Problem 1.19

What spectroscopic terms arise from the configuration 1s²2s²2p⁶3s²3p⁵3d¹?
 
 
 
 

Answer:

This is equivalent to a p¹d¹ (p⁵ is a p¹ "hole") configuration, so treat it this way to simplify things.

L = 2 + 1 = 3, M_L = 3, 2, 1, 0, -1, -2, -3

S = ½ + ½ = 1, M_S = 1, 0, -1

There are no Pauli forbidden states because each electron has a different l quantum number.

There are 10 possible states for d¹ and 6 possible states for p¹, so there are 60 possible states in the mixed configuration. For consistency, always put the p electron first in the microstate.
 

ML\MS	1	0	-1
3	(1+,2+)	(1+,2-)(1-,2+)	(1-,2-)
2	(1+,1+) (0+, 2+)	(1+,1-) (0+, 2-)(1-,1+) (0-, 2+)	(1-,1-) (0-, 2-)
1	(1+,0+) (0+, 1+) (-1+, 2+)	(1+,0-) (0+, 1-) (-1+, 2-)(1-,0+) (0-, 1+) (-1-, 2+)	(1-,0-) (0-, 1-) (-1-, 2-)
0	(1+,-1+) (0+, 0+) (-1+, 1+)	(1+,-1-) (0+, 0-) (-1+, 1-)(1-,-1+) (0-, 0+) (-1-, 1+)	(1-,-1-) (0-, 0-) (-1-, 1-)
-1	(-1+,0+) (0+, -1+) (1+, -2+)	(-1+,0-) (0+, -1-) (1+, -2-)(-1-,0+) (0-, -1+) (1-, -2+)	(-1-,0-) (0-, -1-) (1-, -2-)
-2	(-1+,-1+) (0+, -2+)	(-1+,-1-)(0+, -2-)(-1-,-1+)(0-, -2+)	(-1-,-1-) (0-, -2-)
-3	(-1+,-2+)	(-1+,-2-)(-1-,-2+)	(-1-,-2-)

All 60 microstates are found.

Terms:

M_L = 3, M_S = 1 gives ³F (21 microstates), J = 4, 3, 2

M_L = 2, M_S = 1 gives ³D (15 microstates), J = 3, 2, 1

M_L = 1, M_S = 1 gives ³P (9 microstates), J = 2, 1, 0

M_L = 3, M_S = 0 gives ¹F (7 microstates), J =3

M_L = 2, M_S = 0 gives ¹D (5 microstates), J = 2

M_L = 1, M_S = 0 gives ¹P (3 microstates), J = 1

Net:

³F₄, ³F₃, ³F₂, ³D₃, ³D₂, ³D₁, ³P₂, ³P₁, ³P₀, ¹F₃, ¹D₂, ¹P₁