Match the lanthanoid property given in List-I with lanthanoid ion given in List-II
|
List-I (Lanthanoid Property) |
List-II (Ion) |
|
(A) Strong oxidant |
(I) \( Lu^{3+} \) |
|
(B) Strong Reducing agent |
(II) \( La^{3+} \) |
|
(C) \( [Xe]f^{14} \) configuration |
(III) \( Ce^{4+} \) |
|
(D) largest ionic radii |
(IV) \( Eu^{2+} \) |
Choose the correct answer from the options given below.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
The common oxidation state of the lanthanoids is +3. An ion in another state tends to return to +3: a +4 ion gains an electron and acts as an oxidant, a +2 ion loses one and acts as a reductant. Across the series the ionic radius decreases steadily (lanthanoid contraction).
(A) Strong oxidant — This matches (III) \(Ce^{4+}\). Ce (Z = 58) is \( \mathrm{[Xe]\,4f^{1}\,5d^{1}\,6s^{2}} \); \( \ce{Ce^4+} \) has the noble gas configuration [Xe], but it readily changes to the common +3 state. With \( E^\circ_{(\ce{Ce^4+}/\ce{Ce^3+})} = +1.74\ \mathrm{V} \), it is a strong oxidant.
(B) Strong Reducing agent — This matches (IV) \(Eu^{2+}\). Eu (Z = 63) is \( \mathrm{[Xe]\,4f^{7}\,6s^{2}} \); losing the two 6s electrons gives \( \ce{Eu^2+} \): \( \mathrm{4f^{7}} \). It readily loses one more electron to reach +3, so it is a strong reducing agent.
(C) \([Xe]f^{14}\) configuration — This matches (I) \(Lu^{3+}\). Lu (Z = 71) is \( \mathrm{[Xe]\,4f^{14}\,5d^{1}\,6s^{2}} \); losing the two 6s electrons and the 5d electron gives \( \ce{Lu^3+} \): \( \mathrm{[Xe]\,4f^{14}} \).
(D) largest ionic radii — This matches (II) \(La^{3+}\). La (Z = 57) is \( \mathrm{[Xe]\,5d^{1}\,6s^{2}} \), so \( \ce{La^3+} \) is [Xe]. It stands first in the series, before the contraction begins, so it has the largest radius among the \( \ce{Ln^3+} \) ions.
Shortcut: (A)-(III) appears only in Option (4).