## CHM 501

6. Find all the Russell-Saunders terms for the f2 configuration. Energy order them. For the ground state, find all the J values and energy order these states.

For f2:

L = 6 (ml = 3 for each electron) so ML = 6, 5, 4, … –5, –6

S = 1 (ms = ½ for each electron) do MS = 1, 0, –1

The total number of microstates = 14!/(2!)(12!) = 91

 ML \ MS 1 0 … 6 (3+,3–) 5 (3+,2+) (3+,2–) (3–,2+) 4 (3+,1+) (3+,1–) (3–,1+) (2+,2–) 3 (3+,0+) (2+,1+) (3+,0–) (3–,0+) (2+,1–)(2–,1+) 2 (3+,–1+) (2+,0+) (3+,–1–) (3–,–1+) (2+,0–)(2–,0+) (1+,1–) 1 (3+,–2+) (2+,–1+) (1+,0+) (3+,–2–) (2+,–1–) (1+,0–) (3–,–2+) (2–,–1+) (1–,0+) 0 (3+,–3+) (2+,–2+) (1+,–1+) (3+,–3–) (2+,–2–) (1+,–1–) (3–,–3+) (2–,–2+) (1–,–1+) (0+,0–) …

ML = 5, MS = 1: 3H (33 microstates)

ML = 3, MS = 1: 3F (21 microstates)

ML = 1, MS = 1: 3P (9 microstates)

ML = 6, MS = 0: 1I (13 microstates)

ML = 4, MS = 0: 1G (9 microstates)

ML = 2, MS = 0: 1D (5 microstates)

ML = 0, MS = 0: 1S (1 microstate)

The above are listed in increasing energetic order (as least, that's the prediction).

For the 3H ground state, J = 6, 5, 4. Since the configuration is less than half-filled, the energy order is 3H4 (lowest energy) < 3H5 < 3H6.