JEE (Advanced) · Chemistry, Class 12
d- and f-Block Elements: JEE Advanced PYQ, 2007–2026
d- and f-Block Elements (Chemistry, Class 12) has been asked 12 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.3 questions a paper, 2% of all Chemistry questions. It is the 22nd most-asked of 31 Chemistry chapters. The two 2026 papers had 0 from it. Last checked 28 October 2026.
Questions per year, both papers
Year by year.
| Year | Questions |
|---|---|
| 2007 | 0in 2 papers |
| 2008 | 1in 2 papers |
| 2009 | 1in 2 papers |
| 2010 | 0in 2 papers |
| 2011 | 2in 2 papers |
| 2012 | 0in 2 papers |
| 2013 | 0in 2 papers |
| 2014 | 0in 2 papers |
| 2015 | 1in 2 papers |
| 2016 | 1in 2 papers |
| 2017 | 0in 2 papers |
| 2018 | 0in 2 papers |
| 2019 | 2in 2 papers |
| 2020 | 2in 2 papers |
| 2021 | 0in 2 papers |
| 2022 | 0in 2 papers |
| 2023 | 0in 2 papers |
| 2024 | 0in 2 papers |
| 2025 | 2in 2 papers |
| 2026 | 0in 2 papers |
Each year has two papers, and the number of questions has fallen over time, so compare per paper and expect some drift.
Inside the chapter
Topics asked most.
- Potassium dichromate and permanganate7
- Transition element properties4
- Lanthanoids and actinoids1
Try it
10 d- and f-Block Elements questions to practise.
Our own questions, written in the JEE Advanced pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Chemistry · d- and f-Block Elements
In the laboratory preparation of KMnO₄, green K₂MnO₄ in alkaline solution is oxidised electrolytically at the anode to purple permanganate. What charge is needed to convert 0.10 mol of manganate completely to permanganate at 100% current efficiency? (F = 96500 C mol⁻¹)
Answer: (A)
At the anode, MnO₄²⁻ → MnO₄⁻ + e⁻. Manganese goes from +6 to +7, so n = 1. Charge = n × moles × F = 1 × 0.10 × 96500 = 9650 C.
Q2 · Chemistry · d- and f-Block Elements
Manganese reaches its group oxidation state of +7 in Mn₂O₇ and in MnO₄⁻. With fluorine, however, the highest binary fluoride of manganese is MnF₄. Which statement gives the correct reason for this difference?
Answer: (A)
The highest oxidation states of transition metals are found in their oxides and fluorides, but oxygen stabilises higher states than fluorine does. For manganese, the highest fluoride is MnF₄ whereas the oxide is Mn₂O₇ (Mn +7) and the oxoanion is MnO₄⁻. The reason is that oxygen can form multiple bonds (M=O) using its lone pairs and the metal's d orbitals, but fluorine can form only single M–F bonds. Fluorine is more electronegative and F₂ is the stronger oxidant than O₂, so options based on electronegativity or oxidising power are factually wrong. Ionisation enthalpies depend on the metal alone, not on the ligand.
Q3 · Chemistry · d- and f-Block Elements
Which of the following statements about the chemical behaviour of the actinoid metals is correct?
Answer: (D)
Actinoids are electropositive, highly reactive metals, especially in the finely divided state. They react with boiling water to give a mixture of oxide and hydride and combine with most non-metals at moderate temperatures. Dilute HCl attacks all of them, but HNO₃ affects most of them only slightly because a protective oxide film forms. Alkalis have no action on them, so they are not amphoteric like Al or Zn.
Q4 · Chemistry · d- and f-Block Elements
Lanthanoid ions tend to adopt a stable 4f⁷ (half-filled) configuration where possible. Atomic numbers: Ce = 58, Sm = 62, Eu = 63, Gd = 64, Tb = 65. Atomic ground-state configurations: Sm 4f⁶6s², Eu 4f⁷6s², Gd 4f⁷5d¹6s², Tb 4f⁹6s². Which set contains ions that ALL have the 4f⁷ configuration?
Answer: (B)
Principle: ion configuration = atomic configuration minus the 6s (and 5d) electrons first, then 4f electrons. Eu (4f⁷6s²) → Eu²⁺ is 4f⁷. Gd (4f⁷5d¹6s²) → Gd³⁺ is 4f⁷. Tb (4f⁹6s²) → Tb³⁺ is 4f⁸, and removing one more 4f electron gives Tb⁴⁺ as 4f⁷. Only Eu²⁺, Gd³⁺ and Tb⁴⁺ are all 4f⁷. The other sets include Tb³⁺ (4f⁸), Eu³⁺ (4f⁶) or Sm³⁺ (4f⁵).
Q5 · Chemistry · d- and f-Block Elements
The equivalent mass of KMnO₄ (molar mass 158 g mol⁻¹) depends on the medium in which it acts as an oxidant. What are its equivalent masses in strongly acidic solution (MnO₄⁻ → Mn²⁺) and in neutral solution (MnO₄⁻ → MnO₂), respectively?
Answer: (A)
Equivalent mass = molar mass / number of electrons gained per formula unit. In acid, MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, so n = 5 and the equivalent mass is 158/5 = 31.6. In neutral solution, MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻, so n = 3 and the equivalent mass is 158/3 ≈ 52.7.
Q6 · Chemistry · d- and f-Block Elements
Many Ln³⁺ ions are coloured in aqueous solution because of f–f transitions, and ions whose 4f configurations have the same number of unpaired electrons have similar colours. Atomic numbers: Pr = 59, Nd = 60, Sm = 62, Eu = 63, Gd = 64, Tb = 65, Ho = 67, Tm = 69. Which pair of Ln³⁺ ions is expected to have similar colours?
Answer: (A)
First write the Ln³⁺ configurations. Each Ln³⁺ loses its 6s and 5d/4f electrons to leave 4fⁿ, with n = Z − 57 + 1 − ... which is easiest as n = Z − 57. Pr³⁺ is 4f², Nd³⁺ 4f³, Sm³⁺ 4f⁵, Eu³⁺ 4f⁶, Gd³⁺ 4f⁷, Tb³⁺ 4f⁸, Ho³⁺ 4f¹⁰ and Tm³⁺ 4f¹². Then count unpaired electrons using Hund's rule. In the first half (n ≤ 7) the number of unpaired electrons is n. In the second half (n > 7) it is 14 − n. This gives Pr³⁺ 2, Nd³⁺ 3, Sm³⁺ 5, Eu³⁺ 6, Gd³⁺ 7, Tb³⁺ 6, Ho³⁺ 4 and Tm³⁺ 2. Only Pr³⁺ and Tm³⁺ have equal numbers (2 each), so they are similar in colour; NCERT records both as green. In general, ions with 4fⁿ and 4f¹⁴⁻ⁿ configurations show similar colours.
Q7 · Chemistry · d- and f-Block Elements
In the first transition series the second ionisation enthalpies of chromium (1592 kJ mol⁻¹) and copper (1958 kJ mol⁻¹) are noticeably higher than those of their neighbours manganese (1509 kJ mol⁻¹) and zinc (1733 kJ mol⁻¹). Which of the following best explains this?
Answer: (A)
Ionisation enthalpies across the 3d series generally rise slowly, with irregularities where electron configurations are special. Cr is 3d⁵4s¹ and Cu is 3d¹⁰4s¹. Losing the first electron (4s¹) gives Cr⁺ (3d⁵) and Cu⁺ (3d¹⁰), which have a half-filled and a completely filled d subshell. Removing a second electron means breaking into these stable arrangements, which needs extra energy. Mn⁺ (3d⁵4s¹) and Zn⁺ (3d¹⁰4s¹) lose a 4s electron next, so their second ionisation enthalpies are lower.
Q8 · Chemistry · d- and f-Block Elements
0.72 g of pure ferrous oxalate, FeC₂O₄, is dissolved in dilute H₂SO₄ and titrated warm with 0.020 M KMnO₄. Both Fe²⁺ and C₂O₄²⁻ are oxidised, to Fe³⁺ and CO₂ respectively, and MnO₄⁻ is reduced to Mn²⁺. What volume of the KMnO₄ solution is needed to reach the end point? (Fe = 56, C = 12, O = 16)
Answer: (C)
Principle: at the end point, total electrons lost by the reductant equal total electrons gained by the oxidant. Moles of FeC₂O₄ = 0.72/144 = 0.005 mol (molar mass 56 + 24 + 64 = 144 g/mol). Each formula unit loses 1 electron (Fe²⁺ → Fe³⁺) and 2 electrons (C₂O₄²⁻ → 2CO₂), 3 in total. Electrons lost = 0.005 × 3 = 0.015 mol. In acidic medium each MnO₄⁻ gains 5 electrons (MnO₄⁻ → Mn²⁺), so moles of KMnO₄ = 0.015/5 = 0.003 mol. Volume = 0.003 mol / 0.020 mol L⁻¹ = 0.15 L = 150 mL.
Q9 · Chemistry · d- and f-Block Elements
Acidified K₂Cr₂O₇ and acidified KMnO₄ are each completely reduced by an excess of a suitable reducing agent. What are the spin-only magnetic moments (in BM) of the chromium ion and the manganese ion that remain in solution, respectively?
Answer: (A)
In acidic medium, Cr₂O₇²⁻ → Cr³⁺ and MnO₄⁻ → Mn²⁺. Cr³⁺ is 3d³, with 3 unpaired electrons, so μ = √[3(3+2)] = √15 ≈ 3.87 BM. Mn²⁺ is 3d⁵, with 5 unpaired electrons, so μ = √[5(5+2)] = √35 ≈ 5.92 BM.
Q10 · Chemistry · d- and f-Block Elements
A 1.47 g sample of pure K₂Cr₂O₇ is dissolved in water and made up to 250 mL. A 25.0 mL portion is acidified with dilute H₂SO₄ and treated with excess KI. The iodine liberated is titrated with 0.10 M Na₂S₂O₃ solution, which is oxidised to tetrathionate, S₄O₆²⁻. What volume of the thiosulphate solution is needed to reach the end point? (Molar mass of K₂Cr₂O₇ = 294 g mol⁻¹)
Answer: (A)
Principle: in iodometric estimation, dichromate oxidises I⁻ to I₂ quantitatively, and the I₂ is then titrated with thiosulphate. Moles of K₂Cr₂O₇ in the flask = 1.47/294 = 5.0 × 10⁻³ mol. The 25.0 mL aliquot is one tenth of this, so it holds 5.0 × 10⁻⁴ mol. Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O, so I₂ = 3 × 5.0 × 10⁻⁴ = 1.5 × 10⁻³ mol. I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻, so S₂O₃²⁻ = 3.0 × 10⁻³ mol. Volume = 3.0 × 10⁻³ mol / 0.10 mol L⁻¹ = 0.030 L = 30 mL.
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All 12 real JEE Advanced questions from d- and f-Block Elements.
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About d- and f-Block Elements in JEE Advanced.
How many questions come from d- and f-Block Elements in JEE Advanced?
d- and f-Block Elements (Chemistry, Class 12) has been asked 12 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.3 questions a paper, 2% of all Chemistry questions. It is the 22nd most-asked of 31 Chemistry chapters. The two 2026 papers had 0 from it.
Which topics of d- and f-Block Elements are asked most in JEE Advanced?
Potassium dichromate and permanganate (7), Transition element properties (4), Lanthanoids and actinoids (1), counted over the papers from 2007–2026.
Is d- and f-Block Elements important for JEE Advanced 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 22 of 31 in Chemistry; use that to decide how much time it gets, not whether to study it.
Where can I solve d- and f-Block Elements JEE Advanced previous year questions?
On Nikaalle, free: sign up and practise every past JEE Advanced question from d- and f-Block Elements with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
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