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Chapter 8

Chapter 8. Electron Configuration and Chemical Periodicity. Electron Configuration and Chemical Periodicity. 8.1 Development of the Periodic Table. 8.2 Characteristics of Many-Electron Atoms. 8.3 The Quantum-Mechanical Model and the Periodic Table.

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Chapter 8

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  1. Chapter 8 Electron Configuration and Chemical Periodicity

  2. Electron Configuration and Chemical Periodicity 8.1 Development of the Periodic Table 8.2 Characteristics of Many-Electron Atoms 8.3 The Quantum-Mechanical Model and the Periodic Table 8.4 Trends in Three Key Atomic Properties 8.5 Atomic Structure and Chemical Reactivity

  3. Observing the effect of electron spin. Figure 8.1 The Stern-Gerlach experiment.

  4. principal n positive integers(1,2,3,…) orbital energy (size) angular momentum l integers from 0 to n-1 orbital shape (The l values 0, 1, 2, and 3 correspond to s, p, d, and f orbitals, respectively.) magnetic ml integers from -l to 0 to +l orbital orientation spin ms +1/2 or -1/2 direction of e- spin Table 8.1 Summary of Quantum Numbers of Electrons in Atoms Name Symbol Permitted Values Property

  5. The Effect of Nuclear Charge (Zeffective) The Effect of Electron Repulsions (Shielding) Factors Affecting Atomic Orbital Energies Higher nuclear charge lowers orbital energy (stabilizes the system) by increasing nucleus-electron attractions. Additional electron in the same orbital An additional electron raises the orbital energy through electron-electron repulsions. Additional electrons in inner orbitals Inner electrons shield outer electrons more effectively than do electrons in the same sublevel.

  6. The effect of orbital shape. Figure 8.2

  7. as s,p,d,f Figure 8.3 Order for filling energy sublevels with electrons. Illustrating Orbital Occupancies The electron configuration # of electrons in the sublevel n l The orbital diagram (box or circle)

  8. A vertical orbital diagram for the Li ground state. TA pg. 241 no color-empty light - half-filled dark - filled, spin-paired

  9. PROBLEM: Write a set of quantum numbers for the third electron and a set for the eighth electron of the F atom. 9F 1s 2s 2p n = 2 l = 0 ml = 0 ms= + or -1/2 n = 2 l = 1 ml = -1, 0, or +1 ms= + or -1/2 SAMPLE PROBLEM 8.1 Determining Quantum Numbers from Orbital Diagrams PLAN: Use the orbital diagram to find the third and eighth electrons. SOLUTION: The third electron is in the 2s orbital. Its quantum numbers are The eighth electron is in a 2p orbital. Its quantum numbers are

  10. Table 8.2

  11. Figure 8.4 Condensed ground-state electron configurations in the first three periods.

  12. Table 8.3

  13. Replace w/ Table 8.3 1e

  14. Figure 8.5 A periodic table of partial ground-state electron configurations.

  15. Figure 8.6 The relation between orbital filling and the periodic table.

  16. PROBLEM: Using the periodic table on the inside cover of the text (not Figure 8.12 or Table 8.4), give the full and condensed electrons configurations, partial orbital diagrams showing valence electrons, and number of inner electrons for the following elements: full configuration condensed configuration partial orbital diagram 4s1 3d 4p SAMPLE PROBLEM 8.2 Determining Electron Configuration (a) potassium (K: Z = 19) (b) molybdenum (Mo: Z = 42) (c) lead (Pb: Z = 82) PLAN: Use the atomic number for the number of electrons and the periodic table for the order of filling for electron orbitals. Condensed configurations consist of the preceding noble gas and outer electrons. SOLUTION: (a) for K (Z = 19) 1s22s22p63s23p64s1 [Ar] 4s1 There are 18 inner electrons.

  17. full configuration condensed configuration partial orbital diagram 5s1 4d5 full configuration condensed configuration partial orbital diagram 5p 6s2 6p2 SAMPLE PROBLEM 8.2 continued (b) for Mo (Z = 42) 1s22s22p63s23p64s23d104p65s14d5 [Kr] 5s14d5 There are 36 inner electrons and 6 valence electrons. (c) for Pb (Z = 82) 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2 [Xe] 6s24f145d106p2 There are 78 inner electrons and 4 valence electrons.

  18. Defining metallic and covalent radii. Figure 8.8

  19. Figure 8.9 Atomic radii of the main-group and transition elements.

  20. Periodicity of atomic radius. Figure 8.10

  21. PROBLEM: Using only the periodic table (not Figure 8.15)m rank each set of main group elements in order of decreasing atomic size: SAMPLE PROBLEM 8.3 Ranking Elements by Atomic Size (a) Ca, Mg, Sr (b) K, Ga, Ca (c) Br, Rb, Kr (d) Sr, Ca, Rb PLAN: Elements in the same group increase in size and you go down; elements decrease in size as you go across a period. SOLUTION: (a) Sr > Ca > Mg These elements are in Group 2A(2). (b) K > Ca > Ga These elements are in Period 4. (c) Rb > Br > Kr Rb has a higher energy level and is far to the left. Br is to the left of Kr. (d) Rb > Sr > Ca Ca is one energy level smaller than Rb and Sr. Rb is to the left of Sr.

  22. Periodicity of first ionization energy (IE1). Figure 8.11

  23. First ionization energies of the main-group elements. Figure 8.12

  24. Figure 8.13 The first three ionization energies of beryllium (in MJ/mol). For more data on sequential ionization energies of the elements, go to http://www.webelements.com or click on the button below.

  25. PROBLEM: Using the periodic table only, rank the elements in each of the following sets in order of decreasing IE1: SAMPLE PROBLEM 8.4 Ranking Elements by First Ionization Energy (a) Kr, He, Ar (b) Sb, Te, Sn (c) K, Ca, Rb (d) I, Xe, Cs PLAN: IE decreases as you proceed down in a group; IE increases as you go across a period. SOLUTION: (a) He > Ar > Kr Group 8A(18) - IE decreases down a group. (b) Te > Sb > Sn Period 5 elements - IE increases across a period. (c) Ca > K > Rb Ca is to the right of K; Rb is below K. (d) Xe > I > Cs I is to the left of Xe; Cs is furtther to the left and down one period.

  26. Table 8.4

  27. PROBLEM: Name the Period 3 element with the following ionization energies (in kJ/mol) and write its electron configuration: IE1 IE2 IE3 IE4 IE5 IE6 1012 1903 2910 4956 6278 22,230 SAMPLE PROBLEM 8.5 Identifying an Element from Successive Ionization Energies PLAN: Look for a large increase in energy which indicates that all of the valence electrons have been removed. SOLUTION: The largest increase occurs after IE5, that is, after the 5th valence electron has been removed. Five electrons would mean that the valence configuration is 3s23p3 and the element must be phosphorous, P (Z = 15). The complete electron configuration is 1s22s22p63s23p3.

  28. Electron affinities of the main-group elements. Figure 8.14

  29. Figure 8.15 Trends in three atomic properties.

  30. Trends in metallic behavior. Figure 8.16

  31. The trend in acid-base behavior of element oxides. Figure 8.17

  32. Main-group ions and the noble gas configurations. Figure 8.18

  33. PROBLEM: Using condensed electron configurations, write reactions for the formation of the common ions of the following elements: (a) Iodine (Z = 53) is in Group 7A(17) and will gain one electron to be isoelectronic with Xe: I ([Kr]5s24d105p5) + e- I- ([Kr]5s24d105p6) (b) Potassium (Z = 19) is in Group 1A(1) and will lose one electron to be isoelectronic with Ar: K ([Ar]4s1) K+ ([Ar]) + e- (c) Indium (Z = 49) is in Group 3A(13) and can lose either one electron or three electrons: In ([Kr]5s24d105p1) In+ ([Kr]5s24d10) + e+ In ([Kr]5s24d105p1) In3+([Kr] 4d10) + 3e- SAMPLE PROBLEM 8.6 Writing Electron Configurations of Main-Group Ions (a) Iodine (Z = 53) (b) Potassium (Z = 19) (c) Indium (Z = 49) PLAN: Ions of elements in Groups 1A(1), 2A(2), 6A(16), and 7A(17) are usually isoelectronic with the nearest noble gas. Metals in Groups 3A(13) to 5A(15) can lose their np or ns and np electrons. SOLUTION:

  34. The Period 4 crossover in sublevel energies. Figure 8.19

  35. Apparatus for measuring the magnetic behavior of a sample. Figure 8.20

  36. PROBLEM: Use condensed electron configurations to write the reaction for the formation of each transition metal ion, and predict whether the ion is paramagnetic. (a) Mn2+(Z = 25) Mn([Ar]4s23d5) Mn2+ ([Ar] 3d5) + 2e- (b) Cr3+(Z = 24) Cr([Ar]4s23d6) Cr3+ ([Ar] 3d5) + 3e- (c) Hg2+(Z = 80) Hg([Xe]6s24f145d10) Hg2+ ([Xe] 4f145d10) + 2e- SAMPLE PROBLEM 8.7 Writing Electron Configurations and Predicting Magnetic Behavior of Transition Metal Ions (a) Mn2+(Z = 25) (b) Cr3+(Z = 24) (c) Hg2+(Z = 80) PLAN: Write the electron configuration and remove electrons starting with ns to match the charge on the ion. If the remaining configuration has unpaired electrons, it is paramagnetic. SOLUTION: paramagnetic paramagnetic not paramagnetic (is diamagnetic)

  37. Depicting ionic radius. Figure 8.21

  38. Ionic vs. atomic radii. Figure 8.22

  39. PROBLEM: Rank each set of ions in order of decreasing size, and explain your ranking: SAMPLE PROBLEM 8.8 Ranking Ions by Size (a) Ca2+, Sr2+, Mg2+ (b) K+, S2-, Cl - (c) Au+, Au3+ PLAN: Compare positions in the periodic table, formation of positive and negative ions and changes in size due to gain or loss of electrons. SOLUTION: (a) Sr2+ > Ca2+ > Mg2+ These are members of the same Group (2A/2) and therefore decrease in size going up the group. The ions are isoelectronic; S2- has the smallest Zeff and therefore is the largest while K+ is a cation with a large Zeff and is the smallest. (b) S2- > Cl - > K+ (c) Au+ > Au3+ The higher the + charge, the smaller the ion.

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