Chemistry · d and f Block Elements

NEET (UG) 2024 — Question 93

The pair of lanthanoid ions which are diamagnetic is

Option A:
$\ce{Ce^{4+}}$ and $\ce{Yb^{2+}}$
Correct
Option B:
$\ce{Ce^{3+}}$ and $\ce{Eu^{2+}}$
Option C:
$\ce{Gd^{3+}}$ and $\ce{Eu^{3+}}$
Option D:
$\ce{Pm^{3+}}$ and $\ce{Sm^{3+}}$

Answer: A

Step-by-step solution

Diamagnetic species have all electrons paired. $\ce{Ce^{4+}}$ has $[Xe]4f^0$ configuration, no unpaired electrons. $\ce{Yb^{2+}}$ has $[Xe]4f^{14}$ configuration, all electrons paired. Thus, both are diamagnetic. $\ce{Ce^{3+}}$ has $[Xe]4f^1$ (one unpaired electron). $\ce{Eu^{2+}}$ has $[Xe]4f^7$ (seven unpaired electrons). $\ce{Gd^{3+}}$ has $[Xe]4f^7$ (seven unpaired electrons). $\ce{Eu^{3+}}$ has $[Xe]4f^6$ (six unpaired electrons). $\ce{Pm^{3+}}$ has $[Xe]4f^4$ (four unpaired electrons). $\ce{Sm^{3+}}$ has $[Xe]4f^5$ (five unpaired electrons). Hence, only option A contains both diamagnetic ions.

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Exam: NEET (UG) 2024
Paper: 2024 paper
Subject: Chemistry
Chapter: d and f Block Elements
Topic: Inner transition elements (f-Block elements)

The E^circ value for the Mn^3+ / Mn^2+ couple is more positive than that of Cr^3+ / Cr^2+ or Fe^3+ / Fe^2+ due to change of

Answer: A

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