NCERT Solutions for Class 12 Chemistry Chapter 4: The d- and f-Block Elements

Definition used. A transition element is one which has an incompletely filled d-subshell either in the ground state of the atom or in any of its common oxidation states.

Silver. The ground-state configuration of Ag (Z = 47) is [Kr]4d¹⁰5s¹ — the 4d subshell is completely filled. However, silver exhibits the +2 oxidation state (for example in AgF₂ and in some complexes); in Ag²⁺ the configuration is 4d⁹, i.e. the d-subshell is incompletely filled (one unpaired electron). Therefore silver is regarded as a transition element.

The same reasoning is applied to copper (Cu²⁺, 3d⁹), whereas zinc, cadmium and mercury (only +2 states with d¹⁰) are not transition elements.

Extra depth. The definition of a transition element in terms of any common oxidation state is important: it brings in Cu (Cu²⁺, 3d⁹), Ag (Ag²⁺, 4d⁹) and Au (Au³⁺, 5d⁸) while excluding Zn, Cd and Hg, whose only common ions are d¹⁰. Silver is nevertheless a typical “d-block metal” in many respects — coinage metal, complex formation ([Ag(NH₃)₂]⁺) and catalysis — so the +2 state argument only formalises this.

Conclusion: Silver is a transition element because it shows the +2 oxidation state (4d⁹) with an incompletely filled d-orbital.

Concept. The enthalpy of atomisation measures the strength of metallic bonding. In transition metals the bonding involves both the ns electrons and the unpaired (n−1)d electrons; the greater the number of unpaired d electrons, the stronger the interatomic bonding and the higher Δ_aH° (maxima near the middle of the series: V 515, Cr 397 etc.).

Zinc. Zn has the configuration 3d¹⁰4s². Its 3d subshell is completely filled, so there are no unpaired d electrons and no d electrons take part in metallic bonding; only the two 4s electrons are involved. The metallic bond is therefore weak, giving the lowest enthalpy of atomisation (126 kJ mol⁻¹), as well as a low melting point and relatively high volatility.

Extra depth. Metallic bonding strength tracks the number of unpaired electrons: Δ_aH° climbs to a maximum near vanadium (515 kJ mol⁻¹) and falls to a minimum at Zn (126). The same reason accounts for the low melting point (693 K) and the low boiling point of zinc relative to other 3d metals; Hg, with 5d¹⁰6s² and a still weaker interaction, is a liquid at room temperature.

Conclusion: Zinc has a lowest Δ_aH° because, with a d¹⁰ configuration, no 3d electrons participate in metallic bonding (no unpaired d electrons).

Answer: Manganese (Z = 25) — it exhibits all oxidation states from +2 to +7 (+2, +3, +4, +5, +6, +7).

Reason. Mn has the configuration 3d⁵4s², with the maximum number of unpaired electrons (five d + two s electrons available for bonding). All of them can be lost or shared in compounds, and the successive oxidation states differ by one unit. Examples: Mn(II) in MnCl₂, Mn(III) in Mn₂O₃, Mn(IV) in MnO₂, Mn(V) in K₃MnO₄, Mn(VI) in MnO₄²⁻, Mn(VII) in MnO₄⁻ and Mn₂O₇.

At the beginning (Sc, Ti) there are too few electrons to lose or share, and at the end (Cu, Zn) there are too many d electrons (fewer orbitals available), so the number of oxidation states is smaller.

Extra depth. Mn is the only 3d element showing every oxidation state from +2 to +7, and the maximum states are found in compounds with O or F (MnO₄⁻, Mn₂O₇); Mn(II) is the most stable in acid solution. The reasons for the smaller range at the ends of the series are: too few electrons in Sc and Ti and too many paired d electrons in Cu and Zn.

Conclusion: Manganese shows the largest number of oxidation states (+2 to +7) because of its five unpaired 3d electrons plus two 4s electrons.

Concept. E°(M²⁺/M) depends on the overall energy needed to convert M(s) to M²⁺(aq): atomisation enthalpy (Δ_aH°) + first and second ionisation enthalpies (Δ_iH₁° + Δ_iH₂°) − hydration enthalpy of M²⁺ (Δ_hydH°). A less negative (or positive) E° means that this conversion is energetically unfavourable.

For copper, the very high second ionisation enthalpy (1958 kJ mol⁻¹, because Cu⁺ is a stable 3d¹⁰ ion) together with a high atomisation enthalpy makes the sum 3042 kJ mol⁻¹, and this energy is not balanced by the hydration enthalpy of Cu²⁺ (−2121 kJ mol⁻¹, not sufficiently negative). Hence Cu²⁺ is not stabilised enough, and Cu²⁺ is reduced to Cu easily: E° is positive. Consequently copper does not liberate H₂ from acids; only oxidising acids (HNO₃, hot concentrated H₂SO₄) dissolve it.

Extra depth. The high atomisation and ionisation enthalpies of Cu outweigh its hydration enthalpy, which explains the positive E° and also why copper is a “noble” metal that is not attacked by dilute HCl or H₂SO₄ (no H₂ is liberated). With nitric acid Cu is oxidised by NO₃⁻, not by H⁺: 3Cu + 8HNO₃(dil.) → 3Cu(NO₃)₂ + 2NO + 4H₂O.

Conclusion: Cu has a positive E° because its high Δ_aH° and Δ_iH₁° + Δ_iH₂° are not compensated by the hydration enthalpy of Cu²⁺.

Data (kJ mol⁻¹, Table 4.2).

Explanation. The ionisation enthalpy generally increases along the series because of increasing nuclear charge, but the trend is irregular because the removal of an electron alters the relative energies of the 4s and 3d orbitals, and because different d-configurations have different stability through exchange energy (exchange energy is proportional to the number of pairs of parallel spins in degenerate orbitals; losing it makes ionisation easier, gaining/retaining stability makes it harder). The d⁰, d⁵ and d¹⁰ configurations are exceptionally stable:

• Δ_iH₁°: Cr (653) hardly rises above V (650) because its single 4s¹ electron is removed to give the stable half-filled d⁵ Cr⁺, while Mn (3d⁵4s², 717) shows a sharp rise because of the stable half-filled d⁵ together with the filled 4s²; Zn (3d¹⁰4s²) has the highest value, 906 kJ mol⁻¹, because of its completely filled stable configuration.
• Δ_iH₂°: unusually high for Cr (1592) and Cu (1958) because Cr⁺ (3d⁵) and Cu⁺ (3d¹⁰) have stable half-filled/completely filled subshells which are hard to disrupt; the value for Mn (1509) and Zn (1734) is relatively low as their M⁺ (3d⁵4s¹ and 3d¹⁰4s¹) lose a 4s electron.

Extra depth. Plotting ionisation enthalpies against atomic number gives a zig-zag instead of a smooth rise; the “breaks” correspond to Cr⁺ (d⁵) and Cu⁺ (d¹⁰) for IE₂ and Mn²⁺ (d⁵) and Zn²⁺ (d¹⁰) for IE₃. The same stability argument explains why Mn exhibits a low IE₂ whereas Cr exhibits a high IE₂ — a favourite examination question.

Conclusion: The irregularities arise from the varying stability of the d⁰, d⁵ and d¹⁰ configurations (exchange energy) and from the change in the relative energies of 4s and 3d orbitals on ionisation.

Reason. Oxygen and fluorine are the two smallest and most electronegative elements, so they can oxidise a metal to its highest oxidation state.

• Fluorine stabilises high oxidation states because of the high lattice energy of the fluoride (e.g. CoF₃) or the high bond enthalpy of the covalent fluoride (e.g. VF₅, CrF₆).
• Oxygen can form multiple bonds (pπ–dπ) with the metal atom, e.g. Mn=O in MnO₄⁻ and Mn₂O₇, CrO₄²⁻, VO₄³⁻; therefore oxygen even exceeds fluorine in stabilising high oxidation states (the highest fluoride of Mn is MnF₄ whereas the highest oxide is Mn₂O₇ with Mn(VII)).

Other halogens (Cl, Br, I) are larger, less electronegative and cannot form multiple bonds; moreover they are oxidised by the highest-oxidation-state species.

Extra depth. The tendency of F and O to bring out the highest oxidation state is also seen in compounds of the p-block (e.g. PCl₅, SF₆, IF₇) and has practical importance: KMnO₄ and K₂Cr₂O₇ are powerful oxidants because the metals are in their highest states, and these are present as oxoanions, not as halides. The highest state is often given by fluorides at a lower oxidation number than the oxides (Mn: MnF₄ versus Mn₂O₇).

Conclusion: Oxygen and fluorine, being small and highly electronegative (and oxygen forming multiple bonds), bring the metal to its highest oxidation state.

Answer: Cr²⁺ is the stronger reducing agent.

Reasoning with E° values. E°(Cr³⁺/Cr²⁺) = −0.41 V (negative) whereas E°(Fe³⁺/Fe²⁺) = +0.77 V (positive). A more negative E° means that the lower oxidation state is a stronger reducing agent, i.e. oxidation Cr²⁺ → Cr³⁺ is much more favourable than Fe²⁺ → Fe³⁺.

Reasoning with electronic configuration. Cr²⁺ is 3d⁴ and on oxidation becomes Cr³⁺, 3d³ (t₂g³ — a half-filled t₂g level, particularly stable in aqueous medium). Fe²⁺ is 3d⁶ and on oxidation becomes Fe³⁺, 3d⁵ — half-filled d⁵, but in water the d³ (t₂g³, higher CFSE) configuration is more stable than d⁵.

Cr²⁺ even reduces H⁺: 2Cr²⁺ + 2H⁺ → 2Cr³⁺ + H₂.

Extra depth. The reducing power of the +2 states in the 3d series is V²⁺ > Cr²⁺ > Fe²⁺ — Cr²⁺ is oxidised even by water or H⁺ in acid and so must be handled under an inert atmosphere. Fe²⁺ is a weaker reductant, with a moderate E°(Fe³⁺/Fe²⁺) = +0.77 V, and survives in water, which is why Mohr’s salt is used as a primary standard in redox titrations.

Conclusion: Cr²⁺ (d⁴ → d³) is a stronger reducing agent than Fe²⁺ (d⁶ → d⁵).

Step 1 — configuration. Z = 27 is cobalt: Co = [Ar]3d⁷4s². Co²⁺ = [Ar]3d⁷ (the two 4s electrons are lost first).

Step 2 — unpaired electrons. In the weak-field aqua ion, 3d⁷ is t₂g⁵e_g² (high spin): three unpaired electrons, n = 3.

μ = √[n(n + 2)] = √(3 × 5) = √15 = 3.87 BM

(The observed values for Co²⁺ salts are higher, 4.4–5.2 BM, because the orbital contribution is not completely quenched.)

Extra depth. The spin-only formula gives the minimum value; observed moments for Co²⁺ in octahedral complexes are higher (4.4–5.2 BM) because of orbital contribution, especially for the t₂g⁵e_g² configuration. This is why Table 4.7 lists “calculated” and “observed” moments separately.

Conclusion: The spin-only magnetic moment of Co²⁺ is 3.87 BM.

Reason. Cu⁺ (3d¹⁰) undergoes disproportionation in aqueous solution into Cu²⁺ and Cu:

2Cu⁺(aq) → Cu²⁺(aq) + Cu(s)

The standard potentials are 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 (positive; reaction is spontaneous, K is large).

Why Cu²⁺(aq) is favoured: although the second ionisation enthalpy of copper is high, the hydration enthalpy of Cu²⁺ is much more negative (small size, double charge) than that of Cu⁺, and this more than compensates for IE₂. Hence Cu(II) is the stable ion in water. Cu(I) is stable only in insoluble compounds (Cu₂O, CuCl, CuI) or complexes ([Cu(CN)₂]⁻) .

Extra depth. The disproportionation of Cu⁺ is the reason why copper(I) is stable only in insoluble compounds (Cu₂O, CuCl, CuI) or in complexes with strong ligands (CN⁻, NH₃, thiourea), where the ligand stabilises the d¹⁰ configuration. The reverse (comproportionation) — Cu²⁺ + Cu → 2Cu⁺ — occurs in hot concentrated HCl, giving [CuCl₂]⁻.

Conclusion: Cu⁺ disproportionates (2Cu⁺ → Cu²⁺ + Cu) because the greater hydration enthalpy of Cu²⁺ makes Cu(II) more stable in water.

Concept. Both contractions arise from imperfect shielding: as the nuclear charge increases by one unit from element to element, the added electrons (4f in lanthanoids, 5f in actinoids) do not shield each other completely, so the effective nuclear charge increases and the size decreases.

Reason for the larger actinoid contraction. The 5f electrons shield each other much less effectively than 4f electrons: 5f orbitals are more diffuse (less buried in the atom) and penetrate the inner core less. Therefore, with each unit increase in nuclear charge, the outer electrons are pulled in by a greater net attraction, and the decrease in size from element to element is greater in actinoids than in lanthanoids.

Extra depth. The 4f and 5f orbitals differ in their radial distribution: 5f orbitals extend further from the nucleus, so they are less “inner” and 5f electrons can take part in bonding, giving the actinoids a larger range of oxidation states. Because shielding by 5f electrons is poorer, the effective nuclear charge rises more steeply and the actinoid contraction is greater.

Conclusion: Because of the poorer shielding by 5f electrons (compared with 4f), the actinoid contraction is greater than the lanthanoid contraction.

Rule. For d-block elements the ns electrons are removed first, then (n−1)d; for lanthanoids/actinoids the ns (and any (n−1)d) electrons are removed before the f-electrons when forming M³⁺.

Notes: Ce⁴⁺ and Th⁴⁺ attain noble-gas configurations, which explains their stability; Lu²⁺ loses its two 6s electrons and retains the completely filled 4f¹⁴ subshell plus one 5d electron.

Extra depth. The pattern of loss of electrons — ns first, then (n−1)d — applies to all d-block ions and is the reason that Fe²⁺ is 3d⁶, not 3d⁴4s². In lanthanoids the outermost 6s and the 5d electron (if any) are lost first and the 4f electrons are held tightly, so Ln³⁺ is 4fⁿ.

Conclusion: Cr³⁺ [Ar]3d³; Pm³⁺ [Xe]4f⁴; Cu⁺ [Ar]3d¹⁰; Ce⁴⁺ [Xe]; Co²⁺ [Ar]3d⁷; Lu²⁺ [Xe]4f¹⁴5d¹; Mn²⁺ [Ar]3d⁵; Th⁴⁺ [Rn].

Electronic configurations. Mn²⁺ = [Ar]3d⁵ (half-filled d-subshell, each of the five d orbitals singly occupied with parallel spins — large exchange energy, extra stability). Fe²⁺ = [Ar]3d⁶ (the sixth electron is paired). Oxidation to +3 gives Mn³⁺ = 3d⁴ and Fe³⁺ = 3d⁵.

• Oxidation of Mn²⁺ → Mn³⁺ destroys the stable half-filled d⁵ arrangement (d⁵ → d⁴) and needs a large third ionisation enthalpy (3260 kJ mol⁻¹).
• Oxidation of Fe²⁺ → Fe³⁺ produces the stable half-filled d⁵ configuration (d⁶ → d⁵), requiring a smaller third ionisation enthalpy (2962 kJ mol⁻¹).

The same conclusion is seen in the electrode potentials: E°(Mn³⁺/Mn²⁺) = +1.57 V (Mn³⁺ is a strong oxidant, so Mn²⁺ resists oxidation) whereas E°(Fe³⁺/Fe²⁺) = +0.77 V (Fe²⁺ is oxidised much more easily).

Extra depth. The magnitude of the difference in IE₃ between Mn (3260) and Fe (2962 kJ mol⁻¹) correlates with the standard potentials of M³⁺/M²⁺ (+1.57 V for Mn, +0.77 V for Fe), allowing a direct comparison of thermodynamic data. This explains the green Fe²⁺ turning yellow-brown on exposure to air whereas Mn²⁺ salts (pale pink) are stable in air.

Conclusion: Mn²⁺ (d⁵) is more stable towards oxidation than Fe²⁺ (d⁶) because oxidation of Fe²⁺ gives the stable d⁵ Fe³⁺ whereas oxidation of Mn²⁺ would break the stable d⁵ configuration.

Concept. The +2 state arises by loss of the two 4s electrons, giving M²⁺ with 3dⁿ configuration. As the atomic number increases (Sc → Mn), the nuclear charge increases and the 3d orbitals are stabilised (lower in energy, more tightly held), so the 3d electrons are lost less readily. Therefore, the tendency to lose further electrons to go to +3 and higher states decreases and the +2 state becomes progressively more stable.

Supporting evidence: the successive ionisation enthalpies rise (third IE: Ti 2657, V 2833, Cr 2990, Mn 3260 kJ mol⁻¹), so the removal of the third electron becomes harder; the M³⁺/M²⁺ potential becomes more positive, ending with Mn²⁺ (d⁵), the most stable M²⁺ ion of the first half of the series (Mn³⁺ is a strong oxidant).

Extra depth. The first half of the 3d series is characterised by a rise in the stability of M²⁺ and the second half by the stability of M²⁺ being restricted by the increasingly large IE₃ (Fe → Zn), which is why +3 is not important for Ni, Cu and Zn. The pattern is summarised by E°(M³⁺/M²⁺): Ti −0.37, V −0.26, Cr −0.41, Mn +1.57, Fe +0.77, Co +1.97.

Conclusion: With increasing atomic number the 3d orbitals are stabilised and IE₃ increases, so the +2 state (up to Mn²⁺, d⁵) becomes more and more stable.

Principle. The relative stability of oxidation states is decided to a great extent by the stability of the d⁰, d⁵ (half-filled) and d¹⁰ (completely filled) configurations, together with the numbers of s + d electrons available; but the factor is not the only one (hydration enthalpy, lattice energy, ionisation enthalpies and bond formation also matter).

Limitation. The ionisation enthalpies give only guidance — the problem is very complex. Cu²⁺ (3d⁹) is more stable than Cu⁺ (3d¹⁰) in water because of its much larger hydration enthalpy, which shows that electronic configuration alone does not decide stability.

Extra depth. Electronic configuration is a powerful, but not the only, guide; the stability of an oxidation state in solution also depends on hydration enthalpy and, in solids, on lattice energy (as in CoF₃) — for example Co³⁺ is unstable in water (E° = +1.97 V) although d⁶ is a good configuration, whereas in [Co(NH₃)₆]³⁺ it is stable (see Unit 5).

Conclusion: The electronic configuration (especially d⁰, d⁵, d¹⁰) largely decides the stability of oxidation states, but hydration/lattice enthalpies also contribute.

Method. Add the d electrons to the two 4s electrons to find the maximum oxidation state (sum of s + d electrons, valid up to Mn), and note the common stable states of the corresponding element.

The maximum oxidation state rises to Mn(VII) and then falls abruptly (Fe, Co, Ni, Cu, Zn mostly +2, +3).

Extra depth. The ground-state configurations that are asked about (3d³, 3d⁵, 3d⁸, 3d⁴) correspond to V, Mn, Ni and — for d⁴ — to no neutral atom; recognising this subtle point is a sign of mastery. For Cr the ground state is 3d⁵4s¹, which explains why Cr is more stable as Cr³⁺ (3d³) and Cr(VI) (all six outer electrons used).

Conclusion: 3d³ (V): +5/+4 (also +2, +3); 3d⁵ (Mn): +2 and +7; 3d⁸ (Ni): +2; 3d⁴ occurs only as Cr²⁺/Mn³⁺ ions — for chromium +3 and +6 are the stable states.

Concept. The highest oxidation number equals the group number from Group 3 (Sc) to Group 7 (Mn); oxoanions (MO₄ⁿ⁻ type, tetrahedral) of metals having oxidation state equal to the group number are:

Titanium (Group 4, +4) and Fe (Group 8) do not form such oxoanions in common compounds (ferrate(VI), FeO₄²⁻, would be +6 and is not equal to the group number 8; Ti(IV) is found as TiO²⁺ and TiO₂). Chromate and dichromate are interconvertible by changing pH, with Cr remaining +6.

Extra depth. The tetrahedral oxoanions of the first row (VO₄³⁻, CrO₄²⁻, MnO₄⁻) show the metal in a high oxidation state surrounded by four O atoms; the π bonding between oxygen p orbitals and metal d orbitals shortens the M–O bond and increases stability. Both CrO₄²⁻ (yellow) and MnO₄⁻ (purple) are coloured through charge transfer, not d–d transitions, since the metal has d⁰ configuration.

Conclusion: VO₃⁻/VO₄³⁻ (V +5), CrO₄²⁻/Cr₂O₇²⁻ (Cr +6) and MnO₄⁻ (Mn +7).

Definition. The overall, fairly regular decrease in atomic and ionic radii of the lanthanoids with increasing atomic number, from La to Lu (Ln³⁺ radius falls from 106 pm for La³⁺ to 86 pm for Yb³⁺ — Table 4.9), is called lanthanoid contraction.

Cause. Along the series, the nuclear charge increases by one at each step and the additional electron enters the inner 4f subshell. The shielding of one 4f electron by another is very poor (less than that of one d electron by another, because of the diffuse shape of f orbitals), so the effective nuclear charge increases and pulls in the outer electrons — the entire 4fⁿ shell shrinks.

Consequences.

1. Similar radii of the second and third transition series: the 5d elements have almost the same radii as the 4d elements, in spite of an extra shell (Zr 160 pm, Hf 159 pm).
2. Very similar properties of the pairs Zr–Hf, Nb–Ta, Mo–W, which therefore occur together in nature and are very difficult to separate by chemical methods.
3. The properties of the third-series elements are influenced: high ionisation enthalpies and densities, and the greater stability of higher oxidation states of 5d elements.
4. Gradual change in the properties of lanthanoids (e.g. the basic character of hydroxides decreases and ionic size decreases from La³⁺ to Lu³⁺) makes their separation by ion-exchange methods possible.

Extra depth. The practical effect of the lanthanoid contraction is that the 4d and 5d elements of a group are separated only by special methods (ion exchange, solvent extraction), and that Hf is present in Zr ores at 1–3 %; Zr is used in nuclear reactors only after removing the Hf, which strongly absorbs neutrons. It also explains the unexpectedly high density of the 5d metals (Os 22.6 g cm⁻³).

Conclusion: Lanthanoid contraction is the steady decrease in size of Ln atoms/ions from La to Lu due to poor shielding by 4f electrons; it makes the 4d and 5d elements (Zr/Hf) almost identical in size and properties.

Why “transition”. The d-block elements are placed between the s-block (strongly electropositive, ionic compounds) and the p-block elements (largely covalent); their properties are intermediate (transitional) between those of the two blocks. A transition element is defined as one which has an incompletely filled d-subshell in the neutral atom or in any of its common oxidation states.

Characteristics.

• General configuration (n−1)d¹⁻¹⁰ns¹⁻²; typical metallic properties — high tensile strength, ductility, malleability, high thermal and electrical conductivity, lustre.
• High melting and boiling points and high enthalpies of atomisation (strong bonding by (n−1)d + ns electrons).
• Variable oxidation states differing by one unit (e.g. Mn: +2 to +7).
• Formation of coloured ions (d–d transitions) and of complex compounds.
• Paramagnetism (unpaired d electrons); catalytic properties.
• Formation of interstitial compounds and alloys.
• Smaller variation in ionisation enthalpies, atomic radii etc. across a series.

Elements not regarded as transition elements: Zn, Cd, Hg (and Cn), which have the configuration (n−1)d¹⁰ns² — completely filled d-orbitals in the ground state as well as in their common oxidation states (M²⁺ is d¹⁰). Their properties, such as low melting points and absence of variable valency, differ from those of the typical transition metals.

Extra depth. The word “transition” was introduced because these elements bridge the highly electropositive s-block metals and the electronegative p-block elements, both in position and in chemical behaviour. The group 12 elements (Zn, Cd, Hg) are placed in the d-block but are usually called “post-transition” because their d-orbitals are filled and not used in bonding.

Conclusion: Transition elements have partly filled d-orbitals (atom or ions) and properties intermediate between s- and p-block; Zn, Cd and Hg are not regarded as transition elements.

Because of the small energy difference between the (n−1)d and ns orbitals, the transition elements also show exceptions (Cr 3d⁵4s¹, Cu 3d¹⁰4s¹).

Extra depth. Because the (n−1)d and ns energies are close, the electrons of both subshells can take part in bonding; in main-group elements the inner shells are completely filled and unavailable. The two exceptions to the general (n−1)d^(x)ns² pattern in the 3d series (Cr, Cu) are caused by the extra stability of half-filled and filled d-subshells.

Conclusion: In transition elements the penultimate (n−1)d subshell is being filled [(n−1)d¹⁻¹⁰ns¹⁻²], whereas in non-transition elements the valence s or p orbitals are filled.

General. The most stable and characteristic oxidation state of all lanthanoids is +3 (Ln³⁺ 4fⁿ ions). Besides it, +2 and +4 states also occur occasionally in solution or in solid compounds; their appearance is related to the extra stability of empty (f⁰), half-filled (f⁷) and completely filled (f¹⁴) 4f subshells.

Extra depth. The relative stability of the +2, +3 and +4 states of lanthanoids follows the pattern of ionisation enthalpies: the first three IEs are small, so +3 is easy; the fourth IE is high except when the removal gives f⁰ (Ce⁴⁺) or f⁷ (Tb⁴⁺). In aqueous solution only Ce⁴⁺ and Eu²⁺ are important ions, and they are used as an oxidant and a reductant respectively.

Conclusion: Lanthanoids show +3 mainly, with +2 (Sm, Eu, Tm, Yb) and +4 (Ce, Pr, Nd, Tb, Dy) occasionally.

(i) Paramagnetism. Paramagnetism arises from the presence of unpaired electrons, each with a magnetic moment due to its spin (and orbital) angular momentum. Transition metal atoms and ions have incompletely filled d-orbitals with one or more unpaired electrons (e.g. Fe³⁺ 3d⁵, 5 unpaired; Cu²⁺ 3d⁹, 1 unpaired), so they are attracted into a magnetic field. The spin-only magnetic moment is μ = √[n(n + 2)] BM.

(ii) High enthalpies of atomisation. They have a large number of unpaired d electrons in addition to the ns electrons, and these electrons take part in interatomic metallic bonding; the bonding is therefore strong and a large amount of energy (Δ_aH°) is required to break it (e.g. V 515, Fe 416, Co 425, Ni 430 kJ mol⁻¹ against 126 for Zn). Maxima occur in the middle of each series where the number of unpaired electrons is largest.

(iii) Coloured compounds. In the presence of ligands, the five d-orbitals split into groups of different energies; an electron can be promoted from a lower-energy d orbital to a higher-energy d orbital (d–d transition) by absorbing light of visible-region frequency. The colour seen is the complementary colour of the light absorbed (e.g. Cu²⁺(aq) absorbs red-orange and looks blue). Ions with d⁰ (Sc³⁺, Ti⁴⁺) or d¹⁰ (Zn²⁺, Cu⁺) have no d–d transition and are colourless.

(iv) Catalysis. (a) They can exist in multiple oxidation states, so they can form unstable intermediates and provide a new reaction path of lower activation energy (Fe³⁺ catalyses 2I⁻ + S₂O₈²⁻ → I₂ + 2SO₄²⁻ through Fe³⁺/Fe²⁺). (b) On surfaces they use their d and s electrons to form bonds with reactant molecules, increasing their concentration at the surface and weakening bonds in the reactants (V₂O₅ in the Contact process; Fe in Haber’s process; Ni in hydrogenation). (c) They form complexes.

Extra depth. The four sub-parts are linked: unpaired d electrons lead to paramagnetism, strong metallic bonding and a high Δ_aH°; partially filled d-orbitals give d–d absorption and colour; and the possibility of variable oxidation states and surface bonding underlies catalysis. The magnetic moment of Fe³⁺ (5.92 BM), the colour of Cu²⁺ (blue) and the use of V₂O₅ in H₂SO₄ manufacture are standard illustrations.

Conclusion: Unpaired d electrons explain paramagnetism, strong bonding and high Δ_aH°; d–d transitions explain colour; variable oxidation states and surface bonding explain catalysis.

Definition. Interstitial compounds are those formed when small atoms such as H, C or N are trapped inside the crystal lattice (the voids/interstices) of metals. They are usually non-stoichiometric and are neither typically ionic nor covalent. Examples: TiC, Mn₄N, Fe₃H, VH₀.₅₆, TiH₁.₇ (the formulae do not correspond to the normal oxidation state of the metal).

Why well known for transition metals. The transition metals have crystal lattices (hcp, ccp, bcc) with voids (tetrahedral and octahedral interstices) of suitable size to accommodate the small atoms H, C, N, B; the interstitial atoms also form some bonds with the metal atoms (d-electron participation), which stabilises the structure.

Properties. (i) High melting points, higher than those of the pure metals; (ii) very hard — some borides approach diamond in hardness; (iii) retain metallic conductivity; (iv) chemically inert.

Extra depth. Interstitial compounds such as steel (iron carbide, Fe₃C, in iron) and palladium hydride are the reason why hard, high-melting materials like TiC (m.p. ≈ 3140 °C) and tungsten carbide are used for cutting tools. They retain metallic conductivity because the interstitial atoms occupy voids without breaking the metallic lattice.

Conclusion: Interstitial compounds are formed when small atoms occupy the interstices of the metal lattice; transition metals form them readily because of suitably sized voids and d-orbital bonding.

Transition metals. Variable oxidation states arise from the incomplete filling of d-orbitals — (n−1)d and ns electrons have nearly equal energies and take part in bonding — so the oxidation states differ from each other by unity (e.g. V: +2, +3, +4, +5; Mn: +2, +3, +4, +5, +6, +7; Fe: +2, +3).

Non-transition metals. The variability usually involves the lone/inert-pair electrons, so the oxidation states differ by two units (e.g. Sn: +2, +4; Pb: +2, +4; P: +3, +5; Cl: +1, +3, +5, +7).

Extra depth. The inert-pair effect favours lower oxidation states for the heavier p-block elements (Pb(II) more stable than Pb(IV)) whereas the trend is reversed in the d-block (W(VI) > Cr(VI) in stability), a point often asked. In transition elements oxidation states can also be as low as zero in carbonyls, [Ni(CO)₄] and [Fe(CO)₅].

Conclusion: In transition metals oxidation states differ by one unit and higher states are more stable for heavier members; in non-transition elements they differ by two units and the lower states are favoured by heavier members.

Steps in the preparation of K₂Cr₂O₇ from chromite ore (FeCr₂O₄):

1. Fusion with sodium carbonate in free access of air: 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂ (Fe₂O₃ is insoluble; the yellow sodium chromate is extracted with water and filtered).
2. Acidification of the sodium chromate solution with sulphuric acid, from which orange Na₂Cr₂O₇·2H₂O crystallises: 2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O.
3. Conversion to the potassium salt (Na₂Cr₂O₇ is more soluble than K₂Cr₂O₇) by treating with KCl: Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl. Orange crystals of K₂Cr₂O₇ crystallise out.

Effect of pH. Chromate (CrO₄²⁻, yellow) and dichromate (Cr₂O₇²⁻, orange) are interconvertible in aqueous solution depending on pH; the oxidation state of Cr (+6) is the same in both.

2CrO₄²⁻ + 2H⁺ ⇌ Cr₂O₇²⁻ + H₂O (acidic solution: orange dichromate)

Cr₂O₇²⁻ + 2OH⁻ → 2CrO₄²⁻ + H₂O (alkaline solution: yellow chromate)

So increasing the pH (adding alkali) of an orange dichromate solution converts it into yellow chromate (the orange colour changes to yellow); decreasing pH (acidifying) reverses the change. Structurally, chromate is tetrahedral and dichromate consists of two tetrahedra sharing one corner (Cr–O–Cr angle 126°).

Extra depth. Chromite is the main ore of chromium (also called chrome iron ore, FeO·Cr₂O₃); the oxidising conditions (air/O₂ at high temperature) convert Cr(III) into Cr(VI). The yellow ↔ orange change with pH is a type of acid–base equilibrium between CrO₄²⁻ and Cr₂O₇²⁻ and is used in qualitative analysis; K₂Cr₂O₇ is less soluble than Na₂Cr₂O₇, which is why the potassium salt is crystallised.

Conclusion: K₂Cr₂O₇ is obtained from FeCr₂O₄ by fusion with Na₂CO₃/air, acidification and treatment with KCl; raising the pH turns orange dichromate into yellow chromate.

Oxidising action. In acidic solution dichromate is a strong oxidising agent: the orange Cr₂O₇²⁻ (Cr +6) accepts six electrons and is reduced to green Cr³⁺ (Cr +3):

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

It oxidises iodides to iodine, sulphides to sulphur, tin(II) to tin(IV) and iron(II) to iron(III). The full ionic equation is obtained by adding the dichromate half-reaction to that of the reducing agent (after equalising electrons).

Checks. (i) charge: left −2 + 14 − 6 = +6; right +6 ✔. (ii) left −2 + 14 + 12 = +24; right 6 + 18 = +24 ✔. (iii) left −2 + 8 = +6; right +6 ✔; the 14H⁺ − 6H⁺ released from H₂S leaves 8H⁺ on the left.

Tin(II) as well: 3Sn²⁺ → 3Sn⁴⁺ + 6e⁻, so Cr₂O₇²⁻ + 14H⁺ + 3Sn²⁺ → 2Cr³⁺ + 3Sn⁴⁺ + 7H₂O (charge: left −2 + 14 + 6 = +18; right 6 + 12 = +18 ✔).

Extra depth. Dichromate is a primary standard in volumetric analysis because it is easy to purify and stable; its titration with Fe²⁺ (using diphenylamine as indicator) is a standard method of determining iron. In organic chemistry acidified K₂Cr₂O₇ oxidises primary alcohols to acids and secondary alcohols to ketones; the colour change from orange to green indicates the reaction.

Conclusion: Acidified K₂Cr₂O₇ (E° = 1.33 V) oxidises I⁻, Fe²⁺ and H₂S as per the above equations, being reduced to Cr³⁺.

Preparation of KMnO₄. Pyrolusite (MnO₂) is fused with an alkali (KOH) in the presence of an oxidising agent (atmospheric O₂ or KNO₃) to give dark-green potassium manganate:

2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O

The manganate is then converted into permanganate by (a) electrolytic oxidation in alkaline solution: MnO₄²⁻ → MnO₄⁻ + e⁻ (commercial method), or (b) disproportionation in neutral or acidic solution: 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O. In the laboratory, a Mn²⁺ salt is oxidised with peroxodisulphate: 2Mn²⁺ + 5S₂O₈²⁻ + 8H₂O → 2MnO₄⁻ + 10SO₄²⁻ + 16H⁺. Dark purple crystals separate on concentration.

Oxidising action in acidic solution: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O (E° = +1.52 V); the purple solution is decolourised.

Checks. (i) charge left −1 + 8 + 10 = +17, right +2 + 15 = +17 ✔; (ii) left −2, right +4 − 10 + 4 = −2 ✔ [2Mn²⁺ = +4; 5SO₄²⁻ = −10; 4H⁺ = +4]; (iii) left −2 − 10 + 16 = +4, right +4 ✔.

Titrations of KMnO₄ in the presence of HCl are unsatisfactory because HCl is oxidised to Cl₂; H₂SO₄ is used.

Extra depth. The colour of the permanganate ion (purple) arises from ligand-to-metal charge transfer, not from d–d transitions, because Mn(VII) is d⁰; this makes KMnO₄ act as its own indicator in titrations (the end point is the first permanent pink). It must be stored in dark bottles because it decomposes slowly in light, and its standard solutions are standardised against sodium oxalate.

Conclusion: KMnO₄ is made by alkaline oxidative fusion of MnO₂ followed by electrolytic oxidation (or disproportionation) of K₂MnO₄; in acid it oxidises Fe²⁺, SO₂ and oxalate according to the above equations.

(i) Stability of the +3 state in acid solution. The M³⁺/M²⁺ potential measures the tendency of M³⁺ to be reduced to M²⁺ — the more positive E°, the more readily M³⁺ is reduced and hence the less stable it is.

Stability of +3 state in acid: Cr³⁺ > Fe³⁺ > Mn³⁺

So Fe³⁺ is more stable than Mn³⁺ but less stable than Cr³⁺ (Cr³⁺ is not reduced by water, but Mn³⁺ oxidises water/acid).

(ii) Ease of oxidation of the metals to M²⁺. The more negative the E°(M²⁺/M), the easier is oxidation of the metal:

Ease of oxidation: Mn > Cr > Fe

Extra depth. These E° data are for the acid solution; a more positive E°(M³⁺/M²⁺) means that M³⁺ is a stronger oxidant. The values given (Cr −0.4, Mn +1.5, Fe +0.8 V) show that Mn(III) would oxidise water (E° for O₂/H₂O is 1.23 V) and cannot exist in water without disproportionating, while Cr(III) is the stable +3 state of chromium.

Conclusion: (i) Stability of +3 in acid: Cr³⁺ > Fe³⁺ > Mn³⁺; (ii) ease of oxidation of the metal: Mn > Cr > Fe (iron is oxidised least easily).

Criterion. A transition metal ion is coloured if it has incompletely filled d-orbitals (d¹–d⁹), because an electron can be excited from a lower to a higher d level (d–d transition) by absorbing visible light. Ions with d⁰ or d¹⁰ configuration cannot undergo d–d transitions and are colourless.

Extra depth. The colour of d–d transition depends on the field strength of the ligand, so the same ion can have different colours in different complexes; for example, Cu²⁺ is pale blue in water but deep blue in ammonia solution. Ions such as Zn²⁺ may appear coloured only when the anion itself is coloured, e.g. in charge-transfer compounds.

Conclusion: Ti³⁺, V³⁺, Mn²⁺, Fe³⁺ and Co²⁺ are coloured (partly filled d-orbitals); Cu⁺ (d¹⁰) and Sc³⁺ (d⁰) are colourless.

Data. The stability of M²⁺ is judged from E°(M²⁺/M), E°(M³⁺/M²⁺), the ionisation enthalpies and the d-configuration of M²⁺.

Trend. Sc²⁺ is virtually unknown (Sc is stable only as Sc³⁺, 3d⁰). From Ti to Mn the stability of the +2 state increases (3d orbitals become more stable and the third ionisation enthalpy rises: Ti 2657, V 2833, Cr 2990, Mn 3260 kJ mol⁻¹); Mn²⁺ (3d⁵) is exceptionally stable (E°(Mn³⁺/Mn²⁺) = +1.57 V). Ti²⁺, V²⁺ and Cr²⁺ are strong reducing agents (they liberate H₂ from dilute acid). After Mn, the +2 state is common and stable for Fe, Co, Ni, Cu (Cu²⁺ is more stable than Cu⁺ in water because of hydration enthalpy), and Zn²⁺ (3d¹⁰) is extremely stable.

The general trend of E°(M²⁺/M) towards less negative values reflects the rising sum of IE₁ + IE₂; Mn (d⁵ in Mn²⁺), Zn (d¹⁰) and Ni (highest hydration enthalpy) are more negative than expected.

Extra depth. The plot of E°(M²⁺/M) against atomic number is not a smooth line (Fig. 4.4), which reflects the irregular ionisation and hydration enthalpies. The stability of +2 in the second half of the series is also supported by the large negative hydration enthalpies of the M²⁺ ions (−1866 to −2121 kJ mol⁻¹ in Table 4.4).

Conclusion: Stability of +2 increases from Ti²⁺ to Mn²⁺ (d⁵), remains high for Fe, Co, Ni and Cu, and is greatest for Zn²⁺ (d¹⁰); Ti²⁺, V²⁺, Cr²⁺ are strongly reducing.

Extra depth. The two f-block series should be compared in a table with the points asked: configuration (4f vs 5f), size (lanthanoid vs actinoid contraction), oxidation states (+3 almost exclusively vs +3 to +7) and reactivity. Additionally, the actinoids are radioactive, many are artificial, and their chemistry is less regular — which is why the lanthanoid contraction is more important chemically.

Conclusion: Lanthanoids are less complicated (almost all +3, buried 4f), while actinoids show variable oxidation states, greater contraction, higher reactivity, and radioactivity.

(i) Cr²⁺ (3d⁴) is oxidised to Cr³⁺ (3d³), which has a half-filled t₂g level (t₂g³) — particularly stable (E°(Cr³⁺/Cr²⁺) = −0.41 V), hence Cr²⁺ is strongly reducing. Mn³⁺ (3d⁴) is reduced to Mn²⁺ (3d⁵), a half-filled and exceptionally stable configuration (E°(Mn³⁺/Mn²⁺) = +1.57 V), so Mn³⁺ is a strong oxidising agent.

(ii) In water E°(Co³⁺/Co²⁺) = +1.97 V, so Co²⁺ is hard to oxidise (Co³⁺ would oxidise water). In the presence of strong-field complexing ligands (e.g. NH₃, CN⁻, en) the Co(III) complexes ([Co(NH₃)₆]³⁺, d⁶ with t₂g⁶ — very large crystal field stabilisation energy and low-spin) are much more stable than the Co(II) complexes, so the Co³⁺/Co²⁺ potential falls and Co(II) is easily oxidised (e.g. by air) to Co(III).

(iii) A d¹ ion tends to lose its single d electron to attain the stable d⁰ configuration, or to undergo disproportionation to d⁰ and d² species; hence d¹ ions are unstable. Example: Mn(VI) (3d¹) as MnO₄²⁻ disproportionates in acid, 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O; V(IV) (3d¹) as VO²⁺ is readily oxidised to V(V), and Ti³⁺ (3d¹) is oxidised to Ti⁴⁺ (3d⁰).

Extra depth. The ideas in (i)–(iii) recur throughout inorganic chemistry: half-filled stability (d⁵), strong-field ligand stabilisation (CFSE), and the tendency of d¹ (and d⁹ for Cu²⁺ is less so) to adopt closed-shell configurations. In (ii), the stabilisation of Co(III) by NH₃ and CN⁻ in [Co(NH₃)₆]³⁺ and [Co(CN)₆]³⁻ is a classic illustration that complexation can change relative stabilities of oxidation states.

Conclusion: (i) d⁴ → d³ (Cr²⁺) vs d⁴ → d⁵ (Mn³⁺); (ii) strong-field ligands stabilise Co(III) (d⁶, t₂g⁶); (iii) d¹ ions tend to go to d⁰ or disproportionate.

Definition. A reaction in which the same species is simultaneously oxidised and reduced, i.e. a particular oxidation state of an element becomes less stable relative to one lower and one higher oxidation state and so changes into both, is called disproportionation.

Examples in aqueous solution:

(i) 3Mn^VIO₄²⁻ + 4H⁺ → 2Mn^VIIO₄⁻ + Mn^IVO₂ + 2H₂O

(ii) 2Cu⁺(aq) → Cu²⁺(aq) + Cu(s) (Cu(I) → Cu(II) and Cu(0))

In (i) Mn(VI) goes to Mn(VII) (oxidation) and Mn(IV) (reduction); in (ii) copper(I) is oxidised to copper(II) and reduced to copper metal. Other examples: 2H₂O₂ → 2H₂O + O₂ and Cl₂ + 2OH⁻ → Cl⁻ + OCl⁻ + H₂O.

Extra depth. Disproportionation is the reverse of comproportionation; the thermodynamic condition is that the E° of the reduction couple (to the lower state) exceeds that of the oxidation couple (to the higher state) — for Cu⁺: 0.52 V (Cu⁺/Cu) > 0.16 V (Cu²⁺/Cu⁺). Other common disproportionations are those of hypochlorite (3ClO⁻ → ClO₃⁻ + 2Cl⁻) and of H₂O₂.

Conclusion: Disproportionation = simultaneous oxidation and reduction of the same species, e.g. 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O and 2Cu⁺ → Cu²⁺ + Cu.

Answer: Copper. Cu (3d¹⁰4s¹) readily loses its single 4s electron to give Cu⁺ with the stable, completely filled 3d¹⁰ configuration; the +1 state is therefore frequent in Cu compounds (Cu₂O, CuCl, CuI, Cu₂S).

Other metals of the series (Sc, Ti, V, Cr, Mn…) do not form +1 ions readily, as removal of the 4s electron leaves an unstable configuration (their 4s² is lost together with d electrons to give +2/+3), and the second ionisation enthalpy of the metal is not unusually high. In Cu, the very high second ionisation enthalpy (1958 kJ mol⁻¹) arises from the breaking of the stable d¹⁰ configuration, so Cu(I) is favoured in the solid state, while in water Cu⁺ disproportionates because of the large hydration enthalpy of Cu²⁺.

Extra depth. The only 3d element for which +1 is characteristic is copper: Cu(I) compounds are colourless (d¹⁰ has no d–d transitions) and diamagnetic, in contrast to the blue paramagnetic Cu(II). Silver and gold in the same group show +1 (Ag) and +1/+3 (Au).

Conclusion: Copper shows +1 most frequently, since Cu⁺ has the stable 3d¹⁰ configuration.

Most stable in aqueous solution: Cr³⁺. Cr³⁺ (d³) has the half-filled t₂g level (t₂g³) in an octahedral aqua complex, which gives it a large crystal-field stabilisation energy. E°(Cr³⁺/Cr²⁺) = −0.41 V is negative, so Cr³⁺ is not easily reduced; in contrast, Mn³⁺ (E° = +1.57 V) is a strong oxidant and unstable, Ti³⁺ is easily oxidised to Ti⁴⁺ (d⁰) and V³⁺ can be oxidised to VO²⁺.

Extra depth. The magnetic moments calculated by the spin-only formula (4.90, 3.87, 2.83 and 1.73 BM) can be used to confirm the configurations by experiment; V³⁺ (3d²) and Ti³⁺ (3d¹) ions are used widely as reducing agents. The extra stability of Cr³⁺ is also seen in the inertness of its octahedral complexes, which exchange ligands very slowly.

Conclusion: Number of unpaired electrons: Mn³⁺ 4, Cr³⁺ 3, V³⁺ 2, Ti³⁺ 1; Cr³⁺ is the most stable in aqueous solution.

(i) Basic → amphoteric/acidic with increasing oxidation state. As the oxidation number of the metal increases, its charge/size ratio increases, the polarising power rises, and the bonding in the oxide changes from ionic to covalent. Low-valent oxides are ionic and basic (they accept H⁺: MO + 2H⁺ → M²⁺ + H₂O); high-valent oxides are covalent and acidic (they react with OH⁻ or water to give oxoacids/oxoanions).

For example, Mn₂O₇ is a covalent green oil, whereas MnO is ionic and dissolves in acids. CrO is basic, Cr₂O₃ is amphoteric, and CrO₃ is acidic.

(ii) Highest oxidation state in oxides and fluorides. Oxygen and fluorine are small and the most electronegative atoms, so they can oxidise the metal to its highest state; fluorine stabilises high states through high lattice energy (CoF₃) or high bond enthalpy (VF₅, CrF₆), while oxygen forms multiple (pπ–dπ) bonds with the metal (Mn₂O₇, CrO₃). Examples: Mn(VII) in Mn₂O₇ but only Mn(IV) in MnF₄; V(V) as VF₅/V₂O₅; Cr(VI) as CrF₆/CrO₃.

(iii) Highest oxidation state in oxoanions. Oxygen can form multiple bonds (π bonds by overlap of filled p orbitals of oxygen with empty/partly filled d orbitals of the metal) and has high electronegativity; in oxoanions the metal can thus attain its highest state (more than in halides, where such π-bonding is not possible). Examples: CrO₄²⁻ and Cr₂O₇²⁻ (Cr +6), MnO₄⁻ (Mn +7), VO₄³⁻ (V +5). Mn(VII) is found in MnO₄⁻ and Mn₂O₇ but not in a simple halide.

Extra depth. These three features of transition metal chemistry are linked by the degree of covalency in the M–O bonds: low states are ionic and basic, high states covalent and acidic (Mn₂O₇ is a covalent oil that explodes on heating). The ability of O to form double bonds in oxoanions is shown by the short Mn–O distance in MnO₄⁻.

Conclusion: (i) Ionic character falls and acidic character rises with oxidation state; (ii) and (iii) O and F (small, electronegative, O forms multiple bonds) can stabilise the highest oxidation states.

(i) K₂Cr₂O₇ from chromite ore (FeCr₂O₄):

1. Fusion: powdered chromite is fused with sodium (or potassium) carbonate in free access of air: 4FeCr₂O₄ + 8Na₂CO₃ + 7O₂ → 8Na₂CrO₄ + 2Fe₂O₃ + 8CO₂.
2. Extraction: the yellow solution of sodium chromate is leached out with water and filtered (Fe₂O₃ is left).
3. Acidification with H₂SO₄: 2Na₂CrO₄ + 2H⁺ → Na₂Cr₂O₇ + 2Na⁺ + H₂O (orange Na₂Cr₂O₇·2H₂O).
4. Treatment with KCl: Na₂Cr₂O₇ + 2KCl → K₂Cr₂O₇ + 2NaCl; orange crystals of K₂Cr₂O₇ crystallise out (the potassium salt is less soluble).

(ii) KMnO₄ from pyrolusite ore (MnO₂):

1. Fusion: MnO₂ is fused with KOH in the presence of air or an oxidising agent (KNO₃): 2MnO₂ + 4KOH + O₂ → 2K₂MnO₄ + 2H₂O (dark green potassium manganate).
2. Oxidation of manganate to permanganate: electrolytic oxidation in alkaline solution — MnO₄²⁻ (manganate) → MnO₄⁻ (permanganate) + e⁻ (commercial); or, by disproportionation in neutral/acid solution, 3MnO₄²⁻ + 4H⁺ → 2MnO₄⁻ + MnO₂ + 2H₂O.
3. Crystallisation: the solution is concentrated; dark purple crystals of KMnO₄ separate.

Extra depth. Industrially, sodium dichromate is made on a very large scale (for tanning, pigments and chrome plating) and the potassium salt is obtained from it; the permanganate is made industrially by the electrolytic oxidation of manganate(VI). Pyrolusite (MnO₂) is the most common ore of manganese and is also used in dry cells and as a decolouriser in glass making.

Conclusion: K₂Cr₂O₇: FeCr₂O₄ → Na₂CrO₄ → Na₂Cr₂O₇ → K₂Cr₂O₇; KMnO₄: MnO₂ → K₂MnO₄ → KMnO₄.

Alloys. An alloy is a blend of metals prepared by mixing the components (usually by fusing and cooling). Alloys may be homogeneous solid solutions in which the atoms of one metal are randomly distributed among the atoms of the other; they form readily when the metallic radii of the components are within about 15 % of each other. Alloys are generally harder and have higher melting points than the pure metals; e.g. steel (Fe + C + Cr, V, W, Mo, Mn), brass (Cu–Zn), bronze (Cu–Sn).

Alloy containing lanthanoids: mischmetall — a lanthanoid metal (≈ 95 %) with iron (≈ 5 %) and traces of S, C, Ca and Al.

Uses of mischmetall: a good deal is used in Mg-based alloys to produce bullets, shells and lighter flints. Lanthanoids are also used in alloy steels for plates and pipes.

Extra depth. Besides mischmetall, other alloys of lanthanoids with commercial importance are those used in permanent magnets (SmCo₅, Nd₂Fe₁₄B) and in lighter flints (“misch metal” + iron). The alloy contains about 95 % lanthanoids, 5 % Fe and traces of S, C, Ca and Al as given in the NCERT text.

Conclusion: Mischmetall (Ln ≈ 95 %, Fe ≈ 5 %) is used in Mg-based alloys for bullets, shells and lighter flints.

Inner transition elements. The f-block elements in which the antepenultimate (n−2) f-subshell (4f or 5f) is being filled — the lanthanoids (Ce–Lu, 4f) and the actinoids (Th–Lr, 5f); they are placed separately at the bottom of the periodic table, and together with La and Ac they are 28 elements.

Extra depth. Remember that the f-block is “inner transition” because the filling occurs in the (n−2)f subshell; the elements Ce (58) to Lu (71) and Th (90) to Lr (103) belong to it, and La and Ac are conventionally included for discussion. The elements 104–112 (Rf to Cn) belong to the 6d series and are true d-block elements.

Conclusion: The inner transition elements are those with Z = 59, 95 and 102; Z = 29, 74 and 104 are d-block elements.

Reason. The lanthanoids show practically only the +3 state (occasionally +2, +4), because their 4f orbitals are deeply buried and energetically far from 5d/6s. In the actinoids the 5f, 6d and 7s levels are of comparable energies, and the 5f electrons are not so deeply buried, so these elements show a very wide range of oxidation states; in addition, many are radioactive with short half-lives, which makes the study of their chemistry difficult.

The maximum oxidation state increases from +4 in Th to +5, +6 and +7 in Pa, U and Np and then falls in the succeeding elements; the distribution of oxidation states is uneven and differs for early and late members, so it is unsatisfactory to review their chemistry in terms of oxidation states. In contrast, among lanthanoids all elements show +3 (e.g. Ln³⁺ and Ln₂O₃, Ln(OH)₃). Also, the +3 and +4 ions of actinoids tend to hydrolyse.

Extra depth. The reasons for the variability of the actinoid oxidation states are the comparable energies of 5f, 6d and 7s orbitals and the lesser shielding of the 5f electrons; the unusual stability of +7 for Np and +6 for U (UO₂²⁺, the uranyl ion) illustrates this. The lanthanoids, in contrast, almost always give Ln³⁺ and their chemistry is therefore much more uniform.

Conclusion: Actinoids exhibit a wide and irregular range of oxidation states (+3 to +7), unlike the lanthanoids (+3 almost exclusively), so their chemistry is more complex.

Last actinoid: lawrencium, Lr (Z = 103).

Lr: [Rn] 5f¹⁴ 6d¹ 7s²

Oxidation state. Lr has +3 as its only oxidation state (Table 4.11): loss of the two 7s electrons and the 6d electron gives Lr³⁺ = [Rn]5f¹⁴, the stable completely filled 5f subshell (analogous to Lu³⁺, 4f¹⁴).

Extra depth. Lawrencium is placed with the actinoids but its 6d¹ electron (instead of 5f¹⁵, which is impossible) is part of the pattern in which Lu (4f¹⁴5d¹6s²) is the last lanthanoid; Lr’s +3 ion has 5f¹⁴, a closed subshell. The elements beyond Z = 103 are made in nanogram or smaller amounts and have very short half-lives.

Conclusion: Lawrencium (Z = 103): [Rn]5f¹⁴6d¹7s²; the common oxidation state is +3.

Configuration. Ce (Z = 58) = [Xe]4f¹5d¹6s². Ce³⁺ is formed by loss of 6s² and 5d¹: Ce³⁺ = [Xe]4f¹.

Hund’s rule. The single 4f electron occupies one of the seven degenerate 4f orbitals; there are no other electrons to pair with it, so n = 1 unpaired electron.

μ = √[n(n + 2)] = √(1 × 3) = √3 = 1.73 BM

Extra depth. Cerium’s configuration is [Xe]4f¹5d¹6s² — the 4f and 5d energies are close — and loses the three outer electrons to give Ce³⁺ with a single 4f electron; the fact that Ce⁴⁺ (4f⁰) is also stable makes cerium unique among the lanthanoids as an oxidising agent (E° = +1.74 V). The magnetic moment 1.73 BM is the same as for any one-unpaired-electron ion.

Conclusion: Ce³⁺ = [Xe]4f¹ with one unpaired electron; μ = 1.73 BM.

Correlation. The +4 and +2 states appear when they lead to the extra-stable empty (f⁰), half-filled (f⁷) or completely filled (f¹⁴) 4f subshell: Ce⁴⁺ (f⁰) and Tb⁴⁺ (f⁷) are formed by loss of a fourth electron to reach f⁰/f⁷; Eu²⁺ (f⁷) and Yb²⁺ (f¹⁴) are formed by retention of the extra f electron. The most stable and common +3 state, however, is the aim of all.

Extra depth. The (+4, +2) behaviour among lanthanoids resembles the anomalies in the first ionisation enthalpies and shows that the stability of f⁰, f⁷ and f¹⁴ is general for all electron shells; it also parallels the position of the elements Ba, La, Gd and Lu where 4f⁰, 4f⁷ and 4f¹⁴ appear. In practice only Ce(IV) and Eu(II) are important in solution.

Conclusion: +4: Ce, Pr, Nd, Tb, Dy; +2: Sm, Eu, Tm, Yb — related to the stability of f⁰, f⁷ and f¹⁴ configurations.

Behaviour similar to the lanthanoids is not evident until the second half of the actinoid series, although early actinoids also resemble the lanthanoids in close similarity among themselves and gradual variation in properties which do not involve oxidation-state changes.

Extra depth. The comparison is useful for exams if you highlight that actinoids resemble the lanthanoids only in the second half of the series, while the first half behaves more like d-block metals (U in +6 resembles Mo and W in group 6). Remember that all actinoids are radioactive, so their study needs special facilities.

Conclusion: Lanthanoids (+3, buried 4f, less variation) differ from actinoids (+3 to +7, 5f participating, radioactive, more reactive).

Pm and Md follow the regular filling of 4f/5f (Pm: 4f⁵; Md: 5f¹³); Pa has a 6d electron (5f² 6d¹) as in Table 4.10; Mt (Z = 109) is a 6d (d-block) element — its configuration is the expected one for the 6d series (Table 4.1 lists Mt with 7s² 6d⁷).

Extra depth. The configuration of an element should be written following the aufbau principle but with attention to anomalies; for Pa the 6d¹ electron is present because of the nearness of 5f and 6d, while Md (5f¹³7s²) follows the regular order. The d-block element Mt (Z = 109) is predicted to be 6d⁷7s² and is a transition element.

Conclusion: Z = 61: [Xe]4f⁵6s²; 91: [Rn]5f²6d¹7s²; 101: [Rn]5f¹³7s²; 109: [Rn]5f¹⁴6d⁷7s².

Illustration (Table 4.2 vs. text). Δ_aH° maxima occur near the middle of each series, but the second and third series have greater enthalpies of atomisation than the corresponding 3d elements — explaining their frequent metal–metal bonding. Atomic radii of 3d elements: Ti 147, V 135, Cr 129 pm; Zr 160, Hf 159 pm for the 4d/5d members.

Extra depth. The heavier transition metals form compounds with metal–metal bonds (e.g. Re₂Cl₈²⁻, Mo₂(CH₃COO)₄), have higher oxidation states in simple compounds (OsO₄, RuO₄, WCl₆) and form stable low-spin complexes; the lanthanoid contraction controls the size relationship (Zr ≈ Hf). The reasons are the larger size of the 4d and 5d orbitals, which overlap better with the orbitals of neighbours.

Conclusion: Compared to the 3d series, the 4d/5d metals show more stable higher oxidation states, greater enthalpies of atomisation, higher IE (5d) and almost equal 4d/5d radii because of the lanthanoid contraction.

Concept. In an octahedral field the five 3d orbitals split into a lower set t₂g (d_xy, d_yz, d_zx) and a higher set e_g (d_x²−y², d_z²). Water is a weak-field ligand, so electrons occupy all five orbitals singly (Hund’s rule) before pairing — high-spin configurations. Number of 3d electrons = group-based count after losing the 4s (and then 3d) electrons.

For example, in Fe²⁺ (d⁶) the three t₂g orbitals are singly occupied, then the two e_g orbitals are singly occupied, and the sixth electron pairs in one t₂g orbital: (t₂g)⁴(e_g)². For Ni²⁺ (d⁸) the t₂g set is filled (t₂g⁶) and the e_g orbitals hold one electron each.

Extra depth. The octahedral field splits the 3d orbitals into t₂g and e_g; with the weak-field ligand H₂O, all five orbitals are singly filled before pairing (high spin), so d⁴ to d⁷ ions have the maximum number of unpaired electrons. This is the basis of the magnetic moments in Table 4.7 and the colours in Table 4.8 (d–d transitions between t₂g and e_g).

Conclusion: Number of 3d electrons: Ti²⁺ 2, V²⁺ 3, Cr³⁺ 3, Mn²⁺ 5, Fe²⁺ 6, Fe³⁺ 5, Co²⁺ 7, Ni²⁺ 8, Cu²⁺ 9, occupying t₂g and e_g as tabulated (high spin with H₂O).

The statement is true. The 3d elements differ from the 4d and 5d elements (heavier transition elements) in several respects:

Extra depth. The differences between the first and later transition series are important for the later chapters: the first series forms mainly high-spin octahedral and tetrahedral complexes, whereas 4d and 5d metals form low-spin complexes because of the larger crystal field splitting. The behaviour of Hf and Zr is nearly identical due to the lanthanoid contraction.

Conclusion: First-series elements have lower Δ_aH°, smaller radii, less stable higher oxidation states and are more reactive/electropositive; heavier elements show metal–metal bonding, more stable higher states and similar 4d/5d radii.

Method. Compare the observed magnetic moment with the spin-only value μ = √[n(n + 2)] BM to find the number of unpaired electrons (n): n = 1 → 1.73; n = 2 → 2.83; n = 3 → 3.87; n = 4 → 4.90; n = 5 → 5.92 BM.

Extra depth. Magnetic data distinguish the electron arrangements: the observed moment in [Fe(H₂O)₆]²⁺ (5.3 BM) is somewhat larger than the spin-only value of 4.90 BM due to orbital contribution, and in K₄[Mn(CN)₆] (2.2 BM) the strong-field CN⁻ forces low-spin t₂g⁵ with one unpaired electron. A magnetic moment is thus a simple experimental way of finding the spin state of a complex.

Conclusion: K₄[Mn(CN)₆]: low-spin d⁵, 1 unpaired electron; [Fe(H₂O)₆]²⁺: high-spin d⁶, 4 unpaired electrons; K₂[MnCl₄]: high-spin d⁵, 5 unpaired electrons.