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Students who haven’t passed the safety quiz

Students who haven’t passed the safety quiz. Understanding “periodicity”. Trends within a period what is changing? what is constant? Trends within a group what is changing? what is constant?. 11 Na. 12 Mg. 11 Na. 19 K. First Ionization Energy (first ionization potential)

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Students who haven’t passed the safety quiz

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  1. Students who haven’t passed the safety quiz

  2. Understanding “periodicity” Trends within a period what is changing? what is constant? Trends within a group what is changing? what is constant? 11 Na 12 Mg 11 Na 19 K

  3. First Ionization Energy (first ionization potential) The minimum energy needed to remove the highest-energy (outermost) electron from a neutral atom in the gaseous state, thereby forming a positive ion

  4. Periodicity of First Ionization Energy (IE1) Like Figure 8-18

  5. Trends • Going down a group, first ionization energy decreases. • This trend is explained by understanding that the smaller an atom, the harder it is to remove an electron, so the larger the ionization energy.

  6. Fig. 8.15

  7. Generally, ionization energy increases with atomic number. Ionization energy is proportional to the effective nuclear charge divided by the average distance between the electron and the nucleus. Because the distance between the electron and the nucleus is inversely proportional to the effective nuclear charge, ionization energy is inversely proportional to the square of the effective nuclear charge.

  8. Electrons can be successively removed from an atom. Each successive ionization energy increases, because the electron is removed from a positive ion of increasing charge. • A dramatic increase occurs when the first electron from the noble-gas core is removed.

  9. Trends Going down a group, first ionization energy decreases. This trend is explained by understanding that the smaller an atom, the harder it is to remove an electron, so the larger the ionization energy.

  10. Small deviations occur between Groups IIA and IIIA and between Groups VA and VIA. Examining the valence configurations for these groups helps us to understand these deviations: IIA ns2 IIIA ns2np1 VA ns2np3 VIA ns2np4 Ittakes less energy to remove the np1 electron than the ns2 electron. It takes less energy to remove the np4 electron than the np3 electron.

  11. Refer to a periodic table and arrange the following elements in order of increasing ionization energy: As, Br, Sb. 33 As 35 Br 51 Sb Sb is larger than As. As is larger than Br. Ionization energies: Sb < As < Br

  12. Ranking Elements by First Ionization Energy Problem: Using the Periodic table only, rank the following elements in each of the following sets in order of increasingIE! a) Ar, Ne, Rn b) At, Bi, Po c) Be, Na, Mg d) Cl, K, Ar Plan: Find their relative positions in the periodic table and apply trends! Solution: a. Rn>Ar>Ne b. Bi<Po<At • Na<Mg<Be • K<Cl<Ar

  13. Electrons can be successively removed from an atom. Each successive ionization energy increases, because the electron is removed from a positive ion of increasing charge. A dramatic increase occurs when the first electron from the noble-gas core is removed.

  14. What are the units here???? Fig. 8.16

  15. Left of the line, valence shell electrons are being removed. Right of the line, noble-gas core electrons are being removed.

  16. Identifying Elements by Its Successive Ionization Energies Problem: Given the following series of ionization energies (in kJ/mol) for an element in period 3, name the element and write its electron configuration: IE1 IE2 IE3 IE4 580 1,815 2,740 11,600 Plan: Examine the values to find the largest jump in ionization energy, which occurs after all valence electrons have been removed. Use the periodic table! Solution:

  17. Identifying Elements by Its Successive Ionization Energies Problem: Given the following series of ionization energies (in kJ/mol) for an element in period 3, name the element and write its electron configuration: IE1 IE2 IE3 IE4 580 1,815 2,740 11,600 Plan: Examine the values to find the largest jump in ionization energy, which occurs after all valence electrons have been removed. Use the periodic table! Solution: The largest jump in IE occurs after IE3 so the element has 3 valence electrons thus it is Aluminum ( Al, Z=13), its electron configuration is : 1s2 2s2 2p6 3s2 3p1

  18. Electron affinity (E.A.) • The energy change for the process of adding an electron to a neutral atom in the gaseous state to form a negative ion • A negative energy change (exothermic) indicates a stable anion is formed. The larger the negative number, the more stable the anion. Small negative energies indicate a less stable anion. • A positive energy change (endothermic) indicates the anion is unstable.

  19. The Electron affinity of a molecule or atom is the energy change when an electron is added to the neutral atom to form a negative ion.

  20. Overall periodic trends Note:Electronegativity has similar trend as electron affinity

  21. Reactivity of the Alkali Metals 2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g) Lithium video 2Na(s) + 2H2O(l) 2NaOH(aq) + H2(g) Sodium video 2K(s) + 2H2O(l) 2KOH(aq) + H2(g) Potassium video Trend? 23

  22. More Sodium Reaction Videos 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g) http://www.theodoregray.com/PeriodicTable/ Prepping Na 100 g Na in one piece 150 g Na in small pieces 24

  23. Electronic Configuration Ions Na 1s 2 2s 2 2p 6 3s 1 Na+ 1s 2 2s 2 2p 6 Mg 1s 2 2s 2 2p 6 3s 2 Mg+2 1s 2 2s 2 2p6 Al 1s 2 2s 2 2p 6 3s 2 3p 1 Al+3 1s 2 2s 2 2p 6 O 1s 2 2s 2 2p 4 O- 2 1s 2 2s 2 2p 6 F 1s 2 2s 2 2p 5 F- 1 1s 2 2s 2 2p 6 N 1s 2 2s 2 2p 3 N- 3 1s 2 2s 2 2p 6

  24. Isoelectronic Atoms and Ions H- 1 { He } Li+ Be+2 N- 3 O- 2 F-{ Ne } Na+ Mg+2 Al+3 P- 3 S- 2 Cl- { Ar } K+ Ca+2 Sc+3 Ti+4 As- 3 Se- 2 Br- { Kr } Rb+ Sr+2 Y+3 Zr+4 Sb- 3 Te- 2 I- { Xe } Cs+ Ba+2 La+3 Hf+4

  25. Trends when atoms form chemical bonds Metals tend to lose e-’s Nonmetals tend to gain e-’s Empirical Observation “when forming ionic compounds, elements tend to lose or gain electrons to be more like the nearest noble gas”

  26. Are ions bigger or smaller than atoms? Representative cation Na → Na+ + e- Representative anion F + e-→ F-

  27. Trends in ion size Cations are always smaller than parent atom - decreased e- repulsion (clouds contract) - if emptying valence shell, “n” decreases Anions are always larger than parent atoms - increased e- repulsion (clouds expand) Cations Contract Anions Add

  28. Trends in atom & ion size

  29. Trends in ion size

  30. Ranking Ions According to Size Problem: Rank each set of Ions in order of increasing size. a) K+, Rb+, Na+ b) Na+, O2-, F - c) Fe+2, Fe+3 Plan: We find the position of each element in the periodic table and apply the ideas of size: i) size increases down a group, ii) size decreases across a period but increases from cation to anion. iii) size decreases with increasing positive (or decreasing negative) charge in an isoelectronic series. iv) cations of the same element decreases in size as the charge increases. Solution: a) since K+, Rb+, and Na+ are from the same group (1A), they increase in size down the group:Na+ < K+ < Rb+ b)the ions Na+, O2-, and F- are isoelectronic. O2- has lower Zeff than F-, so it is larger. Na+ is a cation, and has the highest Zeff, so it is smaller: Na+ < F- < O2- c) Fe+2 has a lower charge than Fe+3, so it is larger: Fe+3 < Fe+2

  31. Chapter #9 - Models of Chemical Bonding 9.1) Atomic Properties and Chemical Bonds 9.2) The Ionic Bonding Model 9.3) The Covalent Bonding Model 9.4) Between the Extremes: Electronegativity and Bond Polarity 9.5) An Introduction to Metallic Bonding

  32. Sodium Chloride

  33. Depicting Ion Formation with Orbital Diagrams and Electron Dot Symbols - I Problem: Use orbital diagrams and Lewis structures to show the formation of magnesium and chloride ions from the atoms, and determine the formula of the compound. Plan: Draw the orbital diagrams for Mg and Cl. To reach filled outer levels Mg loses 2 electrons, and Cl will gain 1 electron. Therefore we need two Cl atoms for every Mg atom. Solution: Mg + Mg+2 + 2 Cl- .. 2 Cl . .. .. .. . Cl Cl . .. .. .. .. Mg + Mg+2 + 2 Cl . .. ..

  34. Depicting Ion Formation from Orbital Diagrams and Electron Dot Symbols - II Problem: Use Lewis structures and orbital diagrams to show the formation of potassium and sulfide ions from the atoms, and determine the formula of the compound. Plan: Draw orbital diagrams for K and S. To reach filled outer orbitals, sulfur must gain two electrons, and potassium must lose one electron. Solution: 2 K + 2 K+ + S - 2 S . .. . 2 - . .. .. .. .. .. K . + S 2 K+ + S K

  35. Three Ways of Showing the Formation ofLi+ and F - through Electron Transfer

  36. Lewis Electron-Dot Symbols for Elements in Periods 2 & 3

  37. The Reaction between Na and Br to Form NaBr The Elements The Reaction!

  38. Melting and Boiling Points of Some Ionic Compounds Compound mp( oC) bp( oC) CsBr 636 1300 NaI 661 1304 MgCl2 714 1412 KBr 734 1435 CaCl2 782 >1600 NaCl 801 1413 LiF 845 1676 KF 858 1505 MgO 2852 3600

  39. Figure 9.11: Potential-energy curve for H2.

  40. Covalent Bonding in Hydrogen, H2

  41. Figure 9.10: The electron probability distribution for the H2 molecule.

  42. Covalent bonds http://www.chem1.com/acad/webtext/chembond/cb03.html This is a good overview. animation http://www.chem.ox.ac.uk/vrchemistry/electronsandbonds/intro1.htm

  43. For elements larger than Boron, atoms usually react to develop octets by sharing electrons. H, Li and Be strive to “look” like He. B is an exception to the noble gas paradigm. It’s happy surrounded by 6 electrons so the compound BH3 is stable. Try drawing a Lewis structure for methane.

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