Match the lanthanoid property given in List-I with lanthanoid ion giv…

The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D) (II)

List-I (Lanthanoid Property)	List-II (lon)
(A) Strong oxidant	(III) $Ce^{4+}$
(B) Strong Reducing agent	(IV) $Eu^{2+}$
(C) $[Xe]f^{14}$ configuration	(I) $Lu^{3+}$
(D) largest ionic radii	(II) $La^{3+}$

(A) Strong Oxidant $\rightarrow$ (III) $Ce^{4+}$

Cerium ($Ce$, $Z=58$) has the electronic configuration $[Xe]4f^1 5d^1 6s^2$. The $Ce^{4+}$ ion achieves the stable noble gas configuration of $[Xe]4f^0$. However, the most stable oxidation state for all lanthanoids is $+3$. Therefore, $Ce^{4+}$ has a strong tendency to gain an electron to become $Ce^{3+}$, making it a strong oxidant.

(B) Strong Reducing Agent $\rightarrow$ (IV) $Eu^{2+}$

Europium ($Eu$, $Z=63$) has the configuration $[Xe]4f^7 6s^2$. The $Eu^{2+}$ ion attains a stable half-filled $[Xe]4f^7$ configuration. To reach the more stable $+3$ state characteristic of lanthanoids, it readily loses an electron (gets oxidized), acting as a strong reducing agent.

(C) $[Xe]4f^{14}$ configuration $\rightarrow$ (I) $Lu^{3+}$

Lutetium ($Lu$, $Z=71$) is the last element of the lanthanoid series with the configuration $[Xe]4f^{14} 5d^1 6s^2$. When it forms the trivalent $Lu^{3+}$ ion by losing three electrons, it is left with a completely filled $4f^{14}$ subshell.

(D) Largest ionic radii $\rightarrow$ (II) $La^{3+}$

According to the principle of Lanthanoid Contraction, there is a steady decrease in the size of the tripositive ions ($M^{3+}$) from Lanthanum to Lutetium due to poor shielding by $4f$ electrons. Consequently, $La^{3+}$ (at the start of the series) has the largest ionic radius.