Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 2

${\mathrm{S}}_4^{2+}$ is approximately square planar; $\mathrm{S}{\mathrm{n}}_5^{2-}$ 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 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.