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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]

A regular icosahedron is a Platonic solid with twenty equilateral triangular faces, twelve vertices and thirty edges. Five faces meet at each vertex. Its orientation-preserving rotational symmetry has order sixty: the orbit-stabilizer theorem gives twenty faces times the three rotations stabilizing a chosen face.

The orientation-preserving rotational symmetry group of a regular icosahedron has order sixty. Its conjugacy classes have sizes $1,15,20,12,12$: identity; half-turns about opposite-edge axes; rotations through $\pm 2\pi /3$ about opposite-face axes; and rotations through $\pm 2\pi /5$ and $\pm 4\pi /5$ about opposite-vertex axes, respectively. Conjugation carries an axis to its rotated axis and preserves the rotation angle. The transitive group action on edges, faces and vertices gives the indicated classes, with reversal of the axis identifying the two signs. A normal subgroup must contain whole conjugacy classes; none of the proper nonidentity sums of these class sizes is a divisor of sixty. Thus this group is a simple group.