Q.Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.

The key idea is that certain lanthanoids can achieve a +4 oxidation state when doing so leads to a stable, half-filled, or fully filled $4f$ subshell.

Reasoning:

1. Among lanthanoids, the +3 state is most common. A +4 state is stable only if it results in a noble-gas-like or particularly stable electronic configuration. …

Why Disproportionation and +4 States Matter in Lanthanoids

The lanthanoid series (Ce to Lu, Z = 58–71) typically shows a stable +3 oxidation state. This is because removing three electrons (two from 6s and one from 4f or 5d) leaves a relatively stable configuration. Going to +4 is much harder — it requires removing a fourth electron from the 4f subshell, which is energetically costly.

However, a +4 state becomes favourable when it leads to a particularly stable electronic configuration:

• Empty 4f subshell (4f⁰) — like in Ce⁴⁺
• Half-filled 4f subshell (4f⁷) — like in Tb⁴⁺
• Filled 4f subshell (4f¹⁴) — like in Yb²⁺ (but that's +2, not +4)

The key insight: the +4 state is not common across the series. It appears only where the ionisation energy is offset by the stability gained from a noble-gas-like or half-filled 4f configuration.

Step-by-Step Reasoning

1. Identify the electronic configuration of lanthanoids

   The general outer configuration is $[Xe]4f^{1-14}5d^{0-1}6s^2$. For cerium (Ce, atomic number 58), the configuration is $[Xe]4f^1 5d^1 6s^2$ (NCERT Table 4.9's form; the alternative $[Xe]4f^2 6s^2$ is sometimes quoted because the 4f and 5d orbitals lie very close in energy).

2. Understand what +4 means

   The +4 oxidation state means the atom loses four electrons. For Ce, losing two 6s electrons and two 4f electrons gives Ce⁴⁺ with configuration $[Xe]4f^0$. This is the same electron configuration as xenon — a noble gas — which is exceptionally stable.

3. Compare with other lanthanoids …

Method: Configuration-Based Prediction of Lanthanoid Oxidation States

Steps:

1. Write the lanthanoid's ground-state configuration (general form $[Xe]\,4f^{1-14}\,5d^{0-1}\,6s^2$).
2. Remove electrons stepwise — the $6s$ pair first, then $5d$/$4f$ as needed.
3. Check the resulting ion's $4f$ count: an empty ($4f^0$), half-filled ($4f^7$) or full ($4f^{14}$) subshell confers extra stability.
4. Judge which deviations from +3 are viable: a +4 (or +2) state is well known only where it lands on one of these stable configurations.

Lanthanoid with +4 Oxidation State

Answer: Cerium (Ce)

• Cerium exhibits the +4 oxidation state (as $Ce^{4+}$) in addition to the common +3 state. …

Common Mistakes: The Lanthanoid Well Known for the +4 Oxidation State

The Question

Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.

Correct Answer: Cerium (Ce) — it exhibits both +3 and +4 oxidation states.

🚩 Mistake #1: Naming the wrong element

What students do wrong:

Students often name Europium (Eu) or Ytterbium (Yb) — but these are famous for +2, not +4.

Why this happens:

• Students confuse "stable +4" with "stable +2" in lanthanoids.
• They remember that Eu and Yb deviate from the common +3 state, but forget the direction of deviation.

How to avoid:

• Memorise the two key exceptions clearly:
  • Ce → stable +4 (loses 4 electrons → empty 4f⁰ configuration)
  • Eu → stable +2 (half-filled 4f⁷)
  • Yb → stable +2 (fully filled 4f¹⁴)
• Use a mnemonic: "Ce is +4, Eu and Yb are +2"

🚩 Mistake #2: Confusing lanthanoids with actinoids

What students do wrong:

Students name Uranium (U) or Plutonium (Pu) — these are actinoids, not lanthanoids.

Why this happens:

• Both series are f-block elements, and actinoids show many stable +4 states (Th, U, Pu, etc.).
• Students mix up the two series under exam pressure.

How to avoid:

• Remember: Lanthanoids (4f) → only Ce shows a stable +4 in aqueous chemistry.
• Actinoids (5f) → many show +4, but the question specifically asks for lanthanoid series.

🚩 Mistake #3: Forgetting the reason why Ce shows +4

What students do wrong:

Students just memorise "Ce = +4" without understanding, so they can't apply it to related questions (e.g., disproportionation).

Why this happens:

• Rote learning without connecting to electronic configuration.

How to avoid:

• Understand the electronic basis:
  • Ce atomic number = 58: $[Xe]\,4f^1\,5d^1\,6s^2$
  • Ce⁴⁺ = $[Xe]$ (noble gas configuration, empty 4f)
  • This empty 4f⁰ is exceptionally stable. …