1. Rule for diamagnetism

An ion is diamagnetic if it has no unpaired electrons. So we check the electronic configuration of each ion, especially the $4f$ electrons for lanthanoids.

2. Useful ground-state configurations

• $La(Z=57): [Xe]5d^16s^2$
• $Ce(Z=58): [Xe]4f^15d^16s^2$
• $Yb(Z=70): [Xe]4f^{14}6s^2$
• $Lu(Z=71): [Xe]4f^{14}5d^16s^2$

For cation formation, electrons are removed first from outer orbitals ($6s$, then $5d$), and then from $4f$ if needed.

3. Check each option

Option A: $La^{3+}, Ce^{4+}$

• $La^{3+}$: $La: [Xe]5d^16s^2 \Rightarrow La^{3+} = [Xe]$ No unpaired electrons $\Rightarrow$ diamagnetic.

• $Ce^{4+}$: $Ce: [Xe]4f^15d^16s^2 \Rightarrow Ce^{4+} = [Xe]$ No unpaired electrons $\Rightarrow$ diamagnetic.

So, A is correct.

Option B: $Yb^{2+}, Lu^{3+}$

• $Yb^{2+}$: $Yb: [Xe]4f^{14}6s^2 \Rightarrow Yb^{2+} = [Xe]4f^{14}$ $4f^{14}$ is completely filled $\Rightarrow$ all electrons paired $\Rightarrow$ diamagnetic.

• $Lu^{3+}$: $Lu: [Xe]4f^{14}5d^16s^2 \Rightarrow Lu^{3+} = [Xe]4f^{14}$ Completely filled $4f$ subshell $\Rightarrow$ diamagnetic.

So, B is correct.

Option C: $La^{2+}, Ce^{3+}$

• $La^{2+}$: $La^{2+} = [Xe]5d^1$ One unpaired electron $\Rightarrow$ paramagnetic.

• $Ce^{3+}$: $Ce^{3+} = [Xe]4f^1$ One unpaired electron $\Rightarrow$ paramagnetic.

So, C is incorrect.

Option D: $Yb^{3+}, Lu^{2+}$

• $Yb^{3+}$: $Yb^{3+} = [Xe]4f^{13}$ One unpaired electron $\Rightarrow$ paramagnetic.

• $Lu^{2+}$: $Lu^{2+} = [Xe]4f^{14}5d^1$ One unpaired electron in $5d$ $\Rightarrow$ paramagnetic.

So, D is incorrect.

4. Final answer

The diamagnetic pairs are: $\boxed{A, B}$

5. Comparison with stored correct answer

Stored correct answer: A, B

My derived answer: A, B

They match.