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Inorganic Chemistry

d- and f-Block Elements — Trends, KMnO₄, K₂Cr₂O₇ & Lanthanoids

2 November 202614 pages
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Transition metals are defined by an incompletely filled d subshell, and almost every property this chapter tests (variable oxidation states, colour, magnetism, catalysis) comes straight from those d electrons. Learn why they exist and the property list stops being a list.

What you will be able to do

The big picture

Every property in this chapter traces back to the d electrons.

In one line: A transition element has an incompletely filled d subshell in the atom or in a common ion. Partly filled d orbitals give variable oxidation states, coloured and paramagnetic ions, complexes and catalytic power.
Think of it like this: Think of the d subshell as a five-room hostel. Half-full (one in every room) or full is a stable, settled state. Most of the “odd” behaviour you will meet — the Cr and Cu configurations, Mn²⁺ being stable, Cu⁺ being colourless — is the hostel trying to reach one of those two states.

Electronic configuration

Fill 4s first, remove 4s first.

ElementZConfigurationCommon ions
Sc21[Ar] 3d¹ 4s²Sc³⁺ (d⁰)
Ti22[Ar] 3d² 4s²Ti³⁺, Ti⁴⁺
V23[Ar] 3d³ 4s²V³⁺, V⁵⁺
Cr24[Ar] 3d⁵ 4s¹Cr³⁺ (d³), Cr⁶⁺
Mn25[Ar] 3d⁵ 4s²Mn²⁺ (d⁵), Mn⁷⁺
Fe26[Ar] 3d⁶ 4s²Fe²⁺ (d⁶), Fe³⁺ (d⁵)
Co27[Ar] 3d⁷ 4s²Co²⁺, Co³⁺
Ni28[Ar] 3d⁸ 4s²Ni²⁺ (d⁸)
Cu29[Ar] 3d¹⁰ 4s¹Cu⁺ (d¹⁰), Cu²⁺ (d⁹)
Zn30[Ar] 3d¹⁰ 4s²Zn²⁺ (d¹⁰)
Key idea: When a 3d atom forms an ion, the 4s electrons leave first. Fe is 3d⁶ 4s², so Fe²⁺ is 3d⁶ and Fe³⁺ is 3d⁵ — not 3d⁴ 4s².

Solved example 01 — Is copper a transition element?

Cu has a full 3d¹⁰ subshell. Why is it still called a transition element, while Zn is not?

  1. The definition allows an incompletely filled d subshell in the atom or in a common ion.
  2. Cu²⁺ is 3d⁹ — incomplete. Cu²⁺ is Cu’s most common ion.
  3. Zn’s only common ion, Zn²⁺, is 3d¹⁰ — complete.

Answer: Cu qualifies through Cu²⁺ (3d⁹); Zn never has an incomplete d subshell.

Trends across the 3d series

Small, steady changes — with a few famous exceptions.

Check yourself

Answers at the end.

Oxidation states and E°

How many electrons can a metal give away, and how willingly?

Because 4s and 3d have similar energies, transition metals use both, giving variable oxidation states differing by one. The number of states peaks at Mn (+2 to +7). The highest oxidation state of the d-block is +8 (Os in OsO₄).

CoupleE° / VWhat it tells you
Mn²⁺/Mn−1.18Mn is a fairly strong reducing agent
Zn²⁺/Zn−0.76Stable d¹⁰ Zn²⁺
Fe²⁺/Fe−0.44
Cu²⁺/Cu+0.34The only positive E° in the 3d series: Cu does not release H₂ from dilute acids
Mn³⁺/Mn²⁺+1.57Mn³⁺ is a strong oxidant: Mn²⁺ is d⁵ (half-filled)
Co³⁺/Co²⁺+1.97Co³⁺ is an even stronger oxidant in water
Cr³⁺/Cr²⁺−0.41Cr²⁺ is a reducing agent: Cr³⁺ is d³ (stable half-filled t₂g)
Fe³⁺/Fe²⁺+0.77Fe³⁺ (d⁵) is the more stable ion

Solved example 02 — Mn³⁺ versus Cr³⁺

Explain why Cr²⁺ is reducing while Mn³⁺ is oxidising, though both are d⁴.

  1. Cr²⁺ (d⁴) → Cr³⁺ (d³): losing one electron gives a half-filled t₂g set, which is especially stable. So Cr²⁺ gives electrons away readily.
  2. Mn³⁺ (d⁴) → Mn²⁺ (d⁵): gaining one electron gives a half-filled d⁵ set. So Mn³⁺ takes electrons readily.

Answer: Both ions move toward a half-filled configuration — one by losing, the other by gaining an electron.

Key idea: Cu⁺ disproportionates in water: 2Cu⁺ → Cu²⁺ + Cu. Cu²⁺ is smaller and more highly charged, so its hydration enthalpy is much larger and more than pays for the second ionisation enthalpy.

Colour, magnetism and catalysis

Three properties, one cause: unpaired or partly filled d orbitals.

Magnetism

μ (spin-only) = √[n(n + 2)] BM

n = number of unpaired electrons; BM = Bohr magneton

Unpaired electrons (n)12345
μ / BM1.732.833.874.905.92
Example ionTi³⁺, Cu²⁺V³⁺, Ni²⁺Cr³⁺Fe²⁺Mn²⁺, Fe³⁺

Colour

In a complex or a hydrated ion, the d orbitals split into two energy sets. An electron absorbs visible light to jump between them (a d–d transition), and we see the complementary colour. No d electrons (d⁰) or a full set (d¹⁰) means no d–d transition — colourless: Sc³⁺, Ti⁴⁺, Zn²⁺, Cu⁺.

IonConfigurationColour in water
Ti³⁺3d¹Purple
V³⁺3d²Green
Cr³⁺3d³Violet
Mn²⁺3d⁵Pale pink
Fe²⁺3d⁶Pale green
Fe³⁺3d⁵Yellow
Co²⁺3d⁷Pink
Ni²⁺3d⁸Green
Cu²⁺3d⁹Blue
Zn²⁺3d¹⁰Colourless

Catalysis, interstitial compounds and alloys

Solved example 03 — Spin-only moment

Calculate the spin-only magnetic moment of Fe²⁺ (Z = 26).

  1. Fe is [Ar] 3d⁶ 4s²; Fe²⁺ loses the two 4s electrons: 3d⁶.
  2. In five d orbitals, six electrons give four unpaired (one pair + four singles).
  3. μ = √[4 × 6] = √24.

Answer: μ ≈ 4.90 BM.

Check yourself

Answers at the end.

Potassium permanganate, KMnO₄

Purple, powerful, and the answer depends on the medium.

Preparation

  1. Fuse MnO₂ with KOH in air (or with KNO₃): 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ (green manganate) + 2H₂O.
  2. Oxidise manganate to permanganate — by disproportionation in neutral or acidic solution, 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O, or commercially by electrolytic oxidation in alkaline solution.

Structure: in MnO₄⁻ the Mn is +7 (d⁰) and tetrahedral. Its intense purple colour is not a d–d transition — it comes from charge transfer (an electron moving from O to Mn).

The three media

MediumHalf-equationMn changesElectrons
Acidic (dil. H₂SO₄)MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O+7 → +2 (colourless)5
Neutral / weakly alkalineMnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻+7 → +4 (brown ppt)3
Strongly alkalineMnO₄⁻ + e⁻ → MnO₄²⁻+7 → +6 (green)1
Key idea: Never acidify KMnO₄ with HCl: MnO₄⁻ oxidises Cl⁻ to Cl₂, so some permanganate is wasted and titres come out wrong. Use dilute H₂SO₄. HNO₃ is also avoided — it is an oxidant itself.

Solved example 04 — Permanganate and oxalate

Write the balanced ionic equation for KMnO₄ oxidising oxalate in acid, and find the mole ratio.

  1. Reduction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (× 2).
  2. Oxidation: C₂O₄²⁻ → 2CO₂ + 2e⁻ (× 5).
  3. Add: 2MnO₄⁻ + 5C₂O₄²⁻ + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O.

Answer: MnO₄⁻ : C₂O₄²⁻ = 2 : 5. (For Fe²⁺ the ratio is 1 : 5.)

Potassium dichromate, K₂Cr₂O₇

Orange, stable, and pH decides its form.

Preparation from chromite ore (FeCr₂O₄)

  1. Fuse with Na₂CO₃ in air: 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂.
  2. Acidify the yellow chromate: 2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O.
  3. Add KCl: Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl. The potassium salt is less soluble and crystallises out.
2CrO₄²⁻ (yellow) + 2H⁺ ⇌ Cr₂O₇²⁻ (orange) + H₂O

Add acid → orange; add alkali → yellow. Cr is +6 in both: this is not a redox change.

Structure: chromate is tetrahedral; dichromate is two tetrahedra sharing one oxygen (Cr–O–Cr bridge).

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ (green) + 7H₂O

E° = +1.33 V; Cr goes +6 → +3

Solved example 05 — Dichromate titration

How many moles of Fe²⁺ are oxidised by 0.02 mol of K₂Cr₂O₇ in acid?

  1. Each Cr₂O₇²⁻ accepts 6 electrons (two Cr, each +6 → +3).
  2. Each Fe²⁺ gives 1 electron.
  3. Moles Fe²⁺ = 6 × 0.02.

Answer: 0.12 mol of Fe²⁺.

Lanthanoids

Fourteen elements that look almost alike.

Key idea: Lanthanoid contraction: across the series the 4f electrons shield the growing nuclear charge poorly, so atomic and ionic radii fall steadily from La³⁺ to Lu³⁺. Consequences: 4d and 5d elements of a group have almost the same size (Zr ≈ Hf, Nb ≈ Ta), lanthanoids are hard to separate, and basicity falls along the series.

Solved example 06 — Why Ce⁴⁺ oxidises

Ce⁴⁺ is a good oxidising agent in water. Explain using configurations.

  1. Ce is [Xe] 4f¹ 5d¹ 6s²; Ce⁴⁺ is [Xe] 4f⁰ — a noble-gas core, so it can form.
  2. But the usual, most stable state of a lanthanoid in water is +3.
  3. Ce⁴⁺ therefore accepts an electron to return to Ce³⁺ (E° ≈ +1.74 V).

Answer: It is reduced back to the common +3 state.

Actinoids

Like lanthanoids, but radioactive and far more varied.

FeatureLanthanoids (4f)Actinoids (5f)
Oxidation statesMainly +3 (+2, +4 occasionally)+3 common, but up to +6 (U) and +7 (Np, Pu)
Why4f is deeply buried5f, 6d and 7s are close in energy; 5f extends further out
ContractionLanthanoid contractionActinoid contraction — larger from element to element (poorer 5f shielding)
RadioactivityOnly promethiumAll radioactive
Complex formationWeakStronger

Exam trap gallery

Where marks are lost in this chapter.

#What students writeWhat is actually correctWhy
1Fe²⁺ is [Ar] 3d⁴ 4s².Fe²⁺ is [Ar] 3d⁶.4s is filled first but also emptied first.
2Zn is a transition element.Zn is a d-block element but not a transition element.Zn and Zn²⁺ are both d¹⁰.
3Cu⁺ is blue.Cu⁺ (d¹⁰) is colourless; Cu²⁺ (d⁹) is blue.No d–d transition is possible in d⁰ or d¹⁰.
4KMnO₄ is purple because of d–d transitions.Mn in MnO₄⁻ is d⁰; the colour is charge transfer (O → Mn).The same is true of orange dichromate (Cr +6, d⁰).
5KMnO₄ in neutral solution gives Mn²⁺.Neutral or weakly alkaline: MnO₂ (3 e⁻). Acidic: Mn²⁺ (5 e⁻).Always check the medium before counting electrons.
6Acidify KMnO₄ with HCl.Use dilute H₂SO₄.HCl is oxidised to Cl₂, consuming permanganate.
7Chromate → dichromate is a redox reaction.It is an acid–base equilibrium; Cr stays +6.Only the pH changes.
8Basicity of lanthanoid hydroxides increases from La to Lu.It decreases: La(OH)₃ is the most basic.Smaller Ln³⁺ → more covalent M–OH bond.
9The highest oxidation state in the d-block is +7 (Mn).Os and Ru reach +8 (OsO₄).Mn has the most oxidation states in the 3d series, not the highest overall.
Memory hook: “Four-s first in, four-s first out” — ions lose 4s before 3d. · “Acid five, neutral three, alkali one” — electrons gained by MnO₄⁻ in each medium. · “Fourteen H, six e, two Cr” — Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺. · “Zero or ten, no colour then” — d⁰ and d¹⁰ ions are colourless.

Practice MCQs

Levels: Easy, Moderate, Tough.

Q01. The electronic configuration of Cr is: (Easy)

Q02. Which ion is colourless? (Easy)

Q03. Which 3d metal has a positive E°(M²⁺/M)? (Easy)

Q04. The spin-only magnetic moment of Mn²⁺ is: (Easy)

Q05. In acidic medium, one MnO₄⁻ ion gains: (Easy)

Q06. Which pair has almost the same atomic radius? (Moderate)

Q07. Of these 3d metals, the lowest melting point is shown by: (Moderate)

Q08. KMnO₄ in neutral medium oxidises I⁻ to: (Moderate)

Q09. Moles of Fe²⁺ oxidised by one mole of Cr₂O₇²⁻ in acid: (Moderate)

Q10. Which statement about Cu⁺ in water is correct? (Moderate)

Q11. Which is the strongest oxidising agent in water? (Moderate)

Q12. The most basic hydroxide among these is: (Moderate)

Q13. Which lanthanoid commonly shows +4 oxidation state? (Tough)

Q14. The colour of KMnO₄ is due to: (Tough)

Q15. An ion with μ = 2.83 BM could be: (Tough)

Answer key with explanations

  1. (b) [Ar] 3d⁵ 4s¹ — Half-filled 3d⁵ is extra stable.
  2. (c) Sc³⁺ — Sc³⁺ is d⁰.
  3. (d) Cu — Cu²⁺/Cu = +0.34 V.
  4. (c) 5.92 BM — d⁵, five unpaired: √35.
  5. (c) 5 electrons — +7 → +2.
  6. (b) Zr and Hf — Lanthanoid contraction.
  7. (d) Zn — Zn has no unpaired d electrons for metallic bonding.
  8. (b) IO₃⁻ — Neutral/alkaline gives iodate; acidic gives I₂.
  9. (c) 6 — 6 electrons accepted per dichromate.
  10. (b) It disproportionates to Cu and Cu²⁺ — Large hydration enthalpy of Cu²⁺.
  11. (b) Mn³⁺ — Mn³⁺ → Mn²⁺ gives stable d⁵; E° = +1.57 V.
  12. (d) La(OH)₃ — Largest Ln³⁺ ion.
  13. (c) Ce — Ce⁴⁺ is 4f⁰.
  14. (c) charge transfer from O to Mn — Mn(+7) is d⁰.
  15. (b) Ni²⁺ — Ni²⁺ is d⁸: two unpaired electrons.

Check-yourself answers

One-page revision

Read this page the night before the exam.

Configuration

Trends

Oxidation states

Colour and magnetism

KMnO₄

K₂Cr₂O₇ and f-block

Written by Ram Govind Yadav · Chemistry Educator