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.
 
 
 
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.