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JEE Advanced (IIT Entrance) · Chemistry (JEE & NEET)

d and f Block Elements

Transition and inner transition elements, variable oxidation states, colour, magnetic behaviour and lanthanoid and actinoid chemistry.

Nine concepts. Transition-metal behaviour is one story with many readouts: a partly filled d subshell, close ns and (n-1)d energies, and poor d-shielding. Those three facts explain variable oxidation states, colour, magnetism, nearly flat radii, catalysis and interstitial compounds — then the f-block adds lanthanoid contraction (Zr ≈ Hf) and the actinoids' wider oxidation-state range.

  • JEE Advanced (IIT Entrance)
  • Medium level
  • 9 concepts
  • 5 practice questions

1What counts as a transition element

The d-block occupies groups 3–12: the outer configuration builds as (n-1)d^{1-10}ns^{1-2} (sometimes written with ns^{0-2} when an ion has emptied ns). A transition element, in the IUPAC sense used by NCERT, is stricter — it must have a partly filled (n-1)d subshell in the atom or in a common oxidation state. The ns and (n-1)d levels sit close enough in energy that both can participate in bonding, which is why the same metal opens several oxidation states.

Zinc fails that test: Zn is [Ar]3d^{10}4s^{2} and its common ion Zn^{2+} is [Ar]3d^{10}. With no partly filled d in the atom or the ion, Zn is not a typical transition element — colourless aqua ion, and without the variable oxidation states the rest of the series shows. The same d^{10} reasoning excludes Cd and Hg. Scandium still counts: Sc metal is 3d^{1}4s^{2} even though Sc^{3+} is d^{0}.

Figure. d-electron counts in common ions. Ti³⁺, Fe³⁺ and Cu²⁺ sit in d^{1}–d^{9} and pass the transition-metal test; Zn²⁺ at d^{10} fails it. Bar length is the d count, not unpaired electrons.

Apply the criterion

  1. Place the elementGroups 3–12 are the d-block. That is the position, not yet the transition-metal test.
  2. Write atom and common ionFrom the noble-gas core, place (n-1)d and ns. For the ion, empty ns first, then (n-1)d.
  3. Test partly filled dIf neither the atom nor a stable ion has d^{1}–d^{9}, it is d-block but not a typical transition metal (Zn, Cd, Hg).

Why zinc is not a typical transition metal

Explain why zinc is not regarded as a typical transition element, even though it sits in the d-block.

  • Zn atom[Ar]3d^{10}4s^{2}
  • Remove 4s² → Zn²⁺[Ar]3d^{10}
  • Partly filled d in atom or Zn²⁺?No — both are d^{10}
  • Criterion verdictd-block, not a typical transition element

Pro tip. d-block ≠ transition metal. Zn/Cd/Hg sit in the d-block and fail the partly-filled-d test. Sc³⁺ is d^{0} and colourless, but Sc metal itself has 3d^{1}4s^{2}, so scandium still sits in the transition series by the atom criterion.

Which species has a partly filled d subshell?
  1. Zn²⁺
  2. Cu²⁺
  3. Sc³⁺

Cu²⁺ is 3d^{9}. Zn²⁺ is 3d^{10} and Sc³⁺ is 3d^{0} — neither is partly filled.

2Chromium and copper configurations

Across the 3d series the expected build is 3d^{n}4s^{2}, but chromium and copper break it for extra stability of a half-filled or filled d subshell: Cr=[Ar]3d^{5}4s^{1} and Cu=[Ar]3d^{10}4s^{1}.

The exchange-energy gain from maximising unpaired d^{5} spins, or from a closed d^{10} shell, outweighs the cost of leaving 4s with only one electron. Every other 3d metal keeps 4s^{2} in the ground atom. Mo and Ag show the same pattern in the later series — memorise Cr and Cu first for JEE Main.

Figure. Exchange energy wins: chromium prefers half-filled 3d⁵ with 4s¹ over 3d⁴ 4s²; copper prefers filled 3d¹⁰ with 4s¹ over 3d⁹ 4s². Expected-versus-observed is the exam fork.

Catch the exceptions

  1. Write the naive configCr would be 3d^{4}4s^{2}; Cu would be 3d^{9}4s^{2}.
  2. Move one 4s electron into 3dCr becomes 3d^{5}4s^{1}; Cu becomes 3d^{10}4s^{1}.
  3. Name the stabilityHalf-filled d^{5} and filled d^{10} are the drivers — memorise the two atoms, not a fake third.

Expected versus observed

Write the expected and observed ground-state configurations of Cr and Cu.

  • Cr expected from Aufbau[Ar]3d^{4}4s^{2}
  • Cr observed (d^{5} half-filled)[Ar]3d^{5}4s^{1}
  • Cu expected from Aufbau[Ar]3d^{9}4s^{2}
  • Cu observed (d^{10} filled)[Ar]3d^{10}4s^{1}

Pro tip. When forming cations, still empty 4s before 3d: Cr → Cr⁺ is 3d^{5}, and Cu → Cu²⁺ is 3d^{9}. The atom anomaly does not change the ionisation order.

The ground-state configuration of Cr is
  1. [Ar]3d^{4}4s^{2}
  2. [Ar]3d^{5}4s^{1}
  3. [Ar]3d^{6}4s^{0}

Chromium adopts 3d^{5}4s^{1} for the half-filled d subshell. 3d^{4}4s^{2} is the Aufbau expectation that it leaves.

4Variable oxidation states in the 3d series

Because ns and (n-1)d energies are close, transition metals show several oxidation states by losing different numbers of those electrons. In the 3d series the span is widest at manganese: Mn shows +2 through +7. The highest oxidation state peaks at Mn (+7), then falls again toward Zn. Early in the series the highest OS often equals the count of ns+(n-1)d electrons (Ti +4, V +5, Cr +6, Mn +7); later metals prefer lower states in aqueous chemistry.

Relative stability shifts across the series: the +2 state becomes more common and more stable toward the end (Mn²⁺, Fe²⁺, Co²⁺, Ni²⁺, Cu²⁺), while higher states early on are strong oxidisers when they can fall to a more stable lower OS. Higher oxidation states are usually found in oxides and oxoanions (MnO_{4}^{-}, Cr_{2}O_{7}^{2-}), where oxygen stabilises the metal at high OS. In acid medium those oxoanions are the standard strong oxidisers of this chapter.

Figure. Highest oxidation state across the 3d metals (NCERT ceiling, not the everyday aqueous OS). The bar for Mn is the series maximum at +7; after Mn the ceiling falls toward Zn. Lengths are the OS numbers themselves.

Reading an oxoanion oxidiser

  1. Assign oxygenEach O is −2 in these oxo species.
  2. Balance the chargeMetal OS = total charge − (sum of O contributions).
  3. Place it on the seriesMn(+7) and Cr(+6) sit at the high-OS end of the 3d peak; Fe rarely exceeds +3 in simple aqueous chemistry. In acid, MnO_{4}^{-} gains 5 e⁻ to Mn²⁺ and Cr_{2}O_{7}^{2-} gains 6 e⁻ to Cr³⁺.
High-OS oxo oxidisers (acid medium)
SpeciesMetal OSAcid reductione⁻ gained
MnO₄⁻+7→ Mn²⁺5
Cr₂O₇²⁻+6→ Cr³⁺6
CrO₄²⁻+6Same Cr(+6); alkaline form— (equilibrates with dichromate)
FeO₄²⁻+6Rare; Fe usually stops at +3—

Oxidation state and electrons transferred

Find the oxidation number of Mn in MnO_{4}^{-} and of Cr in Cr_{2}O_{7}^{2-}, then state how many electrons each ion gains when reduced to Mn²⁺ and Cr³⁺ in acid.

  • MnO₄⁻: Mn + 4(−2) = −1Mn = +7
  • Cr₂O₇²⁻: 2Cr + 7(−2) = −2Cr = +6 each
  • Mn(+7) → Mn²⁺ in acidgains 5 e⁻ per Mn
  • 2 Cr(+6) → 2 Cr³⁺ in acidgains 6 e⁻ per dichromate

Pro tip. The +7 in permanganate is the top of the 3d ladder — nothing past Mn reaches it in this series. Acid-medium electron counts (5 for MnO₄⁻, 6 for Cr₂O₇²⁻) are as examinable as the oxidation numbers themselves.

Across the 3d series, the highest oxidation state peaks at
  1. Cr (+6)
  2. Mn (+7)
  3. Fe (+6)

Mn reaches +7 (as in MnO₄⁻). Cr's highest common OS is +6; Fe does not match Mn's +7 in ordinary chemistry.

5Colour from d–d transitions

Many transition-metal ions are coloured because an electron can be promoted between split d orbitals — a d–d transition — by visible light. That requires a partly filled d subshell: between d^{1} and d^{9}. The colour you see is complementary to the wavelengths absorbed; ligands change the splitting, so the same metal ion can look different in different complexes.

Ions with d^{0} (Sc^{3+}, Ti^{4+}) or d^{10} (Zn^{2+}, Cu^{+}) have no d–d promotion available, so simple aqua ions of those counts are colourless. d–d colour is a readout of the partly-filled-d criterion — not a licence to call every empty-d oxoanion colourless, because charge-transfer still paints MnO_{4}^{-} purple and Cr_{2}O_{7}^{2-} orange.

Figure. d–d colour needs a partly filled d set: d¹–d⁹ ions can absorb visible light; d⁰ and d¹⁰ cannot. The bar is the electron count criterion — pigment itself is outside this vocabulary.

Predict colourlessness

  1. Get the d countWrite the ion configuration after removing ns first.
  2. Ask d^{1}–d^{9}?If yes, d–d transitions are possible and the ion is typically coloured in complexes.
  3. If d^{0} or d^{10}No d–d transition — expect colourless aqua ions (Sc^{3+}, Zn^{2+}). Still check for charge-transfer colour in oxoanions.
Colourless versus coloured
Iond countd–d colour
Sc³⁺d^{0}Colourless
Ti³⁺d^{1}Coloured
Fe³⁺d^{5}Coloured
Cu²⁺d^{9}Coloured
Zn²⁺d^{10}Colourless

Which ions are colourless?

Classify Sc^{3+}, Ti^{3+}, Cu^{2+} and Zn^{2+} as coloured or colourless from their d counts.

  • Sc³⁺ = [Ar] → d^{0}colourless
  • Ti³⁺ = [Ar]3d^{1}coloured (d^{1})
  • Cu²⁺ = [Ar]3d^{9}coloured (d^{9})
  • Zn²⁺ = [Ar]3d^{10}colourless

Pro tip. Charge-transfer colour can still appear in some d^{0} oxo species (MnO₄⁻ is purple without a d–d band). For JEE Main stems that name Sc^{3+}/Zn^{2+}, the expected answer is still colourless from the d–d rule.

Which ion is colourless because it is d^{10}?
  1. Cu²⁺
  2. Fe³⁺
  3. Zn²⁺

Zn²⁺ is 3d^{10}. Cu²⁺ is 3d^{9} and Fe³⁺ is 3d^{5} — both partly filled and typically coloured in complexes.

6Spin-only magnetic moment

Paramagnetism of first-row transition-metal ions is estimated by the spin-only formula \mu=\sqrt{n(n+2)} BM, where n is the number of unpaired electrons. More unpaired electrons mean a larger moment and stronger paramagnetism. Orbital contribution is usually neglected for these 3d aqua ions in JEE Main stems.

The fingerprint values are worth knowing cold: \mu\approx 1.73 (1e), 2.83 (2e), 3.87 (3e), 4.90 (4e), 5.92 (5e). Matching a measured \mu to n is a standard JEE move — and it only works after you have the correct ion configuration (empty ns first).

Figure. Spin-only μ against unpaired-electron count. Each bar height is \sqrt{n(n+2)} in BM — the fingerprint list 1.73, 2.83, 3.87, 4.90, 5.92. Fe³⁺ high-spin sits on the n=5 bar.

From ion to μ

  1. Write the ion configEmpty ns before (n-1)d when ionising the atom.
  2. Count unpaired electronsFor high-spin octahedral 3d ions, maximise unpaired d electrons (Hund).
  3. Apply spin-only\mu=\sqrt{n(n+2)} in Bohr magnetons; compare with the 1.73–5.92 fingerprint list.
μ fingerprint
n (unpaired)μ / BMExample ion (high-spin)
11.73Ti³⁺ (d^{1})
22.83V³⁺ (d^{2})
33.87Cr³⁺ (d^{3})
44.90Fe²⁺ (d^{6})
55.92Fe³⁺ / Mn²⁺ (d^{5})

Magnetic moment of Fe³⁺

Calculate the spin-only magnetic moment of Fe^{3+}.

  • Fe atom[Ar]3d^{6}4s^{2}
  • Fe³⁺ (empty 4s, then one 3d)[Ar]3d^{5}
  • Unpaired electrons in high-spin d^{5}n=5
  • \mu=\sqrt{5(5+2)}=\sqrt{35}5.92 BM

Pro tip. Remove 4s electrons first when forming the cation — Fe → Fe³⁺ is 3d^{5}, not a 4s-still-occupied mistake that would give the wrong n. \sqrt{35}=5.916\ldots, reported as 5.92 BM.

A spin-only moment of 3.87 BM corresponds to how many unpaired electrons?
  1. 2
  2. 3
  3. 4

\sqrt{3(5)}=\sqrt{15}\approx 3.87. Two unpaired give 2.83; four give 4.90.

7Catalysis, interstitial compounds and alloys

Three further general characteristics follow from the same electronic and size facts. Transition metals and their compounds are often catalysts because vacant d orbitals and variable oxidation states let them bind reactants, transfer electrons, and release products — classic examples are Fe in Haber's process, V₂O₅ in contact process, and Ni in hydrogenation. Interstitial compounds form when small atoms (H, C, N) occupy voids in the metal lattice; the products are typically hard, high-melting and retain metallic conductivity (steel, TiC, WN).

Alloys form readily because the metallic radii of neighbouring transition metals are similar, so atoms substitute in the lattice without large strain — brass (Cu/Zn), bronze (Cu/Sn) and stainless steels are the exam staples. None of these three properties needs a new electronic rule; they are consequences of partly filled d, small size and the mid-series radius plateau.

Figure. Three uses of the same open d lattice: adsorb and activate reactants (catalysis), host small atoms in voids (interstitial compounds), or swap metal atoms (alloys such as brass). Property follows which guest sits where.

Match property to cause

  1. CatalysisAsk whether vacant d / variable OS / surface adsorption can activate the reactants — that is the reason, not 'transition metals are catalysts' as a memorised label.
  2. InterstitialSmall non-metal in lattice voids → hard, high melting point; stoichiometry need not be simple.
  3. AlloySimilar metallic radii → solid solution or substitutional alloy with little lattice strain.
Property → reason
PropertyPhysical causeExam example
CatalysisVacant d; variable OS; surface adsorptionFe (Haber), V₂O₅ (contact), Ni (hydrogenation)
Interstitial compoundsH/C/N in lattice voidsSteel, TiC — hard, high MP
Alloy formationSimilar metallic radii → easy substitutionBrass (Cu/Zn), bronze (Cu/Sn)
Transition metals form interstitial compounds with H, C or N mainly because
  1. They have very large atomic radii compared with s-block metals
  2. Small non-metal atoms fit in the voids of the metal lattice
  3. Their ions are always d^{10} and colourless

H, C and N are small enough to occupy interstitial voids. The products are characteristically hard and high-melting.

8Lanthanoids and lanthanoid contraction

Lanthanoids are the 4f series (La–Lu). The general outer configuration is [Xe]4f^{1-14}5d^{0-1}6s^{2}. The most stable oxidation state is +3 across the series; notable exceptions that stems love are Ce⁴⁺ (empty f^{0}), Eu²⁺ and Yb²⁺ (half-filled f^{7} and filled f^{14}). Many Ln³⁺ ions are coloured and paramagnetic when f electrons remain unpaired, but f–f bands are often pale compared with d–d colours.

Across the 4f series the ionic radius of Ln³⁺ falls steadily from La³⁺ to Lu³⁺. The 4f electrons shield nuclear charge poorly, so Z_{\mathrm{eff}} on the outer shells rises and the atoms shrink — the lanthanoid contraction. The practical consequence is that 4d and 5d elements of the same group end up nearly the same size: Zr ≈ Hf and Nb ≈ Ta in radius and in much of their chemistry.

Figure. Shannon CN6 Ln³⁺ radius against Z from La (57) to Lu (71), windowed on 80–110 pm so the 17.1 pm drop is the visible fall. Pm is omitted (no stable isotope). The same shrinkage is why Zr ≈ Hf.

Why Zr matches Hf

  1. Fill 4f from La to LuFourteen elements insert between La and Hf in Z; config pattern [Xe]4f^{1-14}5d^{0-1}6s^{2}.
  2. Poor f-shielding shrinks the atomEach added proton is incompletely cancelled; Ln³⁺ radius falls across the series. +3 remains the default OS.
  3. 5d catches 4dThe contraction offsets the usual down-group size increase, so Hf is almost the size of Zr (and Ta of Nb).
Lanthanoid OS anchors
Speciesf count ideaWhy it is stable
Ln³⁺ (general)4f^{n}Default across the series
Ce⁴⁺f^{0}Empty f shell
Eu²⁺f^{7}Half-filled f
Yb²⁺f^{14}Filled f

Size drop La³⁺ → Lu³⁺

Shannon six-coordinate radii are La³⁺ = 103.2 pm and Lu³⁺ = 86.1 pm. How large is the lanthanoid contraction across the series, and what pair of d-block neighbours does it equalise?

  • r(La³⁺) − r(Lu³⁺)103.2 − 86.1 = 17.1 pm
  • CausePoor 4f shielding → rising Z_eff
  • 4d/5d consequenceZr ≈ Hf (also Nb ≈ Ta)
  • Lanthanoid OS check+3 most stable; Ce⁴⁺ / Eu²⁺ / Yb²⁺ are the named exceptions

Pro tip. If a stem links 'nearly identical properties of Zr and Hf' to a cause, the answer is lanthanoid contraction — not inert pair, not diagonal relationship.

Zr and Hf have nearly identical radii mainly because of
  1. The inert-pair effect
  2. Lanthanoid contraction
  3. Actinoid contraction only

Poor 4f shielding shrinks the atoms before Hf, cancelling the usual group size jump so Zr ≈ Hf. Inert pair is a p-block oxidation-state effect.

9Actinoids versus lanthanoids

Actinoids are the 5f series (Ac–Lr). The 5f, 6d and 7s energies lie closer together than their 4f/5d/6s counterparts, so actinoids show a wider range of oxidation states than the lanthanoids — especially the early members U, Np, Pu. All actinoids are radioactive, and the later members increasingly resemble lanthanoids in preferring +3.

There is an actinoid contraction parallel to the lanthanoid one — poor 5f shielding shrinks the ions across the series — but the headline exam contrasts remain: more oxidation states, radioactivity, and the early actinoids' richer redox chemistry. Do not credit actinoid contraction alone for Zr ≈ Hf; that pair is the lanthanoid-contraction story.

Figure. Actinoids stretch oxidation states far past the lanthanoids' stubborn +3, and 5f shielding is even poorer than 4f — so chemistry is messier. The lanthanoid-contraction plot already shows what a contraction looks like; this chart is the three exam contrasts.

Lanthanoid or actinoid stem

  1. Name the series4f → lanthanoids; 5f → actinoids.
  2. Oxidation-state widthLanthanoids are dominated by +3; actinoids open more states because 5f/6d/7s are closer in energy, especially early in the series.
  3. Radioactivity and contractionAll actinoids are radioactive; actinoid contraction is the 5f analogue of lanthanoid contraction — it does not replace lanthanoid contraction as the Zr/Hf cause.
Lanthanoids versus actinoids
FeatureLanthanoids (4f)Actinoids (5f)
Most common OS+3 dominatesWider range; early members vary more
Why OS width differs4f well buried; less bonding role5f/6d/7s closer in energy
RadioactivityMostly stable isotopes in natureAll radioactive
ContractionLanthanoid contraction → Zr ≈ HfActinoid contraction (analogous); later members resemble Ln³⁺
Compared with lanthanoids, actinoids characteristically
  1. Show fewer oxidation states and are non-radioactive
  2. Show more oxidation states and are radioactive
  3. Have no f-contraction analogue

Actinoids open a wider OS range and are radioactive. Actinoid contraction is real and parallels the lanthanoid effect.

Notes

  • Transition metals have partly filled (n-1)d orbitals; their characteristic variable oxidation states arise because ns and (n-1)d energies are close (Mn shows +2 to +7).
  • Colour of their ions comes from d-d electronic transitions, so ions with d^0 (Sc^{3+}) or d^{10} (Zn^{2+}) are colourless.
  • Magnetic behaviour: the spin-only moment \mu=\sqrt{n(n+2)} BM (n = unpaired electrons); more unpaired electrons mean stronger paramagnetism.
  • Lanthanoid contraction: a steady size decrease across the 4f series (poor f-shielding) makes 4d and 5d elements of the same group nearly identical in size (Zr ≈ Hf).
  • Actinoids show more oxidation states, are radioactive, and later members resemble lanthanoids (actinoid contraction).
  • Transition metals show variable oxidation states because (n-1)d and ns electrons have similar energies; the maximum oxidation state rises to the middle of the series (Mn +7) then falls. Higher oxidation states are stabilised by O and F (small, electronegative) as in MnO4-, CrO4^2-, OsO4.
  • Colour arises from d-d transitions (partially filled d orbitals) whose energy depends on the crystal-field splitting; the colour observed is complementary to the wavelength absorbed. d0 (Sc3+, Ti4+) and d10 (Zn2+, Cu+) ions are colourless.
  • Magnetic moment follows the spin-only formula mu = sqrt(n(n+2)) BM where n = unpaired electrons; deviations occur when orbital contribution is significant. This lets you deduce the number of unpaired electrons and hence oxidation state and geometry.
  • The lanthanoid contraction (steady decrease in size across the 4f series due to poor 4f shielding) makes second- and third-row transition metals of the same group nearly identical in size (Zr ~ Hf, Nb ~ Ta), explaining their difficult separation and similar chemistry.
  • Catalytic activity of transition metals and their compounds is linked to variable oxidation states and the ability to form intermediates/adsorb reactants (e.g., V2O5 in contact process, Fe in Haber process, Ni in hydrogenation).

Formulas

  • Spin-only magnetic moment \mu=\sqrt{n(n+2)} BM
  • General configuration: (n-1)d^{1-10}ns^{0-2}
  • Cr=[Ar]3d^5 4s^1,\quad Cu=[Ar]3d^{10}4s^1 (extra stability)
  • Highest oxidation state in the 3d series peaks at Mn (+7)
  • Most stable oxidation state of lanthanoids is +3
  • \mu_{spin-only} = \sqrt{n(n+2)}\,\text{BM}\;(n = \text{unpaired } e^-)
  • \text{CFSE}_{oct} = (-0.4\,n_{t_{2g}} + 0.6\,n_{e_g})\Delta_o + nP
  • 2\,MnO_4^- + 5\,C_2O_4^{2-} + 16\,H^+ \rightarrow 2\,Mn^{2+} + 10\,CO_2 + 8\,H_2O
  • Cr_2O_7^{2-} + 14\,H^+ + 6\,e^- \rightarrow 2\,Cr^{3+} + 7\,H_2O\;(E^{\circ}=+1.33\,V)

Exam traps & shortcuts

  • For colour you need a partially filled d subshell (between d^1 and d^9); d^0 and d^{10} ions are colourless.
  • Unpaired-electron count from spin-only moment: \mu\approx1.73 (1e), 2.83 (2e), 3.87 (3e), 4.90 (4e), 5.92 (5e).
  • Zr/Hf and Nb/Ta have almost identical radii and properties due to the lanthanoid contraction.
  • Use the spin-only formula in reverse: a measured moment of ~5.9 BM means 5 unpaired electrons (high-spin d5, e.g., Mn2+ or Fe3+); ~1.73 BM means 1 unpaired electron.
  • E(standard) for M2+/M shows irregular trends explained by sublimation, ionisation and hydration enthalpies; the unusually low value for Mn2+/Mn and high stability of Mn2+ come from its half-filled d5 configuration.
  • For separating Zr and Hf, exploit tiny differences from lanthanoid contraction (ion exchange/solvent extraction); chemically they are almost identical.

Reference tables

Atom configs that stems treat as fixed facts.

3d configuration anchors
SpeciesConfigurationWhy it matters
Cr[Ar]3d^{5}4s^{1}Half-filled d exception
Cu[Ar]3d^{10}4s^{1}Filled d exception
Fe³⁺[Ar]3d^{5}n=5 → μ=5.92 BM
Zn / Zn²⁺3d^{10}4s^{2} / 3d^{10}Not a typical transition metal

Same d count drives both colour and spin-only μ.

Colour and magnetism quick map
d countd–d colourMax unpaired (high-spin)μ / BM
d^{0}Colourless00
d^{1}–d^{4}Coloured1–41.73–4.90
d^{5}Coloured55.92
d^{6}–d^{9}Coloured4–14.90–1.73
d^{10}Colourless00

Reconstructible from the concepts above.

Formula sheet
RelationReads asWatch for
(n-1)d^{1-10}ns^{1-2}General d-block configPartly filled d in atom or ion → transition metal
Cr 3d^{5}4s^{1}; Cu 3d^{10}4s^{1}Atom exceptionsStill empty 4s first in ions
3d radii nearly flat mid-seriesPoor 3d shielding ≈ rising ZNot lanthanoid contraction
\mu=\sqrt{n(n+2)} BMSpin-only momentFingerprint 1.73…5.92
Highest OS peaks at Mn (+7)3d oxidation-state maximumMnO₄⁻ / Cr₂O₇²⁻ acid e⁻ counts 5 / 6
Lanthanoid contractionLn³⁺ shrinks La → LuZr ≈ Hf; Nb ≈ Ta; Ln OS mostly +3
Actinoids vs lanthanoidsMore OS; all radioactive5f/6d/7s closer in energy

Recap

Read only this the night before.

Transition test
Partly filled (n-1)d in the atom or a common ion. Zn/Cd/Hg fail — d^{10} throughout. d-block ≠ transition metal.
Cr and Cu
Cr is 3d^{5}4s^{1}; Cu is 3d^{10}4s^{1}. Half-filled and filled d win over Aufbau.
Radii and IE
3d radii drop then flatten (poor d-shielding). IE₁ rises irregularly; Zn is highest (3d^{10}4s^{2}).
Oxidation states
Close ns and (n-1)d energies → variable OS. Ceiling peaks at Mn (+7). +2 grows more stable toward Zn. Acid: MnO₄⁻ +5 e⁻, Cr₂O₇²⁻ +6 e⁻.
Colour and magnetism
d–d colour needs d^{1}–d^{9}. Aqua d^{0}/d^{10} colourless; MnO₄⁻ is charge transfer. \mu=\sqrt{n(n+2)}; Fe³⁺ high-spin → 5.92 BM.
Catalysis / interstitial / alloys
Vacant d + variable OS → catalysts. H/C/N in voids → hard interstitials. Similar radii → alloys.
f-block
Ln: [Xe]4f^{1-14}5d^{0-1}6s^{2}, +3 default, contraction → Zr ≈ Hf. Actinoids: more OS, radioactive, actinoid contraction.

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