Which of the following oxidation state is common for all lanthanoids? (i) +2 (ii) +3 (iii) +4 (iv) +5

NCERT Exemplar · Q7

Q. Which of the following oxidation state is common for all lanthanoids?

(i) +2

(ii) +3

(iii) +4

(iv) +5

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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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The most common and stable oxidation state across all lanthanoids is +3. This arises from the electronic configuration of the lanthanoids, where losing two 6s electrons and one 4f electron (or one 5d electron in some cases) leads to a particularly stable half-filled or empty 4f subshell, or simply a lower energy state.

The lanthanoids (elements from Cerium, Z=58, to Lutetium, Z=71) are famous for their remarkably similar chemical properties. This similarity is a direct consequence of the Lanthanide Contraction — the steady decrease in atomic and ionic radii as we move across the series. But the real reason they behave so alike lies in their electronic configurations.

All lanthanoids have the general outer electronic configuration:

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

The 4f orbitals are deeply buried inside the atom, shielded by the 5s and 5p orbitals. This means the 4f electrons are not very involved in bonding. When a lanthanoid forms an ion, the first electrons to go are always the two 6s electrons (they are the outermost). After that, the next easiest electron to remove is either a 4f electron or, in a few cases, a 5d electron.

The +3 state is reached by losing these three electrons. For every single lanthanoid, this +3 ion is either the most stable or one of the most stable states. Let's see why the other options fail.

Why +2 is not common for all:

The +2 state occurs when only the two 6s electrons are lost. This leaves a $4f^n$ configuration. For most lanthanoids, this configuration is not particularly stable. Only a few elements can achieve a stable +2 state, and only under special conditions (e.g., in aqueous solution or solid compounds).
- Eu²⁺ and Yb²⁺ are the most famous exceptions. Eu²⁺ has a half-filled 4f⁷ subshell (very stable), and Yb²⁺ has a fully filled 4f¹⁴ subshell (also very stable).
- Others like Sm²⁺ and Tm²⁺ exist but are strong reducing agents and not common. So +2 is definitely not common for all lanthanoids.
Why +3 is common for all:

This is the sweet spot. Removing two 6s electrons and one 4f (or 5d) electron gives a $4f^{n-1}$ configuration. For the early lanthanoids (like La, Ce, Gd, Lu), this leads to an empty, half-filled, or fully filled 4f subshell — all exceptionally stable. For the others, the +3 state is still the most energetically favourable because the 4f electrons are so tightly bound that removing a fourth electron requires too much energy, while removing only two leaves the ion too large and less stable.

Every lanthanoid forms stable +3 compounds (oxides, halides, etc.). This is the defining oxidation state of the series.
Why +4 is not common for all: …

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Common Mistakes on Lanthanide Contraction & Oxidation States

The Question

Which of the following oxidation state is common for all lanthanoids?

(i) +2 (ii) +3 (iii) +4 (iv) +5

Correct Answer: (ii) +3

Mistake #1: Confusing “common” with “most stable”

What students do: They think of the most stable oxidation state for a few lanthanoids (e.g., Ce⁴⁺, Eu²⁺) and pick +2 or +4.
Why it’s wrong: The question asks for the oxidation state common to all lanthanoids. While +2 and +4 occur for some, only +3 is exhibited by every lanthanoid.
How to avoid: Read carefully — “common for all” means every single element in the series must show that state. Memorise: +3 is universal for lanthanoids.

Mistake #2: Forgetting the +3 stability due to Lanthanide Contraction

What students do: They treat lanthanide contraction as just a size trend, not linking it to oxidation state stability.
Why it’s wrong: Lanthanide contraction causes a steady decrease in ionic radius across the series. This makes the +3 ions (Ln³⁺) more stable because:
- The effective nuclear charge increases.
- The 4f orbitals are deeply buried and shielded.
- Removing more than 3 electrons becomes energetically unfavourable.
How to avoid: Connect the concept: Lanthanide contraction → stable +3 state → common to all.

Mistake #3: Picking +2 because of Eu²⁺ or Yb²⁺

What students do: They recall that Eu and Yb show stable +2 states (half-filled and fully-filled 4f subshells) and assume +2 is common.
Why it’s wrong: Only Eu, Yb, and sometimes Sm show +2. Most lanthanoids do not form stable +2 ions.
How to avoid: Remember the exceptions (Eu²⁺, Yb²⁺) are special cases, not the rule. The rule is +3.

Mistake #4: Picking +4 because of Ce⁴⁺ or Tb⁴⁺

What students do: They recall Ce⁴⁺ (empty 4f) and Tb⁴⁺ (half-filled 4f) and think +4 is widespread.
Why it’s wrong: Only Ce, Pr, and Tb show +4 with any stability. Most lanthanoids do not reach +4. …

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Q.Which of the following oxidation state is common for all lanthanoids?

Concept understanding — Lanthanide Contraction

Lanthanide Contraction: The Intuition

The Precise Statement

Why It Matters

The Numbers (for reference)

Why this formula?

Lanthanide Contraction: Why It Happens

The Core Question

The Formula That Governs It

The Derivation: Step by Step

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

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

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

4. How this shrinks the radius

Method: Electronic Configuration & Stability Analysis

Steps

Common Mistakes on Lanthanide Contraction & Oxidation States

The Question

Mistake #1: Confusing “common” with “most stable”

Mistake #2: Forgetting the +3 stability due to Lanthanide Contraction

Mistake #3: Picking +2 because of Eu²⁺ or Yb²⁺

Mistake #4: Picking +4 because of Ce⁴⁺ or Tb⁴⁺

Q.Which of the following oxidation state is common for all lanthanoids?

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