= Solution
With oxide charge $-2$ and four <thiocyanates> each $-1$, $x-2-4=-2$ gives \b[V(IV), $d^1$]. Counting the terminal oxo as $X_2$ and each <thiocyanate> as X gives
$$
\boxed{N_e=1+4+4(2)=13.}
$$
The expected structure is \b[<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 $\mathrm{M{=}O}$ convention; counting extra oxide pi-donor interactions as additional ligand <electrons> requires a different convention throughout.
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