 Solution

$\mathrm{A}=\mathrm{K}[{\mathrm{B}}_5{\mathrm{H}}_8]$ is nido; $\mathrm{B}=\mathrm{K}[{\mathrm{B}}_6{\mathrm{H}}_{11}]$ and $\mathrm{C}={\mathrm{B}}_6{\mathrm{H}}_{12}$ are arachno. The initial hydride removes an acidic bridging hydrogen as a proton, releasing ${\mathrm{H}}_2$ and retaining the potassium counterion. Addition of one $\mathrm{B}{\mathrm{H}}_3$ unit expands the boron framework; final protonation supplies the neutral hexaborane and KCl. These assignments conserve all the atoms in the displayed sequence, without inserting an unshown dehydrogenation step.

For ${\mathrm{B}}_{\mathrm{n}}{\mathrm{H}}_{\mathrm{m}}^{\mathrm{q}-}$, take one terminal B–H bond per vertex and remove its $2n$ electrons from the valence total. The skeletal electron count is $3n+m+q-2n=n+m+q$. ThusWade's rules assign $n+1$, $n+2$ and $n+3$ pairs to closo, nido and arachno frameworks respectively. A derives from a six-vertex closo octahedron by removal of one vertex, giving a square-pyramidal boron skeleton. B and C have six-vertex open frameworks formally derived from an eight-vertex closo polyhedron by removing two vertices. Their additional hydrogens are distributed over bridging and other open-face positions; counting every hydrogen as an independent terminal B–H vertex would give the wrong classification.

The arachno assignment of $[{\mathrm{B}}_6{\mathrm{H}}_{11}{]}^{-}$ has specific theoretical and NMR support. It should not be replaced by a permanently localized, unrearranged $[{\mathrm{B}}_5{\mathrm{H}}_8{]}^{-}\cdot \mathrm{B}{\mathrm{H}}_3$ picture: the corresponding symmetric adduct participates in fluxional rearrangement rather than being the lowest-energy structure.^[ref]