CHM 501

13. Calculate Z* according to Slater's rules for the first row transition elements, Sc to Zn. Compare these to the Clementi-Raimondi Z* values. Which values compare favorably and which values compare poorly? In those cases where the comparison is poor, suggest a reason.

Answer

Element

Clementi-Raimondi Z*

Slater Z*

Sc

4.632

Z* = 21 – [1(0.35) + 9(0.85) + 10(1.00)] = 3.00

Ti

4.817

Z* = 22 – [1(0.35) + 10(0.85) + 10(1.00)] = 3.15

V

4.981

Z* = 23 – [1(0.35) + 11(0.85) + 10(1.00)] = 3.30

Cr

5.133

Z* = 24 – [13(0.85) + 10(1.00)] = 2.95

Mn

5.283

Z* = 25 – [1(0.35) + 13(0.85) + 10(1.00)] = 3.60

Fe

5.434

Z* = 26 – [1(0.35) + 14(0.85) + 10(1.00)] = 3.75

Co

5.576

Z* = 27 – [1(0.35) + 15(0.85) + 10(1.00)] = 3.90

Ni

5.711

Z* = 28 – [1(0.35) + 16(0.85) + 10(1.00)] = 4.05

Cu

5.858

Z* = 29 – [18(0.85) + 10(1.00)] = 3.70

Zn

5.965

Z* = 30 – [1(0.35) + 18(0.85) + 10(1.00)] = 4.35

The absolute magnitude of the two sets of values is quite different in all cases. However, plotting both sets of Z* values on the same axis shows that they both have the same trend and the same slope, with the Clementi-Raimondi values offset ~1.5 units above the Slater values. The two dips in the plot of the Slater values is a result of the change in electron configuration associated with the half-filled phenomenon. The Clementi-Raimondi determination of Z* does not show this effect, although experimental ionization energies certainly do change around Cr and Cu.