CHM 501

Homework 2

Due September 20, 2001



1. Find all of the term symbols for the s1p1 electron configuration, including J values. Energy order these states.


In this case there are no Pauli forbidden states because each electron is required to have a different l quantum number. Thus, L = 1 and S = 1. The total number of microstates is (6!/1!5!)(2!/1!1!) = 12.

ML\MS

1

0

-1

1

(0s+, 1p+)

(0s+, 1p-) (0s-, 1p+)

(0s-, 1p-)

0

(0s+, 0p+)

(0s+, 0p-) (0s-, 0p+)

(0s-, 0p-)

-1

(0s+, -1p+)

(0s+, -1p-) (0s-, -1p+)

(0s-, -1p-)

Find the states:

ML = 1, MS = 1: 3P (9 states), J = 2, 1, 0

ML = 1, MS = 0: 1P (3 states), J = 1

Since the configuration is less than half-filled, the smallest J value is the lowest energy, so the energy order is 3P0 < 3P1 < 3P2 < 1P1.


2. Use Slater's rules to calculate Z* for Ru, Rh, Pd, Os, Ir, and Pt. What are the periodic trends?


Find the electron configurations and write them according to Slater's prescription. Then use the rules to find the screening constant and then Z*

Ru: (1s)2(2s2p)8(3s3p)8(3d)10(4s4p)8(4d)7(4f)0(5s5p)1
For the 5s valence electron: S = (15)(0.85) + (28)(1.00) = 40.75 so Z* = 44 - 40.75 = 3.25

Rh: (1s)2(2s2p)8(3s3p)8(3d)10(4s4p)8(4d)8(4f)0(5s5p)1
For the 5s valence electron: S = (16)(0.85) + (28)(1.00) = 41.60 so Z* = 45 - 41.60 = 3.40

Pd: (1s)2(2s2p)8(3s3p)8(3d)10(4s4p)8(4d)10(4f)0(5s5p)0
For the 4d valence electron: S = (9)(0.35) + (36)(1.00) = 39.15 so Z* = 46 - 39.15 = 6.85

Os: (1s)2(2s2p)8(3s3p)8(3d)10(4s4p)8(4d)10(4f)14(5s5p)8(5d)6(5f)0(6s6p)2
For the 6s valence electron: S = (1)(0.35) + (14)(0.85) + (60)(1.00) = 72.25 so Z* = 76 - 72.25 = 3.75

Ir: (1s)2(2s2p)8(3s3p)8(3d)10(4s4p)8(4d)10(4f)14(5s5p)8(5d)7(5f)0(6s6p)2
For the 6s valence electron: S = (1)(0.35) + (15)(0.85) + (60)(1.00) = 73.10 so Z* = 77 - 73.10 = 3.90

Pt: (1s)2(2s2p)8(3s3p)8(3d)10(4s4p)8(4d)10(4f)14(5s5p)8(5d)9(5f)0(6s6p)1
For the 6s valence electron: S = (17)(0.85) + (60)(1.00) = 74.45 so Z* = 77 - 74.45 = 3.55

Z* generally increases from left to right on the Periodic Table, but Pt appears to be an exception to this rule. Going down the Periodic Table gives a slight increase in Z* because of the poor shielding abilities of the f electrons, but again Pt appears to be an exception, at least based upon Slater's Rules.