B. E. Douglas, D. H. McDaniel, J. J. Alexander, Concepts and Models of Inorganic Chemistry, 3rd edition, John Wiley & Sons, New York, 1994, page 52.
 
 

Problem 1.23

Use Slater’s rules to calculate the effective nuclear charges for K, Rb, and Cs. Compare these results with those for Cu, Ag, and Au. Comment on the ionization energies of these elements in light of these charges.
 
 

Answer:
 
Element
Electron Configuration
Z* Calculated
Ionization Energy (eV)
K
(1s2)(2s22p6)(3s23p6)(4s1)
19 – [10(1.00) + 8(0.85)] = 2.20
4.341
Rb
(1s2)(2s22p6)(3s23p6)(3d10)(4s24p6)(5s1)
37 – [28(1.00) + 8(0.85)] = 2.20
4.177
Cs
(1s2)(2s22p6)(3s23p6)(3d10)(4s24p6)(4d10)(5s25p6)(6s1)
55 – [46(1.00) + 8(0.85)] = 2.20
3.894
Cu
(1s2)(2s22p6)(3s23p6)(3d10)(4s1)
29 – [10(1.00) + 18(0.85)] = 3.70
7.726
Ag
(1s2)(2s22p6)(3s23p6)(3d10)(4s24p6)(4d10)(5s1)
47 – [28(1.00) + 18(0.85)] = 3.70
7.576
Au
(1s2)(2s22p6)(3s23p6)(3d10)(4s24p6)(4d10)(4f14)(5s25p6)(5d10)(6s1)
79 – [60(1.00) + 18(0.85)] = 3.70
9.225

 

For the Group 1 elements. all of the Z* values are identical and the ionization energies are very similar, so Slater’s rules do an adequate job. For the Group 12 elements, again the Z* values are all the same and larger than found for group 1, consistent with the significantly larger ionization energies. However, Slater’s rules do a poor job of accounting for the large jump in the ionization energy for Au. This is because the rules contribute too much shielding or the 4f electrons.