Chemistry 1B Fall 2013

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Erin Clarke
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1 Chemistry 1B Fall 2013 lectures 5-6 Chapter 12 pp th 7 *( ) 7th 4

2 2012 Nobel in QUANTUM physics 5

3 many-electron atoms and Schrödinger Equation (ppn ) Although the Schrödinger equation for polyelectronic atoms (and molecules) cannot be solved exactly (mathematically), numerical computer calculations give solutions that agree perfectly with experiment but are very complex. We can (and will!!) use the hydrogen-like orbitals as a very good approximation to the exact solutions of the Schrödinger equation for many-electron atoms. 6

4 aims for lectures 5-6: many electron atoms electron spin, the 4 th quantum number and the Pauli exclusion principle effective nuclear charge, Z eff the net attraction for an electron in a many-electron atom knowledge of the principal quantum number (n) and Z eff for and atom s valence electrons leads to an understanding of: E 2s < E 2p E 3p < E 3d E 3d vs E 4s Hund s Rule and electron configuration in many-electron atoms the Aufbau Principle 7

5 new considerations for many-electron atoms (somewhat different order of presentation than Zumdahl how does increased atomic number (Z) and the presence of other electrons affect orbital energies? how does one fill up the available orbitals in many-electron atoms 8

6 the spin quantum number Stern-Gerlach experiment (fig. 8-1 Silberberg) 9

7 Pauli exclusion principle (PEP) (sec 12.10) no two electrons can have the same four quantum numbers: 1s n=1 =0 m = 0 m s =+½ n=1 =0 m = 0 m s = ½ 1s n=1 =0 m = 0 m s =+½ n=1 =0 m = 0 m s =+½ 10

8 many-electron atoms He, Li, Be, B, C, N, O, F, Ne greater nuclear charge (Z) than hydrogen (Z=1) more electrons than hydrogen (one-electron) 11

where we are heading!!! chapter 12 Quantum mechanics describes many-electron atoms by filling hydrogen-like orbitals with the atom s electrons in a manner consistent with the Pauli Exclusion Principle.

9 where we are heading!!! chapter 12 Quantum mechanics describes many-electron atoms by filling hydrogen-like orbitals with the atom s electrons in a manner consistent with the Pauli Exclusion Principle. This description allows us to understand the energies of electrons in atoms and ions, the relative sizes of atoms and ions, and the chemical reactivity and other properties of various elements. 12

10 voila VOILA!!! 13

11 what we will need to figure out How do the energies of the various hydrogenlike orbitals change when we take into account the effects of increased Z and the other electrons present????? in hydrogen E 2s = E 2p in boron why E 2s < E 2p????? 14

12 configurations (for 1s and 2s levels) ground states (preview) E 2s 2s 1s H 1s 1s He 1s 2 1s Li 1s 2 2s 1s Be 1s 2 2s 2 15

13 orbital energies in many-electron atoms Energy dependence on n and Z m e Z Z E J e 18 n ( ) h 0 n n Holds EXACTLY for 1-electon atoms and ions: H, He +, Li 2+,Be 3+,... Here Z is regular nuclear charge with Z=1, 2, 3, 4,... 16

14 remembering Energy and average radius of electron in a hydrogen orbital m e Z Z E J e 18 n ( ) h 0 n n n 2-11 r ( m) Z ionization energy (ionization potential) of an electron in an atom or molecule is energy required to (just) remove an electron from its stationary state (from n ) (HW2 #11). I.E =(E E -E J 18. ) ( ) n Z n n n

15 ionization energy (section problem S3) ionization energy (IE) : the energy required to remove an electron from an atom, ion, or molecule in the gas phase X (g) + IE X + (g) +e (absorbs energy endothermic, + sign for energetics) similar to the work function in the photoelectric effect, except IE refers to gas phase ionization where refers to removal of electron from the solid 18

16 Energy of H 1s vs He + 1s vs Li 2+ 1s (still 1 electron) all 1s electrons (n=1) Z=+1 Z=+2 Z=+3 H (Z=1) He + (Z=2) Li 2+ (Z=3) m e Z Z E J e 18 n h 0 n n increase Z more attraction lower (more negative) energy (ie higher IE) smaller r avg 19

17 ionization energy Silberberg Figure 8.3 Energy of H 1s vs He + 1s vs Li 2+ 1s (still 1 electron) Increase Z more attraction lower energy (ie higher IE) smaller average radius 20

18 Energy of H 1s vs He + 1s (HO 12.1) Z and Ionization Energies IE=( Z J) n 2 2 E Z=1 Z=2 Z (He + 1s) > Z (H1s) 21

19 what happens when other electrons are present: effective nuclear charge (p 546) Important Factoids in Understanding Effective Nuclear Charge Energy dependence on n and Z eff 2 18 eff n 2 E ( J) Z n Z eff and shielding Z eff = Z (shielding of other electrons) Z eff = Z (effect of electron repulsions) 1s electron feels full Z==+2 nuclear attraction electrons in same shell: He + 1s vs He 1s 2 the two 1s electrons shield one another from the Z=+2 nuclear pull 22

20 He + vs He (what positive charge does a 1s electron see?) less attraction (more shielded) 2 nd 1s electron (e - ) Z=+2 Z=+2 He + 1s He + 1s 2 shielding by 2nd electron 1 < Z eff < 2 23

21 He + vs He (what net positive charge does a 1s electron see?) shielding of electron by 2nd 1s electron 1 < Z eff < 2 Z=+2 Z=+2 He + 1s He 1s 2 24

22 Energy of He 1s 2 vs He + 1s ( HO fig 12.2) Z eff and Ionization Energies Z eff (He+ 1s) > Z eff (He 1s 2 ) E IE= J Z eff 1.34 (1.20) Z 2 n IE eff J 12 Z eff =Z=2 this is approximate way to calculate Z eff ; other techniques give slightly different numbers 25

23 energy of He 1s 2 vs He + 1s (same shell shielding) Zumdahl (p ) 2+ e He + 26

24 Silberberg figure 8.4A: energy of He 1s 2 vs He + 1s (same shell shielding) Zumdahl (p ) [ ] 7th 2+ e e Z eff + e actual He atom Zumdahl s hypothetical He atom 2+ e He + 27

25 Z eff : effective nuclear charge Energy dependence on n and Z eff 2 18 eff n 2 E ( J) Z eff and shielding Z eff = Z - shielding of other electrons Z eff =Z (effect of electron repulsions) Z n electrons in different shells Li 2+ 2s vs Li [1s 2 ] 2s shielding the 2s electron 28

26 Z eff for 2s electron in Li 1s 2 2s 2s electron 18 IE experimental = J 2 n IE2s eff 2s Z J Z=+3 shielding by 1s 2 electrons Z eff 1.26 Li 1s 2 2s 29

27 figure 8.4 B (Silb) shielding by inner shell electrons 30

Energy of Li 2+ 2s vs Li [1s 2 ] 2s (HO Fig. 12.3) Z eff (Li 2+ 2s) > Z eff (Li[1s 2 ]2s) IE=0.86 10-18 J Z eff 1.26 E IE=3.8 10-18 J Z eff 1.34 (1.

28 Energy of Li 2+ 2s vs Li [1s 2 ] 2s (HO Fig. 12.3) Z eff (Li 2+ 2s) > Z eff (Li[1s 2 ]2s) IE= J Z eff 1.26 E IE= J Z eff 1.34 (1.20) IE= J Z eff = Z = 3 Z 2 n IE eff J 12 this is approximate way to calculate Z eff ; other techniques give slightly different numbers 31

29 same shell shielding He 1s 2 vs inner shell shielding Li1s 2 2S how would one expect the shielding of 1s electron for another 1s electron [same shell] to compare to the shielding of a 1s electron for a 2s electron [different shells]? shielding of same shell? < > shielding of outer shell He 1s 2 vs Li1s 2 2s How so? [Z eff =Z nucleus -shielding of other electrons] He : Z (1 s) shielding of other 1s electron eff shielding of other 1s electron 0.66 Li : Z (2 s) shielding of two 1s electrons eff 1.74 shielding of each inner shell 1s electron

30 how do the energies of the 2s and 2p orbitals compare in manyelectron atoms? In 1-electron atoms (H-atom) and 1-electron ions, the 2s and 2p orbitals have the. SAME energies Z eff and the effect of penetration of inner shell electron density by electrons in the same shell 33

31 Z eff for 2s vs 2p: 2p and 2s have SAME energy in 1-electon ion penetration by 2s electron of 1s 2 shielding GIVES INCREASED Zeff 2s radial probability 2s electron Z=+3 +3 shielding by 1s 2 Li 1s 2 2s Z eff 1.26 actual 2s electron density (one radial node; one inner maximum in radial probability) 34

32 Z eff for 2s vs 2p no (less) penetration of 1s 2 shielding by 2p electron GIVES relatively smaller Z eff 2p electron Z=+3 +3 shielding by 1s 2 Li 1s 2 2p Z eff for 2s > Z eff for 2p E 2s < E 2p actual 2p electron density (no radial nodes; no inner maxima in radial probability) 35

33 Z eff for 2s vs 2p (handout fig. 12.4) more penetration of inner shell electron density E 2s < E 2p electron see s more +Z and has greater Z eff 36

34 Energy of Li [1s 2 ] 2s vs Li [1s 2 ] 2p (HO Fig. 12.5): penetration Z eff (Li[1s 2 ] 2s) > Z eff (Li[1s 2 ]2p) IE= J Z eff 1.26 IE= J Z eff 1.02 IE= J Z eff = Z= 3 37

35 how do the energies of the 2s and 2p orbitals compare in manyelectron atoms? Configurations and valence-level orbital diagrams Hund s rule 2 nd row aufbau fig 8.8 (Sil) 38

36 3s 3p 3d 4s orbital energy ordering Increasing Z eff due to increasing penetration effects (figure HO 12Z.6); (Z eff ) 3s > (Z eff ) 3p > (Z eff ) 3d (E) 3s < (E) 3p < (E) 3d 4s vs 3d (Z eff vs n) Orbital energy ordering fig 8.6 (Silb) (figure 8.13, Silb) 39

37 configurations Hund s rule Energy ordering Unambiguous (closed shell, 1e, (n-1)e s Ambiguous (e.g. p 2,p 3,p 4, d 2 d 8 ) Examples (periodic table) ground state excited state not allowed configuration transition metal cations exceptions 40

38 unpaired electrons and magnetic properties (Gouy balance) Diamagnetic pushed out of magnet no unpaired electrons Paramagnetic pulled into magnet unpaired electrons Gouy Balance 41

39 the Periodic Table 42

40 end of lectures

41 Z eff and ionization potentials Z eff = Z = 1. IE= J Z eff 1.26 IE= J Z eff 1.26 IE= J Z eff 1.02 IE= J Z eff 1.34 (1.20) IE= J Z eff = Z = 3. Z eff = Z = 2. 44

42 Silberberg figure

43 Silberberg figure

44 Silberberg figure

45 Zumdahl figure

46 Figure HO 8.6: penetration of 3s vs 3p vs 3d; radial nodes = n-l-1 screening by n=2 electrons + 3s: 2 radial nodes 2 inner electron density maxima + 3p: 1 radial node 1 inner maxima 3d: 0 radial nodes 0 inner maxima + less penetration smaller Z eff higher energy more penetration larger Z eff lower Energy 49

47 Figure HO 8.6: penetration of 3s vs 3p vs 3d; radial nodes = n-l-1 screening by n=2 electrons more penetration larger Z 3s: 2 radial nodes eff 2 inner electron density lower maxima Energy + 3p: 1 radial node 1 inner maxima + + 3d: 0 radial nodes 0 inner maxima less penetration smaller Z eff higher energy 50

48 Why B [He] 2s 2 2p 1 is unambiguous ground state configuration Experimental spectrum (states) of B E Only one energy state (ignore 3/2 vs 1/2) 51

49 Why C [He] 2s 2 2p 2 is ambiguous ground state configuration Experimental spectrum (states) of C E Ground energy state (ignore 0 vs 1 vs 2 ) Excited states with same [He] 2s 2 2p 2 configuration 52

50 Why B [He] 2s 2 2p 1 is unambiguous ground state configuration Experimental spectrum (states) of B E Only one energy state (ignore 3/2 vs 1/2) Excited states have excited [He] 2s 1 2p 2 configuration 53

Chemistry 1B Lectures 5-6 Fall 2013

Chemistry 1B Lectures 5-6 Fall 2013 aims for lectures 5-6: many electron atoms Chemistry 1B lectures 5-6 Chapter 12 pp. 557-569 th 7 *(569-571) 7th electron spin, the 4 th quantum number and the Pauli exclusion principle effective nuclear