Q.Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements.

Concept: Lanthanoid Oxidation States — the stability of oxidation states in lanthanoids is governed by the tendency to achieve empty, half-filled, or fully filled $4f$ subshells.

Reasoning:

1. +4 oxidation state occurs when an element can lose electrons to attain a stable $4f^0$ (empty), $4f^7$ (half-filled), or $4f^{14}$ (fully filled) configuration. This is seen in Ce ($4f^1$) → Ce$^{4+}$ ($4f^0$), Pr ($4f^3$) → Pr$^{4+}$ ($4f^1$ — less stable), Nd ($4f^4$) → Nd$^{4+}$ ($4f^2$ — rare), Tb ($4f^9$) → Tb$^{4+}$ ($4f^7$), and Dy ($4f^{10}$) → Dy$^{4+}$ ($4f^8$ — less stable). The most common +4 states are Ce and Tb. …

Why Oxidation States Vary in the Lanthanoids

The lanthanoid series (Ce to Lu, atomic numbers 58–71) is famous for the lanthanoid contraction and for having +3 as the most common oxidation state. But why do some elements deviate to +4 or +2?

The answer lies in the stability of the 4f subshell. The 4f orbitals are buried deep inside the atom, so removing or adding electrons to them costs a lot of energy. However, if by losing or gaining an electron the atom can achieve a noble-gas configuration (like Xe, 4f⁰), a half-filled 4f subshell (4f⁷), or a fully filled 4f subshell (4f¹⁴), the extra stability compensates for the energy cost. This is the same principle that makes Cr and Cu special in the d-block.

Let’s see how this plays out for +4 and +2 states.

Step-by-Step Analysis

1. The +4 Oxidation State: Losing an Extra Electron

The +3 state is the baseline for all lanthanoids. To reach +4, the atom must lose one more electron (from the 4f subshell, since the 6s and 5d are already gone). This is energetically unfavourable unless the resulting 4f configuration is especially stable.

• Cerium (Ce, Z=58):

  Ground state: $[Xe]\,4f^1\,5d^1\,6s^2$ (NCERT Table 4.9's form; the alternative $4f^2\,6s^2$ appears in some sources).

  Ce³⁺: $4f^1$ (one electron in f).

  Ce⁴⁺: $4f^0$ — empty 4f subshell, which is the same as the noble gas Xe. This is extremely stable.

  → Ce is the most common +4 lanthanoid.

• Praseodymium (Pr, Z=59):

  Pr³⁺: $4f^2$.

  Pr⁴⁺: $4f^1$ — not particularly stable, but possible under strong oxidising conditions. The +4 state is less common than Ce⁴⁺.

• Neodymium (Nd, Z=60):

  Nd³⁺: $4f^3$.

  Nd⁴⁺: $4f^2$ — again, no special stability. Nd⁴⁺ exists only in a few compounds (e.g., NdF₄).

• Terbium (Tb, Z=65):

  Tb³⁺: $4f^8$.

  Tb⁴⁺: $4f^7$ — half-filled 4f subshell (7 electrons, each in a separate orbital with parallel spins). This is very stable.

  → Tb⁴⁺ is the second most common +4 lanthanoid, after Ce.

• Dysprosium (Dy, Z=66):

  Dy³⁺: $4f^9$.

  Dy⁴⁺: $4f^8$ — not half-filled, but Dy⁴⁺ is known in some compounds (e.g., DyF₄). It is less stable than Tb⁴⁺.

2. The +2 Oxidation State: Gaining an Electron (or Losing One Less)

To get a +2 state, the atom must lose only two electrons instead of three. This means the +2 ion has one more electron in the 4f subshell than the +3 ion. Again, stability comes from half-filled or full f-shells.

• Europium (Eu, Z=63):

  Eu³⁺: $4f^6$.

  Eu²⁺: $4f^7$ — half-filled 4f subshell. This is exceptionally stable.

  → Eu²⁺ is the most common +2 lanthanoid. It is even stable in water (though it slowly oxidises to Eu³⁺).

• Ytterbium (Yb, Z=70):

  Yb³⁺: $4f^{13}$.

  Yb²⁺: $4f^{14}$ — fully filled 4f subshell. This is also very stable.

  → Yb²⁺ is common, though less so than Eu²⁺.

• Samarium (Sm, Z=62):

  Sm³⁺: $4f^5$.

  Sm²⁺: $4f^6$ — not half-filled or full, but the +2 state is known (e.g., SmI₂ is a famous reducing agent). It is less stable than Eu²⁺ and Yb²⁺.

• Thulium (Tm, Z=69):

  Tm³⁺: $4f^{12}$.

  Tm²⁺: $4f^{13}$ — known in the research literature (e.g., TmI₂) but rare, and not part of NCERT's own +2 list; the exam answer stays Sm, Eu, Yb. …

Method: Electronic Configuration Correlation

This method uses the stability of half-filled and fully-filled f-orbitals to predict oxidation states.

Step 1: Recall the general electronic configuration of lanthanoids

The lanthanoid series (Ce to Lu) has the general outer configuration:

The key is the 4f subshell — its stability depends on how close it is to being empty ($4f^0$), half-filled ($4f^7$), or fully filled ($4f^{14}$).

Step 2: Identify elements showing +4 oxidation state

A +4 oxidation state means the element loses 4 electrons (typically the two 6s electrons and two 4f electrons). This is favoured when the resulting configuration is stable — either half-filled ($4f^7$) or fully filled ($4f^{14}$).

Members with +4 oxidation state: Ce, Pr, Nd, Tb, and Dy (NCERT: Pr, Nd, Tb and Dy exhibit +4 only in their oxides, MO₂; Ce⁴⁺ is the only one stable more broadly)

Step 3: Identify elements showing +2 oxidation state

A +2 oxidation state means losing only the two 6s electrons. This is favoured when the remaining $4f^n$ configuration is stable — again half-filled ($4f^7$) or fully filled ($4f^{14}$).

Here are the common mistakes students make on this topic, along with clear, concept-first corrections.

Mistake 1: Memorising the list without understanding the why

The Mistake: Students simply try to cram the names (Ce, Pr, Tb for +4; Eu, Yb for +2) and forget them under pressure. They often miss Sm for +2, and Nd/Dy for +4.

Why it happens: The question explicitly asks you to correlate with electronic configuration. If you don't connect the oxidation state to stability, you are just memorising a list.

How to avoid it: Always anchor the oxidation state to the stable electronic configuration of the element or its neighbour.

• +4 Oxidation State: Look for elements that can lose 4 electrons to achieve a noble gas configuration (Xe, Rn) or a half-filled f-shell ($f^7$).
  • Ce (Cerium): $[Xe] 4f^1 5d^1 6s^2$ → Ce$^{4+}$ = $[Xe]$ (Noble gas, very stable).
  • Pr (Praseodymium): $[Xe] 4f^3 6s^2$ → Pr$^{4+}$ = $[Xe] 4f^1$ (Not noble gas, but stable).
  • Tb (Terbium): $[Xe] 4f^9 6s^2$ → Tb$^{4+}$ = $[Xe] 4f^7$ (Half-filled f-shell, very stable).
• +2 Oxidation State: Look for elements that can lose 2 electrons to achieve a half-filled f-shell ($f^7$) or a fully-filled f-shell ($f^{14}$).
  • Eu (Europium): $[Xe] 4f^7 6s^2$ → Eu$^{2+}$ = $[Xe] 4f^7$ (Half-filled, very stable).
  • Yb (Ytterbium): $[Xe] 4f^{14} 6s^2$ → Yb$^{2+}$ = $[Xe] 4f^{14}$ (Fully-filled, very stable).
  • Sm (Samarium): $[Xe] 4f^6 6s^2$ → Sm$^{2+}$ = $[Xe] 4f^6$ (Close to half-filled, moderately stable).
  • (Beyond NCERT: Tm$^{2+}$, $[Xe] 4f^{13}$, is known in the research literature — but NCERT's own +2 list is Sm, Eu, Yb, so stick to those in an exam.)

Key takeaway: Don't just list the names. Write the electronic configuration of the ion and state why it is stable (noble gas, $f^7$, or $f^{14}$).

Mistake 2: Confusing the +4 and +2 lists

The Mistake: Writing Ce, Pr, Tb for +2, or Eu, Yb for +4. Or leaving out Nd (Neodymium) and Dy (Dysprosium) from the +4 list entirely — they do show +4, just only in their oxides (NdO$_2$, DyO$_2$), not as freely stable ions in solution the way Ce$^{4+}$ is.

Why it happens: Students try to remember by "first few" or "last few" without the config logic.

How to avoid it: Use the stability rule as a filter.

• For +4: Ask: "Does losing 4 electrons give me a noble gas ($[Xe]$) or a half-filled $f^7$?"
  • Ce$^{4+}$ = $[Xe]$ (noble gas) → Yes (most stable, even in aqueous solution).
  • Pr$^{4+}$ = $[Xe] 4f^1$ (not noble gas, but stable) → Yes (only in oxide, PrO$_2$).
  • Tb$^{4+}$ = $[Xe] 4f^7$ (half-filled) → Yes (second most stable, after Ce).
  • Nd$^{4+}$ = $[Xe] 4f^2$ (not noble gas, not half-filled) → Yes, but only in the oxide NdO$_2$ — weakly stable.
  • Dy$^{4+}$ = $[Xe] 4f^8$ (not noble gas, not half-filled) → Yes, but only in the oxide DyO$_2$ — weakly stable.
• For +2: Ask: "Does losing 2 electrons give me a half-filled $f^7$ or fully-filled $f^{14}$?"
  • Eu$^{2+}$ = $[Xe] 4f^7$ (half-filled) → Yes.
  • Yb$^{2+}$ = $[Xe] 4f^{14}$ (fully-filled) → Yes.
  • Sm$^{2+}$ = $[Xe] 4f^6$ (close to half-filled) → Yes.
  • Tm$^{2+}$ = $[Xe] 4f^{13}$ — known in the literature, but not part of NCERT's list; leave it out of the exam answer.

Memory trick: Ce, Pr, Tb → Can Produce Tetravalent (+4), and Nd, Dy also do so weakly (only as MO$_2$ oxides). Eu, Yb → Easily Yield divalent (+2). (Sm is the "almost" one — NCERT says its behaviour is very much like europium's).

Mistake 3: Forgetting Sm for +2

The Mistake: Only writing Eu and Yb for +2, missing Sm.

Why it happens: Eu$^{2+}$ and Yb$^{2+}$ are the most stable +2 ions, so Sm$^{2+}$ (a strong reducing agent) gets overlooked.

How to avoid it: NCERT's own text says "the behaviour of samarium is very much like europium, exhibiting both +2 and +3 oxidation states" — so the full NCERT answer is Sm, Eu and Yb.

Rule: Write Sm, Eu and Yb for +2, noting Sm$^{2+}$ is the least stable of the three. (Tm$^{2+}$ is known in the research literature, e.g., TmI$_2$, but it is not part of NCERT's list — mention it only as extra knowledge.)

Mistake 4: Writing the wrong electronic configuration for the element

The Mistake: Writing Ce as $[Xe] 4f^2 6s^2$ (wrong) instead of $[Xe] 4f^1 5d^1 6s^2$ (correct).

Why it happens: The 4f and 5d orbitals are very close in energy. Some elements (like Ce, Gd, Lu) have anomalous configurations.

How to avoid it: Memorise the exceptions in the lanthanoid series:

• Ce: $[Xe] 4f^1 5d^1 6s^2$ (not $4f^2$)
• Gd: $[Xe] 4f^7 5d^1 6s^2$ (not $4f^8$)
• Lu: $[Xe] 4f^{14} 5d^1 6s^2$ (not $4f^{15}$)

For the +4 and +2 question, the key ones are Ce and Eu/Yb. Know their exact ground state configurations.

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