Identify the lanthanoid that exhibits zero effective magnetic moment in +3 state.

Step-by-step solution

A species exhibits zero effective magnetic moment only if it contains no unpaired electrons. For lanthanoid ions in the +3 oxidation state, we analyze their electronic configurations to determine which one has no unpaired electrons. Holmium (Ho, atomic number 67): Ground state configuration is [ Xe ] 4f¹¹ 6s^2 , so Ho ³⁺ has [ Xe ] 4f¹⁰ . With 10 electrons in 7 orbitals, there are 4 unpaired electrons according to Hund’s rule. Praseodymium (Pr, atomic number 59): Ground state is [ Xe ] 4f³ 6s^2 , so Pr ³⁺ is [ Xe ]