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Q4.30:
Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element.

Solution :

Step 1: Identification of the Final Actinoid

The actinoid series encompasses the elements from Actinium ($Z = 89$) to Lawrencium ($Z = 103$). The last element in the series is Lawrencium, represented by the symbol $Lr$ and having an atomic number of $Z = 103$.

Step 2: Derivation of Electronic Configuration

To determine the electronic configuration of Lawrencium ($Z = 103$), we apply the Aufbau Principle, the Pauli Exclusion Principle, and Hund’s Rule, while accounting for the noble gas core of Radon ($Rn, Z = 86$).

  • Total electrons to account for: $103 - 86 = 17$ electrons.
  • Distribution follows the sequence: $5f$, $6d$, and $7s$ orbitals.
  • Standard configuration: $[Rn] \ 5f^{14} \ 6d^1 \ 7s^2$.

[Per the Madelung energy-ordering rule, the $5f$ subshell is filled prior to the $6d$ subshell. Once $5f$ reaches stability at 14 electrons, the next electron enters the $6d$ subshell.]

Step 3: Analysis of Oxidation States

The stability of oxidation states in actinoids is governed by the electronic configuration and the relative energy levels of the $5f$, $6d$, and $7s$ orbitals.

Oxidation State Configuration Stability Context
$+3$ $[Rn] \ 5f^{14}$ This is the most stable and predominant oxidation state for Lawrencium. The removal of the three valence electrons ($6d^1$ and $7s^2$) results in a closed-shell $5f^{14}$ configuration, which provides high electronic stability [Per the Shell Model of Atomic Stability].

Visual Representation of Orbital Energy Levels

7s (2e⁻) 6d (1e⁻) 5f (14e⁻) Energy Level Diagram for Lr (Z=103)

Step 4: Theoretical Conclusion

Lawrencium behaves similarly to Lutetium in the lanthanoid series due to the completion of the $f$-subshell. The $5f^{14}$ shell is inert, meaning the electron removal for oxidation primarily involves the $6d$ and $7s$ electrons. The $+3$ state is the primary state observed, as it achieves the stable configuration of $[Rn] \ 5f^{14}$.

Final Solution: The last element is Lawrencium ($Lr, Z=103$). Its electronic configuration is $[Rn] \ 5f^{14} \ 6d^1 \ 7s^2$. Its most stable oxidation state is $+3$, which corresponds to the attainment of the stable $5f^{14}$ closed-shell configuration.


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