Which of the following Lanthanoid ions is diamagnetic? [Atomic number…

The correct answer is Option (4) → \(Yb^{2+}\)

An ion is diamagnetic only when it has no unpaired electrons. For the lanthanoids, the 6s electrons are lost first when the M²⁺ ion forms (for Ce, the 5d electron also), and the number of unpaired 4f electrons decides the magnetic behaviour:

• Ce (Z = 58): [Xe] 4f¹ 5d¹ 6s²; \( \ce{Ce^{2+}} \) is [Xe] 4f² — 2 unpaired electrons, paramagnetic.

• Sm (Z = 62): [Xe] 4f⁶ 6s²; \( \ce{Sm^{2+}} \) is [Xe] 4f⁶ — 6 unpaired electrons, paramagnetic.

• Eu (Z = 63): [Xe] 4f⁷ 6s²; \( \ce{Eu^{2+}} \) is [Xe] 4f⁷ — 7 unpaired electrons, paramagnetic.

• Yb (Z = 70): [Xe] 4f¹⁴ 6s²; \( \ce{Yb^{2+}} \) is [Xe] 4f¹⁴ — all electrons paired, diamagnetic.

So the diamagnetic ion is Yb²⁺.

Why the other options are incorrect

• Option (1) Ce²⁺ — Incorrect. 4f², two unpaired electrons, so it is paramagnetic.

• Option (2) Sm²⁺ — Incorrect. 4f⁶, six unpaired electrons, so it is paramagnetic.

• Option (3) Eu²⁺ — Incorrect. 4f⁷, a half-filled 4f subshell with seven unpaired electrons, so it is strongly paramagnetic.

NCERT (2026-27 edition):
Electronic Configurations – “… Electronic configurations (Ln and Ln²⁺): 58 Cerium, Ce: 4f¹5d¹6s², 4f²; 62 Samarium, Sm: 4f⁶6s², 4f⁶; 63 Europium, Eu: 4f⁷6s², 4f⁷; 70 Ytterbium, Yb: 4f¹⁴6s², 4f¹⁴.”