 Solution

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]