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  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. Bonds Definition Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 3 Bonds are attractive forces that hold atoms together in complex substances. Changes to bonding forces = chemical reactivity Breaking bonds & forming new bonds Covalent Bonds Bonding in molecules Electrons shared between 2 atoms Ionic Bonds Bonding in ionic compounds Electrons are not shared Bonds are electrostatic attractions between oppositely charged atoms/molecules

4. Ionic Bonds Definition Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 4 Na + Cl – Ionic Bonds result from attractive forces between oppositely charged particles: Cation Anion Usually Metal-Nonmetal bonds (Metal = cation, Nonmetal = anion), but also the bond between polyatomic cations & anions. Metal-Nonmetal: - Metals have - Low ionization energies - Easily lose electrons to be stable - Non-metals have very exothermic electron affinities  Formation of lattice stabilizes ions

5. Ionic Bonds Ionic Compounds Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 5 Ionic crystals: Exist in 3-dimensional array of cations and anions called a lattice structure Ionic chemical formulas: Always written as empirical formula Smallest whole number ratio of cation to anion Na + + Cl -  NaCl (s) Attraction between + and – ions in ionic compound.

6. Ionic Bonds Energetics Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 6 For any stable compound to form from its elements – Potential energy of system must be lowered. – Net decrease in energy  H f ° < 0 (negative) Can think of the formation of an ionic bond as an electron transfer:

7. Ionic Bonds Energetics: Hess’s Law Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 7 1. Single step Na(s) + ½Cl 2 (g)  NaCl(s)  H f ° = – 411.1 kJ/mol 2. Stepwise path Na(s)  Na(g)  H f °(Na, g) = 107.8 kJ/mol ½Cl 2 (g)  Cl(g)  H f °(Cl, g) = 121.3 kJ/mol Na(g)  Na + (g) + e – IE(Na) = 495.4 kJ/mol Cl(g) + e –  Cl – (g) EA(Cl) = – 348.8 kJ/mol Na + (g) + Cl – (g)  NaCl(s)  H lattice = – 787 kJ/mol  –––  Na(s) + ½Cl 2 (g)  NaCl(s)  H f ° = – 411 kJ/mol

8. Ionic Bonds Energetics: Hess’s Law Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 8

9. Ionic Bonds Energetics: Lattice Energy Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 9 Amount that PE of system decreases when one mole of solid salt is formed from its gas phase ions. Energy released when ionic lattice forms.

10. Ionic Bonds Energetics: Lattice Energy Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 10

11. Ionic Bonds Energetics: Lattice Energy Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 11 Always  H Lattice = – therefore, exothermic  H Lattice gets more exothermic (larger negative value) as ions of opposite charge in crystal lattice are brought closer together as they wish to be. Ions tightly packed with opposite charged ions next to each other. Any increase in PE due to ionizing atoms is more than met by decrease in PE from formation of crystal lattice. Even for +2 and – 2 ions Therefore, forming ionic solids is an overall exothermic process and they are stable compounds.

12. Ionic Bonds Energetics: Lattice Energy Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 12 Coulomb’s law determines the potential energy of two ions (q 1 and q 2 ) separated by a distance (r), where k is a proportionality constant. The charges on the ions & their size affect lattice energies – As q 1 and q 2 get larger E becomes more negative & potential energy lower – As ionic radii decrease, distance decreases, and E becomes more negative & potential energy decreases

13. Ionic Bonds Cations & Anions Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 13 Metal Left hand side of Periodic Table IE small and positive – Little energy required to remove electrons EA small and negative or positive – Not favorable to attract an electron to it. Least expensive, energy- wise, to form cation Nonmetal Right hand side of Periodic Table IE large and positive – Difficult to remove e – EA large and negative – But easy to add e – – Exothermic—large amount of energy given off – PE of system decreases Least expensive, energy- wise, to form anion

14. Ionic Bonds Ion Electron Configurations Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 14 How electronic structure affects types of ions formed: Na1s 2 2s 2 2p 6 3s 1 = [Ne] 3s 1 Na + 1s 2 2s 2 2p 6 = [Ne] IE 1 = 496 kJ/mol small not too difficult IE 2 = 4563 kJ/mol large ~10 x larger very difficult Can remove first electron, as doesn't cost too much Can’t remove second electron, as can't regain lost energy from lattice Thus, Na 2+ doesn’t form.

15. Ionic Bonds Ion Electron Configurations Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 15 Ca [Ar] 4s 2 Ca 2+ [Ar] IE 1 small = 590 kJ/molnot too difficult IE 2 small = 1140 kJ/molnot too difficult IE 3 large = 4940 kJ/moltoo difficult Can regain by lattice energy ~2000 kJ/mole if +2, –2 charges. But third electron is too hard to remove Can't recoup required energy through lattice formation. Therefore Ca 3+ doesn't form

16. Ionic Bonds Ion Electron Configurations Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 16 Stability of noble gas core above or below the valence electrons effectively limits the number of electrons that metals lose. Ions formed have noble gas electron configuration – True for anions and cations Cl 1s 2 2s 2 2p 6 3s 2 3p 5 = [Ne]3s 2 3p 5 Cl – 1s 2 2s 2 2p 6 3s 2 3p 6 = [Ar] Adding another electron – Requires putting it into next higher n shell Energy cost too high

17. Ionic Bonds Ion Electron Configurations Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 17 O1s 2 2s 2 2p 4 O – 1s 2 2s 2 2p 5 EA 1 = –141 kJ/mol O 2– 1s 2 2s 2 2p 6 = [Ne] EA 2 = +844 kJ/mol EA net = +703 kJ/mol endothermic Energy required to form cation is more than made up for by the increase in  H Lattice caused by higher –2 charge

18. Ionic Bonds Ion Electron Configurations Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 18 Generalization: When ions form… – Atoms of most representative elements (s and p block) – Tend to gain or lose electrons to obtain nearest Noble gas electron configuration – Except He (two electrons), all noble gases have eight electrons in highest n shell Octet Rule Atoms tend to gain or lose electrons until they have achieved outer (valence) shell containing octet of eight electrons

19. Ionic Bonds Octet Rule Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 19 Works well with – Group 1A and 2A metals – Al – Non-metals H and He obey Duet Rule – Limited to 2 electrons in the n = 1 shell Doesn't work with – Transition metals – Post transition metals

20. Ionic Bonds Octet Rule: Example Selenium Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 20 1 2 X X X X X X X X X X 3 4 5 6-1-2 6 valence electrons Apply octet rule: 8 – 6 = 2 Therefore, Selenium will have a -2 charge Se -2

21. Ionic Bonds Transition Metals Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 21 First electrons are lost from outermost s orbital Lose electrons from highest n first, then ℓ e.g.Fe [Ar] 3d 6 4s 2 Fe 2+ [Ar] 3d 6 loses 4s electrons first Fe 3+ [Ar] 3d 5 then loses 3d electrons Extra stability due to half-filled d subshells. Consequences –M 2+ is a common oxidation state as two electrons are removed from the outer ns shell –Ions of larger charge result from loss of d electrons

22. Ionic Bonds Transition Metals Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 22 Not easy to predict which ions form and which are stable But ions with exactly filled or half-filled d subshells are extra stable and therefore tend to form. Mn 2+ [Ar]3d 5 Fe 3+ [Ar]3d 5 Zn 2+ [Ar]3d 10

23. Ionic Bonds Post-Transition Metals Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 23 E.g. Sn [Kr] 4d 10 5s 2 5p 2 Sn 2+ [Kr] 4d 10 5s 2 – Neither has noble gas electron configuration – Have emptied 5p subshell Sn 4+ [Kr] 4d 10 – Does have empty 5s subshell

24. Ionic Bonds Summary Predicting Cation Configurations Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 24 Aufbau Principle: Electron configuration based on “filling” an atom with electrons. Follows order in the periodic table. Consider Bi, whose aufbau configuration is: [Xe] 6s 2 4f 14 5d 10 6p 3. What ions are expected? Rewrite configuration: [Xe] 4f 14 5d 10 6s 2 6p 3 Bi 3+ and Bi 5+ Consider Fe, whose aufbau configuration is: [Ar] 4s 2 3d 6. What ions are expected? Rewrite configuration: [Ar]3d 6 4s 2 Fe 2+ and Fe 3+

25. Ionic Bonds Summary Predicting Anion Configurations Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 25 Non-metals gain electrons to become isoelectronic with next larger noble gas O: [He]2s 2 2p 4 + ? e – → ? N: [He]2s 2 2p 3 + ? e – → ?

26. Ionic Bonds Lewis Symbols (Electron-Dot Symbols) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 26 Electron bookkeeping method Way to keep track of e – ’s Write chemical symbol surrounded by dots for each e – Group #1A2A3A4A Valence e – 1234 e – conf.ns 1 ns 2 ns 2 np 1 ns 2 np 2

27. Ionic Bonds Lewis Symbols (Electron-Dot Symbols) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 27 Group #5A6A7A8A Valence e - 5678 e - conf.ns 2 np 3 ns 2 np 4 ns 2 np 5 ns 2 np 6 For the representative elements Group number = number of valence e – ’s

28. Ionic Bonds Lewis Symbols (Electron-Dot Symbols) Jesperson, Brady, Hyslop. Chemistry: The Molecular Nature of Matter, 6E 28 Can use to diagram electron transfer in ionic bonding

29. Problem Set A 1.Identify the covalent and ionic bonds: a.CaF 2 b.CCl 4 c.NaOH d.NH 4 NO 3 2.Which ionic solid is likely to have the smallest exothermic lattice energy? LiCl; CsCl; NaCl, KCl 3.What ion will form for each element? Draw the elements lewis symbol and write the ion’s electron configuration. a.I b.Rb c.P 4.Predict the ionic compound that will form between the following: a. Aluminum (Al) and Chlorine (Cl) b.Strontium (Sr) and Bromine (Br)