The lanthanide ion having four unpaired electrons is (Given : Atomic numbers of Ce =58 , Nd =60 , Tb =65 and Ho =67 )

Step-by-step solution

The general electronic configuration of lanthanide ions Ln ³⁺ is [ Xe ] 4f^ Z-57 , where Z is the atomic number. For Ho ³⁺ ( Z=67 ): Number of 4f electrons = 67 - 57 = 10 Configuration is [ Xe ] 4f¹⁰ . Since the f -subshell has 7 orbitals, 10 electrons will fill as 3 pairs and 4 unpaired electrons. Number of unpaired electrons = 14 - 10 = 4 . For Nd ³⁺ ( Z=60 ): Number of 4f electrons = 60 - 57 = 3 Configuration is [ Xe ] 4f³ . Number of unpaired electrons = 3 . For Ce ³⁺ ( Z=58 ): Number of 4f electrons = 58 - 57