Question: The most common oxidation state for all lanthanoids is :

Step 1 (Governing Principle): [0.5 Mark]
Lanthanoids (elements from $Z = 57$ to $71$) possess a general electronic configuration of $[Xe] 4f^{0-14} 5d^{0-1} 6s^2$. The stability of the $+3$ oxidation state arises because the energy required to remove three electrons (two from the $6s$ orbital and one from the $5d$ or $4f$ orbital) is relatively low, leading to a stable tripositive ion in aqueous solution and solid compounds.

Step 2 (Step-by-Step Deduction): [0.25 Mark]
While some lanthanoids exhibit $+2$ or $+4$ oxidation states due to extra stability provided by $f^0$, $f^7$, or $f^{14}$ configurations, these are exceptions. For the entire series, the removal of three electrons results in the most energetically favorable state across the board: $$Ln \rightarrow Ln^{3+} + 3e^-$$ This state is the characteristic oxidation state common to all members of the lanthanoid series.

Step 3 (Final Concluded Answer): [0.25 Mark]
The correct option is (c) + 3.

💡 EL10 Concept Insight: Think of lanthanoids like a group of friends who all find it easiest to give away exactly 3 "energy coins" to become stable; even though some might occasionally give away 2 or 4, the "3-coin" trade is the one they all share and use most often.