1. Criterion for diamagnetism

A lanthanoid ion is diamagnetic if it has no unpaired electrons in its $4f$ subshell.

2. Find electronic configurations of the given ions

For lanthanoids, the common neutral configurations are based on $[Xe]4f^n5d^06s^2$ (with some exceptions), and on forming $+3$ ions, electrons are removed first from $6s$ and then $5d$.

• $\mathrm{La}^{3+}$:

  Neutral La: $[Xe]5d^16s^2$

  $\mathrm{La}^{3+} = [Xe]$

  No $4f$ electrons, hence all electrons paired $\Rightarrow$ diamagnetic.

• $\mathrm{Ce}^{4+}$:

  Neutral Ce: $[Xe]4f^15d^16s^2$

  Removing 4 electrons gives:

  $\mathrm{Ce}^{4+} = [Xe]$

  Again, no unpaired electrons $\Rightarrow$ diamagnetic.

• $\mathrm{Nd}^{3+}$:

  Neutral Nd: $[Xe]4f^46s^2$

  $\mathrm{Nd}^{3+} = [Xe]4f^3$

  This has 3 unpaired electrons, so it is paramagnetic.

• $\mathrm{Eu}^{3+}$:

  Neutral Eu: $[Xe]4f^76s^2$

  $\mathrm{Eu}^{3+} = [Xe]4f^6$

  $4f^6$ has several unpaired electrons, so it is paramagnetic.

• $\mathrm{Lu}^{3+}$:

  Neutral Lu: $[Xe]4f^{14}5d^16s^2$

  $\mathrm{Lu}^{3+} = [Xe]4f^{14}$

  A completely filled $4f$ subshell has all electrons paired, so it is diamagnetic.

3. Check each option

• A: $\mathrm{La}^{3+}$ and $\mathrm{Ce}^{4+}$

  Both are diamagnetic. ✅

• B: $\mathrm{Nd}^{3+}$ and $\mathrm{Ce}^{4+}$

  $\mathrm{Nd}^{3+}$ is paramagnetic. ❌

• C: $\mathrm{Lu}^{3+}$ and $\mathrm{Eu}^{3+}$

  $\mathrm{Lu}^{3+}$ is diamagnetic, but $\mathrm{Eu}^{3+}$ is paramagnetic. ❌

• D: $\mathrm{Nd}^{3+}$ and $\mathrm{Eu}^{3+}$

  Both are paramagnetic. ❌

4. Final answer

The diamagnetic lanthanoid ions are:

$\boxed{\mathrm{La}^{3+} \text{ and } \mathrm{Ce}^{4+}}$