Quick Summary
Concept Tested: Lanthanide contraction and exceptions in atomic radii.
Chapter – Subtopic: D and F Block Elements – Lanthanide Series.
Difficulty: ★★★★☆
Estimated Time: 5 minutes
Key Formula: $$r_{\text{Ln}} = r_0 – \Delta r_{\text{contraction}}$$ (where $\Delta r_{\text{contraction}}$ reflects the gradual decrease in radius across the lanthanide series).
Answer: D
One‑line Reason: Europium’s preference for the +2 oxidation state makes its metallic radius larger than expected, giving the order Eu > Ce > Ho > N.
The Question
Arrange the elements in decreasing order of atomic radius:
$\text{Eu},\ \text{Ce},\ \text{Ho},\ \text{N}$
(A) N > Ce > Eu > Ho
(B) Ho > N > Eu > Ce
(C) Ce > Eu > Ho > N
(D) Eu > Ce > Ho > N
Quick Answer
The correct arrangement is Option D: Eu > Ce > Ho > N. Europium (185 pm) is larger than cerium (182 pm) and holmium (177 pm) because Eu readily forms a +2 state, retaining a half‑filled 4f⁷ configuration, which expands its metallic radius. Nitrogen, being a second‑period non‑metal, has a much smaller covalent radius (~71 pm).
Why Other Options Are Incorrect
Option A: Places nitrogen (71 pm) above all lanthanides, ignoring the vast difference between covalent and metallic radii; lanthanide metals are inherently larger than second‑period non‑metals.
Option B: Lists holmium before nitrogen, suggesting Ho > N, which is correct, but the subsequent order Ho > N > Eu > Ce contradicts the actual measured radii (Eu > Ce > Ho) and the special +2 state of Eu.
Option C: Positions cerium larger than europium (Ce > Eu) while ignoring the anomalously large radius of Eu caused by its +2 oxidation state; experimentally Eu (185 pm) > Ce (182 pm).
Video Solution
Video Solution Coming Soon
Understanding the Concept
The lanthanide contraction describes the gradual decrease in atomic and ionic radii across the lanthanide series due to ineffective shielding of the increasing nuclear charge by the 4f electrons. However, exceptions occur for elements like europium and ytterbium, which preferentially exhibit a +2 oxidation state, preserving a half‑filled ($4f^7$) or fully‑filled ($4f^{14}$) subshell. This extra stability leads to a larger metallic radius than predicted by the smooth contraction trend.
Key principle: $$\text{Stability of half‑filled/fully‑filled }4f^n\text{ subshell} \Rightarrow \text{larger metallic radius for Eu}^{2+}\text{ and Yb}^{2+}$$
Detailed Step-by-Step Solution
Step 1: Identify the Nature of Each Element
Nitrogen (N) is a second‑period non‑metal with a covalent radius of about 71 pm. Europium (Eu), Cerium (Ce), and Holmium (Ho) are lanthanide metals with metallic radii measured experimentally as 185 pm, 182 pm, and 177 pm respectively.
Step 2: Apply the Lanthanide Contraction Trend
Across the lanthanide series, the radius generally decreases: $$\text{Ce} > \text{Pr} > \dots > \text{Ho} > \dots > \text{Lu}.$$ This would predict Ce > Ho. However, europium deviates because it prefers a +2 oxidation state, retaining a half‑filled $4f^7$ configuration, which expands its metallic radius.
Step 3: Compare the Measured Radii
Using the given values:
- Eu: 185 pm
- Ce: 182 pm
- Ho: 177 pm
- N: 71 pm
Ordering from largest to smallest gives Eu > Ce > Ho > N, matching Option D.
Final Answer
✅ Option D: Eu > Ce > Ho > N
Essential Formulas for This Topic
$$\Delta r_{\text{contraction}} = k \times (Z_{\text{eff}})$$
where $k$ is a proportionality constant and $Z_{\text{eff}}$ is the effective nuclear charge experienced by the 4f electrons. The formula emphasizes that increasing $Z_{\text{eff}}$ leads to a smaller radius, except when a stable electronic configuration (e.g., $4f^7$ for Eu$^{2+}$) counteracts the contraction.
Common Mistakes to Avoid
Mistake 1: Assuming a Strict Decrease with Atomic Number
Many students think radii must decrease monotonically across the lanthanide series. This ignores known exceptions like Eu and Yb, which adopt a +2 state and show larger radii.
Mistake 2: Treating Nitrogen as Part of the Same Trend
Comparing N’s covalent radius directly with metallic radii of lanthanides leads to incorrect ordering; the two types of radii are not comparable.
Mistake 3: Overlooking Oxidation State Influence
Neglecting the fact that Eu prefers a +2 oxidation state (half‑filled $4f^7$) causes the misconception that Ce > Eu, contrary to experimental data.
Key Concept Summary
- Lanthanide contraction: radii decrease across the series due to poor 4f shielding.
- Europium (and Yb) are exceptions because they favor a +2 oxidation state, preserving a stable $4f^7$ (or $4f^{14}$) configuration.
- Metallic radii of lanthanides are significantly larger than covalent radii of second‑period non‑metals like nitrogen.
- Experimental radii: Eu = 185 pm, Ce = 182 pm, Ho = 177 pm, N ≈ 71 pm.
Golden Rule: Always check for oxidation‑state exceptions when applying periodic trends to the lanthanide series.
Frequently Asked Questions
Q: Why does europium have a larger radius than cerium despite a higher atomic number?
A: Europium commonly forms a +2 oxidation state, retaining a half‑filled $4f^7$ subshell, which is energetically favorable and results in a larger metallic radius than the +3 state expected for most lanthanides.
Q: Can we compare covalent and metallic radii directly?
A: No. Covalent radii (e.g., N ≈ 71 pm) refer to non‑metallic bonding, while metallic radii (e.g., lanthanides) refer to metallic lattices; they are measured differently and cannot be directly juxtaposed without context.
Q: Are there other lanthanides that break the contraction trend?
A: Yes, ytterbium (Yb) also exhibits a +2 oxidation state, leading to a larger radius than expected.
Q: How does the half‑filled $4f^7$ configuration stabilize Eu$^{2+}$?
A: Half‑filled subshells have symmetrical electron distribution, minimizing electron‑electron repulsion and lowering energy, making the +2 state particularly stable for Eu.
Prerequisites to Solve This Question
- Understanding of periodic trends, especially the lanthanide contraction.
- Knowledge of oxidation states and the stability of half‑filled/fully‑filled subshells.
- Familiarity with the difference between covalent and metallic radii.
- Ability to read and interpret experimental radius data.
After Solving This, You Can:
- ✔ Predict atomic size trends for other f‑block elements.
- ✔ Identify exceptions to periodic trends based on electronic configurations.
- ✔ Distinguish between covalent and metallic radii in comparative problems.
- ✔ Apply lanthanide contraction concepts to advanced JEE questions.
Study Tips for This Topic
- Memorize the key exceptions (Eu, Yb) and the reasons behind them.
- Practice comparing radii across different blocks to reinforce the distinction between covalent and metallic sizes.
- Use mnemonic devices for the lanthanide series order to quickly assess trends.
- Review oxidation state stability charts for f‑block elements before attempting trend‑based questions.
Difficulty Rating & Exam Frequency
Difficulty: ★★★★☆
JEE Main Frequency: Moderate – appears in aptitude of periodic trends.
JEE Advanced Frequency: Low – usually as a conceptual twist in lanthanide series questions.
Importance: High for mastering f‑block chemistry and recognizing exceptions to periodic trends.
Related Questions from D And F Block Elements
Written by Nishant Kumar Gupta
Founder of Padho Likho JEE & Senior Chemistry Educator — 12+ Years Experience, Ex-Faculty Allen/Aakash/Narayana.
Last Updated: August 2026
Question Source: JEE Main 2019 PYQ
Topic: D And F Block Elements