Choose the Incorrect statement from the options given below:

The correct answer is Option (3) → Tb⁴⁺ act as reducing agent.

The most stable oxidation state of the lanthanides is +3. Some of them also show +2 or +4, especially when this gives an empty (4f⁰), half-filled (4f⁷) or completely filled (4f¹⁴) f subshell. Such ions still tend to return to the +3 state:

• A +2 ion changes to +3 by losing an electron, so it is a reducing agent.

• A +4 ion changes to +3 by gaining an electron, so it is an oxidising agent.

Terbium (Z = 65) is [Xe]4f⁹6s²; losing the two 6s electrons and two 4f electrons gives \( \ce{Tb^{4+}} \): 4f⁷. This ion tends to gain an electron and become \( \ce{Tb^{3+}} \):

    \( \displaystyle \ce{Tb^4+ + e^- -> Tb^3+} \)

• Option (1) Eu²⁺ act as a reducing agent. — A correct statement. Eu (Z = 63) is [Xe]4f⁷6s²; losing the two 6s electrons gives \( \ce{Eu^{2+}} \): 4f⁷. It readily loses an electron to become \( \ce{Eu^{3+}} \).

• Option (2) Ce⁴⁺ act as an oxidising agent. — A correct statement. Ce (Z = 58) is [Xe]4f¹5d¹6s²; losing all four electrons gives \( \ce{Ce^{4+}} \): 4f⁰. It gains an electron to become \( \ce{Ce^{3+}} \) (E^° = +1.74 V), so it is a good oxidising agent.

• Option (3) Tb⁴⁺ act as reducing agent. — The incorrect statement. \( \ce{Tb^{4+}} \) gains an electron to return to the +3 state, so it acts as an oxidising agent, not a reducing agent.

• Option (4) Yb²⁺ act as a reducing agent. — A correct statement. Yb (Z = 70) is [Xe]4f¹⁴6s²; losing the two 6s electrons gives \( \ce{Yb^{2+}} \): 4f¹⁴, which readily loses an electron to become \( \ce{Yb^{3+}} \).

NCERT (2026-27 edition):
Oxidation States