copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 1 Nature of the Chemical Bond with applications to catalysis, materials science, nanotechnology, surface science, bioinorganic chemistry, and energy William A. Goddard, III, WCU Professor at EEWS-KAIST and Charles and Mary Ferkel Professor of Chemistry, Materials Science, and Applied Physics, California Institute of Technology Course number: KAIST EEWS Room E Hours: Tuesday and Thursday Senior Assistant: Dr. Hyungjun Kim: Manager of Center for Materials Simulation and Design (CMSD) Teaching Assistant: Ms. Ga In Lee: Special assistant: Tod Lecture 7, September 24, 2009 copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 2 Schedule changes There was no lecture on Sept. 22 because of the EEWS conference Goddard will be traveling Oct 2-11 and will not give the lectures scheduled for Oct. 6 and 8 Consequently an extra lecture will be added at 2pm on Wednesday Sept. 30 and another at 2pm Wednesday Oct 14 L8: Sept. 29, as scheduled L9: Sept. 30, new replaces Oct 6 L10: Oct. 1, as scheduled L11: Oct. 13, as scheduled L12: Oct. 14, new replaces Oct 8 L13: Oct. 15, as scheduled copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 3 Last time copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 4 Contour plots of 3s, 3p, 3d hydrogenic orbitals copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 5 The ground state of He atom Put both electrons in 1s orbitals He (1,2) = A [( 1s )( 1s 1s (1) 1s (2) ( 1s = exp(- r) E He = 2( 2 ) 2Z (5/8) where J 1s,1s = (5/8) Applying the variational principle, dE/d = 0 get = (Z 5/16) = E= 2(- 2 ) = - 2 = h 0 Interpretation: two electrons move independently in the orbital 1s exp(- r) which has been adjusted to account for the average shielding due to the other electron in this orbital. On the average this other electron is closer to the nucleus about 31% of the time so that the effective charge seen by each electron is =1.69 The total energy is just the sum of the individual energies. copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 6 Now 3 rd electron to form Li Li (1,2,3) = A [( 1s )( 1s ( 1s by Pauli principle Thus the 3 rd electron must go into 2s or 2p atomic orbital Li (1,2,3) = A [( 1s )( 1s )( 2s Li (1,2,3) = A [( 1s )( 1s )( 2pz (or 2px or 2py) Li + : 1s exp(- r) with = Z = 2.69 R 1s = 1/ = a 0 = 0.2A copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 7 Add 3 rd electron to the 2p orbital Li (1,2,3) = A [( 1s )( 1s )( 2pz (or 2px or 2py) The 2p orbital sees effective charge of Z eff = 3 2 = 1, since it goes to zero at z=0, so that there is no shielding of 1s Get size: R 2p = n 2 /Z eff = 4 a 0 = 2.12A Energy: e 2p = -(Z eff ) 2 /2n 2 = -1/8 h 0 = eV 0.2A 1s 2.12A 2p copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 8 Add the 3 rd electron to the 2s orbital Li (1,2,3) = A [( 1s )( 1s )( 2pz (or 2px or 2py) The 2s orbital must be orthogonal to the 1s, which means that it must have a spherical nodal surface at ~ 0.2A, the size of the 1s orbital. Thus the 2s has a nonzero amplitude at z=0 so that it is not completely shielded by the 1s orbitals. The result is Z eff 2s = 3 1.72 = 1.28 Size: R 2s = n 2 /Z eff = 3.1 a 0 = 1.65A Energy: e 2s = -(Z eff ) 2 /2n 2 = h0 = 5.57 eV 0.2A 1s 2.12A 2s R~0.2A copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 9 Li atom excited states 1s 2s h 0 = -3.4 eV 2p Energy zero h 0 = -5.6 eV h0 = 74.1 eV MO picture State picture (1s) 2 (2s) (1s) 2 (2p) E = 2.2 eV cm nm Ground state 1 st excited state Exper 671 nm E = 1.9 eV copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 10 He, 2 Ne, 10 Ar, 18 Zn, 30 Kr, 36 Aufbau principle for atoms Particularly stable atoms, closed shells Xe, 54 Rn, 86 Uuo, 118 copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 11 Many-electron configurations General aufbau ordering Particularly stable copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 12 General trends along a row of the periodic table As we fill a shell, thus B(2s) 2 (2p) 1 to Ne (2s) 2 (2p) 6 Zeff increases leading to a decrease in the radius ~ n 2 /Zeff And an increase in the IP ~ (Zeff) 2 /2n 2 Example Z eff 2s = 1.28 Li, 1.92 Be, 2.28 B, 2.64 C, 3.00 N, 3.36 O, 4.00 F, 4.64 Ne Thus R(2s Li)/R(2s Ne) ~ 4.64/1.28 = 3.6 copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 13 General trends along a column of the periodic table As we go down a colum Li [He}(2s) to Na [Ne]3s to K [Ar]4s to Rb [Kr]5s to Cs[Xe]6s Things get more complicated The radius ~ n 2 /Zeff And the IP ~ (Zeff) 2 /2n 2 But the Zeff tends to increase, partially compensating for the change in n so that the atomic sizes increase only slowly as we go down the periodic table and The IP decrease only slowly (in eV): 5.39 Li, 5.14 Na, 4.34 K, 4.18 Rb, 3.89 Cs (13.6 H), 17.4 F, 13.0 Cl, 11.8 Br, 10.5 I, 9.5 At 24.5 He, 21.6 Ne, 15.8 Ar, 14.0 Kr,12.1 Xe, 10.7 Rn copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 14 copyright 2009 William A. Goddard III, all rights reservedEEWS Goddard-L04 15 Transition metals; consider [Ar] plus one electron [IP 4s = (Z eff 4s ) 2 /2n 2 = 4.34 eV Z eff 4s = 2.26; 4s