Past exam of the chemistry B course of the University of Cambridge 2022 Paper 2 B8 a i Solution 2026-10-20
Choose along M–N–N. The occupied dinitrogen orbital, with amplitude on the terminal N atoms, overlaps an empty metal orbital of sigma symmetry to give N-to-metal sigma donation. Filled metal and orbitals separately overlap the two empty N orbitals, giving metal-to-N pi backbonding. The lower-energy combination of each interacting pair is bonding and the higher is antibonding; each donation channel starts with two electrons in the occupied fragment orbital. The lower panel of the original orbital diagram shows these three channels.
As for a terminal carbonyl ligand, the donor and acceptor interactions reinforce one another: sigma donation helps make electron density available at the metal, while pi backbonding strengthens M–N and weakens N–N. For conventional electron counting an intact end-on neutral N ligand is an L-type two-electron donor; counting the backbonded electrons a second time as extra donated ligand electrons would double count them.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 2 B8 c iii Solution 2026-10-20
The opposite insertion orientation gives a benzylic metal–carbon bond, . C–Si reductive elimination then gives the branched productA plausible explanation is stabilization of the developing and final benzylic Co–C interaction by the adjacent aromatic pi system, which can outweigh the steric preference for a primary alkyl. That makes an electronically different choice from an ordinary alkyl substituent. This is a mechanistic rationale for the proposed cycle, not proof of its rate-determining step. In the absence of a chiral catalyst or another source of asymmetric induction, the new stereogenic carbon would give a racemic mixture.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 2 B8 c ii Solution 2026-10-20
oxidative addition of an intact Si–H bond initially places H and SiR cis in the metal coordination sphere; this is the local arrangement generated by addition of a sigma bond. hydride and alkene must also attain adjacent sites for intramolecular insertion, and the eventual alkyl and silyl groups must be cis for direct reductive elimination. Subsequent rearrangements should not be confused with the stereochemistry of the bond-addition event itself.
For an alkyl-substituted terminal alkene, insertion puts Co on the terminal carbon and H on the substituted carbon, producing . It avoids the more crowded secondary metal–alkyl arrangement. Replacement of that Co–C bond by C–Si therefore gives the linear product , an Anti-Markovnikov addition outcome. Hydrometallation is syn at the coordinated alkene, although this simple terminal substrate generally does not retain two substituents whose relative stereochemistry could diagnose the complete sequence.
Past exam of the chemistry B course of the University of Cambridge 2022 Paper 2 B8 c i Solution 2026-10-20
Step 1 is oxidative addition of Si–H to Co(I), giving Co(III)–H and Co(III)–SiR. It cleaves the Si–H bond and forms two metal–ligand bonds, increasing the formal metal oxidation state by two. The phosphine dissociation shown before this numbered step makes room for the reaction; alkene binding and phosphine exchange are ligand-substitution events rather than oxidative additions.
Step 2 is migratory insertion of the coordinated alkene into Co–H, equivalently intramolecular hydride migration to the alkene. One alkene carbon gains H and the other gains the Co–C bond. Co remains formally Co(III); a two-electron alkene ligand plus a hydride is replaced by one alkyl ligand, reducing the electron count by two before phosphine reassociation.
Step 3 is C–Si reductive elimination between the alkyl and silyl ligands. It releases the hydrosilylated alkane and regenerates Co(I). The electron count for the drawn sequence is consistent: the one-PMe alkene adduct is 16-electron; oxidative addition gives 18; insertion gives 16; PMe association gives 18; reductive elimination returns the 16-electron catalyst. Because insertion occurs into Co–H, this is the conventional Chalk-Harrod mechanism.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B6 a ii Solution 2026-10-20
cisplatin has square planar molecular geometry and a metal centre. Choose and along the four Pt–donor directions. The strongly antibonding orbital is highest and remains empty; the other four d orbitals are doubly occupied. The square-planar splitting is represented by the conventional qualitative orderThe ordering among the lower occupied levels can depend on the ligand set, and the cis arrangement removes exact fourfold symmetry. The robust result is the filled lower four levels and empty highest level.
satraplatin has approximately octahedral molecular geometry around Pt(IV), . Its large octahedral splitting favours a low-spin complex:All three orbitals are paired. Mixed ligands remove exact degeneracies, but do not change this qualitative occupancy.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B6 b ii Solution Created 2026-10-20 Updated 2026-10-31
cyclohexylamine is expected highest in the spectrochemical series for comparable metal environments. Its alkyl substituent increases the electron-donating ability of nitrogen relative to Ammonia, strengthening the sigma interaction and raising the metal–ligand sigma-antibonding levels. The anionic pi donors tend to raise occupied metal levels and give smaller splittings. Steric effects and the metal can modify the exact ordering, so the comparison is a ligand field trend rather than a universal basicity-to-splitting identity.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B6 b i Solution Created 2026-10-20 Updated 2026-10-31
Ammonia and cyclohexylamine are sigma-donor ligands; chloride and acetate are sigma- and pi-donor ligands. The nitrogen lone pair of each amine points toward the metal and forms the donor interaction. These saturated amines have no low-lying conjugated acceptor system that would make them conventional pi-acceptor ligands.
chloride uses a lone pair for sigma donation and has additional occupied orbitals perpendicular to the bond that can donate with pi symmetry. An oxygen-bound acetate similarly has occupied oxygen-centred orbitals available for sigma and pi donation, although its carboxylate resonance in chemistry changes their energies and distribution. Neither should be classified as a pi acceptor merely because its bonding can be drawn with multiple bonds or resonance in chemistry contributors. None of these four is a standard sigma-donor/pi-acceptor ligand.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B6 c Solution Created 2026-10-20 Updated 2026-10-31
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 B6 d Solution 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 B7 b i Solution 2026-10-20
The neutral product is cis-, and the displaced material is six waters per nickel:The two phosphorus donors of dppe chelate the same metal, forcing adjacent donor positions. The diamagnetism identifies a square-planar, paired Ni(II) centre rather than a tetrahedral high-spin centre. Its valence count is electrons. The reaction is ligand substitution, not reduction of Ni(II).
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B8 a ii Solution 2026-10-20
The chloride complex has Pt(II), , whereas the neutral bis(phosphine) complex has Pt(0), in the neutral-ethene assignment. The lower-valent, phosphine-supported Pt(0) centre back-donates more strongly into ethene , weakening and lengthening its internal C–C bond. This accounts for the longer distance compared with in the Pt(II) complex.
Both are formally 16-electron complexes: for the chloride species and for the phosphine species. Thus the difference is not simply a difference in total valence count. It reflects the energies and occupations of the metal–ligand molecular orbitals. The increased reduction of the ethene ligand gives the Pt(0) complex more metallacyclopropane character; the two descriptions are bonding limits, not distinct products inferred from the distances alone.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B8 a i Solution 2026-10-20
The Dewar-Chatt-Duncanson model has two complementary interactions. The filled alkene orbital donates to an empty metal orbital, and an occupied metal orbital back-donates into the empty alkene orbital. The metal acceptor has the symmetry of the sum of the two ligand lobes facing it; the back-donor has opposite phases at the two carbon positions and matches the antisymmetric combination.Both channels strengthen metal–alkene attachment. Donation removes some electron density from the internal bonding orbital, while back-donation populates its antibonding partner, so coordination generally reduces C–C bond order. The schematic shows orbital phase signs, not positive and negative electric charges.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B8 b ii Solution 2026-10-20
A useful strongly bound limit is Zr(IV), , and Ta(V), , with an aryne dianion in each complex. Charge balance uses two Cp ligands for zirconium, and Cp* plus two methyl ligands for tantalum:Equivalently, an unreduced neutral-aryne description starts from Zr(II) and Ta(III), both , and transfers that pair strongly into the aryne acceptor. These are alternative bookkeeping limits for the same covalent interaction; saying that a formal assignment has no available electrons must not erase the strong metal–carbon bonds in the reduced-ligand description.
Both ligands have therefore lost much of the extra in-plane C–C bonding of free benzyne and approach the same reduced, approximately double-bond-length C1–C2 framework. Their identical C–C distances are consistent with similar aryne reduction in both complexes. The distance is intermediate in a delocalized ring, so it should not be converted into an exact integer bond order.
The M–C distances depend additionally on metal size, effective nuclear attraction and orbital overlap. The high formal charge of Ta(V), the size effects associated with the lanthanide contraction, and effective covalent bonding favour shorter Ta–C contacts than Zr–C contacts. This is consistent with versus . Different coordination envelopes and ancillary ligands also contribute, so this pair is not a controlled measurement of a radius or a metal–carbon bond energy. Similar internal ligand bond order can coexist with different metal–ligand distances.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B8 b i Solution 2026-10-20
The metal interacts with the extra in-plane bonding system of benzyne, not by replacing the ordinary out-of-plane aromatic pi system. Take the ring and metal to lie in the plane, with the metal-to-C–C-midpoint direction along and C1–C2 along . The carbon orbitals directed toward the metal form an occupied symmetric combination and an empty antisymmetric combination.
The occupied in-plane combination donates into an empty metal orbital of matching symmetry. An occupied metal -type orbital overlaps with the in-plane combination, providing back-donation. These are the same symmetry principles as alkene binding, rotated into the plane of the ring. The carbon orbitals perpendicular to the ring continue to participate in its aromatic system. The strained formal triple bond therefore does not supply two independent, ordinary out-of-plane alkene interactions.
Strong population of the in-plane acceptor can be described as reduction of the aryne to a dianionic, metallacyclopropene-like ligand. That description emphasizes two M–C sigma bonds and loss of the extra C–C pi bond. It is particularly useful for early-metal aryne complexes. The lower panels of the orbital schematic distinguish the interacting in-plane orbitals from the perpendicular aromatic set.
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B8 c ii Solution 2026-10-20
The terminal nitrosyls are linear in both proposed electron-precise structures; the formal M–M bond orders are V–V double and Fe–Fe single. Use three-electron terminal linear NO donors, three-electron -NO bridges, and five-electron neutral Cp donors in covalent counting.
For the vanadium dimer the two metal fragments together containA metal–metal bond of order adds to each local count, so gives and 18 electrons per V. For an ionic cross-check, Cp is , each linear terminal NO is conventionally NO, and each bridge is represented as NO. ThusEach V then has electrons.
The iron species is monoanionic, and is a pyrazolate bridge binding through its two nitrogens. This is a four-electron donor overall in ionic counting, or three overall in neutral counting. The carbonyl bridge supplies two electrons overall. The covalent fragment total is consequentlyTherefore gives . In the formal NO description,Each Fe receives electrons. The negative formal oxidation state follows from the chosen nitrosyl assignment and the overall charge; omitting that charge would change the count.
Bent terminal NO ligands would supply fewer neutral-count electrons and demand much higher formal M–M bond orders to attain 18-electron counts. The linear assignments give the ordinary double and single bond solutions. Since NO is a non-innocent ligand, these oxidation states are formal bookkeeping limits, not measured metal charges; actual nitrosyl bending and M–M bond strengths require structural and spectroscopic evidence.[ref]
Principles of structure, bonding, and reactivity for metal nitrosyl complexes by John H. Enemark and Richard D. Feltham. Coordination Chemistry Reviews 13, pp. 339–406; formal nitrosyl assignments and distinction from actual electron distribution
Past exam of the chemistry B course of the University of Cambridge 2023 Paper 2 B8 c i Solution 2026-10-20
A neutral, simple -NO contributes three electrons to the two-metal unit overall. In neutral bookkeeping it is an LX-type bridge: a two-electron donor component plus the odd electron of neutral NO. For a symmetric bridge, this can be apportioned as electrons to each metal in a covalent count; it is not three electrons independently to each metal.
An equivalent ionic description represents the bridge as NO donating two electrons to each metal, four overall. The extra negative ligand charge must then be included when assigning metal oxidation states. Mixing the four-electron ionic donation with a neutral-NO charge assignment would double-count an electron.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 2 B10 c Solution 2026-10-20
The zinc ion is coordinated by protein donors and a water oxygen placed near the reacting C6 center. Zinc acts as a Lewis acid that activates and positions water. Its positive charge withdraws electron density from oxygen, lowers the effective p of coordinated water and facilitates formation of the attacking hydroxide, potentially with assistance from a nearby base. The same metal–oxygen interaction can stabilize developing negative charge as the addition intermediate forms. Zinc stays Zn(II); this is metal ion catalysis, not a zinc redox cycle.
The structure identifies a plausible reaction geometry, but a static X-ray model does not by itself establish the water's protonation state, proton-transfer order or the rate-limiting step. Structural work with bound transition-state analogues independently supports a zinc-containing catalytic center.[ref]
Atomic structure of adenosine deaminase complexed with a transition-state analog: understanding catalysis and immunodeficiency mutations by David K. Wilson and others. 1991, pp. 1278–1284, catalytic zinc and the bound transition-state analogue
An appropriate experimental test would compare native enzyme, a metal-depleted preparation and stoichiometrically zinc-reconstituted enzyme, checking retained fold and active enzyme concentration. Loss and recovery of activity with those controls distinguishes a catalytic metal requirement from nonspecific denaturation. Past exam of the chemistry B course of the University of Cambridge 2024 Paper 2 B6 a Solution 2026-10-20
The carbon end of a carbonyl ligand has a filled molecular orbital with an outward-facing lobe. Its overlap with a vacant metal orbital of the same axial symmetry gives CO-to-metal sigma donation. Occupied metal orbitals can simultaneously overlap with the empty CO antibonding molecular orbitals, giving metal-to-CO pi backbonding. Two perpendicular pi channels are possible; the schematic represents one. Orbital signs indicate phase, not electric charge.
The two interactions reinforce one another: sigma donation transfers density toward the metal, and pi backbonding distributes it into the ligand acceptor orbitals. Both strengthen M–C attachment, while backbonding weakens C–O. Occupation of CO decreases the internal C–O bond order, generally increasing its bond length and lowering its stretching spectroscopic wavenumber. This is why a strongly backbonded carbonyl ligand can have a strong metal–carbon bond and a weakened internal carbon–oxygen bond at the same time.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 2 B6 b iii Solution 2026-10-20
A suitable structure is a triangular Ti metal cluster with one 1,2-Bis(dimethylphosphino)ethane chelate and one terminal chloride at each titanium, and one bridging ligand -Cl spanning each edge. Include a direct Ti–Ti single metal-metal bond along each edge. The last panel distinguishes these direct bonds from the chloride bridges.
The six chlorides have total formal charge , so equivalent centers are Ti(II), . At each titanium the two phosphorus donors supply four electrons, the terminal chloride supplies two, and the two incident chloride bridges each donate a pair in ionic electron counting. Two incident Ti–Ti single bonds supply another two electrons:Each center can contribute one electron to each of its two edge bonds, making three bonds from six metal electrons overall. This is a 14-electron early-metal cluster, not an 18-electron complex. Forcing the 18-electron rule here would invent excessive bond orders. chloride bridges donate to more than one center, so summing local counts is not the same as counting distinct electrons of the whole cluster.
Past exam of the chemistry B course of the University of Cambridge 2024 Paper 2 B6 c Solution 2026-10-20
The principal electronic variable is the net population of metal–metal bonding molecular orbitals relative to antibonding molecular orbitals. For otherwise comparable complexes, a larger bond order generally gives a shorter metal-metal bond; removing an antibonding electron can shorten a bond even though the total electron count falls. Different orbital symmetries also matter: sigma, pi and delta overlaps respond differently to distance and rotation of the metal fragments.
metal size, oxidation state and radial orbital extent set the underlying distance scale. Comparing a first-row metal with a larger heavier congener does not isolate bond order. Increasing positive charge often contracts the metal orbitals, but can also change electron occupancy, ligand binding and electrostatic interactions. The ligand field and donor/acceptor properties of the surrounding ligands in coordination chemistry determine which metal orbitals remain available for direct bonding and how much density instead participates in pi backbonding.
Finally, the geometry can enforce or oppose the preferred separation. Bridging ligands impose bond-angle and bond-length constraints; bulky substituents produce steric hindrance; crystal packing can perturb the observed distance. A bond length is evidence for bonding only in its electronic and structural context. In particular, two bridges can hold metals close without proving a high direct bond order, while a sterically congested formal single bond can be unusually long.




