Arrange the following lanthanoid ions increasing order of their number of 4f electrons:
(A) Ce4+ (Z=58)
(B) Tb4+ (Z=65)
(C) Yb2+ (Z=70)
(D) Eu3+ (Z=63)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (D), (B), (C)
Lanthanoids have the general configuration [Xe]4f1–145d0–16s2. When an ion is formed, the 6s electrons are removed first, then the 5d electron (if present), and only then the 4f electrons. So the number of 4f electrons in an ion is:
\( \displaystyle \text{4f electrons}=(Z-54)-\text{charge on the ion} \)
(54 is the atomic number of xenon, and every electron beyond xenon ends up in 4f once the 6s and 5d electrons are gone.)
(A) \( \ce{Ce^{4+}} \) (Z=58): [Xe]4f15d16s2 → 4 − 4 = 4f0
(D) \( \ce{Eu^{3+}} \) (Z=63): [Xe]4f76s2 → 9 − 3 = 4f6
(B) \( \ce{Tb^{4+}} \) (Z=65): [Xe]4f96s2 → 11 − 4 = 4f7
(C) \( \ce{Yb^{2+}} \) (Z=70): [Xe]4f146s2 → 16 − 2 = 4f14
Shortcut: \( \ce{Ce^{4+}} \) (A) has no 4f electrons and \( \ce{Yb^{2+}} \) (C) a full 4f shell, so the order starts with (A) and ends with (C): Options (1) and (2). \( \ce{Eu^{3+}} \) (6) has fewer than \( \ce{Tb^{4+}} \) (7), which leaves Option (2).
NCERT (2026-27 edition):
Oxidation States – “… In the lanthanoids, La(II) and Ln(III) compounds are predominant species. However, occasionally +2 and +4 ions in solution or in solid compounds are also obtained. This irregularity (as in ionisation enthalpies) arises mainly from the extra stability of empty, half-filled or filled f subshell. Thus, the formation of CeIV is favoured by its noble gas configuration, but it is a strong oxidant reverting to the common +3 state. … Pr, Nd, Tb and Dy also exhibit +4 state but only in oxides, MO2.”