Identification of Pairs of Diamagnetic Lanthanide Ions

To determine which pair(s) of ions are diamagnetic, we analyze the electronic configurations of the given lanthanide ions. A species is diamagnetic when all of its electrons are paired (i.e., the number of unpaired electrons $n = 0$). For $f$-block elements, diamagnetism usually arises when the $4f$ subshell is either completely empty ($4f^0$) or completely filled ($4f^{14}$).

Ground-state electronic configurations of the neutral atoms:

• Lanthanum ($\text{La}$, $Z = 57$): $[\text{Xe}] 5d^1 6s^2$
• Cerium ($\text{Ce}$, $Z = 58$): $[\text{Xe}] 4f^1 5d^1 6s^2$
• Ytterbium ($\text{Yb}$, $Z = 70$): $[\text{Xe}] 4f^{14} 6s^2$
• Lutetium ($\text{Lu}$, $Z = 71$): $[\text{Xe}] 4f^{14} 5d^1 6s^2$

---

Evaluation of Options:

Option (A): $\text{La}^{3+}$ and $\text{Ce}^{4+}$

• $\text{La}^{3+}$: Formed by removing three electrons ($6s^2 5d^1$) from $\text{La}$.
  $\text{La}^{3+}: [\text{Xe}] 4f^0$
  Number of unpaired electrons, $n = 0 \implies \text{Diamagnetic}$.

• $\text{Ce}^{4+}$: Formed by removing four electrons ($6s^2 5d^1 4f^1$) from $\text{Ce}$.
  $\text{Ce}^{4+}: [\text{Xe}] 4f^0$
  Number of unpaired electrons, $n = 0 \implies \text{Diamagnetic}$.

Thus, both ions in Option (A) are diamagnetic.

---

Option (B): $\text{Yb}^{2+}$ and $\text{Lu}^{3+}$

• $\text{Yb}^{2+}$: Formed by removing two electrons ($6s^2$) from $\text{Yb}$.
  $\text{Yb}^{2+}: [\text{Xe}] 4f^{14}$
  The $4f$ subshell is fully filled, so $n = 0 \implies \text{Diamagnetic}$.

• $\text{Lu}^{3+}$: Formed by removing three electrons ($6s^2 5d^1$) from $\text{Lu}$.
  $\text{Lu}^{3+}: [\text{Xe}] 4f^{14}$
  The $4f$ subshell is fully filled, so $n = 0 \implies \text{Diamagnetic}$.

Thus, both ions in Option (B) are diamagnetic.

---

Option (C): $\text{La}^{2+}$ and $\text{Ce}^{3+}$

• $\text{La}^{2+}$: Electronic configuration is $[\text{Xe}] 5d^1$ or $[\text{Xe}] 4f^1$ ($n = 1 \implies \text{Paramagnetic}$).
• $\text{Ce}^{3+}$: Electronic configuration is $[\text{Xe}] 4f^1$ ($n = 1 \implies \text{Paramagnetic}$).

Thus, Option (C) contains paramagnetic ions.

---

Option (D): $\text{Yb}^{3+}$ and $\text{Lu}^{2+}$

• $\text{Yb}^{3+}$: Electronic configuration is $[\text{Xe}] 4f^{13}$ ($n = 1 \implies \text{Paramagnetic}$).
• $\text{Lu}^{2+}$: Electronic configuration is $[\text{Xe}] 4f^{14} 5d^1$ ($n = 1 \implies \text{Paramagnetic}$).

Thus, Option (D) contains paramagnetic ions.

---

Conclusion:

The pair(s) of diamagnetic ions are (A) and (B).