Q.Which of the following lanthanoids show +2 oxidation state besides the characteristic oxidation state +3 of lanthanoids?

The key idea is Lanthanide Contraction and its effect on stability of +2 oxidation states. Due to poor shielding of 4f electrons, atomic/ionic radii shrink across the series, making it harder to remove electrons. However, certain lanthanoids achieve a stable +2 state when it leads to a half-filled (4f$^7$) or fully-filled (4f$^{14}$) configuration.

Reasoning:

1. The characteristic oxidation state for all lanthanoids is +3 (removal of two 6s electrons and one 4f electron).
2. A +2 state is stable only if it results in an especially stable electronic configuration: half-filled (4f$^7$) or fully-filled (4f$^{14}$) 4f subshell. …

The question asks which lanthanoids exhibit a $+2$ oxidation state in addition to the common $+3$ state. This is a classic application of the Lanthanide Contraction and its effect on electronic stability.

The Core Concept: Why +2 Exists

All lanthanoids primarily show a $+3$ oxidation state because losing three electrons ($5d^1 6s^2$ or $4f^{n-1} 5d^1 6s^2$) is energetically favourable. However, a few elements can also form $+2$ ions. The reason is electronic stability.

The $4f$ subshell has a special stability when it is:

• Half-filled ($4f^7$)
• Fully-filled ($4f^{14}$)

If an element can achieve one of these configurations by losing only two electrons (instead of three), the $+2$ state becomes viable. The Lanthanide Contraction (the steady decrease in ionic radii across the series) also makes it harder to remove the third electron from these stable configurations, further favouring the $+2$ state.

Let's check each option:

1. Ce (Cerium, $Z=58$): Electronic configuration: $[Xe]\,4f^1\,5d^1\,6s^2$. To get a stable $4f^0$ (empty) configuration, it loses all three outer electrons to form $Ce^{3+}$ ($4f^1$). It can also lose four electrons to form $Ce^{4+}$ ($4f^0$), which is stable due to the empty subshell. Ce does not show a stable $+2$ state because $Ce^{2+}$ would have $4f^2$ — no special stability.

2. Eu (Europium, $Z=63$): Configuration: $[Xe]\,4f^7\,6s^2$. This is the half-filled $4f^7$ configuration. Losing two electrons ($6s^2$) gives $Eu^{2+}$, which retains the stable $4f^7$ configuration. Losing three electrons would disrupt this stability. Therefore, $Eu^{2+}$ is very stable and well-known. …

Method: Electronic Configuration & Stability of Half/Filled Orbitals

This method uses the Aufbau principle and the extra stability of half-filled and fully-filled 4f orbitals to predict which lanthanoids can exhibit the +2 oxidation state.

Steps

1. Write the general electronic configuration of lanthanoids:

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

2. Identify the +3 oxidation state — all lanthanoids lose the two 6s electrons and one 4f (or sometimes 5d) electron, giving:

   $[Xe]\,4f^{n-1}$ (where $n$ is the number of 4f electrons in the neutral atom).

3. Check for special stability in the +2 state:

   A lanthanoid will show +2 oxidation state if losing only two electrons (the two 6s electrons) results in:

   • A half-filled 4f subshell ($4f^7$), or
   • A fully-filled 4f subshell ($4f^{14}$), or
   • An empty 4f subshell ($4f^0$).

4. Apply to the given options:

   • Ce (Cerium): $[Xe]\,4f^1\,5d^1\,6s^2$

     +3 state: $4f^1$ (not special). +2 state: $4f^2$ — no special stability.

     Ce mainly shows +4 (to reach $4f^0$), not +2.

   • Eu (Europium): $[Xe]\,4f^7\,6s^2$

     +3 state: $4f^6$. +2 state: $4f^7$ — half-filled, extra stable. …

Here are the common mistakes students make on this question, along with how to avoid each one.

Mistake 1: Confusing "Stability" with "Possibility"

Many students think that if an element can show a +2 state in some exotic compound, it counts. The question asks for elements that show a +2 state as a characteristic or common alternative to +3.

• The Trap: Cerium (Ce) can form Ce²⁺ in some solid-state compounds, but it is not stable in aqueous solution or common reactions. Ce prefers +4 (to achieve a noble gas configuration).
• How to Avoid: Focus on aqueous stability and common oxidation states taught in the NCERT syllabus. The +2 state is only stable for Eu, Yb, and Sm (and sometimes Tm).

Mistake 2: Forgetting the "Why" Behind the +2 State

Students often try to memorize the list of elements showing +2 states without understanding the electronic configuration logic. This leads to confusion between Ce and Eu.

• The Concept: The +2 state is stable when it leads to a half-filled (f⁷) or fully-filled (f¹⁴) 4f subshell.
  • Eu²⁺: [Xe] 4f⁷ (half-filled) → Stable
  • Yb²⁺: [Xe] 4f¹⁴ (fully filled) → Stable
  • Ce²⁺: [Xe] 4f² (neither half nor full) → Unstable (Ce prefers Ce⁴⁺: [Xe] 4f⁰, noble gas config)
• How to Avoid: Don't just memorize the list. Write the electronic configuration of the +2 ion. If it ends in f⁷ or f¹⁴, it is likely stable.

Mistake 3: Ignoring the "Lanthanide Contraction" Link

The question is placed under the topic of Lanthanide Contraction, but students often treat it as a separate fact.

• The Connection: Lanthanide contraction causes the similarity in size between:
  • Eu²⁺ and Sr²⁺ (alkaline earth metal)
  • Yb²⁺ and Ca²⁺ This similarity allows Eu and Yb to mimic alkaline earth metals and form stable +2 salts (like EuSO₄, YbCl₂). …