What are the different oxidation states exhibited by the lanthanoids?

Exercises · 4.10

Q. What are the different oxidation states exhibited by the lanthanoids?

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Concept understanding — Lanthanide Contraction

Lanthanide Contraction: The Intuition

Imagine you are walking through a dense forest. With every step forward, you push through thick undergrowth. The deeper you go, the more tired you become — each step feels a little harder, and you find yourself hunching forward, your shoulders pulling inward. That inward pull is exactly what happens inside the lanthanide atoms.

The lanthanides are the 14 elements from cerium ( $Ce$ , atomic number 58) to lutetium ( $Lu$ , atomic number 71). As you move from one element to the next, you add one proton to the nucleus and one electron to the atom. The new electron goes into a 4f orbital — a set of orbitals that are shaped like clover leaves and sit deep inside the atom, close to the nucleus.

Here is the key: 4f orbitals are poorly shielded. They do not spread out far from the nucleus, and they do not block the nuclear charge from pulling on the outer electrons. So when you add a proton, the nucleus gets stronger, and the 4f electrons do almost nothing to stop that extra pull. The result? The entire electron cloud — especially the outermost electrons — gets pulled inward. The atom shrinks.

Note

Shielding is the ability of inner electrons to "block" the outer electrons from feeling the full positive charge of the nucleus. Electrons in s and p orbitals shield well; 4f electrons shield very poorly.

The Precise Statement

Lanthanide contraction is the steady and significant decrease in the atomic and ionic radii of the lanthanide elements as atomic number increases from 58 (Ce) to 71 (Lu).

\text{Atomic radius} \propto \frac{1}{Z_{\text{eff}}}

The total contraction across the entire series is about 15–20 picometers — roughly 10–15% of the initial radius. That is a substantial shrinkage for a single row of the periodic table.

Why It Matters

This contraction has two enormous consequences in chemistry:

Similarity of post-lanthanide elements: After lutetium, the next elements are hafnium ( $Hf$ , 72), tantalum ( $Ta$ , 73), and tungsten ( $W$ , 74). Because the lanthanide contraction has made the atoms so small, these elements have almost identical atomic and ionic radii to their counterparts directly above them in the periodic table — zirconium ( $Zr$ ), niobium ( $Nb$ ), and molybdenum ( $Mo$ ). This is why zirconium and hafnium are chemically almost inseparable — they are the same size.
Difficulty in separating lanthanides: All lanthanide ions ( $Ln^{3+}$ ) have nearly identical chemical properties because their radii change so gradually. Separating them requires hundreds of repeated steps (ion-exchange chromatography, solvent extraction) — a painstaking process that was a major challenge in early nuclear chemistry.

Watch out

A common mistake is to think lanthanide contraction means the atoms get smaller because the 4f orbitals are "full" or because of some repulsion effect. It is purely due to poor shielding of the 4f electrons, which lets the nuclear charge pull everything inward.

The Numbers (for reference)

Element	Atomic Number	Ionic Radius ( $Ln^{3+}$ , pm)
Ce	58	103.4
Pr	59	101.3
Nd	60	99.5

Why this formula?

Lanthanide Contraction: Why It Happens

The Lanthanide Contraction is the steady decrease in atomic and ionic radii of the lanthanide elements (Ce to Lu) as atomic number increases. The key observation: the radii shrink by about 1–2 pm per element, despite adding electrons to the 4f subshell.

The Core Question

Why does adding electrons not increase the size, but instead decrease it?

The Formula That Governs It

The effective nuclear charge ( $Z_{\text{eff}}$ ) experienced by an electron is:

Z_{\text{eff}} = Z - S

Where:

$Z$ = atomic number (protons in nucleus)
$S$ = shielding constant (screening by inner electrons)

The key formula for the trend in ionic radii ( $r$ ) across the lanthanides is:

r \propto \frac{n^2}{Z_{\text{eff}}}

Where $n$ is the principal quantum number of the outermost electron (here, $n=6$ for the 6s orbital).

The Derivation: Step by Step

1. What happens when you add a proton and an electron?

Each lanthanide adds:

+1 proton to the nucleus (increases $Z$ by 1)
+1 electron to the 4f subshell

2. The 4f orbital is "penetrating" but poorly shielding

The 4f orbital has a radial distribution that peaks close to the nucleus (inside the 5s and 5p shells).
However, 4f electrons are very poor at shielding the outer 6s electrons from the nuclear charge.

Why?

The 4f orbital is diffuse and deeply buried — it does not effectively screen the outer electrons because:

Its shape (complex, multi-lobed) means it doesn't occupy the space between the nucleus and the 6s electrons efficiently.
The 4f electrons are inside the 5s/5p shells, so they don't block the nuclear pull on the 6s electrons.

3. The net effect on $Z_{\text{eff}}$

When you add one proton ( $\Delta Z = +1$ ) and one 4f electron ( $\Delta S \approx 0.85$ to 0.95), the change in effective nuclear charge is:

\Delta Z_{\text{eff}} \approx +1 - 0.85 = +0.15 \text{ to } +0.05

Result: $Z_{\text{eff}}$ increases slightly with each element.

4. How this shrinks the radius

From the formula $r \propto \frac{n^2}{Z_{\text{eff}}}$ :

$n$ (the principal quantum number of the 6s orbital) stays constant at 6.
$Z_{\text{eff}}$ increases.
Therefore, $r$ decreases. …

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Lanthanoids show a dominant +3 oxidation state, but also exhibit +2 and +4 states when they lead to a stable (empty, half-filled, or full-filled) 4f subshell. The key is the special stability of empty, half-filled and fully filled 4f configurations.

The lanthanoids (elements Ce through Lu) are famous for their chemical similarity, which arises from the Lanthanide Contraction — the steady decrease in atomic and ionic radii as we move across the series. This contraction happens because the 4f orbitals are poor at shielding the nuclear charge, so each added proton pulls the electron cloud inward. But the real story for oxidation states is about electronic stability.

The 4f subshell can hold 14 electrons. Like all subshells, it is most stable when it is empty ( $4f^0$ ), half-filled ( $4f^7$ ), or fully filled ( $4f^{14}$ ). The +3 state is the default for all lanthanoids because losing three electrons (typically two from the 6s orbital and one from the 4f or 5d orbital) is energetically favourable. However, some lanthanoids can deviate to +2 or +4 if doing so brings them closer to one of these stable configurations.

Let’s break it down systematically.

The default +3 state.

All lanthanoids exhibit the +3 oxidation state. This is because the electronic configuration of a neutral lanthanoid is generally $[Xe]\,4f^n\,6s^2$ (or $[Xe]\,4f^{n-1}\,5d^1\,6s^2$ for a few like La, Ce, Gd, Lu). Removing the two 6s electrons and one 4f (or 5d) electron yields the $Ln^{3+}$ ion with configuration $[Xe]\,4f^{n-1}$ . This is the most common and stable state for all 15 elements.
The +2 state — when it leads to a stable configuration.

A lanthanoid can adopt the +2 state if the resulting $Ln^{2+}$ ion has a particularly stable 4f configuration. This happens for:
- Eu (Europium): Neutral Eu is $[Xe]\,4f^7\,6s^2$ . Losing two electrons gives $Eu^{2+}$ with $4f^7$ — a half-filled subshell. This is very stable.
- Yb (Ytterbium): Neutral Yb is $[Xe]\,4f^{14}\,6s^2$ . Losing two electrons gives $Yb^{2+}$ with $4f^{14}$ — a fully filled subshell. Also very stable.
- Sm (Samarium): also shows +2, less commonly — NCERT notes that samarium's behaviour is very much like europium's, exhibiting both +2 and +3 states. $Sm^{2+}$ has $4f^6$ (one electron short of half-filled); it exists in solid compounds but is a strong reducing agent in solution. ( $Tm^{2+}$ , $4f^{13}$ , is known in the research literature but is not part of NCERT's list.)
Watch out

A common mistake is to think that all lanthanoids can show +2. Only Sm, Eu and Yb do so with any significance (NCERT's list — Tm²⁺ appears only in the research literature). The +2 state for others is extremely unstable or unknown.
The +4 state — when losing more electrons gives stability.

A lanthanoid can adopt the +4 state if the resulting $Ln^{4+}$ ion has a stable configuration. This happens for:
- Ce (Cerium): Neutral Ce is $[Xe]\,4f^1\,5d^1\,6s^2$ (or $4f^2\,6s^2$ ). Losing four electrons gives $Ce^{4+}$ with $4f^0$ — an empty subshell. This is very stable, and $Ce^{4+}$ is a strong oxidising agent.
- Tb (Terbium): Neutral Tb is $[Xe]\,4f^9\,6s^2$ . Losing four electrons gives $Tb^{4+}$ with $4f^7$ — a half-filled subshell. Also stable.
- Pr, Nd and Dy: these also show +4, but only in their oxides, $MO_2$ (NCERT). $Pr^{4+}$ ( $4f^1$ ), $Nd^{4+}$ ( $4f^2$ ) and $Dy^{4+}$ ( $4f^8$ ) reach no special landmark, so their +4 state does not survive outside the oxide lattice.
Tip

Notice the pattern: Ce (+4 → $4f^0$ ), Tb (+4 → $4f^7$ ), Eu (+2 → $4f^7$ ), Yb (+2 → $4f^{14}$ ). The stable configurations are the same ones that govern the magnetic and spectral properties of these ions.
The intermediate +2 and +4 states are not common for all.

For most lanthanoids (e.g., La, Gd, Ho, Er, Lu), the +3 state is the only one observed in aqueous solution. Even where +2 or +4 species exist, they are often too reducing or too oxidising to survive in water, and are stabilised only in solid compounds (e.g., $SmI_2$ , $PrO_2$ ) — cerium(IV) being the notable aqueous exception.
A summary table for quick reference. …

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Oxidation States of Lanthanoids

Method: Electronic Configuration Rule

Step 1 — Know the Default: +3

The lanthanoids' outer configuration is $4f^{1-14}\,5d^{0-1}\,6s^2$ . Losing the two $6s$ electrons plus one $4f$ (or $5d$ ) electron is energetically favourable for every member, so +3 is the characteristic state of the whole series — no special 4f configuration is needed for it.

Step 2 — Oxidation States of Lanthanoids

Lanthanoids exhibit multiple oxidation states, but the most common and stable one is +3.

Oxidation State	Examples	Reason
+3 (most stable)	$La^{3+}, Ce^{3+}, Eu^{3+}, Gd^{3+}, Lu^{3+}$	Removal of the $6s^2$ pair and one $4f$ / $5d$ electron is energetically favourable for every member
+2	$Sm^{2+}, Eu^{2+}, Yb^{2+}$	$Eu^{2+}$ ( $4f^7$ ) and $Yb^{2+}$ ( $4f^{14}$ ) land on stable configurations; NCERT notes Sm behaves very much like Eu
+4	$Ce^{4+}, Pr^{4+}, Nd^{4+}, Tb^{4+}, Dy^{4+}$	$Ce^{4+}$ reaches $4f^0$ and $Tb^{4+}$ reaches $4f^7$ ; Pr, Nd and Dy show +4 only in their oxides, $MO_2$

Step 3 — The Rule for Predicting Oxidation States

"Stability of oxidation states in lanthanoids is governed by the tendency to attain empty ( $f^0$ ), half-filled ( $f^7$ ), or fully-filled ( $f^{14}$ ) 4f subshell."

+3 is universal because removing three electrons (two from $6s$ , one from $4f$ ) is energetically favourable.
+2 and +4 appear only when they lead to a special stability of the 4f configuration.

Step 4 — Quick Mnemonic for Exam …

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Here are the most common mistakes students make when answering questions about lanthanoid oxidation states, along with clear strategies to avoid them.

1. Confusing Lanthanoids with Actinoids

The Mistake:

Students often write that lanthanoids show a wide range of oxidation states (like +3, +4, +5, +6, +7), which is actually true for actinoids (e.g., uranium, neptunium).

Why it happens:

Both series are f-block elements, and textbooks often discuss them together. The similarity in names leads to memory overlap.

How to Avoid:

Remember the rule: Lanthanoids are “+3 specialists”.
The most common and stable oxidation state for all lanthanoids is +3.
Only a few show +2 or +4 (e.g., Eu²⁺, Yb²⁺, Ce⁴⁺, Tb⁴⁺).
Never write +5, +6, or +7 for lanthanoids — those belong to actinoids.

2. Forgetting the Role of Electronic Configuration

The Mistake:

Students list oxidation states without linking them to the stability of half-filled or fully-filled 4f subshells.

Why it happens:

They memorize the list (Eu²⁺, Ce⁴⁺, etc.) but don’t understand why those are stable.

How to Avoid:

Link stability to configuration:
- Eu²⁺ has a half-filled 4f⁷ configuration → extra stable.
- Yb²⁺ has a fully-filled 4f¹⁴ → extra stable.
- Ce⁴⁺ achieves a noble gas configuration (Xe) → stable.
- Tb⁴⁺ reaches half-filled 4f⁷ → stable.
Write the electronic configuration for each ion in your answer — examiners love this.

3. Mixing Up the +2 and +4 Examples

The Mistake:

Students incorrectly claim that all lanthanoids show +2 or +4 states, or they list the wrong elements (e.g., saying Pr³⁺ is common when it’s actually Pr⁴⁺ that is rare).

Why it happens:

The exceptions are few, so students either overgeneralize or misremember the specific elements.

How to Avoid:

Memorize the shortlist:
- +2 state: Only Eu²⁺ and Yb²⁺ are common in aqueous solution. (Sm²⁺ and Tm²⁺ exist but are unstable in water.)
- +4 state: Only Ce⁴⁺ is common. Tb⁴⁺ and Pr⁴⁺ exist but are strong oxidants.
Use a mnemonic:
- For +2: Eu and Yb → “Every Year”
- For +4: Ce, Tb, Pr → “ChTy Party”

4. Applying the Special-Stability Rule to the Wrong State

The Mistake:

Students claim +3 itself is explained by half-filled/full 4f landmarks, or invent extra reasons for it — when most $Ln^{3+}$ ions are not $f^0$ / $f^7$ / $f^{14}$ at all.

Why it happens:

The $f^0$ / $f^7$ / $f^{14}$ rule is so memorable that it gets applied everywhere.

How to Avoid:

+3 is the default for the whole series on energetic grounds — no special configuration needed.
Reserve the landmark argument for the deviations: $Ce^{4+}$ ( $4f^0$ ), $Tb^{4+}$ ( $4f^7$ ), $Eu^{2+}$ ( $4f^7$ ), $Yb^{2+}$ ( $4f^{14}$ ).
In your answer, write: …

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Q.What are the different oxidation states exhibited by the lanthanoids?

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