The trap: Students predict that 5d transition metals (Hf, Ta, W) should be significantly larger than their 4d counterparts (Zr, Nb, Mo) because they sit one period lower. NEET exploits this by asking why 4d and 5d pairs show nearly identical atomic radii and similar chemistry.
What lanthanoid contraction actually is: The 14 lanthanoid elements (Ce to Lu) are filled across the 4f subshell. The 4f electrons shield nuclear charge poorly — their diffuse orbital shape means each added proton is only partially screened. Consequently, effective nuclear charge increases steadily from La (Z = 57) to Lu (Z = 71), shrinking atomic and ionic radii progressively across the series (NCERT Class 12 Chemistry Chapter 4, page 112).
The downstream consequence for 5d elements: By the time you reach Hf (Z = 72), the cumulative contraction across the 14 lanthanoids has offset the expected size increase from adding a new shell. Result: Zr (4d) and Hf (5d) have nearly identical atomic radii (~160 pm). This extends to Nb/Ta and Mo/W pairs — similar size means similar chemistry (lattice energies, bond strengths, coordination behaviour).
Key facts for NEET:
Watch out: Questions may frame the contraction as affecting only lanthanoids themselves. Remember the consequence extends to ALL subsequent elements — the 5d series and even post-lanthanoid p-block elements are affected.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
What is the primary cause of lanthanoid contraction?
Answer: B. Lanthanoid contraction results from the poor shielding ability of 4f electrons, which allows effective nuclear charge to increase across the lanthanoid series, progressively shrinking atomic/ionic radii (NCERT Class 12 Chemistry Chapter 4, page 112).
Why A is wrong: A is wrong because 3d electrons are involved in first-row transition metal chemistry, not in causing the lanthanoid contraction. The contraction occurs due to 4f electron behaviour across Ce–Lu. (trap: confusing d-block shielding with f-block shielding)
Why C is wrong: C is wrong because there is no 'expansion' of 5d — the 5d subshell actually ends up smaller than expected precisely because of the contraction caused by poor 4f shielding. (trap: reversing cause and effect)
Why D is wrong: D is wrong because while relativistic effects exist for heavy elements, NCERT attributes lanthanoid contraction specifically to poor 4f shielding, not relativistic contraction. (trap: importing advanced concepts beyond syllabus)
Which is the most common and stable oxidation state exhibited by lanthanoid elements?
Answer: A. The +3 oxidation state is the most stable and common across the lanthanoid series, as removal of two 6s electrons and one 4f/5d electron consistently gives the most thermodynamically stable configuration (NCERT Class 12 Chemistry Chapter 4, page 112).
Why B is wrong: B is wrong because +2 is shown only by specific lanthanoids like Eu²⁺ (achieves half-filled 4f⁷) and is not the general stable state. (trap: overgeneralising an exception)
Why C is wrong: C is wrong because +4 is shown only by specific lanthanoids like Ce⁴⁺ (achieves empty 4f⁰) and is not the general stable state. (trap: overgeneralising an exception)
Why D is wrong: D is wrong because +5 is not a common oxidation state for any lanthanoid element. No lanthanoid routinely exhibits +5 in its chemistry. (trap: extrapolating transition metal OS range to lanthanoids)
Ce⁴⁺ is stable because it achieves which electronic configuration?
Answer: B. Cerium in +4 state loses all four electrons beyond the Xe core (two 6s + one 5d + one 4f), achieving the stable 4f⁰ configuration equivalent to the xenon noble-gas core (NCERT Class 12 Chemistry Chapter 4, page 112).
Why A is wrong: A is wrong because half-filled 4f⁷ stability explains Eu²⁺ (and Gd³⁺/Tb⁴⁺), not Ce⁴⁺. Ce has only 1-2 f electrons to begin with; it cannot achieve 4f⁷ by losing electrons. (trap: mixing up which lanthanoid achieves which stable configuration)
Why C is wrong: C is wrong because completely filled 4f¹⁴ explains Lu³⁺ stability (or Yb²⁺ stability), not Ce⁴⁺. Ce would need to gain electrons to reach 4f¹⁴, which does not happen in oxidation. (trap: confusing filled vs empty f-shell stability)
Why D is wrong: D is wrong because 5d⁵ is irrelevant here — Ce⁴⁺ has no 5d electrons remaining, and 5d⁵ is a d-block concept (Mn²⁺ analogy). (trap: conflating d-block and f-block stability arguments)
Which pair of elements from the 4d and 5d series has nearly identical atomic radii due to lanthanoid contraction?
Answer: B. Zr (4d, Group 4) and Hf (5d, Group 4) have nearly identical atomic radii (~160 pm) because the lanthanoid contraction offsets the expected size increase from the additional shell in Hf (NCERT Class 12 Chemistry Chapter 4, page 112).
Why A is wrong: A is wrong because Ti (3d) and Zr (4d) are from consecutive d-series but do NOT illustrate lanthanoid contraction — the lanthanoids lie between the 4d and 5d series, not between 3d and 4d. (trap: misidentifying which period transition the contraction affects)
Why C is wrong: C is wrong because while Fe (3d) and Ru (4d) are in the same group, the lanthanoid contraction specifically affects the 4d/5d comparison, not 3d/4d. Fe and Ru do differ significantly in size. (trap: applying the concept to the wrong period pair)
Why D is wrong: D is wrong because Sc (3d) and Y (4d) are again a 3d/4d pair. Lanthanoid contraction operates between periods 5 and 6 (4d and 5d elements), not between periods 4 and 5. (trap: same error as option A)
Eu²⁺ is unusually stable among lanthanoid divalent ions. What is the electronic configuration of Eu²⁺ that explains this stability?
Answer: A. Eu has configuration [Xe] 4f⁷ 6s². Removing two 6s electrons gives Eu²⁺ = [Xe] 4f⁷, a half-filled f-subshell configuration that confers extra stability (NCERT Class 12 Chemistry Chapter 4, page 112).
Why B is wrong: B is wrong because [Xe] 4f⁶ would mean only six 4f electrons — this does not correspond to any special stability (neither half-filled nor fully filled). It would be the configuration of Sm²⁺, which lacks the extra half-filled stability. (trap: miscounting electrons after ionisation)
Why C is wrong: C is wrong because [Xe] 4f⁸ has one electron beyond the half-filled shell and offers no extra stability. It is the configuration of Tb³⁺; Eu²⁺ keeps Eu's seven 4f electrons and does not gain an eighth. (trap: adding instead of removing electrons)
Why D is wrong: D is wrong because [Xe] 4f¹⁴ is the completely filled configuration (Lu³⁺ or Yb²⁺). Eu only has 7 f-electrons in ground state; reaching 4f¹⁴ is impossible by removing electrons. (trap: confusing Eu with late lanthanoids)
As a consequence of lanthanoid contraction, which property of 4d and 5d transition metals in the same group becomes nearly equal?
Answer: C. Lanthanoid contraction makes the atomic and ionic radii of 5d elements nearly equal to their 4d counterparts in the same group. Since similar size leads to similar lattice energies, coordination chemistry, and reactivity, their chemical properties converge (NCERT Class 12 Chemistry Chapter 4, page 112).
Why A is wrong: A is wrong because while ionisation enthalpies are affected, the primary and directly observed consequence is the convergence of atomic radii. IE is a secondary effect, and the question asks for the direct property that becomes 'nearly equal.' (trap: picking a secondary effect over the primary observable)
Why B is wrong: B is wrong because electronegativity is influenced by size but is not the primary property that NCERT highlights as becoming 'nearly equal' — the radii convergence is the fundamental observation from which other property similarities follow. (trap: same reasoning as A)
Why D is wrong: D is wrong because magnetic moments depend on the number of unpaired electrons in the specific ion's d/f configuration, not on atomic radius. Two elements can have identical radii but very different magnetic moments. (trap: conflating size-dependent and electron-configuration-dependent properties)
A student claims: "Since Hf is one full period below Zr, it must have significantly larger atomic radius." What concept directly refutes this prediction?
Answer: D. Lanthanoid contraction — the steady shrinkage across Ce to Lu due to poor 4f shielding — exactly offsets the expected size increase from adding a new shell, making Hf nearly the same size as Zr (NCERT Class 12 Chemistry Chapter 4, page 112).
Why A is wrong: A is wrong because the inert pair effect explains reluctance of 6s² electrons to participate in bonding (e.g., Tl⁺ vs Tl³⁺), not size similarity between 4d/5d elements. (trap: confusing two separate heavy-element phenomena)
Why B is wrong: B is wrong because while relativistic effects exist for very heavy elements, NCERT attributes the 4d/5d size similarity specifically to lanthanoid contraction (poor 4f shielding), not to relativistic shielding. (trap: importing beyond-syllabus physics explanation)
Why C is wrong: C is wrong because 'd-orbital contraction' is not a standard NCERT concept for this context. The contraction relevant here is specifically across the 4f lanthanoid series, not a d-orbital phenomenon. (trap: inventing a plausible-sounding term)
Among the lanthanoid ions Sm³⁺ (4f⁵), Eu³⁺ (4f⁶), Gd³⁺ (4f⁷), and Tb³⁺ (4f⁸), which ion would be LEAST likely to deviate from the +3 oxidation state and show a +2 or +4 state?
Answer: C. Gd³⁺ already has the half-filled 4f⁷ configuration — the most stable f-electron arrangement. It has no thermodynamic incentive to change oxidation state (neither gaining nor losing an f-electron improves stability). Eu³⁺ can reduce to Eu²⁺ (4f⁷), Ce can oxidise to Ce⁴⁺ (4f⁰), and Tb³⁺ can oxidise to Tb⁴⁺ (4f⁷), but Gd³⁺ is already at the stability maximum.
Why A is wrong: A is wrong because Sm³⁺ (4f⁵) can potentially achieve 4f⁶ as Sm²⁺ — while not exceptionally stable, Sm²⁺ does exist. More importantly, Sm³⁺ is NOT at a special stability configuration, so it's not the answer to 'least likely to deviate.' (trap: not recognising that half-filled 4f⁷ is the key stability marker)
Why B is wrong: B is wrong because Eu³⁺ (4f⁶) readily reduces to Eu²⁺ (4f⁷) to achieve the half-filled configuration. This is precisely a deviation from +3 to +2. (trap: confusing 'achieves stability by changing OS' with 'already at stability maximum')
Why D is wrong: D is wrong because Tb³⁺ (4f⁸) can oxidise to Tb⁴⁺ (4f⁷) to achieve the half-filled configuration. This is a known deviation from +3 to +4. (trap: same logic as Eu but in the oxidation direction)
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Pattern: Lanthanoid contraction — effect on 4d/5d element properties (P.CHE.U11.LANTHANOID_CONTRACTION, observed in PYQ 2022, 2025).
Given
The following atomic radii are provided:• Zr (4d, Group 4): 160 pm• Hf (5d, Group 4): 159 pm• Ti (3d, Group 4): 147 pm
Required
Explain why Hf, despite being one period below Zr, has nearly the same atomic radius. Predict whether Hf and Zr would show similar or different chemical properties.
Concept
Lanthanoid contraction: across Ce (Z = 58) to Lu (Z = 71), poor shielding by 4f electrons causes a cumulative decrease in atomic radius (~12–15 pm total). This contraction offsets the expected increase from adding a new electron shell.
Formula/Principle
No quantitative formula — this is a qualitative comparison. The principle: if two elements in the same group have nearly identical atomic/ionic radii, they will exhibit similar lattice energies, bond strengths, and overall chemical behaviour.
Substitution/Application
• Expected: Hf should be ~10–15 pm larger than Zr (one extra shell, analogous to Ti→Zr increase of 13 pm).• Actual: Hf is 159 pm ≈ 160 pm (Zr). The "missing" expansion is absorbed by lanthanoid contraction across the 14 lanthanoid elements.
Calculation
Radius difference: |160 − 159| = 1 pm ≈ negligible.
Compare with Ti→Zr difference: 160 − 147 = 13 pm (a full shell increase without intervening f-block).
Final answer
Lanthanoid contraction explains why Hf and Zr have nearly identical radii (difference ~1 pm vs expected ~13 pm). Consequence: Hf and Zr exhibit remarkably similar chemical properties — they form similar compounds, have similar coordination numbers, and are difficult to separate.
Common trap
Students predict Hf should be "much larger" because it's in period 6. They forget the 14-element lanthanoid contraction that intervenes between the 4d and 5d series. This trap appears as a distractor: "5d elements are always larger than 4d elements in the same group."
Similar NEET-style question
"Nb and Ta have very similar atomic radii. Which of the following best explains this observation?
(A) d-orbital expansion (B) Actinoid contraction (C) Lanthanoid contraction (D) Relativistic effects"
Answer: (C). The same principle applies to all 4d/5d pairs in Groups 4–7.
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Steady decrease in atomic/ionic size across lanthanide series due to imperfect shielding by f-electrons. Causes 4d ≈ 5d size; explains close properties of Zr/Hf, Nb/Ta, Mo/W (4d/5d pairs).
-- NCERT Class 12 Chemistry, Ch. 4, p. 112These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Root cause: concept gap
Lanthanoid contraction (imperfect 4f shielding) makes 5d elements similar in size to 4d. Zr/Hf, Nb/Ta, Mo/W have nearly identical chemistry.
More in The d and f Block Elements: 2 exam traps and mistakes · 3 formulas · 2 question patterns from its other lessons.
4 questions from NEET 2021, 2022, 2024, 2026. Answers verified against NTA official keys.
The pair of lanthanoid ions which are diamagnetic is
Gadolinium has a low value of third ionisation enthalpy because of
The incorrect statement among the following is :
All 15 past-paper questions from The d and f Block Elements →
Recurring question shapes from past papers. Each pattern shows why wrong options look tempting.
predicts large difference due to period shift
Expects 5d to be much larger than 4d
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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