Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B7 a Solution by
Codex 0 2026-10-20
For the usual octahedral Ni(II) hydration-isomer series, distinguish coordinated water from water of crystallization:The first two coordination spheres have no cis/trans choice because they contain zero or one chloride. The last hydration class has two geometrical isomers: the chlorides occupy adjacent or opposite octahedral positions. Thus there are three hydration classes and four structures when cis/trans isomerism is included. Every formula retains two chlorides and six waters overall; the coordination number remains six. The schematic in B6(c) displays the distinct octahedral connectivities. This enumeration states the usual six-coordinate interpretation, rather than treating unspecified changes in coordination number as further members of the same elementary series.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B6 d Solution by
Codex 0 2026-10-20
The more radially extended Pt orbitals give strong metal–ligand overlap and covalent bonding. This helps compensate the energetic cost of reaching Pt(IV), and the large octahedral ligand field splitting strongly stabilizes its low-spin configuration, with a crystal field stabilization energy contribution before separately counting pairing and the full bond energies. Ni bonding is less effective in this respect; Pd compounds also generally favour the very stable M(II) arrangement more strongly than Pt does. The stronger bonding and large ligand-field stabilization available to Pt(IV) make oxidation to more favourable relative to its congeners. This is a comparative trend, not a claim that Ni(IV) or Pd(IV) complexes cannot exist.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B6 c Solution by
Codex 0 2026-10-20
The average Pt oxidation state is , since six bromides balance the two metal atoms. A natural mixed valence assignment is alternating Pt(II), , and Pt(IV), , rather than two equivalent Pt(III) sites. Use neutral square-planar units and octahedral units. At every site place two ammonia and two bromide ligands in an equatorial plane; the Pt(IV) sites have two additional axial bromides pointing toward the adjacent Pt(II) sites.
Along the chain the connectivity isEvery Pt–Br–Pt bridge is asymmetric: short to Pt(IV), long to Pt(II). The latter contact is a secondary axial interaction, consistent with the preference of Pt(II) for a square-planar primary coordination sphere. The Pt(IV) site is smaller and more strongly attracts the bridging bromides. A drawing with all axial distances identical would obscure the distinct local environments.
There are four equatorial bromides and two axial bridging bromides per two-Pt repeat, giving precisely . A direct Pt–Pt bond is not required. This is a proposed structure consistent with the stated composition and two environments; the formula alone does not determine every equatorial cis/trans arrangement or numerical bond length.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B10 a Solution by
Codex 0 2026-10-20
water stabilizes the planar amide more strongly relative to the twisted rotational transition region, increasing the free-energy barrier. nitrogen lone-pair donation into the carbonyl gives the C–N bond partial double-bond character through amide resonance, represented byamide bond rotation between the two amide conformations requires loss of favourable overlap. Near a strongly twisted geometry the nitrogen lone pair is less conjugated with C=O, and the charge-separated resonance in chemistry contribution is reduced. Polar water, a hydrogen bond donor, preferentially stabilizes the planar starting structures, including their polar carbonyl/resonance distribution, compared with this less conjugated rotational region. chloroform provides less of that differential stabilization. What controls the rate is the difference between the starting minimum and the transition state, not the absolute stabilization of either one in isolation.
For a first-order interpretation of the half-lives, the characteristic rate constants areThus the interconversion is 25 times slower in water. With comparable Eyring equation prefactors at the same temperature,which is about at . This temperature is illustrative because none is specified for the measurements.
The two endpoint conformations remain minima separated by a twisted barrier in each solvent. The profiles are schematic, with arbitrary absolute barriers; the half-life ratio supports the relative kinetic comparison. If the quoted half-lives describe relaxation of a reversible cis/trans mixture, , so the ratio directly concerns relaxation rates. Assigning the full difference to one microscopic forward barrier additionally requires comparable equilibrium populations or directional-rate information.
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