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.