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The d and f Block Elements

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Chapter Notes

Overview
This chapter explores the d-block (transition metals) and f-block (inner transition metals) elements, focusing on their electronic configurations, general properties, and trends. The one big idea is that the presence of partially filled d or f orbitals gives these elements unique characteristics like variable oxidation states, coloured ions, complex formation, catalytic activity, and paramagnetism. Students must understand the lanthanoid contraction and the specific preparation and properties of key compounds like K₂Cr₂O₇ and KMnO₄.
Key Concepts (14)
Transition elements (IUPAC definition)
Elements which have incomplete d subshell either in neutral atom or in their ions.
e.g. Scandium (Z=21) has 3d¹ in its ground state, so it is a transition element.
d-block elements
Elements of groups 3-12 in which the d orbitals are progressively filled in each of the four long periods.
e.g. The 3d series includes elements from Sc (Z=21) to Zn (Z=30).
f-block elements
Elements in which 4f and 5f orbitals are progressively filled; placed in a separate panel at the bottom of the periodic table.
e.g. Lanthanoids (Ce to Lu) and actinoids (Th to Lr).
Lanthanoid contraction
The regular decrease in atomic radii of lanthanoids due to imperfect shielding of one 4f electron by another, resulting in similar radii of second and third d-series elements.
e.g. Zr (160 pm) and Hf (159 pm) have nearly identical radii.
Interstitial compounds
Compounds formed when small atoms like H, C, or N are trapped inside the crystal lattices of metals; they are non-stoichiometric, hard, high-melting, and chemically inert.
e.g. TiC, Mn₄N, Fe₃H, VH₀.₅₆.
Disproportionation
A reaction in which a particular oxidation state becomes less stable relative to other oxidation states, one lower and one higher.
e.g. 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O (Mn(VI) disproportionates to Mn(VII) and Mn(IV)).
Spin-only magnetic moment
The magnetic moment calculated using the formula μ = √[n(n+2)] BM, where n is the number of unpaired electrons.
e.g. For Mn²⁺ (d⁵, n=5), μ = √(5×7) = √35 = 5.92 BM.
Paramagnetism
Magnetic property arising from the presence of unpaired electrons; substances are attracted by an applied magnetic field.
e.g. Cu²⁺ (d⁹, one unpaired electron) is paramagnetic with μ ≈ 1.73 BM.
Diamagnetism
Magnetic property where substances are repelled by an applied magnetic field; occurs when all electrons are paired.
e.g. Zn²⁺ (d¹⁰) is diamagnetic.
Ferromagnetism
An extreme form of paramagnetism where substances are very strongly attracted by a magnetic field.
e.g. Iron, cobalt, nickel.
Alloy
A blend of metals prepared by mixing components; may be homogeneous solid solutions.
e.g. Brass (copper-zinc) and bronze (copper-tin).
Oxidation states (variability)
The ability of transition elements to exhibit a variety of oxidation states due to incomplete filling of d orbitals, often differing by unity.
e.g. Manganese exhibits all oxidation states from +2 to +7.
Catalytic activity of transition metals
Ascribed to their ability to adopt multiple oxidation states and form complexes, lowering activation energy.
e.g. Iron(III) catalyses the reaction between iodide and persulphate ions: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂; 2Fe²⁺ + S₂O₈²⁻ → 2Fe³⁺ + 2SO₄²⁻.
Complex compounds
Compounds in which metal ions bind a number of anions or neutral molecules, forming complex species with characteristic properties.
e.g. [Fe(CN)₆]³⁻, [Cu(NH₃)₄]²⁺, [PtCl₄]²⁻.
Formulas & Equations (10)
μ = √[n(n+2)] BM
Reads as: Magnetic moment mu equals square root of n times n plus two, in Bohr magnetons.
Variables: n = number of unpaired electrons; μ = magnetic moment in Bohr magnetons (BM).
Use when: Calculating the spin-only magnetic moment of transition metal ions.
⚠ Students forget to take the square root or incorrectly calculate n for d⁴, d⁵, d⁶, d⁷ configurations (e.g., for d⁶, n=4, not 6).
Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O (E° = 1.33V)
Reads as: Dichromate ion plus 14 hydrogen ions plus 6 electrons gives 2 chromium(III) ions plus 7 water molecules.
Variables: Cr₂O₇²⁻ = dichromate ion; H⁺ = hydrogen ion; e⁻ = electron; Cr³⁺ = chromium(III) ion; H₂O = water.
Use when: Writing the half-reaction for the oxidising action of acidified potassium dichromate.
⚠ Students often forget to balance the charge and atoms, especially the number of H⁺ and H₂O molecules.
2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O
Reads as: Two chromate ions plus two hydrogen ions are in equilibrium with dichromate ion plus water.
Variables: CrO₄²⁻ = chromate ion (yellow); H⁺ = hydrogen ion; Cr₂O₇²⁻ = dichromate ion (orange); H₂O = water.
Use when: Explaining the interconversion of chromate and dichromate in aqueous solution depending on pH.
⚠ Students write the reaction as irreversible or forget that it is pH-dependent.
Cr₂O₇²⁻ + 2OH⁻ ⇌ 2CrO₄²⁻ + H₂O
Reads as: Dichromate ion plus two hydroxide ions are in equilibrium with two chromate ions plus water.
Variables: Cr₂O₇²⁻ = dichromate ion; OH⁻ = hydroxide ion; CrO₄²⁻ = chromate ion; H₂O = water.
Use when: Showing the reverse interconversion in basic medium.
⚠ Students confuse the direction of the reaction with pH change.
2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O
Reads as: Two manganese dioxide plus four potassium hydroxide plus oxygen gives two potassium manganate plus two water.
Variables: MnO₂ = manganese dioxide; KOH = potassium hydroxide; O₂ = oxygen; K₂MnO₄ = potassium manganate; H₂O = water.
Use when: Describing the preparation of potassium manganate by fusion of MnO₂ with KOH and an oxidising agent.
⚠ Students forget to balance the equation or write the wrong product (e.g., KMnO₄ directly).
3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O
Reads as: Three manganate ions plus four hydrogen ions gives two permanganate ions plus manganese dioxide plus two water.
Variables: MnO₄²⁻ = manganate ion (green); H⁺ = hydrogen ion; MnO₄⁻ = permanganate ion (purple); MnO₂ = manganese dioxide; H₂O = water.
Use when: Explaining the disproportionation of manganate to permanganate in acidic or neutral solution.
⚠ Students write the product as Mn²⁺ instead of MnO₂, or forget to balance the equation.
4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂
Reads as: Four iron chromite plus eight sodium carbonate plus seven oxygen gives eight sodium chromate plus two iron(III) oxide plus eight carbon dioxide.
Variables: FeCr₂O₄ = chromite ore; Na₂CO₃ = sodium carbonate; O₂ = oxygen; Na₂CrO₄ = sodium chromate; Fe₂O₃ = iron(III) oxide; CO₂ = carbon dioxide.
Use when: Describing the fusion of chromite ore with sodium carbonate to prepare chromate.
⚠ Students often miss the stoichiometric coefficients or write the wrong product for iron (e.g., FeO instead of Fe₂O₃).
2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O
Reads as: Two sodium chromate plus two hydrogen ions gives sodium dichromate plus two sodium ions plus water.
Variables: Na₂CrO₄ = sodium chromate; H⁺ = hydrogen ion; Na₂Cr₂O₇ = sodium dichromate; Na⁺ = sodium ion; H₂O = water.
Use when: Explaining the conversion of sodium chromate to sodium dichromate by acidification.
⚠ Students forget to include the Na⁺ ions or write the product as K₂Cr₂O₇ directly.
Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl
Reads as: Sodium dichromate plus two potassium chloride gives potassium dichromate plus two sodium chloride.
Variables: Na₂Cr₂O₇ = sodium dichromate; KCl = potassium chloride; K₂Cr₂O₇ = potassium dichromate; NaCl = sodium chloride.
Use when: Describing the preparation of potassium dichromate from sodium dichromate by double displacement.
⚠ Students think this is a precipitation reaction; it is based on the lower solubility of K₂Cr₂O₇.
2Mn²⁺ + 5S₂O₈²⁻ + 8H₂O → 2MnO₄⁻ + 10SO₄²⁻ + 16H⁺
Reads as: Two manganese(II) ions plus five peroxodisulphate ions plus eight water gives two permanganate ions plus ten sulphate ions plus sixteen hydrogen ions.
Variables: Mn²⁺ = manganese(II) ion; S₂O₈²⁻ = peroxodisulphate ion; H₂O = water; MnO₄⁻ = permanganate ion; SO₄²⁻ = sulphate ion; H⁺ = hydrogen ion.
Use when: Describing the laboratory preparation of KMnO₄ by oxidation of Mn²⁺ with peroxodisulphate.
⚠ Students forget to balance the equation or write the wrong oxidation state of sulphur.
How It Works
Preparation of Potassium Dichromate (K₂Cr₂O₇)
  1. Step 1: Fusion of chromite ore (FeCr₂O₄) with sodium carbonate (Na₂CO₃) in free access of air: 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂.
  2. Step 2: The yellow solution of sodium chromate (Na₂CrO₄) is filtered and acidified with sulphuric acid to give orange sodium dichromate (Na₂Cr₂O₇): 2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O.
  3. Step 3: Sodium dichromate is treated with potassium chloride (KCl) to obtain potassium dichromate (K₂Cr₂O₇) by double displacement: Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl.
  4. Step 4: Orange crystals of K₂Cr₂O₇ crystallise out due to its lower solubility compared to Na₂Cr₂O₇.
💡 The final step is a double displacement reaction, not a precipitation; it works because K₂Cr₂O₇ is less soluble than Na₂Cr₂O₇.
Preparation of Potassium Permanganate (KMnO₄)
  1. Step 1: Fusion of MnO₂ with an alkali metal hydroxide (KOH) and an oxidising agent like KNO₃ or air: 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O (produces dark green K₂MnO₄).
  2. Step 2: The manganate(VI) ion (MnO₄²⁻) disproportionates in neutral or acidic solution to give permanganate: 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O.
  3. Step 3: Commercially, it is prepared by alkaline oxidative fusion of MnO₂ followed by electrolytic oxidation of manganate(VI) in alkaline solution.
  4. Step 4: In the laboratory, Mn²⁺ salt is oxidised by peroxodisulphate: 2Mn²⁺ + 5S₂O₈²⁻ + 8H₂O → 2MnO₄⁻ + 10SO₄²⁻ + 16H⁺.
💡 The disproportionation step is key; students often forget that the green manganate ion is an intermediate and that the reaction is pH-dependent.
Interconversion of Chromate and Dichromate
  1. Step 1: In acidic medium, yellow chromate ion (CrO₄²⁻) converts to orange dichromate ion (Cr₂O₇²⁻): 2CrO₄²⁻ + 2H⁺ → Cr₂O₇²⁻ + H₂O.
  2. Step 2: In basic medium, orange dichromate ion converts back to yellow chromate ion: Cr₂O₇²⁻ + 2OH⁻ → 2CrO₄²⁻ + H₂O.
  3. Step 3: The oxidation state of chromium remains +6 in both ions.
💡 The interconversion is reversible and pH-dependent; the colour change (yellow to orange) is a common exam question.
Disproportionation of Manganate(VI) to Permanganate
  1. Step 1: Manganate ion (MnO₄²⁻, green) is stable in alkaline medium.
  2. Step 2: In neutral or acidic solution, it disproportionates: 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ (purple) + MnO₂ (brown) + 2H₂O.
  3. Step 3: Mn(VI) is less stable than Mn(VII) and Mn(IV) under these conditions.
💡 The products are permanganate (purple) and manganese dioxide (brown solid), not Mn²⁺.
Catalytic Action of Fe³⁺ in Iodide-Persulphate Reaction
  1. Step 1: Fe³⁺ oxidises I⁻ to I₂, itself getting reduced to Fe²⁺: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂.
  2. Step 2: Fe²⁺ is then oxidised back to Fe³⁺ by S₂O₈²⁻: 2Fe²⁺ + S₂O₈²⁻ → 2Fe³⁺ + 2SO₄²⁻.
  3. Step 3: Net reaction: 2I⁻ + S₂O₈²⁻ → I₂ + 2SO₄²⁻, catalysed by Fe³⁺/Fe²⁺ cycle.
💡 The catalyst (Fe³⁺) is regenerated; students often forget to show both half-reactions.
Exam Traps — Don't Make These Mistakes
⚠ Zn, Cd, Hg are transition elements
✗ Zinc, cadmium, and mercury are transition elements because they are in the d-block.
→ According to IUPAC, transition elements must have incomplete d subshell in the neutral atom or common ions. Zn, Cd, Hg have d¹⁰ configuration in both ground state and common oxidation states, so they are not transition elements.
✓ Zinc, cadmium, and mercury are not transition elements because they have completely filled d¹⁰ orbitals in their ground state and common oxidation states. However, their chemistry is studied along with transition metals as they are the end members of the 3d, 4d, and 5d series.
⚠ Colour of transition metal ions is due to d-d transitions only
✗ Transition metal ions are coloured because of d-d transitions.
→ While d-d transitions are the primary reason, the statement is incomplete. The colour arises when an electron from a lower energy d orbital is excited to a higher energy d orbital, absorbing light in the visible region. The observed colour is complementary to the absorbed light.
✓ The colour of transition metal ions is due to d-d transitions: an electron from a lower energy d orbital is excited to a higher energy d orbital, absorbing light of a specific frequency in the visible region. The observed colour is the complementary colour of the absorbed light.
⚠ Magnetic moment calculation for d⁶ ions
✗ For Fe²⁺ (d⁶), the number of unpaired electrons is 6, so μ = √(6×8) = √48 = 6.93 BM.
→ In the first row transition metals, the d orbitals split into t₂g and e_g sets. For d⁶ in a weak field (high spin), there are 4 unpaired electrons (t₂g⁴ e_g²), not 6. The spin-only formula uses n = number of unpaired electrons, not total d electrons.
✓ For Fe²⁺ (d⁶, high spin), n = 4 unpaired electrons, so μ = √(4×6) = √24 = 4.90 BM.
⚠ Stability of Cu²⁺ vs Cu⁺ in aqueous solution
✗ Cu⁺ is more stable than Cu²⁺ because it has a d¹⁰ configuration.
→ Although Cu⁺ has a d¹⁰ configuration, Cu²⁺(aq) is more stable in aqueous solution because the much more negative hydration enthalpy (ΔhydH°) of Cu²⁺ more than compensates for the second ionisation enthalpy of Cu. Cu⁺ undergoes disproportionation: 2Cu⁺ → Cu²⁺ + Cu.
✓ Cu²⁺(aq) is more stable than Cu⁺(aq) because the much more negative hydration enthalpy of Cu²⁺ outweighs the second ionisation enthalpy. Cu⁺ is unstable in aqueous solution and disproportionates to Cu²⁺ and Cu.
⚠ Highest oxidation state in oxides vs fluorides
✗ Fluorine stabilises the highest oxidation state better than oxygen.
→ Oxygen stabilises higher oxidation states better than fluorine because oxygen can form multiple bonds (pπ-dπ) to metals, whereas fluorine can only form single bonds. For example, Mn forms Mn₂O₇ (+7) but only MnF₄ (+4).
✓ Oxygen stabilises higher oxidation states better than fluorine due to its ability to form multiple bonds (pπ-dπ) with metals. For example, Mn exhibits +7 in Mn₂O₇ but only +4 in MnF₄.
Compare & Contrast
Electronic configuration and filling of f-orbitals
Lanthanoids
Actinoids
Lanthanoids involve the filling of 4f orbitals (Ce to Lu), while actinoids involve the filling of 5f orbitals (Th to Lr). Actinoids show a wider range of oxidation states due to the smaller energy gap between 5f, 6d, and 7s orbitals.
Structure and colour
Chromate ion (CrO₄²⁻)
Dichromate ion (Cr₂O₇²⁻)
Chromate ion is tetrahedral and yellow in colour, while dichromate ion consists of two tetrahedra sharing one corner (Cr-O-Cr bond angle 126°) and is orange in colour. They are interconvertible by changing pH.
Enthalpy of atomisation and metal-metal bonding
First row transition metals (3d series)
Second and third row transition metals (4d and 5d series)
Second and third row transition metals have greater enthalpies of atomisation than the corresponding first row elements, leading to more frequent metal-metal bonding in compounds of heavy transition metals.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks) — total approximately 12 marks.
Most asked: Electronic configurations of d-block elements (especially exceptions like Cr and Cu) · Lanthanoid contraction and its consequences · Oxidation states of transition metals (especially Mn, Cr, Fe) · Magnetic moment calculations using spin-only formula · Preparation and properties of K₂Cr₂O₇ and KMnO₄ · Colour of transition metal ions · Catalytic properties and complex formation
Don't skip: Explain lanthanoid contraction and its consequences (3-5 marks); Calculate magnetic moment of a given ion (2-3 marks); Describe the preparation of K₂Cr₂O₇ or KMnO₄ with equations (3-5 marks); Why do transition metals show variable oxidation states? (2-3 marks); Compare and contrast chromate and dichromate ions (2-3 marks)
One-Line Revision (read aloud before the exam)
  1. Transition elements are defined by IUPAC as elements with incomplete d subshell in neutral atom or ions.
  2. Zinc, cadmium, and mercury are not transition elements because they have d¹⁰ configuration in ground state and common oxidation states.
  3. The 3d series runs from Sc (Z=21) to Zn (Z=30); 4d from Y to Cd; 5d from La and Hf to Hg; 6d from Ac and Rf to Cn.
  4. Lanthanoids (Ce to Lu) fill 4f orbitals; actinoids (Th to Lr) fill 5f orbitals.
  5. Cr has electronic configuration 3d⁵ 4s¹ and Cu has 3d¹⁰ 4s¹ due to extra stability of half-filled and fully filled d orbitals.
  6. Lanthanoid contraction is the regular decrease in atomic radii of lanthanoids due to imperfect shielding of 4f electrons.
  7. Lanthanoid contraction causes second and third d-series elements (e.g., Zr and Hf) to have nearly identical radii and properties.
  8. Transition metals exhibit variable oxidation states due to incomplete filling of d orbitals; oxidation states often differ by unity.
  9. Manganese shows the largest number of oxidation states (+2 to +7) in the 3d series.
  10. The spin-only magnetic moment formula is μ = √[n(n+2)] BM, where n is the number of unpaired electrons.
  11. Colour of transition metal ions is due to d-d transitions; the observed colour is complementary to the absorbed light.
  12. Cr₂O₇²⁻ (orange) and CrO₄²⁻ (yellow) are interconvertible by changing pH: 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O.
  13. Acidified K₂Cr₂O₇ oxidises I⁻ to I₂, Fe²⁺ to Fe³⁺, H₂S to S, and Sn²⁺ to Sn⁴⁺.
  14. KMnO₄ is prepared by fusion of MnO₂ with KOH and an oxidising agent, followed by disproportionation of manganate.
  15. Oxygen stabilises higher oxidation states better than fluorine due to its ability to form multiple bonds.
  16. Interstitial compounds (e.g., TiC, Fe₃H) are hard, high-melting, chemically inert, and retain metallic conductivity.
  17. Transition metals and their compounds act as catalysts due to their ability to adopt multiple oxidation states and form complexes.
  18. Cu⁺ is unstable in aqueous solution and disproportionates to Cu²⁺ and Cu.
  19. The E° value for Cu²⁺/Cu is positive (+0.34V) because its high enthalpy of atomisation and low hydration enthalpy are not balanced.
  20. Mn²⁺ (d⁵) and Zn²⁺ (d¹⁰) have extra stability due to half-filled and fully filled d subshells, respectively.

MCQs — 10 Questions with Answers

Q1. Which of the following has a completely filled d subshell in its ground state and common oxidation states?

  • A. Scandium (Sc)
  • B. Zinc (Zn) ✓
  • C. Chromium (Cr)
  • D. Manganese (Mn)

Answer: B — Zinc has 3d¹⁰4s² configuration with all d orbitals filled; Sc, Cr, and Mn all have incomplete d subshells.

Q2. The electronic configuration of copper (Cu) is 3d¹⁰4s¹ instead of 3d⁹4s². This is because:

  • A. Half-filled d orbitals are more stable
  • B. Completely filled d orbitals are more stable ✓
  • C. The 4s orbital has higher energy than 3d
  • D. Copper cannot fill all d orbitals

Answer: B — Completely filled d¹⁰ configuration is extra stable due to exchange energy, so Cu prefers 3d¹⁰4s¹ over 3d⁹4s².

Q3. According to IUPAC definition, a transition metal is best defined as:

  • A. An element in the d-block of the periodic table
  • B. A metal with incomplete d subshell in neutral atom or common oxidation states ✓
  • C. An element that forms coloured compounds
  • D. A metal with both d and s electrons

Answer: B — The IUPAC definition specifically requires incomplete d orbitals, which excludes Zn, Cd, and Hg despite their d-block position.

Q4. Chromium has the configuration 3d⁵4s¹. If you were to write it as 3d⁴4s² instead, what would be the main reason this does NOT occur?

  • A. Because 4s electrons cannot enter d orbitals
  • B. Because half-filled d subshell (3d⁵) is more stable than 3d⁴ ✓
  • C. Because 4s orbital has lower energy than 3d
  • D. Because chromium is not a transition metal

Answer: B — The 3d⁵ half-filled configuration has extra stability from exchange energy, making Cr prefer to have one 4s electron rather than two.

Q5. Which property do transition metals exhibit due to their partly filled d orbitals?

  • A. They always show +2 oxidation state
  • B. They form only ionic compounds
  • C. Variable oxidation states and paramagnetism ✓
  • D. They cannot form complexes

Answer: C — Incomplete d orbitals allow multiple oxidation states and unpaired electrons, causing paramagnetism; options A, B, D are false.

Q6. Which of the following statements about d orbitals is INCORRECT?

  • A. d orbitals protrude more to the periphery than s and p orbitals
  • B. d orbitals are more influenced by their surroundings
  • C. d orbitals are completely shielded from external influences ✓
  • D. d orbitals affect the atoms or molecules surrounding them

Answer: C — d orbitals protrude to the periphery and are heavily influenced by surroundings, making option C false; they are exposed, not shielded.

Q7. The first row of transition metals (3d series) goes from Sc to Zn. How many elements have INCOMPLETE d orbitals in their ground state?

  • A. 8
  • B. 9 ✓
  • C. 10
  • D. All 10 elements including Zn

Answer: B — Sc through Cu (9 elements: Sc to Cu) have incomplete d orbitals; Zn (3d¹⁰4s²) has a complete d subshell and is not a true transition metal.

Q8. Both scandium (Sc) and zinc (Zn) are in Group 3 and Group 12 of the d-block respectively. Which statement correctly explains why Sc is a transition metal but Zn is not?

  • A. Sc has fewer total electrons than Zn
  • B. Zn has d¹⁰ configuration while Sc has incomplete d orbitals (3d¹) ✓
  • C. Sc forms more stable compounds than Zn
  • D. Zn is not located in the d-block of the periodic table

Answer: B — Sc (3d¹4s²) has incomplete d orbitals → transition metal; Zn (3d¹⁰4s²) has filled d orbitals → NOT a transition metal by IUPAC definition.

Q9. Calculate the total number of electrons in the d orbital for Manganese (Mn, Z=25) in its ground state.

  • A. 4
  • B. 5 ✓
  • C. 6
  • D. 7

Answer: B — Mn has configuration [Ar]3d⁵4s², so the d orbital contains 5 electrons (3d⁵).

Q10. The lanthanoids (4f series) and actinoids (5f series) are called inner transition metals because their characteristic differentiating electrons enter which orbital?

  • A. d orbitals
  • B. f orbitals ✓
  • C. s orbitals
  • D. p orbitals

Answer: B — f-block elements are defined by progressive filling of f orbitals (4f for lanthanoids, 5f for actinoids), making them inner transition metals.

Flashcards

What is the IUPAC definition of a transition metal?

A transition metal has an incomplete d subshell in its neutral atom or in its common oxidation states.

Why is zinc (Z=30) NOT considered a transition metal despite being in the d-block?

Zinc has a completely filled d¹⁰ configuration in both its ground state and common oxidation states, making it an end member, not a true transition element.

What is the general electronic configuration of transition elements?

The general configuration is (n-1)d¹⁻¹⁰ns¹⁻², where (n-1) is the inner d orbital and ns is the outermost orbital.

Why does chromium have 3d⁵4s¹ instead of 3d⁴4s²?

Half-filled d orbitals (3d⁵) are extra stable due to exchange energy, and the small energy gap between 3d and 4s allows this exception.

Name the four main series of transition metals in order.

The four series are 3d (Sc to Zn), 4d (Y to Cd), 5d (La and Hf to Hg), and 6d (Ac and Rf to Cn).

What are the two series of inner transition metals called?

The 4f series (Ce to Lu) is called lanthanoids, and the 5f series (Th to Lr) is called actinoids.

Why do transition elements show variable oxidation states?

The partly filled d orbitals have similar energies to the ns orbital, allowing electrons from both orbitals to participate in bonding.

What property of d orbitals makes them more influenced by surroundings than s and p orbitals?

d orbitals protrude more to the periphery of the atom, making them more exposed to external influences.

Which elements in Group 12 are studied with transition metals and why?

Zinc, cadmium, and mercury are studied with transition metals because they are end members of the 3d, 4d, and 5d series respectively, despite having complete d¹⁰ configurations.

What is the key difference in property trends between transition and non-transition elements?

Transition elements show greater horizontal (row) similarities than vertical (group) similarities, opposite to the pattern in non-transition elements.

Important Board Questions

Define transition metals according to IUPAC. Give one example of a d-block element that is NOT a transition metal and justify your answer. [2 marks]

IUPAC definition requires incomplete d subshell in neutral atom or common oxidation states. Use Zn (3d¹⁰4s²) as example — complete d means not a transition metal.

Chromium (Cr, Z=24) has electronic configuration [Ar]3d⁵4s¹ instead of [Ar]3d⁴4s². Explain why this anomaly occurs. Also write the configuration of copper (Cu, Z=29) and explain the reason behind its configuration. [5 marks]

Both Cr and Cu show exceptions due to extra stability of half-filled (d⁵) and completely filled (d¹⁰) d orbitals. The small energy gap between (n-1)d and ns allows this. Use exchange energy concept for stability explanation.

Explain how the incompletely filled d orbitals in transition metals lead to four characteristic properties: variable oxidation states, colour formation, complex formation, and paramagnetism. Give one industrial or biological example where any one of these properties is used. [6 marks]

Link incomplete d → unpaired electrons → multiple possible oxidation states, d-electron excitation → colour, d-orbital overlap with ligands → complexes, unpaired d-electrons → paramagnetism. Example: Fe²⁺/Fe³⁺ in redox, KMnO₄ colour in industry, [Cu(NH₃)₄]²⁺ complex, or Fe₃O₄ magnetism.

Assertion-Reason Questions for The d and f Block Elements

Q1. Assertion (A):

Scandium (Z = 21) is classified as a transition element while zinc (Z = 30) is not, even though both are in the d-block.

Reason (R):

Scandium has an incomplete 3d¹ orbital in its ground state, whereas zinc has a complete 3d¹⁰ configuration in both its neutral and common oxidation states.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains why scandium qualifies as a transition element (incomplete d orbitals) while zinc does not (complete d orbitals in all relevant oxidation states). This aligns with the IUPAC definition of transition metals.

Q2. Assertion (A):

Chromium exhibits an anomalous electronic configuration of 3d⁵4s¹ instead of the expected 3d⁴4s².

Reason (R):

The 3d and 4s orbitals have approximately equal energy, and the principal quantum number of 3d is higher than 4s.

Answer: C — A is true but R is false — The assertion is true—Cr does have the anomalous 3d⁵4s¹ configuration. However, the reason is false. The reason should cite the extra stability of half-filled d orbitals (3d⁵), not just energy proximity or principal quantum numbers. The actual reason involves exchange energy and stability from symmetric electron distribution.

Q3. Assertion (A):

Transition metals generally show greater horizontal (period-wise) similarities in their chemical properties than vertical (group-wise) similarities.

Reason (R):

All transition metals in the same period fill electrons into the same (n-1)d subshell, leading to similar electronic configurations and reactivity patterns.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true and correctly related. This is a distinctive feature of transition metals compared to main group elements, where vertical (group) trends dominate. The same d subshell being filled across a period explains the horizontal trend.

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Case-Based Questions (CBQ) for The d and f Block Elements

Case Study:

India's steel industry relies heavily on iron (Fe, Z = 26), a transition metal from the 3d series. Iron exhibits remarkable versatility in its oxidation states, ranging from +2 to +6, although +2 and +3 are most common in biological and industrial applications. The electronic configuration of neutral iron is [Ar]3d⁶4s², and when it loses two electrons to form Fe²⁺, it becomes [Ar]3d⁶. This property allows iron to form stable complexes with various ligands, making it essential in hemoglobin (oxygen transport), cytochrome enzymes (electron transfer), and industrial catalysts (ammonia synthesis via the Haber-Bosch process). The ability to maintain multiple oxidation states and form coloured complexes is a hallmark of transition metals.

Q1. Why does iron lose its 4s electrons before losing 3d electrons to form Fe²⁺, despite having 6 electrons available in the 3d orbitals?

  • A. The 3d orbitals are lower in energy and hold electrons more tightly; the 4s electrons are in the outermost shell and ionize more easily ✓
  • B. The 3d orbitals can hold a maximum of only 6 electrons
  • C. Iron always prefers to lose electrons from the d orbitals
  • D. The 4s orbital is larger and can release electrons faster

Answer: A

Q2. The passage mentions that iron forms coloured complexes. This property is characteristic of transition metals primarily because:

  • A. Their s-block electrons are easily excited by visible light
  • B. Partly filled d orbitals can absorb visible light through d-d electronic transitions, producing colours ✓
  • C. Transition metals reflect all colours of visible light
  • D. Iron atoms are large and scatter light, creating colours

Answer: B

Q3. Based on the electronic configurations provided, which statement about Fe²⁺ ([Ar]3d⁶) is correct?

  • A. Fe²⁺ is not a transition metal ion because it has a complete d subshell
  • B. Fe²⁺ is a transition metal ion with a partially filled 3d subshell containing 6 electrons ✓
  • C. Fe²⁺ cannot form complexes because the d orbitals are not available
  • D. Fe²⁺ has fewer unpaired electrons than the neutral Fe atom

Answer: B

Q4. If iron's configuration is [Ar]3d⁶4s² and it forms Fe³⁺, the resulting configuration would be [Ar]3d⁵. This ion is particularly stable because:

  • A. The 3d⁵ configuration represents a half-filled d subshell with one electron in each orbital, maximizing exchange energy and stability ✓
  • B. Fe³⁺ has lost all its valence electrons
  • C. The 3d⁵ configuration mimics the noble gas configuration
  • D. Five is the maximum number of electrons d orbitals can hold

Answer: A

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Previous Year Questions (PYQ) for The d and f Block Elements

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 56(B) · 1 mark

Which of the following is not a transition metal ?

  • Hg
  • Cu
  • Ag
  • Pt

CBSE 2026 · Set 56(B) · 1 mark

Assertion (A) : Transition metals have high melting points. Reason (R) : Transition metals have incompletely filled d-orbitals.

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.

CBSE 2026 · Set 56(B) · 5 marks

Define transition elements. Mention any two characteristics of transition elements.

CBSE 2026 · Set 56/1/1 · 1 mark

Which of the following is not a transition metal?

  • Sc
  • Ag
  • Hg
  • Cu

CBSE 2026 · Set 56/1/1 · 1 mark

Assertion (A): Actinoids show wide range of oxidation states. Reason (R): Actinoids are radioactive in nature.

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.

CBSE 2026 · Set 56/1/2 · 1 mark

Which of the following is most basic?

  • Mn2O7
  • MnO2
  • MnO
  • Mn2O3

CBSE 2026 · Set 56/1/2 · 1 mark

Assertion (A): Actinoids show irregularities in their electronic configurations. Reason (R): In actinoids 5f, 6d and 7s orbitals are of comparable energies.

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.

CBSE 2026 · Set 56/1/2 · 5 marks

(a) (i) From the given data of E values, answer the following questions: (I) Why E(M²⁺/M) show irregular trend in the above values? (II) Why is E(Cu²⁺/Cu) value exceptionally positive? (III) Why E(Mn²⁺/Mn) value is highly negative? (ii) Write the ionic equations for the oxidising action of potassium permanganate for its reaction with I⁻ in both acidic and alkaline solutions.

CBSE 2026 · Set 56/1/3 · 1 mark

Which of the following transition metals has lowest enthalpy of atomisation ?

  • Cr
  • V
  • Mn
  • Fe

CBSE 2026 · Set 56/1/3 · 1 mark

Assertion (A) : Actinoids show irregularities in their electronic configurations. Reason (R) : Due to varying stabilities of f0, f7 and f14 occupancies of the 5f orbitals.

  • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
  • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
  • Assertion (A) is true, but Reason (R) is false.
  • Assertion (A) is false, but Reason (R) is true.

CBSE 2026 · Set 56/1/3 · 5 marks

(a) (i) From the given data of E values, answer the following questions : (I) Why E M2+/M show irregular trend in the above values ? (II) Why is E Cu2+/Cu value exceptionally positive ? (III) Why E Mn2+ / Mn value is highly negative ?

CBSE 2026 · Set 56/2/1 · 1 mark

Electronic configuration of chromium is:

  • [Ar] 3d4 4s1
  • [Ar] 3d4 4s2
  • [Ar] 3d5 4s1
  • [Ar] 3d5 4s2

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Class 12 Chemistry Formula Sheet

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Class 12 Chemistry Previous Year Questions, Chapter-wise

All 1241 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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