JEE Chemistry — d and f Block Elements Complete Chapter Guide

Chapter Overview & Weightage

d and f Block Elements is a reliable scoring chapter for JEE — it rewards students who memorize smartly rather than understand deeply. That said, the “understanding” part helps you remember the patterns faster.

JEE Main Weightage: 4–5% (1–2 questions per paper)

This chapter almost always shows up in JEE Main. JEE Advanced tests it less frequently but when it does, it combines with coordination compounds. Secure this chapter and you’re looking at guaranteed marks.

Year	JEE Main Questions	Topics Covered
2024	2	KMnO₄ reactions, magnetic moment
2023	1	Lanthanoid contraction consequences
2022	2	K₂Cr₂O₇ preparation, oxidation states
2021	2	Color of transition metal ions, catalytic properties
2020	1	Interstitial compounds, alloy formation

The pattern is clear: reactions of KMnO₄ and K₂Cr₂O₇, magnetic properties, and lanthanoid contraction consequences come up repeatedly. These are your priority zones.

Key Concepts You Must Know

Ranked by how often they appear in PYQs:

Tier 1 — High Frequency (prepare first)

Oxidation states of transition metals — variable oxidation states and which are most stable for Cr, Mn, Fe, Cu
KMnO₄ reactions in acidic, basic, and neutral medium (the products change completely)
K₂Cr₂O₇ — preparation, reactions, oxidizing behavior
Magnetic moment formula and spin-only formula: $\mu = \sqrt{n(n+2)}$ BM
Lanthanoid contraction — causes, consequences (especially on 5d elements)

Tier 2 — Medium Frequency

Color of transition metal ions and the d-d transition explanation
Interstitial compounds and their properties
Catalytic behavior of transition metals (Fe in Haber, V₂O₅ in Contact process)
Actinoid contraction vs lanthanoid contraction

Tier 3 — Low Frequency but Easy Marks

General electronic configurations: (n-1)d¹⁻¹⁰ ns⁰⁻²
Alloy formation conditions
Chromate–dichromate equilibrium

Important Formulas

\mu = \sqrt{n(n+2)} \text{ BM}

where $n$ = number of unpaired electrons.

When to use: Any question asking about magnetic moment of a transition metal ion. First find the electronic configuration, count unpaired electrons, then apply.

JEE trick: For Mn²⁺ (d⁵), $n = 5$ , so $\mu = \sqrt{35} \approx 5.92$ BM — the highest among common ions. This specific value appears in questions.

2\text{CrO}_4^{2-} + 2\text{H}^+ \rightleftharpoons \text{Cr}_2\text{O}_7^{2-} + \text{H}_2\text{O}

When to use: Questions on color changes when acid/base is added to chromate or dichromate solutions. Adding acid shifts right (orange dichromate), adding base shifts left (yellow chromate).

Medium	Mn goes from +7 to	Product
Acidic	+2	MnSO₄ (colorless)
Basic	+6	MnO₄²⁻ (dark green)
Neutral	+4	MnO₂ (brown ppt)

When to use: Whenever KMnO₄ is the oxidizing agent. Identify the medium first — this determines the $n$ -factor and the product.

Step 1: Roasting chromite ore

4\text{FeCr}_2\text{O}_4 + 8\text{Na}_2\text{CO}_3 + 7\text{O}_2 \rightarrow 8\text{Na}_2\text{CrO}_4 + 2\text{Fe}_2\text{O}_3 + 8\text{CO}_2

Step 2: Chromate to dichromate (acidify)

2\text{Na}_2\text{CrO}_4 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{Cr}_2\text{O}_7 + \text{Na}_2\text{SO}_4 + \text{H}_2\text{O}

Step 3: Metathesis

\text{Na}_2\text{Cr}_2\text{O}_7 + 2\text{KCl} \rightarrow \text{K}_2\text{Cr}_2\text{O}_7 + 2\text{NaCl}

Solved Previous Year Questions

PYQ 1 — JEE Main 2024 (Magnetic Moment)

Question: The magnetic moment of a transition metal ion is $\sqrt{35}$ BM. The number of unpaired electrons in this ion is: (a) 3 (b) 4 (c) 5 (d) 6

Solution:

Apply the spin-only formula: $\mu = \sqrt{n(n+2)}$

\sqrt{35} = \sqrt{n(n+2)}

n(n+2) = 35

n^2 + 2n - 35 = 0

(n+7)(n-5) = 0 \Rightarrow n = 5

Answer: (c) 5 unpaired electrons

This corresponds to d⁵ configuration — ions like Mn²⁺ or Fe³⁺.

Memorize these common magnetic moments: Cr³⁺ = $\sqrt{15}$ (3 unpaired), Fe²⁺ = $\sqrt{24}$ (4 unpaired), Mn²⁺ = $\sqrt{35}$ (5 unpaired), Cu²⁺ = $\sqrt{3}$ (1 unpaired). JEE re-tests these values regularly.

PYQ 2 — JEE Main 2022 (KMnO₄ Reactions)

Question: When KMnO₄ reacts with oxalic acid in acidic medium, the oxidation state of Mn changes from: (a) +7 to +4 (b) +7 to +2 (c) +7 to +6 (d) +6 to +2

Solution:

Acidic medium → Mn goes from +7 to +2 (forms MnSO₄).

The reaction:

2\text{KMnO}_4 + 5\text{H}_2\text{C}_2\text{O}_4 + 3\text{H}_2\text{SO}_4 \rightarrow 2\text{MnSO}_4 + 10\text{CO}_2 + \text{K}_2\text{SO}_4 + 8\text{H}_2\text{O}

Mn: +7 → +2 (gain of 5e⁻, gets reduced)
C in oxalic acid: +3 → +4 (loses 1e⁻ per C atom, 2e⁻ per molecule, gets oxidized)

Answer: (b) +7 to +2

Many students write +4 (MnO₂) here — that’s the neutral medium product. In acidic medium, reduction goes all the way to Mn²⁺. The medium question is always asked to catch this confusion.

PYQ 3 — JEE Main 2023 (Lanthanoid Contraction)

Question: Which of the following is NOT a consequence of lanthanoid contraction? (a) Zr and Hf have nearly identical radii (b) 5d elements are larger than expected (c) Lanthanoids can be separated from each other easily (d) Zr and Hf have similar chemical properties

Solution:

Lanthanoid contraction causes:

Size of 5d elements being smaller than expected (4f electrons poorly shield, so nuclear charge pulls 5d shell inward)
Zr (4d) and Hf (5d) having nearly equal atomic radii (~156 pm each)
Zr and Hf having extremely similar chemical properties — hardest pair to separate in the periodic table

Option (b) is wrong — 5d elements are smaller than expected due to poor shielding by 4f electrons, not larger.

Option (c) is the answer: lanthanoids are actually difficult to separate because the size difference between successive lanthanoids is tiny (only ~1 pm).

Answer: (c)

Difficulty Distribution

For JEE Main, this chapter follows a predictable difficulty pattern:

Type	% of Questions	What They Test
Easy	40%	Electronic config, identification of color/oxidation state
Medium	45%	KMnO₄/K₂Cr₂O₇ reactions, magnetic moment calculation
Hard	15%	Combined application: n-factor + redox titration using these reagents

The “hard” questions in this chapter almost always combine it with redox titration or electrochemistry. If you’re preparing for JEE Advanced, practice balancing redox equations involving K₂Cr₂O₇ with variable n-factors carefully.

Expert Strategy

Week 1: Lock down the oxidation states of Cr, Mn, Fe, Co, Ni, Cu. Not just the common ones — know why +2 is stable for Cu (completely filled 3d¹⁰) and why +6 is unstable for Cu.

Week 2: Do all KMnO₄ and K₂Cr₂O₇ reactions. Write them out — don’t just read. The balanced equations matter for n-factor calculations.

Week 3: Lanthanoids and actinoids — this is mostly factual. Use a table: write down electronic config, common oxidation state, and one distinguishing property for La, Ce, Gd, Lu, and then the actinoids Th, U, Pu.

PYQ strategy that works: This chapter has a small syllabus but repeats the same 8–10 concepts every year. Go through JEE Main PYQs from 2019–2024 for this chapter specifically. You’ll see the same question types reappear with slight variations. Once you’ve solved 20 PYQs from this chapter, you’ve essentially seen everything the exam throws at you.

The 80/20 for this chapter: Focus 80% of your time on:

KMnO₄ reactions in three media (with balanced equations)
K₂Cr₂O₇ preparation and reactions
Magnetic moment problems
Lanthanoid contraction consequences

These four topics alone account for ~70% of questions from this chapter in JEE Main over the last five years.

Common Traps

Trap 1: Cr and Cu electronic configurations

You’d expect Cr to be [Ar] 3d⁴ 4s² and Cu to be [Ar] 3d⁹ 4s². But the actual configurations are:

Cr: [Ar] 3d⁵ 4s¹ (half-filled d is extra stable)
Cu: [Ar] 3d¹⁰ 4s¹ (completely filled d is extra stable)

JEE directly asks about which elements show anomalous configurations. The answer is always Cr and Cu (and their heavier counterparts Mo, W, Ag, Au).

Trap 2: Color of Cu²⁺ vs Cu⁺

Cu²⁺ is blue (d⁹, one unpaired electron, d-d transition possible). Cu⁺ is colorless (d¹⁰, no d-d transition). Questions ask about CuSO₄ solution color (blue) vs Cu₂O or CuCl color (colorless/white).

Trap 3: Lanthanoid contraction — confusing cause and effect

The cause of lanthanoid contraction is the poor shielding by 4f electrons (which have a diffuse shape and don’t shield nuclear charge effectively). The effect is decreasing atomic/ionic radius across the lanthanoid series.

A common wrong answer: “increasing nuclear charge” — that’s not the cause, it’s always increasing as we go across a period. The unusual part is the poor shielding that makes the contraction more pronounced than in d-block elements.

Trap 4: Acidic vs oxidizing nature of K₂Cr₂O₇

K₂Cr₂O₇ acts as an oxidizing agent — but it works only in acidic medium. In neutral medium, it doesn’t oxidize effectively. Questions sometimes give a reaction in neutral medium and ask which product forms — the answer is no reaction (or very slow).

Last-minute revision list (30 minutes before the exam):

Cr and Cu anomalous configurations
KMnO₄ products in acidic/basic/neutral
Magnetic moments for Cr³⁺, Fe²⁺, Fe³⁺, Mn²⁺, Cu²⁺
Chromate ⇌ dichromate equilibrium and color change
One consequence of lanthanoid contraction: Zr ≈ Hf in size

These five points cover 60–70% of what JEE Main tests from this chapter.