The lanthanide ion having four unpaired electrons is (Given : Atomic numbers of Ce = 58, Nd = 60, Tb = 65 and Ho = 67)
- Ho3+
- Nd3+
- Ce3+
- Tb3+
Answer
(1) Ho3+
Solution
Electronic configurations of the tripositive lanthanide ions:
- Ce3+ (Z=58): [Xe] 4f1 1 unpaired electron
- Nd3+ (Z=60): [Xe] 4f3 3 unpaired electrons
- Tb3+ (Z=65): [Xe] 4f8 14 - 8 = 6 unpaired electrons
- Ho3+ (Z=67): [Xe] 4f10 14 - 10 = 4 unpaired electrons
Thus, Ho3+ has four unpaired electrons.
Answer: (1)