Select the Lanthanoids which show +2 oxidation state? (A) Yb(B) Ce(C)…

The correct answer is Option (1) → (A) and (C) only

The common oxidation state of the lanthanoids is +3; a +2 ion forms when the atom loses only its two 6s electrons.

• (A) Yb — Correct. Yb (Z = 70) is [Xe]4f¹⁴6s²; losing the two 6s electrons gives \( \ce{Yb^{2+}} \): 4f¹⁴, a completely filled subshell.

• (B) Ce — Incorrect. Ce (Z = 58) is [Xe]4f¹5d¹6s². Besides +3 it shows +4: losing all four outer electrons gives \( \ce{Ce^{4+}} \) with the noble gas configuration [Xe]. It does not form a stable +2 ion.

• (C) Eu — Correct. Eu (Z = 63) is [Xe]4f⁷6s²; losing the two 6s electrons gives \( \ce{Eu^{2+}} \): 4f⁷, a half-filled subshell.

• (D) Ho — Incorrect. Ho (Z = 67) is [Xe]4f¹¹6s²; a +2 ion (4f¹¹) would have no special stability, so holmium shows only the usual +3 state (\( \ce{Ho^{3+}} \): 4f¹⁰).

Note: The NCERT book prints La(II) in the first line of the passage below; it should read La(III).

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 Ce^IV is favoured by its noble gas configuration, but it is a strong oxidant reverting to the common +3 state. … Eu²⁺ is formed by losing the two s electrons and its f⁷ configuration accounts for the formation of this ion. However, Eu²⁺ is a strong reducing agent changing to the common +3 state. Similarly Yb²⁺ which has f¹⁴ configuration is a reductant.”