Past exam of the chemistry B course of the University of Cambridge / 2022 / Paper 2 / B7 / a / iv / Solution

With oxide charge $-2$ and four thiocyanates each $-1$, $x-2-4=-2$ gives V(IV), $d^1$. Counting the terminal oxo as $X_2$ and each thiocyanate as X givesThe expected structure is square-pyramidal molecular geometry, with the short V=O unit at the apex and the four thiocyanate donors in the basal plane. The remaining d electron can occupy an approximately basal $d_{xy}$ orbital that avoids the strongest donor directions and the oxo pi-antibonding orbitals. This is a high-oxidation-state oxovanadium complex; reaching 18 electrons is not its governing stability criterion. The displayed $\mathrm{SCN}$ composition alone does not establish N versus S linkage, and that choice does not change this count. The stated 13-electron result consistently uses a four-electron $\mathrm{M}=\mathrm{O}$ convention; counting extra oxide pi-donor interactions as additional ligand electrons requires a different convention throughout.