Hunds Third Rule - Advanced Quantum Chemistry and Spectroscopy - Lecture Slides, Slides of Chemistry

Download Hunds Third Rule - Advanced Quantum Chemistry and Spectroscopy - Lecture Slides and more Slides Chemistry in PDF only on Docsity! Again as in the H-atom, first- order perturbation theory gives: ( ) ( ) ( )[ ]111 2 ),( ˆˆ),( )( )()*()( +−+−++= ⋅+= ∫ SSLLJJSLE dSLSLEE o o LSnJM o LSnJM o J JJ ζ τψζψ ( )( ) ( )[ ] ( )1 2 121 1 += +−++ =−+ J JJJJEE JJ ξ ζThe energy intervals are: This is the Landé interval rule: the spin-orbit splitting between sequential J levels in a term is proportional to the larger of the J values. Whether the level with the largest value of J lies highest or lowest energy is determined by the sign of ζ. If ζ > 0, the term is said to be regular (as for C), and If ζ < 0, the term is inverted (as for O) Hund’s third rule: If the ground term arises from a configuration for which valence electrons make up a less than half-filled subshells (e.g., C), then the lowest energy term will be regular, whereas If the configuration is more than half-filled subshells (e.g., O), then the lowest energy term will be inverted. docsity.com The Process of Successive Approximation oĤ oĤ soeeo HHH ˆˆˆ ++ 1s22s22p2 1S 1D 3P 2S+1L 2S+1LJ 1S0 1D2 3P2 3P1 3P0 oĤ eeo HH ˆˆ + e.g., the ground state of C docsity.com 49 In Indium (Kr}4d™ 55? 5p 2Piye 5.7864 500 Sn Tin (Kr)4d!5s2 5p? 3Py 7.3439 51 Sb Antimony (Kr}4a!° 552 sp* 4*Sare 8.6084 5200 Te Tellurium (kr) 4a!552 spt SPy 9.0006 53 I Iodine (Kr}4a!9552_ 5p 2Payo 10.4513 Sd Xe Xenon (Kr}4a!° 552 5p® ts 12.1298 55 Cs Cesium (Xe) 6s 25/0 3.8039, 56 Ba Barium (Xe) Gs? 1s) 5.2117 57 La Lanthanum (Xe) Sd 6s? 2Daya 5.5760 58 Ce Cerium (Xepay Sd 6s? 1G4 5.5387 59 Pr Praseodymium = (Xe)4f2_ Gs L ‘Toya 573 60 Nd Neodymium (Xe)4ft Gs? a Ory 5.5250 61 Pm Promethium (Xe)af> 6s B Haya 5.582 62 Sm Samazium (Xe)4f® Gs? iu Ky 5.6437 63 Eu Eurepium (Xe)as7 6s G S72 5.6704 64 Gd Gadolinium (Xe)4s? 5d 6s? : °Do 6.1408 65 Tb Terbinm (Xe)47? Gs ; 8Hysyq 5.8638 66 © Dy Dysprosium (Keay 6s a SIs 5.9389 67 Ho Holmium (Xe)a7" Gs e F570 6.0215 68 Er Erbium (Keay? 6s? 8 He 6.1077 69° Tm Thulium (Xejag 6s? 70 Yb Ytterbium (Keays 6s m1 Lu Lutetium (Xe)47Msd 6s? 72 Hf Hafnium (Xe)4 ssa? 6s? T SF 6.3251 73 Ta Tantahun (Xe)4pMsa® 6s? ro. *Fayo 7.5406 74 W Tungsten (Xe}4 sat 6s? ay 5Do 7.8640 75 Re Rhenium (Xe)apMsa® 6s? noe Sy yo 7.8335 76 Os Osmium (XejasHsa8 6s? Son Spy 3.4382 77) Iridium (Xe)ayMsa? 6s? ioe 4Fyy0 3.9670 78 Pt Platinum (Xe)4sMse? 6s io °Ds 8.9588 79 Au Gold (Ke)ajMsaGs ot 2842 9.2255 80 He Mercury (Xe)4pM5a"Gs? ae 1s) 10.4375 8. TI Thallium (Xe)4pMsal6s? 6p 2Pirg 6.1082 82 Pb Lead (Ke)4sM5a0Gs2 Gp? 7.4167 83 Bi Bismuth (Xe)4fMsa!Gs? Gp? 4S 7.2855 84 Po Polonium (Xe}4fM5d'Gs? Gt SP Bald 85 At Astatine (Ke)4sMsal Gs? Gp" Paro 86 Rn Radon (Xe)4fM5a!6s? 6p? 180 10.7485 87 Fr Francium (Bn) 7s 251 jo 4.0727 88 Ra Radium (Rn) 732 Aso 5.2784 89 Ac Actinium (Rn) 6d 75? 90° Th Thorium (Rn) 6d? 757 91 Pa FProtectinium — (Rn)5f? 6d 7s? A 02 U Uranium (Rn)5f2 6d 75? © 93 Np Neptuninm (Bn)5f4 6d 757 t 94 = Pu Plutonium (Rn)ofo 7s? z 95 Am Americium (in)sf7 7s 96 = =Cm Curium (Rnj5j? Gd 732 A 07 Bk Berkelium. (Bn)5 727s : 98 Cf = Californium (Rn)5 fl? 7s? s 09 Es Einsteinium (Rn)sft 7s? 1000 Fm Fermium (Rnjs ft? 75? 101 9 Md Mendelevium = (Rn)5j#* 7s? 102. No Nobelium (Rnjsft 757 103 Lr Lawreneium (njoj Ts? zpt 104 RF Rutherfordium —_(Rn)5 6d? 757? * The usual LS coupling scheme does not apply for these three elements. note to the NIST table from which this table is taken. See the introductory docsity.com 8. Many electron atoms Need a suitable wave function to describe such N-electron atomic systems, N ≥ 2. Consider perturbation approach. Split Hamiltonian in the following way: VHH ˆˆˆ 0 += ∑ ∑∑ => == = −∇−= N ji ij N i i N i i r eV r Ze m H 1 2 1 2 1 2 2 0 1ˆ 1 2 ˆ rhwhere H0 represents the unperturbed problem of N electrons which do not interact with each other (only with the nucleus) and V is the perturbation which accounts for the electron-electron interaction via Coulomb’s Law docsity.com It is convenient to write: i ii N i i r eZ m H HH 2 2 2 1 0 2 ˆ ˆˆ −∇−= = ∑ = rh This means the unperturbed Hamiltonian can be written as a sum of H-like Hamiltonian operators for each individual electron, I, in the atom. We know the solutions: ( ) ( ) ( ) ( ) ( ) oi n iiii m in siiii ni an eZE orYrRi msmn iEiH i i iiii i iii 2 22 , 2 ,)( , 2 1,,, ˆ −= ×= ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ == = βαϕθφ ρ φφ ρ ρρ l ll ll docsity.com 2 2 2 2 e 22 2 1 e 2 2m2m Ĥ rr1 ZeZe −∇−−∇−= hh e- one e- two If we assume the particles are non-interaction then the resulting Schrodinger Equation is separable and we can write the Hamiltonian and wave function as: 21 ĤĤĤ += )2()1()2,1()( ΨΨ=Ψ=Ψ 21 r,r This a product wave function again is only exact if the electrons are non- interacting docsity.com With this approximation the energy of the system is given by: oo aa 2 e n Z 2 e n ZEEE 2 2 2 22 2 1 2 21 n 2n1 −+−=+= n1 = 1, 2, 3… & n2 = 1,2, 3… How good is this approximation? We can compare the ground state energy of the independent electron helium atom and compare it to that determined experimentally. Eo(H with Z=2) = -54.4 eV Eo(He) = -54.4 - 54.4 = -108.8 eV The true value is -79.0 eV. Thus the independent electron approximation is surprisingly good; i.e., it is in the right order of magnitude. docsity.com Ψ(1,2) versus Ψ(1)Ψ(2) - What’s the Difference? Consider a two electron system. The true many electron wave function should indicate that: The probability of finding the electron at r1, should depend on where the other electron is. 1 r2 r1 21 r2 r1 2E ≠ E Does the product wave function account for this? NO docsity.com Average Shielding Approximation In this approximation, we assume the effect of the other electron is to shield the nuclear charge from the other electron. Consider our electron out here The charge distribution of the other electron ‘shields’ the nuclear charge from the other electron. Within the independent electron approximation, we can therefore do better than just using Z by defining an effective nuclear charge, Zeff oo aa 2 e n Z 2 e n Z E 2 2 2 22 2 1 2 effeff −−= What is the value of Zeff? docsity.com In contrast at the nucleus, the other electron does not screen the nuclear charge at all, so at r2=0 then Zeff = +2e. Out here, the other electron screens out 1e worth of the nuclear charge, so if r2= ∞ then Zeff = +1e Consider both electrons in their ground 1s state. Thus, 1 < Zeff < 2 In fact, it should be a function of r such that Zeff = Zeff (r) docsity.com There are many ways of estimating the effective nuclear charge Zeff from calculation and experimental data. oo aa 2 e n Z 2 e n Z E 2 2 2 22 2 1 2 He effeff −−= For example, given that experimental the first ionization potential of He is 24.6 eV, find an estimate for Zeff eV ao 211.27Hartree1e 2 ==Use: 34.1 2 211.27 2 6.24 eff 2 eff 22 eff =⇒ ⋅ ==∴ Z Z a eZ o A better estimate involving a calculation of <H> yields Zeff2 = 1.6875 Will see that shielding determines many trends associated with the Periodic Table docsity.com