Q32 P - Chemistry: Molecular Nature Of Matter And Change

Question

Only a few lanthanides show an oxidation state other than +3. Two of these, europium (Eu) and terbium (Tb), are found near the middle of the series, and their unusual oxidation states can be associated with a half-filled f subshell.

(a) Write the electron configurations of Eu²⁺, Eu³⁺, and Eu⁴⁺. Why is Eu²⁺ a common ion, whereas Eu⁴⁺ is unknown?

(b) Write the electron configurations of Tb²⁺, Tb³⁺, and Tb⁴⁺. Would you expect Tb to show a +2 or a +4 oxidation state? Explain

Step-by-Step Solution

Answer

a) $E{u}^{2+}:\left[Xe\right]4f^{7}E{u}^{3+}:\left[Xe\right]f{f}^{6}E{u}^{4+}:\left[Xe\right]f{f}^{5}E{u}^{2+}$ is more common ion than Eu³⁺ because it has more stable half-filled 4f subshell.

b) $T{b}^{2+}:\left[Xe\right]4f^{9}T{b}^{3+}:\left[Xe\right]4f^{8}T{b}^{4+}:\left[Xe\right]4f^7+4$ oxidation state rather than +2, because it has more stable half-filled 4f subshell.

Step 1: The electron configuration

The electron configuration is that the distribution of electrons of an atom or molecule (or other physical structure) in atomic or molecular orbitals.

Step 2: Electron configuration of each elements.

Electron configuration for elemental Eu would be: 

1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5d²5p⁶4f⁷5d⁰6s²

Electron configuration for elemental Tb would be:

1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5d²5p⁶4f⁹5d⁰6s² 

a) $E{u}^{2+}:\left[Xe\right]4f^{7}E{u}^{3+}:\left[Xe\right]f{f}^{6}E{u}^{4+}:\left[Xe\right]f{f}^{5}E{u}^{2+}$ is more common ion than Eu³⁺ because it has more stable half-filled 4f subshell.

b) $T{b}^{2+}:\left[Xe\right]4f^{9}T{b}^{3+}:\left[Xe\right]4f^{8}T{b}^{4+}:\left[Xe\right]4f^7$Tb to show +4 oxidation state rather than +2, because it has more stable half-filled subshell.