Chemistry Notes

The d- and f-Block Elements — NEET Notes

The middle two and a half rows of the periodic table — the d-block transition metals and the f-block lanthanoids and actinoids — contain the metals that built civilisation: iron, copper, silver, gold, chromium, manganese, tungsten, uranium. They share a fingerprint: partly filled (n−1)d or n f orbitals. From that single structural fact flow every characteristic property NEET examines — variable oxidation states, intensely coloured ions, paramagnetism, catalytic power, hard high-melting alloys, and the puzzling near-identity of zirconium and hafnium. This chapter is asked 2–3 times every NEET, with K₂Cr₂O₇, KMnO₄, magnetic moment calculations and lanthanoid contraction as perennial favourites. By the end you should be able to write the spin-only formula from memory, manufacture K₂Cr₂O₇ from chromite on paper, and explain why Zr (160 pm) and Hf (159 pm) are inseparable.

Position in the periodic table — four d-series, two f-series

The d-block sits in the broad middle of the periodic table, spanning groups 3 to 12 and flanked by the s- and p-blocks. Across each long period the (n−1)d orbitals of the penultimate shell are filled progressively, generating four horizontal series: 3d (Sc to Zn), 4d (Y to Cd), 5d (La and Hf to Hg) and 6d (Ac and Rf to Cn). The f-block elements — those in which 4f or 5f orbitals are being filled — are pulled out into the panel at the bottom: the lanthanoids (Ce to Lu) and the actinoids (Th to Lr).

IUPAC's modern definition pins the term transition element to a structural criterion: an element whose neutral atom or any of its common ions has an incomplete d sub-shell. By that yardstick Zn, Cd, Hg (and the synthetic Cn) of group 12 do not qualify — both their ground-state atoms and their common +2 ions have a fully filled d¹⁰ configuration. They are studied alongside the transition metals only because they are the end members of the 3d, 4d and 5d series.

Electronic configuration — (n−1)d¹⁻¹⁰ ns⁰⁻² and two anomalies

The outer electronic configuration of a d-block element is written as (n−1)d¹⁻¹⁰ ns⁰⁻². Here the inner d sub-shell may hold anywhere from one to ten electrons and the outermost s sub-shell holds zero, one or two. The (n−1)d and ns sub-shells lie so close in energy that small effects — exchange stabilisation, symmetrical occupation, electron-pair repulsion — can swap an electron between them. Two such swaps appear in the 3d series and are NEET staples.

The two 3d-series anomalies

Chromium (Z = 24)

[Ar] 3d⁵ 4s¹

not [Ar] 3d⁴ 4s²

  • Half-filled 3d⁵ is extra-stable
  • Exchange energy maximised
  • Symmetrical electron distribution
  • Small 3d–4s gap allows the swap

Copper (Z = 29)

[Ar] 3d¹⁰ 4s¹

not [Ar] 3d⁹ 4s²

  • Fully filled 3d¹⁰ is extra-stable
  • All d-electrons paired symmetrically
  • Maximum exchange in 3d set
  • One 4s electron supplied to fill 3d

The same logic gives palladium (4d¹⁰ 5s⁰), Ag (4d¹⁰ 5s¹), Au (5d¹⁰ 6s¹). Two consequences matter for NEET: when these atoms form ions, ns electrons leave before (n−1)d electrons, so Cr⁰ (3d⁵ 4s¹) → Cr²⁺ is 3d⁴ (not 3d³ 4s¹). And the general formula for Zn, Cd, Hg becomes (n−1)d¹⁰ ns² — completely filled in both atom and common +2 ion, which is why they fail the transition-element test.

Five characteristic properties — the d-electron fingerprint

Partly filled d orbitals are the structural reason behind every property NEET expects you to recite for transition metals. The d orbitals extend further into space than s or p of the same shell, so they interact with surroundings and are themselves perturbed by ligands. NCERT highlights five outcomes that come up again and again.

Master rule: all five properties below follow from one fact — a transition metal has unpaired d electrons. Remove the d-electrons (Sc³⁺, Zn²⁺, Cu⁺) and the colour, paramagnetism and catalytic versatility disappear with them.

1 · Variable OS

+2 to +7

Mn shows all seven

Differ by unity (not by two as in p-block). Maximum OS = sum of (n−1)d + ns electrons up to Mn (Mn²⁺O₄⁻ in Mn₂O₇).

PYQ: highest oxide group correspondence

2 · Coloured ions

d–d

visible-region transition

Cu²⁺ blue, Fe²⁺ green, Fe³⁺ yellow. d⁰ (Sc³⁺) and d¹⁰ (Zn²⁺, Cu⁺) — colourless.

NEET trap: d⁰ and d¹⁰ are colourless

3 · Paramagnetism

μ = √n(n+2)

spin-only, in BM

Unpaired electrons give the moment. Pd, Zn²⁺, Cu⁺ — no unpaired — diamagnetic.

PYQ 2020: Cr²⁺ → 4.90 BM

4 · Catalysis

V₂O₅, Fe, Ni

Contact, Haber, hydrogenation

Multiple OS and complex formation let the metal accept and donate electrons in steps. Fe³⁺ catalyses 2I⁻ + S₂O₈²⁻.

5 · Interstitials

TiC, Mn₄N

non-stoichiometric

Small atoms (H, C, N) trapped in metal lattice. Hard, high-melting, retain conductivity, chemically inert.

A sixth, related property is alloy formation. Transition-metal atoms have metallic radii within ~15% of each other, so they form homogeneous solid solutions — chromium, vanadium, tungsten, molybdenum and manganese all alloy with iron to give the steels and stainless steel. Brass (Cu–Zn) and bronze (Cu–Sn) extend the family to non-transition partners.

Variable oxidation states — Mn shows seven

Among transition elements, the elements in or near the middle of each series show the greatest number of oxidation states because they have both unpaired electrons available for sharing and empty orbitals to share into. Manganese exhibits all states from +2 to +7. At the extremes, Sc shows only +3 (too few d-electrons), and Zn only +2 (no d-electron involvement at all). Variable states in the d-block differ from each other by unity, whereas p-block states differ in steps of two — a contrast NEET tests directly.

Down a group, the heavier members favour the higher oxidation state — the opposite of the p-block inert-pair effect. Thus Mo(VI) and W(VI) are more stable than Cr(VI); chromium in dichromate is a strong oxidiser, but MoO₃ and WO₃ are not. The highest oxide of a 3d metal traces the group number up to manganese: Sc₂O₃, TiO₂, V₂O₅, CrO₃, Mn₂O₇ — beyond which the pattern breaks (Fe stops at Fe₂O₃ in stable oxides).

Magnetic moment — the spin-only formula

The d orbital electrons carry a magnetic moment derived from spin angular momentum. For first-row transition-metal complexes the orbital contribution is effectively quenched, so the observed magnetic moment depends only on the count of unpaired electrons. NCERT formalises this as the spin-only formula:

μ = √n(n+2) BM

Spin-only magnetic moment

n = number of unpaired electrons. A single unpaired electron gives 1.73 BM; five unpaired (d⁵ high-spin, e.g. Mn²⁺) give √35 ≈ 5.92 BM. Bohr magneton (BM) is the unit. Tested in NEET 2018 and 2020.

Spin-only magnetic moment — four-step computation

Worked for Cr²⁺ (3d⁴, four unpaired)
  1. Step 1

    Write atomic Z

    For Cr²⁺: Cr is Z = 24, configuration [Ar] 3d⁵ 4s¹. Remove 2 electrons.

    Cr⁰ → Cr²⁺
  2. Step 2

    Lose ns first

    Strip the 4s electron, then one 3d electron. Result: 3d⁴.

    Configuration of ion
  3. Step 3

    Count unpaired

    Hund's rule: four electrons fill four d-orbitals singly. n = 4.

    All parallel spins
  4. Step 4

    Apply formula

    μ = √(4 × 6) = √24 ≈ 4.90 BM. Matches observed.

    NEET 2020 answer

The procedure for any d-block ion: write Z, remove ns electrons before (n−1)d, count the unpaired d-electrons, then plug into μ = √n(n+2). This handles every magnetic-moment PYQ — Co³⁺ (3d⁶, n = 4, μ = √24), Cr³⁺ (3d³, n = 3, μ = √15), Fe³⁺ (3d⁵, n = 5, μ = √35), Ni²⁺ (3d⁸, n = 2, μ = √8) — all tested in NEET 2018 Q.87.

Colour & d–d transitions

Free, gas-phase d orbitals are degenerate — equal in energy. In a complex, the ligands' electric field splits them into two sets separated by an energy gap Δ. An electron from the lower set can absorb a photon of energy hν = Δ — usually in the visible range — and jump to the higher set. The colour seen is the complementary colour of the wavelength absorbed. This is the d–d transition that NCERT lays out in Table 4.8: Ti³⁺ purple (3d¹), V³⁺ green (3d²), Cr³⁺ violet (3d³), Mn²⁺ pink (3d⁵), Fe³⁺ yellow (3d⁵), Cu²⁺ blue (3d⁹).

Ions with no d-electrons (Sc³⁺, Ti⁴⁺, d⁰) or with a fully filled d sub-shell (Zn²⁺, Cu⁺, d¹⁰) cannot undergo d–d transitions and are colourless. NEET 2018 Q.84 hinged on this: of CrO₄²⁻, Cr₂O₇²⁻, MnO₄⁻ and MnO₄²⁻, only MnO₄²⁻ (Mn⁶⁺, 3d¹) shows both d–d transition and paramagnetism — the rest are d⁰. MnO₄⁻ and Cr₂O₇²⁻ are coloured by a different mechanism (charge-transfer, beyond the NEET syllabus), but their d⁰ centres are diamagnetic.

Catalysis, interstitials & alloys

The catalytic prowess of transition metals follows directly from the two features above. Variable oxidation states let the metal accept electrons in one step and donate them in the next, completing the cycle without being consumed. V₂O₅ catalyses the contact process (SO₂ → SO₃), finely divided iron drives Haber's process (N₂ + H₂ → NH₃), and nickel hydrogenates alkenes (the basis of vanaspati). Solid catalysts work because the metal surface forms loose bonds with reactant molecules, raising local concentration and weakening internal bonds — lowering activation energy.

Interstitial compounds arise when small non-metal atoms — H, C, N — slot into the gaps between metal atoms in the crystal lattice. They are non-stoichiometric (TiC, Mn₄N, Fe₃H, VH₀.₅₆, TiH₁.₇), keep the metallic conductivity of the host, and pick up extreme hardness and high melting points. Cementite (Fe₃C) in steel is the most famous example — it is the reason a steel blade outlasts pure iron.

Within a series, the atomic and ionic radii decrease only slightly from left to right because the added d-electron shields the outer ns electrons rather imperfectly — net effective nuclear charge grows, but slowly. Down a group the radius increases from 3d to 4d as expected, but then the 5d radii are nearly identical to the 4d. The cause is the lanthanoid contraction: 14 f-electrons are squeezed in between La and Hf, and 4f shielding is poorer than 3d shielding, so the size shrinks across the whole 4f row and that shrinkage cancels the expected size increase to the 5d series.

Melting points climb sharply across the 3d series, peak at d⁵ (Cr, Mo, W tops the chart with the highest m.p. of any metal at 3683 K), then fall. The cause is metallic bonding: in 3d metals, both 3d and 4s electrons contribute to the lattice; with one unpaired electron per d orbital (d⁵) the bonding electron count is maximised. Mn and Tc are local anomalies — half-filled 3d⁵ 4s² gives stable atoms that bond less strongly than their neighbours. Enthalpies of atomisation track melting points exactly.

Potassium dichromate K₂Cr₂O₇

Potassium dichromate is one of the two inorganic compounds NEET tests at every cycle. Cr is in the +6 oxidation state — its highest. Industrial preparation starts with the chromite ore FeCr₂O₄ and proceeds in three named steps.

K₂Cr₂O₇ from chromite — three-step manufacture

Chromium stays in +6 throughout
  1. Step 1

    Oxidative fusion

    Chromite + Na₂CO₃ + O₂ → Na₂CrO₄ (yellow chromate) + Fe₂O₃ + CO₂. Carried out in free access of air.

    Cr stays +6
  2. Step 2

    Acidify

    Filter; treat with H₂SO₄. Yellow chromate → orange dichromate. 2 CrO₄²⁻ + 2 H⁺ → Cr₂O₇²⁻ + H₂O.

    Crystallise as Na₂Cr₂O₇·2H₂O
  3. Step 3

    Convert to K-salt

    Add KCl. K₂Cr₂O₇ is less soluble than the sodium salt and crystallises out as orange crystals.

    Na₂Cr₂O₇ + 2 KCl → K₂Cr₂O₇ + 2 NaCl

The chromate ion CrO₄²⁻ is tetrahedral; the dichromate ion Cr₂O₇²⁻ is two CrO₄ tetrahedra sharing one corner oxygen with a Cr–O–Cr bond angle of 126°. The two ions are interconvertible in solution by pH — acid favours dichromate, alkali favours chromate. The oxidation state of Cr is +6 in both.

In acidic medium, K₂Cr₂O₇ is a powerful oxidant. The half-reaction is the one to memorise:

This drives the standard list of NEET reactions: iodide → iodine, ferrous → ferric (basis of dichrometry), Sn²⁺ → Sn⁴⁺, H₂S → S (precipitate), and acidified K₂Cr₂O₇ with SO₂ gives green Cr₂(SO₄)₃ — the 2016 PYQ. Potassium dichromate is used as a primary standard in volumetric analysis precisely because it is non-hygroscopic and the orange-to-green colour change is unmistakable.

Potassium permanganate KMnO₄

The other compound NEET keeps returning to is potassium permanganate — manganese in its highest state, +7. NCERT's laboratory preparation is a two-step sequence starting from manganese dioxide.

KMnO₄ — two-step preparation

Mn(IV) → Mn(VI) → Mn(VII)
  1. Step 1

    Alkaline oxidative fusion

    2 MnO₂ + 4 KOH + O₂ → 2 K₂MnO₄ (dark green manganate, Mn⁶⁺) + 2 H₂O.

    KNO₃ may replace O₂
  2. Step 2

    Disproportionation or electrolysis

    Electrolytic oxidation in alkaline solution (commercial), or in neutral / acidic: 3 MnO₄²⁻ + 4 H⁺ → 2 MnO₄⁻ + MnO₂ + 2 H₂O.

    Mn(VI) → Mn(VII)

KMnO₄ forms dark purple, almost-black crystals, isostructural with KClO₄. Both manganate (MnO₄²⁻) and permanganate (MnO₄⁻) ions are tetrahedral; π-bonding involves overlap of oxygen 2p orbitals with manganese d orbitals. The green manganate has one unpaired electron (d¹) and is paramagnetic; the permanganate (d⁰) is diamagnetic but intensely coloured by a charge-transfer transition.

In acidic medium permanganate is the strongest of the common oxidants: it converts oxalate to CO₂, Fe²⁺ to Fe³⁺, nitrite to nitrate and iodide to iodine, with manganese reduced all the way to Mn²⁺ (E° = +1.52 V). In neutral or alkaline medium it stops at MnO₂. Permanganate titrations are never done in HCl — HCl would itself be oxidised to chlorine and corrupt the count.

K₂Cr₂O₇ vs KMnO₄ — oxidising actions in acidic medium

K₂Cr₂O₇ · Cr(VI)

+1.33 V

orange → green Cr³⁺

  • Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
  • Primary standard in volumetry
  • 6Fe²⁺ → 6Fe³⁺ (dichrometry)
  • SO₂ → green Cr₂(SO₄)₃
  • Stable to light; non-hygroscopic
  • Cr(VI) — d⁰, diamagnetic

KMnO₄ · Mn(VII)

+1.52 V

purple → colourless Mn²⁺

  • MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O
  • Self-indicator in titrations
  • 5Fe²⁺ → 5Fe³⁺; oxalate → CO₂
  • Not used with HCl (oxidises to Cl₂)
  • Decomposes in light/heat
  • Mn(VII) — d⁰, diamagnetic

Uses: K₂Cr₂O₇ for leather tanning, primary standard, azo-dye intermediates. KMnO₄ for bleaching wool, cotton, silk; decolouring oils; antiseptic in dilute solution; and as the favourite oxidant of organic chemistry — preparation of carboxylic acids, cleavage of alkenes.

Lanthanoids — the 4f series and lanthanoid contraction

Below the main body of the periodic table sit the fourteen lanthanoids, Ce to Lu, in which the 4f orbitals are filled progressively. Their atomic electronic configurations show 6s² in common, with 4f occupancy variable from 0 (La itself, [Xe] 5d¹ 6s²) to 14 (Lu, [Xe] 4f¹⁴ 5d¹ 6s²). The chemistry of all lanthanoids is overwhelmingly dominated by the +3 oxidation state — every Ln³⁺ ion has a 4fⁿ configuration with n increasing systematically from Ce³⁺ (4f¹) to Lu³⁺ (4f¹⁴).

Occasional +2 and +4 states appear, always near an empty, half-filled or completely filled f sub-shell. Cerium reaches +4 (Ce⁴⁺ = 4f⁰, noble-gas core) and is a strong oxidant; europium drops to +2 (Eu²⁺ = 4f⁷, half-filled, reductant); ytterbium drops to +2 (Yb²⁺ = 4f¹⁴, full, reductant); terbium reaches +4 (Tb⁴⁺ = 4f⁷, half-filled, oxidant). NEET 2016 Q.8 tested precisely this pattern: Gd's stable [Xe] 4f⁷ 5d¹ 6s² preserves the half-filled f, and Eu drops the 5d to keep [Xe] 4f⁷ 6s².

15 elements 4f filling

Lanthanoid contraction — scope & consequence

Across La → Lu, atomic radius shrinks gradually (from ~187 pm to ~173 pm; M³⁺ from 103 pm to 86 pm). Cause: imperfect 4f–4f shielding. Cumulative effect: 5d radii nearly equal corresponding 4d radii — Zr (160 pm) ≈ Hf (159 pm). Asked in NEET 2021.

All lanthanoids are silvery-white, soft, reactive metals that tarnish rapidly in air. They burn in O₂ to give Ln₂O₃, react with halogens to give LnX₃, with hydrogen to give hydrides, and with dilute acids to liberate H₂. Their compounds with f-electron counts other than 0 and 14 are paramagnetic. Applications: mischmetal (≈95% Ln + 5% Fe + traces of S, C, Ca, Al) used in lighter flints and Mg-based alloys; mixed Ln oxides as petroleum-cracking catalysts; individual Ln oxides as phosphors in television and fluorescent screens. Cerium(IV) sulphate is a standard analytical oxidant.

Actinoids — radioactive, variable, contracted

The actinoids span Th to Lr — fourteen elements in which 5f orbitals are filled. Every actinoid is radioactive, which is the first feature that complicates their chemistry: experimental work demands shielded facilities and short-lived isotopes that decay before you finish a measurement. Beyond uranium, the elements are entirely synthetic.

The 5f, 6d and 7s orbitals lie much closer in energy than do 4f, 5d and 6s in the lanthanoids. The consequence is the second distinguishing feature: actinoids show a much wider range of oxidation states than lanthanoids — uranium alone exhibits +3, +4, +5 and +6 in stable compounds; neptunium and plutonium reach +7. NEET 2017 asked exactly this and the canonical answer is "5f, 6d and 7s levels have comparable energies." This contrasts sharply with the lanthanoids, where +3 is the only dominant state.

Like the lanthanoids, the actinoids show contraction along the series — the actinoid contraction — and for the same reason (imperfect f–f shielding). Because 5f electrons shield even more poorly than 4f, the actinoid contraction is slightly larger per element than the lanthanoid contraction. Thorium and uranium are excellent fuel sources for nuclear reactors; americium is used in smoke detectors; plutonium and californium are produced for nuclear weapons and neutron-source applications respectively. The actinoids will continue to expand the periodic table — they remain at the frontier of inorganic chemistry.

NEET PYQ Snapshot

Real NEET previous-year questions — solve before moving on.

NEET 2023

The stability of Cu²⁺ is more than Cu⁺ salts in aqueous solution due to —

  1. second ionisation enthalpy
  2. first ionisation enthalpy
  3. enthalpy of atomisation
  4. hydration energy
Answer: (4) hydration energy

Why: ΔhydH° of Cu²⁺ is −2121 kJ mol⁻¹, much more negative than that of Cu⁺. This more than compensates for the high second ionisation enthalpy of Cu (+1960 kJ mol⁻¹), making Cu²⁺(aq) the more stable species. Hence Cu⁺ disproportionates in water.

NEET 2021

Zr (Z = 40) and Hf (Z = 72) have similar atomic and ionic radii because of:

  1. Having similar chemical properties
  2. Belonging to same group
  3. Diagonal relationship
  4. Lanthanoid contraction
Answer: (4) Lanthanoid contraction

Why: The 14 lanthanoids that intervene between La and Hf squeeze the radius. Imperfect 4f shielding produces a regular contraction across the 4f series, and the cumulative effect makes Hf (159 pm) nearly identical to Zr (160 pm). Same group is part of the story but not the cause.

NEET 2020

The calculated spin-only magnetic moment of Cr²⁺ ion is:

  1. 4.90 BM
  2. 5.92 BM
  3. 2.84 BM
  4. 3.87 BM
Answer: (1) 4.90 BM

Why: Cr is [Ar] 3d⁵ 4s¹. Cr²⁺ is 3d⁴ → four unpaired electrons. μ = √(4×6) = √24 ≈ 4.90 BM.

NEET 2018

Which one of the following ions exhibits d-d transition and paramagnetism as well?

  1. CrO₄²⁻
  2. Cr₂O₇²⁻
  3. MnO₄⁻
  4. MnO₄²⁻
Answer: (4) MnO₄²⁻

Why: In MnO₄²⁻ the manganese is Mn(VI) → 3d¹: one unpaired electron, paramagnetic, capable of d-d transition. CrO₄²⁻ (Cr⁶⁺, d⁰), Cr₂O₇²⁻ (Cr⁶⁺, d⁰) and MnO₄⁻ (Mn⁷⁺, d⁰) all have no d electron — colour comes from charge transfer, and they are diamagnetic.

NEET 2016

Which one of the following statements is correct when SO₂ is passed through acidified K₂Cr₂O₇ solution?

  1. The solution is decolourised
  2. SO₂ is reduced
  3. Green Cr₂(SO₄)₃ is formed
  4. The solution turns blue
Answer: (3) Green Cr₂(SO₄)₃ is formed

Why: Acidified K₂Cr₂O₇ (orange, Cr⁶⁺) is reduced by SO₂ to green Cr³⁺. The reaction is K₂Cr₂O₇ + 3 SO₂ + H₂SO₄ → K₂SO₄ + Cr₂(SO₄)₃ + H₂O. SO₂ is oxidised (not reduced) to SO₄²⁻; the orange solution turns characteristic green.

Expert FAQs

Questions NEET has asked from this chapter, answered straight.

Why is the electronic configuration of chromium [Ar]3d⁵ 4s¹ and not [Ar]3d⁴ 4s²?
The energy gap between the 3d and 4s sub-shells is very small in chromium, and a half-filled 3d⁵ configuration has extra stability due to exchange energy and symmetrical distribution. Promoting one 4s electron into the 3d sub-shell gives the more stable [Ar]3d⁵ 4s¹ ground state. The same logic gives copper the [Ar]3d¹⁰ 4s¹ configuration, with a fully-filled 3d¹⁰ set.
Why are Zn, Cd and Hg not regarded as transition elements?
IUPAC defines a transition metal as an element whose atom or any of its common ions has an incomplete d sub-shell. Zn, Cd and Hg have a fully filled (n−1)d¹⁰ configuration in both the neutral atom and their common +2 ions, so they do not meet the definition. They are studied alongside transition metals because they are the end members of the 3d, 4d and 5d series.
Why do transition metal ions form coloured compounds?
In the presence of a ligand field, the five degenerate d orbitals split into two energy sets. An electron from the lower set can absorb visible light and jump to the higher set — a d-d transition. The colour observed is the complementary colour of the light absorbed. Ions with d⁰ (Sc³⁺, Ti⁴⁺) or d¹⁰ (Zn²⁺, Cu⁺) configurations have no d-d transition possible and are therefore colourless.
What is the spin-only formula for magnetic moment?
μ = √(n(n+2)) Bohr magnetons, where n is the number of unpaired electrons. The formula assumes orbital contribution is quenched, which is a good approximation for first-row transition metal ions. A single unpaired electron gives 1.73 BM; five unpaired electrons (d⁵ high-spin) give √35 ≈ 5.92 BM.
What is lanthanoid contraction and what does it cause?
Lanthanoid contraction is the regular decrease in atomic and ionic radii from lanthanum to lutetium across the 4f series, caused by the imperfect shielding of one 4f electron by another. Its most important consequence is that the 5d transition series members have radii almost identical to their 4d counterparts — for example Zr (160 pm) and Hf (159 pm) — making them difficult to separate.
How is potassium dichromate prepared from chromite ore?
Chromite ore (FeCr₂O₄) is fused with Na₂CO₃ in air to give Na₂CrO₄. The yellow chromate is acidified with H₂SO₄ to crystallise orange Na₂Cr₂O₇. Because Na₂Cr₂O₇ is more soluble, it is treated with KCl to precipitate the less soluble K₂Cr₂O₇ as orange crystals. The chromium stays in the +6 oxidation state throughout.
What is the action of acidified K₂Cr₂O₇ on Fe²⁺ and iodide?
Acidified K₂Cr₂O₇ is a strong oxidising agent (E° = +1.33 V). It oxidises Fe²⁺ to Fe³⁺ (the basis of dichrometry) and iodide to iodine. The dichromate itself is reduced to green Cr³⁺. The balanced equation is: Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻ → 2 Cr³⁺ + 7 H₂O.
Why is the chemistry of actinoids more complicated than that of lanthanoids?
In actinoids the 5f, 6d and 7s orbitals are close in energy, so actinoids show a much wider range of oxidation states than lanthanoids — uranium alone exhibits +3, +4, +5 and +6. Lanthanoids are dominated by the +3 state. In addition, all actinoids are radioactive, which complicates experimental study. Both series do show contraction, however — actinoid contraction is even more pronounced because 5f shielding is poorer than 4f.

Go Deeper

Drill into the subtopics that NEET asks most often.