Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 9

For the CO and NCO cases, the added fragment acts as a Lewis base at boron, rather than becoming a new skeletal vertex. Both have a triangular ${\mathrm{B}}_3$ framework. A useful localized drawing has three $\mathrm{B}{\mathrm{H}}_2$ groups, one B–H–B bridge between two of them, and the donor attached to the third boron. The boron–boron framework includes three-centre two-electron bonds; its edges should not all be read as ordinary independent two-electron single bonds.

For the neutral carbonyl adduct, attach CO through carbon, giving $\mathrm{B}\leftarrow \mathrm{C}\equiv \mathrm{O}$ at the donor-bearing $\mathrm{B}{\mathrm{H}}_2$ site. For the anionic isocyanate adduct, attach $\mathrm{NC}{\mathrm{O}}^{-}$ through nitrogen, giving a B–N=C=O connectivity and the overall $-1$ charge. In both drawings the hydrogen count is $3(2)+1=7$. Weak or asymmetric additional B–H contacts can distort this idealized drawing; a hydrogen-rich borane is not fully characterized by a polygon alone. The carbonyl triborane structure was established in the original borane study.^[ref]

The rhodium fragment is instead a skeletal building block. Its neutral electron count is $9+2(2)+1=14$, so it contributes two skeletal electrons, like a BH vertex. Combining it with ${\mathrm{B}}_3{\mathrm{H}}_7$ gives $3(2)+4+2=12$ skeletal electrons, or six pairs, for four vertices. A consistent low-vertex structure is a tetrahedral $\mathrm{Rh}{\mathrm{B}}_3$ skeleton, with the $\mathrm{RhH}(\mathrm{PP}{\mathrm{h}}_3{)}_2$ fragment capping the boron triangle. Keep the terminal Rh–H ligand distinct from the seven hydrogens on the borane portion. The latter can again be represented by three terminal $\mathrm{B}{\mathrm{H}}_2$ groups and a B–H–B bridge. The total formula then contains eight hydrogens outside the phenyl groups. Treating Rh as a pendant one-centre Lewis base would miss its skeletal role.

Neutral counting assigns five electrons to cyclopentadienyl ligand and two to CO. Thus each $\mathrm{FeCpCO}$ fragment has $8+5+2=15$ valence electrons and contributes three skeletal electrons; each $\mathrm{CoCpCO}$ fragment has sixteen and contributes four:For A the six-pair four-vertex count is the nido count derived from a five-vertex trigonal bipyramid. Removing an axial vertex leaves a tetrahedron: the three equatorial vertices plus the remaining axial vertex. For B the corresponding arachno fragment is obtained by removing both axial vertices from that same parent, leaving a three-metal triangle. Small clusters require specifying the parent and removed vertices; mechanically calling every nido fragment a square pyramid is wrong.

The Fe tetramer is indeed known to have a tetrahedral $\mathrm{F}{\mathrm{e}}_4$ core, with four face-capping CO groups. The carbon atoms of those carbonyls and the iron atoms form interpenetrating tetrahedra, giving a cubane-like Fe–C connectivity. Bridging rather than terminal CO changes the distribution of local donation, not the total electron count above.^[ref]

For a neutral borane ${\mathrm{B}}_{\mathrm{n}}{\mathrm{H}}_{\mathrm{m}}$ with one terminal B–H bond assigned to each vertex, the skeletal electron number is $3n+m-2n=n+m$; an overall charge $q-$ adds $q$ electrons. Suitable twelve-electron correspondences areFor C, four BH contributions plus four additional hydrogens give $8+4=12$ electrons. For D, three BH contributions plus five additional hydrogens and the negative charge give $6+5+1=12$. They map to the tetrahedral and triangular low-vertex skeletons respectively. The neutral formal alternative for D is ${\mathrm{B}}_3{\mathrm{H}}_9$, also with twelve skeletal electrons.

The electron counting correspondence alone must not be read as evidence that the neutral boranes are isolable stable bulk materials. Low-vertex hydrogen-rich boranes can be transient; the triangular octahydrotriborate anion is the established practical three-boron example, whereas neutral ${\mathrm{B}}_3{\mathrm{H}}_9$ is a formal analogy rather than an equally established preparative product. The requested skeletal correspondence should be distinguished from a claim about comparable stability or isolation conditions.^[ref]

For E, each BH supplies two skeletal electrons and each phosphorus vertex supplies three after retaining a lone pair. Hence $10(2)+2(3)=26$ electrons, or thirteen pairs, matching a twelve-vertex closo cluster. Place the two adjacent P vertices on a regular icosahedron. The remaining ten boron vertices fall into four symmetry classes: two sites adjacent to both P vertices, four adjacent to exactly one, two adjacent to neither but opposite the common-neighbour pair, and two opposite the P pair. The populations are $2:4:2:2$, so four boron resonances are expected. Interchanging the identical phosphorus vertices is allowed when determining chemical equivalence.

For F, a ten-vertex bicapped square antiprism offers a consistent closo model. There are eight BH vertices, one B vertex bearing the $\mathrm{NE}{\mathrm{t}}_3$ donor instead of H, and one P vertex. Their skeletal contributions are respectively $16$, $3$ and $3$, giving 22 electrons, eleven pairs. The donor-bearing B contribution follows from B plus the donated lone pair minus the exoskeletal B–N bond, $3+2-2=3$.

One explicit arrangement consistent with six boron signals puts P at one axial cap and the $\mathrm{NE}{\mathrm{t}}_3$ substituent on one B vertex of the nearer square. The remaining mirror plane through that B, the opposite vertex of its square, and the cap axis leaves three B sites individually fixed: the donor-bearing B, its opposite square vertex and the other axial cap. The other six B sites form three exchanged pairs. Thus the nine borons give three singleton environments and three two-site environments, six signals in all. The original compound has a crystallographically characterized positional assignment; the spectrum alone asks for a compatible structure and does not establish a unique positional isomer. A single-crystal X-ray diffraction determination or assigned heteronuclear correlations would distinguish alternatives.^[ref]