4. The d- and f-Block Elements

What you will learn

• Separate d-block position from the transition-element definition.
• Write neutral and ionic configurations and count d electrons.
• Explain physical, redox, optical and magnetic trends with their limits.
• Balance chromium and manganese oxoanion reactions in a stated medium.
• Compare lanthanoids and actinoids.
• Connect applications with relevant properties.

Big idea: closely spaced electron energies create a wide range of useful chemical behaviour.

The middle of the periodic table contains metals that can exchange different numbers of electrons, bind many ligands and respond strongly to their surroundings. A compound’s colour or magnetism can change even when the metal name stays the same.

This chapter links configurations with trends, then studies two important families of oxoanions. Use comparisons carefully: match ionic charge, medium and geometry before drawing a conclusion.

The separation of closely related f-block elements was a long analytical challenge because neighbouring ions behave so similarly. Organising elements by electronic structure helped explain that difficulty and the gradual size changes hidden within otherwise similar chemistry.

The d block is a location in the periodic table. “Transition element” is a definition about electron configuration, so the two expressions are not perfectly interchangeable.

The d-block elements occupy the central groups of the long-form periodic table. Across a transition series, electrons are added to an inner $(n-1)d$ set while the outer shell has principal quantum number $n$. The outer s and inner d energies are close enough for both to influence bonding.

A transition element has an atom with an incomplete d subshell, or can form cations with an incomplete d subshell. Zinc lies in the d block but its usual atom and stable common ion have complete d subshells. That makes it an important boundary case when the stricter transition-element definition is used.

The first familiar series runs from scandium to zinc. Later d series involve more spatially extended orbitals. The f-block rows are usually placed below the main table to keep the diagram compact; chemically they belong within the corresponding long periods. Their position on the printed page does not mean they are separate from the periodic system.

Is every element in the d block automatically a transition element under the incomplete-d definition?

Test the definition rather than the table location. Ask whether the atom or a relevant cation has an incomplete d subshell.

Result: No. Zinc is the standard example of a d-block element that does not meet the usual incomplete-d criterion.

1. What electron-configuration feature is central to the transition-element definition?
2. Why is the f block printed below the main table?

Answers

1. An incomplete d subshell in the atom or an appropriate cation.
2. It makes the table compact; those elements still belong to the long periods.

For ions of the first transition series, remove the outer s electrons before the d electrons. Writing the neutral atom correctly is only the first step.

A useful general form for many d-block atoms is $(n-1)d^{1\text{–}10}ns^{0\text{–}2}$. The close energies permit exceptions to a simple filling pattern. For the first series, chromium is commonly written $[\mathrm{Ar}]3d^54s^1$ and copper $[\mathrm{Ar}]3d^{10}4s^1$. Do not describe these exceptions as proof that one simple rule predicts every atom.

Half-filled and filled subshells can contribute stability, but electron configurations result from the balance of several energies. Avoid a universal explanation that any half-filled subshell must override every other effect. The observed ground state is the fact to explain.

Find the d count of a first-row iron(III) ion.

Iron is in the group whose neutral valence count is eight. Removing three electrons gives the formal d count for this ordinary ionic example. Confirm by starting from the neutral configuration and removing outer s electrons first.

Result: $d^{5}$. For the free-ion high-spin picture this is a half-filled d set; ligand-field splitting can matter in complexes.

For many simple transition-metal ions, group number minus oxidation state gives a convenient formal d count. It is useful bookkeeping for coordination chemistry, not a complete description of where electron density lies in every covalent compound. Keep oxidation-state counting separate from an actual charge-density picture.

1. Give the usual d count of $\mathrm{Co^{3+}}$.
2. Does $\mathrm{Zn^{2+}}$ have an incomplete d subshell?

Answers

1. $d^{6}$.
2. No. It has a complete d subshell.

Many d-block trends become easier when you connect them with three questions: how strongly are atoms bound, which electrons are available, and what stabilises the resulting ion?

Most transition metals are dense, strong metallic conductors with relatively high melting points. Delocalised electrons and contributions from d orbitals often strengthen metallic bonding. These are trends with exceptions, especially near completed d subshells. A trend is not a substitute for the measured property of a particular metal.

Across a series, increasing nuclear charge tends to pull electrons inward. Added d electrons also shield one another, so the size change is more modest than a naive charge-only picture suggests. Ions of the same element usually become smaller as positive charge increases. Compare like oxidation states and like coordination environments.

Ionisation removes an electron from a gaseous species. Successive ionisations become harder, but irregularities reflect the changing stability of initial and final configurations. An observed aqueous electrode potential also contains atomisation and hydration contributions; ionisation enthalpy alone does not determine it.

Close s and d energies permit several oxidation states. Higher states are often stabilised by electronegative ligands such as oxygen or fluorine. The most stable oxidation state depends on the compound and medium. A formal high oxidation number does not mean that an isolated bare ion with that charge exists in the solution.

For the $\mathrm{M^{2+}/M}$ couple, compare the full energetic route from solid metal to solvated ions and electrons. Strong hydration can favour ions, while atomisation and ionisation oppose their formation. Copper’s positive standard reduction potential helps explain why a simple nonoxidising acid does not ordinarily release hydrogen by reacting with copper.

The $\mathrm{M^{3+}/M^{2+}}$ comparison involves different configurations. The half-filled $d^5$ state helps stabilise $\mathrm{Mn^{2+}}$ and $\mathrm{Fe^{3+}}$. Consequently, $\mathrm{Mn^{3+}}$ can be an effective oxidant, whereas the stability of $\mathrm{Fe^{3+}}$ affects its potential differently. Use the tabulated couple and medium for a quantitative comparison.

Strong bonding to oxygen or fluorine can stabilise high formal oxidation states. Oxides and oxoanions are therefore common examples. A compound stable with fluoride need not have a stable chloride analogue in the same oxidation state. Bond energies, lattice effects and the possibility of ligand oxidation all matter.

A more negative standard reduction potential often indicates a stronger reducing metal for the defined couple. Surface passivation can make a thermodynamically favourable reaction slow. Separate equilibrium driving force from the rate controlled by a protective oxide layer.

Unpaired electrons produce paramagnetism. A species with all electrons paired is diamagnetic. The spin-only expression is a useful approximation for many first-row ions when orbital contributions are small; it is not exact for every metal ion.

$$\mu_{\rm so}=\sqrt{n(n+2)}\,\mu_B$$

Estimate the spin-only magnetic moment of a free-ion high-spin iron(III) configuration.

The d count is five. In the unsplit high-spin picture, each d orbital contains one electron. Substitute the unpaired-electron count into the spin-only expression.

Result: $\mu_{\rm so}=5.916\,\mu_B$. A coordination compound needs its ligand field considered before assigning the unpaired count.

In a ligand environment, d orbitals can have different energies. Absorbing visible light can promote an electron between these levels, and the transmitted or reflected light gives the observed colour. Species without available d–d transitions may still be coloured by charge-transfer transitions. Permanganate is a key reminder that “coloured” does not automatically mean an incomplete d subshell.

Small, charged metal ions can accept electron pairs from ligands. A ligand may bind through one or several donor atoms. Coordination changes electronic energies, colour, reactivity and sometimes spin state. The next chapter treats these ideas systematically.

Variable oxidation states support electron-transfer cycles. Metal surfaces can also adsorb reactants, bring them into a useful arrangement and change the activation pathway. Catalysis is not explained merely by saying that a metal “has d electrons”; an actual active site and route are needed.

Small atoms such as carbon, hydrogen or nitrogen can occupy spaces within a metal lattice. The resulting material may be hard and retain metallic conductivity. Such compositions need not have the simple whole-number ratios expected for elementary ionic salts.

Similar atomic sizes allow one metal to replace another in a solid structure. Alloys can have useful combinations of hardness, strength and corrosion behaviour. A metallic solid solution is different from an interstitial compound, although real materials can involve several structural features.

1. Can a d⁰ species be coloured?
2. Why can two ions with the same d count have different magnetic behaviour in complexes?

Answers

1. Yes. Charge-transfer absorption can produce colour even without d–d transitions.
2. Ligand-field splitting and pairing energies can change the number of unpaired electrons.

Chromium and manganese oxoanions combine high oxidation states with useful, medium-dependent redox reactions.

Metal–oxygen compounds range from basic oxides in lower oxidation states to more acidic oxides and oxoanions in higher states. This is a broad trend. Bonding and structure must still be considered for each compound.

Chromate and dichromate are connected by a pH-dependent equilibrium. Acid favours dichromate relative to chromate; adding base favours chromate. The chromium oxidation state remains the same during this acid–base change.

$$2 \mathrm{CrO_{4}}^{2-} + 2 \mathrm{H}^{+} \rightleftharpoons \mathrm{Cr_{2}O_{7}}^{2-} + \mathrm{H_{2}O}$$

Find chromium’s oxidation state in $\mathrm{Cr_2O_7^{2-}}$.

Use oxygen oxidation state $-2$ and require the sum to equal the ion charge: $2x+7(-2)=-2$.

Result: $x=+6$.

Dichromate acts as an oxidant in acid. Balance oxygen with water, hydrogen with hydrogen ions and charge with electrons. A medium stated in a question is part of the chemical information, not an optional annotation.

$$\mathrm{Cr_{2}O_{7}}^{2-} + 14 \mathrm{H}^{+} + 6 e^{-} \longrightarrow 2 \mathrm{Cr}^{3+} + 7 \mathrm{H_{2}O}$$

Industrial preparation of chromate starts from chromium-containing ore under oxidising alkaline conditions. Converting chromate to dichromate is then an acid–base step. Crystallisation and cation choice help isolate the desired salt. The redox step changes chromium’s state; the chromate–dichromate conversion does not.

In $\mathrm{MnO_4^-}$, manganese has oxidation state $+7$. The ion has tetrahedral connectivity. Preparation passes through manganate in alkaline conditions; further oxidation gives permanganate.

$$2 \mathrm{MnO_{2}} + 4 \mathrm{KOH} + \mathrm{O_{2}} \longrightarrow 2 \mathrm{K_{2}MnO_{4}} + 2 \mathrm{H_{2}O}$$

$$\mathrm{MnO_{4}}^{-} + 8 \mathrm{H}^{+} + 5 e^{-} \longrightarrow \mathrm{Mn}^{2+} + 4 \mathrm{H_{2}O}$$

$$\mathrm{MnO_{4}}^{-} + 2 \mathrm{H_{2}O} + 3 e^{-} \longrightarrow \mathrm{MnO_{2}} + 4 \mathrm{OH}^{-}$$

$$\mathrm{MnO_{4}}^{-} + e^{-} \longrightarrow \mathrm{MnO_{4}}^{2-}$$

1. How many electrons does one permanganate ion accept on reduction to Mn²⁺?
2. What structural feature joins the two chromium units in dichromate?

Answers

1. Five.
2. A shared bridging oxygen atom.

The lanthanoids are chemically similar, yet their gradual size change has consequences far beyond that row of the periodic table.

Across the lanthanoid series, the $4f$ set is progressively occupied, with closely spaced configurations involving outer $5d$ and $6s$ orbitals. The trivalent state is especially common because removal of the outer electrons gives a stable ionic framework while the 4f count changes across the series.

The attraction of increasing nuclear charge is not fully screened by added f electrons. Ionic sizes therefore decrease gradually across the series for comparable trivalent ions. This is lanthanoid contraction. Similar sizes of many later d-series pairs are one important consequence.

Because the lanthanoid ions are so similar in charge and size, separating neighbouring elements is difficult. Their hydroxide basicity generally decreases as ionic size decreases and attraction to oxygen increases. State that this is a trend among comparable compounds.

Although the trivalent state dominates, selected elements can also form divalent or tetravalent compounds. The relative stability of empty, half-filled or filled f sets can help interpret examples. Cerium(IV), europium(II) and ytterbium(II) are useful cases, but stability still depends on the compound and medium.

Many lanthanoid ions are coloured and paramagnetic when f electrons remain unpaired. Their internal f electrons are comparatively shielded, which helps make their chemistry more uniform than the first transition series. Lanthanoid metals are reactive and form compounds with electronegative elements.

Why is a gradual change in radius important even if the common oxidation state stays the same?

Charge is only one part of ionic behaviour. A smaller ion has a different charge density and can alter hydration, lattice energies and binding preferences.

Result: The gradual contraction changes physical and chemical properties without requiring a new common charge.

1. What common ionic state helps make lanthanoid chemistry similar?
2. Why is neighbouring-lanthanoid separation difficult?

Answers

1. The trivalent state.
2. Their common ions have very similar charges, sizes and chemical behaviour.

The actinoids share f-block features with the lanthanoids, but their greater oxidation-state variety and radioactivity make the comparison incomplete.

Actinoid filling involves the $5f$ orbitals, with nearby $6d$ and $7s$ levels. The energies and spatial reach of these orbitals can permit more valence involvement than a simple “buried f electrons” picture suggests.

Comparable actinoid ions show a contraction across the series because f electrons incompletely screen increasing nuclear charge. The pattern is not perfectly regular, especially when different oxidation states are compared. Keep charge and coordination environment consistent in a size comparison.

Early actinoids commonly show several oxidation states because the relevant valence orbitals have similar energies. Later members more often favour the trivalent state. Uranium and other early actinoids illustrate how compound formation can stabilise different formal states.

All actinoids are radioactive. Their chemistry therefore includes nuclear stability and radiation considerations in addition to ordinary bonding and redox chemistry. A school-level comparison should describe periodic trends without suggesting that these substances are routine classroom materials.

Two reported actinoid radii come from ions with different charges. Can their difference alone demonstrate the across-series contraction?

Higher ionic charge itself tends to reduce radius. A fair periodic comparison controls that variable.

Result: No. Compare ions in the same oxidation state and similar environments before assigning the difference to contraction.

1. Are all actinoids radioactive?
2. Why are oxidation states often more varied for early actinoids than for lanthanoids?

Answers

1. Yes.
2. The relevant 5f, 6d and outer s orbitals can be close enough in energy to participate differently in bonding.

A useful material is chosen for a property, not just for its place in the periodic table.

Transition metals and their compounds contribute to structural alloys, electrical conductors, catalysts, pigments and electrochemical devices. A metal suitable for one purpose may fail in another because strength, corrosion behaviour, density and chemical reactivity must be considered together.

Catalytic applications exploit an active surface or a cycle of oxidation states. Alloying changes structure and material performance. Magnetic and optical applications depend on electronic states and their environment. These examples show why “contains a transition metal” is a starting point for explanation, not an explanation by itself.

Lanthanoid-containing materials are used in permanent magnets, optical devices and selected catalysts. Mixtures and alloys can have properties that no pure component supplies alone. A material’s performance also depends on processing, purity and microstructure.

A designer needs a metal-based part that resists wear and remains useful in a damp environment. What information should be compared?

Hardness or wear resistance addresses one requirement. Corrosion behaviour in the actual environment addresses another. Density, cost of processing and compatibility with neighbouring materials may also matter.

Result: Choose from measured material properties for the intended environment; a block label or a single high melting point is insufficient.

In an exam answer, connect each use with a mechanism or property: conductivity with mobile electronic charge carriers; catalytic action with an accessible pathway; magnetic behaviour with the electronic state. Avoid presenting a disconnected list of products and element names.

1. What feature can make a transition-metal system useful in a redox catalyst?
2. Why can alloying change practical performance?

Answers

1. Access to suitable oxidation states and a pathway that regenerates the active species.
2. It changes composition and structure, which alter properties such as strength and corrosion behaviour.

Try these original questions without looking at the answers. Show equations, conditions and units when relevant.

4.1 State the incomplete-d definition of a transition element.

4.2 Explain the zinc boundary case.

4.3 Write the usual first-row iron(II) configuration after the argon core.

4.4 Give the formal d count of manganese(II).

4.5 Give the formal d count of nickel(II).

4.6 Which electrons are removed first when forming ordinary first-series cations from their neutral atoms?

4.7 Why do many transition metals show several oxidation states?

4.8 Why are oxygen and fluorine effective at stabilising high oxidation states?

4.9 Why does ionisation enthalpy alone not determine an aqueous metal-ion/metal potential?

4.10 What makes a surface-passivated metal appear less reactive than a potential comparison suggests?

4.11 Estimate spin-only moment for two unpaired electrons.

4.12 Estimate it for three unpaired electrons.

4.13 Estimate it for four unpaired electrons.

4.14 What magnetic classification follows when all electrons are paired?

4.15 Why is colour not a reliable direct count of unpaired d electrons?

4.16 Describe an interstitial compound.

4.17 Distinguish substitutional alloying from interstitial occupation.

4.18 Find chromium oxidation state in chromate, $\mathrm{CrO_4^{2-}}$.

4.19 Does acid-driven chromate-to-dichromate conversion change chromium’s oxidation state?

4.20 How many chromium atoms are present in one dichromate ion?

4.21 Describe dichromate connectivity.

4.22 Which ion is favoured relative to the other on acidifying a chromate/dichromate equilibrium mixture?

4.23 How many electrons does dichromate accept on reduction of both chromium atoms to Cr³⁺?

4.24 Find manganese oxidation state in manganate, $\mathrm{MnO_4^{2-}}$.

4.25 Give the common permanganate reduction product in sufficiently acidic medium.

4.26 Give the common product in neutral or mildly alkaline conditions.

4.27 What changes when permanganate is reduced to manganate?

4.28 Explain why a d⁰ oxoanion can be strongly coloured.

4.29 What is lanthanoid contraction?

4.30 How can contraction affect the later d series?

4.31 Why are trivalent lanthanoid ions difficult to separate?

4.32 Name one familiar divalent-lanthanoid case and one tetravalent case.

4.33 Which f subshell is associated with lanthanoid filling and which with actinoid filling?

4.34 What nuclear property is shared by all actinoids?

4.35 Why is a comparison of radii at different ionic charges unreliable for testing contraction?

4.36 Give a property-based explanation of transition-metal catalysis.

4.37 Why should a material-use answer mention the operating environment?

4.38 Plan a method for predicting magnetism of a transition-metal complex.

Answers

4.1 Its atom has an incomplete d subshell, or it can form cations with an incomplete d subshell.

4.2 It is located in the d block, but its usual atom and common Zn²⁺ ion have complete d subshells.

4.3 $3d^6$.

4.4 $d^{5}$.

4.5 $d^{8}$.

4.6 Outer s electrons are removed before d electrons.

4.7 Their outer s and nearby d electrons can contribute differently to bonding.

4.8 Strong bonding to these electronegative ligands can stabilise high formal metal states.

4.9 Atomisation and hydration, among other contributions, also enter the full energetic comparison.

4.10 A protective film can slow the reaction kinetically.

4.11 $2.828\,\mu_B$.

4.12 $3.873\,\mu_B$.

4.13 $4.899\,\mu_B$.

4.14 Diamagnetic.

4.15 Different ligand fields and charge-transfer transitions can produce colour by different routes.

4.16 Small atoms occupy spaces in a metallic lattice; composition and properties need not resemble a simple ionic salt.

4.17 In substitutional alloying, an atom replaces a lattice atom; an interstitial atom occupies a space between lattice positions.

4.18 $+6$.

4.19 No. Both contain chromium in the same oxidation state.

4.20 $2$.

4.21 Two chromium-centred tetrahedral units share one bridging oxygen.

4.22 Dichromate is favoured relative to chromate.

4.23 $6$.

4.24 $+6$.

4.25 Mn²⁺.

4.26 MnO₂ under the usual school-level conditions.

4.27 Manganese is reduced by one formal oxidation-state unit, accepting one electron per ion.

4.28 Charge-transfer transitions can absorb visible light without a d–d transition.

4.29 A gradual decrease in comparable ionic radii across the lanthanoid series from incomplete shielding by added f electrons.

4.30 It can make corresponding later-series elements unexpectedly similar in size.

4.31 Their similar size and charge give closely similar chemical behaviour.

4.32 Europium(II) and cerium(IV) are useful examples; stability remains compound- and medium-dependent.

4.33 4f for lanthanoids and 5f for actinoids.

4.34 Radioactivity.

4.35 Charge itself changes radius, so it confounds the across-series comparison.

4.36 An accessible surface or oxidation-state cycle can provide a lower-barrier route that regenerates the catalyst.

4.37 Corrosion, surface chemistry and performance depend on the surroundings, not just the pure element.

4.38 Find metal oxidation state and d count, identify geometry and ligand-field strength, assign electron occupancy, then count unpaired electrons before using a suitable magnetic model.

Quick-check answers appear directly after each topic. The selected explanations below show how to check the reasoning, rather than memorise a final line.

What nuclear property is shared by all actinoids?

Radioactivity.

Why is a comparison of radii at different ionic charges unreliable for testing contraction?

Charge itself changes radius, so it confounds the across-series comparison.

Give a property-based explanation of transition-metal catalysis.

An accessible surface or oxidation-state cycle can provide a lower-barrier route that regenerates the catalyst.

Why should a material-use answer mention the operating environment?

Corrosion, surface chemistry and performance depend on the surroundings, not just the pure element.

Plan a method for predicting magnetism of a transition-metal complex.

Find metal oxidation state and d count, identify geometry and ligand-field strength, assign electron occupancy, then count unpaired electrons before using a suitable magnetic model.