Transition Metals and the f-Block: Coordination Chemistry, Crystal Fields, Spectra, Magnetism and Mechanisms

The Transition Metals and the Lanthanides and Actinides sub-headings of Section 2 of the GATE Chemistry (CY) paper. The chapter follows the Transition Metals list in its own order: coordination chemistry — structure and isomerism; the theories of bonding (valence-bond, crystal-field and molecular-orbital); metal–metal multiple bonds; energy-level diagrams in various crystal fields, crystal-field stabilisation energy and the applications of CFT, including Jahn–Teller distortion; electronic spectra with spectroscopic term symbols, selection rules, Orgel and Tanabe–Sugano diagrams, the nephelauxetic effect and the Racah parameter, and charge-transfer spectra; magnetic properties; the Ray–Dutt and Bailar twists; and reaction mechanisms — kinetic and thermodynamic stability, associative and dissociative substitution, and redox reactions. The last section is the lanthanides and actinides: their recovery, periodic properties, and spectral and magnetic properties. Magnetic moments are in Bohr magnetons (BM) and crystal-field energies in units of Δₒ unless stated.

1. Structure and isomerism of coordination compounds

A complex is described by its coordination number and geometry: 2 (linear, Ag(NH₃)₂⁺), 4 (tetrahedral for d¹⁰ and weak-field ions, square planar for d⁸ Pd²⁺, Pt²⁺, Au³⁺ and strong-field Ni²⁺), 5 (trigonal bipyramidal or square pyramidal, close in energy) and 6 (octahedral, by far the commonest). Structural isomers differ in which atoms are bonded: ionisation ([Co(NH₃)₅Br]SO₄ and [Co(NH₃)₅SO₄]Br), hydrate ([Cr(H₂O)₆]Cl₃, violet, and [CrCl(H₂O)₅]Cl₂·H₂O, green), linkage (nitro –NO₂ and nitrito –ONO; S- and N-bound thiocyanate) and coordination isomers ([Co(NH₃)₆][Cr(CN)₆] and [Cr(NH₃)₆][Co(CN)₆]). Stereoisomers differ only in arrangement.

Stereoisomers of common complex types
TypeGeometrical isomersTotal stereoisomersExample
square planar MA₂B₂2 (cis, trans)2[PtCl₂(NH₃)₂]
octahedral MA₄B₂2 (cis, trans)2[Co(NH₃)₄Cl₂]⁺
octahedral MA₃B₃2 (fac, mer)2[Co(NH₃)₃(NO₂)₃]
octahedral M(AA)₂B₂2 (cis is chiral)3[Co(en)₂Cl₂]⁺
octahedral M(AA)₃12 (Δ, Λ)[Co(en)₃]³⁺
octahedral MA₂B₂C₂56 (one pair of enantiomers)[PtCl₂Br₂(NH₃)₂]

2. Bonding theories, crystal-field splitting, CFSE and metal–metal bonds

Valence-bond theory describes octahedral complexes as inner-orbital d²sp³ (low-spin, [Fe(CN)₆]⁴⁻) or outer-orbital sp³d² (high-spin, [FeF₆]³⁻), and square planar ones as dsp²; it explains geometry and magnetism but not colour. Crystal-field theory treats the ligands as point charges. In an octahedral field the d orbitals split into a lower t₂g set (d_xy, d_xz, d_yz, at −0.4Δₒ) and an upper e_g set (d_z², d_x²−y², at +0.6Δₒ), the barycentre preserved. In a tetrahedral field the order inverts — e below t₂ — and Δ_t ≈ (4/9)Δₒ, so tetrahedral complexes are almost always high-spin. In a square planar field d_x²−y² rises far above the rest, which makes d⁸ ions with strong-field ligands diamagnetic and square planar.

The crystal-field stabilisation energy of an octahedral configuration t₂gˣe_gʸ is CFSE = (−0.4x + 0.6y)Δₒ, with a pairing energy P added for each pair forced together beyond the free-ion arrangement. When Δₒ > P a d⁴–d⁷ ion is low-spin, when Δₒ < P high-spin. Δₒ grows with the oxidation state (Co³⁺ > Co²⁺), down a group (4d and 5d complexes are nearly always low-spin) and along the spectrochemical series: I⁻ < Br⁻ < S²⁻ < SCN⁻ (S) < Cl⁻ < F⁻ < OH⁻ < C₂O₄²⁻ < H₂O < NCS⁻ (N) < NH₃ < en < bpy < phen < NO₂⁻ < CN⁻ < CO. The ligand-field (MO) picture explains the order: σ-donors raise e_g*, π-donors (halides, OH⁻) push t₂g up and shrink Δₒ, π-acceptors (CO, CN⁻, phosphines) pull t₂g down and enlarge it. CFSE explains the double-humped curve of hydration enthalpies across the 3d series (minima at d⁰, d⁵ high-spin and d¹⁰), trends in ionic radii, and site preferences in spinels.

The Jahn–Teller theorem: a non-linear molecule in an orbitally degenerate electronic state distorts to remove the degeneracy. The effect is strong when the degeneracy is in the σ-antibonding e_g set — high-spin d⁴ (Cr²⁺, Mn³⁺), low-spin d⁷ and d⁹ (Cu²⁺) — and usually shows as a tetragonal elongation, two long axial bonds and four short equatorial ones; t₂g degeneracies give only weak distortions. Metal–metal multiple bonds arise when d orbitals on adjacent metals overlap: in [Re₂Cl₈]²⁻ d_z² gives a σ bond, d_xz and d_yz two π bonds, and d_xy a δ bond, so the configuration σ²π⁴δ² gives a quadruple bond (bond order 4, Re–Re 224 pm). The δ overlap needs the two ReCl₄ units eclipsed, which is why the ion is eclipsed despite the steric cost.

3. Magnetic properties of complexes

For first-row complexes the orbital angular momentum is largely quenched by the ligand field, and the moment is close to the spin-only value μ_s = √(n(n + 2)) BM for n unpaired electrons (1.73, 2.83, 3.87, 4.90 and 5.92 BM for n = 1 to 5). Orbital contributions survive when the ground term is a T term (for example octahedral high-spin Co²⁺, t₂g⁵e_g², measured 4.7–5.2 BM against 3.87 spin-only), but not for A or E ground terms (Ni²⁺ octahedral, Cr³⁺, Mn²⁺, which match the spin-only value closely). The susceptibility of a magnetically dilute paramagnet follows the Curie law χ = C/T, and μ_eff = 2.828√(χ_M T) BM.

Spin states and moments of representative complexes
Complexdⁿ, spin stateUnpaired e⁻μ_s (BM)
[Ti(H₂O)₆]³⁺d¹11.73
[Cr(H₂O)₆]³⁺d³33.87
[Mn(H₂O)₆]²⁺d⁵ high-spin55.92
[Fe(CN)₆]³⁻d⁵ low-spin11.73
[Fe(H₂O)₆]²⁺d⁶ high-spin44.90
[Co(NH₃)₆]³⁺d⁶ low-spin00
[NiCl₄]²⁻d⁸ tetrahedral22.83
[Ni(CN)₄]²⁻d⁸ square planar00

4. Electronic spectra: terms, selection rules, Orgel and Tanabe–Sugano diagrams, Racah B, charge transfer

The free-ion ground terms of dⁿ are ²D (d¹, d⁹), ³F (d², d⁸), ⁴F (d³, d⁷), ⁵D (d⁴, d⁶) and ⁶S (d⁵). In an octahedral field a D term splits into T₂g + E_g, an F term into A₂g + T₂g + T₁g, and there is a higher P term of the same multiplicity for d², d³, d⁷ and d⁸. Selection rules: the spin rule ΔS = 0 and the Laporte rule (g ↔ u only). d–d transitions in an octahedral complex are g → g and Laporte-forbidden, so they are weak (ε ≈ 1–100 L mol⁻¹ cm⁻¹) and gain intensity only through vibronic coupling; tetrahedral complexes, lacking a centre of symmetry and with d–p mixing, are more intense (ε ≈ 100–1000). Spin-forbidden bands are weaker still, which is why high-spin d⁵ [Mn(H₂O)₆]²⁺, which has no spin-allowed d–d transition at all, is almost colourless.

Orgel diagrams plot the spin-allowed terms against Δ for weak fields: d¹, d⁴, d⁶ and d⁹ (D ground terms) show one band — [Ti(H₂O)₆]³⁺ absorbs at about 20 000 cm⁻¹ (500 nm), which is Δₒ itself — while d², d³, d⁷ and d⁸ (F ground terms) show three; for octahedral d³ Cr³⁺ and d⁸ Ni²⁺ the lowest band ⁴A₂g → ⁴T₂g (or ³A₂g → ³T₂g) equals Δₒ directly. Tanabe–Sugano diagrams plot E/B against Δₒ/B for all terms, including spin-forbidden ones, and show the high-spin to low-spin crossover as a vertical line for d⁴–d⁷. The Racah parameter B measures interelectronic repulsion; in a complex it is smaller than in the free ion because the d electrons spread onto the ligands — the nephelauxetic ("cloud-expanding") effect — with β = B(complex)/B(free ion) < 1 and a nephelauxetic series F⁻ > H₂O > NH₃ > en > Cl⁻ > CN⁻ > Br⁻ > I⁻ (β decreasing, i.e. more covalency to the right).

Charge-transfer bands move an electron between metal and ligand and are Laporte- and spin-allowed, so they are intense (ε ≈ 10³–10⁵). LMCT (ligand to metal) dominates for metals in high oxidation states with filled-π ligands: the intense purple of MnO₄⁻ and yellow of CrO₄²⁻ are O → M transitions in d⁰ ions, and the colours of the silver halides darken as the halide becomes easier to oxidise. MLCT (metal to ligand) occurs for electron-rich metals with low-lying π* ligands, as in the orange [Ru(bpy)₃]²⁺ and [Fe(phen)₃]²⁺.

5. Reaction mechanisms: stability, substitution, the trans effect, twists and electron transfer

Thermodynamic stability (formation constants, the chelate effect) and kinetic lability (the rate of ligand exchange) are independent. Taube called a complex labile if it exchanges ligands within about a minute at room temperature and inert otherwise. [Ni(CN)₄]²⁻ is very stable yet exchanges CN⁻ rapidly; [Co(NH₃)₆]³⁺ is thermodynamically unstable in acid yet survives for days. CFT explains the pattern: octahedral d³ and low-spin d⁴–d⁶ ions (Cr³⁺, Co³⁺, Rh³⁺, Pt⁴⁺) are inert because losing a ligand costs much ligand-field stabilisation; ions with e_g electrons (Cu²⁺, high-spin ions) and d⁰ or d¹⁰ ions are labile.

Substitution mechanisms are classed as dissociative (D), with a lower-coordinate intermediate; associative (A), with a higher-coordinate one; and the concerted interchange paths I_d and I_a. Octahedral substitution is usually dissociative in character: the rate depends little on the entering ligand, and ΔV‡ and ΔS‡ are positive. Base hydrolysis of [Co(NH₃)₅Cl]²⁺ is fast because OH⁻ deprotonates an NH₃ and the amido conjugate base loses Cl⁻ easily (the S_N1CB mechanism). Square planar d⁸ substitution is associative through a five-coordinate intermediate, and the ligand trans to the leaving group governs its rate — the trans effect: CO ≈ CN⁻ ≈ C₂H₄ > PR₃ ≈ H⁻ > CH₃⁻ > I⁻ > Br⁻ > Cl⁻ > NH₃ > H₂O. It is used synthetically: [PtCl₄]²⁻ + 2NH₃ gives cis-[PtCl₂(NH₃)₂] (the second NH₃ replaces a Cl trans to Cl), while [Pt(NH₃)₄]²⁺ + 2Cl⁻ gives the trans isomer.

Tris-chelate complexes M(AA)₃ can racemise without breaking a bond by twist mechanisms through a trigonal-prismatic intermediate: the Bailar (trigonal) twist rotates two opposite triangular faces about the real C₃ axis, and the Ray–Dutt (rhombic) twist rotates about one of the three pseudo-C₃ axes. Electron transfer between complexes is outer-sphere when both coordination shells stay intact and the electron tunnels between them — the rate follows Marcus theory and is fast when the reorganisation energy is small ([Fe(CN)₆]⁴⁻/³⁻) — or inner-sphere when a ligand bridges the two metals. Taube’s proof: [Co(NH₃)₅Cl]²⁺ + [Cr(H₂O)₆]²⁺ gives [CrCl(H₂O)₅]²⁺, with the chloride transferred to chromium, because the substitution-inert Cr(III) product can only have gained Cl while it was still labile Cr(II) bound to the bridge.

6. Lanthanides and actinides

Recovery. The lanthanides occur together in minerals such as monazite (a phosphate, with thorium) and bastnäsite (a fluorocarbonate). After digestion and removal of thorium, cerium is often separated first by oxidation to Ce⁴⁺ and europium by reduction to Eu²⁺. The rest, all Ln³⁺ of nearly identical chemistry, are separated by ion-exchange chromatography or counter-current solvent extraction (tributyl phosphate or organophosphoric acids), which exploit the small, steady fall in ionic radius: with a complexing eluent the smaller, more strongly complexed heavy lanthanides come off a cation-exchange column first.

Periodic properties. The 4f orbitals are buried inside the 5s and 5p shells and shield the nuclear charge poorly, so Ln³⁺ radii shrink steadily from La³⁺ (about 103 pm) to Lu³⁺ (about 86 pm) — the lanthanide contraction. Its consequences: Y³⁺ falls among the heavy lanthanides; the 5d elements that follow have almost the same radii as their 4d congeners (Zr/Hf, Nb/Ta, Mo/W), making them hard to separate; and basicity of the hydroxides falls from La to Lu. +3 is the characteristic oxidation state; Ce⁴⁺ (f⁰), Eu²⁺ (f⁷) and Yb²⁺ (f¹⁴) exist because they reach empty, half-filled or filled f shells. The actinides show a wider range of oxidation states early in the series (U +3 to +6, as in the linear uranyl ion UO₂²⁺; Np and Pu up to +7), because the 5f orbitals extend further and lie closer in energy to 6d and 7s; after americium +3 dominates as for the lanthanides.

Spectra and magnetism. Because the 4f electrons are shielded, ligand fields split f levels by only about 100 cm⁻¹, far less than spin–orbit coupling. So f–f absorption bands are sharp, weak (Laporte-forbidden) and almost independent of the ligand, and emission is line-like — the red of Eu³⁺ and green of Tb³⁺ in phosphors. Magnetic moments follow the free-ion J value, μ_eff = g_J√(J(J + 1)) BM with g_J = 1 + [J(J + 1) + S(S + 1) − L(L + 1)]/[2J(J + 1)], not the spin-only formula. Gd³⁺ (f⁷, ⁸S₇/₂, g = 2) gives 7.94 BM; Nd³⁺ (f³, ⁴I₉/₂) 3.62 BM. Sm³⁺ and Eu³⁺ deviate from this formula because their excited J levels lie within kT and are thermally populated.

Key takeaways

  • MA₄B₂ and MA₃B₃ give two geometrical isomers each; cis-M(AA)₂B₂ is chiral (3 stereoisomers); MA₂B₂C₂ has 5 geometrical and 6 stereoisomers.
  • CFSE = (−0.4 n_t₂g + 0.6 n_eg)Δₒ; Δ_t ≈ 4/9 Δₒ; strong σ-donor/π-acceptor ligands raise Δₒ; Jahn–Teller is strongest for d⁹, high-spin d⁴ and low-spin d⁷; [Re₂Cl₈]²⁻ has a σ²π⁴δ² quadruple bond.
  • μ_s = √(n(n + 2)) BM; orbital contributions survive for T ground terms; lanthanides need μ = g_J√(J(J + 1)).
  • d–d bands are Laporte-forbidden and weak; one band for D-term ions, three for F-term ions; β = B(complex)/B(free) < 1; LMCT and MLCT bands are intense.
  • d³ and low-spin d⁶ are inert; octahedral substitution is dissociative, square planar associative with the trans effect; inner-sphere transfer moves the bridging ligand; Bailar and Ray–Dutt twists racemise without bond breaking.

Practice questions (22)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. What is the crystal-field stabilisation energy of the low-spin complex [Co(NH₃)₆]³⁺, in units of Δₒ, ignoring pairing energy? Give the answer to one decimal place.

    Numerical answer — type the value.

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    Answer: -2.4

    Co³⁺ is d⁶; low-spin puts all six electrons in t₂g: CFSE = 6 × (−0.4)Δₒ = -2.4Δₒ. The high-spin arrangement t₂g⁴e_g² would give only −0.4Δₒ, which is why strong-field NH₃ makes the ion diamagnetic.
  2. What is the CFSE of a high-spin d⁴ ion in an octahedral field, in units of Δₒ, to one decimal place?

    Numerical answer — type the value.

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    Answer: -0.6

    High-spin d⁴ is t₂g³e_g¹: 3(−0.4) + 1(+0.6) = −1.2 + 0.6 = -0.6Δₒ. Low-spin t₂g⁴ would give −1.6Δₒ plus a pairing energy; the single e_g electron is what makes this ion Jahn–Teller active.
  3. What is the CFSE of tetrahedral [NiCl₄]²⁻ (d⁸), in units of Δ_t, to one decimal place?

    Numerical answer — type the value.

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    Answer: -0.8

    In a tetrahedral field e lies at −0.6Δ_t and t₂ at +0.4Δ_t. d⁸ is e⁴t₂⁴: 4(−0.6) + 4(+0.4) = −2.4 + 1.6 = -0.8Δ_t, about −0.36Δₒ. Using the octahedral coefficients the wrong way round gives +0.8 or −1.2.
  4. What is the spin-only magnetic moment of [Fe(H₂O)₆]²⁺ (high-spin), in BM, to two decimal places?

    Numerical answer — type the value.

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    Answer: 4.9

    Fe²⁺ is d⁶; high-spin t₂g⁴e_g² has 4 unpaired electrons, so μ_s = √(4 × 6) = √24 = 4.90 BM. The low-spin cyanide complex [Fe(CN)₆]⁴⁻ has none and is diamagnetic.
  5. What is the spin-only magnetic moment of [Fe(CN)₆]³⁻, in BM, to two decimal places?

    Numerical answer — type the value.

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    Answer: 1.73

    Fe³⁺ is d⁵ and cyanide is a strong-field ligand, so the complex is low-spin t₂g⁵ with one unpaired electron: μ_s = √(1 × 3) = 1.73 BM. High-spin d⁵ would give 5.92 BM, as in [Fe(H₂O)₆]³⁺. The measured value is a little higher because the ²T₂g ground term keeps some orbital moment.
  6. What is the spin-only magnetic moment of [Cr(H₂O)₆]³⁺, in BM, to two decimal places?

    Numerical answer — type the value.

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    Answer: 3.87

    Cr³⁺ is d³, t₂g³ with three unpaired electrons whatever the field strength: μ_s = √(3 × 5) = √15 = 3.87 BM. Its ⁴A₂g ground term has no orbital contribution, so experiment agrees closely.
  7. Which of the following complexes are diamagnetic?

    1. [Ni(CN)₄]²⁻
    2. [NiCl₄]²⁻
    3. [Co(NH₃)₆]³⁺
    4. [CoF₆]³⁻
    Show answer

    Answer: A — [Ni(CN)₄]²⁻; C — [Co(NH₃)₆]³⁺

    Square planar d⁸ [Ni(CN)₄]²⁻ pairs all eight electrons below the high d_x²−y² orbital, and low-spin d⁶ [Co(NH₃)₆]³⁺ fills t₂g⁶. Tetrahedral [NiCl₄]²⁻ has two unpaired electrons and weak-field [CoF₆]³⁻ is high-spin with four.
  8. Nd³⁺ (4f³) has the ground term ⁴I₉/₂. Using μ_eff = g_J√(J(J + 1)) with g_J = 1 + [J(J + 1) + S(S + 1) − L(L + 1)]/[2J(J + 1)], what is its magnetic moment, in BM, to two decimal places?

    Numerical answer — type the value.

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    Answer: 3.62

    S = 3/2, L = 6, J = 9/2: J(J + 1) = 24.75, S(S + 1) = 3.75, L(L + 1) = 42. g_J = 1 + (24.75 + 3.75 − 42)/49.5 = 1 − 0.273 = 0.727, and μ = 0.727 × √24.75 = 0.727 × 4.975 = 3.62 BM. The spin-only formula with three unpaired electrons gives 3.87, which is wrong for an f ion.
  9. What is the magnetic moment of Gd³⁺ (4f⁷, ⁸S₇/₂), in BM, to two decimal places?

    Numerical answer — type the value.

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    Answer: 7.94

    With L = 0, J = S = 7/2 and g_J = 2, so μ = 2√(3.5 × 4.5) = 2√15.75 = 7.94 BM — the same as the spin-only value √(7 × 9), because an S-state ion has no orbital moment to add.
  10. [Ti(H₂O)₆]³⁺ shows a single d–d absorption with maximum at 500 nm. What is Δₒ, in cm⁻¹?

    Numerical answer — type the value.

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    Answer: 20000

    For d¹ the single band ²T₂g → ²E_g equals Δₒ: ν̃ = 1/λ = 1/(500 × 10⁻⁷ cm) = 20 000 cm⁻¹, typed as 20000, about 239 kJ/mol; a d³ or d⁸ ion would need its lowest band, not its most intense one.
  11. How many spin-allowed d–d absorption bands are expected for octahedral [Cr(H₂O)₆]³⁺?

    Numerical answer — type the value.

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    Answer: 3

    d³ has a ⁴F ground term, which splits into ⁴A₂g (ground), ⁴T₂g and ⁴T₁g(F), and there is also ⁴T₁g(P) from the ⁴P term: three quartet transitions from ⁴A₂g. The first equals Δₒ. A D-term ion such as d¹ would show one band.
  12. The Racah parameter B of a first-row metal ion is 1030 cm⁻¹ in the free ion and 790 cm⁻¹ in one of its octahedral complexes. What is the nephelauxetic ratio β, to two decimal places?

    Numerical answer — type the value.

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    Answer: 0.77

    β = B(complex)/B(free ion) = 790/1030 = 0.767, i.e. 0.77. A value below 1 means the d electrons are spread over the ligands and repel each other less; the more covalent the metal–ligand bonding, the smaller β.
  13. Why is the intense purple colour of KMnO₄ not due to a d–d transition?

    1. Mn(VII) is d⁰, so the band is a ligand-to-metal charge transfer
    2. Permanganate is octahedral, so d–d bands are forbidden
    3. Mn(VII) has five unpaired d electrons
    4. The colour comes from K⁺
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    Answer: A — Mn(VII) is d⁰, so the band is a ligand-to-metal charge transfer

    Mn in MnO₄⁻ is +7, with no d electrons, so no d–d transition exists; the band near 530 nm is O 2p → Mn 3d LMCT, allowed and intense (ε ≈ 2400). The ion is tetrahedral, not octahedral, and K⁺ is colourless.
  14. For which ion is the Jahn–Teller distortion in an octahedral complex expected to be strongest?

    1. Cu²⁺ (d⁹)
    2. Ni²⁺ (d⁸)
    3. Cr³⁺ (d³)
    4. high-spin Mn²⁺ (d⁵)
    Show answer

    Answer: A — Cu²⁺ (d⁹)

    d⁹ is t₂g⁶e_g³, degenerate in the strongly σ-antibonding e_g set, so Cu(II) complexes distort strongly, usually by tetragonal elongation. d⁸ (e_g²), d³ (t₂g³) and high-spin d⁵ are orbitally non-degenerate and do not distort.
  15. How many stereoisomers does [Co(en)₂Cl₂]⁺ have (en = ethylenediamine)?

    Numerical answer — type the value.

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    Answer: 3

    The trans isomer has mirror planes and is achiral; the cis isomer has no improper axis and exists as a pair of enantiomers. Total 1 + 2 = 3. Counting only geometrical isomers gives 2, missing the optical pair.
  16. How many stereoisomers, counting enantiomers separately, does an octahedral complex MA₂B₂C₂ have?

    Numerical answer — type the value.

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    Answer: 6

    There are five geometrical arrangements: all three pairs trans; one pair trans and two cis (three ways, choosing which pair is trans); and all three pairs cis. Only the all-cis isomer lacks a mirror plane, so it is a pair of enantiomers: 5 + 1 = 6.
  17. What is the metal–metal bond order in [Re₂Cl₈]²⁻?

    Numerical answer — type the value.

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    Answer: 4

    Each Re(III) is d⁴, giving eight electrons for the σ²π⁴δ² configuration: one σ, two π and one δ bond, bond order 4. The δ bond needs eclipsed ReCl₄ units; in the staggered form the δ overlap vanishes and the order would drop to 3.
  18. Which of the following aqua or ammine complexes are substitution-inert?

    1. [Cr(H₂O)₆]³⁺
    2. [Co(NH₃)₆]³⁺
    3. [Cu(H₂O)₆]²⁺
    4. [Ni(H₂O)₆]²⁺
    Show answer

    Answer: A — [Cr(H₂O)₆]³⁺; B — [Co(NH₃)₆]³⁺

    d³ Cr³⁺ and low-spin d⁶ Co³⁺ have empty e_g orbitals and large ligand-field stabilisation, so losing a ligand is costly: water exchange takes hours or longer. Jahn–Teller Cu²⁺ exchanges in about a nanosecond and Ni²⁺ (with e_g²) in about 10⁻⁵ s, both labile.
  19. Starting from [PtCl₄]²⁻ and adding two equivalents of NH₃ gives mainly which product, and why?

    1. cis-[PtCl₂(NH₃)₂], because Cl⁻ has a greater trans effect than NH₃
    2. trans-[PtCl₂(NH₃)₂], because NH₃ has a greater trans effect than Cl⁻
    3. a 1:1 mixture, because square planar substitution is dissociative
    4. [Pt(NH₃)₄]²⁺, because NH₃ always replaces every chloride
    Show answer

    Answer: A — cis-[PtCl₂(NH₃)₂], because Cl⁻ has a greater trans effect than NH₃

    After the first substitution, [PtCl₃(NH₃)]⁻ has two Cl trans to Cl and one Cl trans to NH₃. The labilised chlorides are those trans to Cl, so the second NH₃ enters cis to the first: cisplatin. The trans isomer is made from [Pt(NH₃)₄]²⁺ and Cl⁻. Square planar substitution is associative.
  20. In the reaction [Co(NH₃)₅Cl]²⁺ + [Cr(H₂O)₆]²⁺ → Co²⁺(aq) + [CrCl(H₂O)₅]²⁺ + 5NH₄⁺, the transfer of chloride to chromium shows that electron transfer is

    1. inner-sphere, through a Co–Cl–Cr bridge
    2. outer-sphere, with both coordination shells intact
    3. dissociative substitution at cobalt
    4. a Bailar twist
    Show answer

    Answer: A — inner-sphere, through a Co–Cl–Cr bridge

    Cr(III) is inert, so Cl could only have joined chromium while it was labile Cr(II), bonded through the chloride to Co(III) during the electron transfer — Taube’s inner-sphere mechanism. An outer-sphere path would leave the chloride on cobalt, which is released as labile Co(II).
  21. The Bailar and Ray–Dutt twists are mechanisms by which tris-chelate complexes M(AA)₃

    1. racemise without breaking any metal–ligand bond, via a trigonal-prismatic transition state
    2. undergo inner-sphere electron transfer
    3. lose one chelate arm and re-coordinate
    4. isomerise from fac to mer
    Show answer

    Answer: A — racemise without breaking any metal–ligand bond, via a trigonal-prismatic transition state

    Twisting two triangular faces relative to each other — about the true C₃ axis (Bailar) or a pseudo-C₃ axis (Ray–Dutt) — passes through a trigonal prism and converts Δ into Λ with every bond intact. Loss of one arm is a different, bond-breaking racemisation path.
  22. Which is a direct consequence of the lanthanide contraction?

    1. Zr and Hf have almost identical atomic radii and chemistry
    2. Lanthanides show many oxidation states from +2 to +7
    3. La(OH)₃ is less basic than Lu(OH)₃
    4. f–f bands are strongly affected by the ligands
    Show answer

    Answer: A — Zr and Hf have almost identical atomic radii and chemistry

    The fourteen-element shrinkage across the 4f series cancels the expected size increase from 4d to 5d, so Hf (after the lanthanides) is almost the same size as Zr. The lanthanides are mostly +3, La(OH)₃ is the more basic hydroxide, and shielded f orbitals make f–f bands nearly ligand-independent.