What does third ionization energy mean?

The third ionization energy is the energy required for: $M^{2+} \rightarrow M^{3+} + e^-$

For lanthanides, this depends strongly on the stability of the electronic configuration of the $M^{2+}$ ion.

Key idea for lanthanides

Lanthanides commonly form $+3$ oxidation state. So, if the $+2$ state is especially stable, then removing the third electron to form $+3$ becomes difficult, and hence the third ionization energy is high.

Find lanthanides with especially stable $+2$ state

Stability in $4f$ series is enhanced for:

• half-filled configuration: $4f^7$
• fully filled configuration: $4f^{14}$

Now check the relevant elements:

• Eu: $[Xe]4f^7 6s^2$

  $Eu^{2+} = [Xe]4f^7$ This is half-filled, hence very stable. So, removing one more electron from $Eu^{2+}$ is difficult. Therefore, third ionization energy is high.

• Yb: $[Xe]4f^{14}6s^2$

  $Yb^{2+} = [Xe]4f^{14}$ This is fully filled, hence very stable. So, removing one more electron from $Yb^{2+}$ is difficult. Therefore, third ionization energy is high.

Check other elements in options

• Gd: $[Xe]4f^7 5d^1 6s^2$ After losing two electrons: $Gd^{2+} \approx [Xe]4f^7 5d^1$ This is not as exceptionally stable as $Eu^{2+}$ or $Yb^{2+}$ for resisting formation of $+3$. In fact, $Gd^{3+} = [Xe]4f^7$ is especially stable, so formation of $+3$ is favored. Hence third ionization energy is not unusually high.

• Dy and Lu similarly do not give the exceptionally stable $+2$ states comparable to $Eu^{2+}$ and $Yb^{2+}$.

Evaluate options

• A: Dy, Gd ❌
• B: Lu, Yb ❌ (Yb yes, Lu no)
• C: Eu, Yb ✅
• D: Eu, Gd ❌ (Eu yes, Gd no)

Final answer

The pair in which both lanthanides have high third ionization energy is: $\boxed{\text{Eu, Yb}}$