Making more complicated atoms starting with He• Pauli Exclusion Principle allows for at most 2 electrons in a spatial energy eigenstate (r) (‘spin-up’, ‘spin-down’)• the nucleus provides Coulomb potential energy of charge Ze• the overall electron wavefunction is (r)sm

• all spin operators commute with all space operators• let’s start to make atoms by filling in electrons• He: n = 1, s state, with 2 electrons in it, [(r) = R10(r)Y00(,)], we put electron #1 into the space state and electron #2 into the same space state and then build the symmetric combination

• therefore the spin state must be antisymmetric: the electrons must be in the singlet state: ground state

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•Knock 1 electron up to n = 2, l = 0 and leave electron 2 where it was, in n = 1, l = 0 (so one is 1s and the second is 2s)• spin states are both TBD at the moment• now, should we build the symmetric space combo or the anti?• the unexcited electron is nearer the nucleus and is said to shield the nuclear charge from the excited electron, thereby reducing the Coulomb attraction with the excited electron, thereby raising (making less negative) the energy

• how else might we do it? Build the antisymmetric space (r)

• that’s not good ground state of He is a spin singlet, symmetric in space (separable!) ),()(),()(

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The first excited state of He

• one refers to this as the paired-spin (up&down) 2-electron state

Parahelium: the three first excited states• in the antisymmetric space combo, the electrons are further apart, so shielding is less successful this is the lower-energy solution• therefore spin is symmetric parahelium comes as triplets• the space wavefunction for parahelium:

• and of course the full wavefunction is to be multiplied by one of the three symmetric two-electron spinors s = 1, m = 1, 0, –1• we will shortly adjust the notation to accommodate both L and S

• in the symmetric space combo the electrons are closer together, so the shielding is more complete, and the excited electron is not bound so tightly… the effect is small but sufficient that the energy is larger• therefore spin is antisymmetric orthohelium comes as a singlet• the space function for orthohelium (multiply by spinor s = 0 m = 0)

Orthohelium: the sole second excited state

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Finding ground states for more-electron atoms• the n = 1 shell is now full with its two electrons ‘up’ & ‘down’ • more precisely, the n = 1 shell is full, as is its l = 0 orbital, with the helium ground state spatial and spin-singlet wavefunction• one can say that both 1s orbitals are occupied• now let L be the total orbital ang mom quantum number and S be the total spin quantum number, so here we have L = 0 and S = 0• the next electron has to go into the 2s orbital but it can choose l = 0 [2s] or l = 1 [2p]. • Which to choose? SHIELDING gives the way!!• in general, the lowest l electrons are closest to the nucleus and so they are not shielded so well, so their energy is lower s, then p• Li and Be thus fill up the 2s orbital• Li is hydrogen-like plus it has the third electron in the 2s orbital• to generate its fully antisymmetric wavefunction we resort to the Slater determinant• for He, the Slater determinant is that of a 2 x 2 matrix

The Slater determinant for He and Li• for helium ground state the determinant captures both terms:

• for lithium ground state we can express full wavefunction fw as a determinant to capture all six terms (put third electron up in 2s):

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The overall antisymmetry is tangled up in spin/space

• the 2s electron was chosen to be up arbitrarily• not at all expressible separably!!• there is a certain ‘cyclicity’ in the spin arrangements• beyond element three it gets rapidly more baroque• notice, though, that if you exchange any coordinate with any other, you get a sign flip: perfect!!• the remaining elements for n = 2 fill up as Hund’s rules govern, getting you to element #10 (Ne) Now the n = 2 shell is full and all three orbitals (1s 2s 2p) are filled and, in addition, are said to have no unpaired spins• thus, the TOTAL spin is zero: S = 0, and the TOTAL L = 0 too!• how did we get there, in more detail?

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Term symbols and spectroscopic notation• for historical reasons we use term symbols to denote the ground state electron configurations: 2S+1LJ

• L is the total orbital angular momentum quantum number, so

• it is denoted by letters S, P, D, F…(confusing!!!)• S is the total spin angular momentum quantum number, so

2)1( magnitude momentumangular orbital LLL

2)1( magnitude momentumangular spin SSS

• the z-component quantum numbers obey the usual rules: range from (e.g. for L) ML = – L, –(L – 1), …., (L – 1), L

similar for z-components of total spin quantum number MS

• multiplicity is 2S+1 which is the number of MS possibilities• finally, J is the sum of spin plus orbital quantum number and it and MJ too obey the same kinds of rules as L and S and ML and MS

• MJ can range from |L – S| up to L + S but not our concern here• the trick is sorting out what J is going to be in terms of L and S

Hund’s rules for filling electron levels• All states must be antisymmetric in exchange of any spin pair!• singlet: spins antiparallel paired (so antisymmetric in spin)• triplet: spins parallel unpaired (so antisymmetric in space)• First rule: other things being equal, the state with highest S has the lowest energy this means that spins tend to fill in by lining up• also called the bus seat rule• physical cause: shielding is worst in this case.. a tricky quantum mechanics problem• Second rule: For a given spin configuration, the state with the

largest L will have the lowest energy• physical cause: electrons orbiting in ‘opposite directions’ tend to bump into each other more often, thus more repulsion energy• Third rule: if a subshell (n, l) is up to and including half filled, then the lowest energy state has J = |L – S|; if it is more than half-filled, the lowest energy state has J = L + S• physical cause: spin-orbit coupling (a magnetic effect)

• in general, empty or filled shells tend to have the simplest term symbols; noble gases always have 1S0 S = L = J = 0 • closed shells (or subshells) will be denoted by the atom and so only the ‘subsequent’ electrons need to be analyzed• both electron configuration (ShellSubshell)# and term symbol• H: (1s) one s electron so S = ½ , L = 0 2S1/2

• He: (1s)2 two (singlet/paired) s electrons: S = 0, L = 0 1S0

• Li: (He)(2s) one s electron so S = ½ , L = 0 2S1/2

• Be: (He)(2s) 2 two (singlet/paired) s electrons: S = 0, L = 0 1S0

• now the fun begins!!• B: (Be)(2p) one p electron: S = ½ , L = 1 2P1/2 [H3]

• C: (Be)(2p)2 two (triplet/unpaired) p electrons : S = 1, L = 1 3P0

• Huh?? Why not L = 2 to obey H2 (would imply m = +1 for both)? But we can’t have both of the p electrons have m = 1, since they are in the same spin state one has m = 1, one has m = 0 to optimize H2 consistent with PEP… so [via H3] J = |L – S| = |1 – 1| = 0

Applying Hund’s rules to simple atoms Z = 1 to 6

• N: (Be)(2p)3 three (unpaired, so spin symmetric) p electrons: S =

3/2 , L = 0 [m = 1, 0, – 1] 4S3/2

• catching on yet? This one was the most extreme..• O: (Be)(2p)4 four (two paired, two not) p electrons: S = 1, L = 1

[m = 1, 0] 3P2 [used version II of H3 here to get J]

• F: (Be)(2p)5 five (four paired, one not) p electrons: S = 1/2 , L = 1

[m = 1] 2P3/2 [used version II of H3 here to get J]

• Ne: (Be)(2p)6 six (all paired) p electrons: S = 0, L = 0 1S0

• we have now filled the second shell all the way up to Z = 10• from here on out to Z = 18 (Ar) we fill 3s shell and then 3p shell just as before and all term symbols are the same.. Ar is closed shell• then, instead of filling 3d subshell, 4s fills next, up to Z = 20 (Ca) which is a closed subshell 1S0

• now the 3d shells start to fill… the ‘transition’ metals have begun, starting with Z = 21 (Sc) and ending with closed-subshell Z = 30 (Zn)• let’s do these 10 and call it a day..

Continuing along the shell n = 2 up to Z = 10

• Sc: (Ca)(3d) one d electron: S = ½ , L = 2 2D3/2 [H3]

• Ti: (Ca)(3d)2 two (triplet/unpaired) d electrons : S = 1, L = 3

3F2 [m = 2, 1 due to PEP, as for carbon]

• V: (Ca)(3d)3 three (all unpaired so spin symmetric) d

electrons: S = 3/2, L = 3 [m = 2, 1, 0] 4F3/2

• the next one (Z = 24, Cr) is very anamalous! one of the 4s electrons moves into the 3d orbital, so we have (here, and for Cu)• Cr: (Ar)(4s)(3d)5 one unpaired s electron and five (all unpaired so spin symmetric) d electrons: S = 3, L = 0 [m = 2, 1,0 –1, –2] 7S3 [used version II of H3 here to get J]

• Mn: (Be)(3d)5 five (all unpaired so spin symmetric) d

electrons: S = 5/2 and L = 0 [why?] 6S5/2 [halfway there!]

• Fe: (Be)(3d)6 six (two paired so spin antisymmetric, four not)• easiest way to deal with it is to work from Ms and add one d electron with opposite spin: S = 2, L = 2 5D4

• Co: (Be)(3d)7 Like Mn but with two opposite d electrons: S =

3/2, L = 3 [m = 2, 1] 4F9/2

Continuing along the shell n = 3 up to Z = 27

• interlude: these large numbers of of unpaired electrons give rise to ferromagnetism (exhibited by Fe, Co and Ni)… the spin’s magnetic moments communicate to nearby spins they line up too whole sample is macroscopically spin-aligned: a huge magnetic dipole!• Ni: (Be)(2p)8 Like Mn but with three opposite d electrons: S =

1, L = 3 [m = 2, 1, 0] 3F4

• Cu: (Ar)(4s)(2p)10 Like Cr so d subshell is closed: S = ½ thanks

to the s electron; L = 0 [m = 2, 1, 0, –1, –2 ] 2S1/2

• Zn: (Be)(2p)10 d subshell is closed: S = 0, L = 0 1S0

• now the 4p subshell start to fill with exactly the same pattern as previously seen for the filling of a p subshell… up to Z = 36 (Kr)• next comes subshell 5s… 4d transitions (with a Cr-like anamoly)… 5p… up to Z = 54 (Xe)• next comes subshell 6s… 4f (one 5d leaks in at first [the lanthanides]… 5d transitions… 6p… up to Z = 86 (Rn)• similar thing for 7s, 5f (the actinides)… stop at Z = 92 (U) [why?]• most horrid term symbol is Neptunium: [Ra](5f)4(6d)1 6L11/2

Most of the rest of the story

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Most of the rest of the story