1. Ionic bonding –evidence for ionic bonding, electron density maps –trends in radii –Born Haber cycles explaining formulae, why AlO is incorrect –polarisation Metallic Bonding Covalency –electron density maps –giant atomic structures –dot-cross diagrams –shapes of molecules VSEPR –electronegativity –polarity of covalent bonds –polarity of molecules Intermolecular forces –trends in physical properties Solutions and dissolving –why certain substances dissolve in particular solvents

2. Melting Points of Period 3 elements

3. Melting Points of Period 2 elements

4. Ionic bonding The electrostatic attraction between oppositely charged ions –Metals, hydrogen and ammonium form positive ions (cations). –Non-metals form negative ions (anions).

5. Evidence for ionic compounds High melting points –strong electrostatic attractions between oppositely charged ions Electrical conductivity only in liquid state or aqueous solution because ions need to move. Coloured ions can be observed migrating to electrodes during electrolysis (e.g. CuCr 2 O 7 ) –green / blue Cu 2+ (aq) moves to cathode –yellow Cr 2 O 7 2- (aq) moves to anode Electron density maps show low electron density between the oppositely charged ions.

6. NaCl (s) Na (s) + ½ Cl 2 (g) Na + (g) + e - + Cl (g) Na + (g) + Cl - (g)  H formation = - 411.2 kJ mol -1 Na (g) + ½ Cl 2 (g)  m1 [Na] = + 496 kJ mol -1 E aff [Cl] = - 348.8 kJ mol -1  H latt (=348.8-121.7-496-107.3-411.2) = - 787.4 kJ mol -1 Enthalpy (H) Born Haber cycle for sodium chloride  H at = + 107.3 kJ mol -1 Na + (g) + e - + ½ Cl 2 (g)  H at [Cl]= + 121.7 kJ mol -1

7. KCl (s) K (s) + ½ Cl 2 (g) K + (g) + e - + Cl (g) K + (g) + Cl - (g)  H formation = - 436.7 kJ mol -1 K (g) + ½ Cl 2 (g)  m1 [K] = + 419 kJ mol -1 E aff [Cl] = - 348.8 kJ mol -1  H latt (=348.8-121.7-419-89.2-436.7) = - 717.8 kJ mol -1 Enthalpy (H) Born Haber cycle for potassium chloride  H at = + 89.2 kJ mol -1 K + (g) + e - + ½ Cl 2 (g)  H at [Cl]= + 121.7 kJ mol -1

8. KBr (s) K (s) + ½ Br 2 (l) K + (g) + e - + Br (g) K + (g) + Br - (g)  H formation = - 393.8 kJ mol -1 K (g) + ½ Br 2 (l) E m1 [K] = + 419 kJ mol -1 E aff [Br] = - 324.6 kJ mol -1  H latt =(324.6-111.9-419-89.2-393.8) = - 689.3 kJ mol -1 Enthalpy (H) Born Haber cycle for potassium bromide  H at [K]= + 89.2 kJ mol -1 K + (g) + e - + ½ Br 2 (l)  H at [Br]= + 111.9 kJ mol -1

9. Li 2 O (s) 2 Li (s) + ½ O 2 (g) 2 Li + (g) + 2 e - + O (g) 2 Li + (g) + e - + O - (g)  H formation = - 597.9 kJ mol -1 2 Li (g) + ½ O 2 (g)  E m1 [Li] = + 1040 kJ mol -1  H latt (=-798+141.1-249.2-1040-318.8-597.6) = - 2862.8 kJ mol -1 Enthalpy (H) Born Haber cycle for lithium oxide  H at = + 318.8 kJ mol -1 2 Li + (g) + 2 e - + ½ O 2 (g)  H at [O]= + 249.2 kJ mol -1 2 Li + (g) + O 2- (g) E aff [O]= - 141.1 kJ mol -1 E aff [O - ] +798 kJ mol -1

10.  H at (Mg) = + 148 kJ mol -1 MgCl 2 (s) Mg (s) + Cl 2 (g) Mg 2+ (g) + 2e - + 2 Cl (g) Mg 2+ (g) + 2 Cl - (g)  H formation =+148+738+1451+243.4-679.6-2526 - 625.2 kJ mol -1 Mg (g) + Cl 2 (g) E m1 [Mg] = + 738 kJ mol -1  aff [Cl] = - 697.6 kJ mol -1  H latt = = - 2526 kJ mol -1 Enthalpy (H) Born Haber cycle for magnesium chloride Mg + (g) + e - + Cl 2 (g)  H at [Cl]= + 243.4 kJ mol -1 Mg 2+ (g) + 2e - + Cl 2 (g)  m2 [Mg] = + 1451 kJ mol -1

11. MgCl (s) Mg (s) + ½ Cl 2 (g) Mg + (g) + e - + Cl (g) Mg + (g) + Cl - (g) Mg (g) + ½ Cl 2 (g)  m1 [Mg] = + 738 kJ mol -1 E aff [Cl] = - 348.8 kJ mol -1  H formation [MgCl (s)] (=+148+738+121.7-248.8-780) = - 21.1 kJ mol -1 Enthalpy (H) Born Haber cycle for MgCl  H at = + 148 kJ mol -1 Mg + (g) + e - + ½ Cl 2 (g)  H at [Cl]= + 121.7 kJ mol -1  H latt [MgCl (s)] = - 780 kJ mol -1

12. MgCl 3 (s) Mg (s) + 3 / 2 Cl 2 (g) Mg 3+ (g) + 3e - + 3 Cl (g) Mg 3+ (g) + 3 Cl - (g)  H formation = +148+9927+366-1047-4500 = + 4894 kJ mol -1 Mg (g) + 3 / 2 Cl 2 (g) E m1 + E m2 +E m3 [Mg] = (738+1451+7738) kJ mol -1 = + 9927 kJ mol -1  aff [Cl] = - 1047 kJ mol -1  H latt = - 4500 kJ mol -1 Enthalpy (H) Born Haber cycle for MgCl 3  H at = + 148 kJ mol -1 Mg 3+ (g) + 3e - + 3 / 2 Cl 2 (g)  H at [ 1 / 2 Cl 2 ]= + 366 kJ mol -1

13. CaI 2 (s) Ca (s) + I 2 (g) Ca 2+ (g) + 2e - + 2 I (g) Ca 2+ (g) + 2 I - (g)  H formation = - 533.5 kJ mol -1 Ca (g) + I 2 (g) E m1 [Ca] = + 590 kJ mol -1  E aff [I]= - 590.8 kJ mol -1  H latt =(590.8-214-1145-590-178.2-533.5) = - 2069.9 kJ mol -1 Enthalpy (H) Born Haber cycle for Calcium iodide  H at [Ca] = + 178.2 kJ mol -1 Ca + (g) + e - + I 2 (g)  H at [I]= + 214 kJ mol -1 Ca 2+ (g) + 2e - + I 2 (g) E m2 [Ca] = + 1145 kJ mol -1

14. MgO (s) Mg (s) + ½ O 2 (g) Mg 2+ (g) + 2 e - + O (g) Mg 2+ (g) + e - + O - (g)  H formation = - 601.7 kJ mol -1 Mg (g) + ½ O 2 (g) E m1 [Mg] = + 738 kJ mol -1 E aff [O]= - 141.1 kJ mol -1  H latt = (-798+141.1-249.2-1451-738-147.7-601.7) = - 3844.5 kJ mol -1 Enthalpy (H) Born Haber cycle for magnesium oxide Mg + (g) + e - + ½ O 2 (g)  H at [O]= + 249.2 kJ mol -1 Mg 2+ (g) + 2 e - + ½ O 2 (g)  m2 [Mg] = + 1451 kJ mol -1 E aff [O - ] +798 kJ mol -1 Mg 2+ (g) + O 2- (g)  H at (Mg) = + 147.7 kJ mol -1

15. BF 3 (s) B (s) + 3 / 2 F 2 (g) B 3+ (g) + 3e - + 3 F (g) B 3+ (g) + 3 F - (g)  H formation = - 1504.1 kJ mol -1 B (g) + 3 / 2 F 2 (g) E m1 + E m2 +E m3 [B] = (578+1817+2745) kJ mol -1 = + 5140 kJ mol -1  aff [F] = - 984 kJ mol -1  H latt (984-237-5140-326.4-1504.1) = - 6223.5 kJ mol -1 Enthalpy (H) Born Haber cycle for Boron fluoride  H at = + 326.4 kJ mol -1 B 3+ (g) + 3e - + 3 / 2 F 2 (g)  H at [F]= + 237 kJ mol -1

16. Al 2 O 3 (s) 2 Al (s) + 3 / 2 O 2 (g) 2 Al 3+ (g) + 6 e - + 3 O (g) 2 Al 3+ (g)+3e - +3O - (g)  H formation = - 1675.7 kJ mol -1 3 E aff [O]= - 423.3 kJ mol -1  H latt = (-2394+423.3-747.6-10280-652.8-1675.7) = - 15 327 kJ mol -1 Enthalpy (H) Born Haber cycle for aluminium oxide  H at [O] = + 747.6 kJ mol -1 2 Al 3+ (g) + 3 e - + 3 / 2 O 2 (g) 3 E aff [O - ] +2394 kJ mol -1 2 Al 3+ (g) + 3 O 2- (g)  H at (Al) = + 652.8 kJ mol -1 2 Al (g) + 3 / 2 O 2 (g) 2(E m1 + E m2 + E m3 )[Al] = 2(578+1817+2745) kJ mol -1 = + 10 280 kJ mol -1

17. B 2 O 3 (s) 2 B (s) + 3 / 2 O 2 (g) 2 B 3+ (g) + 6 e - + 3 O (g) 2 B 3+ (g)+3e - +3O - (g)  H formation = - 1273 kJ mol -1 3 E aff [O]= - 423.3 kJ mol -1  H latt = (-2394+423.3-747.6-13776-1025.4-1273) = - 18 800 kJ mol -1 Enthalpy (H) Born Haber cycle for boron oxide  H at [½ O 2 (g)] = + 747.6 kJ mol -1 2 B 3+ (g) + 3 e - + 3 / 2 O 2 (g) 3 E aff [O - ] +2394 kJ mol -1 2 B 3+ (g) + 3 O 2- (g)  H at (B) = + 1025.4 kJ mol -1 2 B (g) + 3 / 2 O 2 (g) 2(E m1 + E m2 + E m3 )[B] = 2(801+2427+3660) kJ mol -1 = + 13 776 kJ mol -1

18. Lattice energies LiF-1031LiCl-848BeO-4443BeCl 2 -3020 NaF-918NaCl-780MgO-3791MgCl 2 -2526 KF-817KCl-711CaO-3401CaCl 2 -2258 RbF-783RbCl-685SrO-3223SrCl 2 -2156 CsF-747CsCl-661BaO-3054BaCl 2 -2056

19. AlO is not the formula of aluminium oxide Not Al 2+ O 2- Whilst successive ionisation energies of Al increase, E m3 is not especially large. Al 3+ is very much smaller than Al 2+, since its 3rd principal quantum shell is now empty. Consequently, the ions pack more tightly in (Al 3+ ) 2 (O 2- ) 3. Al 3+ also carries a greater charge than Al 2+,, increasing the attraction to O 2- anions. The lattice energy of (Al 3+ ) 2 (O 2- ) 3 is therefore much greater in magnitude than that of Al 2+ O 2-.

20. AlO is not the formula of aluminium oxide Not Al 3+ O 3- O 3- would have a greater radius than O 2- since its extra electron occupies a new principal quantum shell, further from the nucleus and more shielded by inner quantum shells. In spite of the increased charge of the O 3- anion, its large size reduces packing density of the solid. The electron affinity required to form O 3- from O 2- would be large and endothermic. Therefore the lattice energy of Al 3+ O 3- does not make up for the endothermic steps in the Born Haber cycle.

21. Polarisation of the anion X+X+ Y-Y- X+X+ Y-Y-

22. Factors leading to anion polarisation Cation polarizing power increases with –small radius (increasing charge density) –large positive charge (increasing charge density) Anion polarizability increases with –large radius (outer electrons far from nucleus and shielded by inner shells) increasing negative charge increases its size Increasing anion polarisation means increasing covalent character to the bonding –indicated by large difference between theoretical and experimental lattice energies

23. Metallic bonding The attraction between 'positive ions' and a sea of delocalised electrons. Why does the melting point increase across a period, Na<Mg<Al? Electrical and thermal conductivity due to transfer of charge and energy by the movement of delocalised electrons.

24. The covalent bond The attraction between two nuclei and a shared pair of electrons. One electron of the shared pair originating from each atom in a 'standard' covalent bond. Both electrons of the shared pair originate from the same atom in a dative bond. 'Standard' and dative covalent bonds are indistinguishable.

25. HH Hydrogen molecule (H 2 ) H-H

26. Valence Shell Electron Pair Repulsion Sigma bond electron pairs and lone pairs all repel each other around the central atom. The electron pairs move into positions of maximum separation. –2 pairs gives 180  : 3 pairs gives 120  : 4 pairs gives 109.5 , 6 pairs gives 90 . Lone pairs have a greater repulsion than sigma bond pairs. –Each lone pair reduces the expected bond pair - bond pair angle by about 2.5 .

27. Chlorine molecule, Cl 2 Cl Cl-Cl

28. Hydrogen chloride molecule, HCl (g) {not HCl (aq) which is ionic} H-Cl Cl H Why is this covalent?

29. Water molecule, H 2 O HHO O H H 109.5 

30. N Nitrogen molecule, N 2 N N

31. Ammonia, NH 3 HNH H N H H H 107 

32. Methane, CH 4 H H CHH H H H C H C H H H H 109.5 

33. Ethane, C 2 H 6 H C C H H H H H C C H H H C H H H H3CH3C 109.5 

34. Ethene, C 2 H 4 C C H H H H C = C H HH H 121  118 

35. O = C = O Carbon dioxide, CO 2 O O C

36. SHH 104  H2SH2S

37. 109.5  SiHH H H SiH 4

38. CHH O 120  Methanal, HCHO

39. CH N 180  HCN

40. HH 104  O O

41. NHH 109.5  H H + +

42. H O -

43. SOO O O 2-

44. SSO 109  O O 2-

45. Ethene C C H H H H 3  -bond pairs around each C atom repel to positions of maximum separation. trigonal planar C = C H H H H 121  118 

46. Ethene C H H H H H H H H H H H H  bonds shown as lines and wedges or C H H H H

47. Benzene, C 6 H 6 neither but nor

48. Benzene  bonding is delocalised over the whole ring because all 6 p orbitals are coplanar and overlap. –not 3 separate  bonds Benzene is more stable than alkenes and tends to react by substitution rather than addition.

49. Pyrene, C 16 H 10

50. Graphite Flat sheet of C atoms Weak forces between sheets

51. How to draw diamond

52. Diamond

54. The ice structure

55. Dissolving an ionic solid - H O H O H O H O H O H + O H O H O H O H + (aq)

56. Dissolving Forces are broken between the particles within the solute and within the solvent The energy required to break these forces needs to come from new attractive interactions between solvent and solute particles. –If this energy is not supplied, the ‘solute’ does not dissolve.

57. Dissolving (2) Non-polar molecules attract by London forces alone. Ionic solutes have strong electrostatic attractions between the ions. –The energy required to overcome the high lattice energy can often be supplied through hydration of the ions by the highly polar water molecule –Water can also hydrogen bond to polar solutes such as sugars and ethanol.

58. Dissolving (3) Water does not dissolve non-polar solutes. –Hydrogen bonds would need to be broken between solvent water molecules. –This energy is not made up for by London forces between solvent and solute molecules. Non-polar solvents dissolve non-polar solutes. –Little energy is required to break London forces between the solvent molecules. –This energy is supplied by new London forces between solute and solvent molecules.

59. Dissolving (4) Metals dissolve other metals Ionic solids (such as alumina) dissolve in ionic solvents (such as cryolite, Na 3 AlF 6 )