Question

A) Which form(s) of Hg – Hg, Hg+, Hg2+, Hg3+, or Hg4+ – would you predict to

have an unusually large ionization energy compared to other transition metal atoms of the same charge? Why?

B)   What does Slater’s rules state for how Zeff compares for ns and np electrons (e.g. 4s and 4p) within an atom? How would you predict the two Zeff values to compare based on the three factors that contribute to Zeff discussed in class?

Answer 1

A) The electronic configuration of Hg is [Xe]4f¹⁴5d¹⁰6s². The formation of Hg⁺ and Hg²⁺ involves removal of the 6s electrons. The process is energy-consuming; yet not impossible to be supplied by reactions at normal or high temperatures. The third ionization energy of Hg involves the breaking of the complete 5d shell. This is an extremely energy-demanding process and reactions at normal temperatures or elevated temperatures can barely supply the energy. Hence, the third ionization energy of Hg will be unusually high and Hg³⁺, if formed, is formed with utmost difficulty.

B) Slater’s rules for ns and np electrons can be written as

i) An electron in the n shell contributes 0.35 to the shielding constant (except the 1s electrons which contribute 0.30).

ii) Electrons in the (n-1) shell contribute 0.85 each to the shielding constant.

iii) All electrons in the (n-2) shell contribute 1.00 to the shielding constant.

Consider, for example, the effective charge, Z_eff for the 4s electron in Ca (atomic number Z = 20). The electronic configuration of Ca is 1s²2s²2p⁶3s²3p⁶4s². Use Slater’s rules as below:

a) The 4s electron contributes 0.35.

b) We have a total of (2 + 6) = 8 electrons in the 3^rd shell; the total contribution from 8 electrons is (8*0.85) = 6.8.

c) We have a total 10 electrons in the 1^st and 2^nd shells; the total to the shielding constant is (10*1.00) = 10.0

The total shielding constant is S = 0.35 + 6.8 + 10.00 = 17.15.

The effective nuclear charge, Z_eff = Z – S where Z = 20; therefore, Z_eff = 20 – 17.15 = 2.85 (ans).