The d- and f-Block
Elements

The definition and the four configuration anomalies are asked almost every year.

Position in the periodic table

• The d-block occupies groups 3 to 12, between the s- and p-blocks, in four series: 3d (Sc → Zn), 4d (Y → Cd), 5d (La, Hf → Hg), 6d (Ac, Rf → Cn, still incomplete).
• The f-block sits below the main table: lanthanoids Ce–Lu (4f, Z = 58–71) and actinoids Th–Lr (5f, Z = 90–103); both are called inner transition elements.
• General outer configuration of a transition element: $(n-1)d^{1-10}\,ns^{1-2}$.

Definition — and the standing exception

• A transition element has an incompletely filled d subshell either in its ground state or in any of its common oxidation states.
• Zn, Cd, Hg (and Cn) have $\mathrm{(n-1)}d^{10}$ in the atom and in the +2 ion, so they are not regarded as transition metals, although they belong to the d-block. They therefore lack the typical properties — variable oxidation state, colour, paramagnetism, catalysis.

Configuration anomalies

• General lanthanoid configuration: $[\mathrm{Xe}]\,4f^{1-14}\,5d^{0-1}\,6s^{2}$. General actinoid: $[\mathrm{Rn}]\,5f^{1-14}\,6d^{0-1}\,7s^{2}$.
• In ions the ns electrons are always lost first, so the d configuration of an ion is simply $d^{(\text{group no.} - \text{charge})}$ for the 3d series.

Trends are easy; the extremes and the anomalies are what get asked.

Metallic character, hardness, melting point

• All are hard, lustrous, high-melting metals with high thermal and electrical conductivity, because of strong metallic bonding involving unpaired d electrons.
• Melting point rises to a maximum at about d⁵ and then falls. In the 3d series Cr has the highest melting point; the highest in the whole block is W.
• Mn and Tc have anomalously low melting points for their position — the stable half-filled d⁵ configuration makes those electrons less available for metallic bonding.
• Zn, Cd, Hg are soft and low-melting; Hg is a liquid — no unpaired d electrons, hence weak metallic bonding.
• Enthalpy of atomisation is high and follows the same pattern; it is lowest for Zn in the 3d series.

Atomic and ionic radii

• Radius decreases across a series (increasing nuclear charge poorly screened by d electrons), flattens in the middle, then increases slightly at the end (Cu, Zn) as d–d electron repulsion outweighs the nuclear pull.
• Down a group, 3d → 4d shows the usual increase, but 4d and 5d radii are nearly identical because of the lanthanoid contraction.
• Classic consequence: Zr (160 pm) and Hf (159 pm) have virtually the same radius, so their chemistry is almost identical and they are extremely hard to separate. Similarly Nb/Ta and Mo/W.

Ionisation enthalpy and density

• Ionisation enthalpy increases across a series but irregularly, because the removed electron comes from different configurations.
• Cr has an unusually high second ionisation enthalpy (removing an electron from stable 3d⁵) and Cu also has a high second value (removing an electron from stable 3d¹⁰).
• Zn has the highest first ionisation enthalpy of the 3d series (filled 3d¹⁰4s²), and Zn's third ionisation enthalpy is extremely high — which is why Zn never exceeds +2.
• Density increases across a series; Os and Ir are the densest elements known.

Learn the maxima, the single-state elements, and the acidic/basic trend.

• Variable oxidation states differing by one unit (unlike p-block, where they differ by two), because ns and (n−1)d electrons are close in energy.
• Highest oxidation state equals the number of $3d + 4s$ electrons up to Mn, then declines: Sc +3, Ti +4, V +5, Cr +6, Mn +7, after which pairing makes further loss difficult.
• Mn shows the widest range in the 3d series: +2 to +7. The highest oxidation state known in the whole block is +8, shown by Ru and Os (as $\ce{RuO4}$, $\ce{OsO4}$).
• Single-state elements: Sc only +3; Zn only +2 — both are frequently asked.
• High oxidation states are stabilised only by the most electronegative partners: they appear in oxides and fluorides ($\ce{Mn2O7}$, $\ce{CrO3}$, $\ce{VF5}$, $\ce{OsF6}$), never in iodides. Hence $\ce{CuI}$ exists but $\ce{CuI2}$ does not; $\ce{FeI3}$ does not exist.
• Down a group, the higher oxidation state becomes more stable: Cr(VI) is strongly oxidising, but Mo(VI) and W(VI) are stable.
• As the oxidation state rises, bonding becomes more covalent and the oxide more acidic: $\ce{MnO}$ basic → $\ce{Mn2O3}$ → $\ce{MnO2}$ amphoteric → $\ce{Mn2O7}$ acidic.

🎯 Possible oxidation states per 3d element — memorise this

Bold = most common / most stable state. These are the ones NEET matches to configuration questions.

3 patterns to memorise: (1) Sc & Zn = single state (+3 and +2 respectively) — outliers. (2) Mn = widest range +2 → +7. (3) After Mn, max oxidation state drops because 3d pairing kicks in; that's why we never see Fe⁷⁺ or Co⁸⁺ despite the electron count.

One table, four anomalies. Learn the anomalies with their reasons.

• Cu is the only 3d metal with a positive E°(M²⁺/M) = +0.34 V, so it does not liberate H₂ from acids. Reason: its high sublimation and ionisation enthalpies are not compensated by its hydration enthalpy.
• Mn, Ni and Zn have more negative E° than the smooth trend predicts. For Mn it is the stability of the half-filled d⁵ Mn²⁺; for Zn the stability of d¹⁰ Zn²⁺; for Ni it is the unusually high hydration enthalpy.
• E°(Mn³⁺/Mn²⁺) = +1.57 V is very high, so Mn³⁺ is the strongest oxidising agent among 3d M³⁺ ions — it wants to become the stable d⁵ Mn²⁺.
• E°(Cr³⁺/Cr²⁺) = −0.41 V is negative, so Cr²⁺ is the strongest reducing agent — it readily gives an electron to reach the stable d³ Cr³⁺ configuration.
• E°(Co³⁺/Co²⁺) = +1.97 V — Co³⁺ is a powerful oxidant in aqueous solution, though it is stabilised in complexes.
• Ti²⁺ (−1.63 V) is the most negative, so Ti is the most reactive of the series by this measure.
• Overall reactivity: E° values are generally negative and become less negative across the series, so reducing power falls from Ti to Cu.

Guaranteed numerical. The spin-only formula must be automatic.

• $n$ = number of unpaired electrons; 1 BM (Bohr magneton) $= 9.27\times10^{-24}\ \mathrm{J\,T^{-1}}$.
• Paramagnetic when unpaired electrons are present; diamagnetic when $n=0$, i.e. d⁰ and d¹⁰ — so $\ce{Sc^3+}$, $\ce{Ti^4+}$, $\ce{Cu+}$, $\ce{Zn^2+}$ are diamagnetic.
• Magnetic moment rises to a maximum at d⁵ and then falls.

🎯 Worked example — derive μ from an unknown ion

Colour comes from d–d transitions — except in the two ions you will actually be asked about.

• Most hydrated transition metal ions are coloured because a ligand field splits the d orbitals; the ion absorbs a photon of visible light to promote an electron between them (d–d transition) and transmits the complementary colour.
• Ions with d⁰ or d¹⁰ configurations cannot undergo d–d transitions and are therefore colourless: $\ce{Sc^3+}$, $\ce{Ti^4+}$, $\ce{Zn^2+}$, $\ce{Cu+}$, $\ce{Ag+}$, $\ce{Cd^2+}$.

Example-matching questions. Learn catalyst with process, and the interstitial formulae.

Complex formation

• Favoured by small ionic size, high ionic charge, and vacant d orbitals able to accept lone pairs from ligands.
• Examples: $\ce{[Fe(CN)6]^{3-}}$, $\ce{[Cu(NH3)4]^{2+}}$, $\ce{[Ag(CN)2]-}$, $\ce{[Co(NH3)6]^{3+}}$.

Catalytic activity

• Two reasons: variable oxidation states allowing the catalyst to form intermediates and be regenerated, and large surface area providing adsorption sites.

• Other examples to memorise: Ziegler–Natta $\ce{TiCl4 + Al(C2H5)3}$ for polythene; Ni for hydrogenation of oils; Pd in Wacker/Lindlar contexts and Pt in Ostwald's process ($\ce{NH3}$ → $\ce{NO}$); MnO₂ in the decomposition of $\ce{KClO3}$; Cu and ZnO–Cr₂O₃ for methanol.

Interstitial compounds

• Small atoms — H, C, N, B — occupy interstices of the metal lattice.
• Properties: non-stoichiometric and not governed by normal valency; harder than the pure metal; higher melting point; retain metallic conductivity; chemically inert.
• Formulae worth quoting: $\ce{TiC}$, $\ce{Mn4N}$, $\ce{Fe3H}$, $\ce{TiH1.7}$, $\ce{VH0.56}$. Steel and cast iron are interstitial compounds of Fe and C.

Alloy formation

• Transition metals have similar radii, so one can replace another in a lattice — substitutional alloys form readily when radii differ by less than about 15% (Hume-Rothery).
• Examples: ferrous alloys with Cr, V, W, Mo, Mn; brass (Cu–Zn); bronze (Cu–Sn); misch metal (lanthanoid alloy, see §09).

The highest-yield single block in the chapter. Preparation, structure, and medium-dependent behaviour.

Potassium dichromate, K₂Cr₂O₇

Prepared from chromite ore, $\ce{FeCr2O4}$, in three steps:

• Structure: both $\ce{CrO4^2-}$ (yellow) and $\ce{Cr2O7^2-}$ (orange) are built from tetrahedral $\ce{CrO4}$ units; in the dichromate two tetrahedra share one oxygen through a Cr–O–Cr bridge, with the bridging Cr–O bond longer than the terminal ones.

Acid turns yellow chromate to orange dichromate; alkali reverses it. The oxidation state of Cr stays +6 throughout — it is not a redox change.

• Uses: volumetric analysis as a primary standard (unlike KMnO₄), leather tanning, and the chromic acid mixture.

Potassium permanganate, KMnO₄

Prepared from pyrolusite, $\ce{MnO2}$:

• Structure: $\ce{MnO4-}$ is tetrahedral, deep purple, diamagnetic (d⁰); $\ce{MnO4^2-}$ is tetrahedral, green, paramagnetic with one unpaired electron (d¹).

Acidic → 5 electrons, colourless Mn²⁺. Neutral or faintly alkaline → 3 electrons, brown $\ce{MnO2}$ precipitate. Strongly alkaline → 1 electron, green manganate.

Lanthanoid contraction and the non-+3 oxidation states carry the questions.

• Fifteen elements La–Lu; the 4f orbitals fill from Ce to Lu. General configuration $[\mathrm{Xe}]\,4f^{1-14}\,5d^{0-1}\,6s^{2}$.
• Silvery-white, soft, reactive metals; reactivity resembles Ca. $E^\circ(\ce{Ln^3+}/\ce{Ln})$ lies between about −2.2 and −2.4 V, so they are strong reducing agents. They react with water, burn in air to $\ce{Ln2O3}$, and form $\ce{Ln(OH)3}$, $\ce{LnCl3}$, $\ce{LnH2}$, $\ce{Ln2S3}$.
• +3 is the characteristic oxidation state of all of them.

Lanthanoid contraction

• Definition: the steady, small decrease in atomic and ionic radius from La³⁺ to Lu³⁺.
• Cause: each added 4f electron shields the outer electrons from the growing nuclear charge very imperfectly, because 4f orbitals are diffuse and deeply buried, so the effective nuclear charge rises steadily.
• Consequences — a stock question:
  • The chemistry of the lanthanoids is almost identical, making their separation extremely difficult (ion-exchange chromatography is required).
  • Basic strength of the hydroxides falls: $\ce{La(OH)3}$ is the most basic, $\ce{Lu(OH)3}$ the least.
  • The 4d and 5d transition series have almost equal radii — the Zr/Hf, Nb/Ta, Mo/W pairs.
  • Covalent character of the compounds increases slightly from La to Lu.

• Colour and magnetism: most Ln³⁺ ions are coloured (f–f transitions) and paramagnetic; $\ce{La^3+}$ (4f⁰) and $\ce{Lu^3+}$ (4f¹⁴) are colourless and diamagnetic.
• Misch metal — about 95% lanthanoid metal with 5% Fe and traces of S, C, Ca and Al; used in Mg-based alloys for tracer bullets and shells, and in lighter flints. Mixed lanthanoid oxides serve as petroleum cracking catalysts.

The comparison table is the question. Learn the four points of difference.

• Fourteen elements Th–Lr filling 5f; general configuration $[\mathrm{Rn}]\,5f^{1-14}\,6d^{0-1}\,7s^{2}$. All are radioactive. Elements after uranium are synthetic (transuranium).
• Actinoid contraction is greater than lanthanoid contraction, because 5f orbitals shield even more poorly than 4f.
• Wide range of oxidation states: +3 is common to all, but U shows +3 to +6, Np up to +7, and Pu and Am show +3 to +6. Am even shows +2.
• Ionisation enthalpies are lower than those of lanthanoids, so the actinoids are more reactive electropositive metals.

Applications worth a line each

• Fe and steels — construction. Ti — aircraft, low density and corrosion resistance. Cu, Ag, Au — conductors and coinage.
• TiO₂ — white pigment. ZnO — paints. MnO₂ — dry-cell cathode. AgBr — photographic film. Cr — electroplating. U and Pu — nuclear fuel. Th — in the Indian three-stage nuclear programme.

Few in number, so worth a clean sweep — they turn up as matching questions.

Everything quantitative the chapter can ask, in one place.

Twenty questions in the shapes NEET actually uses. Work each one before opening the answer.

1. Calculate the spin-only magnetic moment of $\ce{Mn^2+}$, $\ce{Ni^2+}$ and $\ce{Ti^3+}$.
   Answer

   $\ce{Mn^2+}$ is 3d⁵, $n=5$, $\mu=\sqrt{35}=5.92$ BM. $\ce{Ni^2+}$ is 3d⁸, $n=2$, $\mu=\sqrt{8}=2.84$ BM. $\ce{Ti^3+}$ is 3d¹, $n=1$, $\mu=\sqrt{3}=1.73$ BM.
2. An ion of a 3d metal has $\mu=3.87$ BM. Identify two possible ions.
   Answer

   $\sqrt{n(n+2)}=3.87 \Rightarrow n=3$. Candidates: $\ce{Cr^3+}$ (3d³) and $\ce{Co^2+}$ (3d⁷, high spin).
3. How many moles of $\ce{KMnO4}$ are needed to oxidise one mole of $\ce{Fe^2+}$ in acidic medium? And how many moles of $\ce{K2Cr2O7}$?
   Answer

   $\ce{MnO4-}$ accepts 5 electrons, so $1/5$ mol. $\ce{Cr2O7^2-}$ accepts 6 electrons, so $1/6$ mol.
4. Write the balanced equation for the reaction of acidified $\ce{KMnO4}$ with oxalic acid and state two conditions.
   Answer

   $$\ce{2MnO4- + 16H+ + 5C2O4^2- -> 2Mn^2+ + 10CO2 + 8H2O}$$ Conditions: warm to about 333 K; the reaction is autocatalysed by the $\ce{Mn^2+}$ formed, so it starts slowly and then accelerates.
5. Why is $\ce{Cr^2+}$ reducing while $\ce{Mn^3+}$ is oxidising, when both are d⁴?
   Answer

   $\ce{Cr^2+}$ loses one electron to reach the stable d³ $\ce{Cr^3+}$ configuration, so it acts as a reductant ($E^\circ=-0.41$ V). $\ce{Mn^3+}$ gains one electron to reach the stable half-filled d⁵ $\ce{Mn^2+}$, so it acts as an oxidant ($E^\circ=+1.57$ V).
6. Why is $E^\circ(\ce{Cu^2+/Cu})$ positive, unlike every other 3d metal?
   Answer

   The sum of copper's sublimation and ionisation enthalpies is high, and its hydration enthalpy is not large enough to compensate. The overall process is therefore unfavourable, $E^\circ = +0.34$ V, and Cu does not liberate hydrogen from acids.
7. Explain why $\ce{Cu+}$ is unstable in aqueous solution.
   Answer

   It disproportionates, $\ce{2Cu+ -> Cu^2+ + Cu}$. The hydration enthalpy of the smaller, doubly charged $\ce{Cu^2+}$ is much larger and more than offsets the second ionisation enthalpy of copper.
8. Why is $\ce{MnO4-}$ intensely coloured although Mn is in the d⁰ state?
   Answer

   No d electrons are available, so the colour cannot come from a d–d transition. It arises from ligand-to-metal charge transfer, in which an oxygen lone-pair electron is momentarily transferred to manganese.
9. Give the effect of adding (a) acid and (b) alkali to a yellow solution of $\ce{K2CrO4}$, with equations. Does the oxidation state change?
   Answer

   (a) $\ce{2CrO4^2- + 2H+ <=> Cr2O7^2- + H2O}$ — turns orange. (b) $\ce{Cr2O7^2- + 2OH- <=> 2CrO4^2- + H2O}$ — back to yellow. Chromium stays at +6; this is an acid–base equilibrium, not a redox change.
10. Write the three half-reactions showing how $\ce{KMnO4}$ behaves in acidic, neutral and strongly alkaline media.
    Answer

    $$\ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O}$$$$\ce{MnO4- + 2H2O + 3e- -> MnO2 + 4OH-}$$$$\ce{MnO4- + e- -> MnO4^2-}$$ Manganese ends at +2, +4 and +6 respectively.
11. Why does Mn show the maximum number of oxidation states in the 3d series?
    Answer

    Its configuration is 3d⁵4s². All seven of these electrons can be involved in bonding, so states from +2 up to +7 are accessible; the number of half-filled-to-empty steps available is greatest at manganese.
12. Why do Zr and Hf have almost identical properties?
    Answer

    Lanthanoid contraction cancels the expected size increase from period 5 to period 6, so the two have nearly the same atomic radius (160 and 159 pm) and consequently nearly identical chemistry — which is why they are so hard to separate.
13. Account for the low melting point and low enthalpy of atomisation of zinc.
    Answer

    Zinc has a completely filled 3d¹⁰ subshell, so no d electrons contribute to metallic bonding. Only the 4s electrons participate, giving weak metallic bonds — the same reason mercury is a liquid.
14. Which is a stronger oxidising agent in acidic medium, $\ce{MnO4-}$ or $\ce{Cr2O7^2-}$, and why is only one of them a primary standard?
    Answer

    $\ce{MnO4-}$ is stronger ($E^\circ = 1.52$ V against $1.33$ V). Only $\ce{K2Cr2O7}$ is a primary standard, because $\ce{KMnO4}$ cannot be obtained perfectly pure and decomposes slowly on standing.
15. Why is $\ce{Ce^4+}$ a good oxidising agent, and $\ce{Eu^2+}$ a good reducing agent?
    Answer

    $\ce{Ce^4+}$ readily accepts an electron to become the common $\ce{Ce^3+}$, $E^\circ = +1.74$ V; the 4f⁰ state is reached by losing four electrons, which is unusual, so it does not persist. $\ce{Eu^2+}$ has the stable half-filled 4f⁷ configuration but readily loses an electron to reach the far more usual +3 state, so it reduces other species.
16. Define lanthanoid contraction and give two consequences.
    Answer

    The steady small decrease in size from $\ce{La^3+}$ to $\ce{Lu^3+}$, caused by imperfect shielding by the diffuse 4f electrons. Consequences: the lanthanoids are extremely difficult to separate; basicity falls from $\ce{La(OH)3}$ to $\ce{Lu(OH)3}$; and the 4d and 5d transition series have nearly equal radii.
17. Which lanthanoid ions are colourless and diamagnetic, and why does the spin-only formula fail for the rest?
    Answer

    $\ce{La^3+}$ (4f⁰) and $\ce{Lu^3+}$ (4f¹⁴). For the others the 4f orbitals are shielded by 5s and 5p, so the orbital angular momentum is not quenched and the orbital contribution must be included: $\mu = g\sqrt{J(J+1)}$.
18. Give two reasons why transition metals make good catalysts, with one example each.
    Answer

    Variable oxidation states let the metal form and then release an intermediate — $\ce{Fe^3+}$ catalysing $\ce{2I- + S2O8^2- -> I2 + 2SO4^2-}$. Large surface area provides adsorption sites — finely divided Fe in the Haber process, or $\ce{V2O5}$ in the contact process.
19. What are interstitial compounds? List three properties and two examples.
    Answer

    Compounds formed when small atoms such as H, C, N or B occupy the interstices of a metal lattice. They are non-stoichiometric, harder and higher-melting than the parent metal, retain metallic conductivity and are chemically inert. Examples: $\ce{TiC}$, $\ce{VH0.56}$, $\ce{Fe3H}$, $\ce{Mn4N}$; steel and cast iron are interstitial Fe–C compounds.
20. Complete and balance the preparation of $\ce{K2Cr2O7}$ from chromite ore.
    Answer

    $$\ce{4FeCr2O4 + 8Na2CO3 + 7O2 -> 8Na2CrO4 + 2Fe2O3 + 8CO2}$$$$\ce{2Na2CrO4 + 2H+ -> Na2Cr2O7 + 2Na+ + H2O}$$$$\ce{Na2Cr2O7 + 2KCl -> K2Cr2O7 + 2NaCl}$$

If time is short, revise in this order.

1. $\ce{KMnO4}$ and $\ce{K2Cr2O7}$ — preparation, structure, medium-dependent behaviour, standard reactions.
2. The electrode-potential anomalies with their reasons: Cu, Mn, Zn, and the Mn³⁺/Cr²⁺ pair.
3. Spin-only magnetic moment values for d¹ to d⁵, plus which ions are diamagnetic.
4. Electronic configuration exceptions: Cr, Cu, Mo, Ag, Pd, Pt.
5. Lanthanoid contraction — cause and all consequences.
6. Exceptional lanthanoid oxidation states: Ce⁴⁺, Eu²⁺, Yb²⁺, Tb⁴⁺, Sm²⁺.
7. The lanthanoid–actinoid comparison table.
8. Colours of hydrated 3d ions, and why $\ce{MnO4-}$ is coloured despite being d⁰.
9. Catalysts matched to processes; interstitial compound formulae.
10. Extremes: Cr highest melting point in 3d, Zn lowest enthalpy of atomisation, Os and Ir densest, Ru and Os reach +8.

Each of these is a distractor that has caught candidates before.

• Zn, Cd, Hg are d-block but not transition metals. Conversely Sc is a transition metal even though Sc³⁺ is d⁰, and Cu is one even though the atom is 3d¹⁰.
• The chromate–dichromate interconversion is not a redox reaction — Cr stays at +6.
• $\ce{MnO4-}$ and $\ce{Cr2O7^2-}$ are coloured by charge transfer, not d–d transition; they are also diamagnetic.
• The spin-only formula does not apply to lanthanoids.
• $\ce{KMnO4}$ is not a primary standard; $\ce{K2Cr2O7}$ is.
• The highest oxidation state in the 3d series is +7 (Mn), but the highest in the whole d-block is +8 (Ru, Os).
• Mn has an anomalously low melting point despite sitting at d⁵ — the half-filled shell holds its electrons back from metallic bonding. Cr, not Mn, has the highest melting point in the 3d series.
• Higher oxidation states occur with O and F, never with iodide — hence $\ce{CuI2}$ and $\ce{FeI3}$ do not exist.
• Actinoid contraction is greater than lanthanoid contraction, not smaller.
• Only Pm among the lanthanoids is radioactive; every actinoid is.
• $\ce{Pd}$ is the one element in the block with a vacant outermost s orbital (4d¹⁰5s⁰).
• Ionic radii comparisons must be made at the same oxidation state — comparing Mn²⁺ with Fe³⁺ directly is a trap.