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Q4.20(iii):
Compare the chemistry of actinoids with that of the lanthanoids with special reference to: (iii) oxidation state

Solution :

Step 1: Theoretical Foundation of Oxidation States

The variation in oxidation states between lanthanoids ($Ln$) and actinoids ($An$) is governed by the relative energy gaps between the $nf$, $(n-1)d$, and $ns$ orbitals. In lanthanoids, the $4f$ orbitals are deeply buried and effectively shielded by $5s$ and $5p$ electrons, rendering them less available for chemical bonding [Per Slater’s Rules and Shielding Effect]. Conversely, in actinoids, the $5f$, $6d$, and $7s$ orbitals possess comparable energies, allowing for a broader participation of $5f$ electrons in bond formation.

Step 2: Comparative Analysis of Lanthanoids

Lanthanoids exhibit a dominant $+3$ oxidation state. While some elements display $+2$ or $+4$ states, these are essentially attempts to attain stable $f^0$, $f^7$, or $f^{14}$ electronic configurations [By Hund’s Rule of Maximum Multiplicity].

Feature Lanthanoids ($4f$)
Primary Oxidation State $+3$
Variable States Limited ($+2, +4$)
Cause of Restriction High ionization energy of $4f$ electrons

Step 3: Comparative Analysis of Actinoids

Actinoids exhibit a much greater range of oxidation states. This is attributed to the smaller energy difference between $5f$, $6d$, and $7s$ orbitals. As we move across the series, the $5f$ orbitals become more stable, but in the earlier part of the series (e.g., $Th$ to $Pu$), the energies are sufficiently close to allow the removal of multiple electrons.

Lanthanoids Actinoids +3 +3,+4 +5,+6 +7 Comparison of Oxidation State Diversity

Step 4: Systematic Comparison

  • Lanthanoids: The $+3$ state is the most stable. Stability of $+4$ (e.g., $Ce^{4+}$) or $+2$ (e.g., $Eu^{2+}$) is strictly dependent on the attainment of noble gas configuration or half-filled $f$-orbitals [Electronic Stability Criterion].
  • Actinoids: They show a range of states: $Th(+4)$, $Pa(+5)$, $U(+6)$, $Np(+7)$. The highest oxidation states are observed in the early actinoids where $5f$ electrons are more "exposed" to the nuclear environment compared to $4f$ in lanthanoids.

Step 5: Final Evaluation

The lanthanoids demonstrate a uniform $+3$ state with minimal variation, whereas actinoids demonstrate high variability. This is fundamentally due to the decreasing binding energy of the $nf$ electrons as the principal quantum number increases from $n=4$ to $n=5$, coupled with the decreasing radial contraction of the $f$-orbitals in the actinoid series [Per Relativistic Effects in heavy elements].

Final Solution: Lanthanoids predominantly exhibit a $+3$ oxidation state due to the poor shielding of $4f$ electrons and their deep burial within the atom. Actinoids exhibit a wide range of oxidation states (from $+3$ to $+7$) because the $5f$, $6d$, and $7s$ orbitals are energetically comparable, allowing for the progressive participation of $5f$ electrons in bonding.


More Questions from Class 12 Chemistry The d- and f-Block Elements Exercises


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