Oxidation Number And Oxidation States

To determine the number of unpaired electrons in Lutetium (Lu) in the +3 oxidation state, we first need to understand its electronic configuration. Lutetium has an atomic number of 71. Therefore, its ground state electron configuration is: [Xe] 4f¹⁴ 5d¹ 6s²

Lutetium in the +3 oxidation state means it has lost three electrons. These electrons are removed first from the outermost shell following the order of 6s and then 5d:

• Two electrons are removed from 6s, and one electron is removed from 5d.

Thus, the electron configuration for Lu³⁺ is: [Xe] 4f¹⁴

Next, we look at the 4f sublevel, which is completely filled with 14 electrons. A completely filled sublevel has no unpaired electrons.

Thus, Lutetium in the +3 oxidation state has 0 unpaired electrons.