= Solution
\b[$B=\mathrm{N_3P_3(NHCy)_6}$], with two cyclohexylamino substituents on each original phosphorus. All six P–Cl bonds have been replaced; the six N–H bonds of those substituents remain. Each equivalent of <triethylamine> traps one equivalent of HCl.
\b[$C$ contains three four-membered $\mathrm{P^{V}{-}N(Cy){-}P^{III}(Cl){-}N(Cy)}$ spiro rings], one at each original ring phosphorus. Each PCl$_3$ joins the two geminal NHCy groups: replacement of two of its chlorides forms two new P–N bonds and releases two HCl, consuming both N–H <protons>. The new phosphorus retains one chloride and a <lone pair>, making it three-coordinate P(III); the original six-membered $\mathrm{P_3N_3}$ ring retains P(V). The product formula is $\mathrm{N_3P_3\{(NCy)_2PCl\}_3}$. For each spiro ring, the original phosphorus is bonded to both exocyclic nitrogens, and each of those nitrogens is bonded to the same new PCl center; the braces do not indicate a simple linear substituent.
The three new P–Cl directions permit \b[an all-cis isomer and a cis,trans isomer]. Relative to the central ring, write these as $(+,+,+)$ and $(+,+,-)$, with cyclic relabeling giving equivalent choices. Reversing all signs does not create a third relative configuration. This describes the configuration pattern; puckering and substituent conformations are additional structural details.
\Image[/past-exam-of-the-chemistry-b-course-of-the-university-of-cambridge/2025/paper-2-phosphazenes.png]
{title=Spiro-ring connectivity and the electron-counted diiron and triiron phosphazene products}
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