1. Chem 59-250 Introductory Inorganic Chemistry What is Inorganic Chemistry?

2. Chem 59-250

6. As: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 3

7. Chem 59-250 For more information about these periodic tables visit the site where I obtained the pictures: http://chemlab.pc.maricopa.edu/periodic/default.html

8. Chem 59-250

9. Classes of Inorganic Substances ElementsIonic CompoundsCovalent Compounds Atomic/Molecular Gases Ar, N 2 Simple (binary) NaCl Simple (binary) NH 3, H 2 O, SO 2 Molecular Solids P 4, S 8, C 60 Complex (polyatomic ions) Na 2 (SO 4 ) Complex (polyatomic) As(C 6 H 5 ) 3, organometallic compounds Network Solids diamond, graphite (C  ) “red” phosphorus (P  ) Network ions Mg 3 (Si 2 O 5 )(OH) 2 (talc) Network Solids SiO 2, polymers Solid/Liquid Metals Hg, Ga, Na, Fe, Mg

10. Chem 59-250 Elements Atomic/Molecular Gases Ar, N 2, O 2, Br 2 Molecular Solids P 4, S 8, C 60 Network Solids diamond, graphite (C  ) “red” phosphorus (P  ) Solid/Liquid Metals Hg, Ga, Fe, Na, Mg

11. Chem 59-250 Ionic Compounds Simple (binary) NaCl Complex (polyatomic ions) Na 2 (SO 4 ), Na 2 Mg(SO 4 ) 2 Network ions Mg 3 (Si 4 O 10 )(OH) 2 (talc)

12. Chem 59-250 Covalent Compounds Simple Molecular (binary) NH 3, H 2 O, CO 2, SO 2 Complex Molecular As(C 6 H 5 ) 3, organometallic compounds Network Solids SiO 2, polymers

13. Chem 59-250 Review of Concepts Thermochemistry: Standard state: 298.15 K, 1 atm, unit concentration Enthalpy Change,  H°  H° =  H° products -  H° reactants Entropy Change,  S° Free Energy Change,  G  G =  H - T  S At STP:  G° =  H° - (298.15 K)  S°

14. Chem 59-250 Standard Enthalpy of Formation,  H° f  H° for the formation of a substance from its constituent elements Standard Enthalpy of Fusion,  H° fus Na (s)  Na (l) Standard Enthalpy of Vapourization,  H° vap Br 2(l)  Br 2(g) Standard Enthalpy of Sublimation,  H° sub P 4(s)  P 4(g) Standard Enthalpy of Dissociation,  H° d ½ Cl 2(g)  Cl (g) Standard Enthalpy of Solvation,  H° sol Na + (g)  Na + (aq)

15. Chem 59-250 Ionization Enthalpy,  H° ie The enthalpy change for ionization by loss of electron(s) Na (g)  Na + (g) + e -  H° ie = 502 kJ/mol Al (g)  Al + (g) + e -  H° ie = 578 kJ/mol Al + (g)  Al 2+ (g) + e -  H° ie = 1817 kJ/mol Al 2+ (g)  Al 3+ (g) + e -  H° ie = 2745 kJ/mol Thus: Al (g)  Al 3+ (g) + e -  H° ie = 5140 kJ/mol

16. Chem 59-250

19. Electron Attachment Enthalpy,  H° ea The enthalpy change for the gain of an electron Cl (g) + e -  Cl - (g)  H° ea = -349 kJ/mol O (g) + e -  O - (g)  H° ea = -142 kJ/mol O - (g) + e -  O 2- (g)  H° ea = 844 kJ/mol Electron Affinity, EA = -  H° ea + 5/2 RT EA = -  H° ea

20. Chem 59-250 Not easy to measure so many are missing

21. Chem 59-250 Why should we care about these enthalpies? They will provide us information about the strength of bonding in solids. NaCl (s) Na (s)  Na (g)  Na + (g) ½ Cl 2(g)  Cl (g)  Cl - (g)  H° ea H°dH°d  H° ie  H° sub H°fH°f Lattice Energy, U

22. Chem 59-250 Bond Energy, E A-B Diatomic: H-Cl (g)  H (g) + Cl (g)  H = 431 kJ/mol Polyatomic: H-O-H (g)  H (g) + O-H (g)  H = 497 kJ/mol O-H (g)  H (g) + O (g)  H = 421 kJ/mol Thus: H-O-H (g)  2 H (g) + O (g)  H = 918 kJ/mol Average O-H bond energy = 918 / 2 E O-H = 459 kJ/mol

23. Chem 59-250 H 2 N-NH 2(g)  4 H (g) + 2 N (g)  H = 1724 kJ/mol NH 3(g)  3 H (g) + N (g)  H = 1172 kJ/mol Thus average N-H bond energy = 1172 / 3 E N-H = 391 kJ/mol Since 1724 = 4 E N-H + E N-N We can estimate N-N bond energy to be: 1724 – 4(391) = 160 kJ/mol

24. Chem 59-250 E H-H = 436 kJ/mol E C=O = 745 kJ/mol E C-H = 414 kJ/mol E C-O = 351 kJ/mol E O-H = 464 kJ/mol  H rxn =  E(bonds broken) –  E(bonds formed)  H rxn = (436 + 745) – (414 + 351+ 464) kJ/mol  H rxn = -48 kJ/mol

25. Chem 59-250 Remember that such calculated bond energies can change For H 2 N-NH 2(g) : E N-N = 160 kJ/mol For F 2 N-NF 2(g) : E N-N = 88 kJ/mol For O 2 N-NO 2(g) : E N-N = 57 kJ/mol They are only a rough approximation and predictions must be made cautiously.

26. Chem 59-250 Free Energy Change,  G =  H - T  S At STP:  G° =  H° - (298.15 K)  S° The two factors that determine if a reaction is favourable: If it gives off energy (exothermic)  H =  H products -  H reactants  H < 0 If the system becomes “more disordered”  S =  S products -  S reactants  S > 0 If  G < 0, then reaction is thermodynamically favourable

27. Chem 59-250  G lets us predict where an equilibrium will lie through the relationship:  G = -RT ln K aA + bB + cC + … hH + iI + jJ + … So if  G 1 and equilibrium lies to the right. There are three possible ways that this can happen with respect to  H and  S.

28. Chem 59-250 If both enthalpy and entropy favour the reaction: i.e.  H 0 then  G < 0. S (s) + O 2(g)  SO 2(g)  H° = -292.9 kJ/mol T  S° = 7.5 kJ/mol  G° = -300.4 kJ/mol If enthalpy drives the reaction: i.e.  H |T  S|, then  G < 0. N 2(g) + 3 H 2(g)  2 NH 3(g)  H° = -46.2 kJ/mol T  S° = -29.5 kJ/mol  G° = -16.7 kJ/mol If entropy drives the reaction: i.e.  H > 0 and  S > 0, but |  H| < |T  S|, then  G < 0. NaCl (s)  Na + (aq) + Cl - (aq)  H° = 1.9 kJ/mol T  S° = 4.6 kJ/mol  G° = -2.7 kJ/mol

29. Chem 59-250 How do people obtain these values? Measure change in equilibrium constants with temperature to get  H° using the relationship: Measure the equilibrium constant for the equilibrium, then determine  G° using the relationship ? :  G° = -RT ln K Often not that easy…

30. Chem 59-250 Reduction-Oxidation (RedOx) reactions: Reduction – gain of electrons Oxidation – loss of electrons  E°, the standard potential for an equilibrium, gives access to  G° through the following relationship:  G° = - n F  E° where, n = number of electrons involved F = Faraday’s constant = 96.4867 kJ mol -1 V -1 (e - ) -1 Note: if  G° 0 So favourable reactions must have  E° > 0

31. Chem 59-250 Half-Cell Reduction Potentials Al 3+ (aq) + 3 e -  Al (s)  E° = -1.67 V Sn 4+ (aq) + 2 e -  Sn 2+ (aq)  E° = 0.15 V thus for: 2 Al (s) + 3 Sn 4+ (aq)  2 Al 3+ (aq) + 3 Sn 2+ (aq)  E° = -(-1.67 V) + (0.15 V) = 1.82 V for 6 electrons So:  G° = - n F  E° = - (6 e - ) F (1.82 V) = -1054 kJ/mol

32. Chem 59-250 Oxidation state diagrams (Frost Diagrams) Relative Energy vs. Oxidation State (under certain conditions) Provides: - Relative stability of oxidation states -Energies available or required for RedOx reactions (the slope between reactant and product)

33. Chem 59-250 Oxidation state diagrams (Frost Diagrams) Some important information provided by Frost diagrams:

34. Chem 59-250 The diagram for Mn displays many of these features. Oxidation state diagrams (Frost Diagrams) The most useful aspect of Frost diagrams is that they allow us to predict whether a RedOx reaction will occur for a given pair of reagents and what the outcome of the reaction will be. This is described in the handout.