(a) What is the de Broglie wavelength for a baseball (140 g) moving at 40 m s⁻¹? Comment on your finding. (5 marks) (b)…

(a) de Broglie wavelength of a baseball (5 marks)

λ = h/(mv) = 6.626 × 10⁻³⁴ J s/(0.140 kg × 40 m s⁻¹) = 6.626 × 10⁻³⁴/5.6 = 1.2 × 10⁻³⁴ m.

This wavelength is about 10¹⁹ times smaller than an atomic nucleus (about 10⁻¹⁵ m), so it is far too small to be detected or to cause any diffraction or interference. Wave properties are significant only when the momentum is so small that λ is comparable with the size of the system, as for electrons in atoms; for macroscopic bodies the motion is described accurately by classical mechanics.

(b) Distinguishing cis- and trans-1,2-dichloroethylene by dipole moment (5 marks)

Each C–Cl bond is polar, so each has a bond dipole. In the cis isomer the two C–Cl dipoles lie on the same side of the double bond and their vector sum is non-zero, giving a net dipole moment of about 1.9 D. In the trans isomer they point in opposite directions and cancel exactly, so the molecule is non-polar (μ = 0). Measuring the dipole moment (for example from dielectric constant in solution) therefore identifies the isomer: a non-zero value means cis, and zero means trans. Dipole-dipole attraction also gives the cis isomer the higher boiling point (about 60 °C, against about 48 °C for trans).

(c) Resonance structures of the nitrate ion (5 marks)

The nitrate ion is trigonal planar. Three equivalent Lewis structures can be drawn, each with one N=O double bond and two N–O⁻ single bonds, differing only in which oxygen is double-bonded. In each, the formal charge is +1 on nitrogen, 0 on the double-bonded oxygen and −1 on each of the other two oxygens, so the net charge is −1.

The actual ion is a resonance hybrid of the three: all N–O bonds are equal in length (about 124 pm), with a bond order of 4/3, and the negative charge is delocalised equally over the three oxygens.

(d) Interplanar spacing d₁₁₁ in a simple cubic lattice (5 marks)

For a cubic lattice, d_hkl = a/√(h² + k² + l²).

d₁₁₁ = 200 pm/√(1 + 1 + 1) = 200/1.732 = 115.5 pm.

The (111) planes cut the axes at a, a, a, so they lie at a body-diagonal orientation, and the spacing between two successive planes is the perpendicular distance, one-third of the body diagonal length (√3 a): a/√3. This is the shortest distance between successive (111) planes.

(e) Ferredoxin (5 marks)

Ferredoxins are small iron-sulfur proteins (about 6 to 12 kDa) that act as electron carriers in photosynthesis, nitrogen fixation and metabolism. The iron is bound to inorganic (acid-labile) sulfide and to the cysteine sulfur atoms of the protein, and the Fe centres shuttle between Fe(II) and Fe(III) without change in the structure.

They are classified by the type of iron-sulfur cluster: [2Fe-2S] (two Fe bridged by two S²⁻, as in plant-type ferredoxin), [3Fe-4S], and [4Fe-4S] (cubane-like Fe₄S₄ cluster), with some bacterial ferredoxins carrying two [4Fe-4S] clusters. Rubredoxin, with a single Fe bound by four cysteines, is a related simpler protein.

(f) The chelate effect and the ring size (5 marks)

The chelate effect is the greater stability of a complex containing a polydentate (chelating) ligand than of a similar complex with the corresponding monodentate ligands. For example, [Ni(en)₃]²⁺ is far more stable (log β₃ about 18) than [Ni(NH₃)₆]²⁺ (log β₆ about 9). The reason is mainly entropic: replacing six monodentate ligands by three bidentate ones increases the number of free molecules in solution (four in all), giving a positive ΔS and a more negative ΔG, whereas ΔH is about the same.

Ring size matters. Five-membered rings (for saturated ligands such as en) and six-membered rings (for ligands with delocalised bonds, such as acetylacetonate) are the most stable, since they have the least strain. Three- and four-membered rings are strained, and rings of seven or more have lower stability because of greater loss of freedom and less favourable geometry.

(g) Ferrocene as a sandwich compound (5 marks)

Ferrocene, Fe(η⁵-C₅H₅)₂, has an Fe²⁺ ion held between two parallel cyclopentadienyl rings, like the filling of a sandwich, with all five carbons of each ring bonded equally to iron (Fe-C about 204 pm) and the C-C bonds all equal (about 140 pm). Each ring is C₅H₅⁻ and donates six π electrons, and Fe²⁺ has six d electrons, so the iron has the 18-electron configuration, which explains its great stability. The rings are eclipsed in the gas phase (D₅h) and rotate almost freely (a barrier of only a few kJ mol⁻¹). It is an orange, diamagnetic, air-stable solid that shows aromatic character, for example in Friedel-Crafts acylation.

(h) Ionization isomerism (5 marks)

Ionization isomers have the same formula but differ in which groups are coordinated to the metal and which are counter-ions, so they give different ions in solution. For example, [Co(NH₃)₅SO₄]NO₃ gives NO₃⁻ ions and no precipitate with BaCl₂, whereas [Co(NH₃)₅NO₃]SO₄ gives free SO₄²⁻ and a white BaSO₄ precipitate.

• (i) [Cr(H₂O)₆]Cl₃ and [CrCl(H₂O)₅]Cl₂·H₂O: hydrate (solvate) isomerism, since water is inside the coordination sphere in one and outside as lattice water in the other.
• (ii) [Co(NH₃)₅SO₄]NO₃ and [Co(NH₃)₅NO₃]SO₄: ionization isomerism.
• (iii) [Pt(NH₃)₄][PtCl₆] and [Pt(NH₃)₄Cl₂][PtCl₄]: coordination isomerism, in which ligands are exchanged between the cationic and anionic complexes (here with the metal also in different oxidation states, Pt(II) and Pt(IV)).

(i) Inorganic benzene (5 marks)

Inorganic benzene is borazine (borazole), B₃N₃H₆, which is isoelectronic with benzene (C₆H₆). It is a planar six-membered ring of alternating B and N atoms, each with one H, with sp² hybridised atoms. The B-N bond length (about 144 pm) is between single and double, and the π electrons are partly delocalised, but because N is more electronegative than B the π cloud is polarised towards nitrogen, so borazine is much less aromatic and more reactive than benzene. Unlike benzene it adds HCl, H₂O and other reagents (Cl on B and H on N). It is made by heating diborane with ammonia, or from BCl₃ and NH₄Cl followed by reduction with NaBH₄.

(j) Actinide contraction (5 marks)

The actinide contraction is the steady decrease in atomic and ionic radii on going across the actinides from Ac to Lr (for M³⁺, from about 112 pm for Ac³⁺ to about 90 pm at the end of the series). It is caused by the poor shielding of the nuclear charge by 5f electrons, because f orbitals are diffuse and have complicated shapes: as the nuclear charge rises by one unit and one electron is added to the 5f subshell, the outer electrons feel a larger effective nuclear charge and are drawn in. It is greater than the lanthanide contraction, because 5f orbitals are more diffuse than 4f and shield even less. One consequence is that the later 6d elements have sizes close to their 5d counterparts.