Download 1 Chem. 152 Term Symbols for Atoms with Equivalent Electrons Prof

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Chem. 152
Prof. Anderson
Term Symbols for Atoms with Equivalent Electrons
Equivalent electrons have the same n and l values, so the possiblity exists that they might end up
with all four quantum numbers the same, which is forbidden by the Pauli Principle. In this case
you have to look at all allowable combinations of ML and MS values, and from those values infer
the L and S values for the whole atom.
1. How many states must you identify? Let G = number of spin orbitals available, and N = the
number of electrons to put into those spin orbitals.
The # of allowed states = G! / [N! (G-N)!]
For example, the carbon atom in the ground state has electron configuration 1s2 2s2 2p2.
There are six spin orbitals available for two electrons: 2p+1 α, 2p+1 β, 2p0 α, 2p0 β, 2p-1 α, and
2p-1 β. Therefore G = 6 and N = 2, so
# of allowed states = G! / [N! (G-N)!] = 6! /[2! (6-2)!] = 6! /[2! 4!]
= 6•5•4•3•2•1/[2•1•4•3•2•1]
= 15
2. List all of the 15 possible allowed states. Use ↑ and ↓ for α and β, respectively. (Fill
columns 1-4 below.)
Conf. #
2p+1
2p0
2p-1
ML
MS
↑↓
1
2
0
↑
↑
2
1
1
↑
↓
3
1
0
↓
↑
4
1
0
↓
↓
5
1
-1
↑
↑
6
0
1
↑
↓
7
0
0
↓
↑
8
0
0
↓
↓
9
0
-1
↑↓
10
0
0
↑
↑
11
-1
1
↑
↓
12
-1
0
↓
↑
13
-1
0
↓
↓
14
-1
-1
↑↓
15
-2
0
3. Add the individual electron ml values to get ML for the atom. (Fill column 5 above.)
4. Add the individual electron ms values to get MS for the atom. (Fill column 6 above. For α,
ms = +1/2 and for β, ms = -1/2.)
5. Make a table listing the number of times a given (ML, MS) combination occurs. For
example, the (1, 0) combination occurs twice, in conformations numbered 3 and 4 above.
MS \\ ML
-1
0
1
-2
0
1
0
-1
1
2
1
0
1
3
1
1
1
2
1
2
0
1
0
2
Check to see that the total number of combinations adds up to 15:
row 1: 0 + 1 + 1 + 1 + 0 = 3
row 2: 1 + 2 + 3 + 2 + 1 = 9
row 3: 0 + 1 + 1 + 1 + 0 = 3
15
Check to see the table is symmetric about the center up and down and left and right.
6. Find the (ML, MS) combination which has the largest value of ML. In the table below,
the combination (2, 0) has the largest value of ML. If the largest ML value is 2, there must be
an L quantum number also of 2, and L = 2 means there are five associated ML values: -2, -1,
0, 1, and 2. The highlighted row must have in it five states arising from L = 2 and S = 0 with
(ML, MS) values of (-2, 0), (-1, 0), (0, 0), (1, 0), and (2, 0).
The term symbol with L = 2 and S = 0 has a J value = L+S = 2 + 0 = 2, so the state must be a
D state (L = 2) with multiplicity (2 S + 1) = 2(0) + 1 = 1.
1
D2
Ms \\ ML
-1
0
1
-2
0
1
0
-1
1
2
1
0
1
3
1
1
1
2
1
2
0
1
0
7. Rewrite the Table, eliminating the five combinations already accounted for. There should be
10 (= 15 - 5) states left.
Ms \\ ML
-1
0
1
-2
0
10
0
-1
1
21
1
0
1
32
1
1
1
21
1
2
0
10
0
Look again for the (ML, MS) combination which has the largest value of ML. In the table
above, the combination (1, 1) has the largest value of ML and MS. If the largest ML value is
1, there must be an L quantum number also equal to 1, and L = 1 means there are three
associated ML values: -1, 0, and 1. The same is true for MS: S must be equal to 1 and thus
there must be three associated MS values: -1, 0, and 1. The highlighted area has nine states
arising from L = 1 and S = 1, with (ML, MS) values of (-1, -1), (-1, 0), (-1, 1), (0, -1), (0, 0),
(0, 1), (1, -1), (1, 0), and (1, 1).
The term symbol with L = 1 and S = 1 has J values J = L+S, L+S-1, … |L-S| = 1 + 1,
1 + 1 - 1, …, | 1 - 1| or J = 2, 1, and 0. The states must be P states (L = 1) with multiplicity
(2S + 1) = 2(1) + 1 = 3.
3
P2, 3P1, 3P0
8. After eliminating the 14 (= 5 + 9) combinations already accounted for, there is only 1
(= 15 - 14) state left, which is a state with ML = 0 and MS = 0. This must be a state with L =
0 and S = 0, which means J = 0, also. The term symbol for this state is 1S0.
9. Apply Hund's rules to find the lowest energy state. In this case it is 3P0, having the highest
multiplicity and the lowest J value, because the 2p orbital is less than half filled.