The pair of lanthanides will exceptionally high 3rd ionisation enthalpy than neighbouring elements:
- Lu and Yb
- Eu and Gb
- Eu and Yb
- Dy and Yb
The Correct Option is C
Solution and Explanation
The third ionization energy refers to the energy required to remove the third electron from an atom. Lanthanides exhibit varying ionization energies based on their electronic configurations.
For lanthanides like Eu (Europium) and Yb (Ytterbium), the third ionization energy is high because both elements tend to form stable, highly charged ions after the removal of the first two electrons.
- Europium (\( \text{Eu}^{3+} \)) has a stable \( 4f^6 \) configuration, making the removal of the third electron significantly harder.
- Ytterbium (\( \text{Yb}^{3+} \)) also has a stable \( 4f^{14} \) configuration in its trivalent state, making the third ionization energy high.
Thus, the pair Eu, Yb has high third-ionization energies.
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Concepts Used:
D and F Block Elements
The d-block elements are placed in groups 3-12 and F-block elements with 4f and 5f orbital filled progressively. The general electronic configuration of d block elements and f- block elements are (n-1) d 1-10 ns 1-2 and (n-2) f 1-14 (n-1) d1 ns2 respectively. They are commonly known as transition elements because they exhibit multiple oxidation states because of the d-d transition which is possible by the availability of vacant d orbitals in these elements.
They have variable Oxidation States as well as are good catalysts because they provide a large surface area for the absorption of reaction. They show variable oxidation states to form intermediate with reactants easily. They are mostly lanthanoids and show lanthanoid contraction. Since differentiating electrons enter in an anti-penultimate f subshell. Therefore, these elements are also called inner transition elements.
Read More: The d and f block elements