1. Chapter 8 Atomic Electron Configurations and Chemical Periodicity We know the electronic structure of the hydrogen atom states as determined by the quantum numbers n, l and m. How does this apply to larger atoms? i.e. multiple electron systems How does the electron structure relate to the periodic table ? How does the electron structure relate to the chemical properties of atoms ?

2. Electron Spin and Magnetism Before we can talk about structure we need to learn a bit about the magnetic properties of particles. Recall that electron move the nucleus in orbits corresponding to set angler momentum values Recall, also that when electrons move they generate a magnetic field, B. v B This is analogous to electrical current moving through a loop

3. Electrons in orbit generate magnetic fields, which requires all materials to be magnetic. Is that so?Why?How is this possible? Imagine two electrons in the same orbit moving in opposite directions. B v v B Electron Spin and Magnetism The magnetic fields cancel !! Do electrons occur in pairs in orbitals!!! But not for this reason, since this is not physically correct. Motion of electrons in their orbitals is not responsible for magnetism, even when the electron is unpaired. The net magnetic field averages to zero. Yes!

4. Electron Spin and Magnetism When a beam of atomic hydrogen is passed through a non-uniform magnetic field is splits into two beams This Magnetism is not due to due to orbital motion Another source of magnetism From where? Spin

5. Electron Spin and Magnetism When an external magnetic field is applied the electron will either along or against the field. Being aligned with the field is more stable than against, therefore the up orientation is slightly favored Electron spin is an inherent magnetism associated with it, which has nothing to do with its translational motion. The electron can the thought of as a little magnet More stableLess stable The distribution of up to down depends on strength of the applied magnetic field. B UP (s=1/2)DOWN (s=-1/2) Magnetic field

6. Magnetic Materials Paramagnetic Materials Diamagnetic Materials More electron electrons will align with the field than against the externally applied field. Composed of atoms/molecules containing only paired electrons They are repelled by an externally applied magnetic field. Composed of atoms/molecules with unpaired electrons. The result is a net bulk magnetic field parallel to the applied field, hence an attractive force

7. Ferromagnetic Materials – Have a permanent magnetic field The magnetic field from each atom will add up, as long as the atoms are correctly aligned to give a one strong “bulk’ magnetic field. – i.e Magnets When two electron on separate atoms are close, the field from one will cause the other to align with it since it is more stable Magnetic Materials

8. Pauli Exclusion Principle Fermions - particles have spin ½. electronsprotonsneutrons “Fermions cannot occupy the same space and spin coordinates” This means that no two electrons can have the same quantum numbers, including the spin quantum numbers. Therefore each orbital can only have 2 electrons since there are only two spin states s =1/2 and -1/2. Ex) 1s orbital n = 1, l = 0, m = 0 and s = 1/2 or s = -1/2 1s Orbital

9. Atoms with more than one electron The wavefunctions for multi electron atoms similar to those for the H atom The ground state of such atoms requires that the lowest possible energy wavefunctions be “occupied” box diagram - a simple tool used to add or subtract electrons from the boxes to represent the electron configuration of the element Consider H, He, Li and Be

10. B 5 C 6 N 7 O 8 F 9 Ne 10 1s2p2s Hunds rule Element # e’s Electrons added to each empty orbital in parallel When no new orbitals are available they are paired Maximize spin

11. Electron Configuration A shorthand notation is commonly used to write out the electron configuration of the atoms based on the number of electrons within each subshell It consists of:NUMBERLETTERSUPERSCRIPT (shell i.d.)(subshell)(occupancy)

12. B 5 C 6 N 7 O 8 F 9 Ne 10 1s2p2s Electron Configuration Element # e’s 1s 2 2s 2 2p 1 1s 2 2s 2 2p 2 1s 2 2s 2 2p 3 1s 2 2s 2 2p 4 1s 2 2s 2 2p 5 1s 2 2s 2 2p 6

13. Aufbau order and Energy Levels The sequence of subshells in the electron configurations not exactly same as the energy levels of H The experimental sequence is known as the aufbau order It is a consequence of electron-electron interactions have on the energies of the wavefunctions in all multi-electron atoms Levels in subshells are still degenerate, the subshells are no longer degenerate in each shell, and differ in energy as s < p < d, Some subshells can overlap the levels of a different shell; thus, for example, in neutral atoms 4s lies below 3d

14. Traditional aufbau sequence diagram Instead of filling orbitals in order of increasing n, we should really be filling them in order of increasing n + l n is used as a ‘tiebreaker’ i.e the one with lowest n first Ex) Fluorine 9 e’s 1s 2 2s 2 2p 5 Ex) Scandium 21 e’s 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 1 Ex) Strontium 38 e’s 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2

15. Afbau sequence from Periodic Table s block d block p block f block We can now see that the very origin of the periodic table is the electron configurations of the elements The periodic table can be used to determine the afbau order instead As you increase the # electrons, the block structure indicates the sequence of subshells

16. A more detailed look at the block structure 1 0 2 3 23 4 4 455 5 6 66 6 7

17. The core electrons are represented by the noble gas followed by configuration of the valence electrons. Electron configurations for the larger elements are lengthy to write out. Ex) Ne has an electron configuration of 1s 2 2s 2 2p 6. For Na, we can write either 1s 2 2s 2 2p 6 3s 1 or [Ne]3s 1 Electron Configurations Ex) Sr 38 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 Kr 36 [Kr] 5s 2 Core e’s Valence e’s noble gas notation - the symbol for a noble gas is used as an abbreviation for its electrons.

18. How many core and valence electrons do these atoms have? a) ____core, ____valence c) ____core, ____valence b) ____core, ____valence d) ____core, ____valence Identify the elements with the following electron configurations. a) 1s 2 2s 2 2p 3 c) [Ne]3s 2 3p 3 b) 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 7 d) [Kr]5s 2 4d 5 Exercises N Co P Tc 2 5105 189367

19. Exceptions to the aufbau order Exception to Afbau order result of:Full shell stability Half Shell stability Stability of higher spin state

20. Electron configurations of ions Electron configurations of ions can be determined from that of the neutral atom, i.e. electron configurations predict ions Oxide forming from oxygen : Same electron configuration as neon This rationalizes the kinds of stable ions that are formed for certain elements O = 1s 2 2s 2 2p 4 O 2- = 1s 2 2s 2 2p 6 Ne = 1s 2 2s 2 2p 6 Mg = 1s 2 2s 2 2p 6 3s 2 Mg 2+ = 1s 2 2s 2 2p 6 Magnesium cation from magnesium: Ne = 1s 2 2s 2 2p 6 Same electron configuration as neon

21. Electron configurations of ions Thus, cation electron configuration is obtained by removing electrons in the reverse Aufbau sequence Anion electron configurations are obtained by adding electrons in the usual Aufbau sequence Ions try to achieve: (1) the closest noble gas configuration (2) a pseudo noble gas configuration (closed d or f subshell) (3) a noble gas configuration for everything except d or f electrons Cations always have their electron configurations in the sequence of the H. 1.Li + 2. P 3- 3. Ga 3+ 4. Sn 2+ 5. Sn 4+

22. a)O 2- c) Cl + e) Pb 4+ b) Mg 6- d) Ca + f) Ga 3+ Which of the following ions are likely to form? For those which are not what ion would you expect to form from that element? Exercise O 2- =1s 2 2s 2 2p 6 O = 1s 2 2s 2 2p 4 = Ne Mg = 1s 2 2s 2 2p 6 3s 2 8 12Mg 6- = 1s 2 2s 2 2p 6 3s 2 3p 6 = Ar Cl = 1s 2 2s 2 2p 4 17Cl 1+ = 1s 2 2s 2 2p 3 20 Ca = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 Ca 1+ = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 Pb = [Xe]6s 2 4f 14 5d 10 6p 2 Pb 4+ = [Xe]4f 14 5d 10 82 Ga = [Ar]4s 2 3d 10 4p 1 31Ga 3+ = [Ar]3d 10 a) b) c) d) e) f ) 10 18 16 19 78 28

23. Order of Energy Levels in Ions “Aufbau” energy levels: s below d “Aufbau” energy levels In anion Energy levels in cations Energy levels are lowered due to weakened e-n interactions Energy levels are increased due to enhanced e-n interactions

24. Subtleties of the shell structure of the atom Why does the 4s level in neutral atoms lie below the 3d? The s orbital has a small fraction of its probability density close to the nucleus. 3d orbitals do not have such inner regions, as they only have planar nodes Hence an s electron from a higher shell will sometimes occur at lower energy than a d electron in a lower shell

25. Effective nuclear charge From Li to Ne, nuclear charge increases from 3 to 10 The charge that a 2s or 2p electron feels is different due to the shielding from the electrons in the 1s orbital 2 s orbital penetrate into the 1s orbital and therefore are shielded less on average than d orbitals Note: Shielding effect increases as the number of e’s increase. This is the result of additional shielding from the 2 s and 2 p e’s -1.77 -1.42 -2.78 -3.15 -3.51 -3.87 Note: As Z* increases orbital shrink towards nucleus as e’s are held more tightly dues to stronger electronic interactions.

26. Effect on atomic size Consider the size change from F to I Decrease strongly Increase significantly Increase gently Consider the change in size of the atoms from Li to F Consider the size change from Li to Rb Consider the size change from F to I

27. Size of Atoms and Ions Atomic radius decreases along the period, and increases down the group The radius of an anion is larger than its neutral atom. Removing the electron decreases shielding without changing the charge of the nucleus. Valence electrons of a cations are in a lower energy shell than in the neutral atom, decreasing the ionic radius. Adding the extra electron increases shielding without changing the charge of the nucleus. The radius of a cation is smaller than its neutral atom i.e Z* is smaller. i.e., Z* is larger.

28. Sizes of monatomic ions Anions are larger than cations This is always true across a period of the table Ions in each group of the table get larger in size down the group Isolectronic ions decrease in size across the period, as Z* increases dramatically. Ex) N 3- to F - Na + to Al 3+

29. Trends in ionization energy Increases as Z* increases across each period Decreases down a group since size increases The energy that must be absorbed in order to remove a valence electron from a neutral atom in the gas phase

30. Z* and its effect on size and IE 3+ e-e- e-e- e-e- 9+ e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- e-e- LiF r = 152 pm EA 1 = 520 kJ/mol r = 71 pm EA 1 = 1681 kJ/mol Z*= 1.28 Z*= 5.13 > < <<

31. Periodic distribution of IE 1 values List of the IE 1 in kJ/mol for the elements IE increases across the period IE decreases down the group Z* increases Shielding effect increases i.e Z* decreases

32. Electron affinity Energy released when an element attracts an extra electron into the lowest-energy unoccupied orbital to form an anion For large Z* e’s are held closely to the nucleus therefore e-n interactions will be stronger for an additional electron coming in. Compare Li (Z* = 1.28) with F (Z* = 5.13 ) EA increases in magnitude across period EA decreases in magnitude down the group Negative since energy is released

33. Important Concepts from Chapter 8 Diamagnetic vs. paramagnetic vs. ferromagnetic substances Electron spin Pauli exclusion principle Orbital box diagrams Electron configurations Aufbau order and its exceptions. Predicting ions using electron configurations Core vs. valence electrons Effective nuclear charge Periodic trends (atomic radius, ionization energy, electron affinity, ionic radius)

34. Midterm Test Tuesday, October 10th (the Tuesday after next) at 7:30pm Practice is test on website. Section A Room ???? 90 minute test Chapters: 1, 2, 7, 8

35. Review Chapter 1 Units Significant figures SI System Dimensional Analysis Scientific notation Prefixes and suffixes Propagation through addition and multiplication Precision/Acuracy

36. Chemical & physical properties Dalton’s atomic theory of matter Models of the atom – Rutherford's model Subatomic particles - protons, neutrons, and electrons Elemental Forms Periodic table (groups and periods) – Common properties in each group Elements - names and symbols Atomic number and mass number - # of n’s, p’s, & e’s Isotopes - calculating average atomic mass and percent abundance Avogadro’s number and the mole Chapter 2

37. Chapter 7 Properties of waves - wavelength, frequency, amplitude, speed Electromagnetic spectrum - speed of light Planck’s equation and Planck’s constant Wave-particle duality for light, electrons, etc.) Atomic line spectra: Balmer and Rydberg Series Ground vs. excited states Heisenberg uncertainty principle Bohr and Schrödinger models of the atom Quantum numbers (n, l, ml) Shells (n), subshells (s,p,d,f) Shapes and properties of atomic orbitals –#nodes, # lobes

38. Chapter 8 Diamagnetic vs. paramagnetic vs. ferromagnetic substances Electron spin Pauli exclusion principle Orbital box diagrams Electron configurations Aufbau order and its exceptions. Predicting ions using electron configurations Core vs. valence electrons Effective nuclear charge Periodic trends -atomic radius, ionization energy, electron affinity, ionic radius

39. Sample problem - Chapter 1 90) The smallest repeating unit of a crystal of common salt is a cube with an edge of 0.563 nm. a) What is the volume in nm 3 ? In cm 3 ? V = l*w*h = (0.563 nm) (0.563 nm) (0.563 nm) V = 0.178 nm 3 [(10 2 cm/m)/(10 9 nm/m)] 3 b) The density of NaCl is 2.17 g/cm 3. What is the mass of a single cube? Density =mass/Volume Mass = Density*Volume = (2.17 g/cm 3 )*(1.78*10-22 cm 3 ) = 3.86*10 -22 g = 0.178 nm 3 =1.78*10 -22 cm 3

40. Sample problem - Chapter 1 c) The cube contains four NaCl molecules. What is the mass of a single NaCl molecule? Mass NaCl = Mass cube/(# molecules) = (3.86*10 -22 g)/(4 molecules) = 9.66*10 -23 g/molecule Molecular Mass NaCl = (9.66*10 -23 g)*(6.022*10 -23 g/mol) = 58.2 g/mol (58.4 g/mol) = (9.66*10 -23 g)/(1.661*10 -24 g/u) = 58.2 u

41. Sample problem - Chapter 2 64) When a sample of Phosphorous burns in air P 4 O 10 is formed. One experiment showed that 0.744 g of P formed 1.704 g of P 4 O 10. If the atomic mass of O is assumed to be 16.000 u, compute the atomic mass of P. To make 1 molecule of P 4 O 10 it requires 4 P’s be combined with 10 O’s # P 4 O 10 molecules made = # P consumed/4 = (mass P/atomic mass P)/4 # P 4 O 10 molecules made = mass P 4 O 10 /Molecular mass P 4 O 10 mass P/(4*AM(P)) = mass P 4 O 10 /(4*AM(P)+10*AM(O)) = mass P 4 O 10 /(4*AM(P)+10*AM(O)) mass P/ mass P 4 O 10 = 4*AM(P)/(4*AM(P)+10*AM(O))

42. Let x = AM(P) mass P/ mass P 4 O 10 = 4x/(4x+10*AM(O)) (0.744 g) /( 1.704 g) = 4x/(4x+10*(16.000 u)) 0.437 = 4x/(4x+(160.00 u)) 1.75x+(70.0 u)= 4x 70.0 u = 2.25x x = 31.1 u Recall that AM(O) =16.000 u mass P/ mass P 4 O 10 = 4*AM(P)/(4*AM(P)+10*AM(O)) Sample problem - Chapter 2 = Atomic Mass of P

43. Sample problem - Chapter 3 a) In what group and period is Technetium found 99 Tc is a transition metal at atomic number 43. Period 5Group 7 b) The valence shell of Tc consists of 5s and 4d. List the quantum numbers for electrons in these orbitals. 5sn =5 l = 0 m = 0 4d n =4 l = 2 m = -2,-1,0,1,2 76)The element Technetium is radioactive has to by synthesized and then prepared as NaTcO4. It is used in imaging of brain and thyroid etc.

44. c) Tc emits a  -ray with energy of 0.141 MeV. What is the wavelength and frequency of the  -ray photon? Note: 1 MeV = 10 6 eV where 1 eV = 9.6485*10 4 J/mol Recall that E = h = E/h = 3.41*10 19 1/s (= Hz) Recall that C =  = (2.26*10 -14 J/photon)/(6.62618*10 -34 Js)  C  =(2.998 *10 8 m/s)/(3.41*10 19 1/s) = 8.79*10 -12 m = 8.79 pm E = (0.141*10 6 eV)*(9.6485*10 4 (J/mol)/eV)/(6.022*10 23 photons/mol) = 2.26*10 -14 J/photon

45. Sample problems - Chapter 8 41) Which of the following is not an allowed set of quantum numbers nlmmsms a) 2 0 0 -1 e) 1 1 0 +1/2 c) 2 1 -1 -1/2 b) 3 4 +2 -1/2 f) 5 3 -1 +1/2 g) 2 -1 1 +1/2 d) 2 1 2 +1/2

46. 48)Answer the following questions about elements with electron configurations A = [Ar]4s 2 and B = [Ar]3d 10 4s 2 4p 5 a) Is A a metal, metalloid, or nonmetal?Metal = gr. 2 b) Is B a metal, metalloid, or nonmetal?Nonmetal = gr. 17 c) Which has a larger ionization energy? B d) Which has the larger atomic radius?A e) Which has the largest electron affinity? B f) What charge does A ionize to? +2 g) What charge does B ionize to?