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Electron Configuration and Periodicity: Nuclear Charge and Electron Repulsions

This chapter discusses the effect of nuclear charge (Zeff) and electron repulsions (shielding) on electron orbital energies. It also explores the factors affecting atomic orbital energies and the relationship between electron spin and magnetism.

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Electron Configuration and Periodicity: Nuclear Charge and Electron Repulsions

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  1. Chapter 8 Electron Configuration and Chemical Periodicity .

  2. 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.

  3. ELECTRON SPIN & MAGNETISM Electron Spin: you need to understand magnetism & ionization energy to understand electron spin and quantum number ms. 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 4th Quantum number, ms, 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. Observing the Effect of Electron Spin Figure 8.1 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: IE1 = 1.31 x 106 J/mol for H, 1.68 x 106 J/mol for Fe, and 0.50 x 106 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 Log 5.7 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 IE1 to IE5 in F are almost equal, meaning these 5 e-s are "alike" - And that IE6 = IE7 but > IE1 to IE5, so these two e-s are different “levels” - And that IE8 = IE9 but >>> IE7, 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 IE2 to IE9 are about the same then big jump to IE10 and IE11 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 (IE1)

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

  18. 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

  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 ml ms 1 0 0 +½ 1 0 0 -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. Figure 8.3 Order for filling energy sublevels with electrons as s,p,d,f 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, label with orbital name) 1s 1s

  22. ELECTRONCONFIGURATIONS Hund's Rule: max number of unpaired e-s will occur in ground state Two methods as seen in previous slide:orbital box (next slide) or spectroscopic (spdf) 1s2

  23. A vertical orbital diagram for the Li ground state Figure page 251 no color-empty light - half-filled dark - filled, spin-paired 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. PROBLEM: 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 (b) molybdenum (c) lead SOLUTION: (a) for K 1s22s22p63s23p64s1 [Ar] 4s1 There are 18 inner electrons.

  29. 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 1s22s22p63s23p64s23d104p65s14d5 [Kr] 5s14d5 There are 36 inner electrons and 6 valence electrons. (c) for Pb 1s22s22p63s23p64s23d104p65s24d105p66s24f145d106p2 [Xe] 6s24f145d106p2 There are 78 inner electrons and 4 valence electrons.

  30. ELECTRONCONFIGURATIONS 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]3s23p5 As: [Ar]4s23d104p3

  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. ELECTRONCONFIGURATIONS 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 ml quantum number, 0. A p subshell has three orbitals, so you have to list each one’s ml: -1, 0, +1. A d subshell has ml ranging from -2.,,,0,…+2; etc. • Within each orbital, the ms for the first e- is by definition +½. The second e- is assigned –½.

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

  35. ELECTRONCONFIGURATIONS Make a table of all four quantum numbers for every electron in vanadium. n l ml ms n l ml ms

  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

  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

  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 Cl2 radius is 100 pm Trend: smallest radii are upper right, largest to lower left in general

  45. Defining metallic and covalent radii Figure 8.7

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

  47. Figure 8.9 Periodicity of atomic radius

  48. 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 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.

  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

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