The pair of lanthanides in which both elements have high third - ionization energy is :

Solution

Chemistry ma’am · 3 steps

Write the M²⁺ ion: IE3 is high when the third electron must break a stable 4f set.

1. Where IE3 comes from. The first two electrons leave from 6s, so IE3 depends on where the third electron sits in M²⁺. If removing it breaks a stable 4f⁷ or 4f¹⁴ set, IE3 is high. If it is an outer 5d or 6s electron sitting over a stable f set, IE3 is low.

   Element	M²⁺ configuration	Third electron leaves	IE3
   Eu	[Xe] 4f⁷, half-filled	from 4f⁷, breaks half-filling	high
   Yb	[Xe] 4f¹⁴, full	from 4f¹⁴, breaks full shell	high
   Gd	[Xe] 4f⁷ 5d¹	from 5d, leaves stable 4f⁷	low
   Lu	[Xe] 4f¹⁴ 6s¹	the outer electron, leaves stable 4f¹⁴	low
   Dy	[Xe] 4f¹⁰	from 4f¹⁰, no special stability lost	ordinary

2. Link to +2 state. This is also why Eu and Yb show a fairly stable +2 oxidation state. Their high IE3 makes the step to +3 harder than for the other lanthanides.
3. So. Both elements have a high third ionization energy only in (c): Eu and Yb.

The idea to keep. For lanthanide IE3, write the M²⁺ ion first. Eu²⁺ (4f⁷) and Yb²⁺ (4f¹⁴) hold on to their third electron, which is why both often show +2.

Answer from the NTA final answer key. Solution by PrepWiser.