Assertion (A): The Eu2+ ion is a strong reducing agent.Reason (R): Eu…

The correct answer is Option (3) → (A) is true but (R) is false.

In the lanthanoids the +3 state is the common oxidation state. When a lanthanoid atom forms a cation, the two 6s electrons are lost first; electrons are taken from the 4f level only after that.

Assertion (A): The Eu²⁺ ion is a strong reducing agent. — True. Eu²⁺ readily loses one more electron and changes to the common +3 state, so it reduces other substances.

Reason (R): Europium forms the Eu²⁺ ion by losing two of its 4f electrons. — False. Eu (Z = 63) is [Xe] 4f⁷ 6s² and Eu²⁺ is [Xe] 4f⁷. The ion is formed by losing the two 6s electrons; the 4f⁷ set is kept whole, and this half-filled 4f⁷ configuration is why the Eu²⁺ ion forms.

NCERT (2026-27 edition):
4.5.3 Oxidation States – “…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.”

Table 4.9: Electronic Configurations and Radii of Lanthanum and Lanthanoids (row for europium)

* Only electrons outside [Xe] core are indicated