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Chapter 7 PowerPoint Presentation

Chapter 7

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Chapter 7

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  1. Chapter 7 Ionic Bonding 7.1 Ionic Bonds: Donating and Accepting Electrons 7.2 Energetics of Formation of Ionic Compounds 7.3 Stoichiometry of Ionic Compounds 7.4 Ionic Crystals 7.5 Ionic Radii

  2. Chapter 7 Ionic Bonding (SB p.180) e- e- e- 1:1 ratio of Na+ and Cl- e- Ionic Bonding When a piece of sodium metal is allowed to react with a jar of chlorine gas …... +

  3. Chapter 7 Ionic Bonding (SB p.180) Formation of ionic bond between sodium atom and chlorine atom Cl Na Sodium atom Na 1s22s22p6 Chlorine atom Cl 1s22s22p63s23p5

  4. Chapter 7 Ionic Bonding (SB p.180) - + Chloride ion Cl- 1s22s22p63s23p6 Sodium ion Na+ 1s22s22p6 linked up together by ionic bond Formation of ionic bond between sodium atom and chlorine atom Cl Na

  5. 7.1 Ionic Bonds: Donating and Accepting Electrons (SB p.181) Ionic Bonds: Donating and Accepting Electrons

  6. 7.1 Ionic Bonds: Donating and Accepting Electrons (SB p.181) Donating and Accepting Electrons Ionic bonds are the strong non-directional electrostatic forces of attraction between oppositely charged ions.

  7. 7.1 Ionic Bonds: Donating and Accepting Electrons (SB p.181) – + Internuclear distance Donating and Accepting Electrons

  8. 7.1 Ionic Bonds: Donating and Accepting Electrons (SB p.181) – + – + Donating and Accepting Electrons Cationic radius(r+) Anionic radius(r-) Internuclear distance Internuclear distance = r+ + r-

  9. 7.1 Ionic Bonds: Donating and Accepting Electrons (SB p.182) Electron transfer from a magnesium atom to two chlorine atoms Electron transfer from two lithium atoms to an oxygen atom.

  10. 7.2 Energetics of Formation of Ionic Compounds (SB p. 183)  Hf ø Actually passing through many steps at the molecular level Energetics of Formation of Ionic Compound macroscopiclevel Na(s) + ½Cl2(g)  NaCl(s) microscopic level

  11. 7.2 Energetics of Formation of Ionic Compounds (SB p. 184) ø Standard Enthalpy Change of Atomization (H atom) ø Na(s) Na(g) H atom [Na(s)] = +109 kJ mol-1 1/2 Cl2(g) Cl(g) H atom [1/2Cl2(g)] = +121 kJ mol-1 ø The enthalpy change when one mole of gaseousatoms is formed from its elements in the defined physical state under standard conditions. Questions: Why are the changes endothermic? What type of bond is broken in each case?

  12. 7.2 Energetics of Formation of Ionic Compounds (SB p. 184) ø Ionization Enthalpy (H I.E.) Na(g) Na+(g) + e- H I.E [Na(g)] = +494 kJ mol-1 ø ø Mg(g) Mg+(g) + e-H I.E [Mg(g)] = +736 kJ mol-1 ø Mg+(g) Mg2+(g) + e- H I.E [Na(g)] = +1 450 kJ mol-1 The amount of energy required to remove one mole of valence electrons from one mole of atoms or ions in the gaseous state. Questions: Why are the changes endothermic?

  13. 7.2 Energetics of Formation of Ionic Compounds (SB p. 185) First electron affinity O(g) + e- O-(g) H E.A [O(g)] = - 142 kJ mol-1 Second electron affinity O-(g) + e- O2-(g) H E.A [O(g)] = - 844 kJ mol-1 ø ø ø Electron Affinity (H E.A.) The energy change when one mole of electrons is added to one mole of atoms or ions in the gaseous state. Questions: Why may E.A. have -ve or +ve values?

  14. 7.2 Energetics of Formation of Ionic Compounds (SB p. 185) ø Na+ (g) + Cl-(g) NaCl(s) H lattice[Na+Cl-(s)] – – + + ø Lattice Enthalpy (H L.E.) The energy change when one mole of an ionic crystal is formed from its constituent ions in the gaseous state under standard conditions

  15. 7.2 Energetics of Formation of Ionic Compounds (SB p. 185) ø Na+ (g) + Cl-(g) NaCl(s) H lattice[Na+Cl-(s)] +ve or -ve? – – – – – – + + + + + + Why can’t L.E. be determined directly from experiments? Questions: L.E. can be determined indirectly by either:(1) calculations basing on the knowledge ofelectrostatics in Physics (assuming ions are point charges); or(2) calculations basing on Hess’s Law.

  16. 7.2 Energetics of Formation of Ionic Compounds (SB p. 186) Born-Haber Cycle for the formation of sodium chloride Hatom[Na(s)] HI.E.

  17. 7.2 Energetics of Formation of Ionic Compounds (SB p. 187)

  18. 7.2 Energetics of Formation of Ionic Compounds (SB p. 187) By Hess’s law, ΔHf [NaCl(s)] = ΔHatom[Na(s)] + ΔHI.E.[Na(g)] + ΔHatom[Cl2(g)] + ΔHE.A.[Cl(g)] + ΔHlattice [NaCl(s)] i.e. ΔHf [NaCl(s)] = 109 + 494 + 121 + (-364) +ΔHlattice [NaCl(s)] ΔHlattice [NaCl(s)] = ΔHf [NaCl(s)] +[109 + 494 + 121 + (-364)] = -411 - [109 + 494 + 121 + (-364)] = -711 kJ mol-1 ø ø ø ø ø ø ø ø

  19. 7.3 Stoichiometry of Ionic Compounds (SB p. 189) Stoichiometry of Ionic Compounds Stoichiometry is the simplest ratio of the atoms bonded together in a compound. How can the stoichiometry of an ionic compound be determined?

  20. 7.3 Stoichiometry of Ionic Compounds (SB p. 189) 1 2 In Terms of Electronic Configuration magnesium chloride Example Elements involved Mg(Group II)Cl(Group VII) Ions formed Mg2+Cl- Ratio of ions Chemical formula Mg2+(Cl-)2 or MgCl2

  21. 7.4 Ionic Crystals (SB p. 193) Ionic Crystals Unit cell of NaCl Structure of Sodium Chloride Co-ordination number of Na+ = 6 6:6 co-ordination Co-ordination number of Cl- = 6

  22. 7.4 Ionic Crystals (SB p. 193) A unit cell is the smallest basic portion of the crystal lattice that, when repeatedly stacked together at various directions, can reproduce the entire crystal structure.

  23. 7.4 Ionic Crystals (SB p. 193) corner(Cl-) edge(Na+) face(Cl-) Ionic Crystals Question Determine the number of Na+ and Cl- in a unit cell of sodium chloride respectively.

  24. 7.4 Ionic Crystals (SB p. 193) Diagram showing the two inter-penetrating face-centred cubic structure of Na+ and Cl- ions

  25. 7.4 Ionic Crystals (SB p. 194) Ionic Crystals How to describe the structure? Structure of Caesium Chloride (CsCl) Co-ordination number of Cs+ = 8 8:8 co-ordination Co-ordination number of Cl- = 8

  26. 7.4 Ionic Crystals (SB p. 194) The structure is actually two inter-penetrating simple cubic structure of Cs+ and Cl- ions

  27. 7.4 Ionic Crystals (SB p. 195) Some simple ionic structures

  28. 7.5 Ionic Radii (SB p. 196) X-ray Ionic Radii Photographic plate The technique of X-ray diffraction

  29. 7.5 Ionic Radii (SB p. 196) Electron density map Electron density map found by X-ray diffraction

  30. 7.5 Ionic Radii (SB p. 197) Comparing relative atomic radii of some elements with the ionic radii of the corresponding ions. Size of ion vs size of atom

  31. 7.5 Ionic Radii (SB p. 197) Size of cation < size of atom Reasons:(1) The number of electron shell decreases(2) No. of protons > no. of electrons (p/e ratio increases). The nuclear attraction is more effective to cause a contraction in the electron cloud. Size of anion > size of atom Reasons:(1) Repulsion between newly added electron(s) with other electrons(2) No. of protons < no. of electrons (p/e ratio decreases). The nuclear attraction is less effective and there is an expansion of the electron cloud.

  32. 7.5 Ionic Radii (SB p. 198) Variation of ionic radii of the first 20 elements in the Periodic Table isoelectronic ions Why ionic radius decreases along the isoelectronic series?

  33. 7.5 Ionic Radii (SB p. 198) Isoelectronic ions are ions with the same number of electrons. The following are examples of isoelectronic series: 1. H-, Li+, Be2+, B3+2. N3-, O2-, F-, Na+, Mg2+, Al3+3. P3-, S2-, Cl-, K+, Ca2+

  34. Reason Isoelectronic ions have the same number ofelectrons. An increase in the number of protons implies an increase in the p/e ratio which leads to a contraction of the electron cloud.

  35. The END