Position and electronic configuration of d-block elements
Link to Std 11 : In Std 11 you filled orbitals by the Aufbau principle and met the periodic table blocks; the d-block is where the 3d, 4d and 5d orbitals fill.
Iron, copper, gold and silver are all d-block metals; they are hard, shiny and carry electricity and heat well.
d-block elements are those in which the last electron enters a (n−1)d orbital. Transition elements have an incomplete d subshell either in the atom or in a common oxidation state. General configuration: (n−1)d¹⁻¹⁰ ns¹⁻².
Position and electronic configuration of d-block elements
In the app you can move this figure with your finger.
Open in the appIn the figure : The three d-series (3d, 4d, 5d) lie between groups 3 and 12. The f-block (lanthanoids and actinoids) is placed separately at the bottom.
Rule 1: Configurations of the 3d series — Sc 3d¹4s², Ti 3d²4s², V 3d³4s², Cr 3d⁵4s¹, Mn 3d⁵4s², Fe 3d⁶4s², Co 3d⁷4s², Ni 3d⁸4s², Cu 3d¹⁰4s¹, Zn 3d¹⁰4s². Cr and Cu are exceptions because half-filled and full d subshells are extra stable.
- Easy : Cr (Z = 24) is [Ar] 3d⁵ 4s¹, not 3d⁴4s².
- Medium : Cu (Z = 29) is [Ar] 3d¹⁰ 4s¹, not 3d⁹4s².
- Exam-type : Why is the configuration of Cr different from that of V? The half-filled 3d⁵ arrangement gains extra stability from exchange energy and symmetry.
Rule 2: Configurations of ions — When d-block atoms form cations, the ns electrons are lost first, then the (n−1)d electrons.
- Easy : Fe²⁺ (Z = 26) is [Ar] 3d⁶ (the two 4s electrons go).
- Medium : Mn²⁺ (Z = 25) is [Ar] 3d⁵ — half-filled and stable.
- Exam-type : Cu⁺ is 3d¹⁰ (complete), Cu²⁺ is 3d⁹, so Cu²⁺ has one unpaired electron and is coloured.
Rule 3: Is it a transition element? — Zn, Cd and Hg have the full d¹⁰ configuration in atoms and in their common +2 ions, so they are not usually called transition elements, although they are d-block elements. Sc³⁺ (3d⁰) similarly is not typical, but Sc as an atom has 3d¹.
- Easy : Zn²⁺ = 3d¹⁰ → not a transition metal ion.
- Medium : Cu²⁺ = 3d⁹ → incomplete d subshell, so Cu counts as a transition element.
- Exam-type : Is Sc a transition element? Sc atom has 3d¹ (incomplete), so it is usually included, even though Sc³⁺ is 3d⁰.
Remove the 4s electrons first when making cations from 3d metals.
Writing Cr as 3d⁴4s² and Cu as 3d⁹4s², instead of the correct exceptions.
Memory clue: "Half or full is happy" — d⁵ and d¹⁰ are special; Cr and Cu borrow a 4s electron to get there.
Spot the difference
| Point | Transition element | d-block, non-transition |
|---|---|---|
| Example | Fe, Cu | Zn, Cd, Hg |
| d subshell | Incomplete in atom or ion | Full d¹⁰ in atom and ion |
| Coloured compounds | Usually | Usually colourless |
Practice questions
- 1Write the general electronic configuration of d-block elements.
- 2Write the configuration of Cu (Z = 29).
- 3Name two d-block elements that are not transition elements.
- 4Write the configuration of Fe²⁺ and Fe³⁺ (Z = 26).
- 5Why is Zn not counted as a transition element?
- 6Why is Cr 3d⁵4s¹ rather than 3d⁴4s²?
- 7How many unpaired electrons are there in Mn²⁺ (Z = 25)?
- 8Which has more unpaired electrons: Ni²⁺ (Z = 28) or Co²⁺ (Z = 27)? Show the d configurations.
Show answers (try them yourself first)
- 1(n − 1)d¹⁻¹⁰ ns¹⁻².
- 2[Ar] 3d¹⁰ 4s¹.
- 3Zn and Cd (also Hg).
- 4Fe²⁺: [Ar] 3d⁶; Fe³⁺: [Ar] 3d⁵.
- 5Zn has a full 3d¹⁰ subshell both in the atom and in Zn²⁺, so no incomplete d subshell is present.
- 6The half-filled d⁵ arrangement is more stable because of symmetry and exchange energy.
- 7Mn²⁺ is [Ar] 3d⁵: five unpaired electrons.
- 8Co²⁺ is 3d⁷ (3 unpaired); Ni²⁺ is 3d⁸ (2 unpaired); Co²⁺ has more.
Physical properties and trends in the d-block
Link to Std 11 : In Std 11 you studied how atomic radius and ionisation enthalpy change across a period; the d-block follows a gentler version of those trends.
Tungsten is used for bulb filaments because its melting point is the highest among metals, and iron and copper are strong because their atoms bond strongly.
Transition metals have high melting points, high enthalpy of atomisation, high density and hardness because both ns and (n−1)d electrons take part in metallic bonding. Across a series the atomic radius first decreases and then becomes almost constant. Zn, Cd and Hg are softer and melt at much lower temperatures because their d¹⁰ electrons do not take part in bonding.
In the figure : Radii decrease at first because the nuclear charge rises while the shielding by the added d electrons is not strong. In the middle the two effects balance and the radius stays almost constant; at the end repulsion between the d electrons increases the size a little.
Rule 1: Atomic and ionic sizes — Along the 3d series the radius decreases then stays nearly constant. For the same group, 4d and 5d elements have nearly equal radii because of the lanthanoid contraction (see the f-block section).
- Easy : Zr and Hf have almost identical radii, so they are chemically very alike and hard to separate.
- Medium : Ions with the same charge in a series show a progressive decrease in size.
- Exam-type : Fe²⁺ is smaller than Mn²⁺ since the nuclear charge is greater for the same charge.
Rule 2: Melting points and enthalpy of atomisation — The number of unpaired d electrons that can form bonds increases to the middle of a series and then decreases, so melting point and enthalpy of atomisation peak near the middle (V–Cr–Mn–Fe region) and fall at the end. Mn is lower than expected because of its stable d⁵ configuration.
- Easy : Tungsten (5d) has the highest melting point of all metals.
- Medium : Zn, Cd and Hg have low melting points because there are no unpaired d electrons to bond.
- Exam-type : Why is the melting point of Mn lower than that of Cr and Fe? Its half-filled d⁵ is stable and its d electrons are less available for bonding.
Rule 3: Ionisation enthalpy and electrode potentials — Ionisation enthalpies rise slowly across the series because the nuclear charge increases but the added electrons go to the inner (n−1)d orbital. E°(M²⁺/M) becomes less negative across the series (Cu is +0.34 V); Mn (−1.18 V) and Zn (−0.76 V) are more negative than the trend, because of stable d⁵ and d¹⁰.
- Easy : Cu is the only 3d metal with a positive E°(M²⁺/M), so Cu does not liberate H₂ from dilute acids.
- Medium : Zn is oxidised more easily than Cu: E° (−0.76 V) is more negative than that of Cu (+0.34 V).
- Exam-type : Order of reducing power for Mn (−1.18 V), Fe (−0.44 V), Cu (+0.34 V): Mn > Fe > Cu.
More unpaired d electrons mean stronger metallic bonding, and so higher melting points and atomisation enthalpies.
Saying that Mn has the highest melting point in the 3d series; it is lower than expected because of its stable d⁵.
Memory clue: "Middle is the hardest" — melting points peak near the middle of each series.
Spot the difference
| Point | Middle of a series | End of a series |
|---|---|---|
| Unpaired d electrons | Maximum | Few or none |
| Melting point | High | Low |
| Examples | Cr, W | Zn, Hg |
Practice questions
- 1Name the metal with the highest melting point.
- 2Why do Zn, Cd and Hg melt at low temperature?
- 3Which 3d metal has a positive E°(M²⁺/M)?
- 4Why do the radii of 4d and 5d elements of the same group become nearly equal?
- 5Arrange Mn, Fe and Cu in order of reducing power using E° = −1.18, −0.44, +0.34 V.
- 6Why does the enthalpy of atomisation of the 3d series show a peak in the middle?
- 7Why is E°(Mn²⁺/Mn) (−1.18 V) lower than the trend predicts?
- 8Why can Cu not liberate hydrogen from dilute acid but Zn can?
Show answers (try them yourself first)
- 1Tungsten.
- 2They have no unpaired d electrons (d¹⁰), so metallic bonding is weak.
- 3Copper (+0.34 V).
- 4Because of the lanthanoid contraction: the poorly shielding 4f electrons shrink the size of the 5d elements to match the 4d elements.
- 5Mn > Fe > Cu.
- 6More unpaired d electrons give stronger metallic bonds in the middle; the number drops at the end.
- 7Mn²⁺ has the stable half-filled 3d⁵ configuration, so it is hard to lose the electrons to the metal (the ion is favoured).
- 8E°(Cu²⁺/Cu) is positive so Cu cannot reduce H⁺, whereas E°(Zn²⁺/Zn) is negative, so Zn reduces H⁺ to H₂.
Variable oxidation states and magnetic moment
Link to Std 11 : In Std 11 you assigned oxidation numbers; transition metals have several because the 4s and 3d electrons have almost the same energy.
Manganese compounds in the lab show many colours (purple permanganate, green manganate, pink Mn²⁺), each linked to a different oxidation state.
The ns and (n−1)d electrons are close in energy, so transition metals can lose a different number of electrons, giving variable oxidation states. Successive states differ by one unit (Fe²⁺, Fe³⁺), unlike p-block elements where the difference is two. The spin-only magnetic moment is μ = √[n(n + 2)] BM, where n is the number of unpaired electrons.
In the figure : Filled dots are the more common states. The number of states rises to the middle (Mn has seven) and falls at the end because of fewer unpaired d electrons that can take part.
Rule 1: Variable oxidation states — In the first half of the series the maximum state equals the number of 4s + 3d electrons (Sc +3 to Mn +7). In the second half the maximum state falls (Fe +6 rare, Cu +2, Zn +2). Higher oxidation states occur mainly in fluorides and oxides (oxygen and fluorine stabilise them): Mn₂O₇, MnO₄⁻, CrO₄²⁻.
- Easy : Mn has oxidation states from +2 to +7; in MnO₄⁻ it is +7.
- Medium : Fe commonly shows +2 and +3; Cu shows +1 and +2.
- Exam-type : Why is the highest state found in oxides or fluorides? Oxygen and fluorine are small and highly electronegative and form multiple bonds, stabilising the high states.
Rule 2: Stability of states — Across the series +2 becomes more stable than +3 (Mn²⁺ with d⁵ is very stable, Fe³⁺ with d⁵ is also stable). Cu⁺ disproportionates in water: 2Cu⁺ → Cu²⁺ + Cu, because the hydration energy of Cu²⁺ is high.
- Easy : Fe³⁺ (3d⁵) is more stable than Fe²⁺ (3d⁶) in many cases because of the half-filled d⁵.
- Medium : Cu⁺ is unstable in aqueous solution; it changes to Cu²⁺ and Cu.
- Exam-type : E° check: E°(Cu⁺/Cu) = +0.52 V and E°(Cu²⁺/Cu⁺) = +0.16 V, so for the disproportionation E°cell = 0.52 − 0.16 = 0.36 V, which is positive.
Rule 3: Spin-only magnetic moment — Substances with unpaired electrons are paramagnetic; those with all electrons paired are diamagnetic. μ = √[n(n + 2)] Bohr magnetons (BM). Count n from the d configuration of the ion.
- Easy : Cu²⁺ (d⁹, n = 1): μ = √3 = 1.73 BM; Ni²⁺ (d⁸, n = 2): μ = √8 = 2.83 BM.
- Medium : Cr³⁺ (d³, n = 3): μ = √15 = 3.87 BM; Mn²⁺ (d⁵, n = 5): μ = √35 = 5.92 BM.
- Exam-type : Fe²⁺ (d⁶, n = 4): μ = √24 = 4.9 BM; Zn²⁺ (d¹⁰, n = 0): μ = 0 (diamagnetic).
Higher oxidation states are shown in oxides and fluorides; the +2 state becomes more common towards the end of the series.
Counting the 4s electrons of a cation as still present when writing d configurations.
Memory clue: "n(n+2), then root" — magnetic moment = square root of n(n + 2).
Spot the difference
| Point | Paramagnetic | Diamagnetic |
|---|---|---|
| Unpaired electrons | One or more | None |
| In a magnetic field | Attracted | Slightly repelled |
| Example ion | Cu²⁺, Mn²⁺ | Zn²⁺, Cu⁺ |
Practice questions
- 1Give two oxidation states shown by copper.
- 2Write the formula for the spin-only magnetic moment.
- 3Which ion is diamagnetic: Zn²⁺ or Cu²⁺?
- 4Calculate μ for Mn²⁺ (3d⁵).
- 5Calculate μ for Cr³⁺ (3d³).
- 6Why does Mn show more oxidation states than Zn?
- 7Show that Cu⁺ disproportionates in water. E°(Cu⁺/Cu) = +0.52 V, E°(Cu²⁺/Cu⁺) = +0.16 V.
- 8Why do the highest oxidation states appear in fluorides and oxides?
Show answers (try them yourself first)
- 1+1 and +2.
- 2μ = √[n(n + 2)] BM.
- 3Zn²⁺ (3d¹⁰, no unpaired electrons).
- 4n = 5; μ = √35 = 5.92 BM.
- 5n = 3; μ = √15 = 3.87 BM.
- 6Mn has five unpaired d electrons plus two 4s electrons that can take part; Zn has a full d¹⁰ and loses only the 4s electrons.
- 72Cu⁺ → Cu²⁺ + Cu: E°cell = 0.52 − 0.16 = 0.36 V (positive), so the reaction is spontaneous.
- 8Oxygen and fluorine are small and very electronegative and form multiple bonds with the metal, stabilising the high oxidation state.
Colour, catalysis, complexes and other chemical behaviour
Link to Std 11 : In Std 11 you saw that light of different wavelengths gives different colours; transition metal ions absorb some of the visible light and show the leftover colour.
Copper sulphate crystals are blue, nickel chloride solution is green, and potassium permanganate is deep purple, all because of d electrons.
Most transition metal ions are coloured because a d electron absorbs a part of visible light and jumps to a higher d level (d–d transition). They are also good catalysts, form complexes, make interstitial compounds and alloys.
In the figure : In the field of ligands the five d orbitals split into two sets. A d electron absorbs visible light to go from the lower set (t2g) to the upper set (eg); we see the complementary colour.
Rule 1: Why ions are coloured — Colour needs partly filled d orbitals so that a d–d transition is possible. Ions with d⁰ (Sc³⁺, Ti⁴⁺) or d¹⁰ (Zn²⁺, Cu⁺) are colourless. The colour depends on the metal, its oxidation state and the ligands.
- Easy : Cu²⁺(aq) is blue (d⁹) and Zn²⁺(aq) is colourless (d¹⁰).
- Medium : Ti³⁺ (d¹) is purple, whereas Ti⁴⁺ (d⁰) is colourless.
- Exam-type : Why is Cu⁺ colourless but Cu²⁺ blue? Cu⁺ is 3d¹⁰ (no d–d transition), Cu²⁺ is 3d⁹ (one empty place in d, a transition is possible).
Rule 2: Catalytic behaviour — Transition metals and their compounds catalyse reactions because they can change oxidation state easily and provide a surface on which reactants are adsorbed.
- Easy : Fe is used in the Haber process for ammonia.
- Medium : V₂O₅ is used in the contact process for sulphuric acid, and Ni in hydrogenation of oils.
- Exam-type : Explain: Fe³⁺ catalyses the reaction of I⁻ with S₂O₈²⁻ because it first oxidises I⁻ to I₂ (becoming Fe²⁺) and then Fe²⁺ is oxidised back by S₂O₈²⁻.
Rule 3: Complexes, interstitial compounds, alloys — Small, highly charged ions with empty d orbitals are good at forming complexes (e.g. [Fe(CN)₆]⁴⁻). Interstitial compounds form when small atoms (H, C, N) occupy gaps in the metal lattice (TiC, Fe₃H); they are hard and chemically inert. Alloys form because the metals have similar sizes: brass (Cu–Zn), bronze (Cu–Sn), steel (Fe–C).
- Easy : Steel is stronger than pure iron because of carbon in the lattice and other alloying metals.
- Medium : TiC is extremely hard and is used on cutting tools.
- Exam-type : Why can Fe, Cr, Ni form alloys easily? Their atomic radii differ by less than about 15 %, so they replace each other in the lattice.
Colour needs a partly filled d subshell. d⁰ and d¹⁰ ions are colourless.
Saying that colour arises from the s electrons or from the ligand alone.
Memory clue: "No gap, no colour" — if the d subshell is empty or full, there is no d–d jump and no colour.
Spot the difference
| Point | Coloured | Colourless |
|---|---|---|
| d configuration | d¹ to d⁹ | d⁰ or d¹⁰ |
| Example | Cu²⁺, Ti³⁺, Mn²⁺ | Zn²⁺, Sc³⁺, Cu⁺ |
| Reason | d–d transition possible | No d–d transition |
Practice questions
- 1Name a transition metal used as a catalyst in the Haber process.
- 2Is Zn²⁺ coloured? Give a reason.
- 3Name two alloys of copper.
- 4Explain why Ti⁴⁺ is colourless but Ti³⁺ is purple.
- 5What are interstitial compounds? Give an example.
- 6Why are transition metals good catalysts?
- 7Explain the colour of Cu²⁺ and the lack of colour of Cu⁺.
- 8Why do transition metals form alloys easily?
Show answers (try them yourself first)
- 1Iron.
- 2No; Zn²⁺ is 3d¹⁰ and has no d–d transition.
- 3Brass (Cu–Zn) and bronze (Cu–Sn).
- 4Ti⁴⁺ is d⁰ (no d electron to excite); Ti³⁺ is d¹ and can absorb visible light.
- 5Compounds in which small atoms such as H, C or N occupy interstitial sites in a metal lattice; example TiC.
- 6They have variable oxidation states, which allow intermediate compounds to form, and they provide a surface to adsorb reactants.
- 7Cu²⁺ (d⁹) has a vacancy in the d level and absorbs light (blue); Cu⁺ is d¹⁰ and has no d–d transition, so it is colourless.
- 8Their atomic sizes are similar, so one metal atom can replace another in the lattice.
Potassium dichromate and potassium permanganate
Link to Std 11 : In Std 11 you balanced redox reactions and met oxidising agents; K₂Cr₂O₇ and KMnO₄ are the two best-known transition metal oxidising agents.
A drop of purple KMnO₄ in water decolourises as it oxidises impurities, which is why it is used to purify well water.
K₂Cr₂O₇ is an orange solid made from chromite ore (FeCr₂O₄) via sodium chromate; in alkali it changes to yellow chromate, in acid back to dichromate. KMnO₄ is a dark purple solid made from pyrolusite (MnO₂) via manganate, and is a strong oxidising agent in acidic, neutral and alkaline media.
Potassium dichromate and potassium permanganate
In the app you can move this figure with your finger.
Open in the appIn the figure : Left top: MnO₄⁻ gives Mn²⁺ in acid, MnO₂ in neutral or weakly alkaline medium and MnO₄²⁻ in strong alkali. Bottom: pH decides between yellow chromate and orange dichromate.
Rule 1: Dichromate chemistry — 2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O; adding alkali reverses this. In acid, dichromate is a strong oxidising agent: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O. It oxidises Fe²⁺ to Fe³⁺, I⁻ to I₂, H₂S to S and Sn²⁺ to Sn⁴⁺. Cr(VI) is a hazardous substance.
- Easy : Orange dichromate turns yellow on adding NaOH (chromate forms).
- Medium : Cr₂O₇²⁻ + 6Fe²⁺ + 14H⁺ → 2Cr³⁺ + 6Fe³⁺ + 7H₂O: 1 mol of dichromate oxidises 6 mol of Fe²⁺.
- Exam-type : Mass of K₂Cr₂O₇ (294 g mol⁻¹) needed for 2.78 g FeSO₄·7H₂O (278 g mol⁻¹): moles Fe²⁺ = 0.0100, dichromate = 0.0100 ÷ 6 = 0.00167 mol = 0.49 g.
Rule 2: Permanganate chemistry — Acidic: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O. Neutral or weakly alkaline: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻. Strongly alkaline: MnO₄⁻ + e⁻ → MnO₄²⁻. 1 mol of MnO₄⁻ oxidises 5 mol of Fe²⁺ in acid.
- Easy : A purple solution that becomes colourless in acid shows Mn²⁺ formation.
- Medium : In acid solution, 25.0 mL of 0.020 M KMnO₄ contains 0.0005 mol, which oxidises 5 × 0.0005 = 0.0025 mol Fe²⁺.
- Exam-type : That Fe²⁺ was in 20.0 mL of solution, so its concentration is 0.0025 ÷ 0.0200 = 0.125 M.
Rule 3: Preparation outline and uses — Dichromate: 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂; acidify to Na₂Cr₂O₇, then treat with KCl for K₂Cr₂O₇. Permanganate: MnO₂ fused with KOH and air gives K₂MnO₄ (green), which is oxidised electrolytically or by Cl₂/O₃ to KMnO₄ (purple).
- Easy : Uses of K₂Cr₂O₇: leather tanning, dyes, volumetric analysis.
- Medium : Uses of KMnO₄: bleaching, disinfectant, test for unsaturation (Baeyer's reagent), titrations.
- Exam-type : Why is K₂Cr₂O₇ preferred to Na₂Cr₂O₇ for standard solutions? Na₂Cr₂O₇ is hygroscopic and K₂Cr₂O₇ is not.
In acid: MnO₄⁻ takes 5 electrons (to Mn²⁺), Cr₂O₇²⁻ takes 6 (to 2Cr³⁺).
Writing Mn²⁺ as the product of permanganate in neutral medium; there the product is brown MnO₂.
Memory clue: "Mn five, Cr six" — MnO₄⁻ accepts 5 e⁻, Cr₂O₇²⁻ accepts 6 e⁻ in acid.
Spot the difference
| Point | K₂Cr₂O₇ | KMnO₄ |
|---|---|---|
| Colour | Orange | Dark purple |
| Electrons gained in acid | 6 (to 2Cr³⁺) | 5 (to Mn²⁺) |
| Fe²⁺ oxidised per mole | 6 | 5 |
Practice questions
- 1State the colours of chromate and dichromate ions.
- 2What is the colour of MnO₄²⁻?
- 3Write the half-reaction of KMnO₄ in acid.
- 4How many moles of Fe²⁺ are oxidised by 1 mol of KMnO₄ in acid?
- 525.0 mL of 0.020 M KMnO₄ oxidises Fe²⁺ in 20.0 mL of solution in acid. Find the molarity of Fe²⁺.
- 6What happens to the colour of dichromate on adding NaOH?
- 7How much K₂Cr₂O₇ is needed to oxidise 2.78 g of FeSO₄·7H₂O (278 g mol⁻¹) in acid? (K₂Cr₂O₇ = 294 g mol⁻¹)
- 8Why does KMnO₄ act as a self-indicator in titrations?
Show answers (try them yourself first)
- 1Chromate: yellow; dichromate: orange.
- 2Green.
- 3MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O.
- 45 mol.
- 5n(MnO₄⁻) = 0.020 × 0.0250 = 0.0005 mol; n(Fe²⁺) = 5 × 0.0005 = 0.0025 mol; M = 0.0025 ÷ 0.0200 = 0.125 M.
- 6It turns yellow because chromate ions form.
- 7n(Fe²⁺) = 2.78 ÷ 278 = 0.0100 mol; n(Cr₂O₇²⁻) = 0.0100 ÷ 6 = 0.00167 mol; mass = 0.00167 × 294 = 0.49 g.
- 8The permanganate is purple while the product Mn²⁺ is nearly colourless, so the first excess drop gives a permanent pink colour at the end point.
Lanthanoids and actinoids
Link to Std 11 : In Std 11 you learnt about shielding and periodic trends; the f-block applies the idea of weak shielding by f electrons.
The flint of a lighter is made of mischmetall, an alloy rich in lanthanoids, which gives sparks when struck.
Lanthanoids (Ce to Lu) fill the 4f subshell; actinoids (Th to Lr) fill the 5f subshell. General configuration of lanthanoids: [Xe] 4f¹⁻¹⁴ 5d⁰⁻¹ 6s². The steady decrease in size from La to Lu is the lanthanoid contraction.
In the figure : Each added 4f electron shields the outer electrons only weakly, so the effective nuclear charge rises and the ionic radius shrinks gradually; the total decrease is large, from about 106 pm to 85 pm.
Rule 1: Lanthanoids: properties — +3 is the common oxidation state. Ce shows +4 (Ce⁴⁺ is a strong oxidising agent) and Eu shows +2 (Eu²⁺ is a reducing agent). Lanthanoids are silvery white, soft metals that tarnish in air and are quite reactive. Many of their ions are coloured because of f–f transitions.
- Easy : Ce⁴⁺ has the configuration [Xe], which is stable (noble gas).
- Medium : Eu²⁺ is 4f⁷ (half-filled) and is therefore stable.
- Exam-type : Compare Ce⁴⁺ and Eu²⁺: Ce⁴⁺ is a good oxidising agent (it wants to return to +3) and Eu²⁺ is a good reducing agent (it wants to become +3).
Rule 2: Consequences of lanthanoid contraction — (1) 4d and 5d elements of the same group have almost equal radii (Zr and Hf; Nb and Ta), so they are chemically alike and hard to separate. (2) Basic strength of Ln(OH)₃ decreases from La to Lu. (3) Differences in the properties of Ln³⁺ are small, so they are separated by ion-exchange methods.
- Easy : Zr (160 pm) and Hf (159 pm) are almost the same size.
- Medium : La(OH)₃ is more basic than Lu(OH)₃.
- Exam-type : Why are Nb and Ta so similar? The lanthanoid contraction makes the 5d element Ta almost as small as the 4d element Nb.
Rule 3: Actinoids and uses — Actinoids are radioactive. They show a wider range of oxidation states (+3 to +7) than lanthanoids because the 5f, 6d and 7s energies are close. The 5f electrons shield poorly, giving an actinoid contraction that is larger than the lanthanoid contraction. Uses of lanthanoids: mischmetall (about 95 % lanthanoid metals with iron) in Mg alloys for bullets and shells and in lighter flints; ceria in glass polishing; oxides as phosphors in TV screens.
- Easy : Mischmetall is added to magnesium to make strong alloys.
- Medium : Actinoids show more oxidation states than lanthanoids; U shows +3, +4, +5, +6.
- Exam-type : Which contraction is greater, lanthanoid or actinoid? The actinoid contraction is greater because 5f electrons shield even less than 4f.
f-block: Ln mostly +3, An wider range; both show a contraction because f electrons shield poorly.
Saying that the lanthanoid contraction makes Zr and Hf different; it makes them almost identical.
Memory clue: "Poor shield, small yield" — poor shielding by f electrons shrinks the size.
Spot the difference
| Point | Lanthanoids | Actinoids |
|---|---|---|
| Orbital filled | 4f | 5f |
| Common oxidation state | +3 | +3 to +7 |
| Radioactivity | Not radioactive (except Pm) | All radioactive |
Practice questions
- 1Name the orbital that is filled in lanthanoids.
- 2What is the common oxidation state of lanthanoids?
- 3Name a use of mischmetall.
- 4Define lanthanoid contraction.
- 5Why are Zr and Hf difficult to separate?
- 6Which lanthanoid shows +4 and which shows +2?
- 7Compare the oxidation states of lanthanoids and actinoids.
- 8Explain why La(OH)₃ is more basic than Lu(OH)₃.
Show answers (try them yourself first)
- 14f.
- 2+3.
- 3In lighter flints or in magnesium alloys.
- 4The steady decrease in atomic and ionic radii from La to Lu caused by the poor shielding of 4f electrons.
- 5Their atomic radii and chemical properties are almost the same because of the lanthanoid contraction.
- 6Ce (+4) and Eu (+2).
- 7Lanthanoids mostly show +3 (also +2, +4 for a few); actinoids show a wider range from +3 up to +7 because the 5f, 6d and 7s energies are close.
- 8The size of Ln³⁺ falls from La to Lu, so the hydroxide–ion bond becomes more covalent and the base weaker for the smaller ion.