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Q4.32:
Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements.

Solution :

1. Identification of Lanthanoids with Non-Standard Oxidation States

While the most stable and characteristic oxidation state of lanthanoids ($Ln$) is $+3$, certain elements exhibit stability in the $+4$ or $+2$ states due to the attainment of stable electronic configurations (empty, half-filled, or fully-filled $f$-orbitals).

Oxidation State Lanthanoid Elements
$+4$ Cerium ($Ce$), Praseodymium ($Pr$), Neodymium ($Nd$), Terbium ($Tb$), Dysprosium ($Dy$)
$+2$ Neodymium ($Nd$), Samarium ($Sm$), Europium ($Eu$), Thulium ($Tm$), Ytterbium ($Yb$)

2. Correlation with Electronic Configurations

The stability of these oxidation states is governed by the exchange energy associated with $f$-orbital occupancy [Per Hund’s Rule and Orbital Stability Theory].

  • Cerium ($Ce$, $Z=58$): Electronic configuration $[Xe] 4f^1 5d^1 6s^2$. By losing four electrons, it achieves the stable noble gas configuration of Xenon ($4f^0$), which is highly stable [Ref: Energetic feasibility of noble gas core].
  • Europium ($Eu$, $Z=63$): Electronic configuration $[Xe] 4f^7 6s^2$. By losing two electrons, it attains the $4f^7$ configuration. This half-filled $f$-subshell provides extra stability [Per Exchange Energy Maximization].
  • Ytterbium ($Yb$, $Z=70$): Electronic configuration $[Xe] 4f^{14} 6s^2$. By losing two electrons, it achieves the $4f^{14}$ configuration. This completely filled $f$-subshell offers significant stability [Ref: Symmetry-based stability].
  • Terbium ($Tb$, $Z=65$): Electronic configuration $[Xe] 4f^9 6s^2$. By losing four electrons, it reaches the $4f^7$ configuration, again benefiting from the stability of the half-filled subshell.

3. Visualizing Stability via Orbital Energy Levels

Ce (+4: 4f⁰) Eu (+2: 4f⁷) Yb (+2: 4f¹⁴) Stability Profiles of Lanthanoid Ions

4. Summary of Analytical Findings

The deviation from the common $+3$ oxidation state is not random. It is systematically linked to the energy required to remove electrons versus the energy gained by reaching a stable $4f^0$, $4f^7$, or $4f^{14}$ configuration. Elements like $Ce^{4+}$ act as strong oxidizing agents to return to the $+3$ state, whereas $Eu^{2+}$ and $Yb^{2+}$ act as strong reducing agents to reach the stable $+3$ state.

Final Solution: The lanthanoids exhibiting $+4$ states include $Ce$, $Pr$, $Nd$, $Tb$, and $Dy$, primarily driven by the stability of $4f^0$ and $4f^7$ configurations. Those exhibiting $+2$ states include $Nd$, $Sm$, $Eu$, $Tm$, and $Yb$, driven by the stability of $4f^7$ and $4f^{14}$ configurations.


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