generally structures are compact and symmetrical. General Chemistry: Chapter 10

19. Strategy for Writing Lewis Structures General Chemistry: Chapter 10

20. EXAMPLE10-7 •• •• Step 4: •• Add e- to terminal atoms: O—N—O •• •• •• Writing a Lewis Structure for a Polyatomic Ion. Write the Lewis structure for the nitronium ion, NO2+. Step 1: Total valence e- = 5 + 6 + 6 – 1 = 16 e- Step 2: Identify the central and terminal atoms Step 3: Plausible structure: O—N—O General Chemistry: Chapter 10

21. EXAMPLE 10-7 •• •• O=N=O •• •• •• •• •• O—N—O •• •• •• Step 5: Determine e- left over: 16 – 4 – 12 = 0 Step 6: Use multiple bonds to satisfy octets. + General Chemistry: Chapter 10

22. Practice Problems

23. •• •• O=N=O •• •• 1 FC(O) = 6 - 4 – (4) = 0 2 1 FC(N) = 5 - 0 – (8) = +1 2 Formal Charge These are apparent charges on certain atoms in a Lewis structure that arise when atoms have not contributed equal # of electrons to the covalent bonds joining them FC = # of valence e- in free atom - # of valence lone-pair e- - 1/2( # of bond-pair e-) + General Chemistry: Chapter 10

24. •• •• •• O N O •• 1 FC(O≡) = 6 - 2 – (6) = +1 2 1 FC(O—) = 6 - 6 – (2) = -1 2 1 FC(N) = 5 - 0 – (8) = +1 2 Alternative Lewis Structure + - •• + •• O—N—O •• •• •• •• General Chemistry: Chapter 10

25. - •• + + O≡N—O •• •• •• Alternative Lewis Structures • Sum of FC is the overall charge. • FC should be as small as possible. • Negative FC usually on most electronegative elements. • FC of same sign on adjacent atoms is unlikely. General Chemistry: Chapter 10