JEE Advanced 2026 Paper 2 — Full Mock Test (54 Questions, 180 Marks)
Mathematics · Physics · Chemistry
This MCQ module is based on: D Block Compounds
TOPIC 16 OF 23
D Block Compounds
🎓 Class 12
Chemistry
CBSE
Theory
Ch 4 – The d- and f-Block Elements
⏱ ~14 min
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D Block Compounds
Two Star Chemicals of the d-Block
Two compounds dominate Class 12 d-block chemistry — the orange-red potassium dichromate K₂Cr₂O₇ (Cr in the +6 state) and the deep-purple potassium permanganate KMnO₄ (Mn in the +7 state). Both are powerful oxidising agents, both are used as primary standards in volumetric analysis, and both arise from chromium and manganese ores by very similar industrial routes.
Roadmap of Part 3: oxides and oxoanions of d-metals → preparation, structure, properties and oxidising actions of K₂Cr₂O₇ → preparation, structure, properties and oxidising actions of KMnO₄ in acidic / neutral / alkaline media.
8.14 Oxides and Oxoanions of the 3d Series
Transition metals burn in oxygen at high temperature to form oxides. The maximum oxidation state of the metal in its oxide equals its group number up to Mn:
| Group | 3 | 4 | 5 | 6 | 7 |
|---|---|---|---|---|---|
| Highest oxide | Sc₂O₃ | TiO₂ | V₂O₅ | CrO₃ | Mn₂O₇ |
| Oxidation state | +3 | +4 | +5 | +6 | +7 |
Beyond Mn the maximum drops sharply (Fe₂O₃ is the highest iron oxide). As the metal's oxidation number rises, ionic character drops and acidic character rises: Mn₂O₇ is a covalent green oil that gives HMnO₄ in water; CrO₃ gives H₂CrO₄ and H₂Cr₂O₇.
8.15 Potassium Dichromate, K₂Cr₂O₇
8.15.1 Preparation from Chromite Ore
The starting material is the mineral chromite ore FeCr₂O₄. Three steps convert it to K₂Cr₂O₇.
Step 1 — Oxidative fusion with alkali in air (~1300 K). Sodium chromate is produced.
4 FeCr₂O₄ + 8 Na₂CO₃ + 7 O₂ → 8 Na₂CrO₄ + 2 Fe₂O₃ + 8 CO₂
Step 2 — Acidify the chromate to form sodium dichromate. The yellow chromate (CrO₄²⁻) solution is filtered (Fe₂O₃ removed) and treated with H₂SO₄.
2 Na₂CrO₄ + 2 H⁺ → Na₂Cr₂O₇ + 2 Na⁺ + H₂O
Orange Na₂Cr₂O₇·2H₂O crystallises out.
Step 3 — Convert to less soluble potassium dichromate by metathesis with KCl.
Na₂Cr₂O₇ + 2 KCl → K₂Cr₂O₇ + 2 NaCl
K₂Cr₂O₇ is much less soluble than Na₂Cr₂O₇, so orange-red crystals separate on cooling.
8.15.2 Chromate ⇌ Dichromate Equilibrium
Chromate (CrO₄²⁻) and dichromate (Cr₂O₇²⁻) are interconvertible. Adding acid shifts the equilibrium to dichromate (orange); adding base shifts it back to chromate (yellow). The Cr oxidation number is +6 in both.
2 CrO₄²⁻ + 2 H⁺ ⇌ Cr₂O₇²⁻ + H₂O (yellow → orange)
Cr₂O₇²⁻ + 2 OH⁻ ⇌ 2 CrO₄²⁻ + H₂O (orange → yellow)
8.15.3 Structure of CrO₄²⁻ and Cr₂O₇²⁻
CrO₄²⁻ is a regular tetrahedron: a central Cr surrounded by four O atoms at the corners. Cr₂O₇²⁻ is two such tetrahedra joined at one corner (a shared O atom). The Cr–O–Cr bridge angle is 126°.
8.15.4 Properties & Oxidising Action in Acidic Medium
K₂Cr₂O₇ forms bright orange-red crystals, m.p. 671 K. It is moderately soluble in water and is a stable, weighable solid — making it a perfect primary standard in volumetric analysis. In dilute H₂SO₄ it acts as a powerful oxidiser:
Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻ → 2 Cr³⁺ + 7 H₂O (E° = +1.33 V)
Three Standard Oxidation Reactions (acidic medium)
(a) Iodide → Iodine:
Cr₂O₇²⁻ + 14 H⁺ + 6 I⁻ → 2 Cr³⁺ + 3 I₂ + 7 H₂O
(b) Iron(II) → Iron(III):
Cr₂O₇²⁻ + 14 H⁺ + 6 Fe²⁺ → 2 Cr³⁺ + 6 Fe³⁺ + 7 H₂O
(c) Hydrogen sulphide → Sulphur:
Cr₂O₇²⁻ + 8 H⁺ + 3 H₂S → 2 Cr³⁺ + 3 S↓ + 7 H₂O
Other reductants similarly oxidised: Sn²⁺ → Sn⁴⁺; SO₃²⁻ → SO₄²⁻; alcohols → aldehydes/ketones (used to spot drunk drivers in a "breathalyser" — the orange Cr(VI) turns green Cr(III) in presence of ethanol).
8.15.5 Uses of K₂Cr₂O₇
- Tanning of leather (chrome tanning).
- Primary standard in volumetric analysis (e.g., for Fe²⁺ titrations).
- Manufacture of azo dyes.
- Cleaning glassware (chromic-acid mixture).
- Oxidant in synthetic organic chemistry.
Worked Example 8.8 L3 Apply
Q. Balance the reaction of acidified K₂Cr₂O₇ with FeSO₄ in ionic form and identify the colour change observed.
Cr₂O₇²⁻ (orange) gains 6 electrons per ion; each Fe²⁺ → Fe³⁺ loses 1 electron — so 6 Fe²⁺ are needed.
Cr₂O₇²⁻ + 14 H⁺ + 6 Fe²⁺ → 2 Cr³⁺ + 6 Fe³⁺ + 7 H₂O
Colour change: orange (Cr⁶⁺) → green (Cr³⁺); pale green Fe²⁺ → pale yellow Fe³⁺.
8.16 Potassium Permanganate, KMnO₄
8.16.1 Laboratory and Industrial Preparation
The starting material is pyrolusite ore (MnO₂). Two steps convert MnO₂ to KMnO₄.
Step 1 — Alkaline oxidative fusion with KOH and air (or KNO₃) gives the dark green manganate ion (Mn⁶⁺):
2 MnO₂ + 4 KOH + O₂ → 2 K₂MnO₄ + 2 H₂O
Step 2 — Convert manganate to permanganate. Two routes are used.
Route A — disproportionation (lab method): in neutral or acidic medium,
3 MnO₄²⁻ + 4 H⁺ → 2 MnO₄⁻ + MnO₂ + 2 H₂O
(green) (purple) (brown)
Route B — electrolytic oxidation (industrial method): in alkaline solution, the manganate is oxidised at the anode to permanganate:
MnO₄²⁻ → MnO₄⁻ + e⁻ (anode, alkaline)
Laboratory route from a Mn(II) salt: peroxodisulphate oxidises Mn²⁺ all the way to MnO₄⁻ (Ag⁺ catalysed):
2 Mn²⁺ + 5 S₂O₈²⁻ + 8 H₂O → 2 MnO₄⁻ + 10 SO₄²⁻ + 16 H⁺
8.16.2 Properties
KMnO₄ forms dark purple (almost black) rhombic crystals, isostructural with KClO₄. It is moderately soluble in water (6.4 g per 100 g at 293 K) giving an intensely purple solution. On heating to 513 K it decomposes:
2 KMnO₄ → K₂MnO₄ + MnO₂ + O₂↑
The MnO₄⁻ ion is tetrahedral with strong π-bonding (oxygen 2p with manganese 3d). The intense colour is due to a charge-transfer transition (oxygen → metal); MnO₄⁻ has no unpaired d-electrons, so it is diamagnetic (the slight temperature-dependent paramagnetism comes from molecular-orbital effects beyond NCERT scope). MnO₄²⁻ (manganate, Mn⁶⁺ d¹) is paramagnetic with one unpaired electron.
8.16.3 Structure of MnO₄⁻ Ion
8.16.4 Oxidising Power Depends on pH
Three half-reactions are crucial — and the medium decides the product.
| Medium | Half-reaction | Product | E° / V |
|---|---|---|---|
| Acidic (H⁺) | MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O | Mn²⁺ (pale pink) | +1.52 |
| Neutral / faint alkaline | MnO₄⁻ + 4 H⁺ + 3 e⁻ → MnO₂ + 2 H₂O | MnO₂ (brown solid) | +1.69 |
| Strongly alkaline | MnO₄⁻ + e⁻ → MnO₄²⁻ | MnO₄²⁻ (green) | +0.56 |
Standard Reactions in Acidic Medium
(a) Iodide → Iodine:
2 MnO₄⁻ + 16 H⁺ + 10 I⁻ → 2 Mn²⁺ + 8 H₂O + 5 I₂
(b) Fe(II) (green) → Fe(III) (yellow):
MnO₄⁻ + 8 H⁺ + 5 Fe²⁺ → Mn²⁺ + 4 H₂O + 5 Fe³⁺
(c) Oxalate / oxalic acid → CO₂ (at 333 K):
2 MnO₄⁻ + 16 H⁺ + 5 C₂O₄²⁻ → 2 Mn²⁺ + 8 H₂O + 10 CO₂↑
(d) Hydrogen sulphide → Sulphur:
2 MnO₄⁻ + 16 H⁺ + 5 S²⁻ → 2 Mn²⁺ + 8 H₂O + 5 S↓
(e) SO₃²⁻ / SO₂ → SO₄²⁻:
2 MnO₄⁻ + 6 H⁺ + 5 SO₃²⁻ → 2 Mn²⁺ + 3 H₂O + 5 SO₄²⁻
(f) Nitrite → Nitrate:
2 MnO₄⁻ + 6 H⁺ + 5 NO₂⁻ → 2 Mn²⁺ + 3 H₂O + 5 NO₃⁻
Reactions in Neutral / Faintly Alkaline Medium
(a) Iodide → Iodate:
2 MnO₄⁻ + H₂O + I⁻ → 2 MnO₂ + 2 OH⁻ + IO₃⁻
(b) Thiosulphate → Sulphate:
8 MnO₄⁻ + 3 S₂O₃²⁻ + H₂O → 8 MnO₂ + 6 SO₄²⁻ + 2 OH⁻
(c) Mn²⁺ → MnO₂ (catalysed by ZnSO₄ or ZnO):
2 MnO₄⁻ + 3 Mn²⁺ + 2 H₂O → 5 MnO₂ + 4 H⁺
Why HCl medium is forbidden: KMnO₄ titrations are never performed in HCl because permanganate oxidises Cl⁻ to Cl₂, introducing an error. H₂SO₄ is used instead.
8.16.5 Uses of KMnO₄
- Strong oxidant in synthetic organic chemistry (Baeyer's reagent — alkaline KMnO₄ — tests unsaturation).
- Volumetric analysis (self-indicating in acid; pink end-point).
- Decolourising oils and bleaching wool, cotton, silk.
- Mild antiseptic (dilute solutions for skin disinfection).
- Water purification (oxidises Fe²⁺ and H₂S).
Interactive: KMnO₄ Reaction Predictor by Medium L3 Apply
Choose a reductant and the reaction medium — the simulator returns the balanced ionic equation, the Mn product (Mn²⁺ vs MnO₂ vs MnO₄²⁻) and the visible colour change.
Pick a reductant to begin.
Activity 8.3 — Trace the Path: From Pyrolusite to PermanganateL4 Analyse
Aim: Track the oxidation state of manganese at each step of the KMnO₄ industrial preparation.
Procedure:
- Write the molecular formula of pyrolusite. Determine the oxidation state of Mn.
- Write the formula of the green intermediate K₂MnO₄. Determine the oxidation state of Mn.
- Write the formula of KMnO₄. Determine the oxidation state of Mn.
- Plot the oxidation states across the three stages and identify which oxidant is responsible for each step.
Predict: How many electrons must Mn lose, in total, between MnO₂ and KMnO₄?
MnO₂: Mn = +4 · K₂MnO₄: Mn = +6 · KMnO₄: Mn = +7.
Step 1 (MnO₂ → MnO₄²⁻): Mn loses 2 e⁻; oxidant = O₂ (or KNO₃).
Step 2 (MnO₄²⁻ → MnO₄⁻): Mn loses 1 e⁻; oxidant = electrolytic anode (industrial) or H⁺ via disproportionation (lab).
Total e⁻ lost = 3 per Mn atom from pyrolusite to permanganate.
Worked Example 8.9 L3 Apply
Q. Write the balanced ionic equation for the reaction between acidified KMnO₄ and ferrous oxalate, FeC₂O₄, in solution. (FeC₂O₄ contains both Fe²⁺ and C₂O₄²⁻ as reductants.)
Fe²⁺ → Fe³⁺ loses 1 e⁻; C₂O₄²⁻ → 2 CO₂ loses 2 e⁻; so each FeC₂O₄ unit loses 3 e⁻. Each MnO₄⁻ accepts 5 e⁻.
LCM = 15: take 3 MnO₄⁻ and 5 FeC₂O₄.
3 MnO₄⁻ + 24 H⁺ + 5 FeC₂O₄ → 3 Mn²⁺ + 12 H₂O + 5 Fe³⁺ + 10 CO₂↑
Worked Example 8.10 L4 Analyse
Q (In-text 4.6). Why is the highest oxidation state of a metal exhibited only in its oxide or fluoride?
Both O and F have small atomic size and very high electronegativity. They strip every accessible valence electron off the metal and stabilise the resulting high-charge cation through (i) very high lattice/bond energy and (ii) pπ-dπ multiple bonding (in the case of O, e.g. Mn=O bonds in Mn₂O₇). No other element offers both effects together.
Competency-Based Questions L3 L4
Stimulus: A volumetric analyst is titrating a known amount of FeSO₄ solution against (a) acidified K₂Cr₂O₇ and (b) acidified KMnO₄. He notes the standard electrode potentials are +1.33 V (Cr₂O₇²⁻/Cr³⁺) and +1.52 V (MnO₄⁻/Mn²⁺).
Q1. (MCQ) The bridge angle Cr–O–Cr in the dichromate ion is approximately:
(c) 126° — bent, owing to the lone pairs on the bridging oxygen.
Q2. (MCQ) When acidified KMnO₄ reacts with oxalic acid, the manganese is reduced from:
(c) +7 to +2 — in acidic medium MnO₄⁻ → Mn²⁺ (gain of 5 e⁻).
Q3. (SA) Why is K₂Cr₂O₇ preferred over Na₂Cr₂O₇ as a primary standard?
Na₂Cr₂O₇ is highly hygroscopic — it picks up moisture from air and cannot be weighed accurately. K₂Cr₂O₇ is non-hygroscopic and can be obtained in a pure, dry, weighable form, satisfying the requirements of a primary standard.
Q4. (LA) KMnO₄ titrations are always carried out in dilute H₂SO₄ — never in HCl. Explain.
In HCl medium, MnO₄⁻ also oxidises Cl⁻ to Cl₂: 2 MnO₄⁻ + 16 H⁺ + 10 Cl⁻ → 2 Mn²⁺ + 8 H₂O + 5 Cl₂. This consumes some of the permanganate, giving an over-estimate of the reductant being titrated. H₂SO₄ provides the acidic medium without contributing a reductant, so the stoichiometry is preserved.
Q5. (HOT) Both Cr₂O₇²⁻ and MnO₄⁻ contain the metal in its highest oxidation state, yet MnO₄⁻ is a stronger oxidant. Suggest two reasons.
(i) The reduction potential E°(MnO₄⁻/Mn²⁺) = +1.52 V is higher than E°(Cr₂O₇²⁻/Cr³⁺) = +1.33 V. (ii) The product Mn²⁺ has a stable half-filled d⁵ configuration; the product Cr³⁺ has d³ — although stabilised by a half-filled t₂g, the overall driving force is smaller. Also, the +7 → +2 jump for Mn (5 electrons) is larger than the +6 → +3 jump for Cr (3 electrons), giving more "oxidising power per ion".
Assertion–Reason Questions L4 L5
Choose: A) Both A and R true and R explains A · B) Both true but R does not explain A · C) A true, R false · D) A false, R true.
Assertion (A): The dichromate ion is orange while the chromate ion is yellow.
Reason (R): Adding acid to a yellow chromate solution converts it into orange dichromate.
Answer: B. Both statements are correct, but R describes the equilibrium response to pH; the colour difference itself arises from differences in O→Cr charge-transfer transitions in the two ions, not from the act of adding acid.
Assertion (A): KMnO₄ is diamagnetic at room temperature.
Reason (R): The MnO₄⁻ ion contains Mn(VII) which has a d⁰ configuration — no unpaired electrons.
Answer: A. Mn⁷⁺ has lost all its 3d and 4s electrons → d⁰ → no unpaired electrons → diamagnetic.
Assertion (A): KMnO₄ titrations are conducted in H₂SO₄ rather than HCl.
Reason (R): HCl reacts with KMnO₄ to release Cl₂, introducing error in the titration.
Answer: A. Permanganate's E° (+1.52 V) exceeds Cl₂/Cl⁻ (+1.36 V), so MnO₄⁻ oxidises Cl⁻ to Cl₂.
Frequently Asked Questions - D Block Compounds
What is the main concept covered in D Block Compounds?
In NCERT Class 12 Chemistry Chapter 4 (The d- and f-Block Elements), "D Block Compounds" covers the core chemistry principles and reactions students need for board exam success. The MyAiSchool lesson explains the topic with definitions, structural diagrams, reaction mechanisms, worked examples, and interactive simulations. Key reactions, IUPAC names, and chemical reasoning are highlighted throughout aligned with CBSE 2025-26 syllabus.
How is D Block Compounds useful in real-life or applied chemistry?
Real-life applications of "D Block Compounds" from NCERT Class 12 Chemistry Chapter 4 include drug design, polymer industry, food chemistry, electrochemical cells, fuel cells, dyes/pigments, agrochemicals, and biochemistry. The MyAiSchool lesson links every concept to a tangible industrial or biological example so students see chemistry as a problem-solving framework for the molecular world.
What are the key reactions students should memorize for D Block Compounds?
Key reactions in "D Block Compounds" (NCERT Class 12 Chemistry Chapter 4 The d- and f-Block Elements) are tabulated in the MyAiSchool reaction map. Students should memorize each reaction with its reagent, conditions, mechanism class (SN1/SN2/E1/E2/electrophilic addition/etc), product, and stereochemistry. The Summary section provides a quick-reference reaction chart for last-minute revision.
How does this part connect to other parts of Chapter 4?
NCERT Class 12 Chemistry Chapter 4 (The d- and f-Block Elements) is structured so each part builds chemical understanding sequentially. "D Block Compounds" connects to neighbouring parts via shared functional groups, reaction mechanisms, and structural concepts. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected story rather than disconnected fragments.
What types of CBSE board questions come from D Block Compounds?
CBSE board questions from "D Block Compounds" typically include: (1) 1-mark MCQs on definitions and IUPAC naming, (2) 2-mark short-answer reactions/products, (3) 3-mark mechanism questions, (4) 5-mark long-answer combining mechanism + product + stereochemistry + application. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "D Block Compounds" lesson allows students to explore reaction outcomes, predict products, or compare reaction conditions, with live visual feedback. To use it effectively: (1) try every option/configuration, (2) compare with the analytical reasoning, (3) check IUPAC names and structural correctness, (4) test edge cases from worked examples. The simulation reinforces conceptual intuition that pure mechanism memorisation cannot provide.
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