Question

The Lanthanoid that does not show $+4$ oxidation state is

• A. $\mathrm{Dy}$
• B. $\mathrm{Ce}$
• C. $\mathrm{Tb}$
• D. $\mathrm{Eu}$

Answer 1

The correct option is D

$\mathrm{Eu}$

The explanation for the correct option:

(D) $\mathrm{Eu}$-

• Europium does not demonstrate $+4$ oxidation.
• The electronic configuration of Europium is$\left[\mathrm{Xe}\right]4{\mathrm{f}}^{7}6{\mathrm{s}}^2$.
• After losing 2 electrons, $\mathrm{Eu}$ achieves stability as the f-subshell will be half-filled but can lose 3 electrons also to achieve a +3-oxidation state also.
• For europium, the only oxidation states are +2 and +3.
• Thus, ${\mathrm{Eu}}^{2+}$ is more stable than${\mathrm{Eu}}^{3+}$.

The Explanation for the incorrect options:

(A) $\mathrm{Dy}$-

• The electronic configuration of Dysprosium is$\left[\mathrm{Xe}\right]4{\mathrm{f}}^{10}6{\mathrm{s}}^2$.
• Since they get reduced by accepting electrons, they are good oxidizing agents in $+4$ the oxidation state.
• Dysprosium and its compounds have been used for making laser materials and phosphor activators, and in metal halide lamps.

(B) $\mathrm{Ce}$-

• The electronic configuration of Cerium is$\left[\mathrm{Xe}\right]4{\mathrm{f}}^{1}5{\mathrm{d}}^{1}6{\mathrm{s}}^2$.
• Since they get reduced by accepting electrons, they are good oxidizing agents in $+4$ the oxidation state.
• Cerium The metal is used as a core for the carbon electrodes of arc lamps, and for incandescent mantles for gas lighting.

(C) $\mathrm{Tb}$-

• The electronic configuration of Terbium is$\left[\mathrm{Xe}\right]4{\mathrm{f}}^{9}6{\mathrm{s}}^2$.
• Since they get reduced by accepting electrons, they are good oxidizing agents in $+4$ the oxidation state.
• Terbium is used to dope calcium fluoride, calcium tungstate, and strontium molybdate, all used in solid-state devices.
• It is also used in low-energy lightbulbs and mercury lamps

Hence, the correct option is (D).