Question: The most common and stable oxidation state of a Lanthanoid is:

Step 1 (Governing Principle): [0.5 Mark]

Lanthanoids are the elements of the $4f$-series (from $Z = 57$ to $71$). The electronic configuration of these elements is generally represented as:

$$[Xe] 4f^{0-14} 5d^{0-1} 6s^2$$

The stability of the oxidation state is governed by the removal of electrons from the $6s$ and $5d$ orbitals, followed by the $4f$ orbitals.

Step 2 (Step-by-Step Deduction): [0.25 Mark]

• The $6s^2$ electrons are removed first, resulting in a $+2$ oxidation state.
• The removal of the third electron (from $5d$ or $4f$) leads to the $+3$ oxidation state.
• The $+3$ state is the most stable because the energy required to remove the third electron is relatively low, and the resulting $Ln^{3+}$ ions have a stable configuration that is common to all lanthanoids.

Step 3 (Final Concluded Answer): [0.25 Mark]

The most common and stable oxidation state of a Lanthanoid is +3.

Correct Option: (b) + 3

💡 EL10 Concept Insight: Think of Lanthanoids like people carrying three heavy bags; they find it very easy to drop two bags from their outer shell and one more from their inner shell to become stable, making the $+3$ state their "happy" and most common way to exist!