Which of the following Ln2+ metal ion is diamagnetic?
(Given Atomic number, Yb = 70, Ho = 67, Pm = 61, Dy = 66)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Yb2+
A species is diamagnetic when it has no unpaired electrons. For lanthanoid ions this happens only when the 4f subshell is empty (4f0) or completely filled (4f14); any other 4f count leaves unpaired electrons in the seven 4f orbitals.
The lanthanoids given here have the configuration [Xe]4fn6s2, with n = Z − 56. Forming Ln2+ removes the two 6s electrons, leaving 4fn:
Yb2+: Yb (Z = 70) is [Xe]4f146s2; \( \ce{Yb^{2+}} \) is 4f14, all electrons paired, so it is diamagnetic.
Ho2+: Ho (Z = 67) is [Xe]4f116s2; \( \ce{Ho^{2+}} \) is 4f11, with 3 unpaired electrons.
Pm2+: Pm (Z = 61) is [Xe]4f56s2; \( \ce{Pm^{2+}} \) is 4f5, with 5 unpaired electrons.
Dy2+: Dy (Z = 66) is [Xe]4f106s2; \( \ce{Dy^{2+}} \) is 4f10, with 4 unpaired electrons.
So Yb2+ is the diamagnetic ion.
NCERT (2026-27 edition):
Oxidation States – …
