Out of the following statements, the incorrect statement is : (A) La is actually an element of transition series. (B) Zr and Hf have almost identical atomic radii because of lanthanoid contraction. (C) Ionic radius decreases from $La^{3+}$ to $Lu^{3+}$ ion. (D) Lanthanoids are radioactive in nature.

> [!TLDR] > Lanthanide contraction explains the steady decrease in ionic radii across the lanthanoid series, which in turn causes Zr and Hf to have nearly identical atomic radii. The incorrect statement here is **(D)** — not all lanthanoids are radioactive; only promethium (Pm) is radioactive among them. --- The key to this question is understanding **lanthanide contraction** — the gradual decrease in ionic radii from $La^{3+}$ to $Lu^{3+}$ as you move across the lanthanoid series. This happens because the 4f electrons are poor at shielding the nuclear charge, so the effective nuclear pull on the outer electrons increases steadily. Let’s examine each statement one by one. 1. **Statement (A):** *La is actually an element of transition series.* Lanthanum (La, atomic number 57) is the first member of the lanthanoid series, but it is also placed in the d-block of the periodic table (Group 3). Its electronic configuration is $[Xe]\,5d^1\,6s^2$, which makes it a d-block element. So yes, La is indeed a transition element. This statement is **correct**. 2. **Statement (B):** *Zr and Hf have almost identical atomic radii because of lanthanoid contraction.* Zirconium (Zr, period 5) and hafnium (Hf, period 6) belong to Group 4. Normally, atomic radius increases down a group. But because the 4f orbitals (filled across the lanthanoids) are inserted between Hf and the element above it, the lanthanide contraction causes Hf’s radius to be nearly the same as Zr’s. This is a classic consequence. Statement (B) is **correct**. 3. **Statement (C):** *Ionic radius decreases from $La^{3+}$ to $Lu^{3+}$ ion.* …

Out of the following statements, the incorrect statement is : (A) La is actually an element of transition seri

Question

La^{3+}

CBSECBSE Class XII Board 2025MCQ· 1mImportance★★★★★

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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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Lanthanide contraction explains the steady decrease in ionic radii across the lanthanoid series, which in turn causes Zr and Hf to have nearly identical atomic radii. The incorrect statement here is (D) — not all lanthanoids are radioactive; only promethium (Pm) is radioactive among them.

The key to this question is understanding lanthanide contraction — the gradual decrease in ionic radii from $La^{3+}$ to $Lu^{3+}$ as you move across the lanthanoid series. This happens because the 4f electrons are poor at shielding the nuclear charge, so the effective nuclear pull on the outer electrons increases steadily.

Let’s examine each statement one by one.

Statement (A): La is actually an element of transition series.

Lanthanum (La, atomic number 57) is the first member of the lanthanoid series, but it is also placed in the d-block of the periodic table (Group 3). Its electronic configuration is $[Xe]\,5d^1\,6s^2$ , which makes it a d-block element. So yes, La is indeed a transition element. This statement is correct.
Statement (B): Zr and Hf have almost identical atomic radii because of lanthanoid contraction.

Zirconium (Zr, period 5) and hafnium (Hf, period 6) belong to Group 4. Normally, atomic radius increases down a group. But because the 4f orbitals (filled across the lanthanoids) are inserted between Hf and the element above it, the lanthanide contraction causes Hf’s radius to be nearly the same as Zr’s. This is a classic consequence. Statement (B) is correct.
Statement (C): Ionic radius decreases from $La^{3+}$ to $Lu^{3+}$ ion. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set 56/3/11 markMCQ
$Q. For Questions number 13 to 16, two statements are given — one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer to these questions from the codes (A), (B), (C) and (D) as given below : (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true. Assertion (A) : Separation of Zr and Hf is difficult. Reason (R) : Because Zr and Hf lie in the same group of the periodic table.$
›Reveal solutionSolution
The difficulty in separating Zr and Hf arises from their nearly identical atomic and ionic radii due to lanthanide contraction, not merely from being in the same group. The Assertion is true, but the Reason given is too vague and incomplete to be the correct explanation.

The key to this question lies in understanding why being in the same group makes separation difficult for some elements but not others. Many elements in the same group (e.g., Na and K) are easily separated. The real culprit here is a subtle but powerful effect: the lanthanide contraction.

The Assertion is true. Zirconium (Zr) and Hafnium (Hf) are notoriously difficult to separate. Their chemical properties are so similar that traditional methods like fractional crystallisation or precipitation are extremely inefficient. This is a well-known fact in inorganic chemistry.

The Reason is true, but incomplete. It is a fact that Zr and Hf lie in Group 4 of the periodic table. However, this statement alone does not explain why their separation is difficult. Many elements in the same group (e.g., Oxygen and Sulfur) have vastly different properties and are easily separated.

The missing link: Lanthanide Contraction. The real explanation is the lanthanide contraction. As we move across the lanthanide series (elements 57–71), the 4f orbitals are filled. These orbitals have a poor shielding effect, so the increasing nuclear charge pulls the outer electrons inwards, causing a steady decrease in atomic and ionic radii.

Important

Because of the lanthanide contraction, Hafnium (atomic number 72), which comes after the lanthanide series, has an ionic radius ( $\text{Hf}^{4+} \approx 71 \text{ pm}$ ) that is almost identical to that of Zirconium ( $\text{Zr}^{4+} \approx 72 \text{ pm}$ ), even though Hf is much heavier. …
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CBSE 2025Set 56/4/11 markMCQ
$La^{3+}$
›Reveal solutionSolution
Lanthanide contraction explains the steady decrease in ionic radii across the lanthanoid series, which in turn causes Zr and Hf to have nearly identical atomic radii. The incorrect statement here is (D) — not all lanthanoids are radioactive; only promethium (Pm) is radioactive among them.

The key to this question is understanding lanthanide contraction — the gradual decrease in ionic radii from $La^{3+}$ to $Lu^{3+}$ as you move across the lanthanoid series. This happens because the 4f electrons are poor at shielding the nuclear charge, so the effective nuclear pull on the outer electrons increases steadily.

Let’s examine each statement one by one.

Statement (A): La is actually an element of transition series.

Lanthanum (La, atomic number 57) is the first member of the lanthanoid series, but it is also placed in the d-block of the periodic table (Group 3). Its electronic configuration is $[Xe]\,5d^1\,6s^2$ , which makes it a d-block element. So yes, La is indeed a transition element. This statement is correct.

Statement (B): Zr and Hf have almost identical atomic radii because of lanthanoid contraction.

Zirconium (Zr, period 5) and hafnium (Hf, period 6) belong to Group 4. Normally, atomic radius increases down a group. But because the 4f orbitals (filled across the lanthanoids) are inserted between Hf and the element above it, the lanthanide contraction causes Hf’s radius to be nearly the same as Zr’s. This is a classic consequence. Statement (B) is correct.

Statement (C): Ionic radius decreases from $La^{3+}$ to $Lu^{3+}$ ion. …
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