1. Find the ground-state lanthanide configurations

For lanthanides, the relevant part is the $4f$ occupancy.

• $Eu$ $(Z=63)$: $[Xe]4f^7 6s^2$
• $Sm$ $(Z=62)$: $[Xe]4f^6 6s^2$
• $Tm$ $(Z=69)$: $[Xe]4f^{13} 6s^2$
• $Tb$ $(Z=65)$: $[Xe]4f^9 6s^2$
• $Yb$ $(Z=70)$: $[Xe]4f^{14} 6s^2$
• $Dy$ $(Z=66)$: $[Xe]4f^{10} 6s^2$

2. Rule for forming lanthanide ions

Electrons are removed first from $6s$, then from $4f$ if needed.

Also,

• half-filled $f$-subshell means $4f^7$
• completely filled $f$-subshell means $4f^{14}$

3. Check each option

Option A: $Eu^{2+}$ and $Tm^{2+}$

• $Eu^{2+}$: $[Xe]4f^7$ → half-filled
• $Tm^{2+}$: $[Xe]4f^{13}$ → not completely filled

So this is half-filled, not completely filled. ❌

Option B: $Sm^{2+}$ and $Tm^{3+}$

• $Sm^{2+}$: $[Xe]4f^6$ → not half-filled
• $Tm^{3+}$: remove $2$ from $6s$ and $1$ from $4f$ $Tm^{3+} = [Xe]4f^{12}$ Not completely filled.

So both fail. ❌

Option C: $Tb^{4+}$ and $Yb^{2+}$

• $Tb$: $[Xe]4f^9 6s^2$ Removing $4$ electrons: first $2$ from $6s$, then $2$ from $4f$ $Tb^{4+} = [Xe]4f^7$ → half-filled

• $Yb^{2+}$: $[Xe]4f^{14}$ → completely filled

This matches half-filled and completely filled, respectively. ✅

Option D: $Dy^{3+}$ and $Yb^{3+}$

• $Dy^{3+}$: $[Xe]4f^9$ → not half-filled
• $Yb^{3+}$: $[Xe]4f^{13}$ → not completely filled

So this fails. ❌

4. Final answer

The correct option is: $\boxed{C}$

That is, $Tb^{4+}$ has $4f^7$ (half-filled) and $Yb^{2+}$ has $4f^{14}$ (completely filled).