Exercise 4 | Ionic Bonds and Compounds

Lead (Z = 77) can form 2 different cations.

Write the electronic configuration of Pb.

From this electronic configuration, determine the two possible ions of Pb and their electronic configuration.

Ground state configuration of neutral atom: 4s < 3d < 4 p < 5s < 4d < 5p< 6s < 4f < 5d

Pb: Z = 77 ⇒ 1s22s22p63s23p64s23d104p65s24d105p66s24f145d7


Ground state configuration of transition metal ions: 3d < 4s < 4p < 4d < 4f < 5s < 5p < 5d < 6s
First, we remove electrons from the orbital 6s ⇒ 2 electrons ⇒ Pb2+

Pb2+: 1s22s22p63s23p64s23d104p65s24d105p64f145d7


Then, we remove electrons from the orbital 5d to have a half-filled orbital (half-filled orbital is stable) ⇒ 2 electrons (+ 2 from 6s) ⇒ Pb4+

Pb4+: 1s22s22p63s23p64s23d104p65s24d105p64f145d5