1. Find the ground-state electronic configurations of the relevant lanthanoids

For lanthanoids, the valence configuration is typically: $[Xe] \, 4f^n 5d^{0/1}6s^2$

We only need the number of electrons in the $4f$ subshell after forming the given ions.

2. Evaluate each ion

(A) $\mathrm{Eu}^{2+}$

Europium ($Z=63$) has ground-state configuration: $\mathrm{Eu} : [Xe] \, 4f^7 6s^2$ Forming $\mathrm{Eu}^{2+}$ removes the two $6s$ electrons: $\mathrm{Eu}^{2+} : [Xe] \, 4f^7$ So, this has $4f^7$. ✅

(B) $\mathrm{Gd}^{3+}$

Gadolinium ($Z=64$) has ground-state configuration: $\mathrm{Gd} : [Xe] \, 4f^7 5d^1 6s^2$ Forming $\mathrm{Gd}^{3+}$ removes $6s^2$ and $5d^1$ electrons: $\mathrm{Gd}^{3+} : [Xe] \, 4f^7$ So, this also has $4f^7$. ✅

(C) $\mathrm{Eu}^{3+}$

From $\mathrm{Eu} : [Xe] \, 4f^7 6s^2$ Removing three electrons gives: $\mathrm{Eu}^{3+} : [Xe] \, 4f^6$ So, this is not $4f^7$. ❌

(D) $\mathrm{Tb}^{3+}$

Terbium ($Z=65$) has ground-state configuration approximately: $\mathrm{Tb} : [Xe] \, 4f^9 6s^2$ Forming $\mathrm{Tb}^{3+}$ removes two $6s$ electrons and one $4f$ electron: $\mathrm{Tb}^{3+} : [Xe] \, 4f^8$ So, this is not $4f^7$. ❌

(E) $\mathrm{Sm}^{2+}$

Samarium ($Z=62$) has ground-state configuration: $\mathrm{Sm} : [Xe] \, 4f^6 6s^2$ Forming $\mathrm{Sm}^{2+}$ removes the two $6s$ electrons: $\mathrm{Sm}^{2+} : [Xe] \, 4f^6$ So, this is not $4f^7$. ❌

3. Collect the correct ions

The ions with $4f^7$ configuration are: $\mathrm{Eu}^{2+} \quad \text{and} \quad \mathrm{Gd}^{3+}$ So the correct pair is (A) and (B).

4. Match with the given options

Given options:

• A: (B) and (E) only
• B: (B) and (C) only
• C: (A) and (D) only
• D: (A) and (B) only

Therefore, the correct option is: $\boxed{\text{D}}$

5. Comparison with stored correct answer

Stored correct answer = D

Our derived answer = D

They match.