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Chemistry — Class 12 — L21: d-Block and f-Block Elements

NIOS Code 313 · Module 6 · Chemistry of Elements

Notes extracted from NIOS Chemistry Course (313), Lesson 21 — d-Block and f-Block Elements (313_Chemistry_Eng_Lesson21.pdf). Content covers sections 21.1–21.7.
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Overview — d-Block and f-Block Elements

Between the s- and p-blocks sit the d-block (transition) elements, where the (n−1)d subshell fills. The first series is Sc–Cu (3d). Separately at the bottom of the table, the f-block (inner transition) elements fill 4f (lanthanoids) and 5f (actinoids). This lesson covers definition and configuration, physical and characteristic properties of 3d metals, preparation and redox chemistry of K₂Cr₂O₇ and KMnO₄, lanthanoid contraction and oxidation states, and a comparison of lanthanoids with actinoids.

Section 1: Transition Elements — Definition & Configuration (21.1–21.2)

Transition elements have a partially filled d-subshell in the atom or in a common ion. General configuration: (n−1)d¹–¹⁰ ns¹–². Four series: 3d (Sc–Cu), 4d, 5d, 6d. Cu, Ag, Au count as transition metals because Cu²⁺ (3d⁹), Ag²⁺, Au³⁺ have incomplete d shells. Zn, Cd, Hg have d¹⁰ in atom and ions — not transition elements, though often discussed with the d-block.

First Transition Series (3d) Sc Ti V Cr* Mn Fe Co Ni Cu* Zn *Cr 3d⁵4s¹ · Cu 3d¹⁰4s¹ · Zn not transition (d¹⁰ always) Config: (n−1)d¹–¹⁰ ns¹–² · Partial d required Most occur as oxides/sulphides · Au, Pt free · Al-like abundance varies
3d series highlights configuration exceptions and the Zn boundary of true transition behaviour.

Energy of 3d falls below 4p after Ca, so electrons enter 3d. Half-filled and full d shells are extra stable → Cr and Cu have only one 4s electron. Occurrence: few free (Au, Pt); most as oxides, sulphides, carbonates.

(n−1)d¹–¹⁰ ns¹–²  |  Cr: 3d⁵4s¹  |  Cu: 3d¹⁰4s¹
General transition config · exceptions for half-filled/full d stability

Section 2: Physical Properties (21.3)

Typical metals: high tensile strength, ductility, conductivity, lustre. High m.p./b.p. (usually >1356 K) from small size and strong metallic bonding involving d electrons. High density; maxima around Groups 8–10. Hard (except Zn, Cd, Hg). Atomic radii decrease across the series (poor d shielding raises Z_eff) then rise slightly near the end. 4d and 5d metals of a group are similar in size because of lanthanoid contraction.

Section 3: Characteristic Properties (21.4)

Variable oxidation states: both ns and (n−1)d electrons can bond. After Sc, +2 is very common (loss of 4s). High OS with F and O (Mn +7 in MnO₄⁻, Cr +6 in Cr₂O₇²⁻). Higher OS oxides more acidic (MnO basic; Mn₂O₇ acidic). High-OS compounds are strong oxidants.

Magnetic properties: unpaired electrons → paramagnetism. Spin-only moment:

μ = √[n(n+2)] B.M.
n = number of unpaired electrons · Mn²⁺ (d⁵) → 5.92 BM · Sc³⁺, Ti⁴⁺, Zn²⁺ diamagnetic
Colour & Magnetism of 3d Ions Ti³⁺ violetd¹ · 1.73 BM Ni²⁺ greend⁸ · 2.83 BM Cu²⁺ blued⁹ · 1.73 BM Mn²⁺ paled⁵ · 5.92 BM Colour from d–d transitions · d⁰/d¹⁰ ions colourless & often diamagnetic μ = √[n(n+2)] · more unpaired e⁻ → higher μ
Incomplete d-subshell links colour and paramagnetism of transition-metal ions.

Colour: white light partially absorbed by d–d transitions; complementary colour observed (CuSO₄ blue). d⁰ and d¹⁰ ions usually colourless (Sc³⁺, Ti⁴⁺, Zn²⁺, Cu⁺).

Alloys & interstitial compounds: similar atomic sizes → brass (Cu–Zn), bronze (Cu–Sn), stainless steel. Small H, C, N atoms in lattice voids → hard interstitial compounds (steel hardness from carbon).

Complex formation: small size, high charge, vacant d orbitals accept ligand pairs (L22).

Catalysis: variable OS and surface adsorption. Examples: V₂O₅ (Contact process), Fe (Haber), Ni (hydrogenation), Pd/Cu (Wacker). Fe³⁺ catalyses I⁻ + S₂O₈²⁻ by cycling Fe³⁺/Fe²⁺.

Section 4: Potassium Dichromate (21.5.1)

From chromite FeO·Cr₂O₃: roast with Na₂CO₃ + air (+ CaO for porosity) → Na₂CrO₄ → acidify → Na₂Cr₂O₇ → metathesis with KCl → orange-red K₂Cr₂O₇.

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Acidic dichromate oxidises Fe²⁺, I⁻, SO₂ · Cr(+6) → Cr(+3) · primary standard in volumetric analysis
Dichromate–Chromate Equilibrium Cr₂O₇²⁻ orangeacidic medium 2CrO₄²⁻ yellowalkaline medium Both Cr(+6) · pH switches colour · chromyl chloride test for Cl⁻
Cr₂O₇²⁻ ⇌ 2CrO₄²⁻ + 2H⁺ — orange in acid, yellow in alkali.

With NaCl + conc. H₂SO₄ → red chromyl chloride CrO₂Cl₂ (confirmatory test for chloride). Uses: volumetric oxidant; chrome alum for tanning/dyeing.

Ionic oxidations to memorise: Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O; Cr₂O₇²⁻ + 6I⁻ + 14H⁺ → 2Cr³⁺ + 3I₂ + 7H₂O; Cr₂O₇²⁻ + 3SO₂ + … → Cr³⁺ + SO₄²⁻. Oxidation number of Cr is +6 in both K₂CrO₄ and K₂Cr₂O₇.

Section 5: Potassium Permanganate (21.5.2)

From pyrolusite MnO₂: fuse with KOH/air → green K₂MnO₄ → oxidise (Cl₂, O₃ or anodic oxidation) → purple KMnO₄. Manganate disproportionates in acid: 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O.

MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O  (acid)
Neutral/alkaline: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ · strongest change in acid (5e⁻)

Acidic: oxidises Fe²⁺, SO₂, C₂O₄²⁻. Neutral: Mn²⁺ → MnO₂; H₂S → S. Alkaline: I⁻ → IO₃⁻; ethene → ethylene glycol. Uses: disinfectant (wells, mouthwash), volumetric standard for Fe(II), oxalate, H₂O₂. Heat: 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂.

KMnO₄ — Medium Matters Acidic→ Mn²⁺ (5e⁻)Fe²⁺, SO₂, C₂O₄²⁻ Neutral→ MnO₂ (3e⁻)H₂S, Mn²⁺ Alkaline→ MnO₂ / MnO₄²⁻I⁻ → IO₃⁻ Purple MnO₄⁻ · best oxidising power in acid
Permanganate reduction product depends on pH of the medium.

Section 6: Lanthanoids (21.6)

La–Lu (14 + La): filling of 4f. Config mostly [Xe] 4fⁿ 6s² (La, Gd, Lu have 5d¹). Extremely similar chemically; older name “rare earths” is misleading.

Lanthanoid contraction: 4f electrons shield poorly → gradual decrease in atomic/ionic radii from La³⁺ to Lu³⁺. Consequence: Zr and Hf nearly identical size and chemistry; Hf denser than Zr.

Lanthanoid contraction → r(Zr) ≈ r(Hf)
Poor 4f shielding · 4d and 5d congeners of Groups 4–5 very similar

Oxidation states: characteristic +3. +2/+4 when f⁰ (Ce⁴⁺), f⁷ (Eu²⁺, Tb⁴⁺) or f¹⁴ (Yb²⁺) stabilised. Highly electropositive; compounds largely ionic. More mutual resemblance than d-block metals because +3 dominates.

Section 7: Actinoids (21.7)

Ac–Lr: filling of 5f (with early members mixing 6d). Almost all radioactive (Pm is the only radioactive lanthanoid). +3 common but Th, U, Np, Pu show higher states; Np/Pu up to +7; oxocations UO₂²⁺, PuO₂²⁺. Am²⁺ (f⁷) known.

Ln vs An: both show contraction and +3 prominence; An more radioactive, wider OS range, better complexation, form oxocations, more basic compounds; 5f less effectively screened and closer in energy to 6d than 4f to 5d.

Inner Transition Series Lanthanoids 4f+3 · contraction · few radioactive Actinoids 5fradioactive · high OS · oxocations Both imperfect f-shielding · An chemistry more varied
f-Block comparison: similar +3 chemistry, very different radioactivity and oxidation-state range.

Exam Connections and Chapter Summary

High-yield: definition of transition metal (Zn vs Cu); Cr/Cu configs; μ formula; colour origin; K₂Cr₂O₇ and KMnO₄ preparation outlines and acidic half-reactions; chromate–dichromate equilibrium; chromyl chloride; lanthanoid contraction and Zr/Hf; Ln +3 and special +2/+4; Ln vs An differences.

Connects to L20 (V₂O₅ catalysis), L13 (redox half-cells), L22 (complexes). Practice balancing dichromate and permanganate ionic equations — they appear constantly in volumetric and redox questions.

Intext-style checks: why Cu is transition but Zn is not; μ for V⁴⁺, Cr³⁺, Ni²⁺; coloured vs colourless ions; chromite and pyrolusite formulae; dichromate half-reaction; KMnO₄ in three media; lanthanoid contraction and Zr/Hf; Ln +3; An vs Ln differences. Alloys to name: brass, bronze, stainless steel. Catalysts: Fe, Ni, V₂O₅, PdCl₂.

MCQ Quiz — L21 d-Block and f-Block Elements

0 / 10 correct

Flashcards — L21

1 / 18

Golden Rules — L21 d-Block and f-Block Elements

Most exam-important points from this chapter:

d-Block basics

Partial d in atom/ion. Config (n−1)d ns. Cr/Cu exceptions. Zn not transition. High m.p., density, variable OS.

Colour & magnetism

μ = √[n(n+2)]. Unpaired e⁻ → paramagnetic & often coloured (d–d). d⁰/d¹⁰ colourless/diamagnetic.

K₂Cr₂O₇

From chromite. Acid: 6e⁻ to Cr³⁺. Orange⇌yellow with alkali. Chromyl chloride tests Cl⁻. Volumetric oxidant.

KMnO₄

From MnO₂ via manganate. Acid 5e⁻→Mn²⁺; neutral/alkaline → MnO₂. Disinfectant & volumetric reagent.

f-Block

Ln: 4f, +3, contraction → Zr/Hf similar. An: 5f, radioactive, higher OS, oxocations, more complex chemistry.

(n−1)d¹–¹⁰ ns¹–²
μ = √[n(n+2)] B.M.
Variable OS · coloured ions
Cr₂O₇²⁻ ⇌ 2CrO₄²⁻
Cr₂O₇²⁻ + 14H⁺ + 6e⁻
MnO₄⁻ + 8H⁺ + 5e⁻
K₂Cr₂O₇ · KMnO₄
Lanthanoid contraction
Ln mainly +3

Section 1: d-Block Transition Elements

NIOS Chemistry 313, Module 6 — d-Block and f-Block Elements (sections 21.1–21.7).

Definition & Configuration

Transition element: partial d-subshell in atom or common ion

General config: (n−1)d¹–¹⁰ ns¹–² · First series Sc–Cu (Zn often studied but not truly transition)

Exceptions: Cr [Ar] 3d⁵4s¹ · Cu [Ar] 3d¹⁰4s¹ (half/full d stability)

Cu, Ag, Au are transition (Cu²⁺ 3d⁹) · Zn, Cd, Hg not (d¹⁰ always)

Characteristic Properties

Variable OS: ns + (n−1)d electrons · +2 common after Sc · high OS with F/O

Magnetic: μ = √[n(n+2)] B.M. (spin-only) · n = unpaired e⁻ · Mn²⁺ (d⁵) → 5.92 BM

Colour: d–d transitions · incomplete d · Ti³⁺ violet · Cu²⁺ blue · Sc³⁺/Ti⁴⁺ colourless

Complexes, alloys, interstitial compounds, catalysis (V₂O₅, Fe, Ni)

Exam

Q: μ for Ni²⁺ (d⁸, 2 unpaired)?

√[2×4] = √8

Answer: 2.83 B.M.

Section 2: K₂Cr₂O₇ & KMnO₄

Potassium Dichromate

Ore: chromite FeO·Cr₂O₃ · fuse + Na₂CO₃ + O₂ → Na₂CrO₄ → acidify → Na₂Cr₂O₇ → KCl → K₂Cr₂O₇

Acidic medium: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Oxidises Fe²⁺, I⁻, SO₂ · Chromyl chloride test for Cl⁻ · Cr₂O₇²⁻ ⇌ 2CrO₄²⁻ (orange ⇌ yellow with pH)

Potassium Permanganate

Ore: pyrolusite MnO₂ · fuse KOH/air → K₂MnO₄ → oxidise (Cl₂/O₃/electrolysis) → KMnO₄

Acidic: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O

Neutral/alkaline: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻

Disinfect wells · volumetric Fe²⁺, C₂O₄²⁻, H₂O₂

Section 3: f-Block

Lanthanoids & Actinoids

Ln: Ce–Lu (+ La) · 4f filling · mainly +3 · Eu²⁺, Yb²⁺, Ce⁴⁺ when f⁰/f⁷/f¹⁴

Lanthanoid contraction: poor 4f shielding → radii ↓ La³⁺→Lu³⁺ · Zr ≈ Hf size

An: 5f · almost all radioactive · OS up to +7 (Np, Pu) · form oxocations UO₂²⁺

Section 2: Definitions

Transition element: d-subshell partially filled in atom or common ion.

Spin-only magnetic moment: μ = √[n(n+2)] BM; n = unpaired electrons.

Lanthanoid contraction: Steady decrease in Ln radii due to poor 4f shielding.

Interstitial compound: Small atoms (H, C, N) in metal lattice voids → hard alloys/steel.

Chromyl chloride test: Cl⁻ + K₂Cr₂O₇ + conc. H₂SO₄ → red CrO₂Cl₂ vapours.

Section 3: Visual Map

L21 Map — d & f Block 3d: Sc–Cu properties K₂Cr₂O₇ · KMnO₄ Ln · An · contraction μ = √n(n+2) · coloured dⁿ · variable OS · catalysts · alloys Acid: Cr₂O₇²⁻/6e⁻ · MnO₄⁻/5e⁻ · Ln mostly +3 · Zr≈Hf

Section 5: Q&A (12 Questions)

Q1: Why is Zn not a transition element?

d¹⁰ in atom and common ions — no partial d-subshell.

Q2: Why Cr and Cu have 4s¹ configs?

Extra stability of half-filled 3d⁵ and full 3d¹⁰.

Q3: Magnetic moment of Mn²⁺?

d⁵, n=5 → √[5×7] = √35 ≈ 5.92 BM.

Q4: Why are transition metal ions coloured?

Incomplete d; d–d transitions absorb visible light; complementary colour seen.

Q5: Chromite ore formula?

FeO·Cr₂O₃ (or FeCr₂O₄).

Q6: Half-reaction of dichromate in acid?

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O.

Q7: Effect of alkali on dichromate solution?

Orange Cr₂O₇²⁻ → yellow CrO₄²⁻ (equilibrium shifts).

Q8: Pyrolusite and KMnO₄ preparation outline?

MnO₂ + KOH/air → K₂MnO₄ → oxidise to KMnO₄.

Q9: MnO₄⁻ in acidic medium reduces to?

Mn²⁺ (colourless/pale pink dilute).

Q10: What is lanthanoid contraction? Consequence?

Decrease in size across Ln due to poor 4f shielding; Zr and Hf nearly same size/chemistry.

Q11: Characteristic OS of lanthanoids?

+3; +2/+4 when f⁰, f⁷ or f¹⁴ stabilised.

Q12: One difference Ln vs An?

Actinoids mostly radioactive, higher max OS (up to +7), form oxocations; Ln rarely beyond +4.

Section 6: Tips & Exam Hacks

Memory Aids

  • μ: "Root n times n-plus-two"
  • Dichromate: "6 electrons, 14 H⁺"
  • Permanganate acid: "5 electrons, 8 H⁺"
  • Cr colour: "Orange acid, yellow alkali"
  • Ln: "Always think +3 first"

Exam Tips

  • Balance redox half-reactions carefully
  • Diamagnetic: d⁰ or d¹⁰ (Sc³⁺, Ti⁴⁺, Zn²⁺, Cu⁺)
  • KMnO₄ best oxidant in acid (5e⁻ change)
  • Chromyl chloride confirms Cl⁻
  • Lanthanoid contraction explains 4d/5d similarity

Section 8: Quick Reference

• Config (n−1)d ns · Cr/Cu exceptions · Zn not transition

• μ = √[n(n+2)] · coloured if unpaired d · variable OS · catalysts

• K₂Cr₂O₇ from chromite · Cr₂O₇²⁻ + 14H⁺ + 6e⁻

• KMnO₄ from MnO₂ · MnO₄⁻ + 8H⁺ + 5e⁻ (acid)

• Ln +3 · contraction · An radioactive · high OS · oxocations

Remember: ✓ Partial d required ✓ High OS oxidants ✓ Orange⇌yellow ✓ Zr≈Hf ✓ Ce⁴⁺/Eu²⁺ special

PYQ — Previous Year Questions

Extracted from NIOS Chemistry (313) board exam papers in your PDF. Chapter L21 — d-Block and f-Block Elements only. Use Model Answer for marking points; Explanation for concept clarity.

L21 — d-Block and f-Block Elements

5 question(s) · Sources: 313/MAY/205A, 313/MAY/205B, 313/MAY/205C, 313/TUS/105A

Section A — MCQ / Objective (from papers)

PYQ1. Read the passage given below and answer the following questions : Most of the compounds of d-block elements are coloured or they give coloured solution when dissolved in water. This is generally associated with incomplete (n – 1)d subshell of the transition metal. If red portion of white light is absorbed by a substance, it would appear blue. Identify the incorrect statement from the following : (i) An energy transition of electrons takes place in transition metal ions which absorb some of the energy of visible light. (ii) The colour of the ions is due to the presence of all paired electrons in them. (iii) Blue is the complementary colour of red. (iv) In transition metals, the energy difference between the various d-orbitals is in the same order of magnitude as the energies of radiation of white light. What is the colour of hexahydrated form of ferric ions?

2 marks · Q23 · 313/MAY/205A

Model Answer

State the precise definition from the L21 notes in 1–2 sentences, include formula/example if marks ≥ 2, and avoid extra theory beyond the ask.

Explanation

Definition questions score for accuracy of wording + one supporting point/example. Do not write full chapter summaries.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205A · Q23 · 2 mark(s) · L21.

PYQ2. Read the passage given below and answer the following questions : Most of the compounds of d-block elements are coloured or they give coloured solution when dissolved in water. This is generally associated with incomplete (n – 1)d subshell of the transition metal. If red portion of white light is absorbed by a substance, it would appear blue. Identify the incorrect statement from the following : (i) An energy transition of electrons takes place in transition metal ions which absorb some of the energy of visible light. (ii) The colour of the ions is due to the presence of all paired electrons in them. (iii) Blue is the complementary colour of red. (iv) In transition metals, the energy difference between the various d-orbitals is in the same order of magnitude as the energies of radiation of white light. What is the colour of hexahydrated form of ferric ions?

2 marks · Q27 · 313/MAY/205B

Model Answer

State the precise definition from the L21 notes in 1–2 sentences, include formula/example if marks ≥ 2, and avoid extra theory beyond the ask.

Explanation

Definition questions score for accuracy of wording + one supporting point/example. Do not write full chapter summaries.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205B · Q27 · 2 mark(s) · L21.

PYQ3. Read the passage given below and answer the following questions : Most of the compounds of d-block elements are coloured or they give coloured solution when dissolved in water. This is generally associated with incomplete (n – 1)d subshell of the transition metal. If red portion of white light is absorbed by a substance, it would appear blue. Identify the incorrect statement from the following : (i) An energy transition of electrons takes place in transition metal ions which absorb some of the energy of visible light. (ii) The colour of the ions is due to the presence of all paired electrons in them. (iii) Blue is the complementary colour of red. (iv) In transition metals, the energy difference between the various d-orbitals is in the same order of magnitude as the energies of radiation of white light. What is the colour of hexahydrated form of ferric ions?

2 marks · Q25 · 313/MAY/205C

Model Answer

State the precise definition from the L21 notes in 1–2 sentences, include formula/example if marks ≥ 2, and avoid extra theory beyond the ask.

Explanation

Definition questions score for accuracy of wording + one supporting point/example. Do not write full chapter summaries.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205C · Q25 · 2 mark(s) · L21.

PYQ4. Write True (T) for correct statement and False (F) for incorrect statement (out of four attempt any two) : Copper(I) compounds are white and diamagnetic while copper(II) compounds are coloured and paramagnetic. The common oxidation state of Cu, Ag and Au is +2. Scandium does not exhibit variable oxidation state in its compounds. Among Al, Zn, Mg and Fe, the densest element is Fe. ghr

2 marks · Q23 · 313/TUS/105A

Model Answer

Answer using key concepts from L21 (definitions, equations, and one example where useful). Stay within the suggested word range for a 2-mark NIOS question.

Explanation

Cross-check with L21 notes. Structure: definition/law → working → conclusion. Partial marks for correct equations even if explanation is short.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/TUS/105A · Q23 · 2 mark(s) · L21.

Section B — Short / Long answer (from papers)

PYQ5. Give reason for the following : Transition elements have higher density as compared to s-block elements. Transition metals show high melting and boiling points

2 marks · Q37 · 313/MAY/205A

Model Answer

Present a clear comparison in 2–3 points (definition / structure / property / example). Use a table style in prose: A vs B for each criterion.

Explanation

Comparison answers need parallel points. Award marks for each distinct difference.

How to write for NIOS: Use 30–50 words (VSA) or short objective. Open with definition/equation, then reason, end with conclusion. Paper 313/MAY/205A · Q37 · 2 mark(s) · L21.

Problem Solving — L21 d-Block and f-Block Elements

Six problems spanning this chapter’s NIOS syllabus. Every question is built from the notes and formula sheet: solve with equations first, then read the formal textbook-style write-up, the easy explanation, and the topic in depth (formulas, meaning, exam tips). If the question says draw, a labelled pencil sketch is provided. Explanations open by default.

Question 1 of 6d-block

Define transition elements and give one characteristic property.

(n−1)d ns configuration

Solution — step by step with formulas

  1. d-block elements with incomplete d subshell in atom or common ions; coloured ions / variable OS / catalytic activity.

Final answer: Incomplete d; coloured ions / variable OS

Formulas used in this problem

(n−1)d ns configuration

Textbook formal language

Transition metals show variable oxidation states and complex formation.

Working formulas: (n−1)d ns configuration. State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Middle of the table metals with partly filled d orbitals.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Transition elements

Zn sometimes excluded (d¹⁰).

Linked to chapter notes (L21). Remember: (n−1)d ns configuration. Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (n−1)d ns configuration before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 2 of 6Colour

Why are many transition metal ions coloured?

Solution — step by step with formulas

  1. d–d electronic transitions in ligand field / crystal field (visible light absorbed).

Final answer: d–d transitions

Textbook formal language

Absorption of visible photons promotes d electrons between split levels.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Ion absorbs some colours and we see the complementary colour.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Coloured ions

d⁰ and d¹⁰ often colourless.

Linked to chapter notes (L21). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 3 of 6OS

Why do transition metals show variable oxidation states?

Solution — step by step with formulas

  1. ns and (n−1)d electrons both can participate in bonding.

Final answer: ns and (n−1)d both available

Textbook formal language

Successive ionisation energies allow multiple stable OS.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

They can lose different numbers of electrons fairly easily.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Variable oxidation states

Mn shows OS from +2 to +7.

Linked to chapter notes (L21). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 4 of 6Catalysis

Give one industrial process catalysed by a transition metal/compound.

Solution — step by step with formulas

  1. Haber (Fe); Contact process (V₂O₅); hydrogenation (Ni) etc.

Final answer: e.g. Fe in Haber process

Textbook formal language

Variable OS and surface adsorption enable catalysis.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Metal surface holds reactants and helps them react faster.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Catalytic property

Enzymes also use transition metals in biology.

Linked to chapter notes (L21). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 5 of 6Lanthanoids

What is lanthanoid contraction and one consequence?

Solution — step by step with formulas

  1. Steady decrease in size across La–Lu; similar sizes of 4d/5d pairs; separation difficulties.

Final answer: Size decrease across 4f series

Textbook formal language

Poor shielding by 4f electrons causes gradual contraction.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Atoms get slightly smaller across the f-block.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Lanthanoid contraction

Zr/Hf similar chemistry.

Linked to chapter notes (L21). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.
Question 6 of 6Interstitial

What are interstitial compounds of transition metals?

Solution — step by step with formulas

  1. Small atoms (H,C,N) occupy voids in metal lattice; hard, high m.p.

Final answer: Small atoms in metal voids

Textbook formal language

Non-stoichiometric hard materials important as steels/carbides.

Working formulas: (see solution steps). State the definition or law first (NIOS style), use SI units, and box the final numerical answer with unit.

Easy language (same idea, plain words)

Carbon squeezes into iron lattice—steel properties change.

Read once for the idea, once for the numbers. Write the formula, substitute, then simplify. Check whether you used moles, grams, or litres correctly.

Topic in depth — Interstitial compounds

Retain metallic conductivity often.

Linked to chapter notes (L21). Remember: (see solution steps). Most exam errors are unit mix-ups (g vs mol, mL vs L) or wrong mole ratios from the equation.

Exam tip

Write (see solution steps) before substituting. Keep three significant figures until the end when data allow.

Common mistakes

  • Confusing mass (g) with amount of substance (mol).
  • Forgetting Avogadro’s number unit mol⁻¹ or STP volume 22.7 L mol⁻¹ (1 bar).
  • Using wrong mole ratio from the balanced equation.
  • Mixing up empirical and molecular formulas.