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# TPCS CHEM 639

NMSU

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This 0 page Class Notes was uploaded by Dean Jacobs on Sunday November 1, 2015. The Class Notes belongs to CHEM 639 at New Mexico State University taught by Staff in Fall. Since its upload, it has received 27 views. For similar materials see /class/233226/chem-639-new-mexico-state-university in Chemistry at New Mexico State University.

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Date Created: 11/01/15

Benzene 4 We will solve Schodinger equation forthis molecule by considering only porbitals of six carbons under the Huckel approximation Huckel approximation though quite crude provides very useful results Since a bonding in planar molecules has different symmetry from the 7rbonding corresponding molecular orbitals separate in the Hamiltonian do not have cross offdiagonal element and thus can be solved separately Besides abonding is much stronger than 7rbonding 7r and 7r molecular orbitals lie within the a 7 0 gap Thus HOMO and LUMO orbitals are either of 7r7rtype or nonbonding ntype All that allows considering only pzorbitals in conjugated molecules for treating their spectroscopic features while s and pxpy orbitals will be responsible primarily for a bonding ie molecule s shape Huckel approximation can be summarized with following statements 1 ltp jpj gt 6y a overlap integrals are zero 2 lt p lij gt a diagonal elements equal atomic energies of carbon pz electrons 3 ltp lijj gt neighbors the interaction energy is nonzero only for neighboring carbons Both a and 13 are negative and can be paraterized for a typical conjugated carbon and carboncarbon bond Such a parameterization can be also introduced for hetero atoms 8 O N If we write the secular equation for benzene 74136 9 4a7EZ 7 6132or7li4017E6 0 g o 0 0 13 or 7 E The solutions can be found E a 3 a 7 213 a 7 315 a 213 For an arbitrary cyclic molecule with identical conjugated bonds the solutions can be written asE a 2 cos k and graphically represented as enegries horizontally leveled at the corners of the appropriate polygon symmetrically placed on one of its corners quotGraphical solutionquot to cyclic aromatics V E 0c 2B cos27kn One can see that there are degenerate solutions but symmetries of the appropriate orbitals can be found by finding appropriate eigenfunctions Let s do that analysis for benzene using group symmetry Benzene belongs to the ooint groupD61 When six p7 orbitals are taken as the basis set for a representation of this group we obtain 139 The character matrix is also shown on the top C Z s and 7V swere chosen to go through lli carbon atoms and C 2 sand ad s go in between Dsh E 206 203 C2 3C Z 3C Z i 233 236 7 30d 30V Alg 1 1 1 1 1 1 1 1 A28 1 1 1 1 71 71 1 1 1 1 71 71 Blg 1 71 1 71 1 71 1 71 1 71 1 71 Bz g 1 71 1 71 71 1 1 71 1 71 71 1 Elg 52 Am 1 1 1 1 1 1 71 71 71 71 71 71 AZu 1 1 1 1 71 71 71 71 71 71 1 1 BM 1 71 1 71 1 71 71 1 71 1 71 1 BM 1 71 1 71 71 1 71 1 71 1 1 71 Em 2 1 71 72 0 0 72 71 1 2 0 0 EZu 2 71 71 2 0 0 72 1 1 72 0 0 oo 1quot 6 0 0 0 72 0 0 0 0 76 0 2 1393N360 0074 0000120 4 That representation can be reduced into irreducible for this point group 139 BzgE1g l AZulEZu A bit faster approach is to recognize that D h D6 x C and thus use a smaller group D6 instead D6 E 206 203 C2 30 30 Z A1 1 1 1 1 1 1 AZ 1 1 1 1 71 71 B1 1 71 1 71 1 71 B2 1 71 1 71 71 1 E1 2 1 71 72 0 0 E2 2 71 71 2 0 0 1quot 6 0 0 0 72 0 That representation can be reduced into irreducible for this point group 139 BZ E1 AZ E2 which can be easily assigned g and u subscript based on the required negative character with respect to operation oh in Dsh Thus 139 Bzg Elg l AZu E2 We have not found the eigenfunctions nor energies yet just their symmetries Now the task is easier than before We have two options aTrained eye can draw the nodal planes and assign the molecular orbitals by analysis of their characters with respect to appropriate symmetry operations as shown below prematuraly put energies nextto the MO but note that encreaing number of the nodal planes corresponds to increasing energy of MO Also am with energy or 213 is the lowest because both or and 13 are negative The approach is handy but might be confusing when trying to find those c1 and c2 coefficients to finish the construction 01 I C 01 01 e1 0L 5 1 C1 g c1 c1 01 3901 02 a2U 0L 2B b More straight forward also a bit more tedious approach is by using the projection operator 1sz RVOR Again we ll resort toa smaller group D6 instead of Dsh Then projection operators would look shorter 0 For examplePquot2 ICE1006 10q1100 1 1063 1062 7 10621 7 10622 7 10623 7 1001210 7 100m 7 106218 Using which onpl results in PA2P1P1P2 P6 P3 P5 P4 P1P5 P3 P2 le P6 2P1 l39P2 P3 P4 P5 P6 After normalizing we obtain94a2u p1pz p3 p4 p5 p6 The same way it can be dobe with others 6 Some attention is required for doubly degenerate species The projection operator forEZ PE2 20 71066 7100 71063171063 2002when applied tosayp1 gives PEZPI 2p17pz 7P6 7P5 7m 2P4That gives only one symmetry adapted orbital 95 L271 Pz P6 P5 P3 t 2P4 25 Another has to be found by generating a similar one after applying PE2 say tom PEZPZ 2P2 P3 P1 P6 P4 t 2P5 and making it orthogonal to 95 96 ZPZ P3 P1 P6 P4 l39 2P 95 lt aslzpz lk P1 P6 P4 l39 2P5 gt x Z t m J 2P1 Pz apt Ps 173 2P4t6 ZPZ P3 P1 P6 P4 l39 2P5 P2 P3 P5 Ps AnalogouslyPE 20 1066 10031710q171053 72002and PE Pi 2P1Pz P6 P5 P3 2P4 gt92 ZPI P2 P3 2P4 P5 P6 Z 3 PE PZ 2P2 P3 P1 Ps Pzi 2P5 and I 32 93 2P2 P3 P1P6P4 2P5 2P1P2 P3 2P4 Ps P66 3 3 3 3 3 7 7 7PZ l39EP3 7P5 7P6 ZPz l39P3 P5 P6 BZg 91LP1P2P3P4P5 P6 13 Big 921EQP1Pzip3 2P4 P5 tP6 93P2P3P5 P6 A2 94P1P2P3P4P5P6 Em 95 721217122 P3 l39 2P4 P5 P6 2J3 96P2 P3P5 P6 So we found the eigenfunctions which graphically are represented belowwith shaded areas corresponding to a negative sign and the size of each circle resembling the value of coefficient in the eigenfunction e2u The energies of these MO can be found by evaluating appropriate lt Q lHlQ gt Eng lt 91lHl91gtlt 1Pl Pz P3 Pzi P5 P6lHlJ1 P1 Pz P3 le P5 Ps gt 6 6 r ltP1lHKP1 P2 l39 p3p4p5 Psgt ltleHlP1 P2P3 p4p5p6 gt ltP3lHK P1 Pztm m r psps gt ltP4lHl plpz Pgip4p57 p6 gt ix ltP5lHl plpZlp3 Pzi tPs Ps gt ltP6lHlP1 pZip3p4 P5 Ps gt matri element for highlighted orbitals are zero ltP1lHKP1 P2 fps gt 7 ltpzwp1 7122 123 gt ltp2lHlpz 123 7124 gt e quot ltP4lHKP3 Pzi P5 gt l39 ltP5lHK P4 l39P5 Ps gt ltP6lHKP1P5 Ps gt j 6 7213 new Energies of other orbitals are calculated the same way Eng 916P1 P2P3 P4P5 P6 Eelg 921EQP1P2 P3 2P4 P5P6 93 P2 P3 P5 P6 Ea2u 946P1P2P3P4P5P6 E92u 95 2P1 P2 P3 t2p4 l75 l76 96 P2 P3 175 Ps l b2g 2 b29 0408p1 p2p3 p4p5 p6 e2u ezu 02892p1 pzp32p4p5pe e252 05 pzp3p5pe X e1g 5 e1g 02892p1 p2 p3 2p4 p5p6 e190 05 p2p3 p5 p6 a1u 23 a1u 0408p1 p2p3p4p5p6 We see that MOenergies are symmetric with respect to value of a All planar conjugated hydrocarbons with such property are called alternant Alternant hydrocarbons can be recognized by their ability to have all carbons labeled in two colors alternatingly thus the name ie when each neighbor has a different color The ground state configuration is amze1g4 NIIAlg Note that this ground state energy is lower than that of cyclohexatriene benzene with three localized double bonds The effect of sharing 7relectrons the socalled delocalization or resonance energy equals 213 The lowest excited state configuration is aZuZe1g3eZu1 with energy 213 above the ground state from which following states can be constructed elg x em Bl Bl Emeach of which can be either a singlet or a triplet ie it till fill XI1A1g 1B2ul1B1ul1E1u 3B2u 3B1u39 3E1u Atthis level of comlexity we cannot choose which excited state is the lowest energy IBM 132 or 1E1 all we know is that they each have corresponding triplet of lower energy 331 lt 13 me lt IBM and 3E1 lt 1Em We have to analyse other higher energy oneelectron excited state configurations and find if there are any of the same symmetry as the three under consideration Configurational interaction between configurations of the same symmetry lowers the appropriate lower energy state while increasing the corresponding high energy state energy The nextone electron excited state configurations with energy 313 areazze1g3b2g1 and am1e1g4e2ul Neither of them has a proper symmetry The former makes elg x bzg ezg while the latter a2 x em ezg WeshaH mcmdeevenmghev ane e e mn emted sate mn guvmmnmh enevgwa Yheve san vane menm Wmmpmducesmhqulbq Yhus ammmeweestmes 5u 52and zwnw mm m mn guvmmns Cansequemw n M heme av es energy sate Since in the Dsh group the dipole moment components behave like Em xy and Au z the only dipole allowed transition isE Em filAlg which indeed shows a high extinction coefficient while the lower energy transitions IBM and IBM are electronically forbidden They are observed anyway but due to so call vibronic coupling Before that let s consider electric dipole allowed transitions first The matrix element for the transition moment is Melvle v lt wg39v39l lv eW gtlt wgilulve gtlt W39le gt Mere lt waltzV gtlt wiglvV gt The selection rules in this case are defined by the selection rules for electronic partMeyei and the FranckCondon factor lt 15 luVH gtZ forvibrational part The latterwould be nonzero if that integrand has thetotally symmetric species which means that for totally symmetric vibrations all vibrations are allowed Av 0123 For all other vibrations the totally symmetric integrand appears only for even difference in vibrational number ie when Av 024 In cases like the lowest energy transition in benzene which is electronically forbidden the only way the matrix element for the transition moment would be nonzero is to step back from the BornOppenheimer approximation and consider mixing nuclear vibrational and electronic coordinates so called vibronic coupling The electronic Hamiltonian parametrically depends on on vibrational coordinates He H Z T t H2 39 The excited state wavefunction 1ny becomes accordinly mixed with otherzeroorder electronic states w W0 2k skyE n y n where the degree of mixing ck depends on both the vibronic coupling element lt wng39lufo gt and the 7 1 energy separation between the states 7 The electronic transition moment becomes Mere lt we39lulvEH gtlt IIIOlulvSw gt 2ka lt wElHlWSH gt While the first term 00 transition is zero the other terms might be not Then the intensity of transition would bec me nonzero or as it is often referred to as intensity is borrowed from a neigboring transition That is why B1H transition beeing closer to the allowed E1 is more intense than Bl The energy factor is not the only one responsible for nonzero ck the other part lt wng39lufo gt enforces additional selection rules lt wElH39lvfo gt 0 lt wElQilvfo gt or Av i1i3 5 for vibronicaly active mode the mode for which the symmetry of l39l 2 x 1399 matches that of the ground stateX The rules for othervibrational states would followthe rules of ordinary FranckCondon factor described above In case of benzene vibronically allowed transition should be a2 and em For theglleu state it translates to a reqire for either blg b 1g x Bl or 28 egg x Bl There are 12x3630 vibrational modes in benzene which forthe ground state would be realized in 20 fundamental frequencies 2mg azg a2 2b m 2sz 2bu elg 3em 4e2 g 2e From these modes only ezg satisfy the selection rules v15ezg through v18ezg forvibronic transition By symmetry all four of them can contribute and those that have the highest value of contribute the most CH vibrations do not affect the energies of 7relectrons very much it that should be a mode that shakes the molecular sceleton associated with 7rsystem A likely candidate is 000 in plane bending v18ezg with energy ca 500600 cm 1 606 cm 1 in the ground state and 522 cm 1 in the electronically first excited state Thus the excited vibronic transitions that should be observed are ZIBm18 transition from vibrational ground state in the1A1g toglleu at vibrationally excited state with vquot l at V18 Zg Another possibility is to observe a hotZIBZHlS transition from the vibrationally excited with v l at v18ezg NIIAlg to theglleu and vquot0 at v18ezg The lattertransition is very temperature dependent hot What about combinational transitions Again the other vibrations have to be involved in ring distortion vibrations preferably CC stretching since it affects the Hamiltonian the most and posses appropriate symmetry which can be achived by combining v18ezg with totally symmetric combinations of the ring distortion vibrations such as vza1g 992 cm 1 in the ground state and 923 cm 1 in the electronically first excited state Asa result a series of vibronic transitions 213 2ml81 appears as a famous benzene spectrum near 260 nm Zn 0 0 Transm ttance Transm ttance 37000 38000 39000 40000 Wavemmbercm391 41000 42000 38200 38300 38400 38500 4 Wavenumber cm 38600 38700 Energy digram and optical transitions in naphthalene 8 1 9 6 5104 3 yl Naphthalene belongs to the point groupDZhFor simplicity we can analyze it within DZ group and chose appropriate subscripts later When 10 pz 7 orbitals are taken as the basis set for a representation of this group we obtain 1quot The character matrix is also shown below By custom z axis is chosen parallel to pz orbitals DZ E Czz C20 C206 i 0Xy 0xz 0077 Ag 1 1 1 1 1 1 1 1 31g 1 1 71 71 1 1 71 71 192g 1 71 1 71 1 71 1 71 Egg 1 71 71 1 1 71 71 1 Au 1 1 1 1 71 71 71 71 3m 1 1 71 71 1 71 1 1 32 1 71 1 71 71 1 71 1 33 1 71 71 1 71 1 1 71 139 10 0 72 0 0 710 0 2 That representation can be reduced into the irreducible ones forthis point group 139 2328 333g 2Au 331 Now we can construct symmetry adapted molecular 7r 7orbitals Projection operator for each symmetry species P ZR 1 ROR produces 3281st 10E 710mg10cm71002X1071OWJ1OJXZ71OWZJ lf operating on p1 it results inP525p1 2p17p5 7p8 p4 but obviously there should be another function orthogonal to this one Operation onpz we getPBZEpZ 2pZ 7p6 7p7 p3 Thus we found two symmetry adapted functions forBZgThe same way for others After normalizing we have BZg 91P1 P5 P8P4 H11 0 H12 13 92 P2 P6 P7 P3H21 13 H22 13 33g 93 P1 P4 P5 P8 0 l3 63 94 P2P3 P6P7 3 1 5 0 95 13P9 P10 513 0 13 Au 96P1 P4P5 P8 0 97 P2P3P6 P7 3 1 5 3111 98P1P4P5P8 D 3 13 99 P2P3P6P7 3 15 0 91o 079 P1o 513 0 15 2 lnthe Huckel approximation the matrix elements in Hamiltonian can be found using ltplHlp gt 01 ltp lHlpjneihgbor gt 3 and ltp lHlpjnonneihgbor gt 0For example H11 lt 91lHl91gtlt P1 P5 Ps P4lHlP1P5 Ps P4 gt ltP1lHlP1 gt ltP5lHlP5 gt ltP8lHlP8 gt ltP4lHlP4 D The cross elements are all zeros here because neitherp1p5p8 orp4 neighbor each other Other elements are calculated the same way H12 Hz1lt 91lHl92 gt lt P1 P5 Ps P4lHlPz P6 P7 P3 gt ltP1lHlPz P6 P7 P3 gt ltP5lHlPz P6 P7 P3 gt ltP8lHlPz P6 P7 P3 gt ltP4lHlPz P6 P7 P3 gt ltP1lHle gt ltP5lHlP6 gt ltP8lHlP7 gt ltP4lHlP3 gt I3 Now the secular equation is also reduced from 10x10 to two of 2x2 and two 3x3 328 a eigenvectors 7 Hzx7 7 d7l618 1301713 1 554 lt gt1x7 a0618 1 1 1x 3 13 0 33g 13 113 0 eigenvectors 9111713 J27 0 01713 1 W Ha7 7 d72303 1 7J W Ha7 l a1303 1 715l A a eigenvectors Z Z lt gt 1x 713 7 35 01 7161813 13 1 ats Ha7 5a06l8 1 1x 13 J53 0 BI 3 13 0 eigenvectors lt gta 513 0 owl3 1 y Ha 7 lzx71303 1 7WH J m Ha ma2303 2 1 After normalizing and substituting orbitals we get 3393 Ault2gt B 1 u3 3392 2 B29 Aum B1ult2gt 3391 3291 1 B1u 1au 2b1u 1b3g 11329 1b1u 23030 3b3g 0301 p1 p4 p5pa 0231 pZ p3 p6p7 0461 pg p1 16180 2au 0263p1 p4p5 p8 0425p2 p3p6 p7 13030 23b1u 0400p1p4p5p8 0174p2p3p6p7 0347p9p1 10000 22b3g 0408p2 p3 p6p7 0408p9 p10 06180 22b2g 0425p1p4 p5 p8 0263p2p3 p6 p7 06180 1au 0425p1 p4p5 p8 0263p2 p3p6 p7 10000 2b1u 0408p1p4p5p8 0408p9p10 13030 1b3g 0400p1 p4 p5p8 0174p2 p3 p6p7 0347p9 p10 16180 1b29 0263p1p4 p5 p8 0425p2p3 p6 p7 23030 1b1u 0301 p1p4p5p8 0231 p2p3p6p7 0461p9p10j 23030 3b3g 0301p1 p4 p5p8 0231p2p3 p6p7 0461p9 p1U 16180 2au 0263p1 p4p5 p8 0425p2 p3p6 p7 13030 23b1u 0400p1p4p5p8 0174p2p3p6p7 0347p9p1U 10000 22b3g 0408p2 p3 p6p7 0408p9 p10 06180 szg 0425p1p4 p5 p8 0263p2p3 p6 p7 06180 1au 0425p1 p4p5 p8 0263p2 p3p6 p7 10000 22b1u 0408p1p4p5p8 0408p9p10 13030 1b3g 0400p1 p4 p5p8 0174p2 p3 p6p7 0347p9 p10 1 6180 1b29 0263p1p4 p5 p8 0425p2p3 p6 p7 23030 1b1u 0301p1p4p5p8 0231p2p3p6p7 0461p9p10 So the ground state configuration of naphthalene is lbm Z lbz g 211b3g 212bm Z lau Z Ag What about excitations Low energy singlet excited state configurations are Bl 111m 2 fl12g Z fl13g 2 211m 21sz with energy E 0618 0618 1 2360 Bguzzfjlbm 21113ng Zijlbgg Z2bm Zlau2b3g with energy E 1000 0618 461813 awn1mszj1bzggzij1b3gji jzbmjs 11au22bzg with energy E 06180618 1 6181 B B3u B3u B2u B2u 33u I The two Bguconfigurations have the same energy and higher than that of the BZH but due to their interaction they split into B and 8 Calculation of these new energies is more involved but can be qualitatively analysed as interaction between two transition dipole moments oriented along x axis either with constructive or destructive interference The one with lower energy dipoles are attracting each other has to have lower transition moment dipole moments subtract each other The higher energy one on the other hand has much stronger transition It happens that in naphthalene Bgustate is lower in energy than BZu thus warning that a simplification of the lowest energy transition being that between HOMO and LUMO can fail and often does Multielectron Atoms Chem 639 Spectroscopy Spring 2002 SSmirnov Atomic Units Mass me 1 9109 x10 31kg Charge lel 1602 x 10 19C Angular momentum 1 1055 x10 34Js Permittivity 47m 1 1113 x 103910CZJ391m391 2 me 718 Energy Eh 436x 10 J 27211ev Length a0 0529 x 10 10m km 3 17 Time 1 7 W 242x10 5 Speed 1101 47 as Electrical potential L 1 27211V 471Enan Magnetic dipole 2115 i 1855 gtlt10TZ3JTT1 r3 1 Finestructure constant or 47 C 730x 10 W Hamiltonian fora many electron atom including terms up to spin orbital coupling looks like H 2H Z Z 2 24 LITEH 21 Js k39 1 The last term seems to be the least important because correlation between ljand 5 1 is much smaller if ever exists as compared torEr Thus we need to compare with LIT We saw for hydrogen that the latter is of the order of 1112 10 For havier atoms with atomic number Z lt gT39 fgtzzaZn4 2 and is still much smaller than 1nZ if Z is not very high Coulombic repulsion between the electrons is on the order of 1nZ allowing us to conclude that at least for light atoms Z lt 40 it is safe to assume that perturbation due to electronelectron repulsion is of greater importance ka result for light atoms the total L and S are good quantum numbers and orbital approximation is valid We start with an arbitrary Slater determinant WIIII17 lt3gt with electrons placed into spinorbitals one electron functions 42 Rr Ylm9 p where Ylm9 p are spherical functions for appropriate land m and Rr represents a radial function which would need to be found for a new orbital for example by variational principle For closed shell 4 Atomic Energy Levels Configurations For many practical purposes when exact energies are not of primary concern one can use only one configuration The ground state configurations are fairly well described by the Aufbau principle which tells thatthe order of placing electrons on the orbitals should follow the scheme below The order reflects a combination of increasing energy of atomic orbitals with principal quantum number and increasing effect of shielding for more remote electrons with high azimuthal quantum number Exceptions half filled and completely filled orbitals are more stable causing for example for Crthe ground state configuration to be 4s13d5 rather than 4s23d4 For positive ions the order of orbitals is more hydrogen like shielding is less important thus CuzAr 3d9 since Egd lt E4 Each configuration is usually degenerate ie it has more than one state Terms In multielectron atom all the angular and spin momenta of electrons must add vectorially to make corresponding total angular momentum In light atoms where spinorbit coupling is small the total orbital angular momentum f 27 and total spin 8quot Zquot are relatively good quantum numbers by themselves and the total angular momentum can be calculated as a vector sum 7 3 E 5 That scheme is also called the RusselSanders coupling scheme Without spinorbit coupling the operators H LZLZSZ and Sz commute and thus a simultaneous eigenfunction of all five operators could be labelled by the corresponding eigenvalues 1 InLMLSMS gt 6 Without external fields and spin orbit coupling levels associated with 19 are 2S 1 times degenerate due to the spin differentMS and 2L 1 times degenerate due to different ML ie total degeneracy is g 2S12L1 7 These 2S 12L 1 states of InLMLSMS gt with different projectionsMS and M are label with the term symbol 2 L In the same way as one electron orbitals the values ofL are labelled as What terms arise from a particular configuration Symmetry restrictions empose limitations on how these terms are constructed For example configuration ls12s1 as we saw with He with electrons on different orbitals produces two terms a singlet 1S and a triplet 38 four states altogether Configurations with a completely filled subshell correspond to one state 1S Forthe same reason when reducing a configuration into terms only electrons from incomplete subshells have to be considered Let stake the carbon atom in ls22s22pz configuration kwe conculded above only 2pZ have to be considered We have to consider possible distributions of two p electrons among six spin orbitals in a manner consistent with the Pauli exclusion principle The possible states are often referred to as microstates and in fact correspond to individual Slaterdeterminants For example a microstate l10 means m 1 m 12 mlz 0 mgz 712 The number of microstates in a configuration can be calculated based on degeneracy of each orbital In general total degeneracy of a configuration is 22l 1 per electron and total degeneracy is calculated by multiplying all individual ones If electrons belong to the same orbital then Pauli exclusion principle prevents them from having the same set of quantum numbers Then second electron has only 212lr 1 7 1 options for placing Besides we have to recognize that in multiplying 22111 x 4H1 we count twice the same states Total degeneracy for kelectrons on an l orbital equals quot 21 9 In case of an there are 15 corresponding microstates Slater Determinants for Con guration an g I2ijzt1 k3rkr 1gt0 llrll 71 l071 72 71 Note thatsome microstates such as 11 are impossible due to Pauli principle and thus simultaneousML 2 and MS 1 is impossible as well The microstate l11 I clearly belongsto 1D I 11 I llDML 2 gt while 10 belongsto 3P but neither I10 nor l10 are eigenfunctions of theLZ and L2 The correct combinations can be deduced by application of the lowering operatorLto llDML 2 gt Ll1DML2gt l11 7M L7l1DML 2 gt ML 11111324 1gt ijj11zz l1i 7 1111311 1 7mll 13m 7 1 loi 1121ij 13 7le 1le 71j116 llDML1gt IEE01 l10 le10 7 l10 Similarly the state l3PML 1MS 0 gt l10 is orthogonal to llDML 1MS 0 gt Finishing symmetrization for all the microstates we find that there are 1D g 23 12L 1 5 3Pg 23 12L 1 9 1sg 23 12L 11 Altogether 15 in agreement with the number of microstates Analogous procedure can be used for other configurations The procedure could be somewhat tedeous so here is a table to help in thatjorney Atomic Terms from 2P d1 2D Z 3F d2 3F 3F and d 3 2P 43 d3 ZP 2D 2F 4P 4F 3P d4 13 1F 1G 3F 5D ZP d5 2P 2D 2F 2G 4P 4F 68 Often one needs to know only the lowest energy term Hund proposed a set of empirical rules helping to identify it Hund s rules 1 the term with highest spin multiplicity 2S 1 lies lowest in energy ie forC it is 3P 2 if this rule does not select a unique term then the term with the highestL value lies lowerst in energy H H 0 0 ee 3 in advance the lowest energy state has lowestJ if the subshell is less than halffilled highestJ if the subshell is more than halffilled The order of the higher energy terms and states often reflects the Hund s rules but it is not strict there Pictorial representation method allows a simple visual way to apply the Hund s rules For that purpose 21 1 boxes are drawn for different orbitals each box is labeled with an m and electrons are placed into boxes to maximizeML 2m and MS 2mg Start with the highest m and proceed placing electrons in boxes with spin up When all boxes are filled continue with spins down Examples ie the lowest energy term is 3P ie the lowest energy term is 4F That pictorial approach is also useful in finding all possible terms Identification of unique terms could be simpler if we recognize that each term is 2Si 12L 1 times degenerate and we need to identify only one microstate in the corner say with highestMS and highestML belonging to that term We will place arrows representing electrons with appropriate spins into boxes following Pauli exclusion principle and identify quotcornered microstates ie avoid placing electrons with empty orbitals between them Examples np3 configuration First we start with maximized S and vary L ie 48 term the lowest energy one2S12L 1 4times degenerate This is the only option with electrons having highest total spin Now we have to consider options with pairing them 1 0 71 ML2 MS12 ie 2D term2S12L 1 10 times degenerate ie ZP term2S12L 1 6 times degenerate MS 12 ndZ configuration First we start with maximized S and vary L 39e 3F term the lowest energy one 2S12L 1 21 times degenerate What about microstate 2 1 0 1 2 2 First impression is to assign it to 3D term but it is not I 1G term the lowest energy one 2S 12L 1 9 times degenerate 1D term the lowest energy one 2S 12L 1 5 times degenerate 3P term the lowest energy one 2S12L 1 9 times degenerate 18 term the lowest energy one 2S 12L 1 1 times degenerate A simple way to recognize what terms are possible is based on symmetry considerations Spinorbit coupling and the lowest energy state The effects of spinorbit coupling can be included into Hamiltonian in a general form H o 25ml 3139 10 but it is not very convenient because of the presence of the individual I and E angular momenta For a given term however an equivalent can be derived VVignerEckart theorem H10 mi 539 11 where numerical constant LS is often referred to as the spinorbit coupling This form is only applicable within a single isolated 2ML and assumes no interactions with other terms Now theLZ and S2 no longer commute with H The set of commuting observables now is H JZLZSZJZ rather than H LZLZSZSZ where 339 E E is total angular momentum When g increases the various terms begin to interact and spin orbital coupling in the form 28 is no longer accurate THe individual terms can no longer be considered independently This means that the true wavefunctions are no longer eigenfunctions ofLZand S2 Nevetheless will take them as approximately correct for light atoms That scheme is called the RusselSanders coupling scheme A good set of quantum numbers now is lnLSJMJ gt and the energy of each state now will depend on all of them butMJ From 28 and 29 L3939s39jJAthZ 12 the first energy correction in this new basis set is lt H39SO gt g ltE3939s39 gt TZ JJ 1 7LL 1 7SS 1 13 So each term is now split into states J Depending on the sign of g the lowest energy will be either with highest or lowest J The third Hund s rule state that g is positive when the subshell is less than halffilled and negative when the subshell is more than halffilled Usually g is negative for the halffilled subshell An easy way to explain that sign change is to think of going from a completely filled subshell then lack of electrons can be viewed as holes ie positive signes with corresponding change of sign for g Thus for C 2s22pz we have the lowest energy term 3P split into three states 3P03P13PZ with 3P0 being the ground state For 0 2s22p6 which has the same lowest energy term 3P the ground state is 3P2 instead Notethat Jcan go only fromJmm lL 7 Si toJmax L S Each state 2 L is 2J 1 times degenerate due to different projections ofMJ Complete state specification lists the configuration the term symbol and the J value For example the ground state of C 2st2 3P0 0 1D 1D2 C 1522522p2 3P2 3P1 3P0 H0 H0 Hee H0 HeeHSO The latter degeneracy byMJ can be broken by external electric and magnetic fields

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#### "I signed up to be an Elite Notetaker with 2 of my sorority sisters this semester. We just posted our notes weekly and were each making over $600 per month. I LOVE StudySoup!"

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.