The d- and f-Block Elements

Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Chemistry Class XII, Chapter 4.

Why is copper sulphate blue and zinc sulphate white?

Dissolve a spatula of copper sulphate in water and you get a vivid blue. Do the same with zinc sulphate, which sits right next to copper in the periodic table, and the solution is colourless. Iron(III) gives a yellow solution, manganese(II) a faint pink, and scandium nothing at all. These are not random facts. They all come from one thing: how many electrons sit in the d orbitals of the metal ion. Partly filled d orbitals give colour, magnetism, many oxidation states and a talent for forming complexes. Empty or full d orbitals give none of these. This chapter is that single idea, followed through the periodic table.

Where the d-block sits

The d-block is Groups 3 to 12. Across four long periods the (n−1)d orbitals fill, giving four series. The f-block is the two rows placed below the table, where 4f and 5f orbitals fill.

The general outer configuration is (n−1)d¹⁻¹⁰ ns¹⁻². The ns and (n−1)d levels are close in energy, so electrons can move between them whenever that gives a half filled or fully filled d set, which has extra exchange energy stabilisation.

Who counts as a transition metal. IUPAC defines a transition metal as an element with an incomplete d subshell in the neutral atom or in any of its common ions. Zn, Cd and Hg have d¹⁰ in the atom and in their only common ion (M²⁺), so they are d-block elements but not transition elements. Copper passes the test because Cu²⁺ is d⁹.

Pause and predict: which has more unpaired electrons, a chromium atom or a Cr³⁺ ion?

Check your answer: the atom, with six (3d⁵ 4s¹, all parallel). Cr³⁺ is 3d³ with three. When a transition metal forms an ion, the ns electrons leave first, then d electrons. That is why Fe²⁺ is 3d⁶, not 3d⁴ 4s².

Trends across the 3d series

Moving left to right, nuclear charge rises and electrons enter an inner (n−1)d subshell. d electrons shield the outer 4s electrons poorly, so the pull on the outer shell grows and sizes fall slowly. Almost every trend below comes from this one fact.

The values of ionisation enthalpy are not smooth, and the irregularities are testable. The first ionisation enthalpy of Zn is high (906 kJ mol⁻¹) because it must remove an electron from the very stable 3d¹⁰ 4s² set. The second ionisation enthalpies of Cr (1590) and Cu (1958) are unusually high because the second electron comes out of the stable d⁵ or d¹⁰ core. The third ionisation enthalpy of Mn is high because Mn²⁺ is already a stable d⁵ ion, while that of Fe is lower because Fe²⁺ (d⁶) loses an electron to become d⁵.

Because successive ionisation enthalpies in the d-block are close together, losing more than two electrons is not much harder than losing two. That is what opens the door to many oxidation states.

Oxidation states: the sum of the s and d electrons

A main group element such as Cl shows states 7, 5, 3, 1, separated by two. Transition elements show states that differ by one, because ns and (n−1)d electrons can be lost one at a time. The maximum state of an element up to Mn equals the number of ns plus (n−1)d electrons.

After Mn, the highest state falls, because the d electrons are now held too tightly to be used for bonding. +2 is the lowest common state for the series, from the two 4s electrons, and Sc is the exception with only +3.

Stability of the states. Higher oxidation states are more stable in the left of the series and in the heavier 4d and 5d elements. Lower states are more stable on the right. Compounds in high states are oxidising, and those in low states are reducing, as the table shows.

The case of copper deserves its own line: 2Cu⁺ → Cu²⁺ + Cu in aqueous solution. Cu⁺ looks stable on paper (d¹⁰), but the very large hydration enthalpy of Cu²⁺ more than pays for the high second ionisation enthalpy, so Cu²⁺ wins in water. Cu(I) survives only in insoluble compounds such as Cu₂O and CuCl.

Why fluorine and oxygen reach the highest states. The highest oxidation states appear in fluorides and oxides because F and O are small and very electronegative. Fluorine does it through high lattice and bond energies (VF₅, CrF₆). Oxygen has an extra trick: it forms multiple bonds to the metal. That is why Mn reaches +7 in Mn₂O₇ and MnO₄⁻, while its highest fluoride is only MnF₄. Oxygen also stabilises the oxo-anions VO₄³⁻, CrO₄²⁻ and MnO₄⁻.

The same bonding changes the acid–base character of oxides. The higher the oxidation state, the more acidic the oxide, because the metal holds the oxygen more covalently.

Electrode potentials and reactivity

The standard electrode potential E°(M²⁺/M) of a metal depends on three energies: atomisation, ionisation and hydration. Across the 3d series most metals have negative E° values, so they dissolve in dilute acids and release hydrogen. The values become less negative towards the right because the sum of the first two ionisation enthalpies rises.

Pause and predict: will copper release hydrogen from dilute hydrochloric acid?

Check your answer: no. E°(Cu²⁺/Cu) is positive, so H⁺ cannot oxidise copper. Copper dissolves only in oxidising acids such as nitric acid or hot concentrated sulphuric acid, where the oxidising agent is the anion, not H⁺.

Magnetism: counting unpaired electrons

An ion with unpaired electrons is paramagnetic and is pulled into a magnetic field. An ion with all electrons paired is diamagnetic and is slightly repelled. For 3d ions the orbital motion contributes almost nothing (it is "quenched"), so the spin-only formula works:

$$\mu = \sqrt{n(n+2)}\ \text{BM}$$

where $n$ is the number of unpaired electrons.

For d⁶ to d⁹ you fill the five d orbitals singly first (Hund's rule) and only then pair, so the number unpaired is 10 − (number of d electrons) once you pass d⁵.

Worked example: magnetic moment of an ion. Find the spin-only magnetic moment of Co²⁺ (Z = 27) and of Fe³⁺ (Z = 26), and say which is more paramagnetic.

Fe³⁺ is more paramagnetic. The common wrong step is to count the unpaired electrons of the neutral atom (4 for Fe) or to remove d electrons before s electrons.

Beyond the textbook. For the exam: use $\mu = \sqrt{n(n+2)}$ for every 3d ion. In reality: the spin-only value is an approximation that works best for 3d ions. For the heavier 4d and 5d ions, and for some 3d ions such as Co²⁺, the orbital motion adds to the moment and measured values are higher than the formula predicts. Why it matters: if a question gives you an observed value above the spin-only value, the orbital contribution is the reason, not an extra unpaired electron.

Colour: d–d transitions

In a hydrated ion, the water molecules around the metal split the five d orbitals into a lower and a higher group. An electron in the lower group can absorb a photon of visible light and jump to the higher group. This is a d–d transition. The ion transmits the light that is left, so you see the complementary colour of the light absorbed.

An ion with no d electrons (d⁰) has nothing to excite. An ion with a full d set (d¹⁰) has no empty d orbital to jump into. Either way there is no d–d transition and the ion is colourless.

Beyond the textbook. For the exam: colour needs a d–d transition, so d⁰ and d¹⁰ ions are colourless. In reality: MnO₄⁻ (Mn⁷⁺, d⁰) and CrO₄²⁻ (Cr⁶⁺, d⁰) are intensely coloured even though they have no d electrons. Their colour comes from charge transfer, where an electron jumps from an oxygen lone pair onto the metal. Why it matters: so "d⁰ means colourless" applies to simple aqua ions like Sc³⁺ and Ti⁴⁺, not to oxo-anions of high oxidation states.

Complexes, catalysts, interstitial compounds and alloys

Complex formation. Transition metal ions are small, highly charged and have empty d orbitals that can accept electron pairs from ligands. Examples are [Fe(CN)₆]³⁻, [Cu(NH₃)₄]²⁺ and [PtCl₄]²⁻. The next chapter treats them in full.

Catalysis. A transition metal catalyst works in two ways. It offers a surface to which reactants stick, and it can switch between oxidation states to pass electrons along.

Interstitial compounds. Small atoms such as H, C and N fit into the gaps (voids) of a metal lattice and give compounds such as TiC, Mn₄N, Fe₃H, VH₀.₅₆ and TiH₁.₇. They are non-stoichiometric, very hard, high melting, keep metallic conductivity and are chemically inert. The gaps are filled without changing the metal's crystal structure much, which is why the metal stays metallic.

Alloys. Two metals form an alloy when their metallic radii differ by no more than about 15 percent, so atoms of one can replace the other in the lattice. Transition metals have similar radii, so they alloy readily. Brass is Cu with Zn, bronze is Cu with Sn, and stainless steel is Fe with Cr and Ni. Ferrous alloys use Cr, V, W, Mo and Mn to make steel harder or rust resistant.

Potassium dichromate, K₂Cr₂O₇

Making it. The ore is chromite, FeCr₂O₄. It is roasted with sodium carbonate in excess air:

4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂

The yellow sodium chromate is filtered and acidified to give orange dichromate, which is converted to the potassium salt by adding KCl. The potassium salt is far less soluble than the sodium salt, so orange crystals of K₂Cr₂O₇ separate.

2CrO₄²⁻ + 2H⁺ → Cr₂O₇²⁻ + H₂O, then Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl

Chromate and dichromate switch with pH, without any redox. In both, chromium is +6. In acid the equilibrium goes to orange dichromate. In alkali it goes back to yellow chromate.

Cr₂O₇²⁻ + 2OH⁻ ⇌ 2CrO₄²⁻ + H₂O

Chromate is a single tetrahedron. Dichromate is two tetrahedra sharing one oxygen at a corner, with a Cr–O–Cr angle of about 126°.

Oxidising action in acid. Dichromate is a strong oxidant, and the half reaction is

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O, E° = +1.33 V

The orange solution turns green as Cr³⁺ forms. Typical reactions, all in acid:

Potassium permanganate, KMnO₄

Making it. Pyrolusite, MnO₂, is fused with KOH in air (or with an oxidiser such as KNO₃). The green K₂MnO₄ formed is then oxidised to purple permanganate, electrolytically in alkaline solution.

2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O, then MnO₄²⁻ → MnO₄⁻ + e⁻ at the anode

In the laboratory, Mn²⁺ is oxidised by peroxodisulphate: 2Mn²⁺ + 5S₂O₈²⁻ + 8H₂O → 2MnO₄⁻ + 10SO₄²⁻ + 16H⁺.

Manganate disproportionates in acid. In neutral or acidic solution, manganate (+6) splits into permanganate (+7) and MnO₂ (+4). A disproportionation is a reaction in which one species is both oxidised and reduced.

3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O

Heating. KMnO₄ decomposes at 513 K: 2KMnO₄ → K₂MnO₄ + MnO₂ + O₂.

What permanganate does depends on the medium. In acid the product is almost colourless Mn²⁺ (five electrons taken). In neutral or faintly alkaline solution the product is brown MnO₂ (three electrons taken).

Why not use HCl? A permanganate titration is done in dilute sulphuric acid. In HCl, permanganate oxidises Cl⁻ to Cl₂, so some of the permanganate is used up on the acid and the titre value is wrong. Sulphuric acid is not oxidised by permanganate.

Worked example: a permanganate titration. 25.0 mL of an acidified Fe²⁺ solution needs 20.0 mL of 0.020 M KMnO₄ to reach the pink end point. Find the molarity of Fe²⁺.

The common wrong step is to treat the reaction as 1:1, which gives 0.016 M. Permanganate takes five electrons in acid and iron(II) gives one, so the ratio is 1 : 5. The end point needs no indicator, because the first extra drop of permanganate colours the solution pink.

Pause and predict: a purple permanganate solution is made alkaline and left with a little reducing agent, then it turns green. What is the green species?

Check your answer: manganate, MnO₄²⁻, formed by a single electron gain (Mn from +7 to +6). In strongly alkaline solution the reduction stops at +6.

The lanthanoids

The 14 elements after lanthanum fill the 4f subshell. Their general configuration is [Xe] 4fⁿ 5d⁰⁻¹ 6s². La is [Xe] 5d¹ 6s², Ce is [Xe] 4f¹ 5d¹ 6s² and Gd is [Xe] 4f⁷ 5d¹ 6s² (the half filled 4f⁷ again). They are similar to each other because the 4f electrons are deep inside the atom and take little part in bonding.

Oxidation states. +3 is the common state for all of them. Some show +2 or +4 when that leaves an empty, half filled or full 4f subshell.

Lanthanoid contraction is the steady fall in atomic and ionic size from La to Lu. Ln³⁺ radii drop from about 103 pm (La³⁺) to 86 pm (Lu³⁺). The cause is poor shielding. An f orbital has a diffuse shape, so one 4f electron shields another badly. Each added proton therefore pulls the whole electron cloud in a little further, and these small steps add up over 14 elements.

Consequences of the contraction.

Properties and uses. Lanthanoids are silvery white, soft metals that tarnish in air and burn in halogens. They are strongly electropositive with E°(Ln³⁺/Ln) around −2.2 to −2.4 V, so they react with water to give Ln(OH)₃ and hydrogen: 2Ln + 6H₂O → 2Ln(OH)₃ + 3H₂. They form oxides Ln₂O₃, hydroxides, nitrides, carbides, and hydrides. Many Ln³⁺ ions are coloured through f–f transitions, while La³⁺ and Lu³⁺ (4f⁰ and 4f¹⁴) are colourless and diamagnetic.

Beyond the textbook. For the exam: lanthanoids are used in Mischmetall, as cracking catalysts and as phosphors. In reality: a neodymium–iron–boron alloy makes the strongest permanent magnets known, found in headphones, hard drives and wind turbine generators. Promethium, the only radioactive lanthanoid, has no stable isotope at all. Why it matters: "rare earth" is a misnomer. These elements are not especially rare in the crust. They are hard to separate from one another, for exactly the reason the contraction makes them so alike.

The actinoids

The 14 elements from Th to Lr fill the 5f subshell, with the general configuration [Rn] 5f¹⁻¹⁴ 6d⁰⁻² 7s². All are radioactive, and the elements beyond uranium (Z > 92, the transuranium elements) are all synthetic. The chemistry is harder to organise than that of the lanthanoids for one reason: 5f orbitals are less buried than 4f orbitals. They extend further from the nucleus, so 5f electrons can take part in bonding.

Actinoid metals are reactive and silvery. They react with boiling water to give a mixture of oxide and hydride, combine with most non-metals at moderate temperature, are attacked by HCl, and are only slightly affected by HNO₃ because of a protective oxide layer. Alkalis have no action on them.

Uses worth remembering

Exam traps

Reaction sheet

60-second revision

Where this chapter connects

• Before this: Classification of Elements and Periodicity in Properties gives the periodic trends that this chapter bends for the d-block, and Redox Reactions is the language for every oxidising action above.
• Next: Coordination Compounds explains the colour and magnetism of these ions by crystal field theory, picking up the d–d transitions begun here.
• Related: Electrochemistry turns the E° values here into cell potentials and the Nernst equation.