= Solution
\b[$\mathrm{S_4^{2+}}$ is approximately square planar; $\mathrm{Sn_5^{2-}}$ is a <closo> trigonal bipyramid.] Both have 22 <valence electrons>, $4(6)-2=22$ and $5(4)+2=22$, but their different local orbital requirements prevent inferring identical shapes from this total alone.
For the <sulfur> ring, four S–S <sigma bonds> use eight electrons and four in-plane <lone pairs> use another eight, leaving six electrons in the perpendicular cyclic pi system. The resulting filled bonding and degenerate nonbonding pi levels give a six-pi-electron aromatic description of a four-membered ring, consistent with equivalent S–S links in the ideal square. The out-of-plane p orbitals are retained; this is not a tetrahedral four-vertex <closo> cage.
For <tin>, allocate one mainly s-rich <lone pair> to each atom. The remaining $22-5(2)=12$ electrons form six skeletal pairs, exactly $n+1$ for $n=5$. <Wade's rules> therefore predict a trigonal bipyramid, with three equatorial vertices and two axial vertices. The axial pair is not joined by a polyhedral edge; its bonding is part of the delocalized three-dimensional framework. The contrast is \b[a planar aromatic sigma/pi ring versus a three-dimensional multicentre-bonded <Zintl cluster>], rather than cationic charge alone imposing planarity or anionic charge alone imposing a polyhedron.
\Image[/past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1-sulfur-tin-clusters.png]
{title=Original square-ring and trigonal-bipyramidal models with their different allocations of the same 22 valence electrons}
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