1. Chapter 8 Electron Configuration and Chemical Periodicity.

2. Factors Affecting Atomic Orbital Energies 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. Higher nuclear charge lowers orbital energy (stabilizes the system) by increasing nucleus-electron attractions. The Effect of Nuclear Charge (Z effective ) The Effect of Electron Repulsions (Shielding)

3. ELECTRON SPIN & MAGNETISM Electron Spin: you need to understand magnetism & ionization energy to understand electron spin and quantum number m s. Magnetism and what you already know: earth has a magnetic field, with a north pole and south pole, as do metal magnets Electromagnetism = magnetic field created by electrical current Three definitions for magnetic substances: paramagnetic = attracted to magnetic field diamagnetic = not attracted to magnetic field ferromagnetic = substance that retains its magnetism after being placed in magnetic field (like Fe, Co & Ni)

4. Figure 8.19 Apparatus for measuring the magnetic behavior of a sample.

5. ELECTRON SPIN & MAGNETISM Electrons are spinning charged particles which generate a tiny magnetic field Only two orientations are possible and they are called spin up or spin down, with clockwise being up This is why the 4 th Quantum number, m s, is assigned +½ or -½

6. ELECTRON SPIN Data showed that the He atom was not affected by magnetic field - why? There are 2 e-s present, must be that one is up and one is down, cancelling each others magnetic fields - we say they are "paired" It turns out that only an atom with one or more unpaired e-s exhibits paramagnetism

7. Figure 8.1 Observing the Effect of Electron Spin The Stern-Gerlach experiment.

8. ELECTRON SPIN The lowest energy state for an e- in H is principal level n = 1 There is only one type of orbital at n = 1, because l = 0, which is an s or a spherical orbital For H, the 1 e- is generally in the 1s orbital In He, there are 2 e-s, and it turns out they are both in the 1s orbital & they are paired up, or coupled with one spin up and one spin down Li or Be: 3rd and 4th e-s are in n = 2, but which type of orbital, s or p? FOR THIS WE NEED TO LOOK AT DATA FOR IONIZATION ENERGIES

9. IONIZATION ENERGIES Ionization energy (IE): energy required to take an e- away from an atom First IE removes the e- furthest away from the nucleus Examples: IE 1 = 1.31 x 10 6 J/mol for H, 1.68 x 10 6 J/mol for Fe, and 0.50 x 10 6 J/mol for Na

10. IONIZATION ENERGIES IE in MJ/mol e-#1 2 3 4 5 6 7 8 9 10 11 Na.5 4.6 6.9 9.5 13 17 20 25 29 141 178 Log5.7 6.7 6.8 7.0 7.1 7.2 7.3 7.4 7.9 8.2 8.3 F 1.7 3.4 6.1 8.4 11 15 18 92 106 Log 6.2 6.5 6.8 6.9 7.0 7.2 7.3 8.0 8.03 Na: three energy levels present as shown by big changes in IE (from 1 to 2, 9 to 10) F: two energy levels with slight difference within level - because of s & p orbital differences (7 to 8)

11. IONIZATION ENERGIES If all the IEs are mathematically adjusted for increasing force of attraction between protons and remaining electrons: -then we find that IE 1 to IE 5 in F are almost equal, meaning these 5 e-s are "alike" - And that IE 6 = IE 7 but > IE 1 to IE 5, so these two e-s are different “levels” - And that IE 8 = IE 9 but >>> IE 7, so these two e- levels are really different

12. IONIZATION ENERGIES What does all this mean? IE data define the energy states and orbitals in the atom At n = 1, there's one orbital with e-s that are very hard to remove At n = 2, there are 4 orbitals, but they're different because s has different shape than p

13. IONIZATION ENERGIES Combine the IE data with pairing of e-s into 2e- per orbital from magnetic properties and we determine that: s orbital has up to 2 e-s and each p orbital has up to 2 e-s for a total of 6 If F has 5 e- of similar IE – they must be in p orbitals and they are the easiest to remove so they must be outermost Next 2 e-s are close to same energy, n = 2, but only 2 e-s, so they are in an s orbital The last 2 e-s are very different & have much higher IE, must be close to nucleus, n = 1, 2 e-s in s orbital This data reveals basic e- configuration of F: 1s has 2 e-s, 2s has 2 e-s, 2p has 5 e-s.

14. IONIZATION ENERGIES Now look at Na data: First IE is < 2nd IE but IE 2 to IE 9 are about the same then big jump to IE 10 and IE 11 Means that 1 e- is outermost at n = 3 then 8 e-s in n = 2 (notice slight diff for 2s & 2p) then 2 e-s in n=1

15. IONIZATION ENERGIES Now we look at first IE vs. atomic number Noble gases have high IE All of Group IA atoms have low IE, Group IIA has fairly low IE Transition metals into same IE - these are d e-s Periodic trend of IE - highest at He, lowest at Fr

16. Figure 8.10 Periodicity of first ionization energy (IE 1 )

17. Figure 8.11 First ionization energies of the main-group elements.

18. Table 8.1 Summary of Quantum Numbers of Electrons in Atoms NameSymbolPermitted ValuesProperty principalnpositive integers(1,2,3,…)orbital energy (size) angular momentum l integers from 0 to n-1orbital shape (The l values 0, 1, 2, and 3 correspond to s, p, d, and f orbitals, respectively.) magnetic mlml integers from - l to 0 to + l orbital orientation spin msms +1/2 or -1/2direction of e - spin

19. QUANTUM NUMBERS We know that e-s pair up into two per orbital maximum Pauli Exclusion Principle - a statement of the facts: no two e-s in an atom can have the exact same set of four quantum numbers He 2 e-s: n l m l m s 100+½ 100-½

20. ELECTRON CONFIGURATIONS The best way to figure out quantum numbers is to know electron configuration, so we will do that first! There are several “rules” or physical laws based on data like Ionization Energies If 2p is at higher energy level than 2s, then 3p is higher than 3s Also find that 3d is slightly higher than 3p In multi-electron atoms: - 4s slightly lower energy than 3d, so fill 4s before 3d - always start at 1s, fill in according to increasing energy levels

21. Illustrating Orbital Occupancies The electron configuration n l # of electrons in the sublevel as s,p,d,f The orbital diagram (box or circle, label with orbital name) Order for filling energy sublevels with electrons 1s Figure 8.3 Order for filling energy sublevels with electrons

22. ELECTRON CONFIGURATIONS Hund's Rule: max number of unpaired e-s will occur in ground state Two methods as seen in previous slide: Two methods as seen in previous slide: orbital box (next slide) or spectroscopic (spdf) 1s 2

23. dark - filled, spin-paired light - half-filled no color-empty Figure page 251 A vertical orbital diagram for the Li ground state You will do horizontal orbital box notation.

24. ELECTRON CONFIG USING PERIODIC TABLE Remembering all the rules and the order for filling orbitals looks difficult! It turns out the periodic table is layed out in blocks: s block is groups 1 & 2 p block is groups 13 to 18 d block is transition elements groups 3 - 12 f block is inner transition You can figure out the electron config for the last e- in any element by looking at the periodic table. Then fill in starting from H or nearest noble gas.

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

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

27. Practice Pair up and use a plain (not large) periodic table to do the spdf and orbital box notation for B, Ne and Mg.

28. SAMPLE PROBLEM 8.2Determining Electron Configuration SOLUTION: PROBLEM:Give the full and condensed electrons configurations, partial orbital diagrams showing valence electrons, and number of inner electrons for the following elements: (a) potassium(b) molybdenum(c) lead (a) for K 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 [Ar] 4s 1 4s 1 condensed configuration partial orbital diagram full configuration There are 18 inner electrons. 3d4p

29. SAMPLE PROBLEM 8.2 continued (b) for Mo 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 4d 5 [Kr] 5s 1 4d 5 (c) for Pb [Xe] 6s 2 4f 14 5d 10 6p 2 condensed configuration partial orbital diagram full configuration 5s 1 4d 5 condensed configuration partial orbital diagram full configuration 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 6p 2 There are 36 inner electrons and 6 valence electrons. 6s 2 6p 2 There are 78 inner electrons and 4 valence electrons. 5p

30. ELECTRON CONFIGURATIONS For many e- atoms we can use a shorthand for either method called "noble gas core designation" or condensed version Try for examples: Cl and As Cl: [Ne]3s 2 3p 5 As: [Ar]4s 2 3d 10 4p 3

31. ELECTRON CONFIGURATIONS: VALENCE ELECTRONS Electrons beyond noble gas core are valence electrons: e-s in outermost principal quantum level of atom Practice with Na, As, Mn, and Pu Then determine the principal quantum number of the last electron in each of the above

32. ELECTRON CONFIGURATIONS Hund's Rule: max number of unpaired e-s will occur in ground state Why Cr & Cu don't exactly follow the filling rules: Cr is more stable with 1 e- per orbital including 4s Cu is more stable with full d shell and 1 e- in 4s

33. BACK TO QUANTUM NUMBERS If you have the electron configuration, you have the first two quantum numbers for each electron, n and l. An s subshell has only one orbital with only one m l quantum number, 0. A p subshell has three orbitals, so you have to list each one’s m l : -1, 0, +1. A d subshell has m l ranging from -2.,,,0,…+2; etc. Within each orbital, the m s for the first e- is by definition +½. The second e- is assigned –½.

34. SAMPLE PROBLEM 8.1Determining Quantum Numbers from Orbital Diagrams PLAN: SOLUTION: Use the orbital diagram to find the third and eighth electrons. PROBLEM:Write a set of quantum numbers for the third electron and a set for the eighth electron of the F atom. 9F9F 1s2s2p The third electron is in the 2s orbital. Its quantum numbers are n = l =m l =ms=ms=200+1/2 The eighth electron is in a 2p orbital. Its quantum numbers are 2n = l =m l =ms=ms=1-1/2 So write 2, 1, -1, -1/2 Within 2p, you have -1, 0, +1. Write: 2, 0, 0, +1/2

35. ELECTRON CONFIGURATIONS Make a table of all four quantum numbers for every electron in vanadium. nlml msnlmlmsnlml msnlmlms

36. Table 8.2 in new ed. 8.2

37. Figuree 8.4 Condensed ground-state electron configurations in the first three periods.

38. Table 8.3 in new ed.

39. Table 8.3

40. ELECTRON CONFIGURATIONS Table to remember energy levels IF you don’t have a periodic table handy! 1s 2s 2p 3s 3p 3d 4s 4d 4p 4f 5s 5p 5d 5f (5g) 6s 6p 6d 6f (6g) 7s 7p 7d 7f Follow arrows down and to left to fill in electron configuration.

41. A video on how to write electron configurations and orbital diagrams YouTube - ‪ How to Write Electron Configurations and Orbital Diagrams ‬‏ YouTube - ‪ How to Write Electron Configurations and Orbital Diagrams ‬‏

42. HISTORY OF PERIODIC TABLE Origin is based on "periodic properties" and relative masses Johann Dobereiner grouped triads of elements with similar properties and increasing relative mass In 1864, John Newlands conceived the idea of octaves, since the chem prop's seemed to repeat for every eighth element Current table: Julius Meyer and Dmitri Mendeleev

43. YouTube - ‪ The Periodic Table: Groups and Trends ‬‏ YouTube - ‪ The Periodic Table: Groups and Trends ‬‏

44. PERIODIC TRENDS Trends are the result of atom's e- configuration - # of e-s or really # of protons since its arranged by atomic number Look at Argon's e- density vs. distance from nucleus Not like a billiard ball! Radius is "soft" and is affected by covalent bonding, since it can overlap Cl by itself is 132 pm, but in Cl 2 radius is 100 pm Trend: smallest radii are upper right, largest to lower left in general

45. Figure 8.7 Defining metallic and covalent radii

46. Figure 8.8 Atomic radii of the main- group and transition elements.

47. Figure 8.9 Periodicity of atomic radius

48. SAMPLE PROBLEM 8.3Ranking Elements by Atomic Size SOLUTION: PROBLEM:Using only the periodic table (not Figure 8.15)m rank each set of main group elements in order of decreasing atomic size: (a) Ca, Mg, Sr(b) K, Ga, Ca(c) Br, Rb, Kr(d) Sr, Ca, Rb (a) Sr > Ca > MgThese elements are in Group 2A(2). (b) K > Ca > GaThese elements are in Period 4. (c) Rb > Br > KrRb has a higher energy level and is far to the left. Br is to the left of Kr. (d) Rb > Sr > CaCa is one energy level smaller than Rb and Sr. Rb is to the left of Sr.

49. PERIODIC TRENDS IE: discussed previously, but trend is for highest to upper right, lowest to lower left. Always endothermic process to remove e- Electron Affinity: Trend is most negative EA at F, least likely is Fr, worst are noble gases

50. Table 8.4

51. Figure 8.13 Electron affinities of the main-group elements.

52. Figure 8.14 Trends in three atomic properties.

53. Figure 8.15 Trends in metallic behavior.

54. If you need it: a 15 minute video from the Khan Academy YouTube - ‪ Other Periodic Table Trends ‬‏ YouTube - ‪ Other Periodic Table Trends ‬‏

55. SAMPLE PROBLEM 8.4Ranking Elements by First Ionization Energy SOLUTION: PROBLEM:Using the periodic table only, rank the elements in each of the following sets in order of decreasing IE 1 : (a) Kr, He, Ar(b) Sb, Te, Sn(c) K, Ca, Rb(d) I, Xe, Cs (a) He > Ar > Kr (b) Te > Sb > Sn (c) Ca > K > Rb (d) Xe > I > Cs Group 8A(18) - IE decreases down a group. Period 5 elements - IE increases across a period. Ca is to the right of K; Rb is below K. I is to the left of Xe; Cs is furtther to the left and down one period.

56. SAMPLE PROBLEM 8.5Identifying an Element from Successive Ionization Energies SOLUTION: PROBLEM:Name the Period 3 element with the following ionization energies (in kJ/mol) and write its electron configuration: IE 1 IE 2 IE 3 IE 4 IE 5 IE 6 1012190329104956627822,230 The largest increase occurs after IE 5, that is, after the 5th valence electron has been removed. Five electrons would mean that the valence configuration is 3s 2 3p 3 and the element must be phosphorous, P (Z = 15). The complete electron configuration is 1s 2 2s 2 2p 6 3s 2 3p 3.

57. Atoms to Ions What happens to e- configuration if an atom becomes an ion? Ion size: cations get smaller, anions get bigger More charge is more effect on size ISOELECTRONIC: iso = same, electrons, so same # of electrons We can list isoelectronic series of ions and noble gas atoms that have the same number of electrons.

58. Figure 8.17 Main-group ions and the noble gas configurations. Isoelectronic series: O 2-, F -, Ne, Na +, Mg 2+

59. Atoms to Ions Rank the isoelectronic series on the previous slide by size smallest to largest: Do abbrev. e- config: Mg 2+, Fe 2+ and Fe 3+, Cu 1+ and Cu 2+. Determine which will be paramagnetic, which is more strongly paramagnetic, etc. Then rank them by size.

60. SAMPLE PROBLEM 8.6Writing Electron Configurations of Main-Group Ions SOLUTION: PROBLEM:Using condensed electron configurations, write reactions for the formation of the common ions of the following elements: (a) Iodine(b) Potassium(c) Indium (a) Iodine (Z = 53) is in Group 7A(17) and will gain one electron to be isoelectronic with Xe: I ([Kr]5s 2 4d 10 5p 5 ) + e - I - ([Kr]5s 2 4d 10 5p 6 ) (b) Potassium (Z = 19) is in Group 1A(1) and will lose one electron to be isoelectronic with Ar: K ([Ar]4s 1 ) K + ([Ar]) + e - (c) Indium (Z = 49) is in Group 3A(13) and can lose either one electron or three electrons: In ([Kr]5s 2 4d 10 5p 1 ) In + ([Kr]5s 2 4d 10 ) + e + In ([Kr]5s 2 4d 10 5p 1 ) In 3+ ([Kr] 4d 10 ) + 3e -

61. SAMPLE PROBLEM 8.7Writing Electron Configurations and Predicting Magnetic Behavior of Transition Metal Ions SOLUTION: 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) Mn 2+ (b) Cr 3+ (c) Hg 2+ paramagnetic(a) Mn 2+ (Z = 25) Mn([Ar]4s 2 3d 5 ) Mn 2+ ([Ar] 3d 5 ) + 2e - (b) Cr 3+ (Z = 24) Cr([Ar]4s 1 3d 5 ) Cr 3+ ([Ar] 3d 3 ) + 3e - paramagnetic (c) Hg 2+ (Z = 80) Hg([Xe]6s 2 4f 14 5d 10 ) Hg 2+ ([Xe] 4f 14 5d 10 ) + 2e - not paramagnetic (is diamagnetic)

62. Figure 8.20 Depicting ionic radius.

63. Figure 8.21 Ionic vs. atomic radius.

64. SAMPLE PROBLEM 8.8Ranking Ions by Size SOLUTION: PROBLEM:Rank each set of ions in order of decreasing size, and explain your ranking: (a) Ca 2+, Sr 2+, Mg 2+ (b) K +, S 2-, Cl - (c) Au +, Au 3+ (a) Sr 2+ > Ca 2+ > Mg 2+ (b) S 2- > Cl - > K + These are members of the same Group (2A/2) and therefore decrease in size going up the group. The ions are isoelectronic; S 2- has the smallest Z eff and therefore is the largest while K + is a cation with a large Z eff and is the smallest. (c) Au + > Au 3+ The higher the + charge, the smaller the ion.