Configuration and Terms

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158a Lecture 20 Auf Bau Principle Periodic
Table Periodic Properties
Atomic Configurations
Yielding different atomic states
with
different energies
Fred J. Grieman
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Energy of Atomic Orbitals General Results E
f(n, l) Learned from 1 e- atoms E ? as n ?
ltrgt ? lt1/rgt ?, less
nuclear/electron attraction Learned from
Multi-e- atoms E ? as l ? Perturb.
Theory increased repulsion btwn core e-s
valence e-s as l increases
(e.g., J1s,2pgtJ1s,2s) Variational Theory
decreased screening effect as l
decreases because
penetration
into core via radial distribution
function ?E(n
effect) gt ?E(l effect)
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Configurations states (terms) with different
angular momentum different energies
Add electrons via Auf Bau to get ground
electronic state
4
4f
4d
1S0
2S½
2P3/2,1/2
??
3
4p
2S½
1S0
3d
4s
3p
3s
2
E
2p
2s
1
1s
Shell
n
Z
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Explains Periodic Table
4
4f
4d
3
4p
3d
4s
3p
3s
2
E
2p
2s
1
1s
Shell
n
Z
5
Why not 54.4 ?
?l
e- pair repulsion
?n
Explain Periodic Properties Ionization
Energies
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Configuration and Terms
He results 1s2p ? 1P1 ?
3P2,1,0

Different Energies E(2p) gt E(2s) 1s2s ? 1S0

Esinglet gt Etriplet ?
3S1 (spatial
wavefunction) 1s2 ? 1S0
Given a configuration, how do you determine Terms
(states) possible and their energies?

• Two cases (only consider electrons in open
  subshells)
• Inequivalent electrons (different n and/or l)
• Equivalent electrons (same n and l)

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Inequivalent electrons No Pauli Exclusion

• Steps
• Determine Lmax Lmin ? L Lmax, Lmax-1, , Lmin
• Determine Smax Smin ? S Smax, Smax-1, , Smin
• Determine J from L S

Example He 2p3p configuration (or C 1s22s22p3p
excited state)

• Lmax l1 l2 2, Lmin l1 - l2 0 ? L
  2, 1, 0
• Smax s1 s2 1, Smin ½ - ½ 0 ? S 1,
  0
• All combinations possible because no Pauli
  Exclusion

Create table
Total 36 states
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Equivalent electrons Pauli Exclusion now
restricts possible states

• Steps
• Determine of possible states
• G!/N!(G-N)! G of spin-orbitals, N
  of electrons
• 2) Determine Lmax Lmin from Lz that are Pauli
  allowed
• 3) Determine Smax for Lmax determine states for
  all possible S
• 4) Continue for all possible L
• 5) Continue until you account for all states
• 6) Determine J from L S

Example He 2p2 (same as C 1s22s22p2 in handout)
of possible states 6!/2!(6-2)! 15 states
(In handout reproduced here.)
Lz(max) 2
? Lmax 2, why?
ml 1 0 -1
Lz(min) 0
ml 1 0 -1
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L(max) 2
Smax must 0 because Sz 0
?
ml 1 0 -1
? we have a 1D state J 2 0 2 ? 1D2 which
accounts for 5 states
Lmax 1 1
Smax 1 because Sz(max) 1
ml 1 0 -1
? we have a 3P state J 1 1 2, 2 - 1 1,
1 1 0 ? 3P2,1,0
which accounts for 9 states
Only 1 state left!!! Must be 1S0 which gives us
a total of 15 states
Energies? Can use perturbation theory or
variational theory using proper wave function to
get energies
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Energies of Terms Hunds Rules general rules
based on what weve determined

• For same configuration, highest spin multiplicity
  is most stable.
• Why from what we did before?
• Higher spin multiplicity has
  spatial wave function
• with least
  electron-electron repulsion

II. If same spin multiplicity, highest orbital
angular momentum is most stable. Why?
Electrons moving in same
direction have
less electron-electron repulsion
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Then for He 2p2 or C 1s22s22p2 What would 1D, 3P,
1S order be? E(3P) lt E(1D) lt E(1S) E(3P0) lt
E(3P1) lt E(3P2)
1P1
3P2
3P1
3P0
Spectroscopy Selection Rules ?L ?1 ?S
0 Carbon excited state 1s22s22p3s L 1 S 1
or 0 ? 3P2,1,0 1P1 1P1 ? 1D2 1P1 ?
1S0 allowed
1S0
All forbidden
E
1D2
3P2
3P1
3P0
Test
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