Match the property given in List-I with the element given in List-II…

The correct answer is Option (1) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)

The lanthanoids (Ce to Lu) fill the 4f subshell, and their common oxidation state is +3. A lanthanoid shows a +2 or +4 state when that leaves it with an empty (4f⁰), half-filled (4f⁷) or completely filled (4f¹⁴) 4f subshell, which is extra stable. The 4f count of each ion is worked out from Z: the 6s electrons go first, then a 5d electron if there is one, then 4f electrons.

• (A) lanthanoid, which can show +2 oxidation state — This matches (III) Eu. Eu (Z = 63) is \( \mathrm{[Xe]\,4f^{7}\,6s^{2}} \); losing the two 6s electrons gives \( \ce{Eu^2+} \): \( \mathrm{4f^{7}} \), a half-filled subshell, so \( \ce{Eu^2+} \) is formed (it is a strong reducing agent and goes to +3).

• (B) Radioactive lanthanoid — This matches (I) Pm. Promethium (Z = 61) has no stable isotope; it is the only lanthanoid that is radioactive and is made artificially.

• (C) Lanthanoid which has \(4f^7\) electronic configuration in +3 oxidation state — This matches (IV) Gd. Gd (Z = 64) is \( \mathrm{[Xe]\,4f^{7}\,5d^{1}\,6s^{2}} \); losing the two 6s electrons and the 5d electron gives \( \ce{Gd^3+} \): \( \mathrm{4f^{7}} \).

• (D) Lanthanoid which has \(4f^{14}\) electronic configuration in +3 oxidation state — This matches (II) Lu. Lu (Z = 71) is \( \mathrm{[Xe]\,4f^{14}\,5d^{1}\,6s^{2}} \); losing the two 6s electrons and the 5d electron gives \( \ce{Lu^3+} \): \( \mathrm{4f^{14}} \).

Shortcut: (A)-(III) appears only in Options (1) and (2). Pm is the radioactive one, so (B)-(I) leaves Option (1).