1. The Basics of Chemical Bonding CHAPTER 9 Chemistry: The Molecular Nature of Matter, 6 th edition By Jesperson, Brady, & Hyslop

2. CHAPTER 9: Basics of Chemical Bonding 2 Learning Objectives  … Learning Objectives  Communicate the difference between ionic and covalent bonding.  Predict which ionic compounds have relatively larger lattice energies  Predict ionic compounds  Use the Octet Rule  Familiarity with common covalent molecules: organic molecules  Draw lewis dot structures for covalent molecules  Utilize multiple bonds  Know the exceptions to the octet rule  Predict electronegativity of a bond and overall dipole moment  Recognize and create reasonable resonance structures for molecules

3. 3 Covalent Compounds Form individual separate molecules – Atoms bound by sharing electrons Do not conduct electricity Often low melting point Covalent Bonds Shared pairs of electrons between two atoms Two H atoms come together, why?

4. Covalent Bonds Ionic vs Covalent Bonds Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 4 Ionic Bonds result from electrostatic attraction between a cation and anion: metal-nonmetal (with the exception of NH 4 + and H 3 O + cations). Covalent bonds result from the sharing of electrons between two atoms: nonmetal-nonmetal. Li F Ionic Bonds Covalent Bonds

5. 5 Covalent Bond Attraction of valence electrons of one atom by nucleus of other atom Shifting of electron density As distance between nuclei decreases, probability of finding either electron near either nucleus increases Pulls nuclei closer together

6. 6 Covalent Bond As nuclei get close – Begin to repel each other – Both have high positive charge  Final internuclear distance between two atoms in bond  Balance of attractive and repulsive forces  Bond forms since there is a net attraction

7. 7 Covalent Bond Two quantities characterize this bond Bond Length (bond distance) – Distance between 2 nuclei = r A + r B Bond Energy – Also bond strength – Amount of energy released when bond formed (decreasing PE) or – Amount of energy must put in to “break” bond

8. 8 Lewis Structures Molecular formula drawn with Lewis Symbols Method for diagramming electronic structure of covalent bonds Uses dots to represent electrons Covalent bond – Shared pair of electrons – Each atom shares electrons so has complete octet ns 2 np 6 Noble gas electron configuration Except H which has complete shell with 2 electrons

9. 9 Octet Rule: When atoms form covalent bonds, they tend to share sufficient electrons so as to achieve outer shell having eight electrons – Indicates how all atoms in molecule are attached to one another – Accounts for ALL valence electrons in ALL atoms in molecule Let’s look at some examples Noble Gases: eight valence electrons – Full octet ns 2 np 6 – Stable monatomic gases – Don’t form compounds

10. Covalent Bonds Lewis Dot Structures Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 10 Diatomic Gases: H and Halogens H 2 H· + ·H  H: H or H  H Each H has two electrons through sharing Can write shared pair of electrons as a line (  ) : or  signify a covalent bond

11. Covalent Bonds Lewis Dot Structures Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 11 Many nonmetals form more than one covalent bond Needs 4 electrons Forms 4 bonds Needs 3 electrons Forms 3 bonds Needs 2 electrons Forms 2 bonds methaneammoniawater

12. Covalent Bonds Lewis Dot Structures: Multiple Bonds Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 12 Double Bond Two atoms share two pairs of electrons e.g. CO 2 Triple bond Three pairs of electrons shared between two atoms e.g. N 2

13. Covalent Bonds Lewis Dot Structures Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 13 Diatomic Gases: F2F2 Each F has complete octet Only need to form one bond to complete octet Pairs of electrons not included in covalent bond are called lone pairs Same for rest of halogens: Cl 2, Br 2, I 2

14. Covalent Bonds Lewis Dot Structures Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 14 Diatomic Gases: HF Same for HCl, HBr, HI Molecules are diatomics of atoms that need only one electron to complete octet Separate molecules –Gas in most cases because very weak intermolecular forces

15. Covalent Bonds Lewis Dot Structures: Method Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 15 Step [1] Arrange the atoms next to each other that you think are bonded together. Place H and halogens on the periphery, since they can only form one bond. NH 3 NHHH

16. Covalent Bonds Lewis Dot Structures: Method Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 16 Step [2] Count the valence electrons. The sum gives the total number of e − that must be used in the Lewis structure. For each atom the number of bonds = 8 – valence electrons. Nitrogen has 5 valence electrons, so it will have 8 – 5 = 3 bonds. Hydrogen will have 2-1 = 1 bond. There are 8 total valance electrons NH 3 NHHH

17. Covalent Bonds Lewis Dot Structures: Method Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 17 Step [3] Arrange the electrons around the atoms. Place one bond (two e − ) between every two atoms. Use all remaining electrons to fill octets with lone pairs, beginning with atoms on the periphery. HNHHHNHH 1 lone pair: 2 3 bonds: 6 Total e-8 = total valence e-

18. Covalent Bonds Lewis Dot Structures: Method Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 18 1.Decide how atoms are bonded – Skeletal structure = arrangement of atoms. – Central atom Usually given first Usually least electronegative 2.Count all valence electrons (all atoms) 3.Place two electrons between each pair of atoms – Draw in single bonds 4.Complete octets of terminal atoms (atoms attached to central atom) by adding electrons in pairs 5.Place any remaining electrons on central atom in pairs 6.If central atom does not have octet – Form double bonds – If necessary, form triple bonds

19. Covalent Bonds Lewis Dot Structures: Ex: SiF 4 Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 19 1 Si = 1  4 e – = 4 e – 4 F = 4  7 e – = 28 e – Total = 32 e – single bonds – 8 e – 24 e – F lone pairs – 24 e – 0 e – Skeletal Structure Complete terminal atom octets

20. Covalent Bonds Covalent Compounds: Organic Molecules Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 20 Organic Molecules contain C-H bonds Carbon containing compouns – Exist in large variety – Mostly due to multiple ways in which C can form bonds Functional groups – Groups of atoms with similar bonding – Commonly seen in C compounds Molecules may contain more than one functional group

21. Covalent Bonds Organic Molecules: Carbon Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 21 Alkanes – Hydrocarbons – Only single bonds Isomers – Same molecular formula – Different physical properties – Different connectivity (structure) e.g. CH 4 methane CH 3 CH 3 ethane CH 3 CH 2 CH 3 propane iso-butanebutane

22. Covalent Bonds Organic Molecules: Hydrocarbons Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 22 Alkenes – Contain at least one double bond Alkynes – Contain at least one triple bond ethylene (ethene) butene acetylene (ethyne) butyne

23. Covalent Bonds Organic Molecules: Oxygen Containing Organics Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 23 Alcohols – Replace H with OH Ketones – Replace CH 2 with C=O – Carbonyl group methanolethanol acetone Aldehydes Ketones Carboxylic acids

24. Covalent Bonds Organic Molecules: Oxygen Containing Organics Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 24 Aldehydes – At least one atom attached to C=O is H Organic Acids – Contains carboxyl group – COOH acetaldehyde acetic acid

25. Covalent Bonds Organic Molecules: Nitrogen Containing Organics Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 25 Amines – Derivatives of NH 3 with H’s replaced by alkyl groups methylaminedimethylamine

26. Problem Set B 1.…..